Transcript for:
Comprehensive Overview of ASP Mathematics

okay welcome to this video looking at everything you need for asp mathematics now of course anybody that has studied asp mathematics will know that covering all of the content in a year let alone in seven hours is very challenging now what i have done is i have gone through all of the topics that you can see on the screen and i have covered important aspects and teaching elements to do with all of them so this is a long video but as you will see in a second it is split up so that you can use it effectively now i have gone through every single topic on the asp curriculum so please do make use of this video if it is useful and helpful please do like the video please leave me a comment and don't forget to subscribe to the channel as with all of these videos there is all of the chapters down below and i'm going to show you that in just a second so please do make use of that and don't forget to check in the description for any links to further videos for full lessons on any of these topics so let's have a look at how to use this video so anyone that has followed me from gcse level should hopefully remember this video on the screen where i covered everything for a grade six to nine in your maths gcse exam now this video is going to follow a very very similar format so if you click into the description you will see that every topic is time stamped for you so that you know the name of everything that's on the screen and therefore you can go and you can do some further revision whether that be from a revision guide or whether from a textbook or whether you use some of my lessons which will be next to each topic in the description now if you want another way to scroll through the video as well if you click onto the video you can see in the bottom left it says the name of the topic if you click on that it will bring through all of the chapters down the side and you can just scroll through and you can start and pick up from wherever you left off or you can skip through particular topics if you know that you are comfortable with them so hopefully the way that this video works should be really useful for you again if you have already followed me you will already know how my videos work but just in case you're new to the channel hopefully this just shows you how i set up my videos and how you can go about using it so now that you've seen that let's get started we have a lot to get through [Music] okay so we've got larger coefficients of x here we've got the 2x and the 3x and i'm going to start by expanding these two so let's write that out we've got 2x plus 3 and 3x minus 2. so we know we're going to get the 6x squared here when we do the first two so 2x times 3x gives us the 6x squared and let's have a look at those middle two so we get minus 4x plus 9x and those middle two so minus 4 plus 9 gives us 5x so we're going to have plus 5x minus 4 plus 9. and then we're going to get minus 6 at the end there we go so to save you a little bit of time if you can do this and let's stick that in a bracket i'm going to multiply it by the x minus 4. we go we can do it all from here now so just following the same process as before 6 x squared times x gives us 6 x cubed 6 x squared times negative 4 is negative 24 x squared and then moving on to the underneath 5x times x gives us plus 5x squared and 5x times negative 4 gives us negative 20x and then on to the negative 6 to finish it off negative 6 times x is negative 6x and negative 6 times negative 4 is positive 24 and then we just need to tidy all this up again so we've got the x squareds there and the x piece is here and let's just rewrite all that nice and tidy so 6x cubed negative 24 add 5 is negative 19 so negative 19 x squared negative 20 take away 6 is negative 26 so negative 26x and then plus 24 at the end and there's how to do that one so keep it all really nice and tidy expand whichever two you want to start with tidy it up with your third bracket at the end once it's been simplified and then just go step by step really nice and carefully before you tidy it all up at the end and remember what your answer should look like you will get these cubes to squares the extra numbers sometimes they might cancel each other out but in none of these they have so when we're simplifying an algebraic fraction we're looking for a common factor on the top and the bottom when it comes to algebra to find a factor in algebra we can factorize to find a factor so if we have a look at the top and we just need to factorize both of these now the expression on the top there has an x in both so we can factorize this by x no numbers in both but we can factorize it by x so if we take x out the bracket we would have x and then inside the bracket we'll have x minus sorry plus five okay x plus five if we do the same on the bottom we've got a quadratic on the bottom so you need to know how to factorize quadratics here uh but we've got one and ten or two and five as our options to go in our double bracket and in order to make seven there in the middle we want plus two and plus five and there we go x plus two and x plus five there is a little bit of a hint there in terms of the factors we'd be using because obviously we had x plus five on the top but it's not always going to be the case that we'll get that but obviously so we still need to be able to factorize but there was a little bit of a hint there there was going to be an x plus 5 in one of the brackets now there's an x plus 5 on the top and the bottom so essentially what we can do is we can divide the top and the bottom by x plus 5 which would cancel out these brackets here and if we cancel them out we just need to write what's left we have x on the top and on the bottom we have the x plus two which you can either put in the brackets or leave without the brackets it doesn't matter there but our final answer is x over x plus two all right let's have a look at another one okay so we've got another algebraic fraction here that has already been factorized on the bottom so actually let's have a think about how we'd approach this now we want to cancel off a common bracket or a common factor and the top and the bottom both divide by x plus two so actually if we imagine what this would look like if we were to sort of expand that bracket or just write it as a double bracket we'd have x plus 2 on the top on the bottom it's x plus 2 and x plus 2 as a double bracket there we go right so if we cancel off one of the x plus twos top and bottom we're just going to be careful this time because we can cancel that one off and it'll disappear but not to forget we are dividing by x plus two so when we cancel the one off the top it doesn't just disappear but if we divide it by x plus two it becomes one so writing our final answer we would have one on the top and x plus two on the bottom which again i'll just get rid of the bracket there so one over x plus two so just watch out with these sorts of questions here when the factor is all that there is on the top obviously that does become a one there it doesn't just disappear okay so introducing some negatives same process though we'll stick two x's in one bracket and x in the other as we've got two x squared and again writing down the factors of ten so one and ten or two and five now again one of them is going to double so we could have two and ten or we could have one and twenty are we gonna be able to make nine out of either of those no two and ten we could make eight we could make twelve but we can't make nine and then one and twenty we could make 19 or 21 but we can't make nine so it's not going to be one and 10 which means it's gonna have to be the two and the five gotta be a little bit careful here because we've got negative nine in the middle so we're trying to make negative nine but similar processed when we're factorizing before so let's double one of them so we could have four and five or we could have two and ten there we go just doubling one of them so again um we're trying to make negative nine so it's gonna have to be four and five we can't make nine out of two and ten but we can out of four and five we've just got to figure out okay well which one of these needs to be negative and actually looking at it look we've got a positive ten at the end so they're both gonna have to be negative so it's gonna have to be negative four and negative five so i just wanna make sure that that number above the negative four the two goes in the bracket that's going to be doubled so on the right hand side there we want negative two that's going to allow us to get negative four and on the left bracket there we'll put the negative five and again you can just expand that to just double check so 2x times x is 2x squared 2x times negative 2 is negative 4x negative 5 sum x is negative 5x and then negative 5 times negative 2 so gives us the positive 10 at the end there and tidying that up look these two here do make my negative 9 we get 2x squared minus 9x plus 10 and it matches what we were looking for now for this one here it says simplify this fully and inside the bracket we've got 32x to the power of 5 and then outside it's all to the power of negative two-fifths now obviously when it comes to these negative fractional indices the negative there is gonna do the reciprocal the five on the bottom does a fifth root and the two on the top is a normal power of two and both of these pieces we have to do to that power now when it comes to the 32 we can think about the 32 on its own for the moment and we can do 32 to the power of negative 2 over 5. now we can sort that out and deal with that first now the negative obviously flips it over but before doing that and doing the reciprocal i'm going to do the fifth root of 32. now the fifth root of 32 is 2. okay so 2 times 2 is 4 times 2 again is 8 times 2 again is 16 and then times 2 again for that fifth time gets us to 32. so to start with there we'll do the fifth root i'm just going to write this down here fifth root so that's two then we're going to have to square it because we've got the two on the top so we need to do two squared so that becomes four and then if i do the reciprocal there it becomes 1 over 4 and that's what that negative power does so that 32 there is going to become 1 over 4. so if i write that down to start with i can write down 1 over 4 and then we'll deal with that power now when it comes to the x to the power of five we can obviously multiply the powers when we've got brackets like this so we're going to multiply these powers together so we've got five lots of negative two-fifths and five lots of negative two-fifths will be negative ten-fifths okay just multiplying that numerator by five and negative ten divided by five is negative two there we go so my power is going to turn into negative two so i can write that out here we've got then x to the power of negative two now technically that is the answer that would be fine but there are other ways of writing this as well and obviously we can write this in a slightly different way because x to the power of minus 2 can also be written as i'm thinking about this over here x to the power minus 2 can also be written as 1 over x to the power of 2. remembering that that negative power there does the reciprocal so it's 1 over x squared so i mean we could actually write this in a different way if we imagine x to the power of minus 2 is 1 over x squared we've got a quarter remember these are getting multiplied together so it's a quarter times 1 over x squared and if we multiply those together we get 1 over 4x squared now for this particular question here it does just say simplify fully so we can write it in any of the either of these ways but it might be for whatever reason we might want to just write it as a single fraction so it's always good to be able to know how to write these in different ways so there we go we can also write it as one over four x squared or we could just leave it as a quarter x to the minus two but there we go that's how we're going to approach some of these questions let's have a look at a slightly different one okay so this question doesn't look very nice but you're going to see we're going to treat this in a very similar way to what we did before now again ignoring that x squared for the moment let's just deal with what's in the brackets now everything in the brackets there is to the power of two so we've got the four that we need to do to the power of two and that's easy enough to do four squared is 16 so we've got 16 there in the bracket and then we can multiply these powers that are in brackets as well or times them and if you multiply negative a half by two you get negative one so that's going to be 16 x to the power of negative one and again obviously you can write that working out down to the side you've got minus a half and you're gonna multiply that by two or two over one which comes out as minus two over two which is minus one there we go so there we go that's how we can get our power obviously just thinking about negative a half times two is negative one and now we can obviously put that in a bracket and we can think about that x squared on the outside there we go so obviously when we are multiplying obviously thinking about the fact that the coefficient of this x squared on the front is 1 and then we just multiply this bracket out so 1 times 16 gives us 16 and then we've got the x squared times the x to the power of minus 1 and remember obviously when you're multiplying with powers you add these powers together so we're going to add those two together so x to the power of two add the x to the power of minus one uh obviously just the powers there becomes x to the power of one there we go so 16 x to the power of one and obviously we don't need to write the power of one so we can just write that as 16x and there we go that would be our final answer for that one there we go so something that looks quite nasty actually simplifies down something that's nice and simple for us okay so this question here doesn't look very nice at all but again we're just going to treat it exactly like we have done in the previous questions we're going to deal with the top and then we'll have a look at the fact that it's a fraction so that top bit there can simplify so straight away let's have a look at this so on the top we've got 2 to the power of 3 which is 8. i'm going to multiply those powers okay so 3 lots of a half and then again you can just write that as three over one times a half and you get three over two or three halves there we go so that's eight x to the power of three over two on the bottom nothing simplifies there so we'll just leave that as four x squared there we go and now all we have to do is simplify this down so looking at the numbers we've got 8 divided by 4 8 divided by 4 is 2 and then we are dividing with these powers of x when we divide and we take away the powers so we've got 2x to the power of something we just need to figure out what that is so obviously what we've got there is we've got 3 over 2 and we need to take away 2. so we're going to take away 2. obviously 2 is 2 over 1. if we want to write that in terms of halves we'll have to just change that denominator there just obviously that'll become three over two take away and if we times both of these by two there you go that becomes four halves now when you take those away you get minus a half there we are so just taking away those powers there and you might be okay just doing that without obviously writing all the fraction working out but obviously just to show you there you can just obviously apply a bit of written methods as well but there we go we get 2x and our power is going to be minus a half and again you might be able to just do three over two take away two in your head but there we go so two x to the power of minus a half and again that is a final answer then we can leave it like that but there's another way that we can write this and obviously this is going to be important for thinking about it in reverse as well so x to the power of minus a half well x to the power of a half is the square root of x so that's that'll be the square root of x but it's a negative a half so that'd be one over the reciprocal of that so one over root x so actually what we've got here is we've got and 2 multiplied by 1 over root x there we go and don't forget 2 as a fraction would be 2 over 1. so multiplying the top there we get 2 on the top now multiplying the bottom we get the square root of x so again two different ways of writing your answer you can either write that as two x to the power of minus a half or two over the square root of x either one of those is absolutely fine although i do think it's important to be thinking about how you can write your answers in different ways purely because as i said sometimes they might be going the other way round or you might have a square root on the bottom and have to rewrite it as a power so it is important to be able to write them this way okay here we go so expand and simplify this so slightly different look we've got a number in front of this 2 root 3 here and also this is a slightly odd one because we've got a certain here which isn't fully simplified okay we've got a root 8 there now you don't have to simplify at this point we can see if we can simplify it later on you won't be wrong to simplify it now but i'm just going to leave it and see what happens okay so just to know there are some slight differences in this question but let's have a look 5 times 3 is 15. 5 times negative root 8 is negative five root eight two root two times three remember numbers times numbers thirds times third so it's just two times three which is six so plus six root two and there's no root with the three so just root two and then finishing the last part off is positive times a negative so i've got negative i'm just going to work this out to the side two times the one there is two and root two times root 8 so we get 2 root 16 negative 2 root 16 and root 16's 4 so that's negative 2 times 4 which is negative 8. okay well if we do that to the side here look two times four which is going to be eight so let's just get rid of that working out there and put negative eight okay we can start to tidy this up now we've got 15 take away 8 which we can do but these thirds aren't gonna i'm going to simplify now so what i need to have have a go at doing is first simplifying this because we've got a 5 root 8 there and that can simplify so again i'm just going to do that to the side five lots of root eight so ignoring the five root eight is root four times root two which is two root two and again just remembering we did actually have 5 lots of that so 5 lots of 2 root 2 is 10 root 2. so rather than writing this 5 root 8 here i'm going to write 10 root 2. so i'm going to rewrite this question i've got 15 minus 10 root 2 plus 6 root 2 minus 8. so 15 take away 8 is 7 and then minus 10 lots of root 2 add 6 lots of root two leaves me with negative four lots of root two right there we go so slightly harder there we've got little bits going on where we've gotta just be careful to simplify um but there are little hints there that we you know that we're spotting the root eight to start with that there's going to be some simplifying going on the harder ones here purely because we've got this little three in front of the root two now on the bottom we're gonna take the same process it's gonna times the top and bottom by what's on the bottom there but flipping the sign so i'm going to times top and bottom by five plus three root two and then again five plus three root two so two plus root two on the top multiplied by five plus three root two and then on the bottom we have five minus three root two multiplied by five i put the multiply in there because i'm speaking it as i say it multiplied by five plus three root two and then again just take this nice and slow so two times five is ten two times three root two is six root two root two times five is five root two and root two times three root two is three root four or three times two again you can write that slowly three root four which is three lots of two which is six and there's the top and then the bottom i'm going to apply that little shortcut so five times 5 is 25 and negative 3 root 2 times positive 3 root 2 is negative 3 times 3 is 9 root 2 times root 2 is root 4 so it's minus 9 times 2 which is minus 18. and then we just need to tidy that all up so the 10 plus the 6 on the top and the 16. the 6 root 2 add the 5 root 2 is 11 root 2 and then 25 minus 18. is 7 and then neither of those simplifying nothing on the top divides by 7 there so that would be our final answer for this one so not any negatives on the bottom this time question here we've got a 2x squared so again i'll link in the description the video for obviously factorizing harder quadratics so we'll explain them a little bit but i won't go into as much depth as i did in that video so do check that one out if you're not sure on how to factorize these now in terms of a 2x squared that means in one of our brackets here we're going to have to have a 2x when we're factorizing it so if i set the brackets up we've got a 2x and an x it's all equal to 0 and we just need to figure out where the numbers go now if we've got a 1 at the end our only factors can be 1 and 1 but one of them is going to be doubled okay obviously when we expand our double bracket here we've got a 2x and it's going to double that one in in the other bracket so as they're both ones we know they're both going to be ones here we just need to figure out the symbols now we want minus 1x or minus 1 in the middle so in order to make minus one we're gonna want plus positive one take away two so as this one over here is going to be doubled we'll make that one become our minus two and this will be our positive one here and obviously you can go about expand uh expanding that just to check that it matches but there we go that is going to match we're going to get plus 1 on this expansion here plus 1x and we're going to get minus 2x when we expand that one there that'll give us minus 1 in total so in terms of obviously getting our solutions here it's very similar to what we just looked at if we just set them both equal to zero this one here is just going to become x equals one obviously we haven't got a coefficient there and this one here we're going to flip the sign so x is gonna equal minus one but then we're gonna divide it by the two so minus one over two or minus a half and there's our two solutions x is minus a half and x equals one and again you could obviously draw your sketch of your quadratic there just think about what it looks like it's gonna go through one and minus a half and you could draw that in there we go one and minus a half and that's where our two solutions are now in this one the process isn't any different but we've got something that's going to change here i just want to show you this and it's negative 11 here in front of x so if i write down the values of a b and c we have a equals 3 b equals negative 11 and c equals negative 13. and if i go ahead plugging all these values in i've got to have a look at what changes here so the start of our quadratic formula is minus b but b is already negative so if b is already negative that's going to turn into positive 11. so on the top there we have positive 11. now i don't have to put the plus with it there but i just want to emphasize for you it's going to be plus 11. okay so minus b if b is already negative that's going to become a positive number there so we have 11 plus or minus the square root of and again i've got to be careful here because i've got to do b squared now when i'm squaring a negative number and subbing a negative number in i should always really put it in brackets so negative 11 squared reason behind that and hopefully you've got your calculator in your hand if you type in negative 11 squared without the brackets you're going to get negative 121. and a negative times a negative makes it positive so your calculator is not going to do it in the correct order for you it's going to do the um the power before it does the the minus okay so you just got to be very careful it does the order of operations so if you want to make sure it squares a negative number stick it in a bracket that's very very important when b is negative so we've got b squared minus 4 ac so 4 times 3 times negative 13. okay let's just extend that there we go that's all over 2a to all over 2 times 3. again on this question look that and that was new this time the minus b turning positive and the negative b in the bracket there when you're squaring it so being very very careful with that but there are new bits that you need to be careful on this type of question so let's type this in again we'll do the positive and we'll do the negative there we go and we'll stick these values in so fraction button positive 11 so just 11 plus the square root of in brackets negative 11 squared take away 4 times 3 times negative 13 all over 6 or 2 times 3. press equals and turn it into a decimal you've got point six zero seven two one nine four nine six and again finishing that off by rounding it to three significant figures here again we've got four point six there we go and there's our first solution moving on to our next one back into the calculator takes a little bit of a while to click back there we go top it for a minus and we get minus one point five eight double six zero six eight seven five there we go and then finishing this off rounding it again that becomes negative one point five nine because it's a six there after the eight and there's our two solutions so when it comes to this question here obviously this is what we're going to complete the square for and we're going to find the turning point to start with and have a look at how we apply it for that as well so when we've got this one here now this is our quadratic that we have at the moment 2x squared plus 16x plus 26. now we can only complete the square well i'll say so we can only we we have to get the coefficient of x squared to be one first before going about completing the square so what we're going to do is we're going to factorize it by a factor of 2 okay so that as they all divide by 2 there or just because x squared has a 2 in front of it we're going to take a factor of 2 out so if i factorize this by 2 i get 2 lots of and then dividing everything by 2 we get the 1x squared we get 8x and we get 13 at the end there we go now what i'm going to do is i'm going to basically ignore that that 2 is at the front so all i'm going to do is factorize or complete the square for what's in the middle here okay obviously we've factorized it they've taken out a factor of 2 and now we're going to complete the square for what's inside so normal process for completing the square i am going to halve the coefficient of x so in my bracket i'm going to have x plus 4 obviously halving the 8 there and that is going to be squared obviously remembering why we do that because when we expand that we get the x squared we get the 8x but at the end if we expand it we'd get plus 16. now we don't want plus 16 we want the plus 13 there so we're going to have to take away 3 to make this balance out so i'll have to put minus 3 at the end and now if i did expand and simplify all of that i would get the x squared plus 8x i'd get plus 16 take away the 3 and that'll leave me with the 13 that i'm looking for now obviously we can't forget that that 2 exists that 2 is still at the front there and we have to bring that 2 back in now i'm going to do is i'm going to put all of this in a bracket and rather than using a rounded bracket i'm going to put it in a square bracket just to define obviously it's a different set of brackets there and i'm just going to put that two back at the front all right there we go well we're almost done all i've got to do now is reintroduce that two and that just means expanding out this square bracket by that two and that's easy enough to do we've got two lots of uh the bracket there so i'm just going to write this out let's see what we've got we've got two lots of the x plus 4 squared there we go and then timesing that negative 3 by 2 will leave us with negative 6. and there we go that's that completed in completed square form all we have to do now is obviously like before we identify the turning point from what's in the bracket there so just like before the x coordinate is right here but we flip the sign so that is going to be negative 4 as our x coordinate and our y coordinate is just at the end there the minus six so there we go our y coordinate is minus six there we go and that's the coordinates of our turning point so very very similar to obviously normal completing the square but obviously we just have this extra element there of factorizing out that coefficient of x squared and then reintroducing it at the end and just multiplying that number at the end there that what was normally our y-intercept um or i'm sorry our y-coordinate and actually just multiplying that back out by the factor that we took out at the start so this question we've got 2x squared at the start okay and obviously when we're completing the square here we do have to take in to take that into account and we are going to first complete the square it then says hence solve the equation so just like the last one and given your answer in the form a plus b root three where a and b are integers so let's start with completing the square so we'll take out that factor of two so we have two lots of x squared minus four x minus eight okay and then let's complete the square for what's inside i'm gonna swap colors like i've done before there we go so half that coefficient of x and we have x minus two in bracket squared and that's gonna make plus four at the end and we want minus eight so we're going to have to subtract 12 there to make that balance out reintroducing that too let's see what we get there we go reintroduce the two and we have two lots of x minus two squared minus 24. now for the second part of this question here it says hence solve the equation so we're going to do exactly the same thing we're going to set it equal to zero so if i set this equal to zero and i'll do it separately up here we have two lots of x minus 2 squared minus 24 equals 0. now it's completely up to you in in terms of the steps that you do here but there's a couple of things that we can get rid of from the start either we can get rid of this 2 at the start we could divide everything by 2 or we can add this 24 over to the other side now personally i'd just like to add the 24 over first but it's completely up to you in terms of which order that you do it i'll show you why it doesn't really matter but if i add the 24 over let's have a look we get 2 lots of x minus 2 squared equals 24. now at this point i'm going to divide by 2 so that's going to turn that 24 into a 12. now you could have divided by two at the start it wouldn't obviously affect the zero over here zero would still be zero but you've just got to remember if you do divide by that number you also have to divide the 24 by two because that's going to become 12 on the other side and you've got to make sure you divide everything by two on that left hand side of the equal sign so personally i just like to add it over first i feel like it just helps you not to forget to also divide that by 2 as well so if we go about dividing by 2 now we can get we'll get x minus 2 squared equals 12. and then it's exactly the same as the last question now we can square root both sides and we get x minus 2 equals plus and minus the square root of 12 and then we can add that 2 over to finish it off and if we add the two over let's see we run out of space a little bit here but we get x equals 2 plus or minus the square root of 12. obviously we do just need to have a look and let's just bring this up here we do need to have a look at obviously writing in the form it wants and it wants it as b root 3. so we need to think obviously about that root there and that is root 12 and how does that simplify well let's have a look root 12 the square number that goes into that is 4 so we can write that as the square root of 4 times by the square root of 3 and the square root of 4 is 2 so that's 2 root 3. so finishing this off then we can write it in the form it's asking for and that is the x equals 2 plus and minus 2 root 3. and there is our final answer writing it in that third form that it's asking for in this question now just a very quick discussion of this before you maybe try and have a go but you will notice that the coefficient of x for one is an odd number so when completing the square that's going to form a decimal although you can use a calculator for this question but there is going to be decimals involved and a little hint for you this one doesn't factorize and when a quadratic doesn't factorize you need to use the quadratic formula so this is a calculator question the rest of those could have been non-calculator but this one here is explicitly calculator so a couple of little hints there but i would encourage you to have a go this is our last little question so if you pause the video and have a go otherwise let's have a go going through this one so if i go about completing the square to start with it wants us to draw a sketch showing the coordinates of the turning point and any intersections with the axes so let's take out that factor of minus one so when we do that we get x squared plus three x minus five now if we complete the square for this it's not too difficult particularly as we have a calculator but we get x plus 1.5 you could leave it as a fraction 3 over 2 squared now when you expand that bracket 1.5 squared is 2.25 and we need to get back to negative 5. so you can do that on the calculator if you want if you need to but we're going from 2.25 down to zero and then an extra five so we're gonna have to take away seven point two five to get down to minus five now we need to multiply that minus one back in just like we have done before to put that minus one back on the outside and we get minus one lots of the x plus 1.5 squared minus 7.25 and obviously when we times that minus one in it becomes plus 7.25 so there we go we've got the coordinates of the turning point not too bad there hopefully minus 1.5 the opposite in the bracket and then 7.25 for our y coordinate and again we can solve it from here and if we go about doing that um let's see what we're going to get now we could solve it from there but we should really just solve it from the original quadratic once we've identified the fact that it doesn't factorize we may as well just use the original quadratic because it's not really going to matter which one we use there so if we bring the original quadratic down and we've solved from there we've got our values of a b and c that we're going to have to put into the quadratic formula so a is the number in front of x squared which is minus one b is the number in front of x which is minus three and c is the number at the end or at the start in this case which is the five and if we plug all that into the quadratic formula so obviously you need to know the quadratic formula for this as a while so i'll put that in the description don't worry everything that's going on in this will be linked in the description for you to have a go out i'll try and put them in order for you as well for how they've occurred in the video but if we put those into the quadratic formula let's have a look so we have minus b at the start so that's positive three now when we flip that plus and minus the square root of b squared so negative three in brackets squared take away four times negative one times five and that's all over 2a or minus 2. 2 lots of minus 1. now if we type that into the calculator obviously using the fraction button then just being careful how you sub all these numbers in using the negative button for your negative 1 putting your negative number in bracket there for the b squared for the version with the plus sign we get and i'm gonna round these to two decimal places for this one it doesn't actually say two in this question but for the purpose of our sketch we're not gonna wanna write all these decimals it comes out as minus four point one nine two decimal places and if i go back into my calculator change that plus for a takeaway our version with the takeaway comes out as and again two decimal places 1.19 so there we go um obviously in a question like this we would just leave it as a full decimal if it didn't say how to round it but for the purpose of this little bit of practice we're not going to worry too much about that so obviously we've got a negative and a positive solution again so if we draw our quadratic again we're just going to do a little sketch and it's going to go let's have a look minus four over to one i'm trying to make it as accurate as i can or not not over egging the uh whereabouts is on the graph too much but there we go we've got minus 4.19 over here we've got positive 1.19 we've got a y-intercept which again we can get from our original equation of five and we have our turning point which is just here not the best drawing there by me but minus 1.5 7.25 just about get that on there okay obviously just emphasizing the fact that that turning point is to the left of the y-axis as we have a negative 1.5 as the x-coordinate so there we go real really tiny little sketch there at the bottom but just showing you how this could obviously get a little bit harder obviously with completing the square with an odd number as the coefficient of x but if you have a calculator that's nice and easy enough and then obviously using the quadratic formula as well so it was a pretty tough question to finish on now the discriminant is something that you have met before uh usually obviously from gcse maths you'll have met the quadratic formula and it's something to do with something within the quadratic formula so if we have a look and just have a quick think about the quadratic formula which is minus b plus and minus the square root of b squared minus 4ac all over 2a now normally particularly at gcse level when you've looked at the quadratic formula it's very rare that anything other than a positive number comes up underneath that square root although you will have noticed before maybe you've typed it incorrectly sometimes and you may have found that obviously sometimes it came up with a maths error on the calculator when you type in something under here and you've got to be very careful with the numbers that you type into there but obviously whether depending on whether it's a certain type of number depends on whether we get an answer on our calculator or we don't and that's what we're going to have a look at and actually understanding this bit now that bit on its own underneath the square root that b squared minus 4ac is what we call a discriminant and it determines when we have a quadratic how many solutions we have and whether we even have a solution at all and we're going to have a look at why and how that happens and a few particular questions here and the differences between them but essentially if we forget about the rest of the quadratic formula for the moment we are just concerned with this bit this b squared minus four ac and that is what we call the discriminant okay so we're just going to be having a look at this bit um so i'm just gonna write that up here we've got b squared minus four ac and that is called our discriminant now so grab a piece of paper grab a pen make some notes we're going to get started with these questions here and just have a look at understanding why and how we get our solutions from this little bit under the square root now if you think about the concept of numbers underneath a square root okay we can have three different types of numbers we can have a whole number and let's pick something that actually um square root something like 25 and the square root of 25 would give us two solutions we can get plus five and we can get minus five so we get two solutions when there is a whole number under there sometimes it will be an actual whole number sometimes it might be a decimal okay but we always get two solutions with a whole number we can also have the number zero and if we get the number zero obviously the square root of zero is zero but thinking about the rest of the quadratic formula that would mean we'd only get one solution because obviously in front of the square root there you have the plus and minus and if both the plus and the minus are both zero then both of our solutions are going to be equal or we could have and thinking about when this arises you could get a negative number under there so something like negative three let's just put something random in and you will get no answer on your calculator it would normally say calculator error and that's because we'll get no solutions or in terms of a quadratic when we're solving that we would call that no roots or no real roots okay no roots there we go i'll put that in there so thinking obviously about the concept of the quadratic formula and obviously i will link the video for that in the description if you need to touch up on the quadratic formula but in terms of the type of numbers that go underneath that square root we'll determine what answers we get when we look at these solutions and that's what we're going to have a look at with some of these questions so this first one that i've got on the screen i'm just going to get rid of all of this is looking at a particular quadratic and i have put a picture of this quadratic up here and we'll see obviously that we get these two roots or two solutions here on the x-axis so the equation for this particular quadratic is up here x squared plus four x plus two and in terms of actually figuring out the discriminant of that and we're going to work out the value of the discriminant we just need to plug in the values of a b and c now a is the number in front of x squared hopefully already happy with that and a in this particular case is one b is the number in front of x so b is positive four and c is that number at the end and c is two and if we stick all of these numbers into the discriminant so b squared minus 4ac let's see what we get so we've got 4 squared our value of b take away and then we have 4 times a which is 1 times c which is 2. and if we just work out this calculation here four squared is sixteen take away four times one times two which is eight we get sixteen take away eight and we get the answer eight so the value of our discriminant is eight and if we think about that if it was put into the quadratic formula we would have plus and minus root eight which would give us our two values underneath the square root there again that would give us our two solutions and ultimately determine these two solutions okay so in terms of stating whether this uh obviously without the graph has two roots two equal real roots which we'll discuss in a second no real roots we know that obviously with the positive eight there with the plus and minus square root of eight we would have two roots and we call those two distinct roots okay but i'm just going to put two roots here for this particular example and we can obviously see that from the graph as well but there's two elements there we've got the value of the discriminant and the fact that it has two roots which obviously is also shown on the graph so this question it says 4x minus 5 minus x squared also equals or is equal to q minus x plus p squared where p q and p are integers so obviously that there is highlighting the fact that it can be written in completed square form like that which looks ever slightly different to the way that we've been writing it in the earlier questions but we're going to have a look at how to actually write it like that so it says find the values of p and the values of q and then it starts to ask us some other little bits it says calculate the discriminant and then it also says sketch the curve showing any points of intersection so we'll have a look at those points step by step to start with we're going to have a look at obviously finding the values of p and q and writing it in this completed square form just here so let's write it out so we've got negative x squared and obviously this is a bonus because we've already discussed how to complete the square for this okay but let's have a look minus five so if we take out that factor of minus one to complete the square we'll get positive x squared minus four x plus five and completing the square for that let's obviously halve that coefficient of x so we get x minus two in bracket squared and when we expand that we get plus four we want plus five so we need to add in an extra plus one there and then again not forgetting to reintroduce that minus one because that affects that y coordinate at the end there so minus one and we end up with minus one lots of x minus two squared and then that turns out into minus 1 at the end now obviously it wants it written in this form up here okay q take away those so all that means is putting minus 1 at the start there okay and just shifting this minus one from the end and putting it here at the start and if i write it out like that we have minus one take away the one lots of that bracket so i can get rid of that one there you go and there it is minus one take away x minus two in bracket squared so now it's in the format that we want we can write out what the value of p and what the value of q is so q is at the start so q is the minus one there at the start and p is negative 2 inside the bracket there we go and i always dislike that when they write it like this look they've written plus p just here and actually p's come out as a negative there p is negative 2. so do just watch out for that just because they put plus p there doesn't necessarily mean it's going to come out as a plus or a positive number so the next bit right it says calculate discriminant of this okay now when we're looking at the discriminant and again i've linked the video for that in the description we're looking at the value underneath the square root in the quadratic formula that b squared minus four ac which is called the discriminant and it determines for us how many roots this is going to have so rather than getting us to actually factorize this if it does factorize and finding out whether there are two distinct real roots two equal real roots or no real roots we're actually working out the discriminants okay so if we put these values in then let's see what we've got so obviously um we need to get the values of a b and c from here now i'm going to take it instead from up here where i've rearranged it so the value of a is that one in front of x squared there so a equals the negative one in front of the x squared we've got b which is in front of the x so the positive four so b equals four and then we've got to forget c as well which is the number at the end which is minus five so there we go c equals minus five and if we sub all these values into b squared minus four ac let's see what we get we get four squared take away four times minus one times minus five there we go and if we work that out let's have a look just multiplying all these bits together to start with four times minus one is minus four times minus five is positive 20. so we have 16 take away positive 20 or 16 take 20 which is minus 4. so my discriminant has a value of minus 4. now obviously what we know about the discriminant obviously hopefully you've already checked that video out and you know about the discriminant if we get a negative number that means that means we have no real roots so in terms of this quadratic that we're going to draw it's not going to cross through the x-axis and that's what that tells us and obviously if we were asked to draw this curve and we weren't asked to calculate the discriminant we might be sitting there trying to factorize it we might be putting it into a calculator trying to get in quadratic formula but actually very quickly we can tell using the discriminant whether it does actually have any roots so it's a nice quick way for us to figure that out and obviously this question asks us to do it anyway but just thinking about another type of question if it didn't ask us to calculate the discriminant uh you know it would be a nice quick way just for us to see whether it does factorize or not and that's quite nice about the discriminant as well so if we go about actually drawing this if we want to sketch this curve showing clearly any points of intersection with the axes let's have a think about what it would look like we know that it's going to be an upside down or an n shaped quadratic curve here so if i just draw a little basic axes one thing we do need to get is obviously that y-intercept which we've got up here the y-intercept is -5 so that's going to be down here somewhere so i probably want to draw this a little bit better actually let's get rid of that let's draw it so we can get actually down the bottom there so we know it's going to be -5 we know it doesn't even reach the x-axis there we go and the only thing we need to figure out is whereabouts we're going to put it now we can we can think about where the maximum point is because we've completed the square up here so we've completed the square and we have got from that just here we've got positive two and minus one as our turning point and if we go about actually finding that let's imagine two's there minus one is there so it's around here there we go so if i try and draw this in then we've got it crosses through at minus five we've got a maximum point there and it never touches the x axis and obviously just labeling this on we've got minus five there and that's all we need to label doesn't ask us to label the maximum point there we could put it on if we wanted but there we go there is a sketch of what this curve actually looks like there we got the minus five the y intercept up from our uh equation up here okay well let's write that in y equals minus there we go and we've got the fact of where the turning point was going to be positioned just here from completing the square again and then using the discriminant we figured out there was going to be no roots so we know it wasn't going to get up to the x-axis there so there we go that's how we go about all of these points obviously completing the square to find the turning point to get an idea of whereabouts is going to be positioned uh obviously then uh getting the y intercept from our equation and using the discriminant to figure out whether it was going to touch the x-axis or not okay so on to our next question now again there's a lot of words here so i would advise you pause the video and write this all down and have a good read through but i'm going to assume that you've done that now so we're going to carry on and have a look at this question so it says an arena has 25 000 tickets they know from past events they're only going to sell 10 000 if the ticket costs 30 pounds they also expect to sell an extra thousand tickets for every time the price goes down by one pound and the number of tickets sold given by t can be modeled by this linear equation t equals m minus a thousand p where p is the price of each ticket in pounds there and m is a constant and for starters here it wants us to find the value of m it then goes on to want us to write it in completed square form and use that completed square form to think about that turning point again looking at the maximum amount of money they could charge or they or the most that they could charge in order to maximize their amount of money so obviously i'm hoping that you've written this down you've had a good read so we're going to move everything to the side and here we go now we can get on with answering the question so looking at this one then for starters we've got to find the value of m and we're given this equation t equals m minus a thousand p and it says that let's make sure it's a thousand and it says that m is a constant so we just got to find out what that number is really so we need a value of t and a value of p in order to solve that and if we read our question here we've been given one of those values t is the amount of tickets they expect to sell and we've already been told they know that they're going to sell 10 000 tickets but on the basis that each ticket costs 30 pounds so we have a value of p as well so what we actually have to do to find the value of m is sub those values into our equation so if we put them in t is ten thousand so ten thousand is going to equal m minus one thousand multiplied by which i'll put in brackets thirty so let's just look at that and expand that so 10 000 is going to equal m take away 30 000 and hopefully you can spot straight away there that value is going to have to be 40 000 because 40 000 take away 30 000 will equal ten thousand so for part a i'm going to write this over here so it stays on the screen m is equal to forty thousand there we go so we've got our value of m okay so moving on to part b it says we have the total revenue r can be calculated by multiplying the number of tickets sold by the price of each ticket and this can also be written as r equals p brackets m minus a thousand p this information just there obviously we can expand that that's just given to us in a bracket to start with it wants us to rearrange it into the form and then i want to want it in completed square form so in order to do that we just need to expand that bracket and go about rearranging it into that completed square form and if we do that let's get started and move all of this out of the way first of all we just want to expand that bracket so let's go about expanding the bracket so we'll get r is equal to mp and again you could put the p first but we've got the p at the end on the next piece i'm just going to keep that consistent mp minus 1000 p squared now we already know from part a that m is 40 000 so we can put 40 000 p rather than mp i'm going to get rid of the r equals for the moment just so we can actually look at completing the square for this so at the moment we've got 40 000 p minus 1 000 p squared and notice that we obviously no longer have a constant here so when we complete the square we just need to watch out for that as well so let's rearrange that so that the p squared is first so we have minus a thousand p squared plus the 40 000 p there's a lot of zeros here there we go 40 000 p and to complete the square we need to take out a factor this time of negative a thousand not forgetting you've got your calculator so that's not too bad so we'll start off by taking that factor out so if we do that we've got minus a thousand outside the brackets and in the inside here we've got p squared that's now going to be minus and forty thousand divided by a thousand is forty p there we go so now we just need to complete the square so go about completing the square for that we're going to introduce a new set of brackets so one third minus a thousand and then we'll complete the square so halving minus 40 gives us minus 20 in bracket squared if we were to expand that bracket we would have plus 400 and we obviously want to get back to zero so we just want to minus that 400 and that's going to balance out and complete that square there and then closing off our brackets okay so we've pretty much finished all we actually have to do now is multiply that 1000 back in and if we multiply that back in let's see what we get we get minus a thousand lots of the p minus 20 squared and then negative 1 000 times the negative 400 gives us plus 400 000 there we go so we've got it in completely square form obviously the question wants it written slightly differently though it does want that 400 000 at the start so in order to do that let's get rid of all of this and we'll just write it in the correct way and then we can have a look at part c without all this information everywhere so that is our second to last step we just need to write it in the correct form and that correct form is going to be 400 000 [Music] minus the 1000 p minus 20 squared there we go so that is part b it's in the form that it wants and we could say that a is 400 000 b is 1000 and c is 20. and now we can actually look at part c so let's just label that as part b and here we go part c so it says using your answer to part b or otherwise work out how much the arena should charge for each ticket if they want to make the maximum amount of money and again you don't have to draw a quadratic for this but it fits in nicely just obviously thinking about what a quadratic looks like and it just helps us to visualize it so if we think about this we have t going up here which again always refer back to the question t was given to us at the start and it says where is it t is the amount of tickets sold so the amount of tickets sold is at the top and along the bottom here is going to be our price which is obviously what we're looking at right now the price of the ticket so we already know that in our completed square form we've got our constant right there and we have our 20. so that gives us the coordinates of the turning point so if we think about what this would look like we'd have some form of quadratic we're going to have to sell some tickets um but again we could sell zero tickets if the price was incorrect but if we were to draw a quadratic we don't know exactly what it's going to look like let's just imagine it looked like this and we would be looking at this particular point just here where the amount of tickets is maximized and it's not going to start going back down because we've put the price too high so for this one here um let's have a think we've got 20 there as our x coordinate nothing for not forgetting to flip the sign so 20 and 400 000 for our y coordinate there we go so 400 000 there we go so on the y axis just about fit that in 400 000 just about fit that in and the price down here is going to be 20 and that is where we maximize the amount of profit that we're going to get or they maximize the amount of money for each ticket right at that turning point so the price that we've got there is 20 pounds so the price that we want to sell our tickets for is 20 pounds after that the amount of tickets that we sell is gonna start to go down okay obviously the quadratic isn't going to look exactly like that but it's just a visual example of looking at the turning point and thinking about obviously the the point at which it starts to go back down again after the price has increased too much obviously if we put the price of zero we'd expect to be having a lot of tickets sold so it wouldn't look exactly like that but it's just a nice little visual reminder so there we go that is how to complete these sort of questions and how to just draw a little diagram just to help yourself visualize the turning point so let's label it equation one and two and you've probably noticed in equation two there we've got this negative 5.5 here so we're not going to get a very nice answer here we'll treat you in exactly the same way so we're going to get the y coefficient to be six here so with three and two so let's times the top one by two and the bottom one by three and we'll rewrite them so 8x becomes 16x plus six y equals two times two is four so four times the bottom one by three so nine x minus six y equals negative and let's times that by three we get sixteen 16.5 and now we can just go about adding them together because we've got our different signs there so we're going to add them together so 16 of 9 is 25 so 25x and that's going to equal let's add 4 to negative 16.5 we get negative 12.5 that's not very nice there but 12.5 is half of 25 but if we just go about dividing by 25 what would it look like it would be x equals negative 12.5 over 25 not very nice at all but if you do get this scenario and you don't spot that 12.5 is half of 25 you could times the top and bottom by two just to remove the decimals and we would get negative what would that be 25 over 50 that's a little bit easier to spot there that simplifies down to a half okay so that also equals negative a half which we can write as a decimal if we want as well we can write it as minus 0.5 okay so i'm going to leave it as negative 0.5 it's not very nice one here but we get x equals negative 0.5 now let's sub that in i'm going to sub it into equation 1 just to avoid using that negative 5.5 i'm going to sub it into one let's hope we get a nicer value here so negative 0.5 into into equation one we could do that working out to the side you've got eight lots of negative 0.5 now eight lots of a half is four so eight lots of negative a half is going to be negative four so we get negative four add three y equals two add four to both sides so add this four and you get 3y equals 6 and then dividing by 3 you get y equals 2 as your final answer there now just thinking about what we're actually finding here now this first equation we're going to be having a look at all these sorts of equations here where one is a circle and one is a straight line and if we think about that what that would actually look like without drawing a full graph if we had a circle and a straight line went through it there would be two solutions there would there be one way it crossed over here and another where it crossed over here that is of course unless the straight line is a tangent which could be quite hard for me to draw but if it was a tangent it might just touch the circle at one point there so it could actually only have one solution but the majority of the time these are going to have two solutions now let's get rid of that now in order to solve a simultaneous equation we have to think about eliminating one of the variables so eliminating either the x or the y just like when we have a look at normal simultaneous equations but there's a different process when it comes to these harder ones here so grab a piece of paper grab a pen make some notes we'll have a look at some of these questions now for this first one it says x squared plus y squared equals 13 and the second one says well x equals y take away five or y minus five so essentially if we can substitute which we can we can substitute this y minus five in place of the x because it says x equals y minus five then we can actually eliminate one of the variables there now the way to do that is if i just stick a bracket around this y minus five i'm just going to put that in place of the x up here and let's see what we get so getting rid of the x and putting the y minus 5 would leave us with this we'll have y minus 5 and that is x squared so that's going to be squared plus y squared and that equals 13. now we've got an equation that has no x's in now as long as we've only got one variable in there we can definitely solve it okay but it's going to it's going to form a quadratic okay with y's in but we can actually definitely solve that so if i just go about expanding this bracket to start with obviously i'm going to take my time with the first one here but i might skip a few steps as we move on but that y minus five squared is a double bracket so if we work that out first we've got y minus five y minus five and if we expand that let's see what we get so y times y is y squared then we get minus 5y minus 5y again plus 25. i'm obviously assuming that you are quite comfortable with some of the topics in this in this video here so expanding a double bracket and later on we're going to have factorizing quadratics as well and substitution and a few other bits of math so i'm going to assume that you're quite comfortable with those if you are watching this video if not obviously check out the algebra series in my playlists okay everything's there obviously don't forget those playlists are there because there's quite a lot going on in those now um but if we have a look if we simplify this we get y squared minus 5y minus 5y makes negative 10y and then plus 25. so if i rewrite this equation here let's bring it over here we'll rewrite this but rather than the y minus five squared i'm gonna obviously write that expanded version so we get y squared minus 10y plus 25 and then we've got the plus i've not actually written y squared there let's fix that i should say y squared there plus that y squared and that equals 13. all right there we go right so we can tidy this up because we've got those two y squareds there so we can add those together and simplify that so we'll have 2y squared so 2y squared minus 10y plus 25 equals the 13. now when it comes to solving a quadratic we need to make this equal zero so we need to subtract 13 from both sides so if we get rid of that 13 from both sides let's see what we get we get 2y squared minus 10y take away the 13 from the 25 leaves us with 12 so positive 12 and that equals zero right now we can solve this now we can do this in two ways i can solve it from there or i can have a look at all these numbers now all these numbers all these coefficients of y squared the coefficient of y and the number at the end they're all even so if they're all even we can divide both sides of this equation by two and if i divide everything by two both sides obviously naught divided by two is still naught on the right hand side of the equal sign but if we divide all of these by two we get y squared minus five y plus six equals zero again i've just divided it by two there now this is quite nice and easy for us to factorize so we can factorize this as it is now so six the factors are six we've got one and six or two and three so going into our double bracket there we've got y and y and we just need to get the right pairing here that's gonna equal negative five in the middle and that's gonna be negative 2 and negative 3. now just be careful obviously because you can make negative 5 out of positive 1 negative 6 but we've got a plus 6 at the end there so it's going to have to be negative 2 and negative 3 and that makes minus 5 in the middle and positive 6 at the end right so we've got our two solutions for y i'm going to bring those up to the top in fact i'm going to get rid of all this working out on the left let's get rid of all of this right so we have for y our y minus 2 in the bracket gives us y equals 2 positive 2 and our y minus 3 in the bracket gives us y equals 3. so we've got two of our y coordinates there thinking about obviously what i mentioned at the start that crossover point of the circle in the straight line so we've got y equals 2 and y equals 3. now we just need to find what x equals for both of them so we've got this equation here this second equation i'm going to get rid of those brackets we've got x equals y take away 5. so if x equals y take away 5. then i can find x for both of these coordinates all i have to do is take away 5 from both of them so when y is 2 we get x equals 2 take 5 so x equals negative 3. and for the one on the right there we've got x equals three take away five and y is three so that gives us x equals negative two right so we've got our solutions there so we've got y equals two and x equals negative three and the other one we've got y equals three and x equals negative two obviously thinking about how you could write those as coordinates so the coordinate for this one on the left here would be minus three two and the coordinate for this one the right would be minus 2 3. there we go there's our solution to this simultaneous equation obviously we can pair up the values there we could just say when x is negative 3 y is 2 and when x is negative 2 y equals 3. so it depends on how the question's asking you to write your answer there but we could also write it as coordinates thinking about what that looks graphically okay so looking at a different type of simultaneous equation here only in the way that it looks and it says here obviously solve the simultaneous equations we have y plus four x plus one equals zero and y squared plus five x squared plus two x equals zero so this looks a little bit different to ones we've seen uh obviously previously now again i'm going to take the same approach i want to make one of these equations y equals or x equals and looking at that first one there we can see that obviously there is a coefficient of x we've got a 4x there so i'm going to avoid making x a subject to make enough making a fraction out of this but i'm going to go with making y the subject that's not the nicest here because when i do move those to the other side i get quite a lot of negatives here so if we make that y equals we'll minus the 4x to the other side so we get minus 4x and minus the 1 over as well so we get minus 4x minus 1. now that's not the nicest because that's what i'm going to substitute into the next equation and obviously just bearing in mind here it's a nice one to pick as well because if we had made it x equals we'd have had to have subbed into this 5x squared and would have had to have subbed into the 2x and neither one of those would have been very nice to sub this fraction into so as it stands we're going to substitute it in place of the y squared here so let's go about doing that so obviously our y now is negative 4x minus 1 so we have negative 4x minus 1 and that's all in brackets squared and we're going to add to that 5x squared and we're going to add the 2x to that as well there we go let's just rewrite that because that's a bit messy there we go 5x squared and we're going to add the 2x and it all equals zero so that's quite nice you have to move anything over any point so if we go about expanding this bracket and as it's not as nice a bracket i might just write it out so i can just see it so we've got negative 4x minus 1 and we're going to times that by another negative 4x minus 1. so negative 4x times negative 4x makes positive 16x squared then we have negative 4x times the negative 1 which makes positive 4x and the same below another positive 4x so that makes 8 x's in total so plus eight x and the negative one times negative one makes one and again we're going to add to that the five x squared and the two x there we go and that will equal zero now if we tidy this all up we've got the 16 x squared and the 5x squared and that makes 21x squared we've got the 8x and the 2x and that's going to make 10x and then we've got the plus 1 at the end there we go and that equals 0. that's not the nicest there but in terms of our brackets to make 21 it's unlikely going to be 21 x and x it might be but it's more likely that it's going to be 7x and 3x in order to make that 21x squared and obviously as it's a 1 at the end our only factor options are one and one so it's got to be one and one in both of these and as it turns out look that's going to make a nice 10 in the middle there we've got 7x and a 3x when we expand that so it's 7x plus 1 and 3x plus 1. now again we can get our two solutions from this so x is gonna equal for the first one negative one seventh and for the one on the right there we've got x equals negative a third there we go and there's our two solutions for x now we just need to go about substituting that back in and finding out our y values but again these aren't very nice values to substitute in but we have our equation up here y is equal to negative 4x take away one so y equals negative 4x take away one so we've got negative a seventh so for that first one there we've got when x equals negative one-sevenths y is going to equal negative four multiplied by negative a seventh and then take away one obviously nice and easy if you have a calculator negative four times negative a seventh is positive four sevenths so we have four sevenths take away one and four sevenths take away one more one is seven sevenths so if you take away seven from the four on the top that leaves you with negative three over seven and there's our first one for the next one if we sub in x equals negative a third and just follow the same process then we have to do um y equals negative four multiplied by the negative a third and then again take away one negative four times negative a third is positive four thirds take away one one is three thirds so four thirds take away three thirds leaves us with one third and there's our two solutions so we had when x is negative a seventh y equals negative three sevenths and for the other one there we have when x is minus a third y equals positive a third there okay so there's our two solutions for that question just obviously showing you that obviously if they're written slightly differently if they're not arranged in a way that we've seen before we can still take the same approach and obviously just forming and solving that quadratic and subbing it back into our equation okay so for this question here something slightly different so obviously do make some notes on this like again it's going to be very difficult to draw this particular graph but in terms of the bits of algebra that we're going to look in and look at in this question it would be worth writing down the question and worth writing down part a here with our first equation which we're going to read through and part b with our second equation okay so writing this down making some notes on this because you're going to have a go at this in a sec so it says here the graph of y equals x squared minus 4 x plus 2 is drawn on the grid and again we can see that and again we'll lay we've given that as it's been labeled on the graph there so if we're going to actually go about this just obviously read the question make sure it's all labeled make sure you know what you're looking at so for part a here it says use the graph to find estimates to the solution of the equation and then it says x squared minus 4x plus 2 is equal to 4. so when it says this this means that two equations have actually essentially been set equal to each other and if you've done solving quadratic simultaneous equations which again i'll link in the video you'll know that when you have two equations particularly when one's a quadratic and they're both y equals you can set them equal to each other so instead of this equaling zero this time equals four so if we were to split this up into two equations so for part a here we could say okay well we've got the graph of y equals x squared minus four x plus two which is matched on the left hand side but on the right hand side there we have another equation and that would just be y equals four okay and this is saying where does it equal four okay well obviously if you look at the original equation y is at the start and we're kind of saying where does it equal 4 on the y axis so what we mean by that is we actually have to draw this graph of y equals 4. and if we go to y equals 4 which is on the y axis there we're just going to draw that exact line so the line y equals 4 is a horizontal line that just passes through every single point where y equals 4. and if we were to draw that in i'm going to do it with a slightly dotted line to try and get it a bit straighter again because you need to do this with a ruler and a pencil but there we go you want to draw a nice straight line across and from that we're just going to actually read the coordinates of intersection where it crosses over and that's going to be our estimates so again you want to read this nice and carefully you know already that um you know already that the y coordinates there are four so we just want to give the x coordinates here so the x coordinates and again we just want to read those nice and carefully because we're looking for the solutions to that equation there we go so we can go down read them as carefully as we can and it looks like on the graph there we have minus 0.5 in the middle and we have 4.5 just there so for part a we would say x equals negative 0.5 and the other one x is equal to 4.5 and there would be our solutions to that equation they are the two numbers that you can substitute into that equation to get the answer for now again you are just giving estimates so as obviously you need to draw this as carefully as you can but it might be that your numbers are just slightly different to mine maybe you got minus 0.45 or you're trying to be really really accurate with that so you just need to be very very careful okay um but that is how we're going to go about solving it now part b is a little bit more complicated because if you look at this particular question it doesn't match on the left and there's nothing on the right so we're gonna have to think about what's actually happened here now what actually what's happened is it has been set equal to an equation but it's been rearranged to make it equal zero so what i mean by that is it could have been let's just make it up x squared minus four x plus two maybe they set it equal to i don't know let's go with two x plus three now if it was equal to two x plus three this would be the line that we'd have to draw we'd have to plot the graph of two x plus three and draw it onto our grid but we don't know what it's equal to but if it was equal to that in order for them to make it equal 0 they would minus 2x from both sides and minus 3 from both sides and if you did that you would have x squared if we minus 2x we'd have minus 6x which obviously doesn't equal what it is in the question and if we took away three we would have minus one and that's how it would equal zero so i could get rid of this part and i could imagine that i'd given you that question and you would have to find out what it is that i've taken away to make it equal zero so we're going to try and figure out how we can make the left hand side say x squared minus 4 x plus 2 again so that's the concept of what we have to do here and this is where it does step up and obviously this is the hardest type of question in my opinion that we're going to be looking at so let's have a look now we want it to say x squared minus four x plus two so what would have to be on the other side well at the moment we've got minus five x so to get from minus four x to minus five x we'd have to take away an x from both sides so instead let's add an x to both sides so let's stick an x on the right and if we were to get rid of that now it would be minus 5x on the left so that's the x part done but we also notice there's no number in the equation there that's highlighted and at the moment we've got a 2. so in order to get rid of that 2 we would have to take away 2 from both sides so we need to put a 2 on the right because then if we took away 2 from both sides the number would disappear and we can just test this out so if i take away x from both sides let's see if it matches what we've got in the question we would have x squared minus five x plus two equals two that's a very bad two i was about to draw there let's get rid of that equals two and then for the final step we take away two from both sides and you probably can already see hopefully that we get x squared minus five x equals zero the twos disappear so that does match what we've got so we are correct in saying that we need to draw the graph of x plus two or y equals x plus two so obviously you need to draw the graph of y equals x plus two and in order to do that you need to draw a little table so again up to you where you draw it i'm just going to squeeze in up here i'm going to plot some x values some y values and let's just go with negative 1 0 1 2 and 3. so if it's x plus 2 negative 1 will equal 1 0 will equal two and then we've got a pattern it's going up into so three five seven and again you just need to plot that so let's get rid of all this writing from part a and let's have a go at just finishing off part b here there we go so drawing this in negative one was at one which is just there again being very careful of the axes zero was at two one was at three and you can probably see a pattern here it starts to go up let's cross two of those squares and up one and we can extend that line again you can do it with a ruler and just be very careful when you join those up and there we go perfect so this one here is actually quite nice because it does actually fall on whole number coordinates so this one just here gives us um x equals zero and y equals let's have a look i can't see the number under there i think it's two yeah y equals two and for this one just here we've got x equals and again just trace your finger down x equals five and y equals trace finger across y equals seven and there we go there are two solutions so obviously that one there we were able to give whole numbers again if your line was quite difficult to draw again you would have to give estimates for that but this one was quite nice it gave us whole numbers so that's how we're going to approach these questions that is probably one of the most difficult ones that you could probably do because you actually had to rearrange the equation the concept behind that is quite tricky but i've got a question for you to have a go at so you can have a go at this one you can pause it you can rewind it you can have another look um but here is one for you to have a go at now very quickly before you do have a look at your question it is worth noting that for this particular question because those solutions there were whole numbers you could have actually taken a slightly different approach but it does say to use the graph now for the majority of these questions they're not going to go through whole numbers so this was slightly unique but i thought it was quite nice because it was easy for you to read but of course if you did want to find the solutions to that equation what you could do is factorize it and solve it of course we wouldn't be using the graph and for questions where we are finding estimates of course we wouldn't be able to do that on a non-calculated paper but potentially and this would give answers in third form and things like that and potentially our solution might take a little bit longer to find now for this particular one i know we've given the x and the y coordinate but in order to find the solution of the equation or the solutions of the equation we actually have two solutions our answer would be x equals five and x equals zero so when we're finding the solution to that equation we are not finding the solution necessary to the simultaneous equations so you just need to be careful of the language so for this particular question then it says use set notation to describe the set of values of x for which and then it gives us these two inequalities so we have three x minus five is less than x plus seven and five x is greater than x minus eight so first of all we actually need to go about solving these and then we'll have a think about how we write the set of values that are represented and also how we apply set notation to that so let's go by start by solving them so for the first one there 3x minus 5 is less than x plus 7 we need to subtract x from both sides so we're going to get 2x minus 5 is less than 7 adding 5 to both sides gives us 2x is less than 12 and then solving that to finish it off x is less than 6. so that's our first inequality for the next one we need to again subtract x from both sides so we get 4x is greater than negative 8 and dividing by 4 we get x is greater than negative 2. and there is our second inequality now we need to go about actually thinking how this is drawn on number line now again i mentioned you do need to know about inequalities so again don't forget to check the link in the description if you are not sure on this next step so let's get rid of these inequalities here because these are the two solutions that we need to focus on so we've got x minus six and we've also got the x is greater than negative two so if we draw a little sketch of a number line and it only needs to be a sketch because we're just going to visualize what this actually looks like and how we would go about putting it on a number line if we needed to now it needs to have at least negative two on there so i'm going to write it up to negative three and it needs to go up to six so hopefully i've drawn enough dashes there there we go perfect it goes up to seven so if we were to draw these on the number line we put an open circle as it's less than not less than or equal to above the six and we would point that arrow to the left all the numbers less than six but we've also got this inequality where x has got to be greater than negative two so what i'm going to do is i'm going to put another dot but i'm going to put it above this line and that is greater than negative 2 so that would point to the right so what you can see here is we've actually got an overlap okay or in other words these inequalities on the graph actually intersect so we have to write a set of values for this particular inequality and it's between these two values so as you know and you hopefully remember this from gcse when we're writing a set of values that overlap like this we'll put it in one of these joint inequalities so we are going to write this as and you can use any letter here x is in the question so we'll have x in the middle arrows on either side we have the smaller number on the left negative 2 and the larger number on the right which is 6. and that right there is our final answer in terms of writing the set of values for this for these particular inequalities now the only thing is is that we are also being asked to write this using set notation now set notation just means that we're going to write it in a different way essentially so we're just going to have a look at how we write this now obviously you hopefully have got that written down so i'm going to get rid of this linear scale that we've drawn there our little number line and we're going to look at actually just changing this into set notation now in order to do that when we write these things in set notation we put them in little curly brackets so what we're going to say for this is that for our set and here's our star by curly bracket for all values of x that we're looking at we put a little colon there they are between negative 2 and six and then we close our little curly bracket and that is literally it for set notation now some of these bits never change they're always going to be in a curly bracket and we're always just going to put this little for values of x here at the start and then we write our inequality and that would be it writing this in set notation so very hopefully nice and simple a tiny little step up there from gcse in terms of the way that we write it but it's just a different way of writing the inequality there and that is writing it in using set notation now again the question looks very very similar says use set notation to describe the set of values of x for which and then we have our two inequalities but there is one very big difference this time it says or and if you noticed on the last questions it said the word and so this relates to set notation it gives us a big hint here that we're going to have two different inequalities in this case as it says the word or if you remember from gcse level there is a symbol that can be used instead of the word or and it is the little u for union that we met when we did venn diagrams at gcse level so we are going to be using that symbol when we get to the point in the question of writing our set notation but in terms of actually getting to that point it's very similar to the last one so we're going to solve both of these inequalities and then we're going to draw that on a number line and we're going to have then have a look at how to write our final answer so focusing on the first inequality this one here we are going to get the x's on one side so let's add x to both sides to get rid of that negative x so that would give us one is less than two x minus five at the five would give us 6 is less than 2x and then we would get dividing by 2 3 is less than x now it's quite a bit easier just to swap that over remember if you want to swap the sides of the x and the 3 you just have to move the inequality to be pointing the other way so that would just say x has got to be greater than 3 which is always just a little bit nice to look at so x is greater than 3. on to the second one let's add 3x to both sides so for this one here we have 5 plus 5x is less than 15. take away the 5 from both sides 5x is less than 10 and divide by 5 both sides x is less than 2. right there we go so we've got our two inequalities so again we can get rid of some of this working out we can just focus on our two inequalities there and how we're going to write those in set notation so if we think about this we've got x has got to be greater than three and x has to be less than two and if we draw this on a number line and again we've only got a few numbers there so we'll just go from one to four or something like that or zero one two three four and if we were to draw our inequalities on here again open circles because it is less than or greater than without be equal to we would have x has got to be greater than 3 pointing this way and x has got to be less than 2 pointing this way so as you can see there's no crossover so i'm not going to have to have one of those joint inequalities but we are going to have two separate ones because we do not want to include any numbers between two and three so in order to write this again in terms of our answers we've actually got them both just here we've got x has got to be greater than 3 and x has got to be less than 2. so what we actually have to do is put this straight into our set notation now it looks very similar we still use our curly bracket so we're going to say for the x values where x is and it doesn't matter which one we put first let's just put x is less than two for the start and there we go close our curly bracket and then we come back to this word or so or using our symbol there or okay we can also have the set where the x values are greater than 3. there you go so it looks very very similar to the last one and that's our final answer so this answer if we read it we've got the x x can be less than 2 or x can be greater than three and that's our set notation how we would go about writing it here when there isn't a crossover and there's not an overlap those inequalities on the graph do not intersect at any points okay so this question is slightly different it says find the set of values of x for which both 2x squared minus 5x minus 3 is bigger than nor and 8x minus 7 is less than 5x plus 5. now this one on the left here we have actually just solved that obviously just to save a bit of time here but when we have two different inequalities like this we need to just have a look at the two different solutions and try and come up with a sort of an answer that makes sense here we'll have a look and have a think about what that could be now for that first one there we've already solved that we've got our two solutions so very very quickly obviously when we factorize that we got two x and x and we had one and three and it was minus uh which way was it it was minus three here and plus one we got our solution which was negative a half and positive three and then we wrote our two inequalities so we had x had to be less than negative a half and x had to be greater than three now with our other inequality over here if we go about solving that let's have a look we've got eight x minus seven is less than five x plus five so get rid of the smallest x from both sides we get three x minus seven is less than five add the seven to both sides three x is less than 12 x is less than four so we've got three separate solutions here now what we've got we need to have a look because we've got x is less than minus a half and that's okay that's that fits in with my x uh it's gotta be less than four over there but this one here we've got x has got to be greater than three but over here it says x cannot be less x has got to be less than four so in terms of our solutions here these two are quite important now x uh being less than minus a half that's okay that fits in with this one so we don't need to worry about those two but these two obviously contradict each other we can't have values that are greater than three and less than four apart from the values that are between three and four so what we can do is we can combine these together and we can say okay well we can't be uh we gotta be bigger than three that's okay but it stops at four and that's what that would mean just there that means the x values can be between three and four and also we've got the one over there x is less than uh minus a half so if i get rid of that bit get rid of that bit and that is our final bit of the solution there obviously just thinking about what that means values can be bigger than three but given that one on the right they also have to be less than four so our solution there is okay well they can be between three and four and also less than minus a half over there so there we go you just gotta watch out when you have these two here and thinking about what values actually satisfy those inequalities right okay so three more here on the grid shade the region that satisfies these inequalities so for the first one we've got y is less than three x or y equals three x is the one we're gonna plot so that doesn't have a y intercept so that y intercept is gonna be zero so be careful with my language it does have a y intercept it's just that it's not a whole number though we've just got the zero as the y intercept so we're going to start at zero it's got a gradient of three so i'm going to go across one up three across one up three i'm going to follow that pattern going backwards so left one down three left one down three there we go and that's our first line there y is equal to three x and it's going to be a dotted line as we've not got the extra ink on the inequality it's gonna be uh less than three x there we go so let's just draw that and especially can obviously using a ruler and a pencil now i've got the next one so y equals zero that's an interesting one y equals zero it's actually the x-axis it's just another name for the x-axis so we can go across here that's the line going through all the points where x equals zero sorry y equals zero there we go going across there's our second inequality and our last one here which isn't the nicest we've got four x plus three y equals twelve we need to draw that now again one that looks a bit dodgy so i'm going to apply that same logic i'm going to go when x is zero so we've got zero plus three y equals twelve because four lots of zero is zero there so zero plus three y equals twelve well it's three times a number has to equal twelve so y has to equal four for that one there we go so when x is zero y is four so we just need to find that coordinate so zero four is there and the next one let's do that the other way around so when y is zero what's x going to have to be so we have 4 x plus 3 lots of 0 so 0 equals 12. well 4 times a number has to equal 12 so that has to equal 3 so x has to be 3 there so when y is 0 x is 3 and that is there i'll draw a bit of a bigger cross so we can actually see that there we go so we can draw that in taking that logical approach there not the easiest one for me to draw on the screen here so i'm going to do my best but obviously you will have a ruler and a pencil so we have to do is join those two together and draw a nice straight line there we go going through that all right there we go obviously i've just extended a little bit past the line past the graph there that doesn't really matter let's get rid of that right okay so we've got a basic outline of what it would look like obviously yours would be a bit more accurate here when you're using a ruler and a pencil but that is that last line drawn now you could have taken a different approach for that one again you don't have to take this approach just remember to get that into the form y equals mx plus c we'd have to take away the 4x from both sides so we'd have 3y equals negative 4x plus 12. and then we'd have to divide both sides by three so we'd end up with y equals negative four over three divided by three x plus four there we go and you could actually plot the graph from there just remembering what that means it means there's a y intercept of four which we already have and then that gradient there just means we've got a rise of four and a run of three obviously it's going downwards so we could go along three down four along three down four obviously just reversing the idea of the gradient there but again obviously you want to look into these gradient bits a little bit more and understanding these line equations do check out those videos in the description and look at that coordinate geometry series that's going to really help with understanding that but there we go i like quite like just taking the logical approach and imagining one of the coordinates is zero and finding the other coordinate that matches and then just joining them up with a ruler does work for all of these right so let's get rid of all of these we need to find our region then so let's have a look now we've got a nice little coordinate inside we've got one one there so let's go for that inside the triangle see if it works so 1 1 to x is 1 y is 1. so we've got 1 is less than 3 lots of 1 1 is less than 3. again that's correct so we can take that off the next one we've got 1 is bigger than zero y z y is one sorry so one is bigger than zero that's correct and onto the last one four lots of one is four plus three lots of one which is three so four plus three is less than twelve and that's obviously seven is less than twelve and that's correct as well so we can take that off and there we go that is our region definitely correct there in the middle again this little triangle in the inside right there we go so that is obviously how to draw inequalities on a graph and how to find the shaded region now as i said i've linked in the description link for the worksheet for this particular these next four particular questions so make sure you've got that either printed off or just follow along with me or you can even draw it out yourself if you've got squared paper they're not particularly difficult to draw i do actually tend to draw these myself anyway just to get a good understanding obviously drawing the axes and getting it all plotted to scale okay and making sure that you're aware of that sketch the graph of y equals x cubed minus six x squared plus nine x showing clearly the coordinates of the points where the curve meets the coordinate axes now at the moment obviously we've got the full equation here that needs factorizing first so we have x cubed minus 6x squared plus 9x now these are the nicer versions of the quadratic sorry cubics because these are nice and easy to get a nice little quadratic out of them in order to factorize quite nice and simply if we weren't much more complicated on these we'd have to use the factor theorem and obviously that sort of starts incorporating quite a lot of um quite a lot of math so we get quite a large a large question if we were to do that but obviously check out the factor theorem if you want to have a look at factorizing some of the harder cubics here but we're going to just focus on some of these slightly easier cubics for drawing these as it can get quite complicated as well otherwise but let's have a look now in order to factorize this we can take out a factor of x to start with and when we take excel as there's no number at the end we get inside a nice quadratic here we get x squared minus 6x plus nine okay so that obviously gives us our first solution there we've got x equals zero for that ax on the outside now we can go about uh factorizing this quadratic so obviously if we turn this into a double bracket our factors of nine we can have three and three or one and nine and if we want to make negative six in the middle it's gonna be minus three and minus three so x minus three x minus three there we go and we've got that x on the outside so there is the equation of our graph if we factorize it and then obviously again that gives us our three solutions so we get x equals zero for that one there we get x equals three for this one here and x equals three for this one here and again we know there's no y intercept because we have no number at the end of this equation here so now the y intercept is going to be 0 which we've also proved just via this solution here of x equals 0. so we can go about just drawing it just like we did before now so draw your axes it's positively sloping it goes through zero and it goes through three and it's just going to touch on three as well as we've got those two solutions here of three so in actually just drawing that and not forgetting this isn't the nicest one to be fair because uh we've got um what is it two solutions at three so you just need to watch out obviously the fact that it's sloping up and it's going to go up through the origin oh let's do that again up through the origin down just touching at three and then sloping back up again right there we go so no y intercept twelfth the wind sets of zero and obviously those two solutions on three so just touching on the three point there okay so this question says sketch the graph of y equals and then we have our equation there showing clearly the coordinates of the points where the curve meets the coordinate axes now if you look at the equation that here that we've been given you can see that if we were to expand that out our highest power would be a power of four so this is going to be a quartic graph and when we have a quartic graph there's two different ways that it's going to look we could have a positive quartic graph which is going to have this sort of shape with a little wiggle or we could have a negative quartic graph which is going to have this sort of shape and again those wiggles that we have in the graphs are going to be slightly different depending on our roots so we know this is going to be a negative one so it's going to look like my graph on the right so we know which way it's going to be sloping we just need to figure out those roots so to start with we have the pieces at this beginning so the x at the beginning just here represents a root where x is equal to zero so we know one of our roots already for the second one we have three minus x and again if you're not sure on the root there you can set it equal to zero and solve it so if we say that three minus x is equal to zero and if we add x to the other side we get 3 is equal to x or of course we could write that as x is equal to 3. if we move on to our quadratic there we're going to want to factorize that to find our roots and if we factorize this final quadratic we get in double brackets x plus two and another x plus two now that could be written in a slightly different way of course we could write that as x plus two in brackets squared now when we have that going on that means we have a repeated root which means our curve at some point is going to bounce off the axes so if we go about drawing a sketch of this and we draw an axis we obviously have all our roots we have x is equal to 0 which we can label on there we have x is going to be equal to 3 which we can label over here and we just need to get our root for that x x plus 2 in bracket squared so the root for that is going to be x is equal to negative 2. and of course that's a repeated root so we don't have to draw that twice but we're going to want to label negative 2 over here so that we can make sure this curve looks correct so we know the shape that it's going to be in and it can help to draw the shape just to the side of the graph if that helps just so that you can remember what it's going to look like so for a negative one it is going to have this sort of shape and we're going to need to draw that onto our curve or onto our axes so for this particular one here when we draw this one on the first thing that you're going to notice when we go up is that straight away we have that repeated root so as we come up to the negative 2 it's going to bounce off the axes it's then going to go up through the zero as that's our next route and then it's going to go up a certain distance and it's going to come down through the three so you can see we have that negative quartic shape and again the first thing we had to be aware of was obviously that bounce there for our repeated route but zero and three were distinct roots so it does pass through those points okay so this question says sketch the graph of y equals three over x with a minus one at the end and we'll have a look at that in a sec it says showing any points at which the curve crosses the coordinate axes and writing down the equations of the asymptotes of the curve so let's have a look at this line equation to start with now we know from graph transformations that if um we change the x value it'll move it left and right and if we change the y value which is what this does at the end here it moves it up and down so minus one at the end thinking about a graph transformation and from gcc we'll have sometimes seen it like this we might have seen f of x there we go with a minus one at the end and that there is outside the bracket it affects the y coordinate and minus one it does what we expect it moves it down by one so if we draw a little sketch of this to start with i'm just going to have a look at to start with the graph of y equals 3 over x so y equals 3 over x and what does that look like okay i'm going to draw this on pretty much all these questions and it goes into this area here now if i then do this transformation so i'm going to do this function here i'm going to take away one at the end which moves it down by one we'll have a basic sketch of what that looks like there we go so if i move it down by one then this part here is going to go down below the axis and this part here is also going to move down as well and that's what our graph there's going to look like as a basic sketch now thinking about those asymptotes obviously on the first one here that i've drawn by three over x and let's just label that other one there so this one here is our y equals three over x but with minus one at the end originally as we mentioned on the first bit our asymptote across this way which has the equation y equals zero has now moved down by one so it's going to move down one here there we go and that's going to move down to minus 1. so y is going to equal minus 1 for that asymptote there we go now obviously the graph here has moved downwards it hasn't moved left or right so our original asymptote over here which goes down which has the equation x equals zero has not it's not going to change okay so that is going to stay exactly where it is and we've still got that asymptote there where it never touches the axes there we go so x equals 0 is going to stay exactly the same now there's one more element that we need to have a look at because obviously on our first graph it never crosses over the axis but now we've moved it down this one on the left does cross the axes okay it crosses over just here at that point so we need to find out what is that coordinate just there there are two ways of doing this because it either crosses over the x-axis or the y-axis and at that point in there it crosses over the x-axis and if at a point if a graph ever crosses the x-axis the y-coordinate there is zero so for that point there i know that y is equal to zero and all i need to do is sub that into my equation and solve for what x is going to be so if i put that in obviously we're looking at this equation here y is equal to three over x minus one so if i put zero into there we have zero equals three over x minus one and i just need to solve that so if i add the one over to the other side we get one is equal to three over x there we go so 1 is equal to 3 divided by something now logically we know that the answer for that is 3 3 divided by 3 equals 1 but i also just need to keep going with this and obviously to solve it in a normal way because some of them aren't going to be quite as obvious to see so i can multiply the x over to the other side so i get one x or x equals three there we go so i'm actually just multiplying over that's quite a nice one because it straight away gave me the solution there as x equals three but there we go that's the point at which it crosses over the x-axis so i'll just get rid of this and i'll label that on the graph there on my sketch and i'll put that as three there we go it says here on the same axis sketch the graphs with equations and then we have these two equations to draw here so we've got y equals we have x brackets x plus one brackets x minus four and we also have this other graph y equals two over x plus one it says to indicate all the points where the curves meet the x axis and hence write down the number of real solutions to this equation here and yes as you can see they've set those two equations equal to each other okay so we have the x with our brackets here and we have our other graph 2x 2 over x plus 1 on the right so we're going to have a look obviously doing that and drawing these graphs and seeing how we approach them now first things first we need to be aware of the different types of graphs that we could be looking at drawing obviously forgetting about straight line graphs for the moment just thinking about our curves as both of these are we can have a quadratic a cubic and a reciprocal okay there are other graphs as well but throughout this video they're the main ones that we're going to be focusing on with a little bit of straight line graphs thrown in there as well but with each of these points we're either going to have i'm going to draw these to the side we could have a positive quadratic which makes a u shape or a negative quadratic which makes that n shape we can have a positive cubic which makes this sort of shape or a negative cubic where we have a negative x cubed and then as well as that we can have our reciprocal graphs as well which depending on how we draw our axes we can have our positive reciprocal which has a little while looks a little bit like that or we can have our negative reciprocal which looks just the other way around in those two okay so we're going to be mainly drawing these six graphs here and obviously as i said you need to make sure that you're happy and aware of these um sketches here so obviously do check out sketching quadratics cubics and reciprocals in the videos in the description so let's have a look at this first one then now the first graph there we have three x pieces they're going to be multiplied together there is the x there's the x there and the x there and they're all positive so that's going to make a positive cubic graph now the solutions are given to us that end this particular one here because it's already been factorized so when we've got that one there we've got the first x at the start which gives us one of the x um interceptions and that's going to be x equals zero so i'm just going to write this down we've got x equals zero as our solution for that one we have the x plus one okay and that's going to be x equals negative one okay remember you just flip the signs or you just set it equal to zero remember if there are any difficult ones you can just set it equal to zero so we can say when does x plus one equals zero and then minus the one from both sides we get the x equals negative one which is why you can just flip the signs but don't forget you've got that process there if any more difficult ones come along which they will and then we've got the last one there where x minus four in the bracket and that gives us x equals positive 4. so there's our three solutions and we already know the shape of the graph it's going to be a positive cubic so it's going to look like that so if we start to draw our axes let's just draw it in and i'm not going to actually draw this graph just on just yet but i am going to put these points where i know it's going to cross over so we have minus 1 and 4 and remember it's just a sketch so it doesn't have to be absolutely perfect it's more about where it crosses over and the shape of the graph that we're getting correct here so there we go i know it needs to be around about that big for the moment now we have our other graph there now that one below is a reciprocal graph whenever you're getting a divide by x that's going to form a reciprocal graph now it's a positive x so it's a positive reciprocal so we know in terms of just a little basic idea what it's going to look like it's going to look something like that but this one here also has a plus one on there and hopefully you'll remember that with a plus one at the end that's going to change the y-intercept okay or in other words with not obviously we don't have a y-set with our reciprocal graphs but with a normal graph it changes the y-intercept it moves it up one so in terms of our reciprocal graph here that is going to move up one and if i draw just over the top of my sketch there it just means it's going to be up a little bit higher okay so it's going to cross through the x-axis and that negative region there now in terms of actually drawing this before i draw the sketch again i just want to know where it crosses the x-axis okay so i don't want to have to redraw my sketch twice so let's figure out where it crosses the x-axis and that's easy enough to do because we can just set y equal to zero okay remember that from coordinate geometry if you ever want to find where a graph crosses the x-axis just put y equals zero in so if we sub that in let's do that down here we get zero equals two over x plus one and just remember the reason i'm doing this is because any position on the x-axis the y-coordinate is zero so if we sub that in and now we just need to solve it not the nicest one to solve just because we have a little bit of movement to do it looks like it should be nice and quick but we have to minus one from both sides so we get minus one is equal to two over x multiply both sides by x so we get negative x or negative one x equals two and then divide by negative one and we get x equals negative two there we go so there we are we know it's going to cross over at x equals negative two and i'm going to write that down up there x equals minus two i'm just going to mark that on the graph as well there we go minus two there and now we can go about drawing these in because we know where it's going to cross over all these different points so if i draw the reciprocal in to start with so my reciprocal is going to look something like this and the other one on the right there is also moved up so something like that again it's not perfect okay but you just have to draw these as best you can just extend that axis there and there's our first part of um drawing the reciprocal and i didn't have to draw that one first i could have drawn the cubic first but as i just finished that one we'll draw that one in now the second one that we need to draw here is the cubic and if we said we already know what it looks like we've drawn a little sketch of it up here so it's going to go up through that minus one back down through the zero and then up through the four so if we go up to the minus one back down through the zero and then up through the four and again i can probably just extend that reciprocal graph a little bit again it is only just a sketch so you are going to have these little um situations where you're drawing them where they don't quite meet so you haven't quite drawn it big enough or maybe you've drawn it too small okay so if you do do that obviously just make sure you're always drawing it in a pen um with a pencil and ruler and you can always just rub it out and go again or just draw a second sketch okay it's all about just a little bit of practice with these and just perfecting your graphs as you draw them but there we go there's our sketch if we actually put this onto a graphing calculator we'd probably see that it's not perfect okay and these little dips and um in within that cubic graph could potentially be a lot more stretched out but as it's only a sketch we're not going to stress too much about that and actually we'll have a look at what it looks like in a sec but before we do that it says here hence write down the number of real solutions to the equation and then what they've done here is they've just set them equal to each other now if you remember from gcse if you want to solve two sort of graph equations algebraically all you have to do as long as they're both y equals you just set them equal to each other and find the x coordinate okay solving them simultaneously and then you sub the x coordinate back in to find the y coordinate but this question here is that hence obviously using our diagram there write down the number of real solutions to this equation it doesn't want the solutions exactly just wants to write down how many there are so as we can see on the graph here we've got one here where they cross over and we have another one just there so in total we have two solutions as the graph crosses over the graphs cross over at two points there okay so we would obviously just say two solutions as the graphs cross over at two points i'm just going to write that in there two solutions and there we go okay so that's the end of our first question obviously there was drawing reciprocal graphs there's a little bit of graph transformations there as well with the plus one moving it up up by one and we actually also had drawing a cubic graph and we had some solvent equations as you can see there's a lot of different topics in here even though once you've got the hang of drawing these the topic is actually relatively straightforward okay once you've done quite a few of them but you can see it's quite a lot of topics involved here so do make sure you check out any of those videos on previous topics obviously relating to this but that's our first question and before we move on let's just have a look at what this would look like okay so if you use this app and it's a great little app to use it's called desmos so we actually need to do to have a look at that graph there is just type in the equations for these two graphs and it's a really good tool when you're doing topics like this just for having a look at what these graphs look like so if we type them in we can have y equals x and then what was it was brackets x plus one close brackets and then it was x minus four in the other bracket x minus four there we go and that's going to give us just a little overview or kind of a little look at what it actually looks like if we put our second graph in then which was y equals we had 2 over x so 2 and we'll put divide by x there we go come out of the fraction and then we'll put the plus 1 there and that shows it obviously moving up slightly now it's really good look because obviously we can see that our graph wasn't too far off the reciprocal looked good and the negative part of the graph on the left there looked good but obviously this dip was a lot bigger um a lot smaller on the sketch that i drew okay but as i said it is just a sketch but it's a great little opportunity just to have a look and actually see if your sketch is correct and as we can see there's one crossover point just there and there's another crossover point just there and as well as that we can also see where it crosses through the x-axis we can see the negative two over here that we had we can see these solutions here the minus one the zero and the four and we can pretty much check everything that we had on that question so just a little thing for you to have a look at it's called desmos it's a great little app you can get it on your phone or on an ipad or on a computer um it doesn't really matter what device you use and you should be able to access this on anything so there we go just a little thing to show you along the way let's have a look at our next question so it says the diagram shows them the uh part of the curve with the equation y because f of x and here it is in the diagram the maximum point of the curve is 2 3 as we've already seen write down the coordinates of the maximum point with these different equations i've got quite a lot here all the different scenarios we could actually have a look at so for part a there okay it says f of x minus two so that's inside the bracket so it affects the x coordinate and it does the opposite of what we'd expect so it doesn't subtract two it adds two okay so it just does the opposite so the x coordinate being two there we would add two to the x coordinate and that would make it four and the y coordinate is not going to change so that would stay as three okay so that's the opposite of what we'd expect when we add two to the x coordinate look at the next one let's try to change to a different color here right so f of x minus one now that minus one is outside of the bracket so it's outside the bracket it affects y and it does exactly what we'd expect so minus one we would expect that to subtract one from the y coordinate which it does it just changes the y-intercept actually if you're familiar with your equations of lines and your coordinates and coordinate geometry you should hopefully know that number at the end is the y the y intercept there so it's taking away one from the y intercept essentially just moving the graph down one but we'll have a look at that again a little bit later but minusing one from the y-coordinate the y-coordinate currently is three so if we take one away from that the x-coordinate doesn't change so two but the y-coordinate is also gonna drop down there to two it's gonna subtract one from that okay so that's how we can have different scenarios there with just adding and subtracting numbers and how that affects the x and y coordinate now we're going to look at question c part c there so y equals f of 2x so 2 times x normally when you're timesing by two you'd expect that to double something but as it's inside the bracket affecting x it does the opposite so rather than multiplying by two it divides it by two or halves it there so if we look at the x coordinate the x coordinate is currently two so half of that halving it because it's inside the bracket and changing the x coordinate but does the opposite so half of two is one and the y coordinate remains unchanged and that remains as three there we go so that's that one let's have a look at the other scenario where it's outside the bracket so three lots of f of x for part d here let's have a look at what this does so it's outside the bracket it affects y we're multiplying it by three so it does what we'd expect it makes it three times bigger but it affects the y coordinate this time because it's outside the x bracket so three lots of that y coordinate the y coordinate is currently three so the x coordinate is unchanged and the y coordinate gets multiplied by three and becomes nine there we go so that is our scenarios where we are multiplying by a number so in terms of the x there inside the bracket 2x does the opposite so it divided by 2 but outside the bracket it multiplied the y-coordinate by three okay and on to these negative ones here which don't really follow the same trend but partly what it does so we'll have a look at this one here so we've got minus x in there now it doesn't do the opposite of what we'd expect because the opposite of minus x would just be nothing okay we wouldn't actually change anything at all it just be positive x it'd be the same equation okay but we follow the rest of the rules for this one it's just uh it goes a little bit out the window and we look at our minuses here so when we've got a minus in there all that does is it swaps the symbol of the x coordinate it's still affecting x it's inside the bracket it just doesn't do the opposite this time okay so it just changes the the two as the x coordinate to minus two so the x coordinate is going to change it becomes minus two and the y coordinate remains unchanged stays as three okay so this one is a negative in there that's the only scenario where it doesn't actually do the opposite or anything like that we just have to worry about is it next to the x or is it changing the y and it just flips the sign so the last one here we've got y equals minus f of x and again the minus those outside the brackets so it changes the y coordinate and again it just swaps the sign so instead of being positive 3 there is the y coordinate it's going to be negative 3. so we have positive 2 still and negative 3. okay so again just following all those little rules is it next to x or is it next to the y or outside of the x bracket that tells us which coordinate of x when it comes to adding subtracting multiplying okay we can think about the opposite or the whether it does what we'd expect the only scenario where it doesn't is where we've got this little minus symbol in there and that just changes the the positive or negative symbol of the the corresponding coordinate whether it's x or y whether it's inside the bracket or outside the bracket okay so it says the diagram shows part of a curve of the equation y equals f of x draw a sketch of the graph with these two equations so i'm going to do two on one graph just to save a bit of time here so part a says to draw y equals f of x minus four now we know that x minus four here is going to plus four to all the white towards the x coordinates so add four to x okay now what we've got to do is make sure that we find any whole number coordinates on this graph and they're the key ones so i can see quite a few i'm going to highlight them here so we've got one here we've got one here and we've got a few down the bottom here that go through actual whole number coordinates so they are the coordinates that i need to make sure i move perfectly when i'm drawing my little sketch the rest and we can just do it as accurately as we can so if we're going to add 4 to the x coordinate well let's look at this one here to start with that coordinate there currently is 0 4 so when i add 4 to the x coordinate that is going to become 4 4 okay adding 4 to the zero so 4 4 is the coordinate here and you can see there that's moved it right four spaces and that's what that actually that transformation there actually does okay so when we actually add 4 to the x coordinates they're all going to move right for and therefore it's going to be translated okay to the right okay so we're translating it four to the right now looking at the other one um this one here if i move that four that's going to land on top of that not four coordinates so i can't draw over the top of that that's where that's going to move the one down the bottom here this minus two that if we move that right four that's gonna go to here and then moving the others right four is gonna place them there and there there we go and that is how they would move now if i just get rid of some of these bits here because we're going to do another one in a sec let's get rid of all of them and let's just sketch that in and you've got to just do this nice and neat okay as best you can drawing a nice smooth curve making it look as best as you can like the one on the graph there there we go it's not perfect with my lines there but i'm just moving it right four and showing that it goes through those whole number coordinates let's have a look at part b here we've got y equals f of minus x now minus x we know uh changes the symbol on the x-coordinates so again if we pick a coordinate and let's pick this one to start with okay so this one here let's label this over here that coordinate there is minus two zero okay so if we swap the symbol on the x coordinate there that's going to become positive two zero okay so positive two zero again let's get rid of this first graph positive 2 0 is here there we go so let's have a look at another one and let's pick this coordinate here there we go so that coordinate there is minus one zero and if we change the symbol in front of the x coordinate that becomes positive one zero so positive one zero is there and let's have a look at one more down here on the bottom we might be able to figure out where the rest are going to go afterwards this one here is minus 3 0 and again swapping the x coordinate now that becomes positive 3 0 which is just here so hopefully you can see actually this curve is going to end up in exactly the same place as the last one but what's actually happening here let's have a let's just get rid of all this so we can see it a bit better is something slightly different to the last one because our last one was a transformation it was actually moving right for whereas if you have a look at and i'll try and do them in a different color here this coordinate here the purple one moved to this coordinate here so actually it's not being moved in the same way because as well this coordinate here has moved to this coordinate here what this is actually doing is reflecting the shape in the y axis okay and you might actually be able to see that this coordinate here as well would move two let's make sure i do this carefully now i'm not writing down the coordinates it would move to there and the coordinate which is here on the axis it's not going to move anywhere it's going to stay where it is so actually what's happened is the curve has been reflected but actually all you've got to do is reflect those or change the symbol in front of those x coordinates and again just sketching it in nice and neat so it's quite unique there because oh there we go not very good there make sure you do go through the coordinate points but although there were two different um transformations there two different changes in the function the graph did end up in the same place but there were different transformations one's a reflection and one was a translation okay but as it turned out they ended up in the same place for that one okay so this question says find the equation of the line that passes through the coordinates five seven and three and negative 1. so when we are finding equations of straight lines there are two formulas that we need one is to find the gradient which is that m is equal to y2 minus y1 over x2 minus x1 or the change in y over change in x and the second thing that we're going to need is our equation of a line which is y minus y1 is equal to m brackets x minus x1 of course you can use y equals mx plus c still but this equation is just a little bit quicker but again you can use y equals mx plus c whenever you like so if we plug these values in and we'll have this as our first coordinate and three negative one as our second coordinate okay so if we find our gradient to start with that is going to be y2 minus y1 so we will have negative one take away the first one which is seven over three take away the first one which is five so on the top that comes out as negative eight on the bottom that comes out as negative two so we have a gradient of four so we can just say that the gradient is equal to four we can then put that into our equation of a line so we're just going to substitute that into y minus y1 equals mx minus x1 and if we sub these pieces in and again i'm just going to use the first coordinate there so 5 and 7 and if we substitute those in we get y minus seven is equal to four lots of x minus five and all we have to do is expand this and make it y equals and we will have our line equation so we have y minus seven is going to be equal to 4x minus 20 and then you just need to add 7 to both sides so we get y is equal to 4x minus 13. and there we go we have our line equation done okay so find the equation of a line perpendicular to y equals two minus three x that passes through the point six five so first things first let's get the gradient out of this now it's written in a slightly different way because it's got negative gradient but our gradient there is negative three the number in front of the x let's write that down m equals negative three straight away we're finding a perpendicular line so let's find the perpendicular gradient so mp change it to positive flip it over it becomes a third right so we just need to put a third into our line equation so writing down y equals mx plus c and then let's put this gradient in so we get y equals a third x plus c right from there same process again so we're going to sub this coordinate in 6 5 as that coordinate is on our line so 5 equals y so 5 equals 1 3rd times the x-coordinate which is six plus c right let's expand that little bracket out so we get five equals now one-third times six you can treat this in two different ways you could do a bit of working out for this if you want a few times a third by six remember six is six over one so you can just treat it like normal fractions you get six over three and six over three equals two and when you times a fraction by a whole number it just multiplies the numerator so six times a third is six thirds and six thirds is two okay but you can also just think when it's a fraction times six if it's a nice one a third of six or a third times six is just one third of six so six divided by three equals two so you can do it in two different ways there just depends how confident you are with your fractions here but you can always just treat it in this sense up here so six times a third is six thirds and simplify it down and in this case it does become a whole number so five equals two plus c and then again if we solve that for c subtract two from both sides we get three equals c or c equals three as our value of c there there we go so we've got our y intercept now we can put it back into this line equation and we get y equals one third x plus the three there that we've just worked out okay so this question here says that the straight line l1 has equation 4x minus y equals zero and l2 with the equation 2x plus 3y minus 21 equals 0 intersect at point a work out the area of the triangle aob where b is the point where l2 meets the axes so if we were to draw a sketch of this just to imagine what it looks like which can help with these questions we could imagine what these graphs are going to look like now for the first one that's quite easy to rearrange you get y is equal to 4x so for that first one y is equal to 4x has a y-intercept of zeros and it has quite a high gradient so it looks something like this for the next one again we can make that uh as three y at the start we could make it as um y equals as well but if we just move the 2x and the minus 21 over we'd have that 3y is equal to negative 2x plus 21. now if we divide it all by 3 and have a think about what that equation would be we'd have y is equal to negative two-thirds x plus seven so we could draw that on as well that has got a negative gradient and it has got a y-intercept of seven now it's not very steep but it is negative 2 3 so maybe look something like this and again this is just a sketch and again the y intercept there would be 7 and it was 0 down here so you can actually see the triangle that we're looking at we've got our point b over here that we need to find we have our point a here where they intersect and then you can see the triangle aob is just here that is created via all these lines in the x-axis so if we're going to go about working this out first of all we want to find that point a so the first thing that i would do is i would rearrange that first equation which we've already done and we've got y is equal to 4x now we know what y is equal to we can substitute it into the other equation rather than setting them equal to each other which you could also do as well but we can also just sub it into the other equation so if we sub it into that first equation that we have over there if y is equal to 4x then 2x plus 3 lots of 4y sorry 3 lots of 4x minus 21 is equal to 0. and now we've got an equation that just has x in it but we just need to solve 2x and this is going to equal 12x so that would be 14x which is equal to and add the 21 to the other side 14x is equal to 21. we can now solve that by dividing by 14 and we get x is equal to 1.5 or you could leave that as 3 over 2. so we've got our x coordinate of where they intersect and obviously the equation there is y equals 4x so y is going to equal 4 lots of 1.5 so y is equal to 4 times 1.5 and that is y is equal to six so we have our equator a coordinate where the points intersect and that coordinate is one point five and six and there we go that is the first part of this question we now know the height of the triangle as the y-coordinate there is six the height of that triangle must be six now we just need to know the length of the base so we need to know the coordinate of b and once we've got the coordinate of b we can just do the area of a triangle so moving on to the next stage here we're going to work out the coordinates of b now of course to work out the coordinates of b all we actually need to do here is find the x coordinate we as we already know the y coordinate is zero so if we think about the line equation that that line is meeting down here the line equation was that second one so if we're going to substitute in a value we can just substitute y equals zero so if we do that we get 2x plus three lots of zero and again that'd just be zero so we don't even need to write that you can do if you want but we could just leave it as that becomes zero we're just going to get 2x minus 21 is equal to zero again i'm just going to write here y equals zero so you know that that's where white that y piece has disappeared we can add 21 to both sides so 2x is equal to 21 and then dividing by 2 x is going to equal 10.5 or of course you could leave that as 21 over 2. so the x coordinate at that point b we've got x is equal to 10.5 which means that from 0 to 10.5 the base of our triangle is going to be 10.5 so we've got all the pieces that we need to finish this off an area of a triangle is half base times height so we would just do for area half times 10.5 and multiply that by the height which is six and if we type that into our calculator it gives us a value as a fraction 63 over two or of course you could write that as a decimal and there we go we don't need to give any units as we are looking at the area problem on a coordinate grid so our final answer would just be 63 over 2 and there we go that is the final answer for this one okay so when we're modeling with straight line graphs we just need to take the coordinates from the context of the question so this question here says that container is leaking with water at a constant rate the water remaining is recorded at certain intervals it says at a st at the start so we'll highlight that the depth was 19.1 and after 100 seconds the depth was 6.1 so you can see it's going down over time and it says to do a deduce an equation in the form d equals a t plus b and interpret the meaning of the coefficients of a and b so d in this case is our depth and t is our time and we're going to make an equation using those so d equals a t plus b so in order to do this we obviously need to get the gradient between those points so if we are looking at the points bearing in mind the start time there s will equal zero or we could write that as t equals zero as we are using t so we'll have t is equal to zero we've also got t is equal to 100 seconds here and then we have our other numbers so we have the depth here is equal to 19.1 and we have the depth over here is equal to 6.1 now if we were to draw a little sketch of this we could imagine what it looks like and maybe we would have our depth going up the top which would be the most logical and time down the bottom so the depth starts at 19.1 which we could draw up here and then it is going to go down over time to the point where it's empty so if we think about doing this to get our gradient we're going to do y2 minus y1 over x2 minus x1 so our y coordinates here will be the depth so will do and we can pick the 19.1 if we wanted to start but we'll go with our second coordinate there which is 6.1 so we'll have 6.1 take away 19.1 and that's going to be divided by this the x coordinate the second one which is a hundred so that'd be a hundred take away zero now on the top there that comes out as thirteen or negative thirteen and on the bottom we have 100 so we can simplify that or we could just type into our calculator and that comes out as negative naught point one three so there we go we have our gradient now we can put that into our line equation and again we can use our other coordinates but we know that our gradient is going to be minus 0.13 t so in order to write our equation we could substitute our coordinates into our line equation or for this particular question we can actually get our y-intercept from the graph and you can see here that at 0 seconds it started at 19.1 so if we're going to write our line equation we may as well just take the coordinate from there and write y is equal to and in fact we should keep our letters the same as the question has said because it does say to give it as d is equal to so we would say that the depth is equal to minus 0.13 t plus the 19.1 so there is our linear equation and there we go and we've given it in the format that we've been asked now it does say to interpret the meaning of the coefficients so if we look at the coefficients we have the minus 0.13 where in place of a now that is representing the change in the depth of the water every second as we know it's in seconds so we could write that down the change in depth we could say per second there we go the change in depth per second and we have our y-intercept at the end in place of b and that is the depth at the start okay as given to us in the question so depth at the start so there we go that is part a now part b has asked us something else it says use the model to find the time when the container will be empty in or in other words we want to know when d is equal to zero so for part b if we just put zero in place of d then we've got an equation that we can solve so if we say that 0 is equal to minus 0.13 t plus 19.1 we just need to solve that now the easiest way to solve that would be to add the minus 1.0.13 t to the other side so we would have positive 0.13 t which is equal to 19.1 and then we would just divide both sides by 0.13 and we get a final answer which will give us our time and that time comes out as 146.9 so 146.9 and that is in seconds so we just write seconds at the end of that 146.9 seconds and there we go that would be our final answer for finding the time when the depth is going to be zero or in other words when the container is empty okay so when looking at midpoints and perpendicular bisectors this is quite a common type of question so it says the line segment a b is the diameter of a circle center c where a is negative 1 4 and b is 5 2 says the line l passes through c and is perpendicular to the line a b which is the diameter find the equation of line l so we want to find the perpendicular bisector of this diameter and obviously the perpendicular bisector is going to be perpendicular so we have all our stuff to do with perpendicular gradients so if we start by finding the midpoint of these two we might be able to draw a little sketch of this a little bit better so to find the midpoint you take the coordinates and add them together and divide it by two so if we focus first on the x-coordinate to get the midpoint of the x-coordinate we would do 5 plus negative 1 the two x-coordinates and divide that by 2 and the other coordinate would be 2 plus 4 and again dividing that by 2. so for that particular coordinate there 5 add negative 1 is 4 divided by 2 is 2 and 2 plus 4 is 6 divided by 2 is 3. so we could probably draw an okay sort of sketch and think about what this looks like so a is negative 1 4 so negative 1 4 let's just imagine that's here and then our coordinate in the center is 2 3 so that is to the right obviously and just a little bit lower than the four and then we've got another coordinate which is five two which is to the right again and just a little bit lower so it's going to look something like this now obviously you're not going to be able to draw a perfect sketch of this but you can just have something to help you visualize it so if we imagine it looks something like that although that wasn't very good as my center doesn't look fantastic so you can always get rid of these coordinates and just put them back on so it suits your diagram so let's say it's there there and there so now we have the coordinates of the center so we could also draw in this line a b and let's label that up so that would be a and this would be b and we have our center point there and we are going to find the perpendicular bisector so if we find the or if we draw that in and imagine what it looks like it's going to look something like that and it probably be also worthwhile labeling on that center point two three so how do we go about finding a perpendicular line well first of all we want to find the gradient so if we find the gradient of a b well then we can go about doing that to start with so for finding the gradient y2 minus y1 so that is going to be the y-coordinate of b take away the y-coordinate of one of a sorry over the x-coordinate of b which is five take away negative one the x-coordinate of a so that comes out as negative two over six which is minus a third so our gradient of a b is minus one third which means our perpendicular gradient is going to be our negative reciprocal and that comes out as positive 3. so that is the first thing we need that is the gradient of our perpendicular bisector here so now we can use our line equation so y minus y1 equals mx minus x1 and it is going to have to pass through that coordinate through the center which we can see here 2 and 3. so we would have y minus the y coordinate of 3 is equal to m which is three about to draw the bracket let's get rid of that equals three lots of x minus the x coordinate which is two so if we expand that out we get y minus three is equal to 3x minus 6 and then of course to finish this off we want it just to say y equals so add 3 to both sides there we go and we get y is equal to 3x minus 3 and that is the final line equation and that is the equation of our perpendicular bisector passing through the midpoint of a b in terms of actually the equation of the circle i'm always just going to write that down so i've got x minus a which is the first uh coordinate or the x coordinate within the center squared add to that y minus b which is obviously the y coordinate of the center squared and that's going to equal the radius squared so if we plug these numbers in that we've been given three and one let's just put them into the equation let's see what we get so there we go we've got x minus three squared add y minus 1 squared and that is equal to the radius squared now i'm going to do this in two different ways so if you've seen this before you might have only used one of the ways so i'm going to show you both but in terms of what we did at gcse maybe at this point we'd have imagined these two coordinates okay so if i imagine the point three one which is the center it does always help if you can draw a little diagram of these but if we imagine three one's the center and then negative two five let's have a just have a little think it probably looks something like this there we go and negative two five is somewhere over here let's have a think negative two fives maybe over there somewhere there we go but in terms of looking at these two points we would find the length of the radius by using pythagoras like we've just talked about and we would look at the base or the or the rise and the run there or the change in x and the change in y and then we just use pythagoras to work out the radius so in terms of the base length it goes from negative 2 on the left there over to three which is a length of five and going up it goes from one up to five which is a length of four so you can use pythagoras to work out the length of the radius there and we can just do that quite nice and easy we can do four squared plus five squared which is 16 plus 25 which is 41 and then you can just square root that for the radius so r equals the square root of 41. there we go but obviously we want to know what the radius squared is so when it's in a nice little square root like that we can just take away the square root and it'd be 41. so we could finish off our equation here it'll be x minus 3 squared plus y minus 1 squared equals the radius squared the radius is root 41. so 41. so it says the circle c has equation and then we have the equation here and lots of pieces we've got x squares y squared x y's and a number and it all equals zero says find the coordinates of the center of c and the radius of c and then we've got another question here that says find the coordinates of the points where c crosses the y axis so what i've looked at in a sec but this actual equation here has to be rearranged in order to get into the format that we want obviously we know what the format looks like we've got x minus a in bracket squared and then we've got y minus b in bracket squared and that equals r squared so we have to rearrange this we need to get all those numbers on the right hand side and then we need to get these x parts into brackets and if you hopefully recognize this you should recognize that this looks very similar to completed square form and this is where we're going to use a bit of completing the square to get into this format now if we have a look at the pieces that we have in our equation we've got some x pieces here we've got and i'll highlight this in that color there we've got the x squared and we've got a minus 6x now if we just forget about everything else there we group those together we'll have x squared minus 6x okay now if we were to complete the square on that okay just on that part there we would obviously halve the coefficient of x so it'd be x minus three in bracket squared and obviously if you're not sure on completing the square don't forget i've linked that in the description below so check that out first and then obviously we need to get that back to zero because there's no number after this so to get that back to zero thinking about what that would expand to well negative three times negative three would give us positive nine so i'd also have to take away nine and that obviously that expression there would give us x squared minus 6x so i'm just just writing that in completed square form we also have the y pieces now if we have a look at the y pieces in there let's just highlight those we've got y squared and a plus 2y now if i put those in we've got so i'll put a plus here as well we've got y squared plus 2y now if i complete the square just on that and let's keep the plus there as well we would have the coefficient of y so i'd get y plus one there we go in brackets squared when we expand that that gives us plus one so we have to take that one away to make it balance the expression above the y squared plus two y and that would obviously complete the square on that part so so far what we've done we've completed the square there that's given us a little bit below and we completed the square for the x parts there again for that bit below as well then we've still got that minus 15 so i need to bring that minus 15 back in there we go and it all equals zero so add the minus 15 in there we go and that all equals zero now if we have a look at just the numbers that we've got there we've got a minus nine a minus one and a minus fifteen now if i group those all together and i'll probably skip a few steps on the next question here but if i group all the numbers together and i'll put them all at the end let's see what we've got we've got x minus 3 in bracket squared plus the y plus 1 in bracket squared and then all of those numbers so we've got minus 9 minus one that's minus ten minus another fifteen is minus twenty-five there we go and that all equals zero and obviously just to finish this off we just need to move that twenty-five over to the other side and i'll get it into the equation of a circle the format that we've looked at throughout the video so if we plus 25 to the other side just to finish this off and then we can get the center and the radius from that so we've got x minus 3 in bracket squared plus y plus 1 in bracket squared and that equals 25 now from there we can get the coordinates of the center and we can also get the radius now the easiest part there well i think they're both relatively simple once you've got to this point but it's nice and easy just to square root this number at the end to get the radius so we know that the radius is going to equal the square root of 25 as that is r squared at the end equals the square root of 25 and in this case that becomes a whole number that becomes 5. okay obviously the square root of a number is positive and negative but we can't have a negative radius so it's just going to be 5 there for the length of the radius now in terms of actually getting the equation of sorry the coordinates for the center now obviously we know it's minus a and minus b on these parts here so a is going to have to be that three right there so the center and there we go we know it's going to be three we've taken away three and on this one here we just need to be a little bit more careful because we've got a plus one there so we must have taken away negative one in order for it to turn into positive one and there we go that's our radius and our center taken from that equation okay so there's our two key parts for part a we've got our radius which is five and our center which is three and minus one just need to be careful there when there is a positive number in the bracket like the y plus one that you remember therefore it must have been negative one that we took away in order for it to become positive now the second part here says find the coordinates of the points where c crosses the y axis now in terms of actually thinking about this logically in terms of a coordinate graph if we had a circle let's think about what this would look like so i think the center is at three minus one which is somewhere there and if i was to draw a circle in which i always don't really like doing on the screen here because it's not the easiest to draw but it would look something and it's a really rubbish circle but there we go it looks something like that if i was to draw a basic sketch let's get rid of that center point because that really does not make it look good there we go let's imagine the center there we go three minus one not not to scale but there we go we have these two coordinate points where it crosses through the axes so here and here now the logical thing about this question is that any point on the y axis there the x coordinate down here is always going to be zero so effectively if we want to know where what the y coordinates are where it crosses through the y axis all we have to do is sub in x equals zero let's get rid of that sketch there because it's a little bit of an embarrassing sketch there we go but just to give you a sort of an idea as to what we're actually looking at we're looking at when x equals zero so if i sub that into my equation my equations there i just want to know what does it equal when x equals zero let's put that in so when x equals zero what do we get and i'm subbing it into this equation just here so we get 0 minus 3 in the first bracket so we get negative 3 squared plus we don't know the y coordinates so we've got y plus 1 squared and that equals 25 so we just need to rearrange this now negative 3 squared is 9. so actually i could just think that that's 9 there if i take away 9 from both sides we get y plus 1 in bracket squared equals 16 when you take away that 9. now if i square root both sides obviously we want to get y find out what y is so square root both sides we get y plus 1 equals this plus or minus the square root of 16 so it could be plus and minus four and obviously then we just need to take away one from both sides and we get y equals minus one plus and minus four and that there will give us two values okay because you could have negative one add four so y could equal three or we could have negative one take away four which would equal negative five so we've got three or minus five in terms of where it crosses the y axis they are the two coordinates there so obviously you could draw a bit more of an accurate sketch now now you know where it crosses through the y axis but there we go that's how we're going to go about solving these two parts of the question so for this question looking at intersecting circles with lines it says show that the line x minus y minus 10 equals zero does not intersect with the circle and then we have our circle equation so we've got our two pieces and obviously the process to solve this will be using simultaneous equations so although we looked at simple quadratic simultaneous equations before this question is slightly different as it says this fact here does not intersect and obviously in this relation we are actually looking at the actual circle so the process to do this is substituting that first equation in so we want to make that x or y equals now look at your circle equation because there are two x's in that equation but there's only this one y squared piece so it would be easier for us if we could make it say y equals so if we take our first equation and add y to the other side we get x minus 10 which is equal to y of course you could write that as y equals x minus 10. so if we take that and substitute it into our circle equation just put in a bracket around x minus 10 when we substitute that in we get x squared minus 4x plus and then we have our x minus 10 which is going to be squared and that is equal to 21. so now we have a an equation that is going to be a quadratic that we're going to need to make equal to zero and then to show that it doesn't intersect we're going to have to show that it has no real roots so having a look at this then let's first expand that double bracket and obviously if you are quick at expanding your double brackets you can obviously just write the answer to that which is x squared minus 20x plus 100 so if we write that all out and simplify it we've got 2x squared so 2x squared we've got a minus 4x minus 20x which is going to be minus 24x we have a plus 100 at the end but we're also going to have to minus this 21 to make sure it equals zero so we can simplify that straight away which is 100 take away 21 which is going to be plus 79 and that is now equal to zero now obviously if they did intersect at this point we would want to solve this equation we would want to find our x-coordinates and then we would substitute them back in to find our y-coordinates but this says does not intersect so we will look at the discriminant so for the discriminant we are going to look at b squared minus 4ac and then identify our a b and c so a is 2 b is negative 24 and c is 79. and again it's always worth writing them down a is equal to 2 b is equal to negative 24 and c is equal to 79 okay so putting that into b squared minus 4ac we are going to have negative 24 squared and take away 4 times 2 times 79 and if we type that into our calculator and let's just type that in we get negative 24 squared and the minus 4 times 2 times 79 and the answer comes out as minus 56 so we know that when b squared minus 4ac is less than 0 that less than kind of looks a little bit like my c there but that means it has no real roots okay so because we've got an answer of negative 56 we have no real roots and therefore these do not intersect okay so when looking at tangents to circles we just need to remember that the tangent is going to be perpendicular to the radius so for this question here where it says the point p which is 1 minus 2 that lies on the circle with the center 4 6 and it wants us to find the equation of the tangent to the circle at that point we can start by finding the gradient of the radius now what you can do on a question like this is to draw a little diagram which might help you to visualize it and that's fine to do as well it says the center of the circle is 4 six if we imagine that that is just there and there is a point which is across one minus two let's imagine that's here and obviously you could draw a sketch of the circle again it doesn't have to be perfect but it might help you to visualize it so we are going to find the gradient of this radius that i've just drawn in and that is between the point four six and the other coordinate down there which is one minus so obviously you don't need the actual drawing to do that because we can just use our y2 minus y1 over x2 minus x1 so y2 would be 6 take away minus 2 over 4 take away one so six take away minus two is equal to eight and four take away one is equal to three so the gradient of the tangent is eight over three that means that the gradient of the sorry the gradient of the radius is eight over three now we are going to find the gradient of the tangent because the tangent there is going to have a perpendicular gradient so we are going to find the perpendicular gradient which is the negative reciprocal which is going to be negative three over eight now we can take that and we can put it into the equation of the line given that we know it passes through the point one minus two okay so putting that into the equation of a line we will have y minus y one which is y take away minus two which is going to be y plus two so y plus two is equal to negative three over eight x minus x one and x one is one so x minus one we can now expand that so we have y plus two which is going to be equal to negative three over eight x minus one times minus three over 8 is going to be minus 3 over 8 but it was times a negative so it'd be plus 3 over 8. now we can take away 2 from both sides or what we could do here just to make that actually a little bit easier for us is we could just multiply everything by 8 to remove those fractions and that's fine for us to do so we would have 8y plus 16 is equal to minus 3x plus 3. that makes it a lot easier for us to rearrange that and as now we can just mine a 16 from both sides or we could move everything to one particular side it doesn't really matter but if we bring this up here in fact if we just move everything to the left then we would have 8y plus 3x when we move that 3x over and also minus the 3 over would give us plus 13 and that is equal to 0. of course you could write this as 8y equals or you could make it y equals if you wanted the question might ask you to write it in different ways but it didn't ask us to write it in a particular way so just having it equal to 0 8y plus 3x plus 13 equals 0 is probably the easiest way for us to write that answer okay so when it comes to triangles in circles there's a couple of different methods that you could use now you might be asked to find the center of the circle now this particular question says to show that a b is the diameter of the circle so of course that we could just find the midpoints of the point a and b but if it wasn't the diameter of the circle and we were wanting to find the center then you could actually use the the perpendicular bisectors of the chords so if we looked at the perpendicular bisector of the chord ac we would find the midpoint and then find the perpendicular line equation which would go through the circle like this we could then do the same with c b finding the midpoint and again the perpendicular bisector would cross through like this now those perpendicular bisectors do cross at the center of the circle so if you found the equation of those two lines you could set them equal to each other and of course we've already discussed looking at midpoints of lines and perpendicular bisectors so that additional step there wouldn't be too much work now this particular question is slightly different it's actually telling us to show that a b is the diameter of the circle now because it that is the diameter we know via circle theorems that if it is this angle here must be a right angle and therefore we could use pythagoras to show this proof so in order to use pythagoras we would have to find the lengths of these lines so a to c and c to b we could then use pythagoras to show the length that it matches the length of a to b so if we go about doing that you can see from a to c the length there goes from -8 to minus four now that is a distance change of four so we would do four squared and one to nine is a change of eight so if we did four squared plus eight squared and square rooted our answer that comes out as four root five so the length of that line there is 4 root 5. we can do the same from c to b now that goes from minus 4 to 4 which is a change of 8 and 9 to 5 is a change of 4. so 8 squared plus 4 squared for that length using pythagoras again would give us 4 root 5. so that length there has a length of 4 root 5 which matches the length of a to c so they are our two lengths and if we do pythagoras with those two to find the length of the diameter that should match the length we get when we use this same process so if we just do the same process using pythagoras between these two points then for a to b you can see it goes from -8 to 4 which is a change of 12 and it goes from 1 to 5 which is a change of 4 and if we do that using pythagoras again we get the length for root 10 so that comes out as 4 root 10 and now we can have a look at showing our proof so they are the definite distances if this triangle is right angled then if we were to square both of these sides and square root the answer it would have to match the length of the diameter so if we go about doing that 4 root 5 in bracket squared plus another 4 root 5 in bracket squared and if we square root all of that and we type that into our calculator we get the answer 4 root 10. so as you can see it has matched the length of the diameter therefore it is definitely a right angle triangle and therefore thinking about circle theorems angles made in a semi-circle form that right angle that 90 degree angle that we've shown so it must be a right angle triangle and therefore a to b must be the diameter okay so in this question we've got a slightly harder quadratic on the top and a different sort of quadratic on the bottom so on the bottom there hopefully you recognize that it's a difference of two squares when there's no x piece in the middle there it's because the numbers are the same and they've cancelled each other out so if i start with the bottom i just think it's the easier one there the factors of 16 that are the same are four and four so our difference of two squares there is x plus four and x minus four there we go and that's that factorized on the bottom that gives us that a little helpful hint for the one on the top there which is slightly harder because our factors are four are one and four or two and two now because we've got the little hint here that we you know we know we're going to have a four if it's going to simplify we know it's going to be the one and the four but we just need to have a look and figure out which way it goes around so we've got three x in one bracket and an x and another that'll allow us to get the three x squared again obviously check out my video on factorizing these quadratics or harder quadratics if you're unsure on these but in order to make 11 there we're gonna want the four to get multiplied by three to make twelve so we want plus four that'll make positive 12 and we want to take away one so i'll put the minus one over here in the other bracket and again now we've got it factorized it's nice and easy for us to cancel off these x plus fours and just write what we've got left on the top we've got three x minus one and on the bottom we've got x minus 4 and there we go that is our fraction there fully simplified okay so we've got f of x equals and then we have our cubic and it says use the factor theorem to show that x minus 2 is a factor which we've just done and then hence factorize fully the cubic so let's have a look so the first bit's nice and easy we've just looked at that for that one there x has to equal two so if we sub two in and we'll do this up here we get two cubed minus nine lots of uh two let's call it nine lots of two squared minus nine lots of two squared plus twenty lots of two and then minus twelve i'm just going to type that into the calculator as i've written it so two cubed take away 9 lots of 2 squared let's have a look plus 20 lots of 2 which is 40 and then take away 12. there we go and it equals 0 on the calculator there we are so that all equals 0. so therefore it must be a factor now obviously the new bit that we're going to have a look here is showing that x minus 2 is a factor and using that to factorize it so now we know that x minus 2 is a factor we're going to have a look at how we actually get this cubic so if x minus 2 is a factor there we go we'll have a look what must be in the other bracket and in the other bracket we're going to have the quadratic okay so let's have a look now we know we need to get an x cubed up there so if we need to get an x cubed in this bracket here we're going to have to have an x squared okay thinking back to obviously expanding your triple brackets back from gcse we know that when we multiply this out we've got x times the x squared which gives us the x cube that we want and then we've got the negative 2 times the x squared okay and that would give us negative 2x squared i'm just going to write that down underneath here because at the moment we've got negative 2x squared but whatever piece we put next in the quadratic is also going to give us an x squared in a second so we don't have our completed amount of x squareds or coefficients of x squared just yet so let's have a think now we've got negative 2x squared at the moment and we need to get to negative 9x squared now that's a means i need an extra negative 7x squared so next in my bracket there i'm going to have to put a negative 7 x and that's going to mean that when we expand it now look we get x times the negative 7x that gives me my extra negative 7x squared and it completes my negative 9x squared there we go which is what we're looking for in that cubic now onto the next bit because we also need to do the negative 2 times the second times that negative 7x and that gives us positive 14x now again i'm going to put that down here so plus 14 x because we've not quite completed the x part yet let's get rid of those and let's have a think about that now we need to get to let's have a look plus 20 x so we need to get to plus 20 x and at the moment we're at plus 14 x we need an extra six so that means i need to put a plus six here so we'll put plus six in let's have a look let's do that back in the other color there we are so we need plus six in there and let's have a look at what that gives us so that gives me the extra plus 6x that i was looking for there we are so we've got that plus 6x so that gives me the plus 20x in total that we need and then also we've got the negative 2 times the 6 which gives me the negative 12 at the end there we go so that is definitely factorized at the moment okay obviously we need to factorize this fully though so we have got another step to go about and that's because we need to factorize this quadratic here okay so we need to have a look at this one it's obviously thinking about factorizing quadratics and again i'll link the video for that in the description but we need to factorize this quadratic so for six we can have one and six or we could have two and three and two in order to factorize that let's have a think we need negative seven in the middle and six at the end so that's negative six and negative one so that bracket there is going to factorize into negative six negative one so x minus six and x minus one there we go right okay so obviously we haven't quite finished well we have we just need to actually write the final answer so obviously we've got our first bracket up here x minus two so we've got x minus two and then our two brackets here that we've just got for that quadratic x minus six and x minus one there we go and there that is fully factorized into our three brackets and that is how we go about using the factor theorem okay so this question says the equation x squared plus k minus three in brackets x minus four k equals zero and it has two distinct real roots show that this uh show that k satisfies this inequality and hence find the set of possible values for k now obviously the little hint here on the topic we're going to have to use is when it says has two distinct real roots obviously when we're looking at distinct real roots or roots at all we're looking at the um discriminant okay so when we're looking at this we need to figure out what the discriminant is which is b squared minus four ac and if it has it says it has two distinct real roots we know that b squared minus four ac is going to be bigger than zero okay that's what that means there we're to set that bigger than zero so if we actually plug these values in and see what we have so a is our value in front of x squared so a equals one b is our value in front of x which in this case is a bracket we've got k minus three and c is our value at the end so c is negative 4 k now if i plug these all into b squared minus 4 ac let's see what we get so b squared is going to be our k minus 3 squared so i've got k minus 3 in brackets squared take away 4 times 1 times negative 4k or 4 let's just write that down 4 times 1 times minus 4k is going to equal negative 16 k but we are taking that away so if we're taking away negative 16k that's going to become plus 16k so plus 16k is bigger than zero obviously we just need to expand that bracket there so i'm not going to write it all out i'm just going to expand it as it is so k times k will give us k squared we've got negative 3 k and we've got two of those so minus 6 k and then plus nine and then we've got that plus that 16 k at the end there and that's all bigger than zero and then obviously we've got two pieces we can tidy up we've got the minus 6k the plus 16k and if we tidy that all up we get k squared plus 10k plus nine is bigger than zero and there we go we've got that there what it was asking for show that k satisfies this inequality and there we go it matches so all we have to do to finish this off it says hence find the possible values of k so basically we're just going to solve that quadratic inequality and if we go about factorizing it to start with let's think what we get we've got k in the brackets there we go that's bigger than zero and obviously the factors of nine there we could have one and nine or we could have three and three if we want ten in the middle one and nine is going to work so k plus one and k plus nine so our two solutions we've got k equals negative one and k equals negative nine and it's asking about greater than zero so again we could figure think about this in terms of the quadratic we've got negative nine negative ones we've got two negative roots there let's have a look negative nine negative one and we are looking at the bit above zero so this line and this line so our two solutions we've got k has got to be less than minus nine and k has got to be greater than minus 1 and there we go there's our two solutions there they are the set of possible values k has got to be less than minus 9 and k has got to be greater than negative 1. okay so looking at out some algebraic proof now it says here given that n is a natural number which means that it's going to be positive and then it's going to be odd or even prove that n cubed plus 2 is not divisible by 8. so what we're going to have a look at is we are going to substitute in an even number and an odd number and if we can prove that neither of them are divisible by 8 then that would prove that given our restraints here it is not divisible by 8. so given that n is already in the question rather than using n for my proof i'm going proof i'm going to say that an even number is 2x and i'm going to say an odd number is 2x plus 1. so what i'm going to do is i'm going to substitute them in separately into n cubed plus 2 and hopefully we will show that neither of them are divisible by 8. so it doesn't matter which one we start with i'm just going to start with our even number as it's probably a little bit easier to substitute in and let's just see what we get so for an even number and we'll just highlight that we're doing an even number here we would substitute that in and that would be 2x in brackets cubed plus 2. that's easy enough for us to do because two cubed is going to equal eight so that is going to be eight x cubed plus two now for that particular question there obviously if we were showing it was divisible by eight we would want to factorize it by eight and if you look for this particular one the only way to factorize it by eight would be to put eight outside of the x cubed and then we would have always the plus two at the end so therefore an even number is going to be always two more than a multiple of eight and we'll write that down two more than a multiple of eight there we go and we are now going to move on to an odd number so there we go we have disproved that an even number works if we can also disprove the odd number then that is going to finish off this question and we would have disproved that given these restraints that it has to be a natural number that it could be divisible by eight so if we move on to the odd number which is going to be a little bit longer for us to work out as we're going to have a triple bracket if we substitute that in we're going to have 2x plus 1 and that is going to be cubed and then we're going to have plus 2 at the end so we'll start by expanding a double bracket so if we do 2x plus 1 and we do that squared again you could write your double bracket out but that is going to be equal to 4x squared plus 2x plus 2x so plus 4x plus 1 we now need to expand that by the final 2x plus 1 so if we do that as well times that by another 2x plus 1. so if we do 4x squared times the 2x that gives us 8x cubed 4x squared times the 1 is 4x squared moving on to the 4x times that by 2x is 8x squared times by the 1 is 4x and then onto the 1 times 2x gives us 2x and 1 times 1 gives us 1. now of course all of that also has not forgetting the plus 2 at the end so it's up to you when you add that in you could simplify this but we don't want to forget to put the plus 2 at the end so we may as well just put that in now if we simplify all of that we have 8x cubed if we join together the x squareds there we have 12x squared and adding together the x's we have plus six x and the one and the two makes three so looking at that we can't actually factorize it we could actually take out a factor of two but obviously not out of the full piece we would always have that plus three at the end of course you could actually split the three into two and one but we don't really necessarily need to i mean we could obviously change the plus three into plus two plus one just so that we can factorize it with the plus one remainder but again it doesn't actually matter either way we could factorize it here which would be two lots of four x squared plus sorry four x cubed didn't write that three very well plus six x squared plus three x and then you could have the plus one there but again we're gonna have this plus one on the outside so for this particular question this is always going to be an odd number as we have the plus one at the end so for this one here obviously it's going to be a multiple of two we could write therefore this is going to be always an odd number as we are adding one to an even number and therefore not divisible by eight so there we go what we've actually done is we have subdid an even number and we have subbed in an odd number we have our proof here that this is always going to be two more than multiple of eight for an even number and for an odd number it's always going to be an odd number and therefore not divisible by eight and we have shown our proof for this particular question okay so when we're looking at binomial expansion essentially we're thinking about from gcse level where we've looked at expanding and simplifying a triple bracket we're going to look at sort of how this all sort of interlinks with then higher powers when we have sort of a bracket to the power of 5 or to the power of 8 and how we go about finding those powers using this process now when it comes to binomial expansion it's all sort of interlinked with um sort of the the separate powers when we've looked at brackets now if we think about sort of a nice easy one something like uh x plus five in bracket squared we get a certain pattern that sort of comes out of these and when we expand one of these we get the x squared we get that first piece there's always to that power of 2. then we get the 2 in the middle we get plus 5x plus 5x and we always get these two there and then you get that five at the end which is squared as well to make 25 and then obviously those two in the middle are added together but we get this sort of little pattern and it's all linked to pascal's triangle and pascal's triangle is nice and easy to draw we get a one at the top then you add together uh the numbers either side of it to get the ones below and obviously one at the top has got a zero either side so we just have one and one below that but then when we add these together we get one on the left those two ones add to make two in the middle and then we get one on the end and we can just keep following this process and making this pattern so always one on the outside then the one and two above this next one makes three two and one makes three and one and we can keep on going one three and one makes four three and three make six three and one makes four again and then one at the end and we can keep on going basically with this now when it comes to this double bracket that i've drawn here that is this row here so one two one so we get one of the pieces at the start the x squared we get the two pieces in the middle okay which we've obviously just shown there we got the two in the middle and then we get the one at the end and following that little pattern of one two one now if i come away from that and we have a look at this triple bracket which is what we're gonna have a look at to start with that is obviously linked to the next row down that next row down is one three three one now it's not as easy to spot the pattern with this one um but obviously if we were to expand this all and have a look at all the pieces we would see we get the one at the start we always get an x cubed at the start we get the one number at the end which is going to be that three cubed which we'll have a look in a sec and we get these two lots of the three pieces in the middle i'm going to have a look at those now so if we have a little think when we were to expand this at gcse obviously what we would do is we'd expand two brackets to start with so we'd do x plus three and we're gonna fully expand this i'm gonna do it nice and quick though and x plus three would expand that which gives us x squared plus six x plus nine and then we expand that by another x plus three there we go and we do that we expand this all out we get the x squared times x gives us x cubed the x squared times the 3 gives us 3x squared and then we go on to the 6x and we just keep following this process so 6x times x is 6x squared the 6x times the 3 is 18x 9 times x is 9x and 9 times 3 is 27. okay we simplify this all down so eventually we get x cubed plus and we've got those two lots of the x squareds there which gives us 9x squared the 18x and 9x gives us 27x there we go plus the 27. and we end up with these four pieces here and as you can see in that line we've got those four pieces now it's not quite as clear with the triple bracket here or with the bracket to the power of three as to how we got those two lots of three in the middle but essentially we kind of got and let's have a look we've got the x squared here one two three pieces of x squared and we also got three pieces of x we've got the 18x the 9x the 6x and then obviously we've got the numbers on either end as well we've got the x cubed and the 27. so you can kind of see where this pattern comes from you've got the one lot of the x cube the one lot of the number at the end and then the three of each piece come in on uh coming in the middle there obviously eventually we got two x cubed plus nine x squared plus 27 x plus 27 but we're gonna have a look at how we can actually get to that using binomial expansion and using this process uh within pascal's triangle now obviously we don't want to have to always draw this triangle out every time and thankfully there's a nice little button that we can use to get that on the calculator and if you have your calculator this is a calculator topic it's a button that looks like this and it tends to be that you've got to press shift to get there first but it says n c r and it's normally behind the divide button so you normally have to press shift divide and it gives you this ncr now in order to get these numbers here which was what we're going to use for the first one we have to press obviously it's a power of three so we press three shift that ncr button and you'll get a c and then we press zero for the first one so three c zero and that always gives us a value of one we always get that one for the first one which is just there now for the next one we press three again three c one and we get three c one and that gives us three then we do it again so swap the one but go for two this time so three c two and that gives us let's just have a look we get three again and then for the last one we're gonna do three c3 which gives us one so make sure you know what that that button is in your calculator it's normally just behind the divide so three shift ncr and then whatever number we're looking for we start with zero and then one two three for those next ones so we can actually find out what the pattern's going to be just using this button on our calculator i'm going to be using that throughout the video so this is going to be a very important little button i'm going to be using that every question and i'm mentioning that as we go but as i said you've got to press shift first to get there so let's have a look at how we can actually use that on this question making sure that we can obviously expand this triple bracket using binomial expansion before we start looking at higher powers so if we get rid of this obviously we can get rid of all of this we know what answer we're trying to get to and that's this one here x cubed plus 9x squared plus 27x plus 27 okay so we obviously know that from gcse level that's going to be the answer and let's look at using binomial expansion to get there so once we've got the pattern then and we've got nc 0 is 1 and we've got 1 3 three and one and that's going to be our pattern now what we do is we look at the two pieces within within our bracket so we have an x as the first one and a three is the second one okay let's just actually just highlight different colors we've got an x and we have a three so that first piece and that one there the x is obviously the first piece in our pattern and we have one of those so we have one lot of and that's going to be x and everything here is going to have to be up to a power of 3 so that's going to be x to the power of 3 and we have one of those so all of these little pieces we're going to write now we need to make sure there's always a power of three in all of them now the next one we're going to have three lots of this next piece so we're going to add to that three lots of and the next piece is ever so slightly different now x the first one we're going to drop down that power now because we only have one of those x cubes in there so that x piece is now going to go down to a power of two so one x to the power of two but we need to make sure we balance out these powers and the next piece that we're going to throw in there is that three so we have then a three and that's going to be to the power of one not that we need to write the power of one there okay so all we're going to do is reduce that first piece which in this case is x so that's gone down to a power two and introduce that next piece in the bracket which is a three and we're just going to swap these around now so the x is going to keep reducing and that number the three is going to keep going up in powers until we get to a power of three for that one as well so the next one and again we've got three lots of this next piece so we're gonna add to that three lots of and we have x to the power of one this time that's going down to the power of one and that three is going to go up to a power of two there we go that's our third piece in that pattern done and then on to the last piece we have one of these so we're going to add to this one of now x is going to go down to the power of zero so that's gone and we just have this three at the end which again has to balance out to that power of three so one lot of three cubed and if we expand these all out and see what we get one lot of x cubed is x cubed there we go then we're gonna have three lots of x squared times three well let's just do the 3 times the 3 to start with which is 9 so that's 9 lots of and then we've got x squared with that so 9 lots of x squared then for the next piece let's have a look we've got 3 lots of 3 squared 3 squared is 9 3 times three squared so three times nine is twenty-seven again with that x to the power of one so two plus twenty-seven x and then at the end there we have one lot of three cubed three cubed is twenty-seven so plus twenty-seven at the end and there we go that matches obviously what we've got from expanding our triple bracket at the top we've got x cubed plus 9x squared plus 27x plus 27. so that's all we're going to do for binomial expansion obviously it gets a little bit more complicated as we start to go further down pascal's triangle and start to get larger numbers involved but basically that's all we're going to do we're just going to write down what what the numbers are and obviously in that line of pascal's triangle that we're going to be looking at and obviously just balancing that out uh with our powers of moving through but let's have a look at another one and see how we can apply this to a slightly harder question okay so this question says find the first four terms in ascending power of x of the binomial expansion of three minus a third x to the power of five giving each term its simplest form so let's have a look at this one now we've got a power of five so before we start dealing with the fractions and having a look at how we're going to deal with that we'll find the first four terms in the pattern here so we'll go for five c zero five c one five c two and five c three and just work out what that pattern actually is now obviously we know the first one's one so we go for five c one which is five the next one five c two is ten and the next one five c three is also 10. there we go so that's how many lots of this pattern we're going to have now this time the first piece in our pattern is a three that's okay and the second piece in our pattern is negative a third x now obviously that's okay as well but we're just going to be careful we're typing this into the calculator another way that you could write this here is you could write negative x over three so you could believe it as that as well i might actually write it as negative x over three instead just to sort of um make the terms a little bit easier to write but let's go for this then so we have one lot of the first piece which is three to the power of five so we've only got three more to do now and we're going to add to that we've got five lots off for the next piece let's just highlight these as we go so now we're on this five lots so we've got five lots of and we have three to the power of four and then we've got this negative x over three as well so negative x over three there we go on to the next piece again so we have 10 lots of this one so we've got 10 lots of so we've got 10 lots of and we've got 3 getting down to the power of 3 and then we've got this negative x over 3 which is now being squared and there we go and then for our last piece here again we've got 10 lots of this so we have 10 lots of three now going up going down to a power 2 and we've got the negative x over 3 which is now going to be cubed right so let's deal with all these pieces the first one's okay we've got 3 to the power of 5. work that out nice and easy that's 243. so if 243 is our first one you can always tick these off as you go so that's done the next one five times three to the power four well let's work that out to start with that's 405 so 405 and that's going to be multiplied by negative x over three we'll come back to that and deal with that in a sec onto the next one so we have 10 lots of so 10 lots of 3 cubed so we have 3 cubed which is obviously 27 times that by 10 is 270 so we have plus 270 lots of let's have a look negative x over three well if we square that's going to become positive so we're gonna get x squared on the top and on the bottom three squared is nine so x squared over nine there we go and that's that p start with and onto the next one obviously just being careful with those when you're squaring that remembering that's going to make it positive so we get the x squared on the top positive x squared 9 on the bottom on to the next one 10 times 3 squared is 90 so we've got 90 and that's going to be and we've got negative x over 3 cubed so when we cubed it's going to stay negative so we're going to have negative x cubed on the top and 3 cubed which is 27 on the bottom there we go and that's that's the last piece start with right now simplify all of this then so we've got 243 at the start that's all good and then we've got to be careful on this next bit now negative x over three i just mean obviously we've got we could have had the negative a third there we basically just want to do a third of 405 so if we do a third of 405 we have 405 divided by 3 is 135 and obviously it's negative there so it's going to be negative 135 x there we go on to the next piece we have 9 on the bottom there so we want to divide that 270 by 9. obviously it's positive this time so we'll divide that by 9. 270 divided by 9 is 30 so we get 30x squared onto the next piece 90 divided by 27 is not as nice it gives us a fraction here and gotta be careful because it's negative so that's not going to be plus there it's going to be minus so we get minus 10 over 3 which is what 90 divided by 27 comes out as with an x cubed there we go and that is those four first four terms finished so obviously just be careful on this because obviously sometimes you can get these fractional uh pieces or these fractional coefficients here and that is quite common on these questions so you just need to be nice and careful obviously just watching out for any pieces here that are negative like the two there just making sure that you show those with your negatives here as well when you do have these fractions involved just remember obviously you're just finding a fraction of that coefficient there again it's just like 405 multiplied by negative a third in terms of actually getting these coefficients so here we've got the 135 a third of that and one ninth of 270 gives us the 30. obviously our fractional piece at the end as well there so just being careful there's the first four terms and ascending powers of x and we've simplified each term okay so this question says find the first three terms and descending power of x of the binomial expansion of one plus p x to the power of ten it says where p is a non-zero constant and give each term in its simplest form so obviously p is an actual number that we're going to find and obviously hasn't asked us to find that yet it's just asking us to find those first three terms in the next part of the process though or the next part of the question it says given that in the expansion the coefficient of x squared is nine times the coefficient of x find the value of p so we're going to find that value of p uh you know later on in the question and then it says hence write down the coefficient of x squared and we're just going to follow this step by step and have a look at how to approach this and then you got one two if i go up so find the first three terms then we've got a power of 10 and just bearing in mind our first piece is a one which is always nice and our second piece is this plus px so as we've got a power of 10 we need to do 10 c 0 to start with so we've got 10 c 0 which is 1. we're only finding the first three terms we've got 1 10c1 is 10 and then we have 10c2 which is 45. there we go so that's how many lots of each we're going to have on this particular one here so for this one then we've got one lot of and then that's 1 to the power of 10 the first piece we're going to add to that we've got 10 lots of and it's now 1 to the power of 9 and then px as our second piece and then for the final one we have 45 lots of 1 to the power of 8 and then px to the power of 2. there we go right so if we simplify this down then one to the power of ten gives us one for our first piece ten times one and then times the px gives us ten px so we have ten px for the second piece and then for our final piece there 45 times 1 is 45 and px in bracket squared squares both those letters so we get p squared x squared there we go square in both of those letters so 1 plus 10 px plus 45 p squared x squared and there's our first three terms in ascending powers of x now it says given that in this expansion the coefficient of x squared is nine times the coefficient of x well the coefficient of x squared let's have a look and let's highlight that it's 45 p squared that's what's in front of the x squared and the coefficient of x is 10p so what it's saying is okay if one is nine times the other it's saying if we multiply this coefficient here by nine it would be equal to this coefficient here or in other words 9 times 10 p is going to be equal to 45 p squared or you could do it the other way around we could do 10 p is equal to 45 p squared divided by 9. completely up to which one you do obviously 45 nicely divides by nine so that would be quite a nice way to do it i'm just going to follow this process as i've written it down this way now it doesn't really matter which way it's probably easier actually to divide it by 9 because i think we're going to divide it by 9 anyway in a sec but let's have a look so if we times that by 9 anyway we get 90 p equals 45 p squared and then we just need to solve that as an equation so dividing both sides by p would obviously remove the p from the one side so we'd get 90 equals 45 p just dividing both sides by p there then we could divide both sides by nine or we could actually divide both sides by 15. completely up to you just divide both sides by 45 and that'll tell us what p is so if i divide both sides by 45 just write that down divide by 45 as we've got 45 p there we get 2 equals p or p equals 2. and there we go there's our answer for that part so p equals 2 equals p then it says hence write down the coefficient of x squared well the coefficient of x squared is 45 p squared there we go and we now know that p is 2 so that's 45 multiplied by 2 squared which is 45 times 4 and 45 times 4 is 180 and there we go there's our coefficient of x squared 180 right there we go so there's our final answer so in terms of actually what we did there obviously we had a different piece uh within our bracket we had a px i'm just being careful with that i think the point where you need to be careful with it is obviously this bit here where we get the p squared x squared i think other than that the rest of it was quite nice and simple and then obviously just reading the wording carefully it said one coefficient was nine times the other so obviously multiplying that by nine or dividing the other one by nine there gave us a little equation to solve we got our value of p and then sub that in to find the coefficient of x squared okay so when having a look at binomial estimations we can have questions like this so it says find the first four terms and ascending powers of x of the binomial expansion of and then we have 1 minus x over 4 to the power of 10 and use your expansion to estimate the value of 0.975 again to the power of 10 giving your answer three decimal places so the first thing we want to do is do our binomial expansion so for a power of 10 we'll want to use 10c1 which is equal to 1 we're going to want to use 10 c2 which is equal to 10 10 c3 which is equal to 45 and then 10 c4 for our fourth which is equal to 102 so putting this into our binomial expansion our first piece is one so we have one to the power of 10. our next piece we're going to have plus 10 lots of and then we're going to have 1 to the power of 9 plus oh no sorry multiplied by negative x over four and that is to the power of one we're now going to move on to our next piece so plus 45 lots of one drops down to a power of eight and our negative x over four jumps up to a power of two and then for our final piece and it's quite a large question so let's just try and move some of this out of the way so that we've got enough space there and for our final piece then we're going to have 120 lots of 1 to the power of down to 7 and negative x over 4 is going to go to the power of 3. so we just need to simplify all of this 1 to the power of 10 at the start is going to stay as 1. then we have 10 times 1 times the negative 1 on the top there so that would be 10 over 4 and that is going to come out as 2.5 so it's going to be negative as it is and let's just highlight that the negative x over 4 so rather than being a plus we're going to have a minus so that'd be minus 2.5 and again you can just type that into your calculator but minus 2.5 x for the next piece we have got 45 times 1 times the negative and obviously we need to expand that bracket as well so negative x over 4 squared would come out as x squared over and we can just write this to the side x squared over 16. so we're going to divide by 16 so 45 divided by 16 and if we type that into our calculator gives us plus 2.8125 so two point eight one two five and that's going to be x squared and then on to our final piece x uh minus x over four cubed is going to be negative and we're gonna have x cubed over 4 cubed which is 64. so i'm going to be dividing by 64 on this piece and it's a negative x cubed at the end there so it's going to be a minus 120 times 1 and then divided by the 64 gives us negative 1.875 [Music] and that is x cubed and we're only finding the fourth first four terms here but it does continue so we'll put that it continues there so that is our binomial expansion and now we need to go about solving this estimation so it says in the question here use your expansion expansion to estimate the value of 0.975 to the power of 10 which matches what's in here so what we want is that piece in the bracket the 1 minus x over 4 to be equal to 0.975 so if we set them equal to each other to start with to find out our value of x and then we can actually sub that into our binomial expansion so if we set them equal to each other we get 1 minus x over 4 which has to be equal to as stated in the question 0.975 and we just need to go about solving that and you can do that in two ways we could add the x over 4 to the other side and then subtract the 0.975 from 1 and that leaves us with x over 4 which would be equal to 0.025 again multiplying that by 4 then we would get a value of x which comes out as 0.1 so x is 0.1 we can now use that to substitute it back into our binomial expansion so you can write this down you can write down substitute x equals 0.1 into the expansion but if we just go about substituting that in we've got one at the start and again you could write sub x equals 0.1 but 1 minus 0.1 times 2.5 comes out as 0.25 and sub into the next part so 0.1 sub into x squared gives you naught point naught 1 and then we need to um obviously multiply that by two point eight one two five and that comes out as zero point zero two eight one two five and we've got our last piece there and again you're just typing these all into a calculator take away 1.875 multiplied by 0.1 cubed and that comes out as naught point naught naught just reading off the calculator 1875 there we go and if we simplify all of that uh again just typing all into the calculator it comes out as naught point seven seven six quite a lot of decimals here not point seven seven six two five that comes out as i've just typed in the 0.975 to the power of 10 which is a lot the more decimals so when we sub that into our binomial expansion we get 0.77 now the question here says giving your answer to three decimal places so if we type in 0.975 to the power of 10 into our calculator that gives us the answer and that's where i've got all my decimals from it comes out as 0.776329 there we go there are more decimals there but i'm going to stop there because it only wants it to three decimal places so if we look at these two decimals you can see 0.776 on the top one and 0.776 on the bottom one and we've got a matching amount of decimal places there so they round to the same number so we would give our answer as 0.776 to three decimal places and we've not got much space to write that down but if we just put it over here to the side we would just write naught point seven seven six and just put that that is two three decimal places and there we go that would be our final answer for using estimation with binomial expansions something a little bit different here it says calculate the length of qr now qr is this length over here so let's label that there we go that's qr now at the moment we don't have the opposite angle in fact we don't really have much in this triangle we have the 14 and that is that is actually it but we have got these parallel lines going on okay so hopefully we're gonna be looking at some parallel line rules here so something a little bit different and we have got this right angled triangle spr on the left there now on the right angle triangle we've got uh the length of eight and an angle so obviously we can use socatoa or potentially the sign rule or something like that to work out one of the missing lengths now this one obviously stands out to me because if we can work that one out that gives us another length in our little triangle on the right there so let's think how we could do that so thinking about some trigonometry again some socatoa we've got this side here which is our opposite so we've got the opposite and we're looking for the hypotenuse so we can definitely use socatoe to do this again obviously if you're using formula triangles let's have a think soh so if we're looking for the hypotenuse we're going to do the opposite divided by sine 62. so i can work that out definitely let's call that hypotenuse there let's just call it the question mark to be fair so the question mark is going to equal the opposite which is 8 divided by sine 62. there we go so sine 62. so let's type that in 8 divided by sine 62 and that equals 9.06 let's have a look so that equals 9.06 and there's a few more decimals i'm going to leave it like that this question doesn't actually say let's just go for two decimal places let's go for that there we go we're gonna do two decimal places so let's label that on the diagram so we've got nine point zero six and obviously if i'm doing its two decimal place i should probably keep it a little bit more accurate so it's zero six zero five six there we go i'm going to keep the answer on my calculator anyway and use that now let's look at this right triangle on the right now now i have got two sides and but yeah i'm still missing the angle opposite the one i'm looking for so there must be something to do with the parallel lines here and if we have a look let's just try and highlight this we actually do have an alternate angle here so going along our parallel line it jumps up to here and moves along so we've got an alternate angle there so this particular angle here is alternate with the 62 so that one is also 62 degrees now we have all the information that we need in order to use the cosine rule so opposite that we've got the little a so this would be our big a and then these will be our b and c let's have a look at the one that we've just made there will be our c so we can type that straight into the cosine rule obviously not looking at angles here but we might be looking at angles potentially but this one's looking for a side length so a squared equals b squared so 14 squared plus c squared which isn't that 9.06 and again i'm going to leave that in my calculator so i can use that in a sec and that's squared minus 2 times 14 times c which is this nine point zero six number zero six there we go and that's going to be cos 62. right there we go so let's try typing this in so obviously i've left that answer in my calculator so 14 squared plus answer squared minus 2 times 14 times answer and then cos 62. again you can just write it down and not use the answer button if you want i just quite like just actually using that so let's have a look so let's uh let's write what we get here so a squared equals press equals we get 158.9908 there we go 9908 there are a few more decimals but again we'll leave the answer on the screen so let's square root that so our value of a square root answer and that leaves us with a final value here of 12.609 with a few more decimals but if we do have that two decimal places for this question that would be 12.61 and this is all about centimeters there we go so 12.61 centimeters our final answer so obviously a lot going on in that question we had some socatoa which you know in a different kind of question could be looking for an angle it could be looking for a side length and then we had to apply some cosine rule and obviously there was a little bit of a parallel lines going on in that one as well so something a little bit different with a blend of different topics um but just something to be thinking about how you could apply this topic you know within a different sort of setting or a different sort of diagram that's not just a plain triangle okay so this question says work out the size of angle x give your answer correct the three significant figures now again just thinking about the question look i know we're finding an angle this time but look we have got this pair of opposites here and also we've got a pair of opposites over here so if i just label this up just like before we've got a and big a and b and big b okay but if i was to plug these into my slime like into my slime raw formula like we did before we'd have 6.7 over sine 95 and that would be equal to 5.4 this time over sine x now you can actually use the sine rule like this so obviously we could actually do quite a bit of rearranging here to get that x on its own but obviously uh if what we want to do is we want to get this part on its own this sine x part and if we times that to the other side now it's going to go on to the top of the fraction here and it's not really going to be on its own in that first step so there is a really quick step with the sign rule obviously just to help you with the rearrangement here and that's just to flip them over now the reason we can just flip them over if we think about a little bit of logic here if we have two fractions that are equal to each other let's just pick something easy like one half is equal to two quarters now that is a correct statement there one half is equal to two quarters and if we flip them over we get two over one is equal to four over two two divided by one's two four divided by two is two and two equals two is is again a correct statement so you are able to flip fractions over like this when we're equating them to each other uh obviously they equal different things there but the statements are still correct they still have to equal each other so when we're looking for angles all we're going to do is we're going to flip the fraction and fractions over and use the sine rule upside down so just so to speak so we've got sine a over little a is going to equal sine big b over little b and you're going to see now when we put the numbers in we get to rearrange this in one step just like we did before so if we plug all these values in i've got sine 95 as my big a and that's over the opposite there which is 6.7 and that is equal to sine x i'll put the x in a bracket just like i do with the normal angles and that is opposite the 5.4 okay so now we are actually able to eliminate or get or isolate the sine x on its own by getting rid of this 5.4 on the bottom let's divide by 5.4 again just times both sides by 5.4 so when i times this up and again i'm going to i'm going to write it in the same way as i did before so when it times up to the numerator look i'm going to write it just at the start here and i'll write it down to start with so that would give me 5.4 sine 95 and that's over 6.7 and that now equals sine x there we go so not got the x on its own but it does equal sine x now if i type all of that into the calculator let's see what we get so let's type all of that in so on the top i've got 5.4 sine 95 and on the bottom i've got 6.7 and if i press equals let's see what we get so i get quite a long decimal so we've got 0.802 nine zero three and yarn we're not going to write it all down i'm just going to leave it on my calculator screen for the moment but that is what all of this little bit on the left here equals when i type it all in now obviously that equals sine x it doesn't equal x it's not our answer so that equals sine x and i'll put the x back in a bracket now this would hopefully link through to normal trigonometry when we were looking at socatoe if you want to separate these two parts out and obviously figure out what x is we can do the inverse of sine so i can press that shift sign button and it will do the inverse and that will help me just to figure out what x actually is and not sine x so obviously thinking about your normal trigonometry there press shift and then sign and you get this sign minus one your inverse sign okay and we're going to do the inverse of this value here that's on your calculator so that'll be the inverse sign i'm just going to press the answer button so i'm just going to write ans here rather than that decimal but obviously that is just that decimal there that i'm writing in the bracket you can feel free to write it all in if you want but it seems like a an unnecessary step when we can use our calculator a bit more efficiently here so shift sign ans which is next to the equals button close the bracket and press equals and it gives us the size of the angle which is 53.408 and a few more decimals but again it wants me to round it to three significant figures so i can chop it after the 4 and we get 53.5 there we go and this is about degrees this time so degrees and there's our final answer it's obviously very similar process in terms of the way we use the formula but we have that final step there of using the inverse of sine to find out the size of the angle okay just like with normal trigonometry when we're looking at right angle triangles we still use that inverse of sine when we're looking for an angle okay okay so this question says the area of the triangle is 105 work out the value of x and in this case the value of x is the angle there sitting at the top so if we plug these all into the formula again we've got area equals so 105 equals there you go half a b sine c so let's just call that c and then we've got our a and b here so it's a half times 15 times the 16 there times the sine c and that in this case that is x there so sine x now again this is quite nice i'm going to take the quicker approach i'm going to divide all these pieces straight over a half times 15 times 16. again you could work that out and divide it over but i'm just going to write all the working outs i'm going to 105 i'm going to divide it by a half times 15 times 16 and that's going to equal sine c or sine x in this case there we go so let's type that in so 105 on the top and we've got 0.5 times 15 times 16 on the bottom and it gives us 7 8 or the decimal 0.875 so 0.875 equals sine x and again we want to know the value of x here so just like with all the other trigonometry that we've looked at when he's trying to find the actual x piece here or isolate and find the angle we want to do the inverse of sine so i'm going to leave that 0.875 on the calculator again it's not too difficult to type that in but i'm just going to leave it in because it could be a nasty one i'm going to do shift sign which brings up my sine minus 1 my inverse of sine and i'm going to put my answer in so i'm just going to press the answer button but i'm just going to put 8.875 in there there we go so i want to know what angle gives me that value there for sign so if i type that in so shift sign i'm going to press the answer button and close my bracket and here we go i get the i get my degrees here which is sixty one point zero four four uh nine is a few more decimals but again we are rounding this now to three significant figures as it says up there so that's the zero and again that four's not big enough to round it up so that's going to be 61 points and i'm going to keep the zero in there as it wants three significant figures it's just indicating that it didn't wasn't big enough to round it up there so 61.0 degrees for that question very similar process but when you're finding an angle just like all forms of trigonometry we have to do that inverse at the end so the shift sign to get the inverse to get the angle okay so there we go there's two examples uh just plugging it into the formula dividing everything over to the other side step by step or you can do it all in one jump like i've done on this one and then obviously just remember when you're looking for an angle you have to do that shift sign to get the inverse there to get the angle but work out the bearing from b to c so if we have a look at the bearing from b to c sorry b from c then we're going to be having a look at this angle down here how to get to b from point c so we're going to try and see if we can work out this angle here now we're going to apply a lot of the methods that we did previously in order to get this because um thinking about finding an angle within a non-right-angled triangle we can obviously apply the sine rule or the cosine rule now looking at this if we have a look at this as a triangle so obviously just highlight some of this um we've obviously got this length here this length here we know we can work out this angle here okay this one here let me just put i'll put a little question mark in there there we go we know we can work this one out and then thinking about how we might apply some cosine rule or some sine rule to that or something um how could we actually go about that now we worked out that angle we worked out this length and as long as i have those two i can actually do a bit of uh sine rule okay we can do the opposite pairs here and we'll have an opposite pair there or we could actually use the other opposite pair if once we've if we were to find the other angle potentially but the quickest way is going to be for us to just look at a pair that we can we already have straight away so actually if we go about doing some of those steps that we did previously we should be able to work out any distance or any angle within this triangle let's have a look at actually going about doing that and some of the steps i'm going to do pretty quick here because we have already looked at this particular question in this particular format but let's have a look at how we would go about it so obviously i said it's due east so we obviously knew this was 90 and we'd already done 210 take away the 90 to get that little angle inside the triangle which was 120 degrees there we are so we already know that 120 we then use the cosine rule to get this one here so we did a squared equals let's have a look the other two lengths squared so we had 6.4 squared plus 3.8 squared minus 2 times 6.4 times 3.8 there we are cos 120. there we go okay so i'm going to square root that all in one go just to get the value of a nice and quick so i'm going to do the square root of 6.4 squared plus 3.8 squared minus 2 times 6.4 times 3.8 cos 120 pressing equals there we go and we get our value here which was eight point nine two eight six there we go nine two eight six there's a few more decimals after that i'm just going to leave it as it is anyone's bounced to the nearest degree here but if i label that on there then so let's get rid of the x and we've got eight point nine two eight six a few more decimals there we go it's our little value there so now now that we've got that now we've used the cosine rule we've got a pair of opposites here we are those two and we've got another pair of opposites over here opposite the angle that we're looking for so we can use the sine rule now to work out one of the angles and let's just imagine we could also do the same thing if we wanted to work out this angle over here obviously once we've got one of the angles we can work out all of the angles anyway because it's angles in a triangle but there we go let's get rid of that and we'll focus on the actual angle that we're looking for so when we're looking for an angle we're going to use the you know upside down version of the sine rule so we've got sine a over a equals sine b over b there we go sine b over b so let's just apply all these so let's say this is a and this is my little a and we'll say this is b and this is my little b obviously you can swap and change them if you want so we've got sine x our sine a so sine x over 6.4 what's opposite that there we go and that is equal to sine 120 and that is over what's opposite the 120 the new 8.9286 that we've just worked out right so we can rearrange that we want to times that 6.4 up to the other side up to the top there so we'll have sine x is equal to 6.4 sine 120 all over the 8.9 a there we go so let's work that out using the fraction button so 6.4 sine 120 and that's all over 8.9286 i'm probably going to use the answer button there just to make sure this is super accurate but as i've left it to four decimal places it's not going to make much of an impact on my answer here so we've got our value of sine x so we get sine x equals 0.6207 and again a few more decimals 6 2 0 7 but we want to find the value of x so i'm going to do the inverse of sine so i'm going to get x equals remember we're doing sine minus 1 here so shift sign answer and we get 38.37 degrees there we go 38.37 a few more decimals after that but it wants it to the nearest degree so to the nearest degree x is going to equal 38 degrees there we go so 38 degrees there we are and it does say work out as a bearing so there's one more little step on here let's just highlight that down there there's our last little bit of working out right so now we've got that the angle there is 38 degrees we need to think about this in terms of a bearing now remember the question said the bearing from of b from c now if we look back at c there's quite a lot of drawing here so i might get rid of some of this stuff in a second and we draw our north line in this is where this question gets really quite tricky because obviously we've got the 38 degrees we kind of feel like that question's finished there but it wants the bearing of b from c so we're really what we have to have a look at is the total angle from north okay because it is a bearing not just that angle so we need to know not just the value of x there which is now 38 but the full angle here so if we get rid of that x we'll put our 38 in there as we've worked that out there we go and we also need to work out what the rest of the missing angle is there now there's nothing actually too tricky with working that out but i'm just going to get rid of some of this working so we can kind of see this a little better and we'll have a think about how we're going to actually approach this just this final step here so if we get rid of that we've got all of this information now we know that x is 38 we've worked that out and let's just get rid of some of this writing here get rid of some of that there we go right so hopefully you can see this a little bit more clearly now so in order to find that angle look if we think about one of the earlier steps we looked at in one of the bearings videos and that is just thinking about the angles here within these two north lines within these parallel lines obviously those angles there are co-interior so we can find this angle here if we can find this angle here because those two add up to 180 now that angle there on the right is not too difficult to find because we've got two of the other angles we've got that this one's 90 and we've got this 120 we're already told that the full angle there was 210 so we can find this one here by taking that away from 360. so 360. take away the 210 that's given to us and that leaves us with 150 degrees for that angle there so let's just get rid of some of this working again so if that angle there is 150 there we are that's 150 degrees given that these two here now add up to 180 that means that this little angle here has to be 30. there we go 150 plus the 30 makes them add up to 180 as they are co-interior right so now we've got that that is 30. we can now get the total angle of those two so let's just get rid of some of this again and let's just think about what we've got there there we go so with our full bearing then from north going clockwise from uh to b or sorry bearing of b from c so from c here down to that line with the b on that total angle there is going to be 68 degrees right there we go so the full angle there the full bearing is 68 degrees so obviously our answer is going to be 68 degrees but not forgetting as well as a final step it's got to have three digits so that's going to be zero 68 degrees and not forgetting that in the final step there it's always horrible when you finish off a question like this you've done all this working out all those steps and then you just miss out that one last little mark making sure it's got three digits but that is our final answer oh 68. so there's a lot going on in that question we had a lot of bearings rules then we had cosine rule sine rule and then obviously finding the missing angles using a few more bearings and actually combining that and all and actually making sure you get the final answer in the end of three digits there's a whole lot to think about in that question there and there's a lot going on and a lot of maths that you've got to be thinking outside the box with as well because obviously you've got to be able to think in the moment am i actually what do i need to use here do i need to use the cosine rule do i need to use the sign rule it's always just try and break it down in steps never think about what you're actually trying to find and where you're going with the question so it says here the graph of y equals sine x for x values from minus 270 to plus 270 is shown below and obviously that is shown on the graph so we've got minus 270 all the way over to 270 over here it says on the same axis draw the graph of and we've got y equals 1 minus sine x for values from minus 270 to plus 270. so that's the graph we're going to have to draw there and that doesn't look very nice now this is all sort of uh this question here is kind of very much linked into graph transformations we're actually transforming this graph and we just need to figure out how this graph is transformed but actually we can actually do this on a calculator now this particular graph transformation here has got two different transformations going on and we could rearrange that we don't have to write one minus sine x we could write negative sine x plus one and just sort of rearrange those if we wanted to so we could do it like this we could write minus sine x plus one oh i missed my bracket there let's put that back in there we go plus one and think about what these transformations actually uh represent so that negative in front of the sine x flips the y coordinates from being positive to being negative obviously check out my video and i'll in fact i'll link that in the description below on graph transformations as well because it's probably worth checking out because it does link very strongly into these uh trigonometric graphs so that negative there relates to a flip or a change of the symbols in front of the y coordinates we've then got the plus one at the end and again that's outside the bracket changes the y coordinate it changes the y-intercept it moves the whole graph up by one so you could actually take a graph transformation approach to this you could try flipping it over and then shifting it up one and we could probably do that quite nice and quick i'm gonna do it in two different ways though but just if you are quite keen on your graph transformations you could first flip it over and just think about that by flip oh that's not my pen let's just change that we could take all of these little coordinates here this one this one this one and this one i'm not bothered with the zeros because they'll stay where they are and we could just flip them down change the coordinates there and that would move the coordinates here and we actually end up with and i'll try and draw it there we go we end up with that and then we need to shift it all up by one so take all those coordinates shift them all at one that would shift it at one that would shift at one let's have a look these zero coordinates need shifting up one as well try and shift them all at one it's not the nicest now when you start drawing all over this there we go shift that all at one and it makes this sort of curve here there we go so drawing it using a graph transformation approach let's just get rid of that red one there we go that is where our curve ends up following this graph transformation okay not the nicest though that's quite nasty actually it's a very sort of horribly difficult graph transformation question having to sort of flip it over the x-axis there and then actually shift it at one as well but i'm going to take a little bit of a different approach but just so you know you could actually take that approach there using graph transformations so obviously check that video out if you're not sure on that you can always have a think about how what i've actually done there once you've had a look at that now instead of doing that i'm just going to take this nice calculator approach because this was actually a calculated question as well so rather than doing that i'm just going to plug my numbers into this formula here so i'm going to start with negative 270 so i'm gonna do one take away and i'm gonna do sine minus 270. there we go and that's going to give me my first value so one take away sine and i'll put minus 270 in the bracket and i get the value of zero there we go so that's my zero then i'm just going to change that for the negative 180 so i'm going to put minus 180 i'm just going back into my calculator and changing that value that gives me that value there back in negative 90. there we go that gives me 2. so negative 90 is up here at 2. i'll put 0 in and that gives me 1. there we go put 90 in and that gives me sorry i'll put that in 1 minus 90. yeah that gives me zero there we go then put 180 in the last one nearly done that gives me one where's that gone one is there and then 270. there we go and that gives me two and that is there and again just drawing that up with a nice smooth curve as best you can there we go and there is our transformed graph but again two different methods that you could take to approach that obviously if you're really good with your graph transformations you could have flipped it over first change the y coordinates made them positive negative and the negative one's positive and then obviously shifting it up one doing that transformation with the one at the end changing the y-intercept um or you could just take this approach just plug in the numbers into the calculator plot them all nice and neat and connect them all up together so it's up to you which one you prefer but just a couple of different ways that you can actually approach that question okay so this says here is the graph of y equals cos x between the values minus 180 and 180 and again we can see that on the graph it just stretched from minus 180 to 180. it says on the grid sketched a graph of y equals negative 2 cos x but again between these values okay so taking a little approach to this then i'm just going to take the calculator approach now again you could take a graph transformation approach to this but let's have a look at the end and see what that graph transformation is we'll take that calculator method to start with and just start plugging the values into this formula so we've got y equals negative 2 cos x so on my calculator i'm just going to type in negative 2 because i'm going to start with that first value on the x there which is negative 180 so minus 2 cos and then negative 180 in the bracket and it gives me the value of 2. there we go that equals two and i'm just going to do that same approach for all of them so that two goes up to there and there we go so back into the calculator i'm not going to sort of uh do all the little sub values in between i'm just going to go straight for negative 90 and see where that is so negative 90 ends up at zero there we go so that stays where it is i'm gonna go for zero now so back in negative two cos zero and that gives me a value of negative two and let's go back for 90 negative two cos 90 gives me a value of zero and negative 2 cos 180 gives me a value of 2 there we go so that value of 2 is there and again we just need to join that up with a nice smooth curve there we go and that's our point drawn there so again that's one method that you could approach if i swap to a different color here we'll have a think about the graph transformation as well so negative there flips the coordinates over so i could have actually started by flipping those coordinates over so let's go about doing that we've got this one here which is going to flip to the other side there zero state that one there stays where it is this one here on one flips down to minus one which can just about see it's about there somewhere we've got the one on the axis then which stays where it is and we've got this one down here which flips up above so the minus part okay just like before changes the symbols in front of the y coordinates so we can draw that in okay i'm going to get rid of this in a sec just about draw that in there we go so that's what the negative part does and then you've got the 2 there and again this is all linked to graph transformations but the 2 is going to double the y coordinates okay outside the bracket not affecting x it affects the y coordinate does what we expect two multiplies the y coordinates by two so this coordinate here that's currently at one goes up to two which is there because the zero stays as it is this one here which is at minus one goes down to the minus two which we've got here on our our actual graph again zero stays where it is on the next one and then again finishing it off with this one that one there goes up to the two so you can actually take a graph transformation approach again again i think it is actually a bit harder than just type in on in on the calculator but just so you know there are multiple methods that you could use here there isn't just one way to solve these sorts of questions but there we go that is our final answer there with our nice green line that we've drawn in okay so when it comes to trigonometric identities and equations it is essential that you know the trigonometric graphs so as you can see on the screen i have three of the trigonometric graphs and it's very important that you are able to do a sketch of these these are quite accurate the ones i've done on the screen but you need to be able to draw a sketch of them so you need to know that for y equals sine x it goes in the origin and then we have these waves that go up and over every 90 or up and under every 90. so it goes over the 90 down through the 180 under the 270 and then back through the 360. and this graph here goes on infinitely to the left and right of the y axis but we need to know how to draw a sketch of that and you're going to see why in some later questions for the one below for the cos graph this has just been translated or it's the sine graph which has been translated 90 degrees to the left so you can see instead it starts at 1 on the y axis but then it follows a very similar pattern it goes down through 90 under 180 up through 270 and up and over 360. and again this goes on infinitely to the left and right of the y-axis but it follows this pattern the tan graph is very different it does have these waves and as you can see there are asymptotes at minus 90 and 90 and then each asymptote follows on every 180 degrees and again that is to the right and to the left of these curves so you do need to know these graphs you need to be able to draw basic sketches of them and this is going to be very helpful when we are solving our trigonometric equations as i do use a graphical method again there are other ways of solving trigonometric equations but i do use this graphical method as i do like them and i do think that it's important that you know these anyway so it's always good to be able to use them for further elements so do make sure that you can draw a basic sketch of all of these graphs and you know them very well okay so within this is also some exact trig values now you should know a lot of these exact trig values from gcse level but these are just the basic trig values or the exact trig values now some of these can be written slightly different differently for example 1 over root 3 down here for tan 30 you could actually rationalize that multiplying top and bottom by root 3 and instead you could write it as root 3 over three that's absolutely fine and of course you will have a calculator for all these questions but there are some exact trig values that you should know and also you should be able to then relate these to the trigonometric graphs and we will see on some later questions when we are looking at solving them so for now all you need to do is you need to make sure you know these um exact trig values and you also need to know the method for finding them again i'll link the video in the description for how to find these so if you're not sure on how to find them or how to learn these then obviously you can go and you can check the full video in the description okay so the two main trigonometric identities that we need to know are these two so we'll start with the one on the left we've got sine squared theta plus cos squared theta is always equal or identical to one now this one is super important as is the other one but this one is very important now this can also be rearranged we could minus cos squared theta to the other side which would tell us that sine squared theta is also equal or identical to 1 minus cos squared theta so that is another one of our identities of course we could have also rearranged it by minusing sine squared theta to the other side which would have left us with cos squared theta which is identical to one minus sine squared theta so you need to know that identity and you need to also know these two rearrangements as we'll be using these a lot throughout our trigonometric identities for the one on the right when we're looking at tan theta is equal to sine theta over cos theta we can also do rearrangements of that so if we multiply cos theta to the other side we could also say that tan theta multiplied by cos theta so tan theta cos theta is equal to sine theta and again we could rearrange that by dividing both sides by tan theta and we could say that cos theta is equal to sine theta over tan theta so there are rearrangements of this as well that you need to know but as long as you know the two at the top that i've highlighted you should be perfectly comfortable rearranging them and of course that leaves us with quite a few identities here so really you only have to learn the main two and then obviously how to rearrange them but these are very important i'd make sure that you had these written down on some exam revision flash cards so that you always had these to hand as we're going to be using these for any trigonometric identity we'll be using one or the other and particularly with our trigonometric equations so do make sure you know these very well okay so looking at a trigonometric identity now as you can see here it says prove that 2 sine theta minus cos theta squared plus sine theta plus 2 theta cos theta squared is identical to 5 are always equal to five now this is just one example of lots of different types of questions so of course you do need to practice quite a lot of these but this is a really good one for us to have a look at so what we need to show in this is that for all of this on this left hand side is going to be equal to 5 or is going to always equal 5. and we can go about that by using some of our trig identities now we don't have a tan theta in this one but that's not necessarily to say that it won't appear but obviously we do have our other trig identity that has syn sine and cosine so if we're looking at this one we want to think about how we could change this so that it just becomes one or the other but maybe we'll have a look at expanding it first and seeing what we get so if we expand the first one here that's going to be a double bracket and that would be 2 sine theta minus cos theta and we'll go about expanding that so we would write that twice so 2 sine theta minus cos theta and you can already see that we're going to get some squares involved so if we expand that 2 sine theta times 2 sine sine theta would become 4 sine squared theta and then 2 sine theta times the minus cos theta and we're going to do that twice so that would be minus 4 and you could write either sine or cos first here but four sine theta cos theta again that piece that i've done that's just there i have skipped a step in my expansion i did sine um two sine theta times the minus cos theta and the minus cos theta times the two sine theta so just double that up in one step we now have to do the negative cos theta times the negative cos theta which gives us plus plus cos squared theta now we can move on to our second bracket now we've fully expanded that one and our second bracket is this one here so if we expand that as well that is again two of those brackets so sine theta plus two cos theta and we're going to multiply that by another sine theta plus 2 cos theta okay so if we expand this one sine theta times sine theta gives us sine squared theta then we've got sine theta times the two cos theta and we've got that twice so again that's going to be plus four sine theta cos theta there we go and then we have the 2 cos theta times the 2 cos theta which gives us plus 4 cos squared theta so we've got our two pieces and we need to add them together so we had a plus sign between these two so if we add those two pieces together and let's see what we get so we've got the four sine squared theta and a one sine squared theta here so that would be equal to five sine squared theta we've then got these middle pieces so we've got the negative four sine theta cos theta and the positive four sine theta cos theta so they would just cancel out and then we have the plus cos squared theta and the plus four cos squared theta so that would be equal to 5 cos squared theta now we need to figure out how we're going to make this always equal to 5. so we have two options we could change the sine squared theta into a cos or we could change the cos squared theta into a sine so let's just think about our identity and we'll put one of our rearrangements so let's change one of them we'll say that sine squared theta or we could go in fact let's just go with cos squared theta we'll say that cos squared theta is equal to always equal to one minus sine squared theta and that's going to allow us to change this this cos squared theta here into one minus sine squared theta so if we go about doing that we would still have the five sine squared theta at the start but we would now have five lots of one minus sine squared theta so just replacing the cos squared theta with one minus sine squared theta has now eliminated cos from our identity that we're looking at here so again we could have done it the other way around we could have got rid of sine and replace that with cos using our other rearrangements but it's absolutely fine for this one either way as long as we just have sine so now we could expand that bracket so we would have five sine squared theta and that would be plus five minus five sine squared theta there we go so where are we going to go from here now if we bring that up to the top let's just have a look at what we've got there we have got a and let's just highlight it 5 sine squared theta and a minus 5 sine squared theta which means they're going to cancel out which just leaves us with the plus 5 so we just have the answer 5. so because we've got the answer 5 there we have obviously proved that all of that just simplified down to five so therefore we have proved our identity that the two sine theta minus cos theta squared plus the sine theta plus two cos theta squared is identical to five and we've proved all of that so there we go that's all we had to do obviously we only had to use one of our identities we used this one just here and again you could have done that a slightly different way by rearranging it but again you'd have just got to the answer 5 because you would have had a 5 cos squared theta and a minus five cos squared theta and it would have all just cancelled out and you'd have been left with five okay so moving on to our first trigonometric equation this is going to be quite a tricky one quite a long one and quite tricky to fit all on the screen but i think we'll just manage to fit it in so the first part here says show that 4 sine squared x minus 3 cos squared x is equal to 2 can be written as 7 sine squared x is equal to 5. so the first thing to notice here is that in our second part we only have sine so in our first piece here we're going to want to try and eliminate this cos so if we eliminate that we're going to want to think about which identity would help us to eliminate that and that is that cos squared theta is equal or identical to one minus sine squared theta so if we replace that cos squared theta with one minus sine squared theta it's obviously going to start getting us towards our final answer so if we go about doing that we have 4 sine squared theta or sine squared x sorry because we've got an x in this as we're looking at an equation that is going to be now minus 3 lots of one minus sine squared x there we go and that is equal to two now if we expand that bracket we've got four sine squared x minus 3 that's negative 3 times negative sine squared x so plus 3 sine squared x is equal to 2. there we go now we can add together our sine squared x we've got 4 sine squared x plus 3 sine squared x that adds together to make 7 sine squared x so you can already see we've got something which is matching that 7 sine squared x we also have that minus 3 is equal to 2 and then we could minus 2 from both sides or we could add 3 to the other side now looking at the equation up here what it wants us to be written as it says it's equal to five so we want to add three to the other side which would give us seven sine squared x is equal to five and there we are we have showed that it equals that so that would be our final answer and we have shown that correctly using our trigonometric identity part b here says hence solve four values between 0 and 360 including 0 and 360 that this equation here um is equal to two so we're going to solve this equation and we're going to give our answer to one decimal place or our answers to one decimal place so we'll have a look at this and think about how we'll go about answering this now of course this equation here we have just shown can be written as this equation here now that we would be not be asked to do that unless it was going to be useful another reason it's useful is the second equation that we have shown only has sign in so we're going to be able to use our sine graph to find our solutions and we'll and how we'll go about doing that so at the moment we have 7 sine squared x is equal to 5 and we're going to go about solving it and you'll notice this was a lot of these questions that that part that it's asking you to show it as how it can be written is going to be used in the second part of the question so to go about solving this we want to get what sine x is equal to so if we divide both sides by 7 we'll know what sine squared x is equal to so sine squared x is equal to five over seven now to get what sine x is equal to we just need to square root both sides so we're going to have the sine x is equal to and don't forget when you square root you have plus and minus so plus and minus the square root of five over seven so there we go that's what sine x is equal to and now we can actually use our calculator to do shift sign and find the values there and if we put that into our calculator let's do the positive one to start with so sine minus one of and let me just type it in the square root of five over seven let's have a look let's just write this down so sine minus 1 of 5 over 7 is equal to and we get 57.688 so it wanted it all to one decimal place if so 57.7 degrees if we do the same for our next one so sine minus 1 of negative 5 over 7 and again just go into your calculator sorry the square root of i've just realized i forgot to put the square root symbol there put that in half tightly on the calculator we'll go back we'll put the negative 5 over 7 close our bracket and for this one we get the same answer but the negative version so negative 57.7 degrees so we have actually solved our equation here we've got two answers we've got 57.7 and refer back to our ranges here so it's between naught and 360. but this negative 57.7 falls outside of that so that's not going to be one of our solutions however it is going to be able to help us to find all of our solutions on our sine graph so we're going to use that these answers to see if we've got any other solutions and this is where it comes into play that it helps to be able to draw the sine graph and again this is where you can use different solutions you can use your quadrant method if you would prefer to that's absolutely fine but as i've said before i think it's really important you know these graphs so i always like to discuss them so if we look at the sine graph and we draw a basic sketch of what it looks like if we actually draw that sine starts in the origin obviously it goes up to one and down to minus one and it goes along my uh from 90 180 270 360 and we'll go back a little bit as well as we have that negative value and if i was to draw that in and again it's just a little sketch just so we can visualize it it would look something like that so if we label on our points negative um on the positive side 90 180 and 360 and then we'll have negative 90 and also negative 180 and let's just imagine where these points are so the points that we have we've got 57.7 which would lie between naught and 90 so that would be just there and we've got the negative 57.7 which is between 0 to negative 90 just here so in order to find all of our solutions all we have to do is trace along between naught and 360 and see where these points hit any other points on the curve so we obviously have our 57.7 just here and if we go to the right we obviously hit the curve at another point just here so in order to find that one we can just think about on the graph now 57.7 is 57.7 away from zero so to find that one on the right we would just do 180 and take away 57.7 and if we work that out and turn it into a decimal that comes out as 122.3 and there's one of our other solutions and that's between naught and 360. if we go down to the bottom one at the negative 57.7 and we go across from that we hit the curve here and if we keep on going we hit the curve again here and that is between norton 360. so we have two more solutions just there so to find that one if we add 57.7 to the 180 then that is going to find our next solution so if we do 180 add 57.7 and make sure you type that in correctly and there we go that comes out as this one here 237.7 and again we can do that for the reverse for the one next to it we can do 360. take away 57.7 and that gives us our final answer there of 302.3 now of course there is another point on the graph as i've obviously gone into the negatives there is another point just here but obviously that falls outside of our zero to 360 range so that's not going to be a solution that we can use so the negative 57.7 we can't use but we have got the 57.7 the 122.3 just here on that top part of the curve and then we had also the 237.7 and the 302.3 so we have four answers for this particular question so we would give them we would say that x is going to equal 57.7 and paired with that we've got x is equal to 122.3 on the top part and then down the bottom we had x was equal to 237.7 and matched with that on the right there on the right of that was 302.3 these are all degrees so we would put degrees with all of them and there we go they would be our four answers for this question all given to one decimal place so as you can see very important to know these trigonometric graphs and obviously knowing how to do your inverse of sine and just getting that to the point where it was sine x was equal to a number so that we could actually find that amount in degrees and relate it to the trigonometric graph so that's one type of question and obviously you can have harder ones than this and we're going to have a look at some harder ones as well okay so looking at one of our harder trigonometric equations you can see that obviously doesn't look very nice it says show that 10 sine squared theta minus 7 cos squared theta plus 2 over 3 plus 2 cos theta is always identical to or equal to 4 minus 4 cos theta then it asks us hence solve and we've got a very similar question to what we've looked at before but we're going to start with this first part and show that all of this comes out as four minus cos theta now straight away you can see that in our answer part there that we're looking to show that it's identical to it has a cosine so for all of the pieces that have a sign in or a sine squared theta in we're going to want to turn that into a cos so straight away we'll write down our identity for that and that is that sine squared theta is identical to 1 minus cos squared theta we'll plug that in and see where we get to so if we start by replacing the sine squared theta we have 10 lots of one minus cos squared theta minus 7 cos theta plus 2 and that's all over 3 plus 2 cos theta but we'll just go about simplifying the numerator to start with so the next step for the numerator would be to expand that bracket so we would have 10 minus 10 cos squared theta and that is then minus 7 cos theta plus two so that's our next step and now we're just going to think about whether we can simplify this now the only thing we can really do to simplify this is obviously to add the ten and the two together so we'll add those two together as they are our only like pieces on the top and that would leave us with 12 minus 10 cos squared theta minus 7 cos theta okay there we go so at this point we need to think about whether this factorizes this is just like a trigonometric quadratic if we think back to obviously looking at quadratics we could rewrite this in in terms of x so rather than writing cos i'm just going to replace that with the letter x and rewrite this as a quadratic so if i was to rewrite that it would look like this it would look like 12 minus 10 and that's cos squared so i would have 10x squared minus 7x and i probably want to rearrange that so it looks like a normal negative quadratic which is 12 minus and then put the 7x and then minus 10x squared so this is going to really help if you are struggling to factorize trigonometrics when we have the quadratics and we're trying to actually factorize them you can just write that the cos or sine or whatever we're using in terms of x and then look at it from there so if we go about factorizing that we want to think about those numbers that multiply to make 12 but of course we can actually change this quadratic and um swap back our brackets at the end so just to make it a little bit easier i'm going to write it as positive 10x squared plus seven x minus twelve as this is really nice and quick to factorize our factors here would be one and twelve three and four or two and six and you can obviously see that that is going to be um one of those numbers but we need to think about our brackets now it's unlikely that it's going to be 10 and 1 so we'll go with 2x and 5x and see if it factorizes from there of course if you don't want the factories you could type these into your calculator to get your solutions um but nice and easy for us to factorize we have two times four so it's going to be a it's going to be that one there three and four again take your time with this but two times four would equal eight and three times three um sorry three times five on the other one would be fifteen and fifteen and eight we can make seven with we're trying to make positive seven so we'd want positive here and negative here as that is going to make negative 8 plus 15 which makes plus 7. so we've managed to factorize it all you have to do because we swapped over the negative quadratic and made it positive you just swap these brackets all the pieces in the bracket over so we would have and it doesn't matter when there's a plus but i'll swap it over anyway 3 plus 2x and over here we have 4 minus 5x okay so there are our brackets done but obviously these pieces the x's should have been a cos so we just want to replace that which would mean that our numerator now wouldn't have the x there but would have the cos so our numerator would be 3 plus 2 cos theta and next to that 4 minus 5 cos theta there we go and our denominator if we write that in as well was 3 plus 2 cos theta now if you have a look obviously looking at this fraction here you can see that we have a matching factor on the top and the bottom which is the 3 plus 2 cos theta so they can both be cancelled out which means all we are left with is the 4 minus 5 cos theta so all of that would go down to four minus five cos theta and as you can see that matches what we're being asked to find in the question where we're being asked to show that it was always equal to four minus five cos theta so we've managed to prove that for step one so this would be the first stage of this question obviously you can see that i've run out of a lot of space here so i am going to have to get rid of this working but we're going to have a look now at the second part of the question so we'll get rid of all this working and we will have a look at the second part okay so for the second part of this question it says hence solve between norton 360 including naught but not 360. the equation and then it gives us the original equation or the original pieces that we had there but instead of putting theta they've put x but this time it doesn't say that it equals four minus five cos theta it says that is equal to four plus three sine x now we already know from the first part that all of that is going to be equal to 4 minus cos theta and if we were to write that using the letter x that would just be 4 minus 5 cos x so we know that if both of them are equal to each of one another that four minus five cos x is going to be equal to four plus three sine x and that's going to be a little bit easier for us to use than that fraction and again you're seeing in this question we're using the piece from the start to help us with the second part here so the first thing we're going to want to do is set them equal to each other so if we say that 4 plus it doesn't matter which one we write first let's just go for 4 plus 3 sine x and that is going to be equal to four minus five cos x now again we want to obviously solve this so we want to get this to a point where it only has syn cos or tan and we'll think about how we could maybe get there now the first thing that you might notice here is there's a 4 on both sides so we could minus 4 from both sides and that would just leave us with 3 sine x which is going to be equal to negative 5 cos x so that would be our first step that's gotten rid of a four from both sides what we could now do to uh get the get it so it's just sine x or just cos x and i'm just going to divide both sides by three that's going to remove the three from the start and then we'll think about what we could where we could go with next so if we divide both sides by three that leaves us with sine x equals which is going to be helpful is equal to and i'll just divide the negative five by the three so negative five thirds cos x now at this point here you'd hopefully notice that one of our trig identities is getting gonna get involved and that is if we divide both sides by cos x we're gonna have sine x over cos x now we know from our trigger identity that sine x divided by cos x is equal to tan x so that is going to eliminate our two different trig pieces there because now we're just going to have tan x so if we do that we would have sine x over tan x when we divide by cos x sorry sine x over cos x is going to equal negative 5 over 3. and again you could write in what you've done there we've just divided by cos x there we go and of course sine x over cos x is equal to tan x so we could just rewrite this that tan x is equal to minus five over three so that's going to be very important there because now we're in a position where we can solve it so at this point we can do the inverse of tan and get our solution so for tan minus 1 so we get our angle and that's ten minus one of negative five over three if we type that into our calculator we get the answer and it comes out as minus 59 point zero so we've got our first solution now of course that has fallen outside of our range that we were looking at as we're only looking between 0 and 360. so for this we are going to have to draw our tan graph and we're going to have to visualize where those solutions are so if we get rid of some of this working and we can just draw the graph over here and that's going to help us just to visualize that so here we go let's just get rid of all of this lots of pieces there but we want to draw this tang graph i'm going to draw a nice sketch of that so if we draw some axes again our first wave goes through the origin and then we have our asymptote at 90 and our next one goes up then uh through the 180 and our next asymptote is at 270. so that's going to go up through the center there and i've drawn that very badly so we should probably draw that again again it is just a sketch just to help you to visualize it but it's good to draw it quite well and our next one would start at 270 and it goes up through 360. so we could stop there but of course that does actually carry on but we don't need to draw the rest in because we're only going up to 360. so let's think about where that solution would lie we've got negative 59.0 and that would go between 0 and the negative 90 where are other asymptotes so it's going to be somewhere in the middle let's just imagine it falls just there if we were to trace that down and then of course all we're going to do is go across so we would go across we'll hit the next curve just here and then we're going to go across again and we'll hit the next curve just there again you might know some quick ways of jumping between these 10 values and again if you use any of your quadrants and things like that that's absolutely fine but again as i've explained before i do like to visualize what these graphs look like just to imagine where they're gonna go so for the first one to jump between these two it is just a jump of 180 degrees so if this one here is negative 59 let me just check it on the calculator because it doesn't say in this question how to round it it did come out as negative 59.04 so i'm going to give it to two decimal places just to make it a little bit more accurate so negative 59.04 to jump to the next one there all i need to do is add 180. so if we add 180 to that it comes out as 120 points and then 96 it comes out as and there we go so jumping onto the next one i would go another 180 so plus 180 again and that comes out as on the calculator that comes as 300.96 and of course if we jump to 180 again it would go over our limit which is 360 degrees so these three these solutions here we've got three solutions negative 59.04 which is below zero so we can't have that one but we do have 120.96 and also 300.9 so we do have two solutions there again we would round that as we see appropriate so we could say that x is equal to 120.96 of course we could write to that is equal to or we could say 121 degrees to the nearest degree and we also have x is equal to 300.96 or we could write that that is equal to 301 degrees again just been to the nearest degree there so obviously you can see this one is pretty tough there was quite a lot of steps but you can see a lot of similarity between these trigonometric equations whereby we are isolating it down to either since or 10 we're doing the inverse to find the angle and then we're just sketching it onto the graph so we can see how many solutions we've got given that a equals 3i plus 4j and b equals i minus 2 j find lambda if lambda a minus b is parallel to the vector i plus 3 j now straight away before we get started you can see this time the coefficients of i and j are not equal the coefficients of j in this particular instance right here are three times bigger than the coefficients of i and we're going to bring that concept in in the last step back when we were making them equal to each other we're just going to think about how we're going to approach that so we'll come back to that right at the end so we've got lambda lots of a and a is 3i plus 4j so we've got lambda lots of 3i plus 4j and then we're going to take away from that you got to be careful with this one b and b is i minus two j now as we're taking it away i'm gonna put this one i minus two j into a bracket obviously it doesn't matter if it's a plus between them but when we're taking away as we're taking away both of those we don't wanna forget to take away the negative two and otherwise we might only take away the eye so do be careful when you're doing takeaways particularly when you've got more than one piece that you're taking away you put those pieces in a bracket just to avoid any mistakes there so let's expand this bracket out and then obviously change the signs on that second one so we have 3 lambda i plus 4 lambda j and then we're going to take away i and also take away the negative 2 which makes it plus 2j and there we go that's all of those pieces listed out now again we want to have a look at the coefficients of i and the coefficients of j separately so let's join them together so we've got the 3 lambda i and we are i've also got the minus i so just put minus i and then we've got plus four lambda j plus the two j and if we have a look at the coefficients that we've got here we've got three lambda and also a negative one in front of the other one that'll be hard to highlight that one there when we don't write the one in but we know that that's a one then we can write that in as when we get to it now the coefficients of the other one we've got four lambda being positive again plus two so let's write these out and see what we've got now i'm going to do it like we did before i'm going to change this in a sec because what i'm about to write isn't entirely correct but i'm going to say okay well let's set these equal to each other so we've got three lambda minus one and we'll set that equal to four lambda plus two now technically whatever in there is not mathematically correct and this is where we're gonna have to look back towards this just here and that's because we know that the coefficients of j are three times the size of the coefficients of i and at the moment we've said that they're equal now in order to work around that what we could do is we could say okay well this side here we need to divide that by three you can do that and that's absolutely fine ultimately you're going to get to this next step that we're going to write but we could say okay we need to divide that side by three or what we could say is okay well actually the coefficients of i we're gonna have to multiply by three and if we make the coefficients of i three times bigger they would then match the coefficients of j which are already three times bigger if that makes sense so let's just times this side by three so we'll stick it in a bracket we'll times it by three and then the coefficients would be equal so let's go about times in that by three so the three lambda bill would become nine lambda and the minus one will become minus three so nine lambda take away three has to be equal to four lambda plus two and now we've got a nice easy equation to solve so let's solve this over here so that's minus four lambda from both sides that would leave us with five lambda and let's add three to the other side and that's going to equal five so five lambda equals five so one lambda when we divide by five has to equal one and there's our final answer right there we go so that's how we're going to approach these questions and it's all about looking at the coefficients and making this little equation and making them comparable so if the coefficients or if there is only one letter then we need to eliminate one of the other letters or one of the other directions if the coefficients are equal we can set them equal to each other and solve it and obviously then if the coefficients are different so we have something like i plus three j or maybe you could have f i know five i plus j you would actually have to multiply one of the sides to obviously make them equatable and then obviously you can do your equation from there so this one says we've got three vectors we've got a which is p over minus q we've got b which is q over p and c is seven over four given that a plus two b is equal to c find the values of p and q well first of all let's work out what a plus 2b is so we've got a which is p over minus q and we're going to add to that 2 lots of q over p and it says that that is equal to c which is 7 over 4. now first thing to note here is obviously we have two unknowns we've got a p and we've got a q and there's a little hint in that and where if we do have two unknowns sometimes we may have to look at simultaneous equations there's a little bit of a hint in here that we may have to involve another topic within this question so let's have a look now we need to do two lots of p and q so we've got p on the top and let's just add together that top line there but also times in that that right vector there the two lots of p over q by two at the same time so we have p plus and we're going to have two q's as we're going to multiply that one by two and that's going to equal the top number over there which is seven now if we have a look at the bottom line we've got negative q plus two lots of p which is going to equal four so negative q plus two lots of p is going to be equal to four and there we go we've got two equations both with p and q in and we can use simultaneous equations to solve that now just for the purpose of making this look a little bit nicer what i'm going to do is i'm going to swap over the q and the p just to make sure that they're aligned with our simultaneous equations you certainly don't have to but i think it looks a bit nicer when we line them up in the same way so rather than writing minus q plus 2p i'll write 2p minus q so let's get rid of that and we'll write 2p minus q there we go so we've got our simultaneous equations now in order to solve our simultaneous equation obviously we need a similar numbered coefficient with either ps or q's i tend to like to make the second one the same so i'm going to times the bottom equation by 2 so that we have 2q for both of our equations so times that 1 by 2 and rewrite it so let's have a look when we times that by 2 we get 4p minus 2q and that now equals 8. there we go so we've got our two equations we can forget about the one in the middle in fact let's just get rid of that one okay we lost a bit lost the queue there let's get rid of that there we go and we'll solve it from there let's move this up just above there we go and we have different signs in these simultaneous equations so we'll add them together let's have a look we get 5p the 2 and the minus 2q obviously disappear they cancel each other out and we have 7 plus 8 which is 15 and there we go nice and easy to solve then p is equal to 3. 15 divided by 5. now we've got our value of p just like with normal simultaneous equations we're just going to substitute that back into one of these equations up here so we'll sub it into the top one so if p is 3 we get 3 plus 2q is equal to 7. we can subtract 3 from both sides so 2q is equal to 4 and divide by 2 q equals 2. and there we go solve that question just highlight our two answers there because we've got p equals three and we've got q equals two so something a little bit different there obviously we didn't write it in our normal uh vector vector way with our eyes and our j's just going to show you that we can actually have these questions involved with normal column vectors as well with something a little bit different with simultaneous equations there now with this question here we're going to have a look at the magnitude which is essentially the distance that two vectors actually travel when we combine them together i'm going to have a look at drawing that down and writing a little sketch of them to see what it looks like so as always grab a piece of paper grab a pen make some notes and we're going to start with this question on the screen so it says given that a is 3i plus 4j and b is negative 2i minus 4j and we're going to find two different things here now already you might have noticed there's a little bit of new notation to write and that is this bit here where we have it says find a and it's between these two lines now those two lines just mean the magnitude of whatever's inside it so this means the magnitude of a and if we visualize what a would look like i'm going to have a look at what the magnitude actually is in terms of a visual picture so for a it goes 3i and 4j now that means it goes three across if we just pick any particular direction and then it goes for up obviously not drawing this to scale so that's what our vector looks like now in terms of our starting point we started over here and we finished up there so our vector has actually gone in this direction i'm drawing with that green line and as you can see that forms a right angle triangle now the topics that are going to be involved in this obviously when we're looking at right angle triangles we can have either pythagoras or potentially we could use some socatoa and if we want to find the length of that line which is what the magnitude is in this case the actual distance that we've traveled by going three across and four up we're going to use pythagoras to work that out so we can draw a right angle in if we want but ultimately if we're going to use pythagoras to find this magnitude we just need to go about using pythagoras's theorem so for part a here we have a length of three and a length of four so we're going to do three squared plus four squared and then we're going to square root our answer now rather than doing this in two steps let's just stick our square root right over the top and there we go we can stick that straight into the calculator and we get the answer of five so the magnitude of vector a is five or in other words it travels an actual distance of five units whatever those units may be now for part b it says find the exact value of and exact there giving us a hint that our answer is going to be inserted form which we'll hopefully see in a sec and it wants us to find the magnitude of 2a plus b now it's completely up to you if you want to find two a plus b by writing these in column vector like we went over in part one you can do or you can just use your sort of mathematical logic here to get two a's together so two lots of a well if we deal with the i's to start with we'll have two of these three eyes that gives us 6i and then we've got a b that we're going to add to that and then that we've got negative 2i so we've got 6i take away 2i and that's going to leave us with 4i so it's going to go 4 across then we've got to do the j's so if we do those separately we've got two lots of the four j which is eight j and then we're going to add to that the minus four j over here and that leaves us with four j so we have four i plus 4j so that is our vector 2a plus b of course we're going to find the exact value of the magnitude of that vector so for part b we're going to take the same approach and we're going to use pythagoras again again you could draw a sketch of this four across four up but once you get your head around this hopefully you can avoid potentially having to draw some of these sketches when they are slightly easier like these ones here so in terms of using pythagoras we know that goes four across so we'll do 4 squared it goes 4 up so we're going to do another 4 squared and we're going to square root our answer now 4 squared plus 4 squared is 32 so we have the square root of 32 and to be fair that is the exact value there although when it comes to certs we should always look to simplify them wherever possible and considering you have a calculator for these questions it's easy enough just to type into the calculator and it's for root two there we go it's root 16 root two root 16 goes down to four so four root two there being our final answer for root two so there we go that's what we mean by the magnitude of a vector okay so on to our last little bit here it says in a triangle pqr p to q is four i plus j and p to r is six i minus eight j find the size of the angle q p r in degrees to one decimal place and it also wants us to find the area of the triangle so let's go about drawing this now let's start with p as both these vectors are going from p so let's call this point here p now the first vector there is going four across and one up so let's just imagine that one two three four across one up so let's just imagine it's here there we go and we'll call that point q now to find point r that's going six across and eight down so let's do it again one two three four five six and then one two three four five six seven eight right okay there we go not to scale but just to give us a general idea of what it looks like and that is our point r now let's join these all together so p to q q to r and ah back up to p there we go so this wants us to find the size of angle q p r so q p r is just over here and again let's call that theta there we go and it wants us to find it in degrees to one decimal place now in order to find an angle within a triangle like this as this is not necessarily a right angle triangle we actually don't know we can't really tell just by looking at it like that we would need to use the cosine rule or potentially the sign rule so we're going to involve another topic here and again things like the sine and cosine rule i'll link in the description below just to make sure that you're nice and clued up on those if it's something that's ripped your mind so looking at this then what size is or what sides can we find the length of or in other words the magnitude well we can do that for p to q and we can do it from p to r so let's start with one of them let's go for p to q to start with now here's our uh here's our little notation here we've got four i plus j so we've got a distance of four and one so for p to q and let's write it in proper vector notation let's find the magnitude magnitude of p q so we've got four squared plus 1 squared and we're going to square root that so 4 squared is 16 plus the 1 is 17 so we have the square root of 17. there we go so square root of 17 for our first one for p to r we have 6 i minus 8 j so let's find the magnitude of p to r well that's going to be 6 squared plus the 8 squared and we're going to square root that again okay so let's stop doing these in our head let's get them not typed in so 6 squared plus 8 squared square rooted there we go gives us the answer 10. one that we should know actually there 36 and 64. so there we go there is the length of 10. there we are now have we got enough information to find the angle now if we had the length q to r we could actually use the cosine rule okay but we don't have the length q to r just yet but actually looking at our diagram it's not that difficult to find there's two ways that we could do this we could either take the vector p to r and take away the vector p to q or we could do it vice versa and just be very careful in the direction that we're going in or to be fair we could just use a bit of a logical approach now to get from p to q we've had to go four to the right if you remember drawing those dots and we've got an extra two dots across to get um from q to r so in order to do the vector from q to r we could actually just take this logical approach if we've gone four across from p to q and an extra then two obviously because we've gone six across to r then we know that must have been two i across and we can take the same approach going down from p to q we went up by one but from p to r we went down by eight so if we went up by one to get to q but down by eight to get to r in total that's going to have to go down by nine so minus 9 j now you may not like that logical approach so you don't have to use that you can actually just take one of the vectors and subtract the other one if you have a look and let's just highlight it the vector p to r 6 i minus eight j if we do six i take away the four i in the previous one that leaves us with two i and if we do minus eight j take away the one j that also leaves us with minus nine so you can just take that approach as well but i do think it is worthwhile just thinking about logically why you get the answer and not just simply taking one away from the other and not really understanding why but it is a nice little shortcut there to actually get these vectors when you're trying to find them so there we go we've got enough information now to also find the magnitude of q to r and we can take the same approach just using pythagoras again so from q to r we can do the square root of 2 squared plus 9 squared 2 squared is 4. 9 squared is 81. so that is or the square root of 85. right so we've got all the side lengths now all we need to do is go about actually finding this particular angle so let's get rid of all of this working out now obviously we need our magnitudes there so we'll have to leave those or we can just draw that onto the diagram now root 85 and we'll get rid of all that working out there we go so let's get rid of all of that and just work with our diagram quite a lot working out up to this point but to be fair we've only had to use pythagoras a few times so nothing too bad so to find the size of that angle we obviously need to know the cosine rule and obviously in this particular instance we're looking at the cosine rule for the angle so the cosine rule for the angle is going to be cos theta and you do need to know these well and we're going to do b squared plus c squared minus a squared all over 2 bc or whatever variation we're looking at i'm treating theta as being our big a in this case but let's go for it 10 squared plus we've got root 17 squared which is just 17 but i'm going to write it in anyway just to make sure we don't get confused with any of this and then we're going to take away the root 85 squared which again is just 85 but we'll write it all in as normal and that's all over 2bc so 2 times 10 times root 17. now i'm going to do a little bit of a cheat here before i type that calculation in i am just going to press shift cos and do the inverse straight away so just a little step that i am skipping here rather than writing it all down i'm just going to do the inverse straight away so on the top then 10 squared plus root 17 squared plus root 85 squared all over 2 times 10 times the square root of 17. there we go close brackets on your calculator and we should let's have a look if you type it in correctly we should get 67.16 so there we go our angle to one decimal place is going to be 67 it's 0.16 so 67.2 degrees and again just remember i have skips to step in my writing there just to reduce the amount of working out that's on our screen so there we go there is our angle and the second part is asking us to find the area of the triangle so we've got another bit of trigonometry coming in here we're going to have to use half a b sine c and just as a little tip here i'm going to keep the original answer on my calculator so the 67.16634582 so if you're not going to keep that on a calculator do make sure you have written down the full number there because we don't want that to affect our area and part b so for working out the area we're going to do half a b sine c again we now know the angle here so we can get rid of theta oh not the whole triangle there we might be able to do it there we go and we can put the angle in so 67 in fact let's just write the whole angle in 67 point one six six three four five eight two and now we can do half a b sine c so the area is going to be a half times our two adjacent sides there so 10 times root 17 times sine of this angle 67.16 and the rest of those decimals i'm just going to use the answer button on my calculator so 0.5 times 10 times root 17 times by the answer and there we go obviously typed it in wrong so we're going to have to go right back to the beginning and just re-type that in just to get that at that full angle back in fact no we've written it down so we can actually go ahead and do it and there is the benefit to writing the full angle down if you do type something wrong in your calculator like i have just done then you can go back and you can type it back in without having to worry about getting that full answer back on your calculator so 67.16 six three four five eight two there we go and there's our answer and that comes out as 19. there we go obviously we don't know what the units are here so we could say 19 units squared but i'm not going to worry too much just here when we're practicing on this one but 19 units squared whatever they are okay so let's highlight that off and there is our final answer so just making sure that you're very careful with these obviously we had to involve the cosine rules find angle we had magnitudes we had obviously that bit of logical thinking to find that missing vector there from q to r and then also we had using half a b sine c or our area of a triangle formula as well okay so when it comes to geometric vector proofs a lot of the proofs are very similar to those we've learned at gcse level so i'm going to go through this one quite quickly assuming that you have knowledge of vector proofs from gcse again if you're not sure you can always check out the full video in the description on that but otherwise we're going to have a look at this question so it says o a b is a triangle o to a is a and o to b is b it says the point m divides o to a in the ratio 2 to 1 and m to n is parallel to o to b at the bottom so first things first we should label those on our diagram so o to a and we'll put the direction in is the vector a and o to b is the vector b now what we can do is we can split that up into the ratio so o to a was two to one so this is going to be two thirds a for this part and this part will be one third a now for the first part of the question it says express the vector o to n in terms of a and b so we want to get from o our starting point here all the way over to n so we're going to look at some vectors to start with now moving from o to a is easy enough but a to n is a little bit trickier so we're going to need to look at this and think about the fact that m to n is parallel to o to b and looking at o to b which is only has a b in its vector means that m to n must be a fraction of b so it won't have any a's in that should help us to work it out so the first vector we're going to find here is a to b so if we look at the vector a to b that means we're going to have to go from a down to o and then over to b so that is going to be minus a plus b and just to make that look a little bit neater we'll write that as b minus a just so we reduce our amount of negatives now on to the next part we could think about how we get from a to n now we don't know what the ratio of a to n and n to b is so we're going to have to use some algebra or of course you can use lambda in this instance but i'm going to use the letter x so we're going to say that a to n is some amount of b minus a we're going to find out what x is now if that is the case let's have a look at m to n and think about potentially what that fraction will have to be so to get from m to n we know that that is going to have to be a multiple of b so if we were to expand this bracket we would have to do a to n plus m to a so m to a is already drawn on the diagram that is one third a that gets us from the point m to the point a and then we're going to add to that x lots of b minus x lots of a again that's just the vector above here but just expanded so that would be x b minus x a now we need to just have a look at this look at those a pieces we've got a third a minus xa now obviously in order to eliminate the a's there that means that x has to be one-third so x will have to be one-third and that allows us to put our x in so m to n is going to be third b obviously the a's they've just cancelled each other out so now we have the vector m to n we can get the vector o to n so now we've got m to n we need to think about how we get from a to n okay so that's going to allow us to go from o to a using the vector a and then down from a to n okay so a to n we have already worked out and if we just highlight that so we remember where it was it was this vector here xb minus xa of course we already know that x is a third so we could write that a to n is equal to one third b minus one third a and again let's just write that a little bit clearer there so that is one third b minus one third a okay so that is a to n and obviously we want to get from o to n so o to n if we were to write that in terms of the vectors o to n is going to be equal to o to a plus a to n there we go now we have both of those vectors o to a is in just the letter a so that one is going to be a and we're going to add to that a to n which we've just worked out and a to n is one third b minus one third a there we go and if we simplify all of that that is going to be when let's just highlight what we have we've got the a and minus a third a and that is going to go down to two-thirds a which we can write first so two-thirds a and then we also have the one-third b that's positive one-third b so plus one-third b so there we go that is our vector we have two thirds a plus one third b now we'll highlight that because that's finished but of course you could factorize out a third doesn't ask us to so it is fine to leave it in that form but our final answer there is two thirds a plus one third b now for part b it says show that a to n n to b is in the ratio 1 to 2. so we're going to need a little bit more space for this but in order to do this we're going to think about some of these vectors now if it is in the ratio one to two then we just need to think about some of these um vectors that we've got and we've already worked out the a to n and we'll write that down just up here so we don't lose it because i'm gonna have to rub some of this off the screen but a to n was equal to one-third b minus one-third a so there we go and we can also think about the overall vector of a to b the full length of the line now a to b we wrote down at the very start and that was b minus a and we've got that just over here so we're going to need that vector a to b and we're going to need this vector a to n and then we can think about the other portion of that and why it's going to be in this ratio so if we get rid of some of this working out and there's quite a lot of it so let's just get rid of all of that and now we can go about actually tackling this question so we've got that a to n is one third b minus one third a and the full vector a to b is b minus a now you can already see that a to n if we were to factorize that would be one third of b minus a so if the full line is b minus a a to n is only a third of that then n to b will have to be two thirds of the line so that would be two thirds of b minus a now ultimately what we have shown there is that a to n is a third of the line and enter b is two thirds of the line so as a ratio of a to n n to b that has come out as a fraction which would be one-third to two-thirds and that is obviously can be divided by one third and that is the ratio one to two so there we go we have shown that that would have to be the ratio one to two by looking at the full length of the line and the first part of the line okay so when it comes to modeling with vectors the questions are quite in-depth now some of the other questions that we've looked at are quite similar to this but we're going to have to draw out some bearing diagrams for this so it says a man leaves a starting point o and walks 15 kilometers on a bearing of 120 degrees to reach a so if we imagine that o is just over here by our question and obviously with bearings we need to draw in our north line then we need to think about what 120 degrees would look like so it's a little bit past 90 and it would go down to here we would then reach point a so we should label this up and again draw in another north line so this is going to be point a and we've now got that this is 120 degrees if we move on to our next piece of information it then says from a he walks nine kilometers on a bearing of 240 degrees to point b so if he's standing at a he's going to go 240 which goes past 180 but not past 270 so it's going to go somewhere down here now at the first point it said he worked walked 15 kilometers this one was only 9. so it's not going to go too far but again we are just drawing a sketch so again let's just write that in make sure we have this all labeled nicely and also we're going to want to show that that right there was 240 degrees so we're getting there with our diagram it then says in the final part of this that he then returns directly to o so he's down here at point b and he's going to walk straight back to o so as you can see this has created a triangle we could also put a north line in at b but we'll see how much of this we're going to need in the question now looking at what we have here we want to find the position vector o to a now that can also be phrased in a slightly different way it could say the position vector of a relative to o which just means starting at o and going to a so if we're going to work this out we need to think how we're going to get there so in order to do a position vector we want to know how many how far it's going to cross and how far it's gone down and if we think about that in terms of from o we're going to go across and let's do this in a different color so we can see this it goes across from o over to this north line and then it's going to go down now by doing that i have created a right angle triangle so we can use some trigonometry to figure out these lengths so if i move this 15 kilometers and we just put that above the line as that is going to be in our way a little bit so we'll put that just below this line just here because that was the 15 kilometer line and now we can go about working this out so for that particular line going across we can use some soccer tower now that right angle triangle does split that 120 degree angle as we are going perfectly to the right there then that is going to be 90 just above it which means that just in the triangle here that would be 90 plus 30 which would make that add up to 120. so if we use sokatoa we can work out both of these lengths and we'll write that using then i and j so the length across the top of the triangle we would use cos 4 as we have the hypotenuse and we're looking for the adjacent so for this line just here on the top we would actually do 15 cos 30. again you can slow this down a little bit and think about why it's 15 cos 30 there but hopefully you're happy with your sokatoa onto the one going down this is the opposite so we will use sine as we already have the hypotenuse the hypotenuse again is 15 so that would be 15 sine 30. so the distance is there we could just type into our calculator now of course we're going to write this as a position vector so for part a here we would want to say that it is going to be 15 and then going across was the cos 30 so 15 cos 30 we can put that in a bracket if you want i and then we're going to take away from that as we are going downwards 15 sine 30 so that's going to be 15 sine 30 and that is going to be a j so if we type that into our calculator to work it out 15 cos 30 to one decimal place comes out as 13.0 and that is going to be our vector i and then we're going to go down by and if we type in 15 sine 30 that comes out to one decimal place as 7.5 obviously and that is going to be our vector j so that is the first part of our question there we have found the position vector o to a part b here asks us to work out the magnitude but this time it works it asks us to work out from o to b so if we go over and think about our triangle and think about how we would work that out so if we're trying to work out the distance from o to b we are looking at this line just here and we want to work out the length of that line so to work out the length of that line we could use either sine rule or cosine rule so if we look at what we have in the triangle at the moment can we work out anything in this question so we need to have an angle now what you will see in the green triangle that i have drawn we have already had a 30 degree and 90 degree and therefore this angle in the green triangle would have to be 60 and of course that bearing around a has to fully add up to 360. now there's 240 on the outside 60 just in here so this angle inside the triangle will have to be 60 degrees as well now we can't actually get any of the other angles but do we have any pairs of opposites do we have anything that we can use potentially with the cosine rule now we don't have any pairs of opposites but we do have the angle and we have the 15 and the 9 just next to it so we'll be able to work out the length of that line using the cosine rule so for part b here to get the length of that line o to b which we can call just say ob so ob we'll put that in brackets because that squared is going to be equal to a squared plus b squared so 15 squared plus 9 squared minus 2 a b cos c or depending on which way you write your cosine rule but minus 2 times 15 times 9 cos 60. so if we type that in we get the ob squared is equal to and all of that simplifies down to 171. so there we go to get ob we would just have to square root our answer so ob is going to be the square root of 171 which comes out as and if we do it to one decimal place again it comes out as 13.1 kilometers and there we go that is the length of o to b or in this case we are calling that the magnitude of o to b again they mean the same thing when we're looking at this vector okay so if we move on to part c here it says find the bearing of b from o so if we're looking from o to get down to b we already have part of that angle we have the 120 degrees leading up to the start of our triangle the only thing that we're missing here is this angle here within this triangle so if we call that theta and we'll think about what we could use here so in order to get the angle again we could use sine rule or cosine rule now let's label on this 13.1 actually instead of putting 13.1 let's put the exact value in because we did round part b so that is the square root of 171. now in order to work this out let's just have a quick look and see if we can use the sine rule and for this particular question we have the length opposite the angle we're looking for and opposite that root 171 we have the angle of 60. so for this question here we can use the sine rule so let's use that as it's a little bit quicker so we've got sine theta we're looking for the angle so angle on top sine theta over 9 is going to be equal to let's have a look sine 60 over the square root of 171. now if we rearrange that multiplying both sides by nine we get that sine theta now let's just type that in on the top nine sine 60 over the square root of 171 comes out as 0.5 there's a few more decimals here is six but then there's a few more decimals and in order to work this out now we need to do the inverse so we'll do sine minus one of the zero point five nine again putting the full digits in there using your answer button and it comes out as and let's do it to one decimal place that comes out as 36.6 degrees so we've worked out the angle in the triangle there there's quite a lot going on on the page but that angle just there is going to be 36.6 degrees now obviously the bearing is the overall angle so we need to add that onto the 120 so to get the final bearing we would do 120 and we've kind of run out of space here so i'll put it over to here 120 plus the 36.6 and that comes out as 156.6 degrees and that would be our final answer there for part c okay so on to our final part of this question we've got part d now in order to answer this we're going to need to get rid of a lot of this information but it is very similar to part a so if i get rid of a lot of this information let's have a look okay so now we've got a little bit more space let's think about this diagram now we've got still got quite a lot drawn on the diagram there but if we just draw a little sketch of this triangle just to the side and we'll imagine again this is o this is a and this is b and we can think about obviously what we need to do here now in order to get the position vector from o to b again we're going to need to draw in a right angle triangle so if we go across from o and then down to b again it would form this right angle triangle and all we need to do is think about one of the angles within that triangle and we should hopefully be able to use socatoa again to find our distances so for this particular one here let's not forget we already have the length of the hypotenuse and that was the square root of 171. we then also need to figure out an angle and if you look back in our diagram we actually have this angle here so for that particular angle there obviously our right angle triangle goes just above it but we already know that's an angle of 30 degrees plus we have the 36.6 degrees so the full angle within the triangle just there would have to be 36.6 plus the 30. which comes out as 66.6 so the angle within the triangle and again you could always just draw that triangle separately to the side but what we have is a length of 171 and we have an angle of 66.6 now what i'm going to do for this one i'm just going to round that to 67. you could keep it accurate to be fair and you could write 66.6 either one of those is absolutely fine so in order to work through this again we're going to write our vectors in terms of i and j so to get from o to b we would have to do let's just have a think about this for i we want to work out this length just here now in regards to the angle we have the hypotenuse and we are working out the adjacent so we're going to use cos so we're going to do the hypotenuse which is the square root of cos of the angle which is 66.6 and if we type that in the square root of 171 and then cos 66.6 comes out as 5.19 so that is going to be 5.19 so we'll write this in and we'll just write it straight in as we've done the working out just uh just on the diagram there so for i we would have and let's just put it below where i've drawn the ob we would have five point two if i round that to one decimal place i and then we are going down so we're gonna minus j here to work out j this length just here again we are going to use sine so for sine for that one we can use root 171 sine of the angle so 66.6 and if we type that into our calculator root 171 and all i'm going to do is get rid of cos replace it with sine and press equals and we get 12.00 so for that one there again we'll just write this in so that would be 12.0 to one decimal place and that would go with our j so there is our answer 5.2 i minus 12.0 j of course that could come out slightly differently if you use rounded answers so if you had to use the rounded version of route 171 which came out as 13.1 kilometers or if you use the rounded version of 66.6 and you'd instead use 67 you would get a slightly different answer but that is an overview question that covers a lot of modeling with vectors so when it comes to differentiation essentially what we're actually doing is we are finding uh the an expression that allows us to get the gradient uh a point on a curve okay the gradient of the tangent to the point at a curve so with this particular graph that i've got on the screen this is the graph of y equals x squared okay now if we find any point on this curve so let's go for this one here normally and particularly at gcse level what we'd have done is we would have drawn in a tangent as best as we could and then we'd have worked out the gradient of that by doing change in y over change in x so we'd have drawn a little triangle in we'd have got our rise or over our run our change in y over change in x i'm going to use that to find out an estimate for the gradient now when it comes to differentiation the process is quite simple all we actually do and we write d y over dx which kind of just stands for the difference in y divided by the difference in x like we do at gcc we do change in y over change in x okay but obviously we're actually trying to find the expression for it rather than a numerical value and it's quite simple all that you actually do is you have a look at what the power is and the power here is two we multiply that by the coefficient of of x that's before it now the coefficient of x squared is one at the moment so one times two gives us two okay and then we reduce this power here by one so it's currently a power of two and that just goes down to a power of one so it's two x to the power of one and obviously when we have a power of 1 we don't need to write it so there we go for a nice simple graph like x squared that would be our expression that is our value of d y over dx there 2x and we can use that to find out the gradient at that point on a curve and we're going to look at all of these steps as we move along but here's a bit of an overview on a nice simple one if i wanted to find the gradient at that point there that we've already highlighted that has an x coordinate of two okay so we take two and we'd sub it in and when x equals 2 2x or d y over dx equals 2 lots of 2 or 2 times 2 which is 4. so the gradient at that point would equal 4. now these are all the things we're going to have a look at throughout this video we're going to have a look at obviously differentiating much harder um graph equations than this one and then we're going to have a look at subbing some values in to work out the value of those particular points okay but that's a bit of an overview of what we're going to look at and the process that we're going to go through obviously if you've never seen differentiation before but it's quite nice and simple multiply the coefficient by the power and reduce the power by 1. that is pretty much all we're going to need to do so let's not forget our little formula that we're going to use which is the function of f of x plus h take away the function of x and that's all going to be divided by h so that's the um formula there that you're going to use okay in this particular question our function of x is 5x squared and we're going to show that it goes down to 10x so there we go obviously if you're going to have a go pause the video there and see what you get but otherwise let's carry on with this question okay so this time our function of x is a five x squared so we've got a little bit more working out when we put this piece in here because obviously we've got the first piece to work out which is where we're gonna work out the function of x plus h and if we sub x plus h into five x squared we're gonna get this we get five lots of x plus h and that's going to be squared and then we're going to take away the function of x which in this case is 5x squared and again that's all going to be over h so there's our pieces sub into the equation to the formula there now we just need to simplify it all down and see what we get so if we expand that double bracket x plus h squared we've got five lots of that so we'll sort that out in a sec but if we expand that out we get the same as before we get x squared plus 2xh plus h squared and then we're going to take away 5x squared and again divide it all by h now we can go about multiplying everything in this bracket here by 5. so we end up with 5x squared plus 5 lots of 2xh is 10xh and then 5 lots of h squared will give us 5h squared and then again we've got minus 5x squared at the end and again all over h okay so we've got two pieces that cancel each other out they've got the 5x squared and the minus 5x squared so if we get rid of those what we've got left we'll have 10 xh plus 5h squared all over h we can divide the top and bottom by h there so the h will disappear off the 10x8 so we get 10x one of the h's is going to disappear off the 5h squared so 5h and there we go there's our final piece now again we're going to say as the limit of h approaches zero we're going to change that h for zero so what we end up with is 10x plus five lots of zero which is just plus zero and obviously don't need to write plus zero so we can just get rid of that and we just end up with this 10x piece which is exactly what we're looking for and there we go there's our final answer let's just highlight that and there we go that's how we go about that question part of the process look if we've got y equals five x squared minus three x plus seven find the value of d y over dx when x equals one well that's okay because all we have to do is differentiate and then substitute this value of one in now it doesn't ask it in this question but that is what we're doing we are finding the gradient of the tangent at that point where x equals one but obviously we're not going to have a look at the graphical element or aspect of it just yet we're just going to focus on part of the process and obviously some of these uh just mathematical processes at the moment so for that first one up there times the 5 by 2 gives us 10 so we get 10 x and the power gets reduced to 1. and then for the next bit we have minus 3 and that power of x there is going to disappear so we have 10x minus 3. and let's just label that so d y over dx equals 10 x minus 3. so when x is 1 all we need to do is substitute 1 in so when x equals 1 we are going to get 10 times 1 minus 3 okay which equals 10 minus 3 which equals 7. there we go so that is our value there when x equals 1. there we go so quite nice and simple hopefully okay so this one here does look uh you know a lot nasty than the last one we just looked at and that's primarily because we have this root x here now obviously remember when you have a root there's another way of writing that as a power and that is that the square root of x is also equal to x to the power of a half so don't be put off by a square root it shouldn't throw you at all it's just that we're going to write it in a slightly different way and if we go about actually writing that instead of writing it like that let's write it as 9x to the power of 4 take away 4x to the power of a half divided by x squared there we go so that shouldn't really put us off and now we can obviously get this into a single line because we can look at obviously subtracting this power of x from the one on the top and if we do that let's see what we get we get 9x to the power of 4 take away now technically there's a 1 in front of the x squared so 4 divided by 1 is still 4 and we're still going to have an x and we've got to do a half take away two now obviously just a little thing here obviously you can type that into your calculator but two as a power in terms of halves two is four halves so we're taking away four halves from the one half and that leaves us with negative three halves or three over two so there we go that's obviously now written as a single line obviously just subtracting the powers there when doing that division and now we can go about actually differentiating this so we get d y over dx and that's going to equal and let's have a look 9 times the 4 is 36x reduce that down to a power of 3. and then we have negative 3 halves times negative 4 at the front so that's going to become positive because if we've got negative 4 there and we're going to times by that negative power and that's going to become 12 over 2. 4 times 3 is 12 over 2 and that's going to be x and if we subtract 1 from that power now that's going to go down to minus 5 over 2. there we go just taking away one from three over from negative three over two or in or in essence taking away another one equals two halves okay so taking away another two halves from that power obviously you can type some of these into your calculator but there we go we should be able to hopefully do some of these without a calculator now there's obviously one thing to sort out and that again is that 12 over 2 there and 12 over 2 is a whole number it's 6. so we would write that as 6 there rather than 12 over 2. so if we put that instead and there we go there is that differentiated okay so hopefully just a few little questions there that just show you obviously not to be put off by some of these questions sometimes they don't look the nicest but it's all about getting it into that straight line there and obviously just being very careful with your powers when you multiply them by the coefficient and then reducing it by one okay so when we are finding equations of normals we just need to remember that a normal is perpendicular to the tangent at a particular point so in order to find the equation of the normal we need to find the equation of the gradient of the tangent then we can find the gradient of the normal and then we can find the equation of that line so if we go about this in order to find the gradient of the tangent we'll first differentiate our equation so if we differentiate this equation we're going to find d y by dx now in order to find d y by dx that's quite nice and easy for this particular equation multiplying the x in the coefficient by 2 would give us 2 decreasing the power just gives us a power of 1. then doing the same with the negative 5x over there would just give us negative 5. so we have done differentiating that equation now we can substitute in our coordinates and it asks us to use the coordinate 6 6. so if we put in the value when x equals 6 we get a gradient which i'm going to label as m and our gradient comes out as 12-5 so our gradient is seven for our perpendicular gradient which will be the gradient of the normal we just do our negative reciprocal so the gradient of the normal will be negative one over seven now we have the gradient of the normal we can just put it into the equation of a line so if we use y minus y one so y minus the y coordinate of six is going to be equal to negative one over seven multiplied by x minus x one so x minus six we'll expand that bracket out so we have y minus six is going to be equal to a negative one over seven x plus six times the seventh is six over seven now it doesn't say what form it wants this equation in so it's best that we get rid of the fraction so if we just multiply everything by 7 on the left and the right that will give us quite a nice equation to move around so that would be 7y minus 42 will be equal to negative 1x plus 6. now we may as well just move everything to the left-hand side as it doesn't specify so if we add the x over we'll have 7y and then add the x next to that would be plus x and then at the end we'll go and take away 6 from both sides so we'd have negative 48 and that is equal to zero now of course you can write that equation in any form you'd like for this particular question as it doesn't specify but that's probably the easiest way to do that so when it comes to looking at normals don't forget you have to differentiate to find the gradient of the tangent and then use your negative reciprocal for the gradient of the normal okay so when we're looking at increasing and decreasing functions we're going to want to differentiate the function now instead of writing d y by dx we'll write f dash of x to do our differentiation here just another way of writing d y by dx so if we differentiate this function and we're just looking at this one here we'll start with the 2x cubed so that becomes 6x squared we'll then move on to the negative fifteen x squared which becomes negative thirty x and then moving on to the final piece that just becomes thirty-six so that'll be plus thirty-six at the end and we have found our differentiated function now in order to find when this is increasing which is what this question is asked we're gonna find whereby this differentiated function is greater than or equal to zero so if we just write it as an inequality when is 6x squared minus 30x plus 36 when is that greater than or equal to zero again don't forget that just means when is the gradient going to be bigger than zero or in other words when is it sloping upwards so in order to solve this we need to just solve it as a quadratic inequality this one does simplify so all of those pieces divide by six so we can make it a little bit easier to simplify by doing that and if we divide everything by six we get x squared minus five x plus six and again we're trying to find where that's greater than or equal to zero so if we factorize that we get x minus three and x minus two again in either order and again we're looking at when that is greater than or equal to zero so we've got our two solutions x is equal to three and x is equal to two and then as this is a quadratic inequality we'll just quickly imagine what the curve looks like it goes through two and three so two here three here we are looking at when it is greater than zero so we have this part and this part of the curve and therefore that's going to be when x is less than or equal to 2 and also when x is greater than or equal to 3. so there we go there are our two solutions they are the intervals for when this function is increasing okay so when we're looking at our second derivative we just need to differentiate and then differentiate again so at the moment y is equal to the square root of x minus 3 over x squared now again thinking about indices here we just want to write this in a different way so instead of writing it like that let's change the root x to x to the power of a half and let's change the minus three over x squared to three x to the power of negative two that makes it easier for now us now to differentiate so if we find our first differential again we're just going to multiply the coefficient by the power and decrease the power by one so that becomes nine times a half which we can just write as nine over two x to the power of minus a half when we take away one we've then got negative three times negative two so that becomes plus six and x to the power of negative three so there's our first differential to find our second differential which we write as d squared y over dx squared then we just have to differentiate this once more so nine over two times negative a half becomes negative nine over four again that's to x and if we decrease that power by one it becomes negative three over two we then have 6 times negative 3 which becomes negative 18 and again then that is going to be x to the negative 4 when we decrease the power by 1. so there we go that is called our second differential and we're going to use that to find rates of change but there we go that's all you need to do you just need to differentiate and differentiate again okay so when we are looking for the coordinates of a stationary point on a curve we are going to have a look at the point whereby the gradient is zero so if we were to draw any basic curve and let's just think about a quadratic then we have this point right down the bottom whereby the gradient is equal to zero and if i was to draw a tangent at that point the gradient of that tangent would have a gradient of 0. so that's what we're looking at when we're finding stationary points but of course we have more curves than just a quadratic so when we go about doing this what we are going to do is differentiate the equation and then we will just set it equal to zero and see what coordinates that becomes so we're looking at this equation to start with and then we'll have a look at the nature of the stationary points using our second derivative but to start with let's differentiate this so if we do d y by d x and again we've got lots of differentiation here but that just becomes six x squared uh minus 30x and then plus 24. so again we just need to solve this as we are gonna look when that is equal to zero you could write that all out again i'm just going to put equal zero over here just to avoid writing it down lots of times now the first thing i notice is that everything divides by six so if we divide both sides of this by six it's going to make it easier to factorize so we have x squared minus five x plus four is equal to zero and if we factorize that we get x minus four and x minus one so we've solved that for when it is equal to zero that gives us two x coordinates an x coordinate of 4 and x coordinate of 1. now we just want to find the y coordinate for those as we are finding an actual coordinate here so if we look when x is equal to 4 all we're going to do is substitute 4 into our equation so if we put 4 into there and you can do that on a calculator 2 times 4 cubed minus 15 times 4 squared plus 24 times 4 and then plus 6. and if we substitute that in we get a y value which is equal to negative 10. so our coordinate there is going to be 4 and negative 10. if we do the same for when x is equal to 1 again we're going to get a y value so we're going to want to put x is equal to 1 into our original equation that says y equals now if we substitute that in again doing 2 times 1 cubed take away 15 times 1 squared plus 24 times 1 plus 6 we get a y y-coordinate that is equal to 17 and again that gives us then a coordinate of 1 and 17. so these are the coordinates of the stationary points they are the points at which on our curve there is a gradient of zero now when we're looking at using the second derivative to determine the nature of the stationary points we're going to find whether there is a local minimum point or a local maximum point so first things first we need to find the second derivative so for this one that we've already differentiated the first derivative we're going to find d squared y over dx squared and that becomes let's have a look 2 times 6 is 12 so 12 and then just minus 30. so that's our second derivative we'll just take our x coordinates and substitute them in so when x is equal to 1 the second derivative dx squared minus sorry d squared y over dx squared is equal to negative 18. now if that number is less than zero it is a local maximum point so at the at the coordinate 117 that is a local maximum point and we'll just write that down local maximum point we're going to do the same with the other x coordinates and see what we get for that so when x is equal to and where was the other x coordinate of four when x is equal to four d squared y over d x squared is equal to and that comes out as positive 18. so when it is greater than zero it is a local minimum point so and i should have added over here that this one was less than zero this one is going to be greater than zero so when it's greater than 0 this coordinate here 4 and negative 10 is a local minimum point there we go local minimum points so there we go there's our answers because the second derivative was less than zero when x was one 117 was a local maximum point and for x equals four we got 18 which was greater than zero so four and negative ten was a local minimum point okay so when we're having a look at modeling with differentiation we're going to use a lot of the skills that we've already looked at so this question here says a car is driven from edinburgh to manchester the cost of the journey is in c pounds when the car is driven at a steady speed of v kilometers per hour it says it's modeled here so the cost is equal to 1500 v 1500 over v plus 2 11 or 2 v over 11 plus 60. so this is our model just here and then it asks us to do a few things so the first thing that it asks us to do is find according to the model the value of v that minimizes the cost of the journey so if we want to find the minimum point or the minimum cost of the journey here then we're going to want to differentiate this model and if we differentiate the model we can set it equal to zero to find that point where the gradient is zero and if we imagine what this would look like in terms of a graph if we were to draw a little sketch of this this as it is 1500 over v is a reciprocal graph now we have the plus 60 at the end so it hasn't been raised or translated upwards so it wouldn't be that close to the x-axis but if we were to draw a little sketch it would look something like this and again that's moved up 60 on the left and if we think about where the minimum point is it's going to be as it approaches zero so it'd be somewhere down here and if we were to draw a gradient there then we would have a gradient of zero so that's what we're finding that kind of minimum point there on the reciprocal graph so if we go about actually trying to find that then we just need to differentiate our model so if we differentiate and it's c equals so we will differentiate c with respect to v and to do that first of all we probably want to write this in a slightly different way so let's write that c is equal to one thousand five hundred is v on the bottom on its own so v to the minus one plus and then we have 2 11 v which we could write as just 2 11 v and then plus 60. so if we go ahead and differentiate that negative one times the 1 500 is minus 1 500 v the v is going to go down to a minus 2 plus and then 2 11 times 1 is just 2 over 11. the v goes down to a power of 0 so we end up with just 2 over 11. so this is our differentiated part here and again we want to set that equal to 0. so we should write that as a new line but just to save a bit of space here we'll set it equal to zero just there now we want to go about solving this so we want it to say v equals so if we minus the two elevenths to the other side we'll have one thousand five hundred v to the minus two which is equal to negative two over eleven that's our first step there again it might just help here to rewrite negative one thousand five hundred v to the minus two again you don't really necessarily have to you could take different steps or different approaches here but i'm going to rewrite this back to 1 500 over v squared which is equal to the 2 over 11. so that's equal to a negative 2 over 11 that is technically negative over here and negative 2 over 11 here now what we can do is times both sides by negative 1. that's going to get rid of these negatives so i could just go and erase them but let's write the whole thing out so 1 500 over v squared is equal to positive 2 over 11. so that's the next step now we want to get it to say v squared so if we move the v squared to the right and divide by 2 11 then we'll know what v squared is equal to so multiplying the v squared over 1 500 equals 2 11 v squared and now i can divide by 2 over 11 which just leaves me with v squared being equal to typing that in on your calculator gets you eight thousand two hundred and fifty and now for our final step we just need to write our answer to square root v and square root eight thousand two hundred and fifty which gives us ninety point eight two nine for quite a lot of decimals here eight two nine five and a few more decimals but of course v is talking about a speed as mentioned in the question so speeds we would just give to an appropriate degree of accuracy so v as a speed it would be fine to say 90.8 kilometers per hour so there we go that would be a value of v that minimizes the cost of the journey so now we have that we're going to have a look at finding the minimum the actual minimum cost of the journey okay so for part two where we're finding the minimum cost of the journey we're going to obviously want to keep this minimum speed so if we get rid of everything on the screen but we'll keep that minimum speed and we'll move that up so we can keep that answer okay so i've kept my answer and i've also kept the full rounded decimal so i'm going to keep that answer just up here so that we don't lose that and i've kept the full rounded answer there because that's the number we're going to use if we are going to use this number so putting this into our cost equation there it says that the cost is equal to 1 500 over v so i would do that the cost is equal to 1 500 over my full decimal there 90.8295 and i'm going to add that to 2v over 11 so 2 lots of 90.8295 all over 11 plus 60. so i just need to type that into my calculator and if you type that into your calculator you get a cost that comes out as 93.0289 as our cost so if we round that to the nearest penny as the cost is in pounds that would come out as 93 pounds and three pence and that would be our answer there for the minimum cost of the journey okay so moving on to part b it says prove by using the second derivative that cost is minimized at the speed found in part a i so we don't actually need this cost here for the second part of our question so let's get rid of all of this but we will keep our answer there for part two we'll just keep it up there and we'll get rid of the rest of our working okay so now we've got rid of that we can have a look at part b so if we're going to find the second derivative we do actually need that first derivative so that we can actually use it now the first derivative from earlier in the question let's have a look was dc over dv and i'm going to have to differentiate that again because i forgot what it was but it was negative 1500 v to the minus 2 and then it was plus 2 over 11. so if we're going to find the second derivative so d squared c over dv squared then we're going to want to just do the same again so multiplying negative 1 500 by negative 2 gives us positive 3 000 and that becomes v to the minus 3. 2 over 11 is gone so that is our second derivative now as it is v to the minus three that could be written as 3 000 over v cubed you can write this either way but i do like to rewrite it as the fraction when we can so for this question here the value of v that we got in part a i was the 90.8295 so what we're going to do is substitute that into our second derivative so 3000 divided by 0.9825 cubes there we go and if you type that into your calculator you get the answer and it comes out as zero 0.04003 five there's some more decimals there but of course we're looking here where there is a local minimum or a local maximum point now this number here is greater than zero so we would write therefore it is a minimum cost because as naught point naught four zero zero three five is greater than zero and you could even put therefore it is a local minimum point okay so let's have a look at the last part of this question so part c here says state one limitation of this model now this is where we're going to be thinking about real life and this question here is about driving a car and of course what we worked out is that in order to achieve our minimum cost we would have to drive at 90.8 kilometers per hour for a steady speed for our entire journey now in real life that's not necessarily um possible for us to do so we would write for a limitation it would be very difficult for us to drive at 90.8 kilometers per hour for an entire journey so i'm not going to write that down but just a discussion there that would be the limitation of our model that it would be of near enough impossible for us to drive at a steady speed of 90.8 kilometers per hour for our entire journey when we looked at differentiation we looked at this particular graph and this is the graph of y equals x squared so we had to look at this and we had a look at how to differentiate that and when we differentiate it okay we multiply the coefficient of x there by the power it's okay so when we're differentiating we wrote it like this d y over dx there again we multiply the coefficient of x there which is one so we multiply that by two so we've got two x and then we reduce the power by one so we reduce that power of two down to a power of one and we got this here two x now when we're looking at integrating we're essentially looking at doing the complete opposite here we're just going to reverse this process so instead of multiplying the coefficient and reducing the power we're essentially going to increase the power and divide the coefficient and when we write this out we write it out like this okay and this little symbol here which i'm going to have better in a second but it looks something like that we say to do the opposite that means integrate we put in the bracket what we want to integrate which in this case is 2x and then we say dx after that just means with respect to x so that just means obviously do the opposite that means we integrate 2x with respect to x and we're going to look at how we go about actually doing that so obviously as we've said we're going to increase the power so at the moment that power of 2 2x there the the power of x is a power of 1. so if we increase that it becomes 2x squared and then rather than multiplying by that coefficient we're now going to divide by that sorry rather than multiplying the coefficient by that power we're going to divide the coefficient by that power and the power is 2 so we're going to divide it by 2. and you'll see when we do 2x squared divided by 2 we get back to that original x squared that we had up the top there okay now the problem with this is and obviously this isn't going to be the final answer here the problem with this is that there's a quite a lot of different um like graph equations here that are going to give us 2x when we differentiate it so if i have a look at another example up here we could have y equals and let's just imagine it's x squared plus five now when we differentiate that obviously we multiply the x um squared by the coefficient by 2 and we get 2x so when we differentiate it we get 2x and the 5 then disappears so that gives us 2x as well likewise we could have something like y equals x squared minus 19. when we differentiate that multiply the x squared by 2 reduce the power you get 2x and the 19 disappears so again we get 2x so the problem here with the integration is that we don't actually know what that constant is at the end what that number is at the end and because of that we have to add something onto our answer when we're integrating and that is that we have to put plus c at the end and that is our called our constant of integration again that just obviously says that we don't actually know what the constant is so we always have to put plus c at the end and that's going to be absolutely essential as we move through this video every time we integrate we're going to have to put that plus c at the end because we just don't know what that constant was unless we have any specific values on our graph and obviously with the x squared graph that's quite a nice one but let's just imagine just to sort of give you an overview of everything we're going to be looking at if i imagine that i say okay well on this particular graph that we've just integrated so this x squared plus c we have an x and a y value let's imagine we say okay well y equals five when x equals two okay so if y equals five when x equals two what is that equation going to be well if we sub those values in x squared is going to be four so that would be four plus c and that has to equal five so we know in that case there c would have to equal one and the equation of this line in particular would be x squared plus one or y equals x squared plus one okay so something here that looks a lot more complicated and let's see how we go about this one so we've got these three pieces that we're going to integrate this time so we're going to start off with this 12x cubed so again we're going to increase the power so we have 12 x to the power of 4 and then we're going to divide by that new power so divide by 4. then we're going to add to that we've got 6x so that's going to go up to 6x squared and then we're going to divide by that new power which is 2. and then we're going to take away and this one's not as nice here because we've got a power of two-thirds there but we have 15x we need to figure out what that power's going to be so we need to add one to two-thirds well one in terms of thirds is three-thirds so if we add three-thirds to two-thirds we get five-thirds there we go and that's not a five so let's rewrite that we get five thirds there we go so 15x to the power of five-thirds and then we need to divide by that new power so we're dividing by five-thirds not very nice there there we go but divide by five thirds and then at the end of that plus c okay because we don't know what that constant would have been at the end there we go we just need to simplify this all down so we get these first two look okay we've got 12 divided by four which is three so that's quite nice we get 3x to the power of 4 for the first bit the next bit we've got 6 divided by 2 which is 3 again so 3x squared and then this bit isn't so nice here you've got 15x to the power of 5 thirds divided by 5 thirds obviously if you've got your calculator obviously you can work this out quite nice and easy but don't forget it's not too difficult just to do 15 over one divided by five thirds so times it by the flip version three-fifths and there you go that equals 45 on the top over five which is nine there we go so 15 divided by five thirds is nine just a nice little bit of working out there obviously not always having to rely on your calculator if you've got some nice quick working out that you can do so there we go take away that becomes nine x to the power of five thirds there we go and not forgetting plus c at the end there we go and there we go that's that integrated okay not forgetting that plus c again that's so important obviously something that's obviously new with this even though we're doing the reverse process of differentiating that plus c there is so important to make sure that you're out on the end there we go that's again how we're going to integrate so increase the power by one divide by that new power and then simplify all your terms and just watch out for any sort of complications when you've got fractions in the powers then just being careful with that okay so this question here looks very similar to one we looked at earlier in the video we've got d y over dx equals x squared plus five in bracket squared all over x squared where x cannot equal zero obviously we can't divide by zero so we can't have x is zero on the bottom and it says given that y equals 13 and x equals one find y in terms of x giving each term in its simplest form so obviously you've got a few little bits to sort out here you've got a bracket on the top that needs expanding you'll have to turn this into a single line so dividing all of those pieces by x squared and getting those obviously rewritten but obviously you need to get that into a single line integrate it sub these values in find that value of c so have a go see if you can get get this question done see how far you can get we'll go over the answer in a sec so pause the video there and have a go right okay so we need to sort out what's on the top to start with them so x squared plus five obviously squared means we've got to do a double bracket so we have x squared plus five and another x squared plus five there we go so if we expand this all out we get x to the power of four and then we get plus five x squared plus another five x squared so that's plus ten x squared there we go and then five times five we get plus 25 so that's what we've got on the top and that's all being divided by x squared so obviously we just need to then divide all of these powers by x squared it's actually not too bad if you've obviously got to this point already the first piece there x to the power of 4 divided by x squared while subtract the powers we get x squared there we are the next piece 10x squared divided by x squared it's just 10 so plus 10. and then the last piece there 25 divided by x squared i'll subtract the power from x to the power of zero because x to the power minus two so we get 25 x to the minus two there we go and now it's at a point where we can actually integrate it so if we bring this up and integrate that let's see what we get so x squared well obviously increase the power so we get x cubed divided by three there we are plus 10x increasing the power of x there obviously there's no x before so we'll introduce the x and then for the last piece there if we increase the power it goes to minus one so we get 25 x to the minus one divided by the minus one and then we get plus c at the end right okay so nearly done we obviously just need to simplify this down so we get at the start there we can obviously write it as a third if you want you've got one and one in front of the x cube there so we've got one third x cubed plus 10x and then we have 25 divided by minus 1 so it becomes minus 25 x to the minus 1 plus c and now we can go about subbing these values in so it says over here y equals 13 at the point where x equals one so if we obviously write this as an equation now um let's just get rid of this because we could have put our y equals at the start of that one obviously have decided not to us we were integrating it at that point but now we're looking at it in terms of y equals y in terms of x we can write that out so we've got minus 13 is equal to a third times one cubed so a third times one which is just a third you can put the one in if you want let's just write that in anyway one cubed plus then we've got 10 lots of one which is just 10 take away 25 lots of 1 to the power of minus 1. there we go 1 to the power of minus 1. you can put in the bracket and then plus c right so simplifying this down then we've got minus 13 at the start obviously equals and we have a third times one which is a third plus ten times one which is ten take away 25 and then 1 to the power of anything is 1 so it's just take away 25 plus c obviously we just need to obviously simplify this little bit down and see what we get so we've got a third plus 10 take away 25 right so if we work that out then we've got -13 equals 10 take away 25 is minus 15. add the third to that is minus 14 and two-thirds there we go 14 and two-thirds obviously check that on the calculator but there we go minus 14 and two-thirds plus c so not the nicest here how do we get from minus 14 and two-thirds to minus 13. we're gonna have to add in uh those extra two-thirds to get back to -14 and then add an extra one which is three-thirds so there's five-thirds there that we need to add for c so our value of c is going to be five-thirds there we go that would have to be five thirds there we go obviously you could add 14 and two-thirds the negative 13 there and you'll get your five-thirds or two and or sorry one and two-thirds so we go there's our value of c and obviously to finish this off we just need to plug it back into the original equation up there for our value of c so for our final answer here let's just write this up here we get y equals one-third x cubed plus 10x minus 25 x to the minus one and then plus five thirds there we go and obviously you could write that as a decimal as well if you wanted but obviously when it comes to thirds i tend to just leave it as a fraction rather than having any recurring decimals in there but there we go there's our final answer okay so when having a look at definite integrals let's have a look at the process for going about that so this particular question first of all we need to integrate 2x minus 3 root x now again i'm going to write that in a slightly different way i'm going to write the root x as x to the power of a half so i'm going to write this as 2x minus 3x to the power of a half now to start with all we're going to do is integrate here so if we integrate that we will have and let's just have a look at what we get we get 2 x squared over two and then increasing the power there for the power of a half we get three x to the power of three over two i'm going to divide that by three over two now at this point here you are okay to simplify this so we'll simplify this a little bit now you don't have to because we're going to type this into a calculator anyway when we go about this but we can simplify that so the first piece there if we divide that by 2 we get x squared and for the second piece when we divide that by 3 over 2 we get 2x to the power of 3 over 2. so now we have integrated it what we are going to do is do a definite integral so for this what we are going to do is we are going to substitute the number nine the first number on the top of the integral we are going to substitute the number one in and we are going to subtract them away from each other so when we write this we put our integrated piece into a squared bracket and you will also notice that i haven't put the plus c on the end now you're gonna see here why i haven't put the plus c so we'll come back to that in just a second so the first number we're gonna use is the nine the second number we're gonna use is one and if i go about doing that i'll get something that looks like this so i will put 9 in to start with so that would be 9 squared take away 2 times 9 to the power of 3 over 2. and there we go and then we are going to take away whatever it is when we substitute one in so if i substitute one in we'll get one squared take away two times one to the power of three over two and there we go that is going to work out our answer now you'll notice that we didn't need to put the plus c in because we are going to take these away from each other when doing a definite integral you'll notice that if i had of put a plus c in the first bracket i would be taking away a plus c in the second bracket and therefore they would cancel out anyway so putting that in is kind of pointless so if we work this out and type that into a calculator you will see that i get the answer 28. so our final answer for this is 28. now on a later question you are going to see what we use this for but this was just a practice of going over how to go about doing this so as you can see we integrate it we put it into the squared bracket showing the nine and the one and then we substitute nine in substitute one in and take them away from each other and again if we had different numbers there we would sub those numbers in but again we would just take them away from each other and we're looking to see what our answer is okay so in this question we are going to see what we use definite integrals for and this question here says find the area of the finite region between the curve with equation y equals x brackets x minus 4 squared and the x axis so this is going to show you what we use definite integration for now if we were to draw this curve and think what it looks like it would be a cubic graph and is a positive cubic graph as well now you can see here we get our solutions of x equals zero and we've also got an x equals four solution there and that's a repeated solution so it's going to bounce on the axes so our graph would come up through zero bounce on four and then go back up so if we label that on we've got zero here and four just here and hopefully now you can see that finite region between the curve and the x-axis so if i highlight that we are looking at this region here so when we are using definite integrals we use it to find the region bounded between two of these points so in order to do this we're going to have a look at the equation of the curve we are going to do our definite integrals between 0 and 4 and that will tell us the area that is bounded between those points again it does tell us the area as it is bounded towards the x-axis so we had would have to do something different if it was bounded between the curve and another line but again we will look at that as well so in order to integrate this we'd probably want to expand that equation so if we expand that equation if we do the double bracket first we would get y is equal to x lots of if we expand that double bracket we get x squared minus 8 x plus 16. now we'll multiply it by the x as well so we're going to get y is equal to x cubed minus 8 x squared plus 16 x now we can go about actually integrating this and if we integrate that and again we're going to do it between these two points so we'll write that we're going to do it between four and zero so when we integrate that we get x um let's have a look in fact let's just write down on what we're going to integrate we're going to integrate x cubed minus 8x squared plus 16x and when we integrate it then we can just go ahead and write it in our squared bracket so increasing the power then we get x to the power of 4 over 4. take away 8x to the power of 3 all over 3. just make sure i write that 3 a little bit better and then that is going to be plus 16 x over 2 and of course you could write that as 8x squared sorry 16x squared over 2. you could write that as 8x squared now that's what we're going to be using for our limits here so we'll put that into our squared bracket and we're going to use 4 first and zero second it's quite a nice one when it does have zero it does make the second step quite nice so when we put the numbers in let's have a look at what we get putting four in to start with and again let's just write it in because we're just going to type this in on our calculator so we have 4 to the power of 4 divided by 4 take away 8 times 4 cubed all over 3 and then plus 16 times 4 squared all over two and we're going to take away the second one which is substituting in zero now if you have a look if you substitute zero into the first piece you get zero over four so that's zero the second piece you get zero over three and the third piece you get zero over two so all of that just comes out as zero again you can substitute zero in if you want but just looking at it you can see they all come out as zero so if you type all of that into your calculator we get the answer and it comes out as 64 over 3. and there we go that is our area so the area that we are looking at there is 64 over 3. you could convert that into a mixed number if you want so you could write that as 21 and a third but either of those answers are fine and we could leave our answer in either form so there we go that is our final answer that's what we use this definite integration for we find the two points that are bounded by the curve and the axes and using this we can work out the area underneath okay so when looking at specific areas sometimes we'll be looking at areas that are also under the x-axis and we just have to take something into account when we do this so if we think about what this curve would look like for this particular question we will have um something going under the x-axis so you will see when i expand this bracket and draw if i draw a sketch of it what it would look like so if we draw a little sketch to the side and think about our solutions we've got x equals zero and again we're just looking at this curve here we have got x is equal to one and we have x is equal to negative three so again the question just says to find the finite region between the curve and the x-axis but you will see when i draw this we have a point at negative three we have a point at one and we have a point at zero and when i draw this curve in again just drawing a sketch of it so it's not to scale but it is going to go up through negative three down through zero and then up through 1 and you can see here that we have a finite region above the x-axis here and we have another one below the x-axis here and when we are working this out something's going to change in our answer we'll have a look at that when we get to it but for now we just need to follow very similar steps we're going to integrate our equation and then we are going to use our definite integration between these two points but we're going to do them separately so we're first going to have a look at between negative 3 and 0 and then we're going to have a look between 0 and 1. so first of all let's expand our equation if i expand the first two brackets there or the x by the x minus 1 we get y is equal to x squared minus 1 then we're going to multiply by x plus 3. so if we stick that first part into a bracket and we'll expand that out we get y is equal to x squared times x is x cubed then we get a plus 3x squared and a minus x squared so plus 2x squared and then we have minus 3. so that right there is what we're going to integrate so if we integrate that let's have a look we get and let's just figure this out so increase the power x over 4 x to the power 4 divided by 4 and then we get 2x cubed divided by three and then minus three i've just realized we've forgotten x at the end there's a minus three okay so now we're going to integrate the equation of our curve so first of all let's expand those first two brackets so that would give us y is equal to x squared minus x and we're going to multiply that by the third bracket there or the third piece obviously the first piece wasn't a bracket but we're going to multiply that by x plus 3. if we expand that out we're going to get y is equal to x squared times x is x cubed then we get plus 3x squared minus 2x squared so plus 2x squared and the negative x times 3 is minus 3x so that is our equation expanded now we're going to integrate it so if we integrate that we get x to the power of four over four then we get plus two x cubed over three and then for the final piece there we get minus three x squared over two again we're not going to put our plus c in there because we are going to be using our limits and therefore that plus c is going to disappear anyway so we'll put this in brackets we'll put our dx at the end as we know that is what we are using so when we put our limits in we're going to do this in two parts to start with we will have a look at the first part here and then we will have a look at the second part so if we start by having a look at the first part we are going to put our limits in now so that is between 0 and negative 3 and if we substitute our numbers and we are doing this with respect to the y equation so we'll put y dx here and then we'll substitute our numbers in so when we go about doing this just so that we don't have to rewrite that full equation then we can put zero in to start with which if you have a look just comes out as 0 and then we're going to take away and we need to now substitute negative 3 in so we have negative 3 to the power of 4 all over 4 plus our second fraction 2 times negative 3 cubed all over 3 and then take away 3 times negative 3 squared all over 2 and there we go if we work that out and type that into our calculator for that we get the answer and let's just put this up here so for our first one we get the answer 45 over 4. so that area comes out as 45 over 4 which you can leave like that for the moment and that is our first area moving on to the second one and this time we are going to integrate between the points one and zero let's just change that not there so that this time is between one and zero again so we don't have to write the full equation we'll put y dx and this time when we substitute one in let's have a think we get one to the power of four over four so that's just a quarter plus two times one cubed is just two so it's plus 2 over 3 take away 3 times 1 squared so 3 over 2 there we go and we are going to subtract when we sub 0 in sub 0 into all of it we already know that comes out as zero so that comes out as zero and that comes out as negative seven over twelve now this is the important part and this is where something changes when it's under the axis and area cannot be negative but when it comes out in this form in integration it knows that it's under the axis and it gives us the area as a negative 7 over 12. so all we have to do is we change that area to a positive number so we write that it's positive seven over twelve and that is literally it for this extra element when it is under the axis or under the x-axis so when you get a negative area you know it's under the x-axis and as an area we just write it as the positive version so to finish this question off now we know that the area is positive we just add them together so 45 over 4 plus the positive 7 over 12 and that comes out as you can do this on calculator it comes out 71 over 6. so there we go that is our final answer 71 over six again you could turn it into a mixed number if you want but it's not required and it's not necessary so we're going to leave that as 71 over 6. okay so looking at this question it says the curve c has equation y equals x squared plus 2. it says the line y equals 6 intersects curve c at the points a and b find the area of the finite region bounded by the curve c and the line a b so if we draw a sketch of this to see what it looks like the x squared plus 2 is just an x squared graph that's been translated up by 2 so it would look something like that now if it's been translated by 2 it does cross through the x axis at point 2 just here and that will also help us to draw the line y equals six as we know it has to be above that so if we draw that line in as well it goes across here through the point six and it crosses over here and here so we can call that point a and we can call this one point b so we are looking at the area that is the that is bounded by the curve and the line a b so that is going to be this area just here so again this is above the x-axis but it is bounded by that line y equals 6. so this is an area between a curve and a line now when we go about doing this we just need to think logically about how we would work that out if we can find the coordinates of a and b we can go down and we can imagine where the x coordinate would be just here if we do the same with b we know the x coordinate is going to be just there and that is going to form a rectangle now if we were to work out the area of that rectangle and then subtract the area underneath the curve just here and over here then that would just leave us with the shaded region that is bounded by the curve and the line so to start with we can take many multiple steps here but i'm going to start with solving to find the coordinates where they cross over and we can do that by setting these equations equal to each other so if we take the equation of the curve x squared plus two set it equal to the equation of the line which is six and now we can solve that so if we solve this we get x squared is equal to four and therefore if we square root both sides x has to be equal to plus or minus two so if x is equal to plus or minus two we can see that the x coordinate will be two over here it'll be negative two just over there so now we can work out one of two things we can either work out the area under the curve or we can work out the area of the rectangle probably easier for us to just get the area of the rectangle so we don't forget so for the rectangle there we go that is how that has a base length across here of negative two to two so that is a base length of four and it has a height going up here going from the x axis up to the y equals six line so it has a height of six so the area of our rectangle is going to be four along the bottom multiplied by six which is equal to twenty four so i'm going to underline that so we don't forget about that one now we want to find the area under the curve this is quite nice says it's just between negative two and two so we're going to have to do this twice like on some of the other questions so if we write this out we are going to integrate between two and negative two and the piece we're going to integrate is x squared plus two so we just write x squared plus two dx now we can actually go about integrating that so if we go about writing this out we could put it into our squared brackets and we would say and let's just integrate that now so that would become x cubed over 3 plus and then increasing the power of 2 just makes it 2x and again we'll just write the limits there which is 2 and negative 2. so now we just need to go about actually substituting those numbers in so if we substitute 2 in let's see what we get so for the first one here if we substitute 2 we get 2 cubed divided by 3 plus 2 times 2. for the next one which we're going to take away we've got something very similar looking and that is negative 2 cubed divided by 3. and then that is going to be plus 2 times minus 2. there we go and that is what we need to work out so if we type that into our calculator let's see what we get we get the answer 40 divided by 3 or 40 over 3. now that again could be written as a mixed number but of course you don't need to write it we can just leave it as 40 over 3. so we have the pieces that we need we know that the area of the rectangle was 24 and the area under the curve there bounded by the x-axis is 40 over 3. so we already discussed at the start that to get this area here that is between them we just need to subtract them away from each other so we'll take the area of the curve sorry the area of the rectangle which is 24 and subtract the area under the curve which was 40 over 3 and if we take again just type that into our calculator we get the answer that comes out as 32 over 3 or of course you could write that as a mixed number you could write it as 10 and two-thirds but 32 over 3 is fine so 32 over 3 would be the final answer for the area that's bounded by the line y equals 6 and the curve so something like this 2 to the power of x equals 8. this isn't built to be particularly difficult hopefully you can spot here that 2 to the power of 3 is equal to eight or in other words two times two times two and we could write that we could write two to the power of three is equal to eight now when it comes to logarithms there is another way of writing this now just like when we're looking at these powers here we know this number here this 2 is called the base number and that's going to apply when we look at a logarithm now what we say for a logarithm and this is going to mean exactly the same thing it's just a different way of writing it we're going to say that log with a base of 2 just like here we've got our base number of 2. what log with a base of 2 gives us the answer 8 so log base 2 of 8 and we might say that equals x or something or we're trying to find out what that answer is so what power of 2 gives us the answer 8 and we know that the answer's 3. okay we've just worked that out and that would be the answer there if we wrote log base 2 of 8 our answer would be 3. so all this means here is essentially what power of two what power of two is equal to eight there we go and that is really the fundamentals here behind logarithms and as long as we can understand that concept we can look at it in all different types of ways but all we're going to be looking at here is basically what powers of certain numbers equal particular answers and sort of mixing those and sort of rearranging that in certain ways now we could write this as a sort of not as if not as a formula but we could write this obviously with letters rather than having the numbers here and looking at how obviously we rearrange this so we could say that log base of and let's just call it a with this letter b here is equal to x and what that means in terms of a power and that means that and again the base of a is our main number here or in this term obviously writing a letter at the moment but a to the power of x equals b and that there is obviously something pretty key to write down because that's how we're going to be sort of writing these and writing them in sort of different ways as we move through this video but obviously just thinking about this logically log base two just means the number two to the power of something gives us this answer here okay so what power of two gives us the answer eight and that power there is three so let's think about a different example before we move on i could say something like this obviously rather writing rather than writing it as a power we could say log and let's go for base 3 this time so what power of 3 gives us the answer 9 so we'll go log base 3 of 9 equals x and i might ask you here to find the value of x so if we've got this little problem in front of us we would say okay well 3 to the power of something and we've got obviously got the letter x there is equal to nine so what power of three gives us the answer nine and hopefully that's nice and easy for you to spot we know that it's three squared that equals nine so we would say x equals two or we could say you know log base 3 of 9 equals 2. but that's really the fundamentals here behind what logs mean and how we're going to use them okay so something a little trickier here we've got find the value of x such that log base 4 of a 1 over 64 is equal to x or in other words what power of 4 gives us 1 over 64. well let's write this out so we have the base of 4 so 4 is our base number 4 to the power of something which is our x this time is equal to 1 over 64. so we know from the last one again obviously depending on how comfortable you are with your negative indices we know that in order for it to be a 1 over as the answer it's going to have to be a negative power so we already know part of the power here we know it's going to be 4 to the power of minus something and we need to get to 1 over 64. so what power gets us from 4 to 64 we can work that out just down below let's have a look 4 times 4 is 16 and if we times that by 4 again isn't the nicest one to do in your head but 16 times 2 is 32. double that again we'll times it by 4 and that gets us to 64. so it's 4 times 4 times four or in other words four cubed so it's going to be four to the power of minus three so there we go we know four to the power of minus three gives us one over 64. so we could say log base four of one over 64 is equal to negative 3 the power there or in other words x equals negative 3. and there we go there's our final answer so that's all that this means when we're looking at logs okay so it's just about looking at the the way that the log is written and reading it just like we've recognized before log base and then we have our base number then it gives us our answer 1 over 64 is equal to what power or in other words what power of 4 gets us 1 over 64. okay so as long as you read it like that and sort of understand the way that that writing translates to that wording you should be absolutely fine working through some of these problems okay so this question here will show you how obviously we can use logarithms to actually solve some of these indices problems so we've got here find giving your answer three significant figures the value of x for which 5 to the power of something equals 10. now this is quite nice and this will show you why logarithms so useful because we can actually just type these in straight into the calculator but obviously it does mean that we have to know how to write this power as a logarithm now if we write it out so we're going to have log there we go and then we need to put our base number in so here we've got a base number of five so log base five and our answer is ten so what power of five gives us ten now you can type this straight into your calculator and be able to get our answer straight away so if you have a look on your calculator and everyone has different calculators so you can need to find your button but there is going to be a button on your calculator i'm using a casio class whiz which again i really recommend and i'll link that in the description but just underneath the on button or somewhere on your calculator there's a little button that looks like this and it normally looks like a log and it has like a little colored in square and then a little open square and if you can find that button you'll be able to do these no problem so find that button press that we'll get five in your little base square and then 10 in your larger open square there and type that in and let's have a look we get the answer and i get on my calculator one point four three zero six seven six five five eight which again if we give that to three significant figures chop it after the three and we get one point four three as our answer and there we go it's as easy as that and you can even test that out if you want you can press five to the power of and then put the answer in and you will get the answer 10 obviously if you don't put the rounded answer in but you put the original one in so this question says express log 2 or log base 2 of 4 plus log base 2 of 16 as a single logarithm to base two and a lot of these links here come to our basic understanding of our laws of indices and if i talk about this sort of logically and it will require you to obviously understand what we went over in part one but if we look at log base two of four and we actually calculate that and that means what power of 2 gives us the answer 4. and if we write that down let's go for this here so for this part here what power of 2 gives us the answer 4 and hopefully we're happy to know that is a power of 2. so let's just give that the answer x equals 2 if we obviously incorporate that x into our into our calculation there for the second part we've got log base 2 of 16 and again that means what power of 2 gives us 16. okay and that power of 2 that gives us 16 is 4 so x equals 4 the answer's 4. so really what this question is asking us to do is asking us to add these two numbers together so we have two plus four let's keep it all in the same color there and we know the answer for that is six two plus four is six but obviously it wants it as a single logarithm to base two now in other words and there is a little quick trick to this we don't have to do this for every question you'll see the little trick here but in other words it wants us to write which log with a base 2 has an answer that's going to come out as 6. so log base 2 of what number is equal to 6. and we can reverse that okay so we can think about this as we did in part 1. we can say okay well 2 to the power of 6 is going to equal that number that's going to go with our log there and all we need to do is actually work that out so 2 times 2 is 4 times 2 is 8 times 2 again is 16 times 2 again is 32 and then times 2 again is 64. so x there equals 64. or in other words if we're going to write this log as a single um logarithm to base 2 we would write log base 2 of 64. okay now let's think about how this links to what we've learned already looking at laws of indices okay so if we look at these two numbers that were in our logs 4 and 16 well there's a link between 4 and 16 and the one that we have now 64. and that link is essentially we just have to multiply these numbers together 4 times 16 and that's going to incorporate our first rule here that we're going to look at logarithms let's just get rid of this little bit on the right here and let's have a look at what we're talking about so in other words if we have log and let's just put base x and then a number so let's have it as a for this one and we add to that another log and what is key here is it does have to have the same base and we'll talk about how that links through to our indices as well with another number let's call that b our resultant log okay the log that we're going to get from that is going to be log with the same base base x but it'll be a multiplied by b or a b so we actually have to do we're adding these logs together so long as they have the same base and that's absolutely key we just multiply together those numbers at the end and if we think about how that links back to some of the algebra that we've done at gcse level if we have something like a squared and we multiply that by a to the power of 4 multiplication being involved there we add the powers and we got a to the power of 2 plus 4 which would give us a to the power of 6. and hopefully you can see look those little numbers two and four link back to our two and four here and our six and our six and it kind of all links together back from what we've learned on obviously our laws of indices there okay so that's essentially all we're gonna do when we're adding logs together one thing obviously that's very important here this is when we have the same base number just like when we learnt this with our laws of indices these powers here with our a's can only add together as long as we have the same base letter there and in this case that was the base letter a so a to the power of 2 times a to the power 4 we can add those powers together but not if we change that base number and that rule's going to apply with these logs here and in both of these we had a log base of 2 so we're able to just multiply those numbers to get our final log hopefully nice and simple and obviously you can obviously think about it logically like i did in this first scenario here and working out like that but it's much quicker if we have the same base numbers just to multiply those numbers together at the end so there we go there is our first rule this is your key piece of information to make sure you have written down and that is what we're going to do for adding logs now we're going to have a look at taking away so let's look at that one now okay so you're going to see a lot of links between obviously what we've just looked at and this one here except this time we are taking logs away now again something to note we have the same base number in this case they are bait log base threes and there's going to be a lot of links back to that process that we've done with our powers of indices again now if you think about um again just thinking about our laws of indices if we have something like and let's go with a to the power of four and we divide that by a to the power of one what happened what do you actually do with the powers well not forgetting obviously that we have that power of one in there even though it's not written but when we've done this before we would subtract those powers and that would come out as a to the power of three now if this is something that slipped your mind these laws of indices i'll make sure i'll link the video for that in the description so you can check that one out and make sure that you sort of clued up on your laws of indices there but that sort of idea there is going to link through the powers in this case were subtracted and that was when we were doing a division this time here in our logs we are subtracting those logs and you might be able to guess that we're going to end up dividing these numbers 81 divided by 3 the opposite of what we just did but let's have a look at how that works with our logical approach so log base 3 of 81 that means what power of 3 gives us the answer 81 and that's going to be 3 to the power of 4 so in this case x equals 4 well the answer to that log is 4. for our second log there we've got and this is always an interesting little log as well we've got 3 to the power of something equals 3. so log base 3 of 3 is 1 and that little relationship there is going to apply to any log if we have log base 4 of 4 that's going to be 1 log base 5 of 5. any log base that has the same number after it our power there or our answer there is always going to be one so we've got our two numbers we've got four and one and it's as simple as four take away one and that would give us our final answer which for this one comes out as three and obviously we're not trying to work out that answer there we're trying to know as a single logarithm to base three and if we think about that we are trying to work out therefore that which log with a base of three okay what's the answer if the power is three and that means what base of three okay with a power of three equals x there we go or in other words 3 to the power of 3 equals that x value there so we know that 3 cubed is equal to 27 and there we go that's the number that needs to go with our log so if we write that out as a single log to the base of 3 we have log base 3 of 27 is going to be our final answer and let's just have a look at how that all links through so again thinking about those original numbers we had 81 and 3 and 81 divided by 3 is 27 so it follows that kind of idea that we were talking about just here when we were dividing with powers looking at our laws of indices and we ended up subtracting them it's kind of like the opposite of that when we're subtracting our logs we can divide those numbers at the end of the log so log base 3 of 27 so hopefully that's okay it's just the opposite of when we were adding okay but i just wanted to show you there the logical wave and why it works and obviously the link between that and our laws of indices that we've looked at previously so again two little notes one is the logs have to be the same base when we're adding we can times the numbers at the end when we're subtracting we can divide the numbers at the end but just making sure that we write this down and thinking about this for any kind of logs here as long as we have log and i'm going to use the same letters here base x of a and we take away log with the same base of base x of b our resultant log there we get log base x of a over b i'm going to put that a over b there in a bracket so there we go that is adding and subtracting logs we've got one more thing to have a look at before we look at our final questions here so let's have a look at that one now so for our final little rule on logarithms we're going to look at something like this and it says express 2 log base 2 of eight as a single logarithm to base two and this links nicely into our one of our other little rules of indices that we've looked at in the past and if we think about something like this where we've had a and let's just go for a cubed to the power of two and we have something like that what is it we do with the powers well hopefully we remember we multiply the powers so something like that we would do a to the power of three times two which would end up resulting in a to the power of six now this is obviously when we are we multiply the powers and it's when we have sort of a power to another power in the bracket just like that and that's going to link through quite nicely to this now this is at the moment is not a single logarithm because what we have here is two lots of log base two of eight okay we've got this little two at the start or in other words we're multiplying that by two and what we can do is we can manipulate this logarithm we can write it in a slightly different way and i'll show you why it works so at the moment we've obviously got it written as 2 log base 2 of 8. now all we do is we can take this number at the start listen this number 2 here and we can put it as a power at the end of the logarithm or in other words i'm going to write it here we can say that it's also equal to log base 2 of 8 squared and we know of that obviously we can write that as an actual number we can write that out and we'll write that out in a second but we're going to have a look at why this works at the moment obviously 8 squared is 64. but we'll have a look at that just now now if we look at this first logarithm so it's 2 log base 2 of 8. now at the moment forgetting about the 2 for the moment because we'll look at two lots of it in a second but just this log base two of eight that means what power of two gives us the answer eight and hopefully we know that that is actually three isn't it okay so we can say x equals three two cubed equals three but we also have two lots of that okay so not forgetting about the fact we have the two there we've got two lots of x equals three or two lots of three and if we write that out two times three equals six so our answer just there would be six now let's have a look at our other one over here we have log base two of eight squared or in other words what power of two is going to give us the answer to eight squared which we already know or we've established is 64. not the nicest one there but the power of 2 that gives us 64. and you're happy to check this the answer is 6. we get x equals 6. so both of these give us the answer six and that's just going to show you obviously that they mean the same thing essentially but again this is just another way that we're able to manipulate the way that we write logs so this number at the front can be taken as a power at the end okay so we've moved that to just to being a power here with the eight and obviously we could do that vice versa as well if i have if i had originally log base two of eight squared i could bring that two to the front okay i could just put that at the front like we've done just there okay so again this is just another way that we're able to write logs and this is important as you're going to see on the final question that i'm going to show you because the only time we can add and subtract logs is when they are written as a single logarithm so having it written this first way that we have in the question is not very good to us if we're looking to do some calculations with it we always want to bring that number back up to the power so that's something very important to make sure that you've got there as well okay but it's just another way of writing the logs and if we were to kind of write it out we could say something like this a log base x of b can also be written as log base x of b to the power of a and again vice versa so there's another little rule that you're going to want to have written down but hopefully just with a little bit of practice they'll sort of um ingrain into your memory anyway but this is our final little piece that i needed to show you so we're gonna have a look at a final question whereby we first have to change the logs and then apply some of the other rules that we've looked at but hopefully this here shows you obviously how to go about that i'm not forgetting here we need to get our final answer down so our final answer instead would have been log base 2 not of 8 squared but of 64. and there's our final answer okay so let's highlight that off and now we're going to have a look at putting this all together okay so on to our next question it says solve the equation 2 log base 5 of x take away log base 5 of 3x equals 2. so slightly different we don't have a log on the right hand side of the equation here but we do have some other little bits to deal with and again just taking note these logs do have the same base so we can apply all of these little rules but this log on the left is not a single logarithm we have 2 log base 5 of x so in order to do our subtraction rule here we're going to have to use our little power rule and we're going to have to move that two up to the x or actually just our rearrangement of logs there potentially bringing in our power rule later on so in order to get that log on the left to become a single log we can bring that to at the front up to the power with x and again thinking about that from some of our earlier videos and do make sure you check those out first if any of these little bits of the logs are sort of going past you so let's bring that two up so that would be become log base 5 of x squared rather than 2 log base 5 of x take away log base 5 of 3x and that is equal to 2. now they are both single logarithms with the same base we can apply our subtraction rule like we did before so we're going to divide those pieces so we're going to have here log base 5 of x squared over the 3x and again you can put that in brackets if you want put that in a bracket and that is equal to two okay so we're gonna have to deal with this log now as we only have the log on the one side we can't just cancel out from both sides we're going to use our power rule here i'm thinking about what this means and it's a little bit more abstract that when we have just a numerical value here but we have what power of 5 gives us the answer x squared over 3x and it tells us here that that answer is 2. so in other words 5 to the power of 2 is equal to this x squared over 3x what power of 5 gives us that and it tells us the power is 2. so that's 5 squared so in other words if we write that out in order to remove our log here we can write that x squared over 3x is equal to 5 squared and obviously we don't need to write 5 squared there we can actually remove that and write the answer as 25. so let's just get rid of that obviously you leave that for your working but let's just turn that into 25 as we know 5 squared is 25. all right so from here we can actually form our equation now oh well we've actually already formed our equation but we can get it into a singular line so the first thing we need to do is we need to remove this 3x at the moment it's being divided by 3x so let's multiply that over to the other side and once we do that let's bring this up here we get x squared is equal i'm not quite drawing the 2 in there there we go is equal to 75 x and there we go now obviously we have a quadratic as we have an x squared in there so we want to get everything on the same side so equals zero so that's minus 75 x to the other side so we get x squared minus 75 x equals zero there we go and just like with any quadratic we need to factorize it this quadratic here has no number at the end so in order to factorize that we're going to want to put it into a single bracket so let's take out a factor of x so we get x brackets x minus 75 and that equals zero there we go so we've got our two solutions here we've got two possible solutions we've got the x on the outside which gives us x equals zero and we've got the bracket there which gives us x equals positive seventy-five now we've got two solutions and when we've got two solutions we definitely need to check these out now when we're looking at a log we can't have log base of zero okay we can't have log base of zero because that means what power of whatever the base is in this case five what power of five gives you the answer zero and there is no number that you can put to a power of five or power of anything to get the answer zero just like you can't get a negative number you can get very small numbers fractions you can get very large numbers but you can't get negatives and zeros so zero this one here if we think about that in terms of substituting it back into either of our logs here if we were to put it into this log on the right we would have log base five of three lots of zero which would be zero and that's not possible okay we can't have that so zero here is not going to be included we can have 75 though if we were to put 75 in there we'd get some very large numbers okay we get 225 for the one on the right and 75 squared on the one on the left so 75 would be fine it would give us a positive whole number so x equals 75 in this case is our final answer for that question and there we go let's have a look at our next one okay so when having a look at exponential graphs it's important that we're able to sketch them so this question says on the same axis sketch the graphs of y equals 2 to the power of x and y equals a half to the power of x so if we draw a little sketch of some axes we can think about what these would look like now for the first one we will sketch y equals 2 to the power of x now this is an increasing function here so it would be sloping upwards and again with an exponential function the graph approaches but never touches zero so this would be our y equals two to the power of x and now we'll have a look at y equals a half to the power of x so for y equals a half to the power of x that is a decreasing function so it would be sloping downwards that would look very similar to this but it would just be sloping the other way and as you can see it is decreasing and again getting closer and closer to it approaches but never touches zero so this one here would be y equals one half to the power of x there we go put that in a bracket and to the power of x now as with all exponential functions there are a couple of things that you need to be aware of so for these exponentials here if we were to put in an x value of 0 which is what's on the y axis there we would get a y coordinate just here of one anything to the power of zero is one so that's something to be aware of another thing that we should be aware of as well is that we have an asymptote now this here never actually touches the x-axis so the x-axis is our asymptote now for this one here the equation of the x-axis is y equals zero so our asymptote of these graphs and we'll just label that there a symptote there we go is y is equal to zero of course that would change if we put a graph transformation in something like two to the power of x plus five would move the graph upwards by five which would move our y-intercept and would also change our asymptotes but there we go that is looking at sketching exponential graphs so when it comes to differentiating an exponential there is just a few rules that we need to know so this question here says to differentiate y equals 2 e to the power of 3x with respect to x now if we are going to differentiate this all we actually have to do is do it in a similar way to how we differentiate but just one little thing changes so when we do the differential so d y over d x of this exponential we multiply the coefficient of e by the coefficient of the x so that means we are just going to multiply the 2 by the 3 this time which makes it 6. we still have e and the power stays the same so that is just still going to be a power of 3x so if at any point you have to differentiate an exponential this is all that you need to do multiply the coefficient of e by the coefficient of the x in the power and then keep the power the same so differentiating an exponential is quite nice and simple okay so when we are solving an equation with an exponential and where this question here says to solve the equation e to the power of two x plus three is equal to 7 and we're going to give our answer in exact form and this is when we start to use natural logarithms so in order to remove the e at the start we take the natural logarithm of both sides now the natural logarithm is just the inverse of our exponential there so if we write this out e to the power of 2x plus 3 is equal to 7 when we take our natural log of both sides the e disappears we get two x plus three and that is going to be equal to the natural log of seven so it's quite nice and easy to get rid of the exponential all you need to do is take the natural log of both sides and essentially it just drops the power down now we can just solve it like a normal equation so we would take away 3 from both sides which would leave us with 2x is equal to the natural log of 7 take away 3 and then we would divide both sides by 2 so x is equal to the natural log of seven minus three all divided by two of course that could be written in a different way we could put that divided by two in front of both so we could say a half of log seven minus a half of three which is just could just be written as three over two so either one of those would be fine as our answer i tend to write my answer like this most of the time so i do prefer that one on the right but both of those answers there are fine and we have given our answer an exact form as we have left the natural log within our answer okay so another type of equation where we're going to be using natural logs so for this particular one here you can see what we actually have is an exponential quadratic here and we're just going to have a look at factorizing it in order to find our solutions so it says solve that equation e to the power of 2x minus e to the power of x plus 12 is equal to zero and again giving your answers in exact form so for this question here it does give us a bit of a hint that we're going to be using natural logs and to start with we just need to factorize it so at the start of both of our brackets we will have e to the power of x as when you expand that e to the power of x multiplied by e to the power of x would equal e to the power of 2x and that is all equal to zero of course if it wasn't equal to zero you may just have to rearrange these first but once we've got it into that form we can now factorize so looking at the twelve there we could have some factors we're trying to make eight in the middle so two and six is going to work so in our bracket we'll have minus 2 and minus 6 and that would equate to negative 8 e to the power of x in the center there so we have two solutions for the first one we have e to the power of x has to be equal to positive 2 and here we have e to the power of x has to be equal to positive six now before we just go ahead and solve that just as a side note when we have this here e to the power of x has to be positive okay e to the power of x must be positive and that means if we do get an answer which says e to the power of x is equal to negative three for example we would have to discard that answer and we would just use the one but this one here we have both being positive e to the power of x equals two and e to the power of x equals six so these are both going to be solutions and it does give you a little bit of a hint here as it says the word answers so for this one we will now just take the natural log of both sides so for the first one on the left there x would equal the natural log of 2 and x would equal the natural log of 6 and again they are both in exact form so we would leave our answers like that so e x equals the natural log of two and the natural log of six okay so when we are having a look at modeling with exponentials the questions do get quite tricky so it says here the value of a car and that is in v pounds can be modeled by the equation v is equal to 15 700 e to the power of minus 0.25 t plus 2 300 where t is a real number and t is greater than or equal to zero where the age of the car is given to us in t years always worth highlighting that the age of the car is t years just so we don't forget about that so it says here using the model find the initial value of the car now the initial value of the car would be found when t is equal to zero so all we need to do to find the initial value of the car is substitute t equals zero into our equation and see what we get now if we do substitute t equals zero we would have fifteen thousand seven hundred e to the power of and negative 0.25 t times 0 is just going to equal 0. now anything to the power of 0 is 1 so that is just going to be 15 700. we then have to add 2 hundred to that and that means we just have fifteen thousand seven hundred plus the two thousand three hundred which gives us a final answer of and it's in pounds so we would say eighteen thousand pounds and that would be the initial value of the car so we'll just underline that as that's our first answer on to part b it says given the model predicts that the value of the car is decreasing at a rate of 500 pounds per year at the instant where the time is equal to capital t show that 3 000 and let's just highlight this 3925 e to the power of negative 0.25 capital t is equal to 500. so in order to look at this we would want to differentiate that as we are looking at the rate of change or the gradient there when it is going down by 500. now also just taking note here if it is going down by 500 pounds per year that would be a negative 500 as our gradient so in order to find this we're just going to want to differentiate it to start with so we're going to differentiate v and we're going to do that with respect to t so if we differentiate dv by dt and we are going to be doing this at the instant where t is equal to capital t so we're just going to replace the little t with big t in our differential so if we go about doing that here let's have a look at what we get so we're going to multiply 15 700 by negative 0.25 and if we type that into our calculator we get negative 3925 e to the power of and don't forget when you're differentiating an exponential the power is not going to change so it will be to the power of negative 0.25 and it says that little t is equal to big t so big t now we have got that all we need to do is think about setting this equal to the gradient that we've been told and that is that it's going down by 500 so if we set that equal to 500 we would have negative 3925 e to the power of negative 0.25 t and that is equal to negative 500. now we've got one step in order to make it look like how we've been asked to make it look like and that is to divide both sides by negative one well that would remove the negative from both sides so our final answer would be positive 3925e to the power of negative 0.25 t and that is going to be equal to positive 500. that is exactly what the question has asked us to show so we have finished off our answer and we have shown what the question was asking for okay so on to part two it says here hence find the age of the car at this instant giving your answer in years and months to the nearest month so for this particular part of the question we have our equation to solve we've just got that 3925 e to the negative 0.25 t is equal to 500. so we want to know the value of t at this instant so the first thing that we can do and if we carry on from our part one and we'll just label this as part one part i and then we'll carry on here with part two so the first thing that we're going to do is divide by the 3925 and if we go about doing that then we will get and let's just have a look we will have if we type it into our calculator e to the power of negative 0.25 t and in fact we don't really need to type it in because we could just write it as 500 over 3925 let's actually put the correct number there 3900 3925 so for the next step we want to remove the exponential so if we take the natural log of both sides we will have that negative naught point two five t is equal to the natural log of and we'll put this in brackets 500 over 3925 [Music] we now have one more step to get our value of t and that is to divide by negative 0.25 and if we do that we'll write that over here divide by negative not .25 and we get a value of t that is equal to and what would it be let's type in our calculator so the natural log of 500 over 3925 and then divide that by negative 0.25 and we get the answer 8.242 a few more decimals here so we'll leave it at 8.242 and if we look at that there that is in years and remember it said that in the question and we highlighted that at the start so that is eight and a quarter or almost eight and a quarter years and it does say give it to the nearest month so a quarter of a year would be equal to three months so our answer for this here would be eight years there we go three months and that would be our final answer for part two of part b so eight years three months and we'll just highlight that and that part is finished now for part c it says the model predicts that the value of the car approaches but and does not fall below a state the value of a now in order to do this we need to think about what the exponential graph actually looks like so if we do this and we can just about fit it in at the top here if i get rid of t equals zero if we draw a small sketch of an exponential graph up here and we think about what it would look like we know it is decreasing in value so it would come down something like this now we would have an asymptote just up here and if we draw this in and think about the reason i have drawn it up at the top there the reason i have drawn it up at the top is because our exponential has this plus 2 300 just at the end now that means it has been translated upwards and this value here would be 2 300. so as it has moved up from 2300 that means it is never going to approach well it's going to approach but it is never going to fall below 2 300. so the value of a to finish this question would be three 2 pounds and we'll write this down the bottom or highlight that as our final answer and there we go and that is looking at modeling with exponentials okay well done if you have made it this far it means that you've managed to sit through seven hours of this video and seven hours of me talking so i do certainly congratulate you on that and i do hope this was useful and helpful if it was please do like the video please leave me a comment don't forget to subscribe to the channel but otherwise i will have to see you for the next one [Music] you