okay welcome back this is Mr Hassan's math Channel I'm now answering question number seven from the January 2024 International a level Ed Excel pure mathematics P1 paper and this question here is all about graph sketching part A anyway is and we're asked to sketch the graph of the curve with equation y = 4x - K where K is a positive constant we need to show on our sketch the coordinates of any points where C cuts the coordinate axes and the equation of the vertical ASM toote to C the vertical ASM toote to the curve now when you want to sketch the graph of any curve the first thing you need to determine is what what is its basic shape all right so you should have you should be able to recognize from the equation of a curve its basic shape okay so you have straight line graphs which have got the form y = mx + C where you have X without any power of um anything but one okay no no squared no cubed not in the denominator um you know multiplied by a constant sometimes and a constant added to it that's a straight line graph have that type of shape okay then you have equations of the form y = ax^2 + BX + C which is called a quadratic type of curve okay where the highest power is X squ there's no X in the denominator and it has either this type of shape here or this type of shape here which it's called a quadratic it's a parabola shape has either a minimum or a maximum depending on the value of this a okay then you have also cubic types of curve which are like y = ax Cub plus bx^ 2 plus CX plus d where the highest power is X cubed there's no X in the denominator again and it has either this type of shape up down up or this type of shape down up down something like this okay there's variations of it you could have points of inflection and so on you know there's different variations of it but basically that's the basic kind of shape that you'll have and um you know the the shape is determined whether it's this this way or that way by the value of a then you have what are called reciprocal curves which have the kind of um look like this yal something overx okay where X is in the denominator now this type of shape of graph has some sort of a discontinuity it's called a reciprocal curve the certain values of X and Y can't take okay um so this is a reciprocal type of curve and whenever you have X in the denominator you end up with a reciprocal curve and we can see here that this is definitely going to be a reciprocal type of curve this type of curve here where there's X in the denominator and that's you know something important for us to know now when you ever you have a reciprocal curve you have to work out what the asmp tootes are What are the values that cannot go into this function so for example into this into this equation so the first thing we look at is we see what value makes the denominator zero the denominator of any fraction cannot be zero it will be undefined so the value of x which makes the denominator zero okay will be a value that is going to be where the vertical ASM toote is X will not be able to reach that value so we look at the denominator and we equate it to zero so when x - k equal 0 we can say x is equal to K so xal K is the ASM toote this is one of the ASM tootes this is one of the ASM tootes this is the vertical ASM toote okay the vertical ASM toote which is what we had to uh write the equation so xal K is the vertical ASM toote and we can see also here that if I make y equals z if I put y equals 0 what's going to happen you'll have 0 = 4x - K when you try to solve that you get 0 = 4 which is a contradiction so that means y cannot equal zero so that's the horizontal ASM toote okay so you have a vertical asmt and a horizontal ASM toote now um if for example there was a number added to this fraction on the outside like for example one then yal 1 would be the the hor the horizontal ASM toote because when yal 1 then you'll get 4x - k equal 0 again and then you won't be able to solve it so whatever numers added to the whole fraction that y equals that will be the horizontal ASM toote okay so that's the way to work out the horizontal ASM toote in this case there's nothing added to it so y equals zero is the horizontal asop okay so now we have enough information to be able to make a sketch of this curve so I'm going to do that now so we have our axis Y axis the x-axis so xals K now K is a positive constant so x equal K will be this is the y- axis this is the x-axis xal K will be on the positive side of the axis so let's just um do that here so xal K would be say somewhere over here okay so that's xal K that's the vertic the hor the vertical ASM toote and you have another ASM toote as I mentioned which is going to be y equal Z which we're not actually told to show but that's that's those are the ASM tootes so we know that the um you know this is basically going to be of this shape over here let me just make that a bit clearer a bit better as a curve this side is going to go like this sorry it's not very neat it's a bit difficult with this pen here okay that's a bit better I think something like that it's a bit jaggedy but should try and make it smoother than that um now on this side we know that definitely there's the ASM toote going to be over here so it's going to definitely going to hit the y axis somewhere down here and we can actually calculate that part okay so I I need to write that that point down because it says where it crosses the the coordinate axis it doesn't cross the x-axis by the y- axis it crosses there how do we know when something crosses the y- axis well we know the equation is y = 4x - K so it crosses the y axis as with any other function when x = 0 cuz the equation of the y- axis is xal 0 so we're going to have y = 4 over 0 - K so y = -4 over K so th that's the coordinates of the um is 0 and Min - 4 over K let me put it on that side because there's nothing written on that side so that's Z - 4 over K that's the point where it crosses the y- axis so that is the sketch of this curve u y = 4 4x - K okay so showing get okay so we written down the coordinates of the or the equation of the curve and the coordinates of the point where it crosses the axis so that's how you can draw the curve so you should know that y = 1 /x will look like this okay if it was negative 1 /x it would look something like this in these two parts that's all okay um so there we have the answer to this question so we should understand how to sketch these curves uh reciprocal curves quadratic curves linear curves cubic curves and exponential curves I think that comes more in P2 but that's basically how we deal with that um question so now for part two of the question I think I'm going to have to use some of this okay so for Part B then it says given that the straight line with equation Y = 9 - x y = 9 - x does not cross or touch C find the range of values of K okay so remember this is y equal x over um Sorry 4 over 4 over x - K okay and this that's the point K there so given that the straight line with equation y y = 9 - x does not cross or touch C so 9 - x would be um a straight line which goes through N9 on the x axis and the Y AIS when X is zero Y is 9 when Y is z x is 9 so it's going to be like a straight line going something like this okay go through nine on both axes okay okay so we got to find the values of K such that this curve does not touch nor cross uh the curve so basically what we got to do is we've got to solve simultaneously the equation y equal uh 4x - K and Y = 9 - x and show that U find the value of K for which there will be no solution to this equation that there will be no solution to this the simultaneous equation that we form okay so when we solve a pair of simultaneous equations what we do is we substitute one of the equations into the other so what I'm going to do is I'm going to write this as uh 9 - x = 4x - K okay so I'm just replace this y with what y equals from the other equation that's always the easiest or the the most sensible way to think about it when it comes to other types of questions where you have to substitute sh um into like where equations of circles you get confused if you just say equate the two equations that works here but not in every case so think about it in terms of substituting instead of Y in one equation about y equals in the other and that will make things um okay for you so now we can solve this by we can um multiply both sides by x - K so we have 9 - x * x - k equals 4 when we expand this bracket we're going to have basically 9x - 9 K - x^2 plus KX = 4 so we have a quadratic okay now let me write the quadratic so we have the X squ as positive makes life easier X2 and you're going to have um + 9x - KX okay soorry yes that goes on this side and then we're going to have um yeah plus 9x we're doing sorry let me do that again so we're going to have zero on this side you have x^2 you're going to have Min - 9x and Min - KX sorry - 9x and - KX and then we're going to have + 9 K and + 4 okay I've just all of these change signs when I put them on this side okay you just subtract 9x from both sides you subtract 9k you add 9k to both sides you add x s to both sides and you take away KX on both sides so it end up with this so we can take out the minus um X from here so we have - x * 9 + K okay and here we're going to have + 9 k + 4 all right so here we have x^2 minus in Brackets 9 + 9 + k sorry 9 + KX you got to be careful about writing KS and X's they look very similar sometimes - 9 + KX + 9 k + 4 so equals z equals zero of course okay because it's an equation so we know that this is a quadratic equation and uh we know that when you have a quadratic equation and the discriminant b^2 - for AC is positive you have two solutions so if I wanted to find the values of K such that uh the line cuts the curve in two places I would solve um I would I would rewrite the discriminant well I'll take the discriminant from here and put them put that into this inequality and I'd find where there are two solutions if I wanted to show where the line is a tangent to the curve then I would equate this to zero because there will only be one solution okay b^2 - 4 A C equals um when it's equal to zero that means a line is a tangent to the curve so it'll give me the places where the line is a tangent to the curve if I put um b^2 - 4 a cal0 that will that will show me the points where there would be only one solution that would be where the line is a tangent to the curve so it would give me two answers for K there'll be two values of K for which this would have only one solution one one is a tangent over here the other one is a tangent over there okay that would be the case where it touches the curve and doesn't cut through it but we want to find the case where B ^2 - 4 a c is less than zero when the discriminant is less than zero then there's no solution to this quadratic to this quadratic equation which means there's no real solution no real solution okay when there's no real solution that means the line will not touch nor intersect with the Curve and that's what we want to find does not cross or touch the curve so we want to find the case where b s - 4 a c is less than zero so in this case our a is 1 a 2 remember it's the equation of quadratic ax^2 + BX plus C okay so that that's the quadratic equation so a is the coefficient of x s which is 1 B is the coefficient of x which is - 9 + K okay - 9 + K and C is the constant which is 9 k + 4 so we want this to be true um where we want B ^2 - 4 a c to be less than Zer and we got to find the values of K for which this inequality is true so what we're going to do now is we're going to substitute these values in so when you put this in here the minus sign will cancel out you're just left with 9 + K all 2 right because a minus time minus is going to be positive - 4 * a which is 1 * C which is 9 k + 4 and we want this to be less than zero so it look look like we're going to have a quadratic inequality to solve so that's going to be 81 plus you expand this you're going to have 9k twice which is 18 k + k^ 2 you're going to have -4 * 9 which is - 36 K and - 4 * + 4 which is - 16 is less than Z so now if we simplify this we got k^ 2 - 18 K and we got Min - 16 + 81 so 81 - 16 81 - 16 gives us 65 so we're going to have + 65 is less than Zer so we got to find the values of K for which this inequality is true okay so remember this represents the discriminant of the um you know of the quadratic this represents the values of K for which the discriminant uh you know value is given so we want to find when the discriminant is negative okay so when we're solving this inequality we're basically finding the values of K for which the discriminant is less than zero when the discriminant is below the this axis here the discriminant will have a negative value and therefore it will be um such that that those values of K will give us no solution for the quadratic equation that we formed so now uh what we have to do is solve this inequality so the first thing is to find what called the crital critical values which is find when does when does this actually equal zero so we have k^2 - 18 k + 65 = 0 so the first thing here is to factorize this I think that this factorized with with 15 and three uh sorry uh with 15 and um 15 * 3 no that's 45 um 65 13 and five 13 and five that's the one 13 * 5 13 * 5 gives us 65 and 13 + 5 is 18 yes so they're both negative so K minus 13 and K minus 5 okay that will give us um 65 as a product and that would give us - 18 as a sum so that's the right value so k = 13 and K = 5 are the two critical values so if I just just adjust this a little bit to make our diagram a bit more realistic here so now now I know that this looks like if you were to to sketch something in this form it's going to be um this type of shape so it goes through the zero at five and 13 and it's going to have this type of it's going to look something like this so we can see that that the discriminant is less than zero for the values of what when K is between five and 13 okay if the value of K is between 5 and 13 the discriminant will have a negative value okay so it says find the values of range of values of K good all right so when K is between these values then you'll have no solution to the equation okay and uh you know the curve and the line will never touch so to find where a curve or two functions intersect we solve the equation of the two simultaneously so this is y = 9 - x when you solve these simultaneously that tells you the values of X where they intersect okay now so we want to we we have this unknown which is K so we solve them simultaneously by replacing the Y in one of them with y equals in the other and we end up with a quadratic equation which has K in it all right so we want to find the values of K for which this has no solution which is when the discriminant is negative so we set up our you know um discriminant b^2 - 4 a c now B is - 9 - 9 + k and a is 1 and C is 9 k + 4 we put them into this inequality for the discriminant and then we solve the inequality by first you know we end up with a quadratic expression which we're going to solve this quadratic inequality so we we find where does this equal zero that's those are the places where the discriminant will equal zero those would be the two x values where you would have a tangent to the curve okay the tangent to the curve would occur when x = 5 and xal 13 okay somewhere over there all right but we want to find where there's no solution when the line basically is between those values when the line is basically between those two values okay so when the line is between those two values that's when you're going to have um basically um no solution there will be no solution k equal 13 and k equal 5 okay so there's the answer to that question here I hope that was clear all right um was that the end of it I think that must be the end of it yep next is question number eight okay so that completes this question other questions from this particular paper um can be found in the playlist that will appear at the top of the screen over here um other questions from the topic of this is all to do with graphs and equations I guess you'll find the playlist over here for that and you can subscribe to the channel by clicking on this link and the video that's linked over here tells you how to use my channel to find what you're looking for thank you for watching and see you soon