can someone please confirm if you can see my screen thank you all right good okay now first of all I'm sorry for starting really late uh I'm thinking to go back to our early time start from hopefully next week onward is already uh happening around 5 50 these days next week that I I think it's going to be around six so what I can do is I can start my paper one class uh maybe around 4 pm and then have the stats class before motherab as well and then we could start paper three right after another it could be you know sometime around 6 30. sorry did I say start it would start the paper three class around 6 30 6 35 hopefully from next week onward okay right now uh Maghrib is around 5 555 so what happens is I start my S1 class after that and then that takes time uh so the P3 class is going forward and forward because of that but by next week I think I can move my S1 class before to to a time before Maghrib and then we could start start P3 earlier hopefully by around 6 30 inshallah okay 6 36 to 35 that's that was the time that we started from initially will gradually I'll go back to that time hopefully from next week at some point next weekend well actually that's the plan uh now let me just start this other session as well here all right so we were doing Vlogs yesterday and we talked about linear law in our last class okay now today we are going to do a couple of more practice questions on this and then we'll also learn how to draw a graphs of log functions and exponential functions right so those those are the two things that we are going to be doing today we will be doing a couple of more practice questions on linear law because we did not do many yesterday and then we will learn how to draw graphs of E power something or Ln of some kind right so that's the second thing and that's going to come okay all right let's start with this then so we were doing linear dot remember what linear law is we try to convert whatever function we're given to a form like this y equals MX plus C they will always tell us what they want on the vertical axis and what they want on the horizontal latched that will always be given so they often use terminology like this y against X so whatever is written again before y that means that is what is supposed to be on the vertical axis and whatever is after against that is supposed to be what is on the horizontal axis so the subject of the equation that you make has to be this thing that that comes before the word against we did some examples in this we also learned how to prove that something can be written in a straight line form that's what we did and we did one Quest practice question from the question from the worksheet as well now let's do a couple more before we move on to the graphs thing okay so this is uh paper three logs worksheet all right let's do this this is question number eight from the worksheet let's try this first all right question number eight from the worksheet it says the variables X and Y satisfy the equation x raised to power n times y equals c where n and C are constants when x equals 1.1 y equals 5.2 and when x equals 3.2 when x equals 3.2 y equals 1.05 so you're given a point here you're doing 2.0 and part in the first part it says find the values of N and C now what is the equation that we've given the equation is X power n y equals c now what can we do here we could start by taking log on both sides what we're looking for is the values of N and C okay one idea is actually to just input these values directly in the function that's also possible just input x equals 1.1 here and Y equals 5.2 here you get one equation in X and Y in n and C and then you could input these two values of X and Y in this equation that will give you another equation in terms of N and c will have two equations two unknowns you can solve them simultaneously to find the values of N and C that's one idea but then in the first second part you see what you're asked to do you're asked to explain why the graph of l and y against Ln X is a straight line now for that you will have to do all of that process in which you are converting to m x plus C form right so you'll have to do that that whole thing anyway so what you can do is what you can do is you can just do just convert to this form in the first part as well okay so rather than having to do all of that work in the second part in any case you can just do the first part using that method as well so that the second part becomes easier after that okay so what we do is we start by converting this to a straight time form how is that right now we could take log in both sides let's do that Ln of X raised to power n y that equals Ln of C now on the left side we could apply the multiplication law and break it down like this actually L and X raised to power n plus Ln of Y on the other side you've got Ln of C now make l and y the subject take everything else to the other side so that becomes minus Ln X raised to the power n plus Ln of c n comes down here and it becomes minus n into Ln X Plus Ln of C and now this is the straight line form and the straight line form is such that in place of Y you've got Ln y in place of X you've got Ln X so the gradient is going to be minus n and The Intercept is going to be Ln of C that's the intercept vertical intercept all right so this is what we have figured out until this point now we are given two points what are those two points let me write those down here 1.10 and 5.20 that's one point remember these are values for X and Y similarly we've got another point that point is 3.20 and 1.05 and these are the values of X and Y now this is where you have to be very careful with the access that we're using you see the graph that you're going to have in this case is going to be Ln y against Ln X so on the vertical axis you're going to have l and y so the graph whatever the graph looks like the axis are going to be Ln y and Ln X so the points that you have on this graph what is their form going to be it's going to be lnx for the horizontal coordinate and L and Y for the vertical coordinate so if you were to input values in this function if you were to input values in this function you will have to convert these two things to Ln X and Ln y first of all these are not points on this graph these points do not lie on this particular graph on this graph points have to look like this this is the form that you need to have for points in this graph all right now how do you do that how do you do that conversion you have to first convert these points to Ln X and Ln y so take log of both of these numbers take log of 1.1 and see what that turns out to be and N of 1.1 consumer tell me what that is 0.0953 0.0953 so that's this first thing Ln of 5.2 this is 1.649 [Music] then Ln of 3.2 this is 1.163 and then Ln of 1.05 that is 0.0487 so these are actually the points on that graph this is Ln X and in Ln Y and this is Ln X and lny once you have done this conversion once you've done this conversion now you can use these points to find the gradient and the intercept and see what data and RW okay now how do you find the gradient from this gradient is going to be Y2 minus y1 that's 0.0487 minus 1.649 divided by 1.163 minus 0.0953 this is going to give you the gradient of this curve okay let's see what this turns out to be foreign this is minus 1.49 8 and that could be rounded off to minus 1.50 so the value of M here is minus 1.50 so that's one thing that we have figured out all right now let's write that down we know that the value of n there should be so since gradient you know just m is equal to minus 1.50 if you compare the two n is equal to minus n so we put this equal to minus n what does this give you now n equals 1.50 that is one value that you have to figure out once you've figured this out the rest is relatively easier because now you've got l and y equals minus n you know the value of n now that was minus 1.5 0 times Ln X Plus Ln of c and now you can input any of these two points here in place of l and y and L and X so you've got one point here for instance in which you've got these values in place of l and y you could invert 1.649 on the other side you've got Ln X in case of Ln X dividend for 0.0953 and then you can solve this equation to find the value of C from this all right what do we get from this uh this is going to give us 1.79 for Ln of c and then you can find the value of C by taking or it must divide it the other round first what lnc equals 1.79 remember the base of Ln is e that goes to the other side and becomes the base of 1.79 so that's e raised to the power 1.79 and C turns out to be equal to pretty close to Six you can say that 6.0 so you have the value of N and you have the value of C from this does that make sense to everyone now there's actually another way of doing this now I only do it did it this way because now the second part is just a one-liner since you've already converted everything to this form but if you think that's too complicated this method another possibility could be to just input values directly that's also possible the equation is X raised to power n times y equals c the points that you're given are 1.1 and 5.2 that's 1.1.1 and 5.2 and the second one is 3.2 and 1.05 so if you did not have this second part you would you could actually just do it directly and that might be easier than the first method like we did just input these values these are values for X and Y right now if you don't convert it to that straight line form you just input those values of X and Y directly into this now one value of x is 1.1 let's import that that gives us 1.1 raise to power n into 5.2 and this equals c this is one equation the other equation is three three point two raised to the power n whole whole multiplied by 1.05 and that equals c you got these two equations and I just solved these two equations simultaneously both have C on one side so you could say just equate both of them right so that gives you 1.1 whole raised to the power n into 5.2 that equals 3.2 raised to the power n into 1.05 take the N terms take the terms that have n and n in them on to the same side that's 1.1 raise to power n divided by 3.2 raised to the power n on the other side you have 1.05 divided by 5.2 and now you could say since the power on both the numerator and the denominator is and you can write it like this 1.1 over 3.2 whole raised to the power n that equals 1.05 over 5.2 now if you want you could convert it into decimals and that's going to be okay if you were to do that uh on the left side you have 1.1 over 3.2 that gives us 0.3437 raised to power n on the other side 1.05 over 5.2 that gives us 0.2019 all right now just take log in both sides Ln of 0.34 3 7 raised to the power n equals Ln of 0.2019 the power comes down there error becomes n l n of 0.3437 equals Ln of 0.2019 and you can get the value of n from this this to the other side gets divided the value of n will turn out to be 1.498 from here as well right that's what we got here 1.498 which is 1.5 0 essentially that's the value of n that we get now once you have the value of n put this back in any one of those two equations and find the corresponding value for C C equals 1.1 raised to the power n into 5.2 this is one equation you could input the value of n in this that gives you 1.1 raised to power 1.5 in fact it's better to use the more accurate value 1.498 into 5.2 evaluate this and see what you get you'll see that it also runs off to 6.0 okay that's the value of C that you have this would be another way of doing this I think this is easier but now for the second part what do you have to do it says explain why the graph of l and y against Ln X is a straight line what would you have to do for that we'll just need to convert your equation to this form so if you haven't done that in the first part you'll do that in the second part then convert it to this form such that you have l and y on the left side and Ln X on the right side like this and then you say this is y equals m x plus C form this is important you have to write it down it's y y equals MX plus C form that is why it's a straight line okay you could also mention if you want in places of Y but l and y in place of X you've got this but that's not as important you have to State this for sure then it's y equals MX plus C point that is why it's a straight line after converting into this form first all right so that's the second part if you've done this in the first part already then this becomes easier and now it's your choice whichever method you would want to use in the scripts all right any questions on this no sir okay let's just quickly do one more question on this and then we start graphs of exponential and log functions now I'm taking one question from a recent paper uh you can actually open worksheet first of all this was the worksheet okay for three long okay remember I also told you to try this question this is question number three from the worksheet now this question says two variables quantities X and Y are related by the equation y equals ax power n so there are two variables X and Y they are related like this y equals a into X raised to the power n where A and N are constant the diagram shows the result of plotting l and y against Ln X for four pairs of values of X and I use the diagram to estimate the values of A and N now in this case we'll have to use the diagram to find the gradient and the vertical intercept right and then convert it to that to that form it will look something like this Ln y equals n l n x plus Ln a and you can wait to a phone like this n will be the gradient and Ln a will be the vertical intercept and you'll find that gradient and vertical intercept from the diagram now since they follow this form and it says that l and y against Ln X is a straight line what you can do is you can draw a line through these points and you'll see they'll pass that line will pass through all of these points this line will pass through all of these points okay if you divide using a ruler this will pass to all of these points you'll use this line to find the gradient take any two points and then this thing to find the y-intercept and find those values right this is one problem one type of problem so you have an equation given and you are told that X and Y is satisfy this equation now let's consider a slightly different statement and this is from another paper just copy this question from somewhere uh I think it should be here yeah so this is from October November 17. paper three one let's try this one now it's face it's kind of similar to that but there is a difference there's a slight difference in this case it says two variable quantities X and Y are believed to satisfy an equation of this part where C and A are constants and then an experiment is produced an experiment produced four pairs of values of X Y the table below shows the corresponding values for X and Y now rather than points they have driven plotting the points they've shown the points here in the form of a table and now it says by plotting l and y against Ln X for these four pairs of values and drawing a suitable straight line estimate the values of c and a now it's slightly different from the first from the earlier example that I showed you because in this example it says they are related by this equation and there's no experiment so if you were to join these points they will fall on an exact straight line all of them will pass through um the line will pass through all of these points but this statement that you see here in this question in this case it says they are believed to satisfy an equation but they're not sure and they're conducting an experiment to basically confirm if they satisfy that equation or not okay and whenever there's an experiment there are possibilities of Errors right so those of you study Sciences particularly you would know whenever you conduct an experiment is human error so all the points are not going to fall on an exact straight line there will be variation okay and that is okay so you just need to be aware that there is going to be variation so the line that you have to draw in this case is the suitable line that they're talking about that line is going to be the best fit line okay so in this case we'll have to draw the best for tonight now let's do this question quickly this is the relation that will be given y equals C into a raised to the power of x this is the equation C and A constants an experiment is producing these values we need to find the values of c and a what we have to do is plot l and y against X let's plot that thing first so what do we have in the vertical axis the l and y and the horizontal axis we've got X let's plot those points 0.9 and 1.7 you see the values that are that are given they're already in this form X and Ln y so you don't have to do any calculations you just have to plot them 0.9 and 1.7 where is that point going to be it's going to be somewhere here this is the first point 0.9 and 1.7 then you have 1.6 and 1.9 that's going to be here then 2.4 and 2.3 and then you have 3.2 and 2.6 all right these are the points that you have and you see they are not lying on a on a straight line in this case they're not all in all in a straight line so what you do now is you draw a best red line now remember how used to draw a best fit lines the lines should be such that it's close to most of the points right it's it's passing through uh it's it's passing through these points such that the distance of that line from these points is very minimal that's one thing it should pass through uh it should it should be very close to these points that's one thing and the second thing is you should have roughly the same number of points above and below the line it it's not uh uh a very strict rule number above and below the nine okay so now considering that keeping that in mind you could draw a Best Bet line here okay so let me try so this could be one possible line right so of course there's going to be chances of error here and that's okay but that's one possible best fit line that you could draw here okay you could slightly move it above if you wanted in this case because yeah so now you have two points above the line two below the line somewhat and that's your best print line because this is one possible line that you could draw now you have to use this line to find the values of c and a now you know what you what you have in the vertical axis and the vertical axis you've got l and y so this is your if I were to write that form y equals MX plus c y equals MX plus C this is the form that we need in place of Y we need Ln of Y in place of X we need X so we need to convert this equation to that form how does that work we take log in both sides that gives us Ln y equals Ln of C into a raised to the power x now we split it up and it becomes Ln y equals Ln of C plus Ln of a raised to power x rearrange this and bring X down here that's going to be X Ln a plus Ln of C I can write it like this Ln y equals Ln of a whole multiplied by X so the coefficient of x is Ln a plus Ln of C compared to the standard form y equals m x plus C you got X in place of X that's what we wanted in place of Y so what's the gradient of that line gradient of the line is Ln of a and the vertical intercept is Ln of C now you find those two things things from the graph find the gradient how do you find the gradient you are not going to use the original points for finding the gradient keep that in mind you cannot use the original points you have to use points that lie on this line all right you have to use points on this line not the original Point especially if they're not on the line okay let's take two points on this best red line so you could take two points which are relatively easier to read let's say I'm I take this point here this is 3.9 and 2.9 so you should take points at a distance from each other as well this should not be too close to each other that's an important thing to remember there should not be too close to each other this is one point two point three point nine and two point nine take another point on the line you could take this for instance are let's say this one this point is 0.5 and 1.5 Now find the gradient read the y-intercept from here as well what's the y-intercept vertical intercept that is 0.3 let's find the gradient of this line gradient is 2.9 minus 1.5 divided by 3.9 minus 0.5 what do we get from this the gradient 2.9 minus 1.5 divided by 3.4 the gradient turns out to be 0.41 8 for me this is the gradient 0.4118 okay now we can use those things here we can say gradients value is 0.4118 and that gradient should equal Ln of a we can find the value of L and A from this Ln a equals 0.4118 so a is going to equal e raised to the power 0.4118 now you want your answer connect to two significant figures keep that in mind so when you write down this answer e raised to the power 0.4118 you can see what that turns out to be e raised to power 0.4118 that is one point five one all right that's the value of a we wanted two significant figures so this would become 1.5 now what about the vertical intercept we know from the graph the vertical intercept is 0.3 so what we say is let's equate Ln of C that's 0.3 and that's going to give us the value of C which is e raised to power 0.3 in this case which is one point three four nine for me if we were to round this off I'm sorry uh this number was 1.3 not 0.3 this was 1.3 right so from the graph that's actually 1.3 not 0.3 so it's 1.3 here so e raised to the power 1.3 let's see what that turns out to be this is three point six seven if you were divided off to two significant figures that would be 3.7 this is the value of C that we were looking for all right so this is how we find the values of A and C given this information any questions on this any problems anyone foreign by the way was there any lag during class at any point no sir okay that's good honor just give me a minute and then we'll start the graphs thing okay is it okay if we spend 20 25 more minutes on graph now logs and no not logs Ln and E power x cross lnfx and E bar X graphs this is what we want to do do next now desktop is actually tested with iteration right so it's tested with iteration but since it's it's relevant to logs let's do it right now so in this we want to learn how to draw log function exponential functions now keep one thing in mind first of all the graphs of these functions so the shape of the graph you can just take points and you know try to draw that I'll just tell you first of all if you have a graph like this y equals E power x you can take some points plot them you'll figure out the graph of this function looks something like this you can do it like you usually used to do take points plot them and see what you get the graph is going to turn out to be something like this okay this is the graph of E power x function now an important point on this graph is when the value of x is zero because when the value of x is 0 the value of 1 here is going to be one and the value of y here is going to be 1. so when you have the power equal to zero the value of the function is one so this is an important point that you would show in this graph and the rest of the graph would look like this okay so that's the shape of this graph that is to keep in mind is exponential function it's increasing like this it's increasing exponentially right that's why it's called an exponential function it's basically increasing like this that's one thing uh that's the shape of this graph and another graph is y equals Ln of x this graph looks something like this in the shape of this graph is something like this that goes this way and in this case when the value of x is one that means when the number inside here is equal to 1 you get 0 as the result okay in this case this is an important point that you have and these are the graphs this function looks like now if you notice these two functions are actually inverse of each other these are inverse of each other so they're actually reflections of each other in the line Y equals X as well right so remember inverse functions are reflections of each other in the line Y equals X so these are actually inverse graphs how can you know that you've got this function y equals E power x how do you find inverse make X the subject if we were to make X subject for that you'll have to take log on both sides that gives you L and E raised to power x let's say the original function was f of x that was e raised to power x we converted it to Y that became that became y equals Z power x and then you take login both sides and on the right side you get X into L and E which is just X so this is x equal to Ln y f inverse of X turns out to be Ln of X so if the function is E power x its inverse is Ln X and also the other variant if you add this function Ln X its inverse would be E power x all right so these two functions are actually inverse of each other so they're graphs you can see look kind of the opposite as well E power x is going like this and L in X is reflection of the art graph and the line Y equals X okay so that's the that's the first thing you have to know the general shape of the of these original graphs E power x looks like this and Ln X looks like this that's the first line now the problem I would be you would not only get functions like these you'll have transformation the transformations of these functions so for example you could have something like this y equals 2 e raised to power minus X or you could have y equals minus into Ln of X plus 2 something like this how do you draw these graphs or we could have plus three outside as well let's say how are you going to draw draw these graphs now the easiest way for the cells that you take some points take some points on on these functions plot them and then then just then just join those points that you get but for that you need to have at least some idea of the shape of the graph all right at least some idea of the shape of the graph so you should know what possible shapes you could get for these graphs so these are the original shapes but remember when you do Transformations you would have Reflections you have a graph like this as well you could have a graph like this as well or a graph like this as well so there are transformations that are possible for this function as well similarly for log as well you could have a graph like this you could have a graph that is going like this or like this so there are many different possibilities in general you would see roughly the shapes are going to look like this they're going to be four possible shapes roughly it's going to look either like this okay either like this or like this or like this or like this these are the four possible shapes that you will be seeing in these graphs for both exponential function and log functions all right these are the four possible shapes roughly that you will be getting so you have to keep these in mind that is going to be one of these four shapes all right so this these are the only possibilities that you have it's one of these four possible shapes now how do we for the graph which points do we take on uh on the spot for these functions that is the key thing now so we have to figure out a way to take points smartly so that we use those important points at a minimum so for example in this case I told you this is an important point in this case this is an important point now let me write that down here whenever you are drawing the graph of a function like this e raised to power something it is to power something you have anything outside being multiplied or being added that's okay you could have that whenever you're drawing graph or function like this or function like this a e raised to the power of something plus minus B okay what you have to do is there is there is one important point that you're going to be using here for sure and that is when the power is equal to zero so the important Point here is going to be important point is then power is equal to zero so whatever is in this box when that is equal to zero that is an important point now it's not always going to be x equals 0 because you could have something else in the power as well so for instance you would have a graph like this you could have y equals you could have a function like this y equals 2 e raised to the power power x minus two you want to draw the graph of this function let's say what you're going to do in this case is you're going to put the power equal to 0 x minus 2 equal to 0. what does that give you x equals 2 this is an important point on the graph of this function all right this is an important point so you have to show this graph show this point when you have x equals to what's the corresponding value for y in this case when you put x equals to e raised to the power 0 that's 1 into 2 the Y value in this case turns out to be 2 as well that's 2 into 1 which is 2. this is an important point now two and two and what you do now is you take one point in fact two points on either side of this you've got the important point in the middle this point you have that in the middle here and then you draw a table such that you have two points to the left of it and two points to the right of it okay and that is going to help you draw the graph of the function why are we using this point well at this point if you look at these graphs the shape of the graph actually starts changing rapidly after this so until one until this point it was pretty flat after it crosses one it starts Rising again like it starts Rising rapidly like this on the other hand the log function this is the important for the log function this is an important point the graph is really steep until this point and then it kind of changes the gradient rapidly and it becomes flatter after after this point so it's kind of the point after which the behavior of the graph you know starts changing quickly all right so for this the point is where the power is going from negative side to positive side okay and this is then the number inside is equal to is basically either less than or greater than one we'll talk about that in a moment for E power x what we're doing is for E power is something what we're doing is we're taking this as an important point what point when the power equals equals zero now once you've taken this point you take one point on either side of this in fact two side two points each on both sides two points each on both sides take one point to the left one take two points to the left two points that to the right okay so let's try for instance in this case you could try I don't know minus two let's try that if I input minus two here at minus two I get two into e raised to power minus 2 and that gives me 0.27 okay sorry when X is minus 2 The Power should be minus 4 in fact this should give me zero point zero four instead so it's a pretty small number right if I input 0 there add 0 I get 0.27 and then at 2 I have to let's take some points to the right as well let's say I take 4 here and 6 here what do they give me two e raised to the power 4 minus two that's two this is 14 point something 14.8 we have power 6 uh x x x equal to 6 that is giving me one and nine now if you take this Middle Point and then two points on both sides you're going to figure out what shape it's going to be so you know there are four possible chips we saw that we saw those shapes above these are the four possible chips it's going to be one of these shapes roughly okay so what you do is you plot these points you have two two so let's actually keep the scale a bit small because we have got a very big number we can't you don't necessarily have to show that but it's just showing us the trajectory of the graph that the graph is going to go very high at this point right so it's just sketching you don't have to plot exact numbers you can just do it slightly roughly as well and that's okay on the x-axis we've got minus we've got minus 2 0 2 4 6 and so on what point do we have at -2 it's something very close to 0. at 0 again it's slightly above but still it's a very small number then at 2 we've got two let's say this point here is 2. so you've got one point here and then at four it becomes a pretty large value that's going to 14.8 let's say so let's show that here so roughly now the graph is going to go like this now you should know the shape of the grid that's all that matters this is what the graph looks like that's the graph of the function that you have to draw it's 2 e raised to the power x minus 2. this is the shape of the graph okay now it's not going to touch the x axis here it's just going to stay close to the x axis like this and never actually touch the scrap okay there's actually a name that we've brought to this this thing is called an asymptote for E power something graphs this asymptote is always going to be horizontal so your graph will be get becoming horizontal at some point so here the graph is becoming you know almost horizontal like this okay so you have a horizontal you have a horizontal asymptote here this is what the graph looks like okay so this is the way for this is the method for drawing graphs of exponential functions you take one important Point what's that important point when power is equal to zero when power is equal to zero whatever value of x you get take one two points to the left and play two points to the right of it plot those points then draw that graph okay now there's a function on the on your calculator as well called the table function that really helps in these graphs you can actually draw this whole table on your calculator without having to write down all of these values on paper and that makes things a bit more efficient uh we will learn that later when we do iteration we'll have to draw a lot of graphs there so we will learn how to use that table function at that point uh I think we haven't done that yet right uh table function in any other topic yeah we haven't so we will do that in iteration okay if you want to learn yourself it's called the table functional calculator it actually makes these tables for you directly so you can just have a look at those values from the calculator plot them directly here and you will get your graph so you won't have to do this working here manually anyway so that's the exponential graph that's how we think about it how about what about the log graphs Allen graphs let's say you've got a function you've got a function that looks like this y equals minus 2 into Ln of X plus four you got this lunch you want to draw the graph of this function how do we think about this now for these graphs we've got two important points in fact the important point is only one actually but I'll show you I tell you what the other thing uh means when I write that down so it's Ln of let me draw a general form actually the first axis that's its Ln of something a times some constant multiplied by Ln of something we don't know what that is then plus minus something outside it could be anything of this one the important point is going to be when this thing in the Box whatever that is is equal to one that is the important point okay and in this case you will have a vertical asymptote you see this was the original log of Ln X it becomes vertical on this side this y-axis here is the asymptote of this graph with which the curve you know keeps getting closer but never actually touches it this is the vertical asymptote of this graph Allen Glass have vertical asymptotes exponential graph these have horizontal asymptotes now in in the log graphs you can also find the line that is actually going to be the asymptote how do you do that for the all you have to do is put this thing in the Box whatever that is sorry you put this thing in the Box equal to zero why because that is when the log function becomes undefined right so if there was any positive value inside log the function will be defined but if it's equal to 0 or anything negative the function becomes undefined 0 is the cutoff point where the function you know starts becoming undefined so this value is going to give you the asymptote now let's take an example let's say we've got something like this y equals minus 2 into Ln of X plus 4 that was the function that we wrote earlier I think right minus 2 into Ln of X Plus 4. how do you draw the graph of this function what you do is you take an important Point what's that important point the important point is when you put the thing inside Ln equal to one so put that equal to 1. X plus 4 equals one what's the value of x that you get from this x equals minus 3 this is an important point that you have to show on the graph what's the corresponding value of y on this if you put x equal to minus 3 y will turn out to be 0 in this case if you had some number being added outside it could be something other than 0 as well and that's okay that's also possible Right you could have by something else as well in this case it's turning out to be 0 because when we put x equals minus 3 here this thing becomes 0 and that multiplied to minus 2 gives us 0. so the value of y is zero so the important point that we have therefore is minus 3 and 0. so when we draw a table this is the point that we take in the middle this is the point that we take in the middle this is minus three and zero now we have to take two points each and both sides you could actually do with less than that as well if you have you know if you figure out the shape that's also fine but let's just keep this for now minus three and zero this is the point that is the important point we are keeping this in the middle here minus 3 and 0. now once we're done with that you can also find the asymptote how do you find the asymptote for the SM code you put this thing inside the log equal to zero so what we have inside the log we've got X plus 4. if you put this equal to 0 that gives you x equals minus 4. this is your SM dot this line x equals minus 4 you're going to have your asymptote here your graph cannot cross this line all right now let's take a couple of points on both sides of this and then we'll draw a graph using these points and the asymptote let's take points to the left now remember we just said x equals minus 4 is going to be the asymptote that means the graph cannot cross that point if if you were to input any value here that was less than -4 the graph would turn out to be kind of the function will turn out to be undefined so you can only take values which are greater than minus four so you could write for instance uh minus 3.9 you could try minus 3.5 on the right side you could try for instance what happens at 0 you could write what happens at five so in log function what you do is you take values on one side that are closer to each other on the asymptote side and the other side that are slightly further away from each other like this but you'll realize you can actually figure out the shape of the graph pretty easily and that's all that matters at the end since it's a sketch the actual points do not matter where is the point minus three zero that may be somewhere here let's say this point is -3 and the y coordinate is zero this is one point that you have all right where's the SM dot the asymptote is at x equals minus Force that's going to be here and now what you do is you draw a vertical line here this is your sm20 plus graph your God cannot go to this line This is the line x equals minus four this is your SM door now just this bunch is actually going to tell you that the graph either looks like this encode is like this this is one possibility right or like this because you could either have a symptom like this or the or like what we did and the first case these are the only two possibilities now if you are able to figure this out only one or two point two more points are going to be enough for you if not you can take more and then figure out the shape let's try and figure this out now if we were to input minus 3.9 in that function let's see what that tells us minus 3.9 I would have minus 2 into Ln of 1 and that is of minus 3.9 plus four and that gives me 4.6 so at minus 3.9 I'm getting 4.6 at minus 3.5 I'm getting 1.4 add 0 I'm getting minus 2.77 and at 5 I'm getting minus 4.4 you can see the shape of the graph now so to towards the left of three the points are 1.4 4.6 so if I would deployed these here let's say this is two this is uh four minus 3.9 and 4.6 that's going to be somewhere here then minus 3.5 and 1.4 that may be somewhere here so you can see how the graphics you can see how the graph looks like on this side it's going to go upwards this way look at something like this and just by looking at that you can actually figure out the shape of the graph on the right side as well if you remember the possible shapes sorry remember the possible chips it's one of these four possible chips once you've figured out what the left hand side looks like the part on the right it should go like this right you could confirm that using those points as well just plot those points at zero required minus 2.77 so we have one point here and then at five five maybe somewhere here let's say you've got this point which is minus 4.4 and that may be somewhere here but you don't even need to have all of those points once you have this important point and the asymptote the rest is going to be fairly easy to figure out so this is how you draw the graph the log functions graph now two things that you need to remember important points how do you find those important points for exponential functions you put the power equal to zero that is how you get the important point and the exponential function right and for log functions you get the important point when you put the part inside the the log equal to one that is when you get the important point when you put that equal to 0 you get the asymptote all right so just use those important points and the asymptotes take a couple of points on either side and just use the fact that the graph is going to be one of these four possible shapes and you'll get your graph okay I hope that makes sense if you have any quick questions let me know otherwise that's it for today that completes this topic as well you have to remember this method when we when we reach iteration after a month and a half inshallah because this is going to be important at that point uh we will be doing graphs questions with that topic all right because that's that's where it is actually tested all right so that's it if you have any quick questions let me know otherwise that's it I'll see you again tomorrow thank you