Yes So I think last class we started with the graphs So today I'll start with the property of log. I'm sure you would have already done it in your ad maths Yeah, I think so just a quick revision of properties of Logarithms So starting with the first property log of m To the base a log of n to the base a And that's gonna be log of mm m into the base. Yeah provided these all logs are defined Okay Now how does this property come about it comes about from the very basic? Exponential laws for example if I call this as p and if I call this as q So when you say log of m to the base a is p m becomes a to the power p Yeah, correct. And this becomes n becomes a to the power of q Yeah, now when you multiply mn when you multiply these two Mm-hmm. You get mn is equal to a to the power of p plus q Which is yeah, which is as good as saying log mn to the base a is p plus q Correct, and that's nothing but log of m to the base a plus log of n to the base a Oh, yeah, okay. Got it. Yeah And this can be extended to any number of logs. So if you have log m log n log P log q like that. It'll keep on getting multiplied over here. Okay. Yeah, this property is very useful in Performing calculations in doing Calculations in doing calculations So in J and other exams probably calculator is not allowed and many people prefer using log tables for doing the calculations complex calculations in chemistry and physics in physics you will have modern physics which involves a lot of calculations and They're really dealing with a lot of numbers which are having You know higher powers and a lot of decimal places Okay, so this property is useful in that regard next property is log m to the base a minus log n to the base a I whose log m by n is log m by n to the base a Again, the proof is similar to what we had done in the previous one. So I'm not going to do the proof here log of m to the power n to the base a is Is n log m to the base a Can you prove this quickly? Log it and log m to the base a so How would you prove that? Okay? Okay, so, oh, yeah, okay first Do log a Okay Second, okay, so let's say log m to the base a is X Okay, so m is a to the power x Okay, correct. So m to the power n would be what? And power n would be a to the power x whole to the power n which is nothing but a to the power n x correct Okay, which is as good as saying log of m to the power n to the base a is nx Okay, yeah, okay x here is log m to the base a so you can replace it over here So it becomes log m to the base a Oh, and hence the property Okay, again these two properties are useful in Calculations If you have to raise a number to a power Then this property number three will be useful and if you have to divide two numbers then property number two will be useful Yeah Going to the next property property number four log of m to the base n Can be written as log m to the base a By log n to the base a this is called the change of base property This is called the change of base property. This type of property is useful in answering questions like What is log let's say 64 to the base of 16 Yeah, okay now if you start counting powers on 16 16 to the power 0 16 to the power 1 16 to the power 2 Exceed 64 right so we cannot find a power like this because we only know how to you know calculate Powers of natural number manually else. We have to use a calculator. So in this case is what we do we try to find a number Which whose power contains 64 also and whose power contains 16 also Yeah, four is such a number whose power will contain 64 also 16 also So what you do is I'll choose a suitable a here it is four And I will do this calculation So when I do that this numerator becomes three denominator becomes two That means log 64 to the base 16 is 3 by 2 that means 64 can be obtained by raising 16 to the power of Okay Again, so this is just a verification that we did now. Can you prove this property? um This Try it out Okay, fine um so log n Okay, I'll uh, I'll equate it to something say P then n to the power p equals m Okay, then okay if n to the power p equals m and here you have log a log m base a by log n base a okay, and uh Okay, and then um, I'll okay if I have this then I'll be To the power p by Do you need my help in? Let me see. Try it out. Try it out So Okay, so basically I just I did uh Enter power p equals m and then uh, whatever m was there. I substituted n to the power p So now, uh, it'll be like p log n base a by log n base a Okay, and then I think I can cancel out log n base a and log n base a and that gives you p but P So p was the same thing as log m uh to the base n so I guess that proves it Okay So I have some different approach. Let's say log m to the base a is x Okay, and log n to the base a is y Okay, so from here I can say m is a to the power x Correct from here. I can say n is a to the power y correct. Yeah Now m which is a to the power x Can I write it as a to the power y to the power of x by y? Okay, wait so m m equals a to the power x Yeah, and a to the power x I have rewritten as I have rephrased it as a to the power y Raise to the power x by y Okay, okay, I can find so which makes m is n to the power x by y Correct because a to the power y this term was actually your n Yeah, yeah, okay, so that means log of m to the base n will be x by y Correct and x itself was log m to the base a and y itself was log n to the base b Okay, but yeah easy. So this was the critical step in this problem. Okay, but what I did was I'm right? Yeah. Yeah, that was also correct Okay log m to the base n is Reciprocal of log n to the base m Okay Can you prove this? Reciprocal of log n to the base m Okay Into the piece m So if log m to the base n equals something Then n to the power x equals m Okay, if n to the power x equals m Am I getting somewhere? You can use the previous property itself Oh, is it the previous property? Okay. Yeah, it actually means you have to prove this this n to This is one Oh, okay, okay fine Okay, I'm sorry I have to do that then log and then you can change the base Huh, you can change the base over here. Yeah, it's the base to oh a okay You can do this Yeah, okay, this gets cancelled this gets cancelled this gets cancelled you get a one which is your RHS Okay, I got it Correct. Yeah Okay, next property property number six A to the power log m to the base A A to the power log m to the base a is m itself Yeah, and this is a very very widely used property and you'll find this property used so many places in your class 11 Okay That means if this number and this number are the same Then the answer would be this number That's okay. Yeah, I'll just do a quick verification. Let's say four Race to the power log 16 to the base four So you can do this Race to the power log 16 to the base four Then if you calculate the power it will is actually two which gives you 16 itself. So this number comes as an answer Got it. Yeah, can you prove this? Okay No hint this property comes from the very basic definition of log itself Yeah So I did Because x and a to the power x So log of What's that going to be that's going to be a um data power x equals So basically uh for this log uh m base a I simply equated a log m base a to a variable Okay, like x and I got a to the power x equals m Okay, let me write what you did Okay, and then when I substituted a to power x equals m and that you know in the power log m base n You got log a a to power x. So that becomes a to power x which we already said was m. Yes, correct So this x you have to replace it back with this term. Yeah, and you got the answer That's why it comes on the very basic definition of log Correct. So if somebody asks you what is e to the power ln x. What is ln by the way? ln natural log ln means to the base e lg means to the base 10. Yeah This is mostly used widely used in calculus This is using calculus Okay This is used in calculations Okay, so this term will be x itself. Yeah, so what is e to the power ln x cube? Lv x cube x cube itself, okay. Yeah next property is A to the power log Uh, b to the base, let's say n Is same as b to the power log a to the base n That means a and b can swap their positions and the value will not change Okay, so I'll I'll just do a quick verification of this. Let's say our two to the power log Let's say a 16 to the base of four Yeah, okay. What is this answer? Um Which is going to four, okay Even if you swap the positions that means if you do 16 log four to the log two to the base four This becomes 16 to the power of Ha, which is again four. So these two are equal So this is just a quick verification. Can you prove this? Okay, uh log b log a to the base n So I have to prove it and how do I prove it exactly I am going to um Equate that to x. I'm going to try it that way then n to the power x is when I give me b Okay, so uh I'm I created log b to the base m to x. Okay, that gives n to the power x. Um, that equals b Okay, that equals b then um Okay, then what do I do? Um So I first I put that over our b's power n to the power x Um b to the power x. Yeah Okay, so a log n base v equals b log Probably it gives b to the power x. Okay, and we know that b to the power x equals n to the power x So it's not going to help me. Let's see. Um n to the power x hold to the power x Okay, what am I doing? No, no, it's okay. Let me see. Wait, wait, wait, wait No, this is going to be Am I getting anywhere? I don't know if I'm getting anywhere. No, no, one second Why if I do I know you can help me out Okay See uh, you can say let log a to the power n b x And log b to the power n b y. Okay Okay, so a is n to the power x b is n to the power y Okay Okay Yeah Okay Now what does this property saying this property is saying a to the power of log b to the base n that means n to the power x to the power y Correct and b to the power log a to the base n means n to the power y to the power of x They both are same. That's correct. Actually because this is also n to the power x y and this is also n to the power x y Oh, oh nice Yeah, so hence Okay, nice Well, now we are going to focus more on graphs. So we have seen the property so I think the shifting and Uh, shifting of the graph to the left and the right. I've already told you as well as up and down also Correct. Have you also done changing the sign of x and y also? Yeah, I've done reflection right reflection about x axis when you change the sign of y reflection about y axis when you change the sign of x So a quick question on this will take up. Uh, Let's say plot Plot three minus y in log log x plus three to the base of half x plus three to the base of half Okay, so, uh, I have to give you like a general structure, right? Yeah, I mean roughly how would it look like? Uh, you just draw it on your notebook and let me know once you're done Okay, just do a step by step. Don't be uh, don't jump any step. Okay So step number one you would have drawn the graph of log x to the base half Yeah, and I think it intersects the Um, the x axis at one comma zero. Yeah, it comes it comes from top down like this. Yeah Okay, and then when it has to be x plus three then it ships to the left by three units So that means that it ships, um One minus two it becomes minus two comma zero there Right. So second step is when you change your x with x plus three and as you rightly said It is going to move three units to the left. Yeah So the graph would not appear like this To y, uh, three is added if three is added then it's just a question. Arushi. What would be the asymptote here? Uh, the asymptote. Oh, it shipped uh over there. It was x equals zero X equals minus three x equal to minus or you can say x plus three is equal to three. Yeah, okay Now what do you want to do? Do you want to change the sign of y first or do you want to do y plus three first? Okay, I'll do y plus three first. I did that so, um, if I do y plus three first then, uh, it'll ship to the, uh, it'll ship Down by three units. Absolutely. Absolutely. So it'll ship down Minus two comma minus three So it'll ship down by three units. So it'll be Coming like this Yeah, minus two comma minus three will be the point at which, um, it cuts Yeah, where does it cut the x-axis? Oh, it cuts the x-axis or just, um, it Um, okay, it's pushed where it cuts the x-axis and y-axis both Okay, um, so initially it was cutting at one comma zero, but then it moved there And now it moves like three units to the left and now you can directly find from the equation itself Let's say you want to know where it cuts the x-axis put your y as zero Okay, so when you do that you do that you get this Okay Yeah, and this is like saying half to the power one-third half to the power of three Minus three that is going to be one eighth minus three. This will be your coordinates, which is actually minus 23 by eight Okay, so it's kind of that way. Okay, fine. It'll cut at minus 23 by eight comma zero So this is to know where it cuts the x-axis. Now. How do I know where it cuts the y-axis? It'll be minus two comma minus three. So put x as zero then. Yeah, okay So when you put x as zero It becomes y plus three is equal to log three to the base of half So why becomes log three to the base of half minus three? Yeah, this will be zero comma log three to the base half minus three point Okay, got it. Okay. And yes now what next I have to do remember I have to get three minus y over here Correct. So I will change my sign of y. That's my fourth step You change the sign of y you reflect it across the x-axis. Absolutely. We'll reflect it about the x-axis So when you reflect it about the x-axis How would the graph look like? Okay, so you have that point minus equal minus three and if you have to reflect that about The x-axis then the y-coordinate should stay the same and it goes up It would be two comma minus three So this point will remain the same So this will be reflected like this and this entire part this entire part as you can see I'm running my pen on it Yeah, this entire part will be this part Correct. So now this point will become zero comma three minus log three to the base half Okay, yeah, and this one still remains the same 23 by eight comma zero Yeah, let's check on jujibra whether we are getting the same result or not. Okay Yeah, so our function was three minus y is equal to log one by two of x plus three Okay, this side graphic issue. That's why it's not showing complete Yeah, uh, don't think that's stopping there. It's going down and out. Okay So this point if I try to find out it will be minus 23 by eight That's near about this minus 2.86 I just got up this slide graphic issues. Huh, okay Okay, right Yeah Now let's talk about exponential graphs exponential functions Okay, now exponential functions are functions of the nature a to the power x Where a is a positive quantity not equal to one Okay, fine. Yeah, if you see this it is almost the same Property which log base also used to follow, right? Yeah When I used to define log x to the base a I used to say that a should be greater than zero and a cannot be one It cannot be one. Yeah. So almost the same thing is followed here as well Yeah, okay If your a is negative what happens for fractional powers, you'll start getting imaginary values of f of x Correct And if a is one it'll actually become a straight line. So that is treated as a constant function not an exponential function When I say a is greater than one and not equal to zero. It means there are two possibilities One where your a is greater than one And other when your a is between zero and one Okay, right So for both both these scenarios the graph is slightly different Okay So I'll I'll I'll draw the graph for both of them separately Let's take a case where a is greater than one. Tell me any value of a greater than one Two two So let's say I want to draw the graph of two to the power of x Yeah Now as you can see Two to the power x if you take value of x as zero y will be one Correct. So the function will be cutting the y axis at zero comma one Got it, correct As you increase the value of x as one you will get a value of two Yeah for two you'll get a value of four Yeah, three you'll get a value of eight like that Okay, correct for minus one you will get a half Yeah Minus two you can one four like that. So if you connect these Dots you would realize that the graph of this function would look like this Yeah, it increases and remember here Your x axis that is y equal to zero would be acting like an sm tote Okay, fine. That means an exponential function will never ever touch The y axis and hence it will never go negative as well Okay, yeah an exponential function can never be negative remember that Yeah So if somebody says solve this equation three to the power x is equal to minus two That's not possible Because it can never be negative If at all you have to solve three to the power x is equal to zero then x has to be negative infinity That's also, you know very difficult to achieve Yeah, okay What about When your a is between zero to one in this also I'll ask you some questions just like I asked you in log What happens if I ask you y is equal to four to the power x? Okay, that would be y equals two to the power two x right that means will it be more bent like this Or will it be Less bent like this So is the answer the white graph or the green graph? Okay, okay, so uh, so wait, so this would be the same thing as y equals two to the power two x So if you put in a value of x say one Should give you like, uh, okay Um So here at x equals um at x equals zero it was another y axis it got at all of them. It got that Yeah So if I put um x as x as one for example Then y will be four But if I put x for for the first one if I put x as one then it was two Right, so we are we are when it's one it's four and over there when it was one it was two So, um, for example, this is x of our one. Okay. Yeah, um, so you're right So it is two for this and it should be four for the other one, right? A white one so it has to be the white one It has to be the white one and another thing that I would like to highlight over here I think I didn't tell you before If you have the graph of This known to you correct and if you want to draw the graph of something like let's say kx Where k is greater than one? Okay, then this graph This graph would be the graph of this more bend towards y axis more bend towards y axis Okay Okay, so if it's like y equals f of x and I'll give you an example very simple example to relate Let's say y is equal to x Okay, you know the graph of this is a straight line like this Yeah Correct, and if I ask you y is equal to uh four x Okay, then the graph of this will be more bent towards the y axis and become like this Okay, so as to say that it will shrink along the y axis Okay, okay So when you were when you were doing this analysis that uh four to the power x was Uh two to the power of two x then I got reminded of this because here Now you're changing your x with kx where k is greater than one So your your graph will shrink more towards y axis so it'll bend like the white one Okay, got it. Honestly. Yeah So higher the value of the base the more bending it will have towards the y axis Yeah Now let's go on to the uh Base being between zero to one. So let's choose half to the power x Now given that you know y equal to two to the power x graph Can you produce the graph of y is equal to half to the power x without re-plotting it? Yeah without re-plotting it if you know the graph of y equals to the power x Oh, okay. Uh, when we did the logarithm graph, um, it was the mirror image of this So that also be the case here Uh, see here are you see half to the power x is as good as two to the power negative x Yeah, okay. You are changing the sign of x Yeah, so if you're changing the sign of x means you will reflect this graph about y axis Yeah, so won't the graph appear like this then Oh, yeah, correct. Got it. Yeah Again, it will cut it at the same place zero comma one. So this will be the graph of half Uh to the power of x Okay, got it. Okay What will happen if I ask you the graph of one fourth to the power of x Will it bend more will it be like this? Well, it goes one y equals um two to the power minus Two x Um, it should bend more towards the y-axis, right? Absolutely. Now you've got the trick Yeah, so this is as good as saying y is equal to two to the power of minus two x So you are replacing your x with two x that means bending towards y axis or shrinking towards y axis should happen Okay Now another important Fact is the relation between a to the power x And log x to the base a these two graphs are actually related How they're related, let me show you on a diagram how they're related. You will be able to identify Okay, if I draw if I choose a common base, let's say I choose a base as two Okay, then I draw y is equal to two to the power x. Let me draw it with a yellow pen So y is equal to two to the power x will be like this Okay Okay, and let me draw y is equal to log x to the base of two So that graph will be like this Okay, do you see some kind of uh similarity between these two graphs? Yeah, they do look like mirror images of each other about which line uh y equals x Absolutely correct. So this is zero comma one then this is one comma zero And as you rightly identified, they are mirror images about the line y equal to x Yeah Mirror images about this line y equal to x Such functions which are mirror images of each other about y equal to x are called inverses of each other Yeah So if you want to find inverse of any function you just have to swap their positions of x and y Yeah So if I swap the position of x and y in this you'll see that I end up getting this Let's try this out. So x is log y to the base two That means y is equal to two to the power x. So I got this function Okay Here also you stop the position of x and y you'll end up getting this function So let's see x is equal to two to the power y. So y is equal to log x to the base two So we got this function back. Okay. Yes. So in inverse case You can get the other function just by swapping the position of x and y which brings me to a very important rule that if you replace or if you swap if x and y are swapped x and y are swapped The graph gets reflected About y equal to x line That's reflected About the y equal to x line Okay, is that fine? Yeah, okay. Yeah. So with this rule in mind Can we plot the graph of x equal to y square minus four y plus three? Okay, yeah So x equal to y square minus four y plus three So if x and y are swapped the graph gets reflected about the y equals x axis x line, sorry, okay, so Okay, so first if I Swap x and y here then it'll be x square minus four x plus three then Okay, if I okay, I need to first plot this and then reflected about the y equals x line So just include it and let's check whether we get the right answer So if I have to plot this then I'm gonna Just try this meanwhile. I'll have some water and be back. Yeah, I'm just doing rough sketches Why who's x for x square minus four x plus three rough? So it has like roots at one and three and then what x is at two comma minus one Then I have to reflect it about y equals x and that's the problem right reflecting Imagining the reflection is slightly problematic for many of us. So I So basically I think that you would have to just wait for the main Points we did just like a like swap x and y for example For the for the inverse the vertex will probably be at x equals minus one and y equals two Maybe like that. You go that way. Okay. Let's let's try it out. See first of all, you would have done this graph because this is easy to plot correct? Yes So when you plot this as you rightly pointed out, this is a factorizable and even if it is not factorizable We know how to complete the square and do it right So let me try that. So this is x minus two whole square Minus one correct. Yeah, so it's y plus one is equal to x minus two square which clearly indicates that The y equal to x square parabola has to be shifted two units to the right And one unit is down right so it it'll appear like this So at this point that is the vertex which we call it will be two comma minus one now When you have to swap your x with y and y with x you need to actually imagine the reflection of this graph about y equal to x line which actually is difficult for many of us including me when I was preparing uh as a 12th grader so What I did I devised a mechanism to do that Okay, and I shared that with you. You'll not find that in any book Okay, so what I did first I reflected this graph about the x axis Which is quite easy to do because reflection about x axis is the easiest you can do in any graph. Yeah Is that fine? So first step is first step is what I did. I reflected it about The x axis Right and then I drew this on a piece of paper and I just turned the paper 90 degree anti clockwise Okay, 90 degree anti clockwise So if you turn the paper 90 degree anti clockwise you end up getting the desired graph So basically the graph of it would look like this Right, you can try plotting it on a piece of paper and just turn the piece of paper 90 degree anti clockwise Okay, so this this point will now become This point will now become the vertex Okay, okay Okay, so is that is the point going to be the same thing as minus one comma two Yeah minus one comma two Basically, these two will stop their positions. Yeah So, I mean if you need to reflect it about that point, uh, is is it like is it possible to simply like swap Swap the positions of x and y for all the points and do it That is actually that's that's actually what should be done But we it it takes a lot of time swapping all the points Okay, this is the mechanism which I device. Let's check whether we get the same answer when you plot it on gvg bra Okay So we had the graph as x is equal to y square minus four y You see that oh, yeah So this point is the vertex over here, which is at one comma a minus one comma two Okay, now, how does this work that time when I was doing it? I could not figure it out, but somehow this trick was working for me in all the cases Okay, but later on when I was in my graduation, then I realized it actually comes all from the Transformation of graphs. So when I reflected it about the x axis What I'm doing is let's say this is the graph which I want to reflect about y equal to x line So first step was when I did it when I did its reflection about x axis I did minus y is equal to f of x correct. Yeah, this is what I obtained from step number one Correct. Okay. Now when I rotated it 90 degree anti clockwise I realized that I'm making the x axis as the y axis and I'm making y axis as the negative x axis correct Oh, so I'm making x axis as the y axis and I'm making y as the negative x axis That's the meaning of rotating 90 degrees anti clockwise So in this graph if you replace your y with negative x And x with y We automatically see that our x and y get swapped Okay, and that's why my methodology actually worked in all the cases Oh nice Okay, so this is something which you'll not find in books and Many cases in especially like chapters inverse trigonometry and all you would be required to do the reflection about y equal to x line And in those chapters this method will be very very helpful Okay Like For that I would suggest you do it on a very small piece of paper. Just draw a graph like this Okay, and just turn the paper 90 degree like this Uh, I mean I use my eraser actually On the eraser, I drew a rough diagram and then I just tilted it and I retested back on my ansescript Oh Okay, can you suggest me an alternate way to do this? An ultimate way to do this that means first I reflected about x axis and rotated it 90 degree anti clockwise You can also do it by another way also. Can you suggest me that if at all you can think of Of course one way one way is imagining the reflection about y equal to x but assuming that that is difficult for me to do Yeah Uh, any other similar way to do it. Okay, fine. I find one is that you could simply swap them But uh, the other way, what would it be? Okay, um, if I first reflected it across, um, The y axis where I get the same thing No, if I started about the y axis and it would bring me to y equals negative x Then Go ahead Absolutely brilliant awesome Probably you're the first person who could answer this So first reflected about y axis. Yeah About y axis When you're doing that you're actually doing you're actually changing the sign of x. So you're doing this actually Okay, okay. And then when you said reflected clockwise 90 degree Clockwise 90 degree that means you're making y as x and x as minus y, right? Yeah, y is x and x is minus y So basically you end up swapping the position of x and y Okay, position got swapped We'll take few more questions on this Uh plot x is equal to e to the power of three minus Why? Okay, I have a question. Um, when you when you do reflected like that, right? You don't get a function anymore, right? It just it does not become a function because it doesn't pass the yes In the in the previous case. It is no longer a function. That's correct. Good observation. Yeah Yeah, that's what's relation That would be a relation but not a function because it doesn't uh satisfy the vertical line test Got it So after plot x equals e to the power three minus y So my first the first thing I'm going to do is going to be to swap them How three minus x Hmm This is the same thing as y equals e to the power minus x plus three Yeah, so if you have e to the y equals e to the power x um, that's going to be like the first the bone structure Then I would plot it. Okay um, so when x is zero Well when x is zero y is e What is the value of e? um But wait no wait when x is zero y is not e. What am I saying? I say zero y is one Yeah, so if you cut the y axis the original graph the will cut the y axis at y equals one Okay, you'll go like that and um, why don't we cut the x axis? But we'll never oh it okay. Yeah Whenever you have this exponential graph like e to the power x it's never going to it's never going to touch the x axis, right? Uh Whenever you have like e to the power x e to the power x will never touch the x axis. That's right. Yeah So now I have e to y equals e to the power minus x plus three. So basically I'm going to move move it Three three units to the left So if I move it three units to the left my point on and the y axis which was zero comma one will Will be if I move it three into the left, it will be minus three comma one Okay, and then I have to reflect this across the Okay, minus three comma one Then I have to reflect this across okay minus x. So this will be reflected across the y axis If it is reflected across the y axis then it would be um three comma one Okay, now again, I'm done with that But now I have to plot I have to reflect this across x equals y axis So I have that drawn there Now if I have to reflect this across x equals y axis Then my that point is basically going to be at one comma three Right. Are you done with the graph? Yeah Okay, the first thing that you would like to do is plot this graph, correct? Yeah, so for this we have to start with y is equal to e to the power x So the graph would look like this Yeah, since e is greater than one it will follow e is actually 2.71828 like that It's an irrational number just like pipe. Yeah So it's greater than one so it has to be of this shape Now what do you want to do next do you want to change your x with x plus three or do you want to change your x with minus x The graph like three into the left. So I did x plus three. Okay. Never mind. We do that So if you do that it gets shifted three units to the To the left if it gets three units shifted three units to the left The graph will appear like this Yeah I think it'll cut at a higher value, right because this is shifting to the left, right? So a higher point will come and cut it Yeah, at zero it will be at higher position. Yeah Because when you put x as zero y will be e cube e cube will be almost like a tish kind of figure a tish This will be zero comma e cube Now next step is changing x with minus x Yeah, he will reflect it across y axis. Many reflected about the y axis So that will become something like this Okay And then we have to swap the position of x and y So we have to swap the position of x and y that means we have to imagine the reflection of this graph About y equal to x line and again as I already discussed with you you can follow the approach which I told you So first reflect it about x axis which is like this Okay, and then rotate it 90 degree 90 degree anti clockwise direction. So how will it appear? Um, it will be just It will appear like this Correct Okay Yeah, I got it. This point would be e cube comma zero Okay, fine. Yeah, because you should shift it uh swap the point stuff. Yes. Yes Let's see on jujibra itself so we have x is equal to e to the power of Minus one You see that Yeah, yeah. Oh, okay Correct e cube is almost no e cube is much e cube is almost near to 22. I think because 2.718 To the power of three will be approximately lesser than 27 Yeah, so that's why I got it. That's what we have shown over here. Yeah Let's try this one. Uh x is log to the base of two uh, let's say five minus three y Three y. Yeah, okay and second So the relation is y equals e to the power x equals and so y equals e to the power x and basically that I spoke of y equals log x to the base a does I have any relation here? y equals e to the power x Then y equals log x Okay, let me first I will swap it So y equals log Yeah, now I've got this fine, so Um, if I have I'm going to plot log x to the base two the base two Then what would happen? Okay, how would my graph be? My graph would sort of be like cut the x-axis at x equals has to be Minus okay first negative of that would be reflection across y-axis direction of y-axis should be less And this reflection across y-axis then it could move to minus one comma Zero okay Now, uh look towards the y-axis Then usual minus one by three comma zero Okay, then after that this has this graph is basically moved like five units to the left So if it is moved five units to the left, then it would be Like minus five minus one by three Okay, which is minus five point three to move like here You shift it After that I have to just reflect it across the y equals x-axis Okay My defense y equals x-axis now it will be um simply swap And it'll be like zero comma one one by three. Okay done done. Okay. So let's discuss this So first of all you would start with the graph of log x to the base two which is going to be like this Yeah, then what I'm going to do next I'm going to change my x with x plus five Or you can say five plus x Okay, that means I'm shifting this to the left By five minutes. So my graph would appear like this. I'm going a miniature version So this point will now cut at minus four comma zero Okay, and this will become the asymptote. That's x equal to minus five will become the asymptote. Okay Next what I'm going to do is I'm going to change x with minus x So that's going to make it five minus x means I'm reflecting the graph about y-axis Yeah, so if I do that My graph would appear like this My graph will appear like this Okay, so now this point will become four comma zero for me Yeah, and this will be the asymptote Okay, now I'm going to make x as three x Which means shrinking the graph more towards the y-axis. So basically The points will not change but this graph will shrink like this Or the point won't change. Okay, this point won't change because you're shrinking about y-axis But this point will become four by three comma. Sorry Uh, yeah four by three comma zero Okay, because it is getting one third shrink. This means one third shrink is happening. Yeah, okay Getting the point Yeah, and you can also check that when you put x as four by three your y will become zero Okay, got it right because log of Five minus four by three into three will be log of one and log of one to any base is going to be zero Okay, which clearly indicates it is passing through four by three comma zero Okay, now I have to swap the position of x and y Okay, so I have to swap the position of x and y means the fourth step is I have to take the reflection of this graph about y equal to x line So that would become First what did you do first? What should we do? Uh, I reflected about yeah reflected about the x-axis.  All right, so this is the final answer now plot this one log Uh, sorry Let's say x is equal to two minus three minus y to the power of five I'll simply swap it and x to the power of five origin, right? x to the power five would be uh Like this it'll be more flat like this of the x cube x cube is more like this So first take y equals x to the power of five That is at zero comma zero Okay, and then to that you're adding three. So we're going to shift to the uh to the left by three units It's going to be Isn't that I should just Why do that? I need to go on Yes Done? Yeah done Okay So let's discuss this. Uh, it'll be first y is equal to Two minus three minus x to the power of five Yeah Okay So we can write this as um First of all, we'll start with x to the power five graphs Okay x to the power five we have already discussed. It's going to be like this First thing we'll do is uh, we'll first rephrase it as y minus two is equal to uh negative three minus x to the power five which is as good as saying y minus two is x minus three to the power five Yeah, so what I did was I shifted as a negative sign to the y part so I could reflect it So in this case, you will have to shift it three units to the right and two units up Correct So it will be like this. Uh, just be careful Just be careful whether it's passing through origin or not origin Will not pass when x is zero y will be negative number so it'll pass like this Yeah, okay So this point will be two comma three Correct. Yeah Now, uh, next what you have to do is we have to start the position of x and y Yeah So to start the position of x and y means we have to take the reflection of this about y equal to x line So for that, we first have to draw it Draw its mirror image like this So the mirror image would be Like this about And then rotate it Rotate it 90 degree and take clockwise Yeah, so it will appear like this So this point will come to minus two comma minus three. Sorry, uh Just yeah, I think this point This point will You are actually trying to turn it This graph has a bend actually that's why it is slightly difficult to imagine the reflection Yeah Oh, you're reflecting this about x axis, right? So this is the graph This graph is not The reflection I'll redraw it The reflection would be like this This point will become Two comma minus three Okay, but isn't it I'm not rotating it 90 degrees anti clockwise Yeah So if I rotate it 90 degree anti clockwise it will now appear like this This will be the Three comma two It's just two comma three. I don't know how where I went wrong. I can't tell you what I did Swap those two Got two to that side so it became y minus two equals minus three Minus of three minus next to power five Okay Then our negative sign that was on that side I mean minus two with the negative sign again minus y plus two equals three minus x to the power five so uh here Okay, so then what I did was I contracted the graphs zero comma zero Then um, I shifted it three units to the left to the left by it so became Minus three comma zero. I should have changed the left actually looks like this actually Yeah, yeah, because this part this part when you turn it 90 degree it will be like this So this this face This face will be this face And this face will be like this Okay Are you getting such kind of a Uh point is uh, two comma three. I mean the reflected point is two comma three I don't know. I mean so, okay, so I did that and I moved it like three units to the to the left Then after that, uh, what did I do? I Idea is clear how it works Yeah, yeah So we'll try one more, uh plot Plot x is equal to y minus one whole cube Plus three You can do the steps in any order, right? You still get the same answer Yeah So first of all we'll do y is equal to x minus one whole cube plus three Yeah So it's as we were saying y minus three is x minus one whole cube Yeah So the graph of x cube will get shifted One unit to the right and three units up Yeah So one is to the right and three units up means the graph. Yeah, the graph will come like this And now I have to reflect this graph about y equal to x line Okay, so it's a three comma one. Yeah, so its reflection would be like this So this point will become three comma one Is that fine? Yeah, this is how you got it? Yeah Now the next thing that we're going to discuss is functions which are Uh Of the nature one by x One by x square one by x cube like that Okay So let's say if I ask you plot the graph of y is equal to one by x Yeah, okay So initially if somebody doesn't know the graph he will choose some points of x and then plot it Yeah Now if you choose a value of x which is very very small positive number, let's say x is 0.0000001 What would be the value of y be? My value of y will be it'll also uh, it'll be it'll be larger It'll be very large value. It will be almost plus infinity Correct So the graph will come from plus infinity like this Oh, that's right And as you keep increasing the value of x it'll fall down Okay And as x goes to plus infinity As x goes to plus infinity y value will almost die to zero but never become zero Yeah In a similar way if your x is slightly negative value Let's say negative 0.0000001 Then y will almost go to negative infinity, correct? Yeah In the neighborhood here it'll go to negative infinity. So it'll come back like this Okay Okay, so it really graph like this. So both are these both the arms of this graph are these two arms Yeah, okay This graph is called a rectangular hyperbola when we do the chapter in conic sections We'll study more about rectangular hyperbola You'll find these graphs also in physics in isothermal curves isothermal curves The graph of p versus t p v curve p v graph All the transformation that we have learned so far would be applicable to this I remember it has two asymptotes one is a horizontal asymptote Mm-hmm one's vertical This equation is y is equal to zero and one is vertical x is equal to zero, okay Now plot y is equal to one by x minus three graph So you're changing x with x minus three, right? Yeah for three units to the right Right, so how would it look like? Now this vertical asymptote will shift three units to the right So think as if now your vertical asymptote is like this It has come x equal to three Yeah Okay, now this has become x equal to three line. So your graph will be like this Yeah Okay, let's see that on on geojibra So this was the graph of y is equal to one by x Yeah expand it So this is the graph of one by x minus three as you can see it has shifted three units to the right You see that? Yeah, I'll just show you the asymptote x equal to three Okay, okay if I ask you to plot the graph for Oh, so it's actually tending towards x equal three, but it's never x equals three because if it does reach x equal three Then it becomes not defined when it is an infinity infinity So asymptote is what asymptote is a line which touches the curve at infinity Okay, okay. Yeah Now plot this graph Okay, that's just you have to just move this like one absolutely. So basically we have to now make the graph like this This line is x equal to three This line is y is equal to minus one Okay Understood how it works next is Functions of this type one by x squared again the story is same But in this case a slight difference would be when you're very close to zero. Let's say on the positive side of zero y will become infinitely big And so would be the case when you're left to zero So again y will become infinitely big that means the graph will come from infinity and go like this on both the sides of the graph Okay It's very obvious that this curve will be symmetrical about y axis because the sign the power on x is even actually Oh, okay. So even if you do y y is equal to one by negative x whole square the graph doesn't change So we are thinking about y axis doesn't change the figure So such such a thing will only happen when the function is already symmetrical about y axis Okay Okay So here it has again two asymptotes one is your x is equal to zero and other is y equal to zero So this is a horizontal. This is a vertical Okay, so plot y is equal to one by x plus two whole square plus three Y equals one by x plus two whole square plus three So here We'll go move up by three units So we'll be like at y equals three there'll be an asymptote Also, there'll be an asymptote at Yeah, x equals minus two What would be the final Shape of the graph B It's always advisable that you make the asymptotes first Okay Horizontal asymptote will be y equal to three vertical asymptote will be x equal to minus two correct So you just have to shape the graph near about these two lines. So basically it'll be this So both the arms will move simultaneously. Many people do a mistake. They just move one of the arms or The other arm both have to move together Okay Now there's not much of a difference in the graph of one by x and one by x cube Is just that the steepness of these two arms will change So can you tell me how would the steepness change? Let's say this is the graph of one by x Then how would be the graph of one by x cube B? Let's say this is point one This is point minus one. Okay Okay, so when x equals Okay, so when x equals minus one When x equals minus one y will also be minus one Okay, so it'll also be minus one so Be here and then when x equals one y will also be one So it'll come there Um Okay, then when x equal It'll be one by eight. It'll be one by eight. Uh, so won't that be that'll be like much smaller Be like 0.125 Okay, just a question when x belongs to this interval minus one to one, which is greater one by x or one by x cube Man Okay, wait, oh one by x will be one. Um, and wait, so when x equals minus one One by x will be minus one Do we say that one and minus one but let's say somewhere in between let's say point five if you take Okay, which will be greater one by x or one by x cube Okay, so one by zero point five cube is eight or one by zero point five. Okay, it'll be one by x cube Okay, so the graph will be one by x cube graph will be higher in this zone so Let in this zone The graph of one by x two will be higher But the moment you cross one, let's say you become a two the graph will fall down like this Okay When greater than one or lesser than minus one, uh, one by x greater. Yes So basically here the same story will be repeated here also Lesser lesser lesser and here it'll be more Okay, I'll show you about the graphs on the same y is equal to one by x And y is equal to one by x cube You see that change Red graph is the graph of one by x cube green graph is the graph of one by x Okay, so after this point one You can see that red graph is below the green graph Yeah, and between minus one to one or between zero to one you can say the red graph is higher than the green graph Okay, but yeah, okay. So what will happen if I do one by x to the power five? Can you predict what will happen? Yeah, okay, so uh It was by Between minus one and one one by x cube greater than one by x So basically one by x greater than one by x cube within the same uh domain like between x Lesson one and greater than minus one And minus one one by x cube will be greater Did you see this happening? Yeah Oh, yeah, okay So basically what is happening is the high the power of x if you keep on increasing by odd numbers For example one to the power x to the power five On this side it'll become higher but on the other side will become lower Okay, so It will become more like this Get in the point Got it. Yeah. Now I'll come to that What if I have Yeah, what if I have y is equal to one by x to the power four? Okay Fine I'll check first between minus one and one If I check between and then y was one one by x squared Okay, so if I check, uh minus one and one is for example, if I take 0.5 as usual and 0.5 To the power four would be okay. That would give you 16 Um, but then one by zero point five Square that gives me four Okay, so between again between minus one and one uh Between minus one and one uh one by x to the power four is greater than one by But then when x is greater than one or when x is lesser than minus one Then uh one by x square would be greater than one by x of alpha Yes, absolutely So this is how the graph of one by x this is yellow one is the graph of y is equal to one by x to the power four Yeah, white one is the graph of y is equal to one by x squared So again all the shifting concepts will be equally valid here as well So let's draw the graph of y is equal to Three minus y is equal to one by two minus x to the power of five Okay One by two minus x to the power five as in this way you have shifted two units to the left See you can also reframe it rephrase it as y minus three is equal to one by x minus two to the power five Listen to ship there and then uh y shifts three units up. Yes. So let's discuss this So x to the power five will be Yeah, first of all the vertical and the horizontal asymptotes so x minus two means the vertical asymptote will shift add x equal to two And y minus three means vertical asymptote will shift to y equal to three. Okay Okay, and your graph will be positioned like this Understood Yeah, great. Now I'm going to talk about special functions So we have already seen polynomial functions of various degrees What I take cubic We also saw reciprocal functions one by x one by x square etc We also saw log function exponential function or now time has come that I introduce some special functions to you which will be Ask in your class 11 And the first one being the mod function which you call as the modulus function Okay, what is modulus function the simplest of all modulus function is mod of x Right, what are the meaning of mod mod means absolute value this function gives the absolute value of the input Okay, if I say mod of minus three point two It'll give you three point two. Yeah. Okay, so this function returns This function returns The absolute value of x that means without the sign into consideration. That's the magnitude part Okay, so if I say 3.2 modulus, it'll just return 3.2 Okay, so if I have to define this function in a simpler way without using these parallel lines I would define this function like this This function would return x to you if x is already a positive quantity Correct. Yeah, for example, when I put 3.2 it returned 3.2 to me because 3.2 was already a positive quantity Yeah, but when it is a negative quantity, this will return negative x to you Correct So when x is already a negative quantity, it will return negative of the x value that is negative of negative 3.2 Which is 3.2 Okay, so if I have to plot such a function on the graph I will plot it like this in the interval zero to infinity It will follow the line y equal to x Okay, which is going to be a line like this Yeah, but I cannot go down because it is only valid for x greater than equal to zero Okay, yeah Now for negative values, that means from zero to negative infinity. I will plot y equal to negative x Okay, that would be like this Okay, so it'll look like a v and remember these angles will be 45 45 45 45 Such functions are also given A name of piecewise functions What's the piecewise function piecewise function is a function which is defined in pieces Right, so most most of the function that you know human beings themselves are piecewise functions We behave differently in different situations Correct. So it's like this is the situation and this is how it will behave Okay, this is the situation and this is how it will behave So when your x is greater than equal to zero, it will behave like a line y equal to x Yeah And when x is less than zero, it will behave like the line y equal to negative x So when it's greater than zero and x is greater than or equal to zero it'll behave as y equals x and otherwise We behave as y equals minus x Absolutely Okay And again, all the concepts that we have talked about Transformations will be valid on this as well. So if I ask you Plot the graph of y is equal to mod of x minus two Y is mod of x minus two Okay, if it's mod of x minus two then it will shift two units to the right So the uh, you know that the point where it touches the x axis will be at x equals two Okay, so what will happen to this v Huh, what will happen to this v graph? It will simply shift to the right by two units. Yes, it will shift to the right by two units. So it will be like this So this point will become x equal to two Yeah, so this will be y equals the right hand part of the graph will be y equals x minus two and the left hand part will be y equals minus x plus two Absolutely, this will be y is equal to Minus of x minus two you can say or you can also say two minus x Okay So in light of this you can actually redefine Redefine means writing without mod the way I've written over here. This is called redefining the function Yeah, redefining means Defining the function without mod into place Okay So mod of x minus two is defined as x minus two when x is greater than equal to two And is defined as two minus x when x is less than two Okay, in general in general mod of x minus a is redefined as x minus a when x is greater than a Greater than equal to a and a minus x When x is less than a Okay, so I'll give you a quick question over here There's a function f of x which is Made up of these three functions addition of mod x plus one mod x minus two mod x minus three Okay question is redefine f of x So if I need to do that then one x plus one can be written as So one can be written as minus x minus one and x minus one Now here you have to be very be very careful about the intervals where you are redefining the function Okay In this case see The way to proceed this is first of all see What is the critical point for this mod function minus one night? Yeah, because left to minus one right to minus one it will behave differently correct Okay, yeah, what are the critical point for this mod function? Okay, two critical point for this mod function three Okay, now what I will do is I will try to redefine this function in these four zones Okay So assuming your x is less than minus one Okay, this guy will behave as negative x plus one is whatever x is less than minus one It will be negative of x plus one. Okay The best way to figure that out was is uh choose any value less than minus one. Let's say minus two Okay, what is minus two plus one mod? Oh minus. Oh, okay. I'll just be If this is a negative quantity just put the answer of this as Negative of x plus one. That's it. Okay. That's the best way to solve this problem in a faster way Okay, okay.  It will be redefined as minus three x plus four Okay, fine. Okay So if you need to redefine this, okay, fine. Got it. Fine. I understand So let me just write this in brackets so that we don't have to This is okay. Now what about I've already reached the end of the sheet. Okay. What about minus one to two? Minus one to two. Okay. We'll start from here Take any value between minus one and two. Let's say zero itself Okay, zero plus one is a positive quantity. So this will remain x plus one itself Yeah Zero minus two is a negative quantity. So it will become minus of x minus two Okay, let me correct. So it's in his sign. Um, yes See the approach is the value that you put over here zero If it gives you a negative value, this is a negative value Then you change the function of the function become negative. Yeah, there's a negative over here Okay, fine. That's a trick that I'm telling you to do it in a faster way Okay, okay, fine Now when you put a zero over here, what do you get zero minus three negative quantity? So it'll become negative of x minus two. Okay. Yeah minus x plus three. Yeah If you simplify it, it'll become uh, I'm just writing it in brackets. It will be uh negative x Or two plus three plus four. That's just six Mm-hmm Then there's no space. I'm moving to the right. Mm-hmm. What about two to three zone? Okay, let me write the function again over here. It has got hidden So mod of x minus. What was the function mod of x plus one? Yeah mod of x minus two mod of x minus Yeah Two to three any value take two point five two point five. This is positive. So it'll become x plus one only correct This is also positive. So it'll become x minus two But this will be negative. So right negative x minus three Okay, got it. It'll simplify to be Uh, um, let's do let's do okay. What about the interval? Three to infinity three to infinity or greater than three Again number like four all of them will be positive x plus one x minus two x minus three correct So that will become uh three x minus four four Okay, so this is the redefinition of the function. So now if I ask plot the graph of this function, how will you plot? Okay, so First I said, uh, okay. I can look at these points minus one Two and three Yeah, so when you're less than minus one, how is the function behaving the function is behaving like a line y is equal to minus three x plus fours look at this Okay, correct. Yeah, it will be a line having a negative slope and cutting the graph at four. So like this I'm sorry I don't have to draw the full graph That is three minus one and uh, what is it zero? No, less than minus one. I have to stop here Correct at minus one exactly the function will have a value of Uh, minus three into minus one, which is three three plus four, which is going to be seven Okay, so this guy is going to be at minus one comma seven Okay In the interval minus one to two Now what about minus one you can include minus one here or you should include minus one here. Where should you include? Um, I think uh the first one right you can include at both the places doesn't matter because when you put a minus one here It still gets a seven. Okay. This will also give you a seven. This will also give you a seven Okay So minus x plus six will now be a graph with a slightly lesser negative slope Okay, and it will Go like this cutting the x-axis at six. This point will be zero comma six Okay And it will stop at two And what's the value at which it should stop at two? Minus two plus six, which is four four this point will be two comma four Okay, now two to three it will be x plus two. So two to three it will be x plus two will be like this Correct. Okay, and at three it will stop at three comma five Okay And greater than three it will be three x minus four. So it will be a line like this Okay, so we have got four different lines No one two three four. Yeah four different lines This line is y is equal to minus three x plus four minus three Y is equal to minus x plus six minus x plus six This line is y is equal to x plus two Y is x plus two and that is y equals x plus three x minus Four Okay So let me show you on jiu-jitsu as well. Okay In jiu-jitsu the command for this is abs abs is for mod. Absolutely. Okay. Absolutely So abs x plus one plus abs x minus two First abs x minus three Oh Because they're one two three four like four different ratings one two three four And it's okay. So this part of the line this part of the lines That's y is equal to minus three x plus four This part of the line Is y is equal to minus x plus six Okay This part of the line Is y is equal to x plus two. Yeah And again this part of the line Is y is equal to three x minus Question for you now Redefine mod two x minus three plus mod x plus four plus mod x plus four. Okay. I really define this Then first i'm going to uh, so the points at which it touches is is um If we were just uh, like at two x minus c and x plus four for two x minus Let's see the point at which it will cut the um the x axis If you first plot the critical point Critical points is the point where this is going to be zero. When is this going to be zero at x equal to three by two? All right three by two correct. Oh, yeah, um Four minus four. Okay. I should it's correct. So the critical point for this will be a minus four So you have three regions now one That is below minus four Second region is between minus four and three by two and third region is this Okay So if it's going to be there first I have to I'm going to first obtain the consideration in vision one Then um for values of x Lessor than minus four I'll have to check So when values of x are lesser than minus four, I'm going to substitute that in um in there So if I substitute that in there then So you have to get into a minus four Then minus four to three by two Yeah, and greater than three by two and we have also learned that you can include both the points and both the positions Not a Minus four. I have to first put that inside the function, right? So two, um, so it's less than minus four Take minus Five in just this part the inside part Yeah, what do you get? minus 13 Yeah, the negative quantities will be um three minus two x. Yeah or minus of two x minus Yeah Then if I put in um If I put in negative five in that the minus five plus four which is again negative Just minus one Uh, so if it's minus one then what happens, um, then it has to be like this. It's going to be minus x minus four Okay, so here it's going to be um Minus one minus three x you can you can calculate it here itself. So it'll be minus three x minus one Okay, what about this four two minus four to three by two Okay, I'll take zero then if I take zero then what'll happen? It'll be minus three Okay, the first will be negative. So it'll be three minus two x Um plus and then um the next one will be positive. So it'll be x plus four So it'll be seven minus x That'll be minus of two x minus three Plus x plus four correct Yeah Then well x is greater than or equal to three by two Uh, I can take any value. I'll take in two if I take in two then what'll happen Two into two four minus three is positive one. So it'll be two x minus three plus um Two plus four which is six again positives x plus four will be three x minus one Yeah So first plot the critical values that's minus four T by two will be here So how will it behave before minus four? Before minus four It's very important to know this point actually Yeah, so if you put in minus four there there'll be uh, 12 minus 11 So it's minus four comma 11 Okay Yeah, what about this interval minus four to three by two? Minus four three by two it'll have a negative gradient again And at the point three by two over to be that it'll be seven minus three by two seven minus one point five Which is five point five At this point will be three by two comma 11 by two Yeah Then again The whole content between three by two and infinity It'll have a positive gradient and um Yeah Let's check this What was my function? A mod of two x minus three Plus mod of x plus four So this line is what we found that as a three x minus. What was that three x minus This is seven minus x. Yeah, as you can see y intercept is also seven Yeah, this is finally three x plus one Now, uh, something very interesting that you would note about it When I asked you to plot the graph of mod x minus two You actually uh plotted it like this So we shifted it to units to the right Yeah So this point becomes two comma zero. Okay There's a way to do this in a faster way In fact, I can give you mod of any function that I would like then the approach is what I'm going to tell you next So for that you just first draw y equal to f of x graph For example in this I would draw y is equal to x minus two graph Okay, that would be a line like this Correct. Yeah, this is y equal to x minus two graph. Okay. Now which the part of the graph which is already above the x axis Don't disturb it. Let the part be as it is But the part which is below the x axis you reflected about the x axis you reflected about the x axis That means you make it like this And it is this part And it is this part Okay Okay So first draw this and second step Just then you would yeah, it would be the negative It would be negative of that because yeah, it would be minus y, right? Yeah, so for all these values your y was negative, right? Yeah So model was to actually send it to the positive side. So something like minus one got sent to plus one over it Okay. Okay. So the rule is second step. Don't disturb the part which is already above x axis and Reflect the part Which is below the x axis to above the x axis Okay So for example, if I ask you plot y is equal to mod of x square minus three x plus two Ui equals mod of x square minus 3x plus two How will you do this? Okay, how would I do this? Fine. So first, um, I would have to um, okay. So first, um, let me okay. I'll first draw the original, uh, graph And then I reflect it right reflect the part which is below the x axis above the x axis and don't disturb The one that is already above it. Yes. So first I'd have to, uh, plot this So first plot without the mod? So without the mod, yes Without the mod, if I plot it, then what'll happen? It'll be minus b by two a It will be a parabola like this Two, one by two So let me just like write out like the important points in So that'll be So the vertex should be at 1.5 comma 0.2 minus 0.25 um minus 0.25 then um Okay, then when does it cut it out? It cuts at one Now what I'd have to do is I'd simply have to reflect this above So this part is already above right these two parts These two parts are only above the x axis. So don't disturb them Yeah So if I have to reflect it there then my vertex is going to have to become Uh, minus one by four right so it's going to have to become one 1.5 No, no, no. Uh, yeah x will be the same, but y will become 1.5 one by four. Yeah, so it'll be it'll be like this So now this point will become 1.5 comma one by four Okay, yeah understood So only the part which is below you just have to reflect it above what is already above don't disturb them Yeah, let's check it out So y is equal to abs x square Minus 3x plus 2 You see the small part Yeah, okay Fine, so I think you will be able to solve a few more questions on the worksheet which I sent Yeah, yeah, I will be able to do that do as much as many as you can Okay, so in the next class, which is beyond sunday, I guess Yeah sunday, I'll talk about more other transformations Okay, and some more special functions like gia function fractional part function all those things Okay, okay, then thank you. Bye. Bye. Thank you. Bye. Bye. See you. See you