 Hi, I'm Zor. Welcome to Unizor education. I would like to talk about graphs of exponential functions. We did learn about what exactly exponent is, how to raise some number into some power. We did talk about properties of exponential functions. And now it's time to use these properties and these definitions to basically chart. How the exponential functions look like on the graph. Now, before that I would like actually to encourage you to go again through the lecture about the graphs in general. And I will remind certain properties of the graphs relative to these particular problems which I'm going to present right now, which are related to exponential functions. So, very, very briefly, for details I do refer you to the lecture about the graphs. So, if you have a function, what happens with its graph if we will do certain transformations with argument or the function itself? Well, the easiest for instance is what happens with this graph if I add h to the function? Well, basically the whole graph moves up by a. Well, if a is negative it's obviously down. And the form and the shape of the graph actually doesn't really change at all. Now, what happens if instead of adding to the function I will add to the argument? Okay, in this case the graph goes to the left by a or if you wish by minus a to the right. Again, I refer you to the lecture about graphs, it's very easy to prove this, but in any case in this case if a is positive the whole graph moves to the left, if a is negative to the right by absolute value of a. Now, next transformation, what happens with the graph if I multiply an argument by a? Again, it's very easy to prove that if a is positive number, let's say the a is greater than one, then the graph is basically squeezed towards the vertical y-axis. If a is positive but less than one then it's basically is stretching and the ratio is a. If a is greater than one it's squeezed by a, if a is less than one it stretches by one over a. If a is negative it's basically two transformations. First you do what is required to do for the negative thing. For instance, if a is equal to minus one, now how this graph looks relative to this one? Well, it's basically a symmetry. You transfer graph from this way to this way and that's it, because whatever was happening with a positive x now is happening with a negative x. So, we covered this one and finally what if you multiply the function? Well, then again this is a vertical transformation. Whatever you do with a function is always vertical, whatever you do with argument is always horizontal. So, in this case it's stretching or squeezing by a depending on whether a is positive or negative and whether it's greater than one or less than one. If it's greater than one it's just stretching. If it's positive but less than one it's squeezing by one over a. If it's negative you first reverse the graph this way and then do the stretching or squeezing. And finally what if the situation is complex one, which is this. So, you do some linear transformation on argument and linear transformation on the function. Well, this actually you have to do step by step. So, let's transform the graph from this to this in steps. So, first we do this, you know how to do this. Then you do this, we need plus C so it's this. So, first this is a horizontal stretching or squeezing. Then this is a horizontal shift by C over B. Then you multiply it by A and then you add D to the result. Which means you're stretching vertically and then you're shifting vertically. So, these are sequences of steps which you have to do to transform from this to this. So, we'll do it step by step. And now, let me just go straight to the problems related to graphs of different exponential functions. Now, this was a short refreshing course on how the graphs are transformed. And let me repeat again, go to the graphs lecture to get into the details why it's happening the way how I have described. Now, let's go to these six different graphs and I will use the properties which were described in the previous lecture. So, the function y is equal to 2 to the power of x plus 1, x, y, 0. Now, all exponential functions and in this particular case it's 2 to the power of x are going through the point 0, 1. Because if x is equal to 0, 2 to the power of x is equal to 1. 2 to the power of x is rising to infinity as x goes to plus infinity and goes to asymptotically to 0 when x goes to negative infinity. So, this is function 2 to the power of x. Now, how can I transform it to plus 1? Well, very easily it used to be the x axis as an asymptotic boundary for this particular graph. So now, since we are shifting everything up by 1, this would be the next asymptote. And the whole graph will be instead of going through 0, 1, it will go to 0, 2. So, the graph will be like this. The whole graph is shifted up. So, this is y is equal to 2 to the power of x plus 1. So, there is only one transformation. We started with 2 to the power of x which is obvious how it looks. And then, just shifted it by 1 upwards. That's it. Next, y is equal to 1 half to the x plus 1 minus 2. Okay, so let's do it 1 by 1. First, we do 1 half to the power of x that happens in this particular case. Now, this is the case when the base is less than 1. Which means that the function instead of going this way, it goes this way. So, this would be the graph and this is the point 1. Okay, so this is y is equal to 1 half to the power of x. Next, what we have to do is we have to do instead of to the power of x, we have to raise it to the power of x plus 1. So, we add 1 to the argument of this function. Now, if we add 1, the graph shifts left by this way. So, instead of intersecting the y axis at point 1, so instead of going through point 0, 1, it will go through point minus 1, 1. However, the asymptotic behavior as x is increasing remains the same. So, this will be basically the same thing. By the way, just as a verification, in case x is equal to 0, the function is equal to 1 half to the power of 1, which is 1 half. And this is the function y is equal to 1 half to the power of x plus 1. So, we have shifted the graph to the left by 1 unit. So, whatever was 0, 1 becomes minus 1, 1. Now, the last thing is we have to subtract 2 from the function, which means the function moves down by 2 units. Well, this is 1 unit, so this will be minus 1 and this will be minus 2. So, this point will go to this point. The x-axis as an asymptotic line will actually be x-axis shifted down. So, this will be our asymptotic line. Now, the function used to intersect the y-axis at point 1 half. So, we shifted down by 2, so it will be here, which is minus 1 and a half, right? And probably, it would be interesting to know when this function actually intersects the x-axis because the behavior will be like this, right? So, this is one interesting point. We have already calculated it. When x is equal to 0, it's 1 half minus 2, which is minus 1 and a half. Now, when does the function equals to 0? That's interesting. When exactly it crosses the x-axis when y is equal to 0? Well, if y is equal to 0, let's just consider 1 half to the x plus 1 should be minus 2 equal to 0. So, 1 half x plus 1 equals to 2. So, when minus 1 to some degree is equal to 2 when the degree is equal to minus 1, right? 1 half to the minus 1 is equal to 2. Now, when minus 1 is x plus 1, when x is equal to minus 2, right? So, this is actually at point minus 2. This basically ends the construction of this graph. So, it goes through the point minus 1 minus 1. That's what used to be this. Crosses the y-axis at minus 1 and a half and crosses x-axis at minus 2. That's it. Next. Okay, y is equal to 3.5 to the power of 2x plus 1 plus 2. Well, this is as complex as it can be, really. I mean, the linear manipulation with an argument and some linear manipulation with the function itself. All right, so let's do it step by step. First, we start with 3.5 to the power of x. This is something which we start with, right? So, what's the graph of this function? Well, the base is greater than 1, which means it's increasing. So, if this is point, it goes this way. So, this is 3.5 to the power of x. That's easy. Next. Next, we do 3.5 to the power of 2x. Now, what does it mean that I'm multiplying argument by 2? It means that the graph actually should be squeezed towards y-axis because if I squeeze it by the factor of 2, then multiplying my argument by 2, I will return it back to the place where it was and that's how my value of the function would be exactly the same. So, the graph would be squeezed. Now, just as a reference point, if my argument is equal to 1, my original function is equal to, well, that's obviously not to the scale. Let me just do it slightly differently so it will be a little bit better for the scaling. Now, let's consider that 1 is here, 1 is here. Now, this is 2, this is 3, this is 4. So, 3.5 would be here. So, the function actually goes this way. So, as x is equal to 1, my function is equal to 3.5 to the power of 1, which is 3.5 here. But now, as I'm squeezing it towards this particular, towards the y-axis, now it will be in the 1 half, the same value, 3.5. So, it will be this way and this also will be. So, on the left, for the negative x, the new graph would be below the old one. On the right, when the x is positive, the new graph would be steeper and closer to the y-axis. Okay, so that's 3.5 to the power of 2x. Now, I have this. Well, I can do it in two different ways. I can always represent 3.5 to the power of 2x plus 1 as 3.5 to the power of 2x times 3.5 to the power of 1 because the powers, exponents are added together, right? So, I can actually multiply the function by 3.5 or alternatively, I can do it differently. I can do it this way. That's also 2x plus 1, right? Now, both approaches, 1 means I'm multiplying the function by 3.5, which means everything goes to the right of the 0. It will be steeper by 3.5 times and on the left, it will be closer to the 0 because it's less than 1 by 3.5 factor. So, that's my multiplication by 3.5. But adding to the argument means that I have to really shift the whole graph to the left by one half. And interestingly enough, we are supposed to get exactly the same graph, right? So, how... Now, I don't want to prove that this is exactly the same. I'll just basically build both cases separately. So, in one case, let me just do it on a different scale. I'm multiplying by the factor of 3.5, which means this point will go to somewhere here, this point will go here, so it will be something like this. Now, if I'm shifting by one half, it means I'm shifting to the left by one half. And that also would be something similar to this, right? So, basically the result should be exactly the same thing and we shouldn't really worry about how we will approach it because it's still the same. So, in any case, if I will shift it, let's say, by one half to the left, my new graph would be something like this. This is minus one half, it will be one, but everything else would be something like this. Well, I'm not sure that it will probably be something like this, rather. Again, I don't want to put too much artistic talent into this, but basically the graph of 3.5 to the power of 2x should be shifted by one half to the left. Now, just as a reference point, you can check what exactly would be the point where the graph is intersecting the y-axis. That happens when x is equal to zero and the value of this, if x is equal to zero, would be two times one half is one, so 3.5 to the first degree, so it's still be 3.5. So, the graph would intersect here. So, again, my artistic talent is challenged here. It would be something like this. Now, on the different note, if I have this represented as this, if x is equal to zero, this is one and the result is also 3.5. Obviously, that's how it's supposed to be. In any case, when x is equal to zero, y would be equal to 3.5 and the graph would look like this. Finally, class 2, what does that mean? It means that this asymptotic line should really be shifted to this position up by two and the whole graph, obviously, it used to cross the y-axis at 3.5. Now it would be 5.5 somewhere here, so the graph would look something like this. What's most important in all these manipulations is to realize how to do step-by-step transformation of the original graph, which is 3.5 to the power of x, into whatever the graph you have. This is just an illustration of the steps necessary and I will repeat it for a couple of other graphs as well. I'm sure you will understand the point. Next, one-quarter to the power of minus x minus one. We can do it differently, but let me do it straight. First, I will draw one-quarter to the power of x, which is this, and then I will do multiplication by minus one in the argument. Now multiplication by minus one means, as I was saying before, reflection relative to the y-axis, so actually the graph would look now like this. And minus one, which means we go down by one, so the symptom will be here. It will be something like this. So it will go to zero, zero, because this x is equal to zero. This is one, one minus one, zero, so it will go through this. So this is basically the way how the graph would look. And this is a symptom. Let's do it slightly differently. We know that negative exponent means one over the corresponding positive exponent. So one over x, one, one-fourth at the power of minus x is the same as one over, one over four to the power of x, which is four to the power of x. So I reverse it. So it's basically four x, and four x really looks like this. I mean four to the power of x. And four to the power of x really looks like this. So instead of doing this, we can always do this, which basically means exactly the same thing. All these transformations are leading to exactly the same thing, because after we draw four to the power of x, we just have to subtract one, which means exactly like this. That's it. And in the future, again, if you have something like a negative exponent, you can do both things. You can transform the function first and then draw the graph. Or you can draw the graph and then reflect it symmetrically relative to the violence. It doesn't matter. Okay. Three to the power of minus x plus four. Three to the power of minus x plus four minus four. Okay. I will introduce even more transformations in this particular case, because I can actually, before starting thinking about how the graph actually looks like, I can always transform it in the following way. It's three to the power of minus x times three to the fourth minus four is equal to... Three to the fourth is h1 times... Now, three to the power of minus x, I can always put three to the power of minus two to the power of x minus four, because if I'm raising something into one power and then to another, that's the multiplication of exponents, right? Which means h1 times three to the power of minus two is one over three to the power of two, which is one nines to the power of x minus four. Now, why did they do all this? Well, because this seems to be a little easier than this. That's all. So I have to do two transformations. My original graph would be one nines over to the power of x, which is this. This is one. Since one nines is a base less than one, the graph would look this way asymptotically approaching the x-axis as x is increasing to infinity. Now I have to multiply it by h1, which means instead of going to the point zero one, it will go through point zero h1, so it would be something like this. But it will still be approaching zero, because even if you multiply by h1, small value remains still small and increasingly closer to the x-axis. And now I have to shift it by four, which means my asymptotical line would be here, minus four, and everything else is the same. Now, for some kind of exercise, you can always find out what exactly are these points of intersection. This is when the function has an argument zero, which means this is zero, so this is one, so it's h1 minus four, so it's 77, which means it goes this way, higher than that. It's positive, from h1 down to 77. Now, the point where it crosses the x-axis is when it's equal to zero. So if I will equate it to zero, three to the power minus x plus four minus four is equal to zero, so three to the minus two x plus four is equal to four, and it's kind of difficult to calculate, but what if this is three, for instance? If it's three, then this is supposed to be one, so x should be equal to three seconds. If it's equal to, let's say, nine, it would be two, so x should be equal to one. So for four, it would be somewhere between three seconds, with one and a half and one, so wherever this point is, it should be something like one point, and four or something like this. Anyway, this is the behavior of the graph. Don't pay attention that this is just one and a half or something like this, and this is 77, this is the scaling issue, and obviously I'm not drawing to the scale in this particular case. What's important is that I have made this particular transformation or simplification, if you wish, before charging this particular graph. At the same time, I could have done it in a little bit more classical way, if you wish, which means I would do this, three minus two x minus two minus four, so I would start with three to the power of x, and this is this. Then I would multiply it by minus two, which means I would reverse it, I would symmetrically reflect it relative to the y-axis, which is this, and then stretch it vertically by two, so it would be something like this. Then I would subtract two from the x, which means the whole graph should be shifted to the right by two, so it would be something like this, and then down by four, so it would be like this. More or less the same shape, and again you can calculate all the points in the intersection with x's, but that's the same graph just differently obtained through different transformation steps. Okay. And the last problem, these are all exercises of basically the same thing, minus two x minus two minus two. Okay. Let's do it in a more, I would say, classical approach. So first we do one-half to the power of x, which is this. This is one, this is zero. Then, so this is one-half to the power of x. Next is one-half to the power of minus two x. So it's reflecting relative to the y-axis because it's minus, which is this, and then stretching it by the factor of two, which means this. Next is one-half to the power of minus two x plus one. Right? To get minus two here I have to, which means the whole graph shifts to the left by one unit. So it would be something like this. And finally minus two, which means the whole graph goes there by two units. So it would be minus two, would be something like this. And at point zero it's equal to zero minus two, so it's reverse, it's four minus two, two, so it would be something like this. So it should be something like this. And where it crosses the x-axis, that's when the y is equal to zero, let's calculate what it is. If this is equal to zero, it means two is equal to one-half to the power of minus two x plus one, or two is equal to two to the power of x plus one, this is one, so it's x is equal to zero. So it's interesting. No, it can't be zero. It cannot be zero. No, it's minus right. So, yes. So which means two x plus one is equal to one, so x plus one is equal to one-half, x is equal to minus one-half. So this point is minus one. So that's basically it. I was trying to show how from the first basic graph which you have, you can transform it into any other form which has some linear combination of arguments and the functions. It's not easy, it just takes some practice and if you would like, you can just do it yourself many times. In any case, what's important to remember always is for exponential graph, the basic most important point is if base is greater than one, it's increasing this way and goes down to zero to the left. And if base is less than one, that's the other way around. So it goes to zero when x goes to plus infinity and it goes to infinity when x goes to minus infinity. Everything else basically is a consequence from this. That's it for today. Thank you very much and good luck.