 In our previous video, we talked about exponential functions, how they look, right? We saw that the most general exponential function will look like y equals some c times a to the x plus one, plus k, excuse me. That's the most general form right there. And how adjusting the base and the scalar and the shift affects the graph here. We wanna kind of reverse this process now. Let's not come up with the equation, see what the graph looks like. What if we know what the graph looks like? Can we come up with an equation, right? So let's suppose we have an exponential function with no transformations in it whatsoever, it's just f of x equals a to the x. How do we determine what the base is supposed to be? Well, if we know a point, like say it goes through the points two comma nine, this would tell us that we come over two and we have to come up nine, one, two, three, four, five, six, seven, eight, nine, right? We have this point right here. So we have to go through here. We know it also goes through the point zero comma one. And two comma nine. So we know this. So we kind of, we essentially know what the function's gonna look like. It would have to look something like this. Whoops. Make sure that you don't cross the x-axis because it's asymptotic to it. So this is what our function's gonna look like. So we know the graph, but what does the function, what does the formula look like? Now, if two nine is a point on the graph that we see right here, that means this is a solution to this equation, right? We can look at f of two. And in one instance, f of two is gonna look like a squared. But in another instance, f of two is gonna equal nine. So we get that nine equals a squared. If we take the square root of both sides, we end up with a equals three. Now, when you take square roots, you typically think of plus or minus three. But remember, when it comes to these bases, the a, we have a stipulation, a has to be positive and a can equal one. And therefore that tells us whoops, we don't want the negative value. We actually just want a equals three. That is the function in play here was f of x equals three to the x. All right, so that can help us out with the base. Let's try that again. Let's consider the function g of x equals eight to the x. So just a exponential function without transformations in it whatsoever, okay? What do we do in order to build this function right here? Well, again, since there's no transformations, the standard y-intercept is gonna be one. We also have the point three comma one eighth. So that's gonna be like super teeny. So we get this point right here. And so we can anticipate that the graph is gonna do something like this. This one is gonna be a decay model. Notice how it's decreasing over time. So roughly speaking, this is what the graph g would look like. But just like the last example, we can use this point. We can plug it into our formula right here. And so g of three tells us that a cubed is gonna equal one eighth. So taking the cube root of both sides, we end up with a is equal to, well, the cube root of one eighth, which is the same thing as the cube root of one over the cube root of eight, which gives us one half. So that's the base there. And like we saw in the previous video, if your base is less than one, that's a decay model. If your base is greater than one, you're gonna have a growth model. So this function would be g of x is equal to one half to the x power. So we need a point. We need a point on the graph to determine the base of the function. What if, on the other hand, we do have some transformation in play here? What if we have just a stretch? So h of x looks like c times a to the x. Well, because there's a stretch, there's two parameters we don't know. We don't know c and we don't know a. So we're gonna need two points of data to help us out here. Now, if I could pick any point of data, the y-intercept is gonna be my favorite point to use here. So what we see here is the y-intercept is gonna equal three. We also know the other point is gonna look like three to the 24 or three common 24. I can't fit that on the picture. So let me, let me relabel the picture as like three. I think I can get away with this. Three, six, nine, 12, 15. Then we get 18 and then 24. I could fit that on there. Okay, so we can take the point three common 24 and we can also take the point zero, three. So that actually would move it here. This is a challenge you often see with exponential functions that because they grow so rapidly, you have to sometimes change the scale in order to fit them into a reasonable picture of any kind. So connecting the dots, we would get something like this. So this is the picture we anticipate. So how do we determine this thing? Well, like I said, the y-intercept is the place to get started with because when you look at this function, like all the previous ones, the horizontal asymptote was the x-axis. That only is gonna change if we start shifting the graph up and down. So I want you to examine what's the distance? What's the distance here between the asymptote and the function with the y-intercept? I should be saying what's the distance here between these values? Because it really turns out this is gonna help us determine what the parameter c here is going on here. So I'll come back to that into a little bit, but if we were to plug in, if you plug in your y-intercept into this function right here, what you see is the following. You see that h of zero is equal to c times a to the zero, which should equal three. Now what is a to the zero? That's always gonna equal one irrelevant of what the base a is because as long as a is positive and not equal to one, we know a to the zero is gonna be one. Heck, it could equal one and that would still be true. So we get that c is actually equal to three. So knowing the y-intercept is extremely important because it helped us figure out that c-value very quickly, the initial value, okay? And notice the distance between the y-intercept and the horizontal asymptote actually was three units, right? Because remember, this is y equals three. So from there, we now learn that our function has the form h of x is equal to three times a to the x. Well, if we use the other point, that can help us determine what a is supposed to be. So we see that h of three is equal to three times a cubed. That should equal 24. In which case, if we divide both sides by three, that would then give us a cubed is equal to eight. And then taking the cube root of both sides, we end up with a equals two. And so in the end, our function h of x looks like three times two to the x. So we were able to figure out both the vertical stretch and the base of this exponential function using the points provided. All right, so I wanna do one more example of such a thing here, where here the graph of the function is given to us. And we wanna come up with a formula to match up here. Now I'm gonna help us out here. I'm gonna give us the base. So this function is supposed to be, it has a base of one half, but we don't know the constant it got stretched by. And we don't know yet the shift. And so let's determine how this comes apart here. So our function does pass through the point zero, one and one half, okay? So the first thing I want you to do is when you're trying to graph an exponential function or in this case, come up with the formula of the graph sort of given, your first target is going to be the plus K. That's who you should go after first. And that's because the shift is actually the most obvious of all of these parameters here because the location of the horizontal asymptote is gonna give you this shifting value right here. Because for an untransformed exponential function, like if you have y equals a to the x, your horizontal asymptote, your horizontal asymptote is gonna happen at the x-axis. So if you shifted things, whoop, you go up by a factor of K, that takes the x-axis and then it moves it up as well. And so the location of the horizontal asymptote tells you the shift. So notice our horizontal asymptote is right here, it's at the location y equals three, that's the K value. So we see that K is gonna equal three, which that's quite helpful here. So then rewriting our equation, we have f of x equals C times one half to the x plus three. All right, now we have to look at, we have to look at the C value right here, for which we could try to plug in a value, which hey, if you know the y-intercept, we can plug that in, plug it in the y-intercept, you would see that f of zero, because it worked great for us last time, f of zero is equal to C times one half to the zero plus three. Okay, well the nice thing about the y-intercept is when you take a base to the zero power, it disappears, poof, it disappears. And so you end up with C times one plus three, which is just C plus three. But the y-intercept, according to the graph, is supposed to equal one, right? So you get one, and so then solving for C, C is gonna equal one minus three, which is equal to negative two, all right? And so this is the observation I was alluding to toward earlier. Look at the distance between the y-intercept and the horizontal asymptote. How far apart are they? They are two units away from each other, but for your standard exponential function, you have the asymptote, and the graph should be sitting above it, but ours is sitting below it, because of the reflection, okay? So you wanna take the distance from the y-intercept to the asymptote, and that will give you this number right here. So we actually wanna think of this as a negative distance, because our graph sits below its horizontal asymptote, thus agreeing with the negative two we found right here. And so what does this now tell me about my function? We actually have it. We have f of x equals negative two times one half to the x plus three. And so we are able to figure out this function right here. Now you'll notice that on this example, I provided to you the base, but on the other previous examples, I didn't provide to you the base. We had to solve for it. So really, if we wanted to do the whole enchilada, we'd be given a function like f of x equals c times a to the x plus three, plus k, right? And by the steps we took here, we are able to determine the shift just by the location of the asymptote. We're able to find the scalar c by looking at the difference between the y-intercept and the asymptote. We haven't yet used anything that would actually be given by the one half here. So that's where actually this third point, the second point, excuse me, comes into play here. If you were trying to compute that, you would plug in the point one comma two, you would get f of one is equal to two, but it's also equal to c, which we know to be negative one half, times a, a to the first right, plus three for which, okay, what's going on here? You could subtract three from both sides. You're going to get negative one is equal to negative two a, right? Divide both sides by two and you get a equals one half. So we could have actually figured out the base. I didn't have to tell it to you. So it's like secret, secret, secret. Oh, we found out that it's supposed to be a one half. We can solve for it. That's because we used the value a equals one, all right? But what if we wanna use a different point? Like we put the different point over here where two equals whatever it turned out to be. It would be just like we did before. You'd have to, you have a square here. You have to take the square root. And so we can put all of these ingredients together. If we know basically two points on the graph, particularly if we know the y-intercept, we know the location of the horizontal asymptote and we have any other point on the graph, then we can take our function y equals c a to the x plus k and we can come up with a formula that'll match that exponential growth. And this can be done for any exponential function under the sun.