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Transformation of an Exponential Function / Example 14.4

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Hi! My name is Rebecca Muller. During this video, we're going to look at exponential functions. Now, you're already somewhat familiar with functions that have exponents, but those are typically of the form-- something like g of x equals x squared. Now, in this format, notice that the base is actually a variable, and your exponent is a constant. We're going to switch that when we go to exponential functions. We're going to look at functions of the form f of x equals something like 2 to the x, where now the base is our value 2. That's going to be a constant. And the variable is going to be in the exponent. So let's look at a definition now of an exponential function. The exponential function f with base b is written as f of x equals b to the x power, where b is greater than 0; b does not equal 1; and x is any real number. So a couple of things to point out with this definition. We're only going to allow our base to be positive numbers. And that's so that we don't have things that are undefined or that it's switching from being a positive number to a negative number. Think about a base like negative 2. If you square it, you end up with a positive 4, but if you cube it, you end up with a negative 8. We don't want that to happen. So we're going to restrict our conversation to only those that have a base greater than 0. And why won't we let the base equal 1? Well, that's because if the base is equal to 1, 1 to basically any power is 1. So that's not going to be an exponential function. That's going to be that horizontal line that we've seen previously. It's a linear function. So let's look at the specific things that we want to consider during this video. First, we want to look at evaluating exponential functions. And then we're going to look at graphing exponential functions. And then we're going to look at some applications dealing with money. Those are the interest applications. We're going to begin by reviewing some properties of exponents, and we'll do that in the midst of evaluating exponential functions. So the function I want to start with is f of x equals 4 to the x power. So to evaluate, we're just simply going to substitute in values for x and then see where it takes us. What are the values for y that are going to correspond to that? Let's start with f of 3. That's going to equal 4 to the 3rd power. And 4 cubed is equal to 64. Next, let's look at f of 0. That's going to equal 4 to the 0 power. And you may recall that a number raised to the 0 power is equal to 1, as long as the base is not 0. So this is going to equal 1. Let's look at f of negative 2. We'll have 4 to the negative 2 power. And so again, reviewing exponent properties, we know that when we raise to a negative exponent power, that means that we're going to take the reciprocal of the base raised to that power. That is, this is 1 divided by 4 squared. So we lose the negative sign when we place that in the denominator. And we can evaluate that as 1/16. And if we look at f of 1/2, this is going to equal our base 4 raised to the 1/2 power. And again, let's review the properties. When we raise a base to the 1/2 power, that's the same as taking the square root of the base. So this is going to equal the square root of 4, which is going to give us back the value of 2. Now let's move on looking at graphs of exponential functions. As our first example, we're going to look at f of x equals 2 to the x power. Now, because we don't know what this graph looks like, sensibly that means that we can-- to start with a table of values. And I'm just going to draw a T chart here. Let's substitute some values like negative 2, negative 1, 0, 1, 2 to start with. So substituting in the value of negative 2, that's going to give us 2 to the negative 2 power. That's going to be the same as 1 divided by 2 squared, which is 1/4. Let's go ahead and plot that point. So we're at negative 2, 1/4-- about right there. By substituting the value of negative 1, we're looking at 2 to the negative 1 power, which is the same as 1/2. Let's plot that point. So negative 1, 1/2. Substituting in the value of 0 for the exponent, we end up with 2 to the 0, power which is 1. Substituting in 1, 2 to the first power is 2. And then substituting in 2, we end up with 2 squared, which is 4. Now I'm going to do one more just to point out something. If I look at what happens when I substitute in the value of 3, 2 cubed is going to equal 8. So what we're noticing is that we're doubling up on values here as we end up going to the next integer value. So we have 3 and then 8. And now we're going to connect the dots in order to see what this picture is showing us. So, once I get past values of x equal to 0, that is, once I get to positive values of x, what we notice is-- that our graph is going to be rising pretty rapidly and heading off toward infinity. Because the values, as we continue making them larger, will make the expression larger altogether-- 2 to the 4th will be 16; 2 to the 5th would be 32, and so on. When we're coming down this way, notice that it looks like we're getting closer and closer. And the question could be, well, does it ever end up crossing the x-axis? Well, in order to do that, you'd end up have having to have 2 raised to this power equal to 0. And something that you want to remember is-- if you have a positive number raised to a power, the result will always be positive. So there's no way I can end up getting 0 back. So again, positive number raised to a power is always positive. So thinking about it, we are going to continue getting closer and closer to the x-axis, but we're never going to end up hitting that x-axis. I mean, you could think about what happens if I end up with something like 2 to the negative 10? That's going to be 1 over 2 to the 10th power, which is going to be a extremely small number close to 0. So, you've seen that kind of look before when a graph gets closer, and closer, and closer to a line, but it never hits it. That's called asymptotic behavior. And what we can see is that we have a horizontal asymptote. And that's going to be concurrent with the x-axis. So we have a horizontal asymptote, and its equation is going to equal y equals 0. Let's go ahead and also point out the domain and the range for this function. Now, for the domain, we can look back at the equation and note that we're just raising to an exponent power. And so we have no restrictions on what an exponent can be. The domain is going to be all real numbers. You can also see that by looking at the graph. If you go from left to right, you can see you're picking up all x values. So our domain can be written as negative infinity to infinity. For our range, that's going to be our y value. We can go back to the idea of a positive number raised to a power is always positive. And we can also see it from the graph. We end up picking up any value above 0 for our y values. So our range is going to be from 0 to infinity. And notice that we cannot include 0 in part of the range. So we have now a description of what a exponential function looks like in general. We're always going to have this horizontal asymptote. We're always going to have the other side of the graph heading off toward infinity. Now, like a lot of other graphs that we've worked with, at this point we want to look at what happens if we talk about transforming those graphs by using translations, or reflections, or stretches, and shifting, and all that kind of jazz. So let's look at an example now. We're going to start off with the graph that we had a few moments ago. That's f of x equals 2 to the x power. And we have it depicted here. Let's look at, what do you think would happen if we're now going to look at f of x equals 2 to the negative x? Well, recalling what that negative is going to do with the x, you may recall that what happens is it's a reflection across the y-axis. So we're going to take the picture that we have here, and we're going to have a reflection, which means we're going to flip it across the y-axis in order to get a result. This is what it looks like. Notice that the graph that we started with, the 2 to the x, is an increasing function. It's always increasing as we move from left to right. The second graph, f of x equals 2 to the negative x, is a decreasing function. Now we could also write 2 to the negative x as 1/2 raised to the x power. And so it turns out that if our base is going to be greater than 1, we're going to end up with an increasing function on its basic format. If our base was between 0 and 1, we would end up with a decreasing format for our graph. Let's now move on. Let's look at f of x equals 3 to the x as our starting point. So you'll notice that this one is a little bit steeper than the one we saw previously. Because for instance, if I substitute in the value of 1, instead of getting back 2 I'm going to get back 3. And what about this transformation-- f of x equals 3 to the x plus 1 power minus 2? Let's discuss what's going to go on. We know that the plus 1 is going to end up being a horizontal shift to the left. The minus 2 is going to end up being a vertical shift down. Now, not only does it shift points, but it also is going to shift any asymptotes that we might have in the graph. And of course, we have an asymptote right on the y-axis. So the shift to the left isn't going to bother it, but the shift down is going to. It's going to affect it, and we're going to end up with a new horizontal asymptote. This is what that graph is going to look like. We see the horizontal asymptote at y equals negative 2. And then we have our shift over to the left. And that's going to take place in order to connect the dots. And you end up with the new graph. And now a third example of that. We're going to start again with our f of x equals 2 to the x power. This graph is going to now be a new function. What we're going to have-- negative 2 to the x plus 1. So what does the negative 2 in front of the 2 to the x do? That's going to be a reflection across the x-axis. And then we're going to add 1, which means it's going to be a vertical shift up. So again, let's think about what happens to the horizontal asymptote. It used to be on the x-axis. The reflection across the x-axis is not going to change it, but then the shift up will. Here's a picture of new graph. We're going to have, again, the horizontal asymptote will be x-- at our y value-- equal to 1. And we have that graph that's been reflected across. And so we see it's going to be upside down from what we were expecting. It's time for a quick quiz. Which of the following is the correct graph of the function f(x) equals negative 5 to the x plus 2. We're given three graphs in A, B and C. All of them have a characteristic shape of an exponential graph. In part A, it appears that we have a horizontal asymptotes at y equals 2, we also have the graph decreasing for all values and we have a y intercept of 3. In part B, we're given a graph which is increasing for all values of x it looks like we have a horizontal asymptotes at y equals negative 2 and a y intercept at negative 1. In part C, we have a graph that is decreasing for all values of x it looks like we have a horizontal asymptotes at y equals 2 and we have a y intercept at positive one. Choose from A, B or C. You're correct, the answer is C. Sorry the correct graph is given in C. In our function f of x equals negative 5 to the x plus 2. First of all we know that we have the negative sign in front of 5 to the x that indicates a reflection across the x-axis the plus 2 at the end indicates a shifting upward of 2 units. The graph that has been reflected across the x-axis and then shifted up to units yields in asymptotes at why equals 2 and the graph will be decreasing for all values of x. One of the most familiar applications dealing with exponential functions comes in the form of looking at interest. That is, we're looking at money, and it's being compounded over time. So we're going to develop a formula for that now. You need to understand what compound interest is all about. Let's take a scenario that looks something like this. We're going to start with $100. I'm just going to keep my numbers really easy to work with. We're going to say that we're able to go to a bank and get an interest rate of 8% annually. And we're going to look at this being compound in quarterly, which means 4 times a year. Now, let's see how that plays out when we're really looking at what happens to the money. So we're going to have the number of quarters that we've been discussing. And then we're going to have the amount that we end up with. So, at the beginning, at time 0-- if you want to think of it like that-- we're going to have just the amount that we're going to invest. That's our value of 100. At the end of the first quarter, are they going to give you the full 8%? Well, no. That's what happens over an entire year. So it should make sense to you that at the end of the first quarter, they're going to give you a quarter of the interest. And that means that they're going to give you back the total amount, the $100. Plus you're going to end up getting an extra amount, which is your interest. And that extra amount is going to be the 8% divided into 4 equal parts. And in this case, you can easily calculate that-- 8% divided into 4 equal parts is going to be 2%. Now, from dealing with problems earlier in your career, you know that when we're dealing with percentages, we end up not using the percent in our calculations. We instead change it to its decimal format. So I'm going to change that to 0.02, and that's being multiplied times 100. So this, again, is our original amount plus an extra 2% of our original amount. Well, I'm going to do a little bit of work with this, because we want to develop a formula. Notice we have two terms here, and both terms have 100 in them. So I'm going to factor out the value of 100, leaving me with 1 plus 0.2. Now, if I really wanted to figure out what that amount is, I can go ahead and calculate it-- 2% of 100 is easy to calculate. That's $2. Add it to the 100, and we end up with $102. Now let's consider what happens in the second quarter. Well, in the second quarter, we start off with the amount of $102. And we're going to now get another 2%. But that's going to end up being 2% of $102. I'm going to do similar work to what I did previously. I'm going to notice I have a 102 in both of these. That can be factored out-- 102 multiplied-- times 1 plus 0.02. But before we go any further, let's think about where the 102 came from. If I go back to the second to last step here, we notice that 102 was equal to 100 times 1 plus 0.02. So I'm going to replace it here. It's going to be 100 multiplied times 1 plus 0.02. Now remember, that part is just the 102. And now we're multiplying it times another 1 plus 0.02, which means I could write it as squared. And that amount, if I plug into a calculator and figure it, out is $104.04. OK. Let's continue going. I now want to consider what happens in quarter number 3. And we'll figure out what the amount is that I end up with. So in quarter 3-- same scenario. I start off with $104.04. I'm going to add to that an extra 2% of the $104.04. I can factor out that amount that repeats, leaving me with 1 plus 0.02. The $104.04, that's equivalent to the 100 times 1 plus 0.02 squared. So I have 100 times 1 plus 0.02. Now I'm going to go ahead and write it in. It's squared. That's this part. Let's multiply it times another 1 plus 0.02. And now I can rewrite this as 100 times 1 plus 0.02 cubed. What do you think is going to happen in the 4th quarter? Well, we're going to start with what we had previously. We're going to end up having to add another 2%. We're going to then be able to replace what we had with the expression that we had from the prior step. And all together, what we end up with is our $100 times 1 plus 0.02. Only this time, we'll have multiplied-- times that 4 times. So if we generalize this to the number of quarters that we're working with, we can see that the amount we end up with, our A, is going to equal the 100-- the amount we start with-- multiplied-- times-- this 1 plus 0.02. And the exponent power is going to be a variable. That exponent power changed each time, dependent on how many quarters we were working on. So we can think about this as the number of quarters that we were dealing with in that expression. So with this scenario, we're getting a generalized version. Now, what if we take it a little bit-- a step further? In other words, what if it wasn't $100. What if it was just some amount that we were investing. Well, the term we use for that is called the principle. And we're going to use a capital P to stand for the principle. That's the initial amount that you're going to invest. The value of 1 is there, because we want to get back the initial amount. We want to get back the principle. Plus, how did we come up at this 0.02? If you recall, we took our interest rate, which I wrote as 8%. But notice that that is equivalent to the decimal format. And what I did was I divided it by the number of times I was compounding per year. When we had the compounding quarterly, we knew that was 4 times a year. Let me just add a variable to this. I'm going to call that amount a value n. And we're going to say n equals 4. So to get our 2% here, we ended up taking our r value of 8% and dividing it into 4 equal parts. So it was r divided by n. And then how many quarters do we end up with? Well, here we only looked at 1 full year. And we had 4 quarters in that frame. What if it had been 2 years? It would have end up being the number per year, which was our value n, times the number of years. We would have had 8 times that we had quarters. And so the exponent power here is going to end up being the number of times it's compounding per year-- that's n-- times the number of years, which is t, giving us the formula that we're going to use for interest compounded n times a year. So let's wrap up what we just found in the last example. We have a formula now for interest compounded n times a year. A equals p times 1 plus r over n quantity to the nt power. The A stands for the amount that we end up with. The p is the principle. That's the amount we start with. The r is our interest rate, and we need to use it in decimal format in the formula; n stands for the number of times it's compounded per year. The t is our time. And in this formula the time needs to always be in years. So now it's going to be your chance to try one of these problems. I wanted you to try the formula yourself. And here's the problem we want to consider. A parent invests $2,500 at the birth of a child. The annual interest rate is 2.5% compounded monthly. If the money is left alone until the child's 18th birthday and then cashed in to use for a college fund, how much money will she receive? And then answer the second question. How much money was made off of the original investment? So take some time now to look at the formula. Figure out what all of the variables are that you are given in the formula and what it is you're trying to find. And then solve the formula. Try it now. Pause the tape. All right. Let's see how you did. When we were given a problem like this, the key is, first of all, figuring out what formula we're using. But of course, that part was easy, because we had just finished with the formula. So we know that we're going to be looking at the formula A equals p times 1 plus r over n, to the nt power. If you were reading this, and you didn't have the prior example to go with, how would you recognize that this is the formula? It has to do with the fact that we're told we have an interest rate that is compounded monthly, which means n times a year. That's exactly what the formula is for. Next, we want to identify all of the variables that we're given. So I'm just going to start reading through the problem. "A parent invests $2,500 at the birth of a child." Since that is the amount of investment, that's our principal amount. So our p value in our formula is going to equal 2,500. Continue reading. "The annual interest rate is 2.5%." Let's just stop right there. We're given the annual interest rate. The rate is 2.5%. But remember, I need to put it in decimal format. And so this is per 100, which means that you're going to end up moving your decimal point 2 places to the left. So our r value is 0.025. Continuing on-- it's compounded how many times a year? Well, it's compounded monthly. So that means that the value of it that's compounded per year is going to be 12. That's our value for n-- n equals 12. And finally, if the money is left alone until the child's 18th birthday, that means that the time for this investment is going to equal 18. We're asked to figure out, how much money will she receive at the end of that investment period? Which means that we're asked to find the value of A. So, once you take the problem that's given to you, then you want to come up with the formula and write down all of the pertinent facts. Now it's just a matter of substituting those facts into the formula and then solving. So we're going to have our value of A equals-- in place of p we'll have $2,500. That's going to be 1 plus-- we'll put in our r value-- 0.025; divided by n, which is 12. And that's raised to the 12 times 18th power. Now, if you're using a calculator, which of course you're going to in order to evaluate this, what I did here was to emphasize the fact that these two values in the exponent need to be multiplied together. And so be cautious when you're using a calculator-- that you have parentheses around this. Or even evaluate what 12 times 18 is to start with before you put it as your exponent. If you've used your calculator to come up with a solution, you should have come up with our A value as approximately equal to $3,918.95. Now, it's an approximation. But when we're dealing with money, we're always going to take that to the nearest cent. So our answer will be that the amount she ends up with is $3,918.95. The second part of the question is just asking you to compare this to the amount that was invested. So how much was made of the investment? Well, if I take away the principle of $2,500, that's going to leave us with $1,418.95. So in that 18 year period, this is the amount of money that was made of the investment. Now let's look at this formula and think about what would happen if the value of n is becoming larger and larger. For instance, what if I, instead of compounding monthly-- which is where n is equal to 12-- I'm compounding daily? That would be n equals to 365, perhaps. And sometimes when you're dealing with financial groups, they want it to be 360 for a year. But nonetheless, it's a large number. What if it was compounded hourly? Then you could take 365 and multiply times 24-- very large number. What if you wanted to compound it every minute? You'd have to multiply that amount times 60. So you can see that this denominator is getting extremely large. Now, there's a concept of compounding continuously. Well, what that means is that my value of n is actually going toward infinity. So I have n heading toward infinity. Well, you can't substitute in the value of infinity into this formula. And in fact, you can see, therefore, we need a new formula. However, we're going to base our new formula off of our old one. So follow along, and I'll show you how it develops. We're going to look at the graphs of y equals 2 to the x and y equals 3 to the x. Now we can see that they both go through the point 0, 1. We can also see that the graph of 3 to the x is steeper than the graph of 2 to the x. One way to measure that is to look at the fact that we have tangent lines-- or in other words, a line that is best representative of the graph-- right at the value of 0, 1, for both of these. So we're going to zoom in, and then I'm going to show you what the tangent lines look like. Now, the tangent lines are now, more obviously, showing you that the graph of 2 to the x is not as steep as the graph of 3 to the x. Because it's a straight line, I can find the slope of that straight line. And so it turns out that for the graph of y equals 2 to the x, the slope is somewhere around 0.7-- 7/10. For the graph of 3 to the x, the graph-- the slope, that is-- is somewhere around 1.1. Well, somewhere in between a slope of 0.7 and a slope of 1.1 is a slope that is exactly equal to 1. It's going to have bearing on what we're going to do in a moment. It turns out that that number-- that base that's going to give us a slope of exactly 1-- is going to be a number between 2 and 3. And since we have a slope of 0.7 at 2 to the x and at point 1.1 at 3 to the x, then it would make sense it's going to be a little bit closer to the value of 3. The number, it turns out, is very close to 2.7. And you may have had experience with this before, or maybe not. It's a number that we call e. And so again, when we're looking at an exponential function with a base raised to the x power, the slope of exactly 1 occurs when we're dealing with y equals e to the x. Now the e is used for a guy named Euler, OK-- E-U-L-E-R. It's also-- if you're unfamiliar with it-- the good thing to think about is it's a number like the number pi. You got used to that when you were dealing with circles somewhere early on. And pi, you know, is around 3.14. Well, that's how you want to start thinking about e; e is a number similar that. It's irrational, and it's around 2.7. And so that gives you something to go with. It's important to us, mainly because of this connection. It turns out that as-- that "if," that is-- let me put "if." But if my value of x is close to 0, then we can see that the graphs of e to the x and the graph of that tangent line are basically going to be the same. They're going to be on top of each other the closer and closer you get. What is going to be the equation of that tangent line? Well, it has a slope of 1, and it has a y-intercept of 1. So its equation is going to be x plus 1-- 1x plus 1. I'm going to rewrite this a little bit differently. I'm just going to say e to the x is approximately equal to 1 plus x. And now I'm going to come back to my formula that I have over here, and we're going to deal with that in this context. If I come back over to my formula, I'm going to make a substitution of a variable. I'm going to let x equal r over n. Now, if x is equal to r over n, and n is going to infinity, what does that tell me about x? That my x value has to be getting closer and closer to 0. Because if my denominator is getting very big, my fraction overall is getting close to 0. Ooh. So now what does that look like? A equals p times 1 plus x to the nt power. Hey-- 1 plus x-- it's approximately equal to e to the x whenever I have x close to 0, and it actually equals e to the x when x equals 0. So I'm going to replace my 1 plus x with my e to the x. Now, let's just simplify somewhat. We know that when we have e raised to a power and then raised to another power, we can rewrite that. And I'm going to move ahead and move over here. We can rewrite that by multiplying our exponent powers. So I'm going to have A equals p times e to the x, times n, times t. I still have an n in that formula. But remember, at the beginning, we let x equal r over n. So now substituting back out, we have p times e to the r over n, times n, times t. And look what happens to the n. It's gone. We have A equals p times e to the rt power. And this is the formula that we're going to use whenever we have interest compounded continuously. So you see that it has a relationship to the formula that we developed earlier for interest compounded n times a year. But whenever we're compounding continuously, we cannot use a value for n. You know, I like to call it the PERT formula, because it's easy to remember-- p times e raised to the rt power is going to give us our value for A whenever we are compounding continuously. Now, let's look at what would be different about the investment that those parents made if instead of compounding monthly, you were able to find an investment that was compounding continuously. So instead of using the other formula, we're going to use our PERT formula-- A equals p times e to the rt power. We're still trying to find the value of A. The p is still going to equal 2,500. The r value is still going to be 2.5%. And we're still going to use the decimal format to substitute into the equation. Our T value is still going to be 18 years. So now we're going to substitute into our given value and find A equals 2,500, times e raised to the power, which is going to be the product of 0.025 and the value 18. Using a calculator, we can figure out what this is going to equal. Now, if you've never worked with a calculator with the e key, a lot of times it's going to be above a key that's going to be labeled with an ln. So you might see something like this on your calculator. This one might say ln. And above here you might see something that says e to the x power. So you might have to press second, ln key, in order to end up with e to a power. Using that calculator, you come up with a value for A, which is approximately equal to 3,920.78. Now, when we did this with it compounding monthly, you may recall that the amount that the parents ended up having-- or the child ended up having, I should say-- for the college fund was $3,918.95. Notice that this amount is going to be slightly more. And of course, nothing else has changed except for the way it was being compounded per year. So the rule of thumb is that the more times you can compound per year, the better off for an investment. Of course, if you're being charged interest, it's worse for you if they're compounding more often. So now it's time for you to try some problems dealing with exponential functions on your own. Good luck. Time now for another quick quiz. Which formula would be appropriate to use to solve the following problem. Fred Flintstone invest two stones (worth $500) with Slade Bank at 5% compounded semiannually. What will be the value of his investment at the end of one year? Should we use the formula given in part A, which is A equals P times the quantity 1 plus r divided by n raised to the nt power with the value n equal to 2. Should we use the formula given in B. A equals P times e to the rt power with the value of t equal to one half or should we use the formula given in C. A equals P times the quantity one plus r raised to the t power with the value of r equal to one half. Your correct we should use the formula given in A. Sorry the correct formula is given to us in answer A. In this problem we're told that we're looking at an investment that is going to be compounded semiannually. This means we want to use the formula A equals P times quantity 1 plus r over n raised the nt power with n stands for the number of times is being compounded per year. Semi annually indicates that would be twice a year so n equal 2 is going to be the correct result. B is the formula we use if it was going to be compounded continuously, C would be a formula we could use if we knew that we were compounding annually but in either case we wouldn't have our r value equal to one half.