Constructing exponential functions from tables
About this video
The video finds an exponential function D(x) = A(R)^x + K from a table. Because consecutive values do not share a common ratio, there must be a constant K, so the narrator looks at the differences between consecutive values (1536, 384, 96, 24), which share the ratio 1/4. Two equations from D(-1) = -9 and D(-2) = -33 then give A = -2 and K = -1, so D(x) = -2(1/4)^x - 1.
You'll be able to
- test whether a table shows a simple exponential by checking for a common ratio.
- use the differences between consecutive outputs to find the common ratio when there is a constant K.
- set up and solve two equations to find A and K.
Key words
- common ratio
- The number you multiply by each time x goes up by 1 in an exponential function.
- vertical shift (K)
- A constant added to an exponential that moves its graph up or down.
- delta
- The change between two successive values, found by subtracting.
- system of equations
- Two or more equations solved together to find more than one unknown.
Transcript
- [Instructor] We are told the table represents the exponential function D. Fair enough. What I wanna do in this video is try to figure out what D of x actually looks like in the form where it's equal to some constant A times our common ratio R to the x power, and then potentially some constant K.
So if you feel inspired, definitely pause this video and have a go at that. Okay, now let's do this together. So the first thing I would like to do when I'm dealing with something that could be some type of an exponential, we'll see is it a very simple exponential function.
Maybe K is zero and it's just A times R to the x. Well, if that's the case, I'll just have a straight up common ratio between every successive value of D, every time I increment x by one. So let's see if that happens. And I'll start with the low numbers here. Let's see.
To go from negative 33 to negative 9, we definitely increased our x by one. When we increased our x by one, what did we multiply, or what could we have multiplied? Well, if you take negative 33 and you multiply that times negative 9 over negative 33, you would get negative 9.
Negative 33 times this, that cancels out with a denominator, you'd get negative 9. This is the same thing as negative 9 over 33 is the same thing as 3 over 11. All right. Now when we go from negative 3 to negative 2, we're also incrementing x by one. Let's see if we have that same common ratio.
So here we would've to multiply by negative 33 over negative 129. And let's see, if we divide the numerator by 11, we get 3. But if you divide negative 129 by 11, you're not going to get 11. So we don't have a just straight up common ratio here. So there must be some K factor here.
This isn't just a simple exponential function. There's some constant, and to figure out in that situation what the common ratio could be, or what it is, we need to do something a little bit different. One way to think about it, just to visualize it a little bit, what's going on, if we were a simple exponential function, let me draw some axes here just to visualize what we're about to do.
So that's the x-axis, that's the y-axis. A simple exponential function would look something like this if you think about its graph. We just figured out that it's not that, there's going to be some K component. K could be positive, it could be negative. But just for visualization purposes, let's assume that K is positive.
If K is positive, it's essentially going to shift a simple exponential up like this where this delta is equal to K. And so to figure out the common ratio, we just have to think about every time we increment x by one, how much has our y, or in this case D of y equals D of x...
How much does D change every time we increment by one? How much does D change? And what's the common ratio on that change? How much more do we change the second time than the first time? And that is going to be our common ratio. If we didn't have a K factor, we could have just done it right from the get-go.
But now what I'm gonna do is take the delta between successive points and then figure out what I have to multiply to get there. And that will be our common ratio. We'll still have to figure out A and K, but it gives us a good start. So let's do that. And I'll start with these lower values here.
And actually we still have to validate that this is even an exponential. So let's see, if I am going from... let me get my calculator out just to make this a little bit simpler. I could hand do everything here, but that will take a while. So if to go from negative 2,049 to negative 513, what I did, I could just say 513 negative minus 2,049 negative.
So that would just be adding that. That equals 1,536. So this first delta right over here is plus 1,536. Now what's going to be the next one? Get the calculator out again. And we are going to get, if we have 129 negative, and I could do the same thing, I could just add 513 to that.
That's the same thing as subtracting negative 513. So plus 513 is equal to 384. So that is going to be plus 384. And then when we add one in x, again, this is going to be negative 33 plus 129. So this is gonna be plus 96. And then last but not least, when we go from negative 33 to negative 9, that is going to be plus 24.
So now let's see if there's a common ratio amongst these deltas. So I can immediately see to go from 96 to 24 that is multiplying by one-fourth, or you could even say dividing by 4. But we're trying to find, what do we multiply every time? Let's see, to go from 384, let me get the calculator out just for sake of speed.
And let me see these other common ratios. Let me see if every time I can just divide by 4. Whoops. So if I get 1,536 divided by 4, that is equal to 384. If I divide by 4 again, that is equal to 96. And of course if I divide by 4 again, that is equal to 24. So we can feel very good that this is an exponential indeed.
We have a common ratio of one-fourth. So in this little mystery hunt for these different variables, for these different coefficients or these different unknowns, I should say, we already know what R is. So we can say that D of x is equal to A times one-fourth to the x power.
And maybe I could write it like that. And then of course we have A plus K. So how are we going to figure out A and K? Well, we can substitute some of these x values. We know the corresponding D of x, and then if we have two unknowns, we need essentially two different equations to solve for those two unknowns.
So let's start with some of the simpler numbers here. I could use any of them, but let's use the simpler ones. D of negative 1 is going to be A times one-fourth to the negative 1 plus K, which is the same thing as... one-fourth to negative 1 is just 4. So it's 4A plus K.
And we know that when x is negative 1, D of negative 1 is equal to negative 9. So this is a nice clean equation. And now let's do D of negative 2. D of negative 2 is equal to A times one-fourth to the negative 2 power. Lemme make this very clear. Plus K, which is equal to...
one-fourth to the negative 2 power is going to be 16. So we get 16A plus K is equal to D of negative 2 is equal to negative 33. This is another nice equation. Now we have two equations with two unknowns. So let's see if we can solve this. So let's see. What I will do is let's subtract the second equation from the first equation and let's see what we get.
So 4A minus 16A is going to be negative 12A. K minus K, those cancel out. So that just gives you zero is equal to negative 9 minus 33 is positive 24. So there we get A is equal to... if we divide both sides by negative 12, A is equal to negative 2. Very nice. Now we can go and substitute back into either one of these to solve for K.
Let's do into this first one, 4 times negative 2 plus K is equal to negative 9. So that's negative 8 plus K is equal to negative 9. Or if we add 8 to both sides, we get K is equal to negative 9 plus 8, which is equal to negative 1. And now we are in the home stretch.
We just have to write it out. We deserve a little bit of a drum roll. We get D of x is equal to... A is negative 2. So I'll write negative 2 times... maybe I'll write it this way: one-fourth to the x power, plus K. So instead of saying plus negative 1, I'll just write minus 1.
And there you have it. We have figured out this exponential.
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