Graphing exponential growth & decay | Mathematics I | High School Math | Khan Academy

This is from the graph basic exponential functions on KH Academy, and they ask us to graph the following exponential function. They give us the function ( H(x) = 27 \cdot \left(\frac{1}{3}\right)^x ). So our initial value is 27, and ( \frac{1}{3} ) is our common ratio. It's written in kind of standard exponential form. They give us this little graphing tool where we can define these two points, and we can also define, uh, we can define a horizontal asymptote to construct our function. These three things are enough to graph an exponential if we know that it is an exponential function.

So let's think about it a little bit. The easiest thing that I could think of is, well, let's think about its initial value. Its initial value is going to be when ( x = 0 ). ( H(0) = 27 \cdot \left(\frac{1}{3}\right)^0 ), which is just 1, and so you're just left with ( 27 \cdot 1 ) or just 27. That's why we call this number here, when you write it in this form, you call this the initial value. So when ( x ) is equal to 0, ( H(0) = 27 ), and we're graphing ( y = H(x) ).

Now let's graph another point. So let's think about it a little bit. When ( x = 1 ), what is ( H(1) )? It's going to be ( \left(\frac{1}{3}\right)^1 ), which is just ( \frac{1}{3} ), and so ( \frac{1}{3} \cdot 27 ) is going to be 9. So when ( x = 1 ), ( H(1) = 9 ), and we can verify that.

Now let's just think about the asymptote. So what's going to happen here when ( x ) becomes really, really, really, really, really big? Well, if I take ( \left(\frac{1}{3}\right) ) to like a really large exponent, say to the 10th power, or to the 100th power, or to the 1000th power, this thing right over here is going to start approaching zero as ( x ) becomes much, much, much larger. So something that is approaching 0 times 27, well, that's going to approach 0 as well. So we're going to have a horizontal asymptote at 0.

You can verify that this works for more than just the two points we thought about. When ( x = 2 ), this is telling us that the graph ( y = H(x) ) goes through the point (2, 3). So ( H(2) ) should be equal to 3. You can verify that that is indeed the case. If ( x = 2 ), ( \left(\frac{1}{3}\right)^2 ) is ( \frac{1}{9} ), and ( \frac{1}{9} \cdot 27 = 3 ). We see that right over here when ( x = 2 ), ( H(2) = 3 ).

So I feel pretty good about that. Let's do another one of these. So graph the following exponential function. Same logic: when ( x = 0 ), the ( G(z) ) is just going to boil down to that initial value. So let me scroll down. The initial value is -30.

Now let's think about when ( x = 1 ). When ( x = 1 ), ( 2^1 ) is just 2, and so ( 2 \cdot (-30) = -60 ). So when ( x = 1 ), the value of the graph is -60.

Now let's think about this asymptote, where that should sit. So let's think about what happens when ( x ) becomes really, really, really, really, really negative. When ( x ) is really negative, ( 2^{-1} ) is ( \frac{1}{2} ), ( 2^{-2} ) is ( \frac{1}{4} ), and ( 2^{-3} ) is ( \frac{1}{8} ). As you get larger and larger negative values, or in another way, as ( x ) becomes more and more negative, ( 2 ) to that power is going to approach zero.

So (-30 \cdot) something approaching zero is going to approach zero. So this asymptote is in the right place. Our horizontal asymptote, as ( x ) approaches negative infinity, as we move further and further to the left, the value of the function is going to approach zero. We can see it kind of approaches zero from below. We can see that it approaches zero below because we already looked at the initial value, and we used that common ratio to find one point. Hopefully, you found that interesting.