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Section 12.2 Graphs and End Behavior of Exponential Functions

Now that we have looked at the formula for exponential functions, we will spend this section exploring the graphs, including the end behavior

Looking at the two graphs, we can see that the end behavior is different from what we've seen before: it has a horizontal asymptote on one side, while the other blows up to \(\infty\text{.}\)

Example 12.21.

Suppose we want to find the end behavior of \(f(x)=7(1.56)^x\text{.}\) Since \(b=1.56\gt 1\text{,}\) we can see that this is exponential growth. Based on the shape of the exponential growth graph, we have the following end behavior:

  • As \(x\rightarrow -\infty\text{,}\) \(y\rightarrow 0\text{.}\)

  • As \(x\rightarrow \infty\text{,}\) \(y\rightarrow \infty\text{.}\)

Checkpoint 12.22.

Find the end behavior of the function \(f(x)=\frac{1}{2}(0.01)^x\text{.}\)

Answer.
  • As \(x\rightarrow -\infty\text{,}\) \(y\rightarrow \infty\text{.}\)

  • As \(x\rightarrow \infty\text{,}\) \(y\rightarrow 0\text{.}\)

Solution.

Since \(b=0.01\lt 1\text{,}\) we can see that this is exponential decay. Based on the shape of the exponential decay graph, we have the following end behavior:

  • As \(x\rightarrow -\infty\text{,}\) \(y\rightarrow \infty\text{.}\)

  • As \(x\rightarrow \infty\text{,}\) \(y\rightarrow 0\text{.}\)

Example 12.23.

Suppose we want to find the end behavior of \(f(x)=7(1.56)^{x+3}-2\text{.}\) Since \(b=1.56\gt 1\text{,}\) we can see that this is exponential growth. So, we know what the shape of the graph looks like, but notice that we have some transformations here. The \(+3\) is a horizontal shift, which won't affect our end behavior, so we can ignore that. However, the \(-2\) shifts the whole graph down. That won't affect the side that goes to \(\infty\) (since \(\infty-2\) is still \(\infty\)), but it will move our horizontal asympote down 2. So, we have the following end behavior:

  • As \(x\rightarrow -\infty\text{,}\) \(y\rightarrow -2\text{.}\)

  • As \(x\rightarrow \infty\text{,}\) \(y\rightarrow \infty\text{.}\)

Checkpoint 12.24.

Find the end behavior of the function \(f(x)=\frac{1}{2}(0.01)^{x-4}+9\text{.}\)

Answer.
  • As \(x\rightarrow -\infty\text{,}\) \(y\rightarrow \infty\text{.}\)

  • As \(x\rightarrow \infty\text{,}\) \(y\rightarrow 9\text{.}\)

Solution.

Since \(b=0.01\lt 1\text{,}\) we can see that this is exponential decay. So, we know what the shape of the graph looks like, but notice that we have some transformations here. The \(-4\) is a horizontal shift, which won't affect our end behavior, so we can ignore that. However, the \(+9\) shifts the whole graph up. That won't affect the side that goes to \(\infty\) (since \(\infty+9\) is still \(\infty\)), but it will move our horizontal asympote up 9. So, we have the following end behavior:

  • As \(x\rightarrow -\infty\text{,}\) \(y\rightarrow \infty\text{.}\)

  • As \(x\rightarrow \infty\text{,}\) \(y\rightarrow 9\text{.}\)