Q. 90 - Precalculus Enhanced with Graphing Utilities

Question

Find an exponential function with horizontal asymptote $y=-3$ whose graph contains the points $(0,-2)$ and $(-2,1)$.

Step-by-Step Solution

Answer

The exponential functions are ${\left(\frac{1}{2}\right)}^x-3$ and ${(-\frac{1}{2})}^x-3$.

Step 1. Using given information.

Consider the given horizontal asymptote $y=-3$.

And given the two points on the exponential function are $(0,-2)$ and $(-2,1)$.

If the horizontal asymptote is $y=0$, then the exponential function is $y=a^x$.

If the horizontal asymptote is $y=-3$, then the exponential function is $y=a^x-3$.

Step 2. Finding exponential function.

At $(-2,1)$, the exponential function is

$$\begin{gathered} 1=a^{-2}-3 \\ 1+3=a^{-2} \\ {a}^{-2}=4 \\ \frac{1}{a^2}=4 \\ {a}^2=\frac{1}{4} \\ a=\pm \frac{1}{2} \end{gathered}$$

So the exponential functions are $y={\left(\frac{1}{2}\right)}^x-3$ and $y={(-\frac{1}{2})}^x-3$.