Answer
$y=0.5^x-3$
Work Step by Step
The general formula for an exponential function: $Ca^x+b=y$.
The horizontal asymptote is $b$.
Hence $b=-3$.
Thus, the tentative equation is $y=Ca^x-3$.
We use the two given points to find the values of $C$ and $a$.
Using $(0,-2)$ gives:
$y=Ca^x-3\\
-2=C\cdot a^0 -3\\
-2=C \cdot 1 -3\\
-2=C - 3\\
-2+3=C\\
1=C$
Thus, the tentative equation is no $y=1 \cdot a^x -3 \longrightarrow y=a^x-3$.
Use $(-2, 1)$ to obtain:
$y=a^x-3\\
1=a^{-2}-3\\
1=\frac{1}{a^2} - 3\\
1+3=\frac{1}{a^2}\\
4=\frac{1}{a^2}\\
4\cdot a^2=1\\
4a^2=1\\
a^2=\frac{1}{4}\\
\sqrt{a^2} = \pm\sqrt{\frac{1}{4}}\\
a=\pm \frac{1}{2}$
Since $a$ cannot be negative, then $a=\frac{1}{2}=0.5$
Thus, the equation is $y=0.5^x-3$.
