Answer
$f(x)= \frac{1}{5}\cdot\left(3\right)^x$
Work Step by Step
Let $f(x)=a b^x$. We know from the graph that $f(-1)=a b^{-1}=\frac{1}{15}$ and $f(2)=a b^{2}=\frac{9}{5}$. So
$$
\begin{aligned}
\frac{a b^2}{a b^{-1}} & =\frac{9/5}{1/15} \\
b^3 & =27 \\
b & =\frac{27}{3}
b= 3
\end{aligned}
$$ and
$$ \begin{aligned}
& a b^{-1}=\frac{1}{15} \\
& a=b\cdot \frac{1}{15} \\
&=3\cdot \frac{1}{15} \\
& =\frac{1}{5}.
\end{aligned}
$$ Hence $$f(x)= \frac{1}{5}\cdot\left(3\right)^x$$
