Answer
$
f(x)=\sqrt{\frac{3}{5}}\cdot\left(\frac{1}{2}\sqrt{\frac{3}{5}}\right)^x
$
Work Step by Step
We want to have $f(x)=a b^{x}$. Give that the points $(-1,2),(1,0.3)$ lie on the curve of $f(x)$. We must have $f(-1)=a b^{-1}= 2$ and $f(1)=a b^{1}= 0.30 $
$$
\begin{aligned}
\frac{a b^{1}}{a b^{-1}} & =\frac{f(1)}{f(-1)} \\
b^{2} & =\frac{0.30}{2} \\
b & =\frac{1}{2}\left(\sqrt{\frac{3}{5}}\right)\end{aligned}
$$ Determine $a$:
$$
\begin{aligned}
ab^{-1}& =2 \\
a & =2b= 2\cdot\frac{1}{2}\left(\sqrt{\frac{3}{5}}\right) \\
a & =\sqrt{\frac{3}{5}}
\end{aligned}
$$ It follows that
$$
f(x)=\sqrt{\frac{3}{5}}\cdot\left(\frac{1}{2}\sqrt{\frac{3}{5}}\right)^x
$$
