Answer
$f(x)=\frac{2^{29/6}}{5}((0.25)^{5/6})^x$
Work Step by Step
We want to have $f(x)=a b^x$.
The graph shows that the points $(2.3,0.4),(3.5,0.10)$ lie on the curve of $f(x)$.
We must have $f(2.3)=a b^{2.3}=0.40$ and $f(3.5)=a b^{3.5}=0.10$. This gives
$$
\begin{aligned}
\frac{a b^{3.5}}{a b^{2.3}} & =\frac{f(3.5)}{f(2.3)} \\
b^{1.2} & =\frac{0.1}{0.4} \\
b & =(0.25)^{1 / 1.2}=(0.25)^{5/6}\approx 0.31498
\end{aligned}
$$ We determine $a$:
$$
\begin{aligned}
a b^{2.3}& =0.40\\
a & =\frac{0.40}{b^{2.3}} \\
a & =\frac{0.40}{((0.25)^{5/6})^{2.3}}\\
&=\frac{0.40}{(0.25)^{23/12}}\\
&=0.40(2^{23/6})\\
&=\frac{2^{29/6}}{5}\\
&\approx 5.70175
\end{aligned}
$$
It follows that
$$
f(x)=\frac{2^{29/6}}{5}((0.25)^{5/6})^x
$$
