Answer
a) $f(x)=\frac{31}{8} x+\frac{49}{4}$
b) $f(x)=5\cdot (2)^x$
Work Step by Step
a) Let $f(x)=b+m x$, where $m$, the slope, is given by:
$$
m=\frac{\Delta y}{\Delta x}=\frac{f(2)-f(-3)}{(2)-(-3)}=\frac{20-\frac{5}{8}}{5}=\frac{\frac{155}{8}}{5}=\frac{31}{8}
$$
and
$$
\begin{aligned}
& 20=b+m(2) \\
& 20=b+\left(\frac{31}{8}\right)(2) \\
& 20=b+\frac{31}{4} \\
& b=20-\frac{31}{4}=\frac{49}{4} .
\end{aligned}
$$
Hence
$$
f(x)=\frac{31}{8} x+\frac{49}{4}
$$
b) Let $f(x)=a b^x$. We know that $f(2)=a b^2= 20$ and $f(-3)=a b^{-3}=\frac{5}{8}$. So
$$
\begin{aligned}
& \frac{f(2)}{f(-3)}=\frac{a b^2}{a b^{-3}}=\frac{20}{\frac{5}{8}} \\
& b^5=20 \cdot \frac{8}{5}=32 \\
& b=2 .
\end{aligned}
$$
and $f(2)=a 2^2= 20$ or
$$
\begin{aligned}
20 & =a(2)^2 \\
20 & =4 a \\
a & =5 .
\end{aligned}
$$
Hence
$f(x)=5\cdot (2)^x$
