Answer
$g(x)=\left\{
\begin{array}{cc}
x-1,& \quad \text{if }x\lt 3\\
-x+5, & \quad \text{if }x \geq 3
\end{array}
\right.$
Work Step by Step
The absolute value function $g(x)=a|x-h|+k$ can be written as a piecewise function.
$g(x)=\left\{
\begin{array}{cc}
a[-(x-h)]+k,& \quad \text{if }x-h\lt 0 \\
a(x-h)+k, & \quad \text{if }x-h \geq 0
\end{array}
\right.$
Given $y=g(x)=-|x-3|+2$
$\implies a=-1, h=3$ and $k=2$
Then, $g(x)$ as a piecewise function is
$g(x)=\left\{
\begin{array}{cc}
-1[-(x-3)]+2,& \quad \text{if }x-3\lt 0 \\
-1(x-3)+2, & \quad \text{if }x-3 \geq 0
\end{array}
\right.$
Simplifying, we get
$g(x)=\left\{
\begin{array}{cc}
x-1,& \quad \text{if }x\lt 3\\
-x+5, & \quad \text{if }x \geq 3
\end{array}
\right.$