Answer
$5;\dfrac{9}{2}; 4; 1; 0$
Work Step by Step
We are given the function:
$f(x)=\begin{cases}
-\dfrac{1}{2}x+4,x\leq 0\\
(x-2)^2,x\gt 0
\end{cases}$
Compute $f(x)$ for $x=-2,-1,0,1,2$:
$f(-2)=-\dfrac{1}{2}(-2)+4=1+4=5$
$f(-1)=-\dfrac{1}{2}(-1)+4=\dfrac{1}{2}+4=\dfrac{9}{2}$
$f(0)=-\dfrac{1}{2}(0)+4=0+4=4$
$f(1)=(1-2)^2=1$
$f(2)=(2-2)^2=0$
Place the values in a table:
