# Chapter 2 - 2.2 - Functions - 2.2 Exercises - Page 183: 33

$5;\dfrac{9}{2}; 4; 1; 0$

#### 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:

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