Answer
Absolute maximum $f(3)=-3$ and minimum $f(-2)=-\frac{19}{3}$. See graph.
Work Step by Step
Step 1. Given the function $f(x)=\frac{2}{3}x-5, -2\leq x\leq 3$, we have $f(-2)=\frac{2}{3}(-2)-5=-\frac{19}{3}, f(3)=\frac{2}{3}(3)-5=-3$, and we can identify the absolute maximum as $f(3)=-3$ and minimum as $f(-2)=-\frac{19}{3}$ on the given interval.
Step 2. See graph.