Answer
$\frac{17}{2}$
Work Step by Step
Given $f(x)=-\frac{2}{3}x^2+6x-5$, we have $a=-\frac{2}{3}\lt0, b=6$, thus there is a maximum at $x=-\frac{b}{2a}=-\frac{6}{2(-\frac{2}{3})}=\frac{9}{2}$ with $f(\frac{9}{2})=-\frac{2}{3}(\frac{9}{2})^2+6(\frac{9}{2})-5=\frac{17}{2}$
