Answer
$8.5.$
Work Step by Step
Let's compare $f(x)=-\frac{2}{3}x^2+6x-5$ to $f(x)=ax^2+bx+c$. We can see that $a=-\frac{2}{3}, b=6, c=-4$. $a\lt0$, hence the graph opens down, hence its vertex is a maximum. The maximum value is at $x=-\frac{b}{2a}=-\frac{6}{2\cdot(-\frac{2}{3})}=4.5.$ Hence the maximum value is $f(4.5)=-\frac{2}{3}(4.5)^2+6(4.5)-5=-13.5+27-5=8.5.$
