Answer
$\frac{17}{2}.$
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 $a=-\frac{2}{3}, b=6, c=-5.$
The maximum value is at $x=-\frac{b}{2a}$. (From Chapter 3-3. Page 142.) Hence here:$-\dfrac{6}{2\cdot-\frac{2}{3}}=\frac{9}{2}.$
Hence the maximum value is $f(\frac{9}{2})=-\frac{2}{3}(\frac{9}{2})^2+6(\frac{9}{2})-5=-\frac{2}{3}\cdot\frac{81}{4}+27-5=-\frac{27}{2}+22=\frac{17}{2}.$
