#### Answer

The graph has a maximum value of $18$.

#### Work Step by Step

RECALL:
(1) The graph of the quadratic function $f(x)=ax^2=bx+c$ opens:
(i) upward when $a \gt0$ and has its vertex as its minimum; or
(ii) downward when $a\lt0$ and has its vertex as its maximum.
(2) The vertex of the quadratic function $f(x)=ax^2=bx+c$ is at $\left(-\frac{b}{2a}, f(-\frac{b}{2a})\right)$
Let's compare $f(x)=-2x^2+12x$ to $f(x)=ax^2+bx+c$.
We can see that $a=-2$, $b=12$, $c=0$.
$a\gt0$, hence the graph opens down, hence it's vertex is a maximum.
The maximum value is at $x=-\frac{b}{2a}=-\frac{12}{2\cdot (-2)}=3.$
Thus, the maximum value is $f(3)=-2(3)^2+12(3)=18.$