#### Answer

maximum: $12$

#### Work Step by Step

Let's compare $f(x)=-x^2+8x-4$ to $f(x)=ax^2+bx+c$.
We can see that a=-1, b=8, c=-4.
$a\lt0$, hence the graph opens down, thus it's vertex is a maximum.
The maximum value is at $x=-\frac{b}{2a}=-\frac{8}{2\cdot(-1)}=4.$
Hence the maximum value is $f(2)=-4^2+8(4)-4=12.$