Answer
maximum: $16$
Work Step by Step
Let's compare $f(x)=-3x^2+12x+4$ to $f(x)=ax^2+bx+c$.
We can see that $a=-3, b=12, c=4$.
$a\lt0$, hence the graph opens down, thus it's vertex is a maximum.
The maximum value is at $x=-\frac{b}{2a}=-\frac{12}{2\cdot(-3)}=2.$
Hence the maximum value is $f(2)=-3(2)^2+12(2)+4=16.$
