Answer
The function has a maximum value.
The maximum value is $-1$.
Work Step by Step
Comparing $f(x)=-4x^{2}+4x-2$ with $ax^{2}+bx+c$, we see that $a=-4$ and $b=4$.
As $a\lt0$, the parabola opens down and the function has a maximum value. The maximum value is the y-coordinate of the vertex.
The x-coordinate of the vertex is given by
$x=-\frac{b}{2a}=-\frac{4}{2(-4)}=\frac{1}{2}$
Evaluating the function at $x=\frac{1}{2}$, we get the y-coordinate of the vertex as
$f(\frac{1}{2})=-4(\frac{1}{2})^{2}+4(\frac{1}{2})-2$
$=-1$
