Answer
The function has a maximum value.
The maximum value is $\frac{133}{2}$.
Work Step by Step
Comparing $y=-\frac{1}{2}x^{2}-11x+6$ with $ax^{2}+bx+c$, we see that $a=-\frac{1}{2}$ and $b=-11$.
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
$x=-\frac{b}{2a}=-\frac{-11}{2(-\frac{1}{2})}=-11$
Evaluating the function at $x=-11$, we get the y-coordinate of the vertex as
$y=-\frac{1}{2}(-11)^{2}-11(-11)+6=\frac{133}{2}$
