Answer
The function has a maximum value.
Maximum value is $10$.
Work Step by Step
Comparing $h(x)=-\frac{1}{4}x^{2}+3x+1$ with $ax^{2}+bx+c$, we see that $a=-\frac{1}{4}$ and $b=3$.
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
$x=-\frac{b}{2a}=-\frac{3}{2(-\frac{1}{4})}=6$
Evaluating the function at $x=6$, we get the y-coordinate of the vertex as
$h(6)=-\frac{1}{4}(6)^{2}+3(6)+1=10$
