Answer
Please see the graph.
Work Step by Step
$h(x) = -(x-3)^2+2$
A graph of the form $a(x-h)^2+k$ has the vertex $(h,k)$. Thus, this graph has its vertex at $(3,2)$. Since the coefficient of the $x^2$ term is negative, we know the graph opens down.
We pick $x=0$ and $x=5$ to find two other points for the graph.
$x=0$
$h(x) = -(x-3)^2+2$
$h(x) = -(x-3)^2+2$
$h(0) = -(0-3)^2+2$
$h(0) = -(3^2)+2$
$h(0)= -9+2$
$h(0)=-7$
$x=5$
$h(x) = -(x-3)^2+2$
$h(5) = -(5-3)^2+2$
$h(5) = -(2^2)+2$
$h(5) = -4+2$
$h(5)=-2$
$(0, -7)$ and $(5,-2)$ are also on the graph, as well as the vertex of $(3,2)$.
