Answer
See graph
Work Step by Step
Finding the vertex of the graph of the parabola:
$$x=-\frac{b}{2a}=-\frac{10}{2(-1)}=5$$ $$h(5)=-(5)^2+10(5)-21=4$$
Thus, the vertex is at point $(5,4)$.
Finding the $x$-intercepts, set $h(x)=0$:
$$0=-x^2+10x-21$$ $$0=x^2-10x+21$$ $$0=(x-3)(x-7)$$ $$x-3=0$$ $$x=3$$ $$x-7=0$$ $$x=7$$
Thus, two $x$intercepts are at $(3,0)$ and $(7,0)$.
Plotting the three points, and drawing a curve on the points, the sketch of the graph of the function is as shown.
