Answer
See the graph
Work Step by Step
Given: $y=-(x-3)(x-7)$
$y=-x^2+10x-21$
The coefficients are $a=-1, b=10$, and $c=-21$. Because $a \lt0$, the parabola opens down.
Find the vertex:
$x=\frac{-b}{2a}=\frac{-(10)}{2.(-1)}=5$
then $y=-(5-3)(5-7)=4$
The vertex is $(5,4)$.
Draw the axis of symmetry $x=5$.
The y-intercept is $-8$. Plot the point $(0,-21)$. Then reflect this point in the axis of symmetry to plot another point, $(10,-21)$.
Evaluate the function for another value of x, such as $x=3$.
$y=-(3-3)(3-7)=0$
Plot the point $(3,0)$ and its reflection $(7,0)$ in the axis of symmetry.
Draw a parabola through the plotted points.
