Question 963559
In vertex form,
{{{y=a(x+3)^2+16}}}
Use the y-intercept,
{{{-20=a(0+3)^2+16}}}
{{{-20=9a+16}}}
{{{9a=-36}}}
{{{a=-4}}}
So then,
{{{y=-4(x+3)^2+16}}}
For x-intercepts, set {{{y=0}}} and solve for {{{x}}},
{{{0=-4(x+3)^2+16}}}
{{{4(x+3)^2=16}}}
{{{(x+3)^2=4}}}
{{{x+3=0 +- 2}}}
{{{x=3 +- 2}}}
{{{x=1}}} and {{{x=5}}}
(1,0) and (5,0)