Question 389266
{{{y=a(x+h)^2+k}}}
Using the vertex,
{{{y=a(x-1)^2-108}}}
Using the y-intercept,
{{{-105=a(0-1)^2-108}}}
{{{a=3}}}
So then,
{{{y=3(x-1)^2-108}}}
Find {{{x}}} when {{{y=0}}}
{{{3(x-1)^2-108=0}}}
{{{3(x-1)^2=108}}}
{{{(x-1)^2=36}}}
{{{(x-1)=0 +- 6}}}
{{{x=1 +- 6}}}
{{{highlight(x=-5)}}} and {{{highlight(x=7)}}}