Question 978905
In vertex form,
{{{y=a(x-(-4))^2+2}}}
{{{y=a(x+4)^2+2}}}
So then solving for the y-intercept,
{{{-30=a(0+4)^2+2}}}
{{{-30=16a+2}}}
{{{-32=16a}}}
{{{a=-2}}}
So then,
{{{y=-2(x+4)^2+2}}}
Solving for the x-intercept,
{{{0=-2(x+4)^2+2}}}
{{{-2=-2(x+4)^2}}}
{{{1=(x+4)^2}}}
{{{x+4=1}}} and {{{x+4=-1}}}
{{{x=-3}}} and {{{x=-5}}}
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*[illustration V10.JPG].