Question 1181909


use vertex form of the equation for a parabola

{{{y=a(x-h)^2+k}}}

given

x-intercepts at {{{-3}}} and {{{1}}}=>({{{-3}}},{{{0}}}) and ({{{1}}},{{{0}}})
vertex point at ({{{- 1}}}, {{{-8}}})=>{{{ h=-1}}}, {{{k=-8}}}

{{{y=a(x-(-1))^2+(-8)}}}
{{{y=a(x+1)^2-8}}}........use x-intercept to calculate {{{a}}}
{{{0=a(-3+1)^2-8}}}
{{{0=4a-8}}}
{{{4a=8}}}
{{{a=2}}}

your equation is

{{{y=2(x+1)^2-8}}}


{{{ graph( 600, 600, -10, 10, -10, 10, 2(x+1)^2-8) }}}