Question 1135281


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

if the vertex at ({{{-3}}}, {{{5}}})=>{{{h=-3}}} and {{{k=5}}}

{{{y=a(x-(-3))^2+5}}}

{{{y=a(x+3)^2+5}}}

 and if passes through the point ({{{0}}},{{{ -14}}}) =>{{{x=0}}} and {{{y=-14}}}

{{{-14=a(0+3)^2+5}}}....solve or {{{a}}}

{{{-14=a(3)^2+5}}}

{{{-14-5=9a}}}

{{{a=-19/9}}}





your equation is: {{{y=(-19/9)(x+3)^2+5}}}


{{{drawing( 600, 600, -10,10, -20, 10,
circle(0,-14,.12), locate(0,-14,p(0,-14)),
circle(-3,5,.12), locate(-3,5,V(-3,5)),
 graph( 600, 600, -10,10, -20, 10, (-19/9)(x+3)^2+5)) }}}