Question 1176633


{{{f (x) = a(x - h)^2 + k}}}, where ({{{h}}}, {{{k}}}) is the vertex of the parabola 

if given  vertex at the origin ​(​{{{0}}},​{{{0}}}) then {{{h=0}}} and{{{k=0}}}

 {{{f (x) = a(x - 0)^2 + 0}}}
 {{{f (x) = ax^2 }}}


if passes through the point ​({{{-6}}}​,{{{-3}}}) we have

{{{-3= a(-6)^2 }}}
{{{-3= 36a }}}
{{{a=-3/36}}}
{{{a=-1/12}}}

and your equation is:

{{{f (x) = -(1/12)x^2 }}}


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-6,-3,.12),locate(-6,-3,p(-6,-3)),
 graph( 600, 600, -10, 10, -10, 10, -(1/12)x^2)) }}}