Question 1162554

standard form: 

{{{(y-k)^2 =4p(x-h)}}} where {{{h}}} and {{{k}}} are coordinates of vertex

since given  vertex at ({{{-3}}}, {{{-3}}}), we have

{{{(y-(-3))^2 =4p(x-(-3))}}} 

{{{(y+3)^2 =4p(x+3)}}} .........eq.1


since passes through ({{{0}}},{{{ 0}}}), we have 

{{{(y+3)^2 =4p(x+3)}}}
{{{(0+3)^2 =4p(0+3)}}}
{{{9 =12p}}} 
{{{p=9/12}}} 
{{{p=3/4}}} 

so, your equation is:

{{{(y+3)^2 =4(3/4)(x+3)}}}

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


{{{drawing( 600, 600, -10, 10, -10, 10, 
circle(-3,-3,.12),circle(0,0,.13),
locate(-3,-3,v(-3,-3)),locate(0.2,0.5,p(0,0)),

 graph( 600, 600, -10, 10, -10, 10, sqrt(3(x+3))-3,-sqrt( 3(x+3))-3)) }}}