Question 1162553



standard form: 

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

since given  vertex at ({{{-2}}}, {{{4}}}), we have

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

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


since passes through ({{{0}}},{{{ -4}}}), we have 

{{{(y-4)^2 =4p(x+2)}}}
{{{(-4-4)^2 =4p(0+2)}}}
{{{64=8p}}} 
{{{p=64/8}}} 
{{{p=8}}} 

so, your equation is:

{{{(y-4)^2 =4*8(x+2)}}}

{{{(y-4)^2 =32(x+2)}}}


{{{drawing( 600, 600, -10, 10, -10, 10, 
circle(-2,4,.12),circle(0,-4,.12),
locate(-2,4,v(-2,4)),locate(0,-4,p(0,-4)),
 graph( 600, 600, -10, 10, -10, 10, sqrt(32(x+2))+4,-sqrt( 32(x+2))+4)) }}}