Question 1162555

given: Vertex at ({{{-4}}}, {{{-2}}}); passes through ({{{-3}}}, {{{0}}}) and ({{{9/4}}}, {{{3}}})



standard form: 


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

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

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

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


since passes through ({{{-3}}},{{{ 0}}}), we have 

{{{(0-(-2))^2 =4p(-3-(-4))}}} 
{{{(0+2)^2 =4p(-3+4)}}} 
{{{4 =4p(1)}}} 
{{{p=4/4}}} 
{{{p=1}}} 

so, your equation is:

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


and since passes through  ({{{9/4}}}, {{{3}}}), check if true:

{{{(3+2)^2 =4(9/4+4)}}}

{{{5^2 =4(9/4)+16}}}

{{{25 =9+16}}}

{{{25 =25}}}-> true


{{{drawing( 600, 600, -10, 10, -10, 10, 
circle(-4,-2,.12),circle(-3,0,.12),circle(9/4,3,.12),
locate(-4,-2,v(-4,-2)),locate(-3,-0.3,p(-3,0)),locate(9/4,3,p(9/4,3)),
 graph( 600, 600, -10, 10, -10, 10, sqrt(4(x+4))-2,-sqrt( 4(x+4))-2)) }}}