Question 1190857


There is more than one solution. 

Note that ({{{-3}}},{{{-1}}}) is on the line {{{y=x+2}}}:
{{{-1=-3+2}}}
{{{-1=-1}}}->true

so the vertex can be ({{{-3}}},{{{-1}}}). 

It doesn’t have to be there, but that is an easy solution.

use vertex formula

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

Length of latus rectum is {{{6}}}, so the leading coefficient is ±{{{1/6}}}

 substitute the coordinates the vertex and the leading coefficient

{{{y = (1/6)(x+3)^2-1 }}}

or 

{{{y = -(1/6)(x+3)^2-1}}}


graph: {{{y = (1/6)(x+3)^2-1 }}}

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


graph: {{{y = -(1/6)(x+3)^2-1 }}}

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