Question 1187363

There is more than one solution. 

Note that ({{{-3}}},{{{-1}}}) is on the line {{{y=x+2}}}, so the vertex can be ({{{h}}},{{{k}}})=({{{-3}}},{{{-1}}}). 
It doesn’t have to be there, but that is an easy solution.

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

using vertex form:

{{{y=a(x-h)+k}}..........plug in vertex coorinates and the leading coefficient

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


see the graph: 

first solution: {{{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 )) }}}



second solution: {{{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 )) }}}