Question 966396
the vertex of this parabola is at ({{{-3}}}, {{{-1}}})
when the y-value is {{{0}}}, the x-value is {{{4}}} 

so, you are given two points that lie on a parabola and we can use them to find equation

since one point is a vertex, use vertex form:

{{{y=a(x-h)^2+k}}} where {{{h=-3}}} and {{{k=-1}}}

{{{y=a(x-(-3))^2-1}}} 
{{{y=a(x+3)^2-1}}} 
{{{y=a(x^2+6x+9)-1}}} 
{{{y=ax^2+6ax+9a-1}}} 

use the y-value is {{{0}}}, the x-value is {{{4}}} 
{{{0=a*4^2+6*4+9*4-1}}} 
{{{0 = 49a-1}}} 
{{{1=49a}}} 
{{{1/49=a}}} the coefficient of the squared term in the parabola's equation 

and your equation is:

{{{y=(1/49)(x+3)^2-1}}} 


{{{drawing( 600, 600, -20, 20, -10, 20,
circle(-3,-1,.15),locate(-3,-1.3,V(-3,-1)),
circle(4,0,.15),locate(4,0.7,p(4,0)),
 graph( 600, 600, -20, 20, -10, 20, (1/49)(x+3)^2-1)) }}}