Question 605216
When the equation is of the form
{{{ ax^2 + bx + c }}}, the vertex is at
{{{ x[v] = -b/(2a) }}}
given:
{{{ -b/(2a) = -2 }}}
{{{ -b = -4a }}}
{{{ b = 4a }}}
and 
{{{ y[v] = -4 }}}
{{{ -4 = a*(-2)^2 + b*(-2) + c }}}
{{{ -4 = 4a - 2b + c }}}
and, from above,
{{{ -4 = 4a - 2*(4a) + c }}}
{{{ -4 = 4a - 8a + c }}}
{{{ c = 4a - 4 }}}
So far, I have
{{{ y = ax^2 + (4a)*x + 4a - 4 }}}
and this must go through (0,0), so
{{{ 0 = a*0 + (4a)*0 + 4a - 4 }}}
{{{ 4a = 4 }}}
{{{ a = 1 }}}
{{{ b = 4a }}}
{{{ b = 4*1 }}}
{{{ b = 4 }}}
{{{ c = 4a - 4 }}}
{{{ c = 4*1 - 4 }}}
{{{ c = 0 }}}
So,  the equation is:
{{{ y = x^2 + 4x }}}
here's a plot:
{{{ graph( 400, 400, -6, 10, -6, 10, x^2 + 4x ) }}}