Question 880112
The zeros appear to be -3, and 1, and vertex (-1,-4).


Using a factored form, {{{y=a(x+3)(x-1)}}} and you can use the vertex to find a.  Solving symbolically for a gives {{{a=y/((x+3)(x-1))}}}; substitute the values of the vertex:
{{{a=-4/((-1+3)(-1-1))}}}
{{{a=-4/(2(-2))}}}
{{{a=1}}}


The equation is {{{y=1(x+3)(x-1)}}}; so although you SAY you want standard form, you showed the general form.  If you want general form, then just perform the multiplication.  {{{highlight(y=x^2+2x-3)}}}, GENERAL FORM.


Did you really want standard form, as {{{y=a(x-h)^2+k}}} ?