Question 1028757
The "parent equation" for simple parabolas is 

{{{y=x^2}}}


Using vertex form


{{{y-k=a(x-h)^2}}} where h,k is the vertex, we have:


{{{y-1=a(x+1)^2}}}


now, we have a standard parabola, shifted to get your vertex. But, if you enter 0 for X, you get 2, not the intersection 3 you need. This is where 'a' comes in. It will leave the vertex alone, but transform the shape to 'pull up' the parabola. We want (0,3) to be a solution.

{{{3-1=a(0+1)^2}}}  


{{{2=a(1)^2}}}   so a=2


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


and you rearrange to


{{{y-1=2x^2+2x+1}}}


{{{y-1=2x^2+4x+2}}}


{{{y=2x^2+4x+3}}}