Answer :
Given:
Vertex at (-1,-2)
Goes through point (0,1)
The vertex form of a quadratic equation is defined as
[tex]\begin{gathered} y=a(x-h)^2+k \\ \text{where} \\ (h,k)\text{ is the vertex of the equation} \end{gathered}[/tex]Substitute (h,k) with the vertex (-1,-2)
[tex]\begin{gathered} y=a(x-h)^2+k \\ y=a(x-(-1))^2+(-2) \\ y=a(x+1)^2-2 \end{gathered}[/tex]Now we solve for a, by substituting (x,y) with (0,1) to the vertex form of the previous equation.
[tex]\begin{gathered} y=a(x+1)^2-2 \\ 1=a(0+1)^2-2_{} \\ 1=a(1)^2-2 \\ 1=a(1)-2 \\ 1=a-2 \\ 1+2=a \\ a=3 \end{gathered}[/tex]Putting it together with the vertex form we have
[tex]y=3(x+1)^2-2[/tex]Expand the equation, and change it into standard form
[tex]\begin{gathered} y=3(x+1)^2-2 \\ y=3(x^2+2x+1)-2 \\ y=3x^2+6x+3-2 \end{gathered}[/tex]Simplify further, and the equation of the parabola in standard form is
[tex]y=3x^2+6x+1[/tex]