Answer:
[tex]y=2x^2+8x+11[/tex]
Explanation:
A quadratic equation in vertex form is generally given as;
[tex]y=a(x-h)^2+k[/tex]
where (h, k) is the coordinate of the vertex.
Given the coordinate (-2, 3), we'll have that;
h = -2
k = 3
Given a y-intercept of 11 and we know that at the y-intercept x = 0.
Substituting the above values into the vertex form equation and solving for a, we'll have;
[tex]\begin{gathered} 11=a\lbrack0-(-2)\rbrack^2+3 \\ 11=4a+3 \\ 4a=8 \\ a=\frac{8}{4} \\ a=2 \end{gathered}[/tex]
Substituting a = 2, h = -2 and k = 3 into the vertex form equation and taking it to standard form, we'll have;
[tex]\begin{gathered} y=2(x+2)^2+3 \\ y=2(x^2+4x+4)+3 \\ y=2x^2+8x+8+3 \\ y=2x^2+8x+11 \end{gathered}[/tex]
