Answer:
[tex]\large\boxed{y=2(x+3)^2-6}[/tex]
Step-by-step explanation:
The vertex form of a parabola f(x) = ax² + bx + c:
[tex]y=a(x-h)^2+k[/tex]
(h, k) - vertex
[tex]h=\dfrac{-b}{2a}\\\\k=f(h)[/tex]
We have the equation:
[tex]y=2x^2+12x+14\\\\a=2,\ b=12,\ c=14\\\\h=\dfrac{-12}{2(2)}=\dfrac{-12}{4}=-3\\\\k=f(-3)=2(-3)^2+12(-3)+12=2(9)-36+12=18-36+12=-6[/tex]
Finally:
[tex]y=2(x-(-3))^2+(-6)=2(x+3)^2-6[/tex]
