Answer:
[tex]\large\boxed{y=3(x-2)^2-8}[/tex]
Step-by-step explanation:
The vertex form of an equation of a parabola y = ax² + bx + c:
[tex]f(x)=a(x-h)^2+k[/tex]
(h, k) - vertex
[tex]h=\dfrac{-b}{2a},\ k=f(h)[/tex]
We have the equation:
[tex]y=3x^2-12x+4\\\\a=3,\ b=-12,\ c=4[/tex]
Substitute:
[tex]h=\dfrac{-(-12)}{2(3)}=\dfrac{12}{6}=2\\\\k=f(2)=3(2^2)-12(2)+4=3(4)-24+4=12-24+4=-8[/tex]
Finally:
[tex]y=3(x-2)^2+(-8)=3(x-2)^2-8[/tex]
