Question

Rewrite the quadratic function in vertex form. Y=3x^2-12x+4

Answer 1

Answer:

$\large\boxed{y=3(x-2)^2-8}$

Step-by-step explanation:

The vertex form of an equation of a parabola y = ax² + bx + c:

$f(x)=a(x-h)^2+k$

(h, k) - vertex

$h=\dfrac{-b}{2a},\ k=f(h)$

We have the equation:

$y=3x^2-12x+4\\\\a=3,\ b=-12,\ c=4$

Substitute:

$h=\dfrac{-(-12)}{2(3)}=\dfrac{12}{6}=2\\\\k=f(2)=3(2^2)-12(2)+4=3(4)-24+4=12-24+4=-8$

Finally:

$y=3(x-2)^2+(-8)=3(x-2)^2-8$