Question

Find the axis of symmetry of 2x^2 + 12x + 16 Show your work

Answer 1

The vertex form:

$y=a(x-h)^2+k\\\\(h,\ k)-vertex$

The axis of symetry is x = h.

$y=ax^2+bx+c\to h=\dfrac{-b}{2a}$

We have

$y=2x^2+12x+16\to a=2,\ b=12,\ c=16$

Substitute:

$h=\dfrac{-12}{(2)(2)}=\dfrac{-12}{4}=-3$

Answer: x = -3