Question

Explain how to find the vertex of 2x^2+12x+16

Answer 1

The vertex form:

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

Use $(a+b)^2=a^2+2ab+b^2\qquad(*)$

$2x^2+12x+16=2(x^2+6x)+16=2(x^2+2(x)(3))+16\\\\=2(\underbrace{x^2+2(x)(3)+3^2}_{(*)}-3^2)+16=2((x+3)^2-9)+16\\\\=2(x+3)^2+(2)(-9)+16=2(x+3)^2-18+16\\\\=\boxed{2(x-(-3))^2-2}\to h=-3,\ k=-2$

Answer: (-3, -2).

Other method:

$f(x)=ax^2+bx+c\\\\vertex:(h,\ k)\\\\h=\dfrac{-b}{2a},\ k=f(h)$

We have

$f(x)=2x^2+12x+16\\\\a=2,\ b=12,\ c=16$

Substitute:

$h=\dfrac{-12}{(2)(2)}=\dfrac{-12}{4}=-3\\\\k=f(-3)=2(-3)^2+12(-3)+16=2(9)-36+16=18-36+16=-2$

Answer: (-3, -2).