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=(-b)/(2a),\ k=f(h)`

We have

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

Substitute:

`h=(-12)/((2)(2))=(-12)/(4)=-3\\\\k=f(-3)=2(-3)^2+12(-3)+16=2(9)-36+16=18-36+16=-2`

Answer: (-3, -2).