Question

What is the vertex form of f(x)?
What is the minimum value of f(x)?

f(x)=x^2-6x+16

Answer 1

The vertex form of
`f(x)=ax^2+bx+c`

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

where:
`h=(-b)/(2a);\ k=f(h)`

We have:
`f(x)=x^2-6x+16\to a=1;\ b=-6;\ c=16`

substitute:

`h=(-(-6))/(2\cdot1)=(6)/(2)=3\\\\k=f(3)=3^2-6\cdot3+16=9-18+16=7`
The vertex form of f(x):

`f(x)=(x-3)^2+7`
The value of minimum is equal k.
Therefore:
`y_(min)=7`