Question

What are the coordinates of the vertex of the function f(x) = x2 -12x +5?

Answer 1

`\huge\boxed{(6;\ -31)}`

Explanation:

METHOD 1:

Let:
`f(x)=ax^2+bx+c`.

The coordinates of the vertex:

`(h;\ k)\to h=(-b)/(2a);\ k=f(h)=(-(b^2-4ac))/(4a)`

We have

`f(x)=x^2-12x+5\to a=1;\ b=-12;\ c=5`

Substitute:

`h=(-(-12))/(2(1))=(12)/(2)=6\\\\k=f(6)=6^2-12(6)+5=36-72+5=-31`

METHOD 2:

The vertex form of an equation of a quadratic function:

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

We have:

`f(x)=x^2-12x+5\to a=1`

Complete to the square
`(a\pm b)^2=a^2\pm2ab+b^2`

`x^2-12x+5=x^2-\underbrace{2(x)(6)}_(12x)+5=\underbrace{x^2-2(x)(6)+6^2}_(a^2-2ab+b^2)-6^2+5\\\\=\underbrace{(x-6)^2}_((a-b)^2)-36+5=(x-6)^2-31\\\\h=6;\ k=-31\to(6;\ -31)`