Question

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

Answer 1

Answer:

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

Step-by-step explanation:

METHOD 1:

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

The coordinates of the vertex:

$(h;\ k)\to h=\dfrac{-b}{2a};\ k=f(h)=\dfrac{-(b^2-4ac)}{4a}$

We have

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

Substitute:

$h=\dfrac{-(-12)}{2(1)}=\dfrac{12}{2}=6\\\\k=f(6)=6^2-12(6)+5=36-72+5=-31$

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 $(a\pm b)^2=a^2\pm2ab+b^2$

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