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

User Lukap
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1 Answer

3 votes

Answer:


\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)

User Meez
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