How do you write the equation in vertex form y=x^2-6x+11?

1 Answer
Mar 2, 2018

1(x-3)^2+2

Explanation:

A quadratic equation, for reference, is:

ax^2+bx+c=0

Vertex form of a quadratic:

a(x-h)^2+k

where the vertex is (h,k)

Here we can convert x^2-6x+11 into vertex form by taking the perfect square of x^2-6x:

x^2-6x+9+11-9

(to find a perfect square, we halve the coefficient of bx, which is b, and then square the answer)

Here, we halved 6, which equals 3. Then, we squared 3, which equals 9.

When we add 9, remember that we must also subtract 9 at the end to keep the equation correct.

(x^2-6x+9)+2

Here is a perfect square. All we need to do is halve the coefficient, add it with x, and square the whole thing.

(x^2-6x+9)+2=(x-3)^2+2

To finish it in the vertex form:

1(x-(+3))^2+2

The vertex is (h,k), so here the vertex will be:

(3,2)

Thus, solved.