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

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Answer 1 · 2018-03-02T13:50:04

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

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.