What is the vertex form of #y=x^2-2x+6#?

1 Answer
Aviv S.
Mar 2, 2018

In vertex form, the parabola's equation is #y=(x-1)^2+5#.

Explanation:

To convert a parabola in standard form to vertex form, you have to make a squared binomial term (i.e. #(x-1)^2# or #(x+6)^2#).

These squared binomial terms -- take #(x-1)^2#, for example -- (almost) always expand to have #x^2#, #x#, and constant terms. #(x-1)^2# expands to be #x^2-2x+1#.

In our parabola:

#y=x^2-2x+6#

We have a part that looks similar to the expression we wrote before: #x^2-2x+1#. If we rewrite our parabola, we can "undo" this squared binomial term, like this:

#y=x^2-2x+6#

#color(white)y=color(red)(x^2-2x+1)+5#

#color(white)y=color(red)((x-1)^2)+5#

This is our parabola in vertex form. Here's its graph:

graph{(x-1)^2+5 [-12, 13.7, 0, 13.12]}

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