# What is the vertex form of y= x^2 -6x+8 ?

Feb 26, 2016

$y = {\left(x - 3\right)}^{2} + \left(- 1\right)$

#### Explanation:

The general vertex form is
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - a\right)}^{2} + b$ for a parabola with vertex at $\left(a , b\right)$

To convert $y = {x}^{2} - 6 x + 8$ into vertex form, perform the process called "completing the square":

For a squared binomial ${\left(x + k\right)}^{2} = \textcolor{b l u e}{{x}^{2} + 2 k x} + {k}^{2}$
So if $\textcolor{b l u e}{{x}^{2} - 6 x}$ are the first two terms of an expanded squared binomial, then $k = - 3$ and the third term must be ${k}^{2} = 9$

We can add $9$ to the given expression to "complete the square", but we we also need to subtract $9$ so that the value of the expression stays the same.

$y = {x}^{2} - 6 x \textcolor{red}{+ 9} + 8 \textcolor{red}{- 9}$

$y = {\left(x - 3\right)}^{2} - 1$
or, in explicit vertex form:
$y = 1 {\left(x - 3\right)}^{2} + \left(- 1\right)$

Typically I leave the value $m$ off when it is $1$ (the default anyway) but find that writing the constant term as $+ \left(- 1\right)$ helps me remember that the $y$ coordinate of the vertex is $\left(- 1\right)$