# What is the vertex form of y= x^2-6x+5?

Dec 17, 2015

$y = {\left(x - 3\right)}^{2} + \left(- 4\right)$ with vertex at $\left(3 , - 4\right)$

#### Explanation:

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

Given $y = {x}^{2} - 6 x + 5$

We can "complete the square"
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} - 6 x \textcolor{red}{+ {3}^{2}} + 5 \textcolor{red}{- {3}^{2}}$

$\textcolor{w h i t e}{\text{XXX}} y = {\left(x - 3\right)}^{2} - 4$

Dec 17, 2015

$y = {\left(x - 3\right)}^{2} - 4$

#### Explanation:

To find the vertex form of the equation, we have to complete the square:

$y = {x}^{2} - 6 x + 5$

$y = \left({x}^{2} - 6 x\right) + 5$

When completing the square, we must make sure the bracketed polynomial is a trinomial. So $c$ is ${\left(\frac{b}{2}\right)}^{2}$.
$y = \left({x}^{2} - 6 x + {\left(\frac{6}{2}\right)}^{2} - {\left(\frac{6}{2}\right)}^{2}\right) + 5$

$y = \left({x}^{2} - 6 x + {\left(3\right)}^{2} - {\left(3\right)}^{2}\right) + 5$

$y = \left({x}^{2} - 6 x + 9 - 9\right) + 5$

Multiply $- 9$ by the $a$ value of $1$ to bring $- 9$ outside of the brackets.
$y = \left({x}^{2} - 6 x + 9\right) + 5 - \left(9 \cdot 1\right)$

$y = {\left(x - 3\right)}^{2} + 5 - \left(9\right)$

$y = {\left(x - 3\right)}^{2} - 4$

$\therefore$, the vertex form is $y = {\left(x - 3\right)}^{2} - 4$.