# How do you write the Vertex form equation of the parabola y=x^2-6x+5?

Dec 5, 2016

#### Explanation:

Given $y = {x}^{2} - 6 x + 5 \text{ [1]}$

The vertex form of a parabola of this type is:

$y = a {\left(x - h\right)}^{2} + k \text{ [2]}$

where $\left(h , k\right)$ is the vertex and "a" is the coefficient of the ${x}^{2}$ term in standard form.

The first step is to add 0 to equation [1] in the form $a {h}^{2} - a {h}^{2}$. Because $a = 1$, we add ${h}^{2} - {h}^{2}$ to equation [1]:

$y = {x}^{2} - 6 x + {h}^{2} - {h}^{2} + 5 \text{ [3]}$

If we expand the square in equation [2], ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$, we can see that $- 2 h x$ must equal $- 6 x$ in equation [3]. Write this equation:

$- 2 h x = - 6 x$

Solve for h:

$h = 3$

Substitute ${\left(x - h\right)}^{2}$ for ${x}^{2} - 6 x + {h}^{2}$ in equation [3]:

$y = {\left(x - h\right)}^{2} - {h}^{2} + 5 \text{ [4]}$

Substitute 3 for every h:

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

Simplify the constant terms:

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

This is the vertex form.