# How do you write the Vertex form equation of the parabola y=x^2+4x-7?

Sep 29, 2016

$y = {\left(x - \left(- 2\right)\right)}^{2} + \left(- 11\right)$

#### Explanation:

First recall the general vertex form:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$

Given
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} + 4 x - 7$

The required value of $\textcolor{g r e e n}{m}$ is clearly $\textcolor{g r e e n}{1}$ and we can ignore it at this point.

We are trying to get a squared binomial: ${\left(x - \textcolor{red}{a}\right)}^{2}$

Completing the square in the given equation:
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} + 4 x \textcolor{m a \ge n t a}{+ {2}^{2}} - 7 \textcolor{m a \ge n t a}{- {2}^{2}}$

$\textcolor{w h i t e}{\text{XXX}} y = {\left(x + 2\right)}^{2} - 11$

Into proper vertex form:
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{1} {\left(x - \textcolor{red}{\left(- 2\right)}\right)}^{2} + \textcolor{b l u e}{\left(- 11\right)}$

graph{x^2+4x-7 [-10.84, 9.16, -12.92, -2.92]}