# How do you write the vertex form equation of the parabola (x+5)(x+4)?

Apr 24, 2016

$y = {\left(x - \left(- \frac{9}{2}\right)\right)}^{2} + \left(- \frac{1}{4}\right)$

#### Explanation:

The general vertex form for a parabola (with a vertical axis of symmetry) is
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$ ($\textcolor{g r e e n}{m}$ can be thought of as a parameter that effects the "spread" of the parabola).

Given:
$\textcolor{w h i t e}{\text{XXX}} y = \left(x + 5\right) \left(x + 4\right)$

This can be rewritten as:
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} + 9 x + 20$

Completing the square:
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} + 9 x \textcolor{b r o w n}{+ {\left(\frac{9}{2}\right)}^{2}} + 20 \textcolor{b r o w n}{- {\left(\frac{9}{2}\right)}^{2}}$

$\textcolor{w h i t e}{\text{XXX}} y = {\left(x + \frac{9}{2}\right)}^{2} + \left(\frac{80}{4} - \frac{81}{4}\right)$

$\textcolor{w h i t e}{\text{XXX")y=color(green)(1)(x-color(red)(""(-9/2)))^2+color(blue)(} \left(- \frac{1}{4}\right)}$
with vertex at $\left(\textcolor{red}{- \frac{9}{2}} , \textcolor{b l u e}{- \frac{1}{4}}\right)$

(Note that when $\textcolor{g r e e n}{m} = 1$ it is often omitted.)

graph{(x+5)(x+4) [-5.95, -0.474, -1.327, 1.41]}