# How do you find the vertex of the parabola y=1/2(x+1)(x-5)?

Feb 13, 2016

vertex: $\left(2 , - 4 \frac{1}{2}\right)$

#### Explanation:

Our objective will be to transform the given equation into vertex form:
$\textcolor{w h i t e}{\text{XXX}} y = m {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$ with vertex at $\left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$
If $y = \frac{1}{2} \left(x + 1\right) \left(x - 5\right)$
then
$\textcolor{w h i t e}{\text{XXX}} 2 y = {x}^{2} - 4 x - 5$

$\textcolor{w h i t e}{\text{XXX}} 2 y = \left({x}^{2} - 4 x \textcolor{g r e e n}{+ 4}\right) - 5 \textcolor{g r e e n}{- 4}$ (completing the square)

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

$\textcolor{w h i t e}{\text{XXX")y=1/2(x-color(red)(2))^2+color(blue)("("-9/2")}}$

which is the vertex form with vertex at $\left(\textcolor{red}{2} , \textcolor{b l u e}{\text{("-9/2")}}\right)$

Here's the graph to show that this result is reasonable.
graph{1/2*(x^2-4x-5) [-5.086, 7.4, -5.96, 0.28]}