# What is the vertex form of y= x^2+x/2-4 ?

Dec 21, 2015

$y = 1 {\left(x - \left(- \frac{1}{4}\right)\right)}^{2} + \left(- 4 \frac{1}{16}\right)$

#### Explanation:

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

Complete the square:
$\textcolor{w h i t e}{\text{XXX}} y = {x}^{2} + \frac{1}{2} x \textcolor{g r e e n}{+ {\left(\frac{1}{4}\right)}^{2}} - 4 \textcolor{g r e e n}{- {\left(\frac{1}{4}\right)}^{2}}$

Re-write as a squared binomial plus a simplified constant:
$\textcolor{w h i t e}{\text{XXX}} y = {\left(x + \frac{1}{4}\right)}^{2} - 4 \frac{1}{16}$

Complete vertex form is $y = m {\left(x - a\right)}^{2} + b$
so we adjust signs to get this form (an include the default value for $m$)
$\textcolor{w h i t e}{\text{XXX}} y = 1 {\left(x - \left(- \frac{1}{4}\right)\right)}^{2} + \left(- 4 \frac{1}{16}\right)$
which has its vertex at $\left(- \frac{1}{4} , - 4 \frac{1}{16}\right)$
graph{x^2+x/2-4 [-3.813, 6.054, -4.736, 0.196]}