# What is the vertex form of y=4x^2+x-6?

##### 1 Answer
Dec 27, 2015

$y = 4 {\left(x - \left(- \frac{1}{8}\right)\right)}^{2} + \left(- \frac{97}{16}\right)$

#### Explanation:

To find the vertex form of a quadratic equation we use a process called completing the square.

Our goal is the form $y = a {\left(x - h\right)}^{2} + k$ where $\left(h , k\right)$ is the vertex. Proceeding, we have

$4 {x}^{2} + x - 6 = 4 \left({x}^{2} + \frac{1}{4} x\right) - 6$

$= 4 \left({x}^{2} + \frac{1}{4} x + \frac{1}{64} - \frac{1}{64}\right) - 6$

$= 4 \left({x}^{2} + \frac{1}{4} x + \frac{1}{64}\right) - \frac{4}{64} - 6$

$= 4 {\left(x + \frac{1}{8}\right)}^{2} - \frac{97}{16}$

$= 4 {\left(x - \left(- \frac{1}{8}\right)\right)}^{2} + \left(- \frac{97}{16}\right)$

Thus the vertex form is
$y = 4 {\left(x - \left(- \frac{1}{8}\right)\right)}^{2} + \left(- \frac{97}{16}\right)$

and the vertex is at $\left(- \frac{1}{8} , - \frac{97}{16}\right)$