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

Aug 6, 2016

$y = 4 {\left(x - \frac{1}{8}\right)}^{2} - \frac{65}{16}$

#### Explanation:

x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{1}{8.}$
y-coordinate of vertex:
$y \left(\frac{1}{8}\right) = 4 \left(\frac{1}{64}\right) - \frac{1}{8} - 4 = \frac{1}{16} - \frac{1}{8} - 4 = - \frac{65}{16}$
$V e r t e x \left(\frac{1}{8} , - \frac{65}{16}\right)$
Vertex form of y:
$y = 4 {\left(x - \frac{1}{8}\right)}^{2} - \frac{65}{16}$