# What is the vertex form of y= x^2 – 7x – 4?

Jul 18, 2017

$y = {x}^{2} - 7 x - 4$

use completing a square

$y = {\left(x - \frac{7}{2}\right)}^{2} - {\left(- \frac{7}{2}\right)}^{2} - 4$

$y = {\left(x - \frac{7}{2}\right)}^{2} - \frac{49}{4} - \frac{16}{4}$

$y = {\left(x - \frac{7}{2}\right)}^{2} - \frac{65}{4}$

since ${\left(x - \frac{7}{2}\right)}^{2}$ is a +ve value, therefore it has a minimum vertex $\frac{7}{2}$ at $\left(\frac{7}{2} , - \frac{65}{4}\right)$