# What is the vertex of  y= (x-3)^2-5x^2-x-1?

Nov 10, 2017

The vertex is at $\left(- \frac{7}{8} , \frac{177}{16}\right)$

#### Explanation:

The equation given is a quadratic $y = a {x}^{2} + b x + c$

The vertex is at $\left(h , k\right)$ where $h = - \frac{b}{2 a}$

First expand the equation
$y = {x}^{2} - 6 x + 9 - 5 {x}^{2} - x - 1$

Simplify
$y = - 4 {x}^{2} - 7 x + 8$

the $x$ value of the vertex is $\frac{7}{-} 8$ or $- \frac{7}{8}$

plug the value for h back into the equation to get k

$y = - 4 \cdot - \frac{7}{8} \cdot - \frac{7}{8} - 7 \cdot - \frac{7}{8} + 8$ = $\frac{177}{16}$

The vertex is at $\left(- \frac{7}{8} , \frac{177}{16}\right)$