# What is the vertex of  y= (x-3)^2-5x^2-2x-9?

May 9, 2016

We will first have to expand and simplify the expression, before completing the square.

#### Explanation:

$y = {x}^{2} - 6 x + 9 - 5 {x}^{2} - 2 x - 9$

$y = - 4 {x}^{2} - 8 x$

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

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

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

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

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

As in any quadratic function of the form $y = a {\left(x - p\right)}^{2} + q$, the vertex is situated at the point $\left(p , q\right)$.

Thus, the vertex is a $\left(- 1 , 4\right)$

Hopefully this helps!