# What is the vertex of  y= (x-2)^2-3x^2-4x-4?

Oct 13, 2017

$\left(- 2 , 8\right)$

#### Explanation:

The formula for the x-value of the vertex of a quadratic is:

$\frac{- b}{2 a} = \text{x-value of the vertex}$

To get our $a$ and $b$, it's easiest to have your quadratic in standard form, and to get that, work your quadratic all the way out and simplify, getting you:

$y = {x}^{2} - 4 x + 4 - 3 {x}^{2} - 4 x - 4$

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

In this case, you have no $c$ term, but it doesn't really affect anything. Plug in your $a$ and $b$ into the vertex formula:

$\frac{- \left(- 8\right)}{2 \left(- 2\right)} = \text{x-value of the vertex}$

$\text{x-value of the vertex} = - 2$

Now plug your newly found $\text{x-value}$ back into your quadratic to solve for its $\text{y-value}$, which gives you:

$y = - 2 {\left(- 2\right)}^{2} - 8 \left(- 2\right)$

$y = 8$

Concluding that the coordinates of the vertex of this quadratic are:

$\left(- 2 , 8\right)$