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

Aug 21, 2016

Vertex is at the origin $\left(0 , 0\right)$

#### Explanation:

This is a somewhat unusual format for a parabola! Simplify first to see what we are working with..

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

What does an equation tell us about the parabola?

The standard form is $y = \textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{m a \ge n t a}{c}$

$\textcolor{red}{a}$ changes the shape of the parabola - whether it is narrow or wide, or open upwards or downwards.

$\textcolor{b l u e}{b} x$ moves the parabola to the left or right

$\textcolor{m a \ge n t a}{c}$ gives the y-intercept. It moves the parabola up or down.

In $y = - 2 {x}^{2}$ there is no x-term, and $c = 0$

This means that the parabola has not moved to the left or right, nor has it moved up or down, although it is 'upside-down' with a maximum TP.

Its vertex is at the origin $\left(0 , 0\right)$

Changing it to vertex form will give $y = - 2 {\left(x + 0\right)}^{2} + 0$
graph{-2x^2 [-4.92, 5.08, -3.86, 1.14]}