# What is the vertex form of y=x^4+2x^2-2?

Feb 4, 2016

I think the best we can do is
$\textcolor{w h i t e}{\text{XXX}} y = {\left({x}^{2} + 1\right)}^{2} - 3$

#### Explanation:

A reference to "the vertex form" implies that we are dealing with a parabolic equation; $y = {x}^{4} + 2 {x}^{2} - 2$ is not a parabolic equation.

We could replace ${x}^{2}$ with $z$, so
$\textcolor{w h i t e}{\text{XXX}} y = {z}^{2} + 2 z - 2$

color(white)("XXX")y=(z^2+2zcolor(red)(+1)-2color(red)(-1)
$\textcolor{w h i t e}{\text{XXX}} y = {\left(z + 1\right)}^{2} - 3$
which is a parabolic equation with the vertex at $\left(\textcolor{b l u e}{z} , y\right) = \left(- 1 , - 3\right)$

Replacing $z$ with ${x}^{2}$ gives
$\textcolor{w h i t e}{\text{XXX}} y = \left({x}^{2} + 1\right) - 3$
but as noted, this is not a parabola (see graph below)
graph{x^4-2x^2-2 [-6.783, 7.26, -4.515, 2.505]}