# What is the vertex of  y= 2|x|^2 – 4x+1 ?

Feb 16, 2018

${y}_{\text{vertex}} = \left(1 , - 1\right)$

#### Explanation:

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

First note that ${\left\mid x \right\mid}^{2} = {x}^{2}$

Hence, $y = 2 {x}^{2} - 4 x + 1$

$y$ is a parabolic function of the form $y = a {x}^{2} + b x + c$ which has a vertex at $x = - \frac{b}{2 a}$

$x = - \frac{- 4}{2 \cdot 2} = 1$

$y \left(1\right) = 2 - 4 + 1 = - 1$

Hence, ${y}_{\text{vertex}} = \left(1 , - 1\right)$

We can see this result from the graph of $y$ below:

graph{2abs(x)^2-4x+1 [-5.55, 6.936, -2.45, 3.796]}