# How do you find the axis of symmetry, and the maximum or minimum value of the function y = 2x^2 - 4x + 1?

Jan 20, 2017

The axis is $x = 1$
The minimum value is $y = - 1$

#### Explanation:

Let`s complete the squares

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

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

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

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

This is the vertex form of the equation

The axis of symmetry is $x = 1$

The vertex is $\left(1 , - 1\right)$

The minimum value is $y = - 1$

graph{(y-(2x^2-4x+1))(y-100x+100)((x-1)^2+(y+1)^2-0.002)=0 [-5.546, 5.55, -2.773, 2.774]}