# What is the axis of symmetry and vertex for the graph f(x)=2x^2-4x+1?

Aug 2, 2017

vertex at $\left(x , y\right) = \left(1 , - 1\right)$
axis of symmetry: $x = 1$

#### Explanation:

We will convert the given equation into "vertex form"
$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{m} {\left(x - \textcolor{red}{a}\right)}^{2} + \textcolor{b l u e}{b}$
where
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{m}$ is a factor related to the horizontal spread of the parabola; and
$\textcolor{w h i t e}{\text{XXX}} \left(\textcolor{red}{a} , \textcolor{b l u e}{b}\right)$ is the $\left(x , y\right)$ coordinate of the vertex.

Given:
$\textcolor{w h i t e}{\text{XXX}} y = 2 {x}^{2} - 4 x + 1$

$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{2} \left({x}^{2} - 2 x\right) + 1$

$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{2} \left({x}^{2} - 2 x + \textcolor{m a \ge n t a}{1}\right) + 1 - \left(\textcolor{g r e e n}{2} \times \textcolor{m a \ge n t a}{1}\right)$

$\textcolor{w h i t e}{\text{XXX}} y = \textcolor{g r e e n}{2} {\left(x - \textcolor{red}{1}\right)}^{2} + \textcolor{b l u e}{\left(- 1\right)}$

The vertex form with vertex at $\left(\textcolor{red}{1} , \textcolor{b l u e}{- 1}\right)$

Since this equation is of the form of a parabola in "standard position"
the axis of symmetry is a vertical line passing though the vertex, namely:
$\textcolor{w h i t e}{\text{XXX}} x = \textcolor{red}{1}$ 