# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation f(x)=2x^2-4x+1?

Jun 12, 2015

Rewrite $f \left(x\right) = 2 {x}^{2} - 4 x + 1$ in vertex form to determine vertex is at $\left(1 , - 1\right)$ and the axis of symmetry $x = 1$.
The $f \left(x\right)$ intercept is at $f \left(x\right) = 1$; the x intercepts are at $x = 2 \pm \frac{\sqrt{2}}{2}$

#### Explanation:

The vertex form of a quadratic is
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = m {\left(x - a\right)}^{2} + b$ with a vertex at $\left(a , b\right)$

$f \left(x\right) = 2 {x}^{2} - 4 x + 1$
extract $m$
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = 2 \left({x}^{2} - 2 x\right) + 1$
complete the square
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = 2 \left({x}^{2} - 2 x + 1\right) + 1 - 2$
$\textcolor{w h i t e}{\text{XXXX}}$$f \left(x\right) = 2 {\left(x - 1\right)}^{2} + \left(- 1\right)$ with a vertex at $\left(1 , - 1\right)$

The given equation is a parabola in standard orientation, so the axis of symmetry is a vertical line through the vertex; that is $x = 1$

The $f \left(x\right)$ and $x$ intercepts are determined by solving the equations for $f \left(x\right)$ with $x = 0$ and for $x$ with $f \left(x\right) = 0$ respectively. (Use the quadratic formula to determine the $x$ solution).