# How do find the vertex and axis of symmetry for a quadratic equation y=4x^2-2x+2?

Jun 7, 2015

The general form of a parabolic equation ins vertex form is:
$\textcolor{w h i t e}{\text{XXXX}}$$y = m {\left(x - a\right)}^{2} + b$ where the vertex is at $\left(a , b\right)$

$y = 4 {x}^{2} - 2 x + 2$
can be converted into that form (most easily by a completion of the squares process)
$\textcolor{w h i t e}{\text{XXXX}}$$y = 4 \left({x}^{2} - \frac{1}{2} x\right) + 2$

$\textcolor{w h i t e}{\text{XXXX}}$$y = 4 \left({x}^{2} - \frac{1}{2} x + {\left(\frac{1}{4}\right)}^{2}\right) + 2 - 4 {\left(\frac{1}{4}\right)}^{2}$

$\textcolor{w h i t e}{\text{XXXX}}$$y = 4 {\left(x - \frac{1}{4}\right)}^{2} + \frac{7}{4}$

The vertex is at $\left(\frac{1}{4} , \frac{7}{4}\right)$