# How do you write y=3x^2-6x+1 into vertex form?

May 3, 2015

Vertex form for a quadratic is
$y = m \left(x - a\right) + b$
(where $\left(a , b\right)$ is the vertex)

$y = 3 {x}^{2} - 6 x + 1$

$y = 3 \left({x}^{2} - 2 x\right) + 1 \text{ extracting the "m" constant}$

$y = 3 \left({x}^{2} - 2 x + 1\right) - 3 + 1 \text{ [completing the square](http://socratic.org/algebra/quadratic-equations-and-functions/completing-the-square-1)}$

$y = 3 {\left(x - 1\right)}^{2} - 2 \text{ simplification}$

$y = 3 {\left(x - 1\right)}^{2} + \left(- 2\right) \text{ into completely vertex form}$