# What is the vertex form of y = 12x^2 - 6x + 8?

Nov 30, 2015

$y = 12 {\left(x + \frac{1}{4}\right)}^{2} + \frac{29}{4}$

#### Explanation:

You can get this equation into vertex form by completing the square

First, factor out the coefficient of the largest power of x:
$y = 12 \left({x}^{2} - \frac{1}{2} x\right) + 8$

then take half of the coefficient of the x to the first power and square it
$\frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \rightarrow {\frac{1}{4}}^{2} = \frac{1}{16}$

add and subtract the number you just found within the parenthesis
$y = 12 \left({x}^{2} + \frac{1}{2} x + \frac{1}{16} - \frac{1}{16}\right) + 8$

take the negative $\frac{1}{16}$ out of the parenthesis
$y = 12 \left({x}^{2} + \frac{1}{2} x + \frac{1}{16}\right) - \frac{3}{4} + 8$

factor and simplify
$y = 12 {\left(x + \frac{1}{4}\right)}^{2} + \frac{29}{4}$ $\leftarrow$ answer