# What is the vertex form of y= 4x^2 + 10x + 6 ?

May 13, 2016

$y = 4 {\left(x - \left(- \frac{5}{4}\right)\right)}^{2} + \left(- \frac{1}{4}\right)$

#### Explanation:

$y = 4 {x}^{2} + 10 x + 6$

$= 4 \left({x}^{2} + \frac{5}{2} x + \frac{3}{2}\right)$

$= 4 \left({x}^{2} + 2 \left(x\right) \left(\frac{5}{4}\right) + {\left(\frac{5}{4}\right)}^{2} - {\left(\frac{5}{4}\right)}^{2} + \frac{6}{4}\right)$

$= 4 \left({\left(x + \frac{5}{4}\right)}^{2} - {\left(\frac{5}{4}\right)}^{2} + \frac{6}{4}\right)$

$= 4 {\left(x + \frac{5}{4}\right)}^{2} - \frac{25}{4} + \frac{24}{4}$

$= 4 {\left(x + \frac{5}{4}\right)}^{2} - \frac{1}{4}$

So:

$y = 4 {\left(x + \frac{5}{4}\right)}^{2} - \frac{1}{4}$

Or we can write:

$y = 4 {\left(x - \left(- \frac{5}{4}\right)\right)}^{2} + \left(- \frac{1}{4}\right)$

This is in strict vertex form:

$y = a {\left(x - h\right)}^{2} + k$

with multiplier $a = 4$ and vertex $\left(h , k\right) = \left(- \frac{5}{4} , - \frac{1}{4}\right)$