# What is the vertex form of  y= (9x-6)(3x+2)+4x^2+5x?

Apr 25, 2017

$y = 31 {\left(x + \frac{5}{62}\right)}^{2} - \frac{1513}{124}$

#### Explanation:

$y = \left(9 x - 6\right) \left(3 x + 2\right) + 4 {x}^{2} + 5 x$

= $27 {x}^{2} + 18 x - 18 x - 12 + 4 {x}^{2} + 5 x$

= $31 {x}^{2} + 5 x - 12$

= $31 \left({x}^{2} + \frac{5}{31} x\right) - 12$

= $31 \left({x}^{2} + 2 \times \frac{5}{62} \times x + {\left(\frac{5}{62}\right)}^{2} - {\left(\frac{5}{62}\right)}^{2}\right) - 12$

= $31 {\left(x + \frac{5}{62}\right)}^{2} - 31 {\left(\frac{5}{62}\right)}^{2} - 12$

= $31 {\left(x + \frac{5}{62}\right)}^{2} - \frac{25}{124} - 12$

or $y = 31 {\left(x + \frac{5}{62}\right)}^{2} - 12 \frac{25}{124}$

i.e. $y = 31 {\left(x + \frac{5}{62}\right)}^{2} - \frac{1513}{124}$

and vertex is $\left(- \frac{5}{62} , - 12 \frac{25}{124}\right)$

graph{y=31(x+5/62)^2-1513/124 [-3, 3, -20, 20]}