# What is the vertex form of y=9x^2+14x+12?

May 31, 2017

$y = 9 {\left(x + \frac{7}{9}\right)}^{2} + \frac{59}{12}$

#### Explanation:

A quadratic is written in the form $y = a {x}^{2} + b x + c$

Vertex form is known as $y = a {\left(x + b\right)}^{2} + c ,$ giving the vertex as $\left(- b , c\right)$

It is useful to be able to change a quadratic expression into the form $a {\left(x + b\right)}^{2} + c$. The process is by completing the square.

$y = 9 {x}^{2} + 14 x + 12 \text{ } \leftarrow$ the coefficient of ${x}^{2}$ must be $1$

$y = 9 \left({x}^{2} + \frac{14}{9} x + \frac{12}{9}\right)$

To make a square of a binomial, you need to add on $\textcolor{b l u e}{{\left(\frac{b}{2}\right)}^{2}}$
It is also subtracted so that the value of the expression is not changed. $\textcolor{b l u e}{{\left(\frac{b}{2}\right)}^{2} - {\left(\frac{b}{2}\right)}^{2} = 0}$

$y = 9 \left({x}^{2} + \frac{14}{9} x \textcolor{b l u e}{+ {\left(\frac{7}{9}\right)}^{2} - {\left(\frac{7}{9}\right)}^{2}} + \frac{12}{9}\right)$

$y = 9 \left(\textcolor{red}{\left({x}^{2} + \frac{14}{9} x + {\left(\frac{7}{9}\right)}^{2}\right)} + \textcolor{g r e e n}{\left(- \frac{49}{81} + \frac{12}{9}\right)}\right)$

$y = 9 \left(\textcolor{red}{{\left(x + \frac{7}{9}\right)}^{2} + \textcolor{g r e e n}{\left(- \frac{49}{81} \frac{12}{9}\right)}}\right)$

$y = 9 {\left(x + \frac{7}{9}\right)}^{2} + 9 \left(- \frac{49}{81} + \frac{108}{81}\right)$

y = 9(x+7/9)^2 + 9(59/108))

$y = 9 {\left(x + \frac{7}{9}\right)}^{2} + \frac{59}{12}$