# What is the vertex form of y=(3x+1)(x+2)+ 2?

Feb 28, 2017

Vertex form is $y = 3 {\left(x + \frac{7}{6}\right)}^{2} - \frac{1}{12}$ and vertex is $\left(- \frac{7}{6} , - \frac{1}{12}\right)$

#### Explanation:

Vertex form of quadratic equation is $y = a {\left(x - h\right)}^{2} + k$, with $\left(h , k\right)$ as vertex.

To convert $y = \left(3 x + 1\right) \left(x + 2\right) + 2$, what we need is to expand and then convert part containing $x$ into a complete square and leave remaining constant as $k$. The process is as shown below.

$y = \left(3 x + 1\right) \left(x + 2\right) + 2$

= $3 x \times x + 3 x \times 2 + 1 \times x + 1 \times 2 + 2$

= $3 {x}^{2} + 6 x + x + 2 + 2$

= $3 {x}^{2} + 7 x + 4$

= $3 \left({x}^{2} + \frac{7}{3} x\right) + 4$

= $3 \left(\textcolor{b l u e}{{x}^{2}} + 2 \times \textcolor{b l u e}{x} \times \textcolor{red}{\frac{7}{6}} + \textcolor{red}{{\left(\frac{7}{6}\right)}^{2}}\right) - 3 \times {\left(\frac{7}{6}\right)}^{2} + 4$

= $3 {\left(x + \frac{7}{6}\right)}^{2} - \frac{\cancel{3} \times 49}{{\cancel{36}}^{12}} + 4$

= $3 {\left(x + \frac{7}{6}\right)}^{2} - \frac{49}{12} + \frac{48}{12}$

= $3 {\left(x + \frac{7}{6}\right)}^{2} - \frac{1}{12}$

i.e. $y = 3 {\left(x + \frac{7}{6}\right)}^{2} - \frac{1}{12}$ and vertex is $\left(- \frac{7}{6} , - \frac{1}{12}\right)$
graph{(3x+1)(x+2)+2 [-2.402, 0.098, -0.54, 0.71]}