# How do you write a quadratic function in intercept form who has x intercepts -2/3, 1/6 and passes through points (7/3, 13/2)?

Jun 25, 2017

$y = \frac{1}{18} \left(3 x + 2\right) \left(6 x - 1\right)$

#### Explanation:

$x = - \frac{2}{3} \text{ is a root}$

$\Rightarrow 3 x = - 2 \Rightarrow 3 x + 2 \leftarrow \text{ is a factor}$

$x = \frac{1}{6} \text{ is a root}$

$\Rightarrow 6 x = 1 \Rightarrow 6 x - 1 \leftarrow \text{ is a factor}$

$\Rightarrow y = a \left(3 x + 2\right) \left(6 x - 1\right) \leftarrow \textcolor{red}{\text{ in intercept form}}$

$\text{to find a substitute " (7/3,13/2)" into the equation}$

$\frac{13}{2} = a \left(9\right) \left(13\right) = 117 a$

$\Rightarrow a = \frac{1}{18}$

$\Rightarrow y = \frac{1}{18} \left(3 x + 2\right) \left(6 x - 1\right) \text{ is the equation}$