# How do you find the vertex and intercepts for y = 3(x - 2) (x + 2)?

Aug 2, 2018

$\text{vertex "=(0,-12)," intercepts } x = \pm 2$

#### Explanation:

$\text{to find the x-intercepts set y = 0}$

$3 \left(x - 2\right) \left(x + 2\right) = 0$

$\text{equate each factor to zero and solve for x}$

$x - 2 = 0 \Rightarrow x = 2$

$x + 2 = 0 \Rightarrow x = - 2$

$\text{to find the y-intercept set x = 0}$

$y = 3 \left(- 2\right) \left(+ 2\right) = - 12$

$\text{The vertex lies on the line of symmetry which is}$
$\text{positioned midway between the x-intercepts}$

${x}_{\text{vertex}} = \frac{+ 2 - 2}{2} = 0$

${y}_{\text{vertex}} = 3 \left(- 2\right) \left(+ 2\right) = - 12$

$\textcolor{m a \ge n t a}{\text{vertex }} = \left(0 , - 12\right)$
graph{3(x-2)(x+2) [-40, 40, -20, 20]}