# What is the vertex form of y= (x + 2)(x - 6)?

May 14, 2016

$y = {\left(x - 2\right)}^{2} - 16$

or if you want it in exactly $y = a {\left(x - h\right)}^{2} + k$ form:

$y = 1 {\left(x - 2\right)}^{2} + \left(- 16\right)$

#### Explanation:

$y = \left(x + 2\right) \left(x - 6\right) = {\left(x - 2\right)}^{2} - 16$

How did I get that?

Since we have factors $\left(x + 2\right)$ and $\left(x - 6\right)$ this parabola will intercept the $x$ axis at $\left(- 2 , 0\right)$ and $\left(6 , 0\right)$.

Then the vertical axis of symmetry must lie half way between these, making it the line $x = 2$, which is also the $x$ coordinate of the vertex.

So we must have:

$y = {\left(x - 2\right)}^{2} + c$

for some constant $c$.

Then putting $x = 2$ in the original equation, we find:

$\left(x + 2\right) \left(x - 6\right) = \left(2 + 2\right) \left(2 - 6\right) = 4 \cdot \left(- 4\right) = - 16$

So $c = - 16$

The vertex of the parabola is at $\left(2 , - 16\right)$