# How do you find the vertex of y= (x-3)(x+3)?

Jul 15, 2015

Construct the vertex using part geometry and part algebra to find $\left(0 , - 9\right)$...

#### Explanation:

This is a vertical parabola that cuts the $x$-axis at $\left(- 3 , 0\right)$ and $\left(3 , 0\right)$.

Since parabolas are symmetrical, its axis must be midway between these, namely the vertical line $x = 0$.

The axis intersects the parabola at the vertex, so substitute $x = 0$ into the equation to get:

$y = \left(0 - 3\right) \left(0 + 3\right) = - 3 \cdot 3 = - 9$

that is the point $\left(0 , - 9\right)$.

graph{(x-3)(x+3) [-19.91, 20.09, -12.08, 7.92]}

Jul 15, 2015

Multiply out to find $y = \left(x - 3\right) \left(x + 3\right) = {x}^{2} - 9$

${x}^{2} \ge 0$ for all $x$, with minimum when $x = 0$,

giving $y = {x}^{2} - 9 = {0}^{2} - 9 = - 9$, that is $\left(0 , - 9\right)$

#### Explanation:

$\left(x - 3\right) \left(x + 3\right)$ is recognisable as a factorisation of ${x}^{2} - {3}^{2} = {x}^{2} - 9$

(use the difference of squares identity: ${a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)$)

So $y = \left(x - 3\right) \left(x + 3\right) = {x}^{2} - 9$

This takes minimum value when ${x}^{2} = 0$, that is when $x = 0$

When $x = 0$, we have $y = {x}^{2} - 9 = 0 - 9 = - 9$

So the vertex is at $\left(0 , - 9\right)$