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

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Answer 1 · 2015-07-15T12:54:21

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

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

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:

$y = (0-3)(0+3) = -3 * 3 = -9$

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

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

Answer 2 · 2015-07-15T13:01:11

Multiply out to find $y = (x-3)(x+3) = x^2-9$

$x^2 >= 0$ for all $x$ , with minimum when $x=0$ ,

giving $y = x^2-9 = 0^2-9 = -9$ , that is $(0, -9)$

$(x-3)(x+3)$ is recognisable as a factorisation of $x^2-3^2 = x^2-9$

(use the difference of squares identity: $a^2-b^2 = (a-b)(a+b)$ )

So $y = (x-3)(x+3) = x^2-9$

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 $(0, -9)$

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