What is the vertices y=(x+3)(x-3) ?

The vertices y=(x+3)(x-3) is Minimum (0,-9)

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vertices y=(x+3)(x-3)

vertices

y = (x + 3) (x - 3)

Minimum (0, - 9)

Solution steps

y = (x + 3) (x - 3)

y = (x + 3) (x - 3)

The parabola parameters are:

a = 1, m = - 3, n = 3

x_{v} = \frac{m + n}{2}

x_{v} = \frac{( - 3 ) + 3}{2}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 9

Therefore the parabola vertex is

(0, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, - 9)

f (x) = - 3 x^{2} - 18 x - 25

y = 18 x^{2} - 14 x + 9

f (x) = - 3 x^{2} + 18 x - 8

y = x^{2} - 6 x + 8

y = (x - 1)^{2} - 2