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)

\frac{x + 6}{x - 3}

f (x) = x^{4} - 32 x^{2} + 4

- 2 - x

f (x) = \frac{- 4 x + 20}{x ^{2} - 9 x + 20}

f (x) = \frac{x ^{2}}{x - 9}