What is the vertices-x^2-6x-9 ?

The vertices-x^2-6x-9 is Maximum (-3,0)

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vertices-x^2-6x-9

vertices

- x^{2} - 6 x - 9

Maximum (- 3, 0)

Solution steps

y = - x^{2} - 6 x - 9

The parabola parameters are:

a = - 1, b = - 6, c = - 9

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{( - 6 )}{2 ( - 1 )}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(- 3, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 3, 0)

f (x) = x^{2} - 5

y = x^{2} + 3 x - 5

y = x^{2} + 3 x + 2

y = 4 x^{2} - 16 x + 22

x^{2} + 8 x - 12 y + 88 = 0