What is the vertices f(x)=-6x^2-6x-6 ?

The vertices f(x)=-6x^2-6x-6 is Maximum (-1/2 ,-9/2)

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vertices f(x)=-6x^2-6x-6

vertices

f (x) = - 6 x^{2} - 6 x - 6

Maximum (- \frac{1}{2}, - \frac{9}{2})

Solution steps

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

The parabola parameters are:

a = - 6, b = - 6, c = - 6

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

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

Simplify

- \frac{- 6}{2 ( - 6 )} : - \frac{1}{2}

x_{v} = - \frac{1}{2}

Plug in

x_{v} = - \frac{1}{2}

to find the

y_{v}

value

y_{v} = - \frac{9}{2}

Therefore the parabola vertex is

(- \frac{1}{2}, - \frac{9}{2})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 6

Maximum (- \frac{1}{2}, - \frac{9}{2})

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

y = (2 x + 5)^{2}

f (x) = - 9 x^{2} + 9 x + 1

+ x^{2} + 4 x - 1

y^{2} - 12 y - 4 x = - 28

What is the vertices f(x)=-6x^2-6x-6 ?
The vertices f(x)=-6x^2-6x-6 is Maximum (-1/2 ,-9/2)