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

The vertices f(x)=-2x^2-2x-2 is Maximum (-1/2 ,-3/2)

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

vertices

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

Maximum (- \frac{1}{2}, - \frac{3}{2})

Solution steps

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

The parabola parameters are:

a = - 2, b = - 2, c = - 2

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

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

Simplify

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

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

Plug in

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

to find the

y_{v}

value

y_{v} = - \frac{3}{2}

Therefore the parabola vertex is

(- \frac{1}{2}, - \frac{3}{2})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (- \frac{1}{2}, - \frac{3}{2})

f (x) = - 3 x^{2} + 5 x

(x - 3)^{2} = 8 (y - 5)

(x - 2) = 3 (y - 5)^{2}

y = \frac{x ^{2} x ^{1}}{x}

X^{2}

What is the vertices f(x)=-2x^2-2x-2 ?
The vertices f(x)=-2x^2-2x-2 is Maximum (-1/2 ,-3/2)