What is the vertices-(x^2)/2-2x ?

The vertices-(x^2)/2-2x is Maximum (-2,2)

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vertices-(x^2)/2-2x

vertices

- \frac{x ^{2}}{2} - 2 x

Maximum (- 2, 2)

Solution steps

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

The parabola parameters are:

a = - \frac{1}{2}, b = - 2, c = 0

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

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

Simplify

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

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(- 2, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{2}

Maximum (- 2, 2)

y = - x^{2} + 5

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

f (x) = 3 x^{2} - 2 x - 1

7 x^{2} - 40 x - 25

f (x) = (x - 3)^{2} - 1