What is the vertices-x^2-2x+8 ?

The vertices-x^2-2x+8 is Maximum (-1,9)

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vertices-x^2-2x+8

vertices

- x^{2} - 2 x + 8

Maximum (- 1, 9)

Solution steps

y = - x^{2} - 2 x + 8

The parabola parameters are:

a = - 1, b = - 2, c = 8

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

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

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 9

Therefore the parabola vertex is

(- 1, 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 1, 9)

y = 4 x^{2} - 8

f (x) = 2 x^{2} - 12 x

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

f (x) = 2 x^{2} - 8 x - 5

f (x) = 2 x^{2} - 8 x + 6