What is the vertices f(x)=-x^2-2x+8 ?

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

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vertices f(x)=-x^2-2x+8

vertices

f (x) = - 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)

- x^{2} + 6 x - 5

y = x^{2} - 9 x + 20

y = x^{2} - 11 x + 13

y = 3 x^{2} - 12 x + 10

f (x) = - 2 x^{2} - 4 x + 13

What is the vertices f(x)=-x^2-2x+8 ?
The vertices f(x)=-x^2-2x+8 is Maximum (-1,9)