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

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

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 = x^{2} + 2 x - 1

y = - 2 x^{2} - 4 x + 6

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

x^{2} + 5 x - 1

f (x) = 2 x^{2} - 4 x - 1