What is the vertices-x^2+6x+8 ?

The vertices-x^2+6x+8 is Maximum (3,17)

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vertices-x^2+6x+8

vertices

- x^{2} + 6 x + 8

Maximum (3, 17)

Solution steps

y = - x^{2} + 6 x + 8

The parabola parameters are:

a = - 1, b = 6, c = 8

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

x_{v} = - \frac{6}{2 ( - 1 )}

Simplify

- \frac{6}{2 ( - 1 )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 17

Therefore the parabola vertex is

(3, 17)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (3, 17)

y = 6 x^{2} - 29 x - 5

f (x) = 2 x^{2} + 7 x + 3

x = - y^{2}

f (x) = - x^{2} - 8 x - 7

4 x^{2} - 3 x + 5