What is the vertices-2x^2+8x-1 ?

The vertices-2x^2+8x-1 is Maximum (2,7)

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vertices-2x^2+8x-1

vertices

- 2 x^{2} + 8 x - 1

Maximum (2, 7)

Solution steps

y = - 2 x^{2} + 8 x - 1

The parabola parameters are:

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

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

x_{v} = - \frac{8}{2 ( - 2 )}

Simplify

- \frac{8}{2 ( - 2 )} : 2

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = 7

Therefore the parabola vertex is

(2, 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (2, 7)

y = x^{2} + 3 x

y = (x - 2)^{2} + 1

x = - 3 y^{2} + 5

f (x) = x^{2} + 10 x + 6

- 3 x^{2} - 12 x - 9