What is the vertices f(x)=-2x^2+8x-11 ?

The vertices f(x)=-2x^2+8x-11 is Maximum (2,-3)

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vertices f(x)=-2x^2+8x-11

vertices

f (x) = - 2 x^{2} + 8 x - 11

Maximum (2, - 3)

Solution steps

y = - 2 x^{2} + 8 x - 11

The parabola parameters are:

a = - 2, b = 8, c = - 11

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} = - 3

Therefore the parabola vertex is

(2, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (2, - 3)

f (x) = - 3 x^{2} + 4 x + 4

f (x) = 2 (x - 5) (x - 1)

- x^{2} + 4 x - 6

y = \frac{6 x}{7} \cdot \frac{4 x}{3}

y^{2} - 6 y - 4 x + 5 = 0

What is the vertices f(x)=-2x^2+8x-11 ?
The vertices f(x)=-2x^2+8x-11 is Maximum (2,-3)