What is the vertices f(x)=-x^2+6x-11 ?

The vertices f(x)=-x^2+6x-11 is Maximum (3,-2)

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vertices f(x)=-x^2+6x-11

vertices

f (x) = - x^{2} + 6 x - 11

Maximum (3, - 2)

Solution steps

y = - x^{2} + 6 x - 11

The parabola parameters are:

a = - 1, b = 6, c = - 11

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} = - 2

Therefore the parabola vertex is

(3, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (3, - 2)

y^{2} = - 20 x

2 x^{2} - 6 x + 15

(x + 1)^{2} = - 6 (y - 4)

x^{2} - 2 x - 4 y + 13 = 0

5 x^{2} - 10

What is the vertices f(x)=-x^2+6x-11 ?
The vertices f(x)=-x^2+6x-11 is Maximum (3,-2)