What is the vertices y=-x^2+18x-82 ?

The vertices y=-x^2+18x-82 is Maximum (9,-1)

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vertices y=-x^2+18x-82

vertices

y = - x^{2} + 18 x - 82

Maximum (9, - 1)

Solution steps

y = - x^{2} + 18 x - 82

The parabola parameters are:

a = - 1, b = 18, c = - 82

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

x_{v} = - \frac{18}{2 ( - 1 )}

Simplify

- \frac{18}{2 ( - 1 )} : 9

x_{v} = 9

Plug in

x_{v} = 9

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(9, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (9, - 1)

+ 2 (x + 1)^{2} - 3

f (x) = x^{2} - 4 x - 21

f (x) = - 3 x^{2} - 2 x + 1

y = 3 (x - 3)^{2} - 1

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

What is the vertices y=-x^2+18x-82 ?
The vertices y=-x^2+18x-82 is Maximum (9,-1)