What is the vertices y=-x^2+6x-1 ?

The vertices y=-x^2+6x-1 is Maximum (3,8)

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vertices y=-x^2+6x-1

vertices

y = - x^{2} + 6 x - 1

Maximum (3, 8)

Solution steps

y = - x^{2} + 6 x - 1

The parabola parameters are:

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

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} = 8

Therefore the parabola vertex is

(3, 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (3, 8)

- (x - 30)^{2} + 40

f (x) = x^{2} + 3 x - 10

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

x^{2} - 6 x + 5 y = 34

f (x) = 5 x^{2} + 1