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

The vertices y=-x^2-6x+1 is Maximum (-3,10)

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

vertices

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

Maximum (- 3, 10)

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

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = 10

Therefore the parabola vertex is

(- 3, 10)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 3, 10)

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

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

- 4 x^{2} - 12 x - 8

y = x^{2} - 5 x - 14

x^{2} + 14 x - 8 y + 65 = 0