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

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

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vertices y=x^2-6x-1

vertices

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

Minimum (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 \cdot 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

Minimum (3, - 10)

y = x^{2} - 5 x - 24

2 x^{2} + 5 x - 12

f (x) = - x^{2} - 4 x + 12

f (x) = 2 x^{2} - 12 x + 10

18 x - 9 x^{2} - 9