What is the vertices f(x)=x^2-6x-1 ?

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

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vertices f(x)=x^2-6x-1

vertices

f (x) = 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^{2} + 6 y + 8 x + 25 = 0

9 y^{2} - 49 x^{2} = 441

\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1

y^{2} + 6 y - 2 x + 13 = 0

x^{2} + y^{2} - 25 = 0

What is the vertices f(x)=x^2-6x-1 ?
The vertices f(x)=x^2-6x-1 is Minimum (3,-10)