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

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

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vertices y=x^2-2x-1

vertices

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

Minimum (1, - 2)

Solution steps

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

The parabola parameters are:

a = 1, b = - 2, c = - 1

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

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

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(1, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, - 2)

3 x^{2} - 2 x + 1

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

x (2 x + 3)

- x^{2} + 6 x + 8

y = 6 x^{2} - 29 x - 5