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

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

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

vertices

f (x) = 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)

f (x) = x^{2} - 4 x + 5

y^{2} = 8 x

x \leq y^{2}

\frac{1}{2} x^{2} + 8 x + 5

y = 2 x^{2} - 12 x + 3

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