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

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

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

vertices

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

Minimum (1, - 1)

Solution steps

y = x^{2} - 2 x

The parabola parameters are:

a = 1, b = - 2, c = 0

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} = - 1

Therefore the parabola vertex is

(1, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (1, - 1)

\frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1

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

f (x) = (x - 5) (x + 1)

x^{2} + (y - 4)^{2} = 4

x^{2} - 6 x + 5 y = - 34