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

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

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vertices y=x^2-2x

vertices

y = 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)

f (x) = 3 x^{2} + 6 x + 4

f (x) = 5 x^{2}

f (x) = x^{2} - 6 x + 9

x^{2} + x + 1

y = 2 x^{2} + 5 x - 12