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

The vertices f(x)=x^2-x is Minimum (1/2 ,-1/4)

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vertices f(x)=x^2-x

vertices

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

Minimum (\frac{1}{2}, - \frac{1}{4})

Solution steps

y = x^{2} - x

The parabola parameters are:

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

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

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

Simplify

x_{v} = \frac{1}{2}

Plug in

x_{v} = \frac{1}{2}

to find the

y_{v}

value

y_{v} = - \frac{1}{4}

Therefore the parabola vertex is

(\frac{1}{2}, - \frac{1}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (\frac{1}{2}, - \frac{1}{4})

x^{2} - 8 x + 12

X^{2} - 5 X - 24

y = x^{2} + 8 x + 7

4 t^{2} + 1

f (x) = \frac{1}{2} (x - 1)^{2}

What is the vertices f(x)=x^2-x ?
The vertices f(x)=x^2-x is Minimum (1/2 ,-1/4)