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

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

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vertices y=x^2-x

vertices

y = 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})

f (x) = 6 x^{2} - 1

f (x) = - 2 x^{2} + 16 x - 31

f (x) = - x^{2} - 6 x - 10

y = 2 x (x - 4) + 7

x = y^{2} + 2 y - 1 - (2 y^{2} - 3 y + 6)