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

The vertices y=2x^2-x is Minimum (1/4 ,-1/8)

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vertices y=2x^2-x

vertices

y = 2 x^{2} - x

Minimum (\frac{1}{4}, - \frac{1}{8})

Solution steps

y = 2 x^{2} - x

The parabola parameters are:

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

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

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

Simplify

x_{v} = \frac{1}{4}

Plug in

x_{v} = \frac{1}{4}

to find the

y_{v}

value

y_{v} = - \frac{1}{8}

Therefore the parabola vertex is

(\frac{1}{4}, - \frac{1}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (\frac{1}{4}, - \frac{1}{8})

- x^{2} + 14 x - 45

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

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

y = x^{2} + 16 x + 71

f (x) = 2 x^{2} - 12 x + 1