What is the vertices y=2x^2-x-1 ?

The vertices y=2x^2-x-1 is Minimum (1/4 ,-9/8)

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vertices y=2x^2-x-1

vertices

y = 2 x^{2} - x - 1

Minimum (\frac{1}{4}, - \frac{9}{8})

Solution steps

y = 2 x^{2} - x - 1

The parabola parameters are:

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

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{9}{8}

Therefore the parabola vertex is

(\frac{1}{4}, - \frac{9}{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{9}{8})

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

y^{2} - 12 y + 4 x + 4 = 0

2 x - y^{2} = 4

f (x) = - 2 x^{2} + 5

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

What is the vertices y=2x^2-x-1 ?
The vertices y=2x^2-x-1 is Minimum (1/4 ,-9/8)