What is the vertices f(x)=2x^2-x-6 ?

The vertices f(x)=2x^2-x-6 is Minimum (1/4 ,-49/8)

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vertices f(x)=2x^2-x-6

vertices

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

Minimum (\frac{1}{4}, - \frac{49}{8})

Solution steps

y = 2 x^{2} - x - 6

The parabola parameters are:

a = 2, b = - 1, c = - 6

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{49}{8}

Therefore the parabola vertex is

(\frac{1}{4}, - \frac{49}{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{49}{8})

x^{2} - 4 y = 0

y^{2} = 4 x

3 x^{2} + 12 x + 4 y^{2} - 40 y + 100 = 0

(x + 2)^{2} + (y - 4)^{2} = 25

9 x^{2} - 4 y^{2} - 72 x = 0

What is the vertices f(x)=2x^2-x-6 ?
The vertices f(x)=2x^2-x-6 is Minimum (1/4 ,-49/8)