What is the vertices f(x)=6x^2-x+1 ?

The vertices f(x)=6x^2-x+1 is Minimum (1/12 , 23/24)

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vertices f(x)=6x^2-x+1

vertices

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

Minimum (\frac{1}{12}, \frac{23}{24})

Solution steps

y = 6 x^{2} - x + 1

The parabola parameters are:

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

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

x_{v} = - \frac{( - 1 )}{2 \cdot 6}

Simplify

x_{v} = \frac{1}{12}

Plug in

x_{v} = \frac{1}{12}

to find the

y_{v}

value

y_{v} = \frac{23}{24}

Therefore the parabola vertex is

(\frac{1}{12}, \frac{23}{24})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 6

Minimum (\frac{1}{12}, \frac{23}{24})

y = 2 x^{2} + 10 x + 12

x^{2} + 4 x - 21

x^{2} + 4 x - 12

- x^{2} + 2 x + 3

2 x^{2} + 4 x - 1

What is the vertices f(x)=6x^2-x+1 ?
The vertices f(x)=6x^2-x+1 is Minimum (1/12 , 23/24)