What is the vertices f(x)=7x^2-x+1 ?

The vertices f(x)=7x^2-x+1 is Minimum (1/14 , 27/28)

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vertices f(x)=7x^2-x+1

vertices

f (x) = 7 x^{2} - x + 1

Minimum (\frac{1}{14}, \frac{27}{28})

Solution steps

y = 7 x^{2} - x + 1

The parabola parameters are:

a = 7, b = - 1, c = 1

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

x_{v} = - \frac{( - 1 )}{2 \cdot 7}

Simplify

x_{v} = \frac{1}{14}

Plug in

x_{v} = \frac{1}{14}

to find the

y_{v}

value

y_{v} = \frac{27}{28}

Therefore the parabola vertex is

(\frac{1}{14}, \frac{27}{28})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 7

Minimum (\frac{1}{14}, \frac{27}{28})

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

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

- x^{2} - 6 x - 9

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

y = x^{2} + 3 x - 5

What is the vertices f(x)=7x^2-x+1 ?
The vertices f(x)=7x^2-x+1 is Minimum (1/14 , 27/28)