What is the vertices f(x)=9x^2-x+1 ?

The vertices f(x)=9x^2-x+1 is Minimum (1/18 , 35/36)

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vertices f(x)=9x^2-x+1

vertices

f (x) = 9 x^{2} - x + 1

Minimum (\frac{1}{18}, \frac{35}{36})

Solution steps

y = 9 x^{2} - x + 1

The parabola parameters are:

a = 9, b = - 1, c = 1

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

x_{v} = - \frac{( - 1 )}{2 \cdot 9}

Simplify

x_{v} = \frac{1}{18}

Plug in

x_{v} = \frac{1}{18}

to find the

y_{v}

value

y_{v} = \frac{35}{36}

Therefore the parabola vertex is

(\frac{1}{18}, \frac{35}{36})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 9

Minimum (\frac{1}{18}, \frac{35}{36})

f (x) = - 2 x^{2} - 4 x + 6

y = s^{2} + 2 s + 2

f (x) = 3 x^{2} + 2 x - 1

x^{2} - 5 y = 0

y = 2 x^{2} - 8 x

What is the vertices f(x)=9x^2-x+1 ?
The vertices f(x)=9x^2-x+1 is Minimum (1/18 , 35/36)