What is the vertices f(x)=3x^2-x+1 ?

The vertices f(x)=3x^2-x+1 is Minimum (1/6 , 11/12)

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vertices f(x)=3x^2-x+1

vertices

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

Minimum (\frac{1}{6}, \frac{11}{12})

Solution steps

y = 3 x^{2} - x + 1

The parabola parameters are:

a = 3, b = - 1, c = 1

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

x_{v} = - \frac{( - 1 )}{2 \cdot 3}

Simplify

x_{v} = \frac{1}{6}

Plug in

x_{v} = \frac{1}{6}

to find the

y_{v}

value

y_{v} = \frac{11}{12}

Therefore the parabola vertex is

(\frac{1}{6}, \frac{11}{12})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (\frac{1}{6}, \frac{11}{12})

y = - 2 x^{2} + 5

f (x) = - x^{2} - 2 x + 15

- 2.318 + 0.2356 x - 0.002674 x^{2}

y = 4 x^{2} + 5 x - 7

y = x^{2} + x + 8

What is the vertices f(x)=3x^2-x+1 ?
The vertices f(x)=3x^2-x+1 is Minimum (1/6 , 11/12)