What is the vertices 3x^2-x+5 ?

The vertices 3x^2-x+5 is Minimum (1/6 , 59/12)

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vertices 3x^2-x+5

vertices

3 x^{2} - x + 5

Minimum (\frac{1}{6}, \frac{59}{12})

Solution steps

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

The parabola parameters are:

a = 3, b = - 1, c = 5

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{59}{12}

Therefore the parabola vertex is

(\frac{1}{6}, \frac{59}{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{59}{12})

y = x^{2} - 8 x + 17

y = 2 x^{2} - 4 x + 15

4 (x + 6) + 3 (2 + x + 3 x^{2}) - 2 x^{2}

f (x) = - x^{2} + 6 x - 11

y^{2} = - 20 x