What is the vertices y=3x^2-x-2 ?

The vertices y=3x^2-x-2 is Minimum (1/6 ,-25/12)

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vertices y=3x^2-x-2

vertices

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

Minimum (\frac{1}{6}, - \frac{25}{12})

Solution steps

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

The parabola parameters are:

a = 3, b = - 1, c = - 2

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{25}{12}

Therefore the parabola vertex is

(\frac{1}{6}, - \frac{25}{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{25}{12})

y = x^{2} + 16 x + 71

f (x) = 2 x^{2} - 12 x + 1

f (x) = x^{2} + 9 x + 8

8 x^{2} + 12 x + 3

- 5 x^{2} + 118 x + 7

What is the vertices y=3x^2-x-2 ?
The vertices y=3x^2-x-2 is Minimum (1/6 ,-25/12)