What is the vertices y=3x^2+x-1 ?

The vertices y=3x^2+x-1 is Minimum (-1/6 ,-13/12)

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vertices y=3x^2+x-1

vertices

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

Minimum (- \frac{1}{6}, - \frac{13}{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{13}{12}

Therefore the parabola vertex is

(- \frac{1}{6}, - \frac{13}{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{13}{12})

28 x^{2} + 47 x + 15

y^{2} - 8 y + 4 x = 0

4 y^{2} + 9 y^{2} + 32 x - 18 y + 37 = 0

x^{2} + 3 x + 2

y = + (x - 3)^{2}

What is the vertices y=3x^2+x-1 ?
The vertices y=3x^2+x-1 is Minimum (-1/6 ,-13/12)