What is the vertices 3x^2+2x+1 ?

The vertices 3x^2+2x+1 is Minimum (-1/3 , 2/3)

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vertices 3x^2+2x+1

vertices

3 x^{2} + 2 x + 1

Minimum (- \frac{1}{3}, \frac{2}{3})

Solution steps

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

The parabola parameters are:

a = 3, b = 2, c = 1

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

x_{v} = - \frac{2}{2 \cdot 3}

Simplify

- \frac{2}{2 \cdot 3} : - \frac{1}{3}

x_{v} = - \frac{1}{3}

Plug in

x_{v} = - \frac{1}{3}

to find the

y_{v}

value

y_{v} = \frac{2}{3}

Therefore the parabola vertex is

(- \frac{1}{3}, \frac{2}{3})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- \frac{1}{3}, \frac{2}{3})

3 x^{2} + 8 y = 0

3 x^{2} + 2 x - 3

2 x^{2} + 4 x + 1

- x^{2} + 2 x + 8

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