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

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

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

vertices

f (x) = 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})

f (x) = 9 x^{2} - x + 1

f (x) = - 2 x^{2} - 4 x + 6

y = s^{2} + 2 s + 2

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

x^{2} - 5 y = 0

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