What is the vertices f(x)=3x^2+6x+1 ?

The vertices f(x)=3x^2+6x+1 is Minimum (-1,-2)

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vertices f(x)=3x^2+6x+1

vertices

f (x) = 3 x^{2} + 6 x + 1

Minimum (- 1, - 2)

Solution steps

y = 3 x^{2} + 6 x + 1

The parabola parameters are:

a = 3, b = 6, c = 1

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

x_{v} = - \frac{6}{2 \cdot 3}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(- 1, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- 1, - 2)

y = x^{2} + 6 x + 5

f (x) = 3 x^{2} + 6 x - 1

t^{2} + 2

y = - x^{2} - 4 x + 1

x^{2} - 3 x + 5

What is the vertices f(x)=3x^2+6x+1 ?
The vertices f(x)=3x^2+6x+1 is Minimum (-1,-2)