What is the vertices f(x)=3x^2+6x+4 ?

The vertices f(x)=3x^2+6x+4 is Minimum (-1,1)

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vertices f(x)=3x^2+6x+4

vertices

f (x) = 3 x^{2} + 6 x + 4

Minimum (- 1, 1)

Solution steps

y = 3 x^{2} + 6 x + 4

The parabola parameters are:

a = 3, b = 6, c = 4

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} = 1

Therefore the parabola vertex is

(- 1, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- 1, 1)

f (x) = 5 x^{2}

f (x) = x^{2} - 6 x + 9

x^{2} + x + 1

y = 2 x^{2} + 5 x - 12

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

What is the vertices f(x)=3x^2+6x+4 ?
The vertices f(x)=3x^2+6x+4 is Minimum (-1,1)