What is the vertices 3x^2+6x+2 ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 3x^2+6x+2

vertices

3 x^{2} + 6 x + 2

Minimum (- 1, - 1)

Solution steps

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

The parabola parameters are:

a = 3, b = 6, c = 2

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)

2 x^{2} - 3 x

f (x) = 4 (x - 2)^{2} + 11

f (x) = (2 x + 2) (x - 5)

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

y = 4 - (x - 3)^{2}