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

The vertices y=3x^2+2x+6 is Minimum (-1/3 , 17/3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=3x^2+2x+6

vertices

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

Minimum (- \frac{1}{3}, \frac{17}{3})

Solution steps

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

The parabola parameters are:

a = 3, b = 2, c = 6

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{17}{3}

Therefore the parabola vertex is

(- \frac{1}{3}, \frac{17}{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{17}{3})

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

f (x) = - x^{2} + 4 x + 12

y^{2} = 12 x

y = \frac{x ^{2} x ^{1}}{x}

(x - 3)^{2} = 8 (y - 5)

What is the vertices y=3x^2+2x+6 ?
The vertices y=3x^2+2x+6 is Minimum (-1/3 , 17/3)