What is the vertices-3x^2+2x+1 ?

The vertices-3x^2+2x+1 is Maximum (1/3 , 4/3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

vertices-3x^2+2x+1

vertices

Maximum (\frac{1}{3}, \frac{4}{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 ( - 3 )}

Simplify

- \frac{2}{2 ( - 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{4}{3}

Therefore the parabola vertex is

(\frac{1}{3}, \frac{4}{3})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (\frac{1}{3}, \frac{4}{3})