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

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

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-1

vertices

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

Minimum (- \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 \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{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

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

x^{2} - x

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

- (x - 30)^{2} + 40

f (x) = x^{2} + 3 x - 10

f (x) = x^{2} + 6 x + 11

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