What is the vertices y=3x^2+6x-12 ?

The vertices y=3x^2+6x-12 is Minimum (-1,-15)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=3x^2+6x-12

vertices

y = 3 x^{2} + 6 x - 12

Minimum (- 1, - 15)

Solution steps

y = 3 x^{2} + 6 x - 12

The parabola parameters are:

a = 3, b = 6, c = - 12

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} = - 15

Therefore the parabola vertex is

(- 1, - 15)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (- 1, - 15)

y = \frac{1}{2} x - \frac{3}{2} (4.2)

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

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

\frac{x ^{2} - 7}{x - 3}

Y (x) = - 1 x + 7, (5, - 8)

What is the vertices y=3x^2+6x-12 ?
The vertices y=3x^2+6x-12 is Minimum (-1,-15)