What is the vertices x^2+4x-12 ?

The vertices x^2+4x-12 is Minimum (-2,-16)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices x^2+4x-12

vertices

x^{2} + 4 x - 12

Minimum (- 2, - 16)

Solution steps

y = x^{2} + 4 x - 12

The parabola parameters are:

a = 1, b = 4, c = - 12

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{4}{2 \cdot 1}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = - 16

Therefore the parabola vertex is

(- 2, - 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 16)

- x^{2} + 2 x + 3

2 x^{2} + 4 x - 1

(x + 2)^{2} = - 12 (y - 5)

2 x^{2} - 8 x + 6

f (x) = x^{2} + 14 x - 72