What is the vertices f(x)=(x+2)^2-4 ?

The vertices f(x)=(x+2)^2-4 is Minimum (-2,-4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=(x+2)^2-4

vertices

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

Minimum (- 2, - 4)

Solution steps

y = (x + 2)^{2} - 4

Rewrite

y = (x + 2)^{2} - 4

in the form

y = a x^{2} + b x + c

y = x^{2} + 4 x

The parabola parameters are:

a = 1, b = 4, c = 0

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} = - 4

Therefore the parabola vertex is

(- 2, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 4)

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

(x - 1)^{2} = - 4 (y - 2)

f (x) = \frac{1}{3} x^{2} + 2 x + 1

4 x^{2} - 8 x + 2 y = 5

y = 2 x^{2} + 5 x - 7

What is the vertices f(x)=(x+2)^2-4 ?
The vertices f(x)=(x+2)^2-4 is Minimum (-2,-4)