What is area y^2=4x+4,y=4x+16 ?

The answer to area y^2=4x+4,y=4x+16 is No Solution

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area y^2=4x+4,y=4x+16

area

y^{2} = 4 x + 4, y = 4 x + 16

No Solution

Solution steps

Isolate

y

for

y^{2} = 4 x + 4 : y = 2 x + 1, y = - 2 x + 1

f_{1} (x) = 2 x + 1

f_{2} (x) = - 2 x + 1

To find the intersection points solve

f_{i} (x) = f_{j} (x)

2 x + 1 = - 2 x + 1 : x = - 1

2 x + 1 = 4 x + 16 :

No Solution for

x \in R

- 2 x + 1 = 4 x + 16 :

No Solution for

x \in R

Curves not bounded

= No Solution

\frac{d}{d x} (\frac{x - 3}{3 + sin ( x )})

y = \frac{3}{x}, at P (0.5, 6)

\frac{d y}{d x} = - 3 x y^{3}

x \to 1 lim (x^{2} - 3)

4 y^{^{''}} - 4 y^{^{'}} + 2 y = 0

What is area y^2=4x+4,y=4x+16 ?
The answer to area y^2=4x+4,y=4x+16 is No Solution