Please-i-need-someones-help-on-this-How-do-i-find-an-Asymptote-to-a-curve-and-also-how-find-a-general-solution-for-a-differential-equation-

Question Number 63983 by Rio Michael last updated on 11/Jul/19

P l e a s e i n e e d s o m e o n e s h e l p o n t h i s

H o w d o i f i n d a n A s y m p t o t e t o a c u r v e ?

a n d a l s o h o w f i n d a g e n e r a l s o l u t i o n f o r a d i f f e r e n t i a l

e q u a t i o n .

Answered by MJS last updated on 12/Jul/19

asymptotes

f (x) : y = \frac{1}{x}

is not defined for x = 0 \Rightarrow the line x = 0 is an

asymptote

now {let}^{'} s look at the inverse (solve y = \frac{1}{x} for x)

\bar{f} (y) : x = \frac{1}{y}

similar as above, this is not defined for y = 0 \Rightarrow

\Rightarrow the line y = 0 is also an asymptote

f (x) : y = \frac{x - 1}{x + 1}

not defined for x + 1 = 0 \Rightarrow asymptote x = - 1

\bar{f} (y) : x = \frac{y + 1}{1 - y}

not defined for 1 - y = 0 \Rightarrow asymptote y = 1

f (x) : y = \frac{x + 1}{(x - 1) (x + 2)}

not defined for x + 2 = 0 and x - 1 = 0 \Rightarrow 2 asymptoted

x = - 2 and x = 1

\bar{f} (y) : x = - \frac{y - 1 \pm \sqrt{9 y^{2} + 2 y + 1}}{2 y}

not defined for 2 y = 0 \Rightarrow asymptote y = 0

f (x) : y = \frac{(x - 1) (x + 2)}{x + 1}

not defined for x + 1 = 0 \Rightarrow asymptote x = - 1

\bar{f} (y) : x = \frac{y - 1 \pm \sqrt{y^{2} + 2 y + 9}}{2}

defined \forall y \in R \Rightarrow no asymptote

but

f (x) : y = \frac{(x - 1) (x + 2)}{x + 1} = x - \frac{2}{x + 1} \Rightarrow we have

another asymptote y = x

same here :

f (x) : y = \frac{3 x^{2} + x + 2}{2 x - 5 l 1} = \frac{3}{2} x + \frac{5}{4} + \frac{13}{4 (2 x - 1)}

not defined for 2 x - 1 = 0 \Rightarrow asymptote x = \frac{1}{2}

plus asymptote y = \frac{3}{2} x + \frac{5}{4}

\bar{f} (y) : x = \frac{2 y - 1 \pm \sqrt{4 y^{2} - 16 y - 23}}{6}

this is defined \forall y \in R, although x \notin R for some

values of y . \sqrt{4 y^{2} - 16 y - 23} might is not always

a real number, but {it}^{'} s always \in C . this is

different from terms like \frac{term}{x + a} which are not

defined for x + a = 0 in any set if numbers

we can also have asymptotic curves

f (x) : y = \frac{(x + 1) (x + 2) (x + 3)}{x} = x^{2} + 6 x + 11 + \frac{6}{x}

has an asymptote x = 0 plus the asymptotic

curve y = x^{2} + 6 x + 11

Commented by Rio Michael last updated on 12/Jul/19

G o d b l e s s y o u s i r .