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

$$Pleaseineedsomeoneshelponthis$$$$HowdoifindanAsymptotetoacurve?$$$$andalsohowfindageneralsolutionforadifferential$$$$equation.$$$$$$

Answered by MJS last updated on 12/Jul/19

$$\mathrm{asymptotes}$$$$$$$$f\left(x\right):y=\frac{1}{x}$$$$\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}x=0\Rightarrow \mathrm{the}\mathrm{line}x=0\mathrm{is}\mathrm{an}$$$$\mathrm{asymptote}$$$$\mathrm{now}{\mathrm{let}}^{\prime }\mathrm{s}\mathrm{look}\mathrm{at}\mathrm{the}\mathrm{inverse}(\mathrm{solve}y=\frac{1}{x}\mathrm{for}x)$$$$\overline{f}\left(y\right):x=\frac{1}{y}$$$$\mathrm{similar}\mathrm{as}\mathrm{above},\mathrm{this}\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}y=0\Rightarrow$$$$\Rightarrow \mathrm{the}\mathrm{line}y=0\mathrm{is}\mathrm{also}\mathrm{an}\mathrm{asymptote}$$$$$$$$f\left(x\right):y=\frac{x-1}{x+1}$$$$\mathrm{not}\mathrm{defined}\mathrm{for}x+1=0\Rightarrow \mathrm{asymptote}x=-1$$$$\overline{f}\left(y\right):x=\frac{y+1}{1-y}$$$$\mathrm{not}\mathrm{defined}\mathrm{for}1-y=0\Rightarrow \mathrm{asymptote}y=1$$$$$$$$f\left(x\right):y=\frac{x+1}{(x-1)(x+2)}$$$$\mathrm{not}\mathrm{defined}\mathrm{for}x+2=0\mathrm{and}x-1=0\Rightarrow 2\mathrm{asymptoted}$$$$x=-2\mathrm{and}x=1$$$$\overline{f}\left(y\right):x=-\frac{y-1\pm \sqrt{9y^2+2y+1}}{2y}$$$$\mathrm{not}\mathrm{defined}\mathrm{for}2y=0\Rightarrow \mathrm{asymptote}y=0$$$$$$$$f\left(x\right):y=\frac{(x-1)(x+2)}{x+1}$$$$\mathrm{not}\mathrm{defined}\mathrm{for}x+1=0\Rightarrow \mathrm{asymptote}x=-1$$$$\overline{f}\left(y\right):x=\frac{y-1\pm \sqrt{y^2+2y+9}}{2}$$$$\mathrm{defined}\mathrm{\forall }y\in \mathbb{R}\Rightarrow \mathrm{no}\mathrm{asymptote}$$$$\mathrm{but}$$$$f\left(x\right):y=\frac{(x-1)(x+2)}{x+1}=x-\frac{2}{x+1}\Rightarrow \mathrm{we}\mathrm{have}$$$$\mathrm{another}\mathrm{asymptote}y=x$$$$$$$$\mathrm{same}\mathrm{here}:$$$$f\left(x\right):y=\frac{3x^2+x+2}{2x-5\mathrm{l}1}=\frac{3}{2}x+\frac{5}{4}+\frac{13}{4(2x-1)}$$$$\mathrm{not}\mathrm{defined}\mathrm{for}2x-1=0\Rightarrow \mathrm{asymptote}x=\frac{1}{2}$$$$\mathrm{plus}\mathrm{asymptote}y=\frac{3}{2}x+\frac{5}{4}$$$$\overline{f}\left(y\right):x=\frac{2y-1\pm \sqrt{4y^2-16y-23}}{6}$$$$\mathrm{this}\mathrm{is}\mathrm{defined}\mathrm{\forall }y\in \mathbb{R},\mathrm{although}x\notin \mathbb{R}\mathrm{for}\mathrm{some}$$$$\mathrm{values}\mathrm{of}y.\sqrt{4y^2-16y-23}\mathrm{might}\mathrm{is}\mathrm{not}\mathrm{always}$$$$\mathrm{a}\mathrm{real}\mathrm{number},\mathrm{but}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{always}\in \mathbb{C}.\mathrm{this}\mathrm{is}$$$$\mathrm{different}\mathrm{from}\mathrm{terms}\mathrm{like}\frac{\mathrm{term}}{x+a}\mathrm{which}\mathrm{are}\mathrm{not}$$$$\mathrm{defined}\mathrm{for}x+a=0\mathrm{in}\mathrm{any}\mathrm{set}\mathrm{if}\mathrm{numbers}$$$$$$$$\mathrm{we}\mathrm{can}\mathrm{also}\mathrm{have}\mathrm{asymptotic}\mathrm{curves}$$$$f\left(x\right):y=\frac{(x+1)(x+2)(x+3)}{x}=x^2+6x+11+\frac{6}{x}$$$$\mathrm{has}\mathrm{an}\mathrm{asymptote}x=0\mathrm{plus}\mathrm{the}\mathrm{asymptotic}$$$$\mathrm{curve}y=x^2+6x+11$$

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

$$Godblessyousir.$$

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