anonymous
  • anonymous
Identify the asympatote
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\[y=1/x+1-2\]
anonymous
  • anonymous
@Kreshnik Im sorry
anonymous
  • anonymous
...@daja2fly I'm sorry too! but I'm not going to help you !

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More answers

anonymous
  • anonymous
you know it better ! bye .
anonymous
  • anonymous
@Mindy1234 do u no or not
Callisto
  • Callisto
when 1/(x+1) = undefined, then x=? is the vertical asymptote, if i remember it correctly...
anonymous
  • anonymous
0
Callisto
  • Callisto
0?
anonymous
  • anonymous
whats your question
Callisto
  • Callisto
i didn't ask anything...
anonymous
  • anonymous
so how do i figure this out
Callisto
  • Callisto
when 1/(x+1) = undefined, x+1 =0 , solve x and you'll get the vertical asymptote
anonymous
  • anonymous
ok
anonymous
  • anonymous
-1
anonymous
  • anonymous
-1 callisto
Callisto
  • Callisto
x=-1 it should be..
anonymous
  • anonymous
hello
anonymous
  • anonymous
\[\Large y=\frac{1}{x+1}-2=\frac{-2x-1}{x+1}\] Horizontal asymtote... \[\LARGE y=\lim_{x\to \infty}\frac{-2x-1}{x+1}=\lim_{x\to\infty}\frac{\cfrac{-2x}{x}-\cfrac{1}{x}}{\cfrac{x}{x}+\cfrac1x}\] \[\LARGE =\lim_{x\to\infty}\frac{-2-0}{1+0}=-2\] \[\LARGE \boxed{y=-2}\] Vertical Asymptote... \[\LARGE x=-1\] Not vertical , not horizontal, something in the middle asymptote (I don't know how to say it...not american...but I guess you know what I mean.) \[\LARGE y=kx+b\] \[\Large k=\lim_{x\to \infty}\frac{f(x)}{x}=\lim_{x\to\infty}\cfrac{\cfrac{-2x-1}{x+1}}{x}\] \[\Large \lim_{x\to \infty}\frac{-2x-1}{x^2+x}=0\] So \[\LARGE k=0\] and there's no asymtote like this When I was about to finish it... google crashed and I was about to brake my PC.. (here I wrote it again. hope you understand it HOW TO DO !) . @daja2fly

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