anonymous
  • anonymous
Find the vertical asymptotes, if any, of the graph of the rational function.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
f(x)= \[\frac{ x }{ x^2 +1 }\]
jim_thompson5910
  • jim_thompson5910
solve x^2 + 1 = 0 for x
jim_thompson5910
  • jim_thompson5910
tell me what you get

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anonymous
  • anonymous
Would it just be x^2 = -1 ?
jim_thompson5910
  • jim_thompson5910
good so far, what's next?
anonymous
  • anonymous
there will be no vertical asymptote
anonymous
  • anonymous
Hmm, Im not sure? Do I replace the variable with -1?
jim_thompson5910
  • jim_thompson5910
take the square root of both sides x = sqrt(-1) or x = -sqrt(-1) but there's a problem, you can't take the square root of -1 and get a real number
jim_thompson5910
  • jim_thompson5910
so x^2 + 1 = 0 has no real solutions
jim_thompson5910
  • jim_thompson5910
leading to the fact that \[\Large \frac{ x }{ x^2 +1 }\] has no vertical asymptotes
anonymous
  • anonymous
Ah, ok. Thank you.! Just out of curiosity, what happened to the x on top of the fraction?
jim_thompson5910
  • jim_thompson5910
the numerator doesn't play any role in finding the vertical asymptotes unless you can make it cancel with something in the denominator
jim_thompson5910
  • jim_thompson5910
if you had something like x -------- x^2+x the fraction would simplify to 1 ________ x + 1 and this would be a case where the numerator plays a role
anonymous
  • anonymous
So basically the numerator was pointless in this equation?
jim_thompson5910
  • jim_thompson5910
pretty much
jim_thompson5910
  • jim_thompson5910
the basic thing is to simplify as much as possible (which couldn't be done in this case) then look at the denominator only
anonymous
  • anonymous
Got it, thank you again.!
jim_thompson5910
  • jim_thompson5910
sure thing

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