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gjhfdfg

  • 2 years ago

Find the vertical asymptotes, if any, of the graph of the rational function.

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  1. gjhfdfg
    • 2 years ago
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    f(x)= \[\frac{ x }{ x^2 +1 }\]

  2. jim_thompson5910
    • 2 years ago
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    solve x^2 + 1 = 0 for x

  3. jim_thompson5910
    • 2 years ago
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    tell me what you get

  4. gjhfdfg
    • 2 years ago
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    Would it just be x^2 = -1 ?

  5. jim_thompson5910
    • 2 years ago
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    good so far, what's next?

  6. k.rajabhishek
    • 2 years ago
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    there will be no vertical asymptote

  7. gjhfdfg
    • 2 years ago
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    Hmm, Im not sure? Do I replace the variable with -1?

  8. jim_thompson5910
    • 2 years ago
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    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

  9. jim_thompson5910
    • 2 years ago
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    so x^2 + 1 = 0 has no real solutions

  10. jim_thompson5910
    • 2 years ago
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    leading to the fact that \[\Large \frac{ x }{ x^2 +1 }\] has no vertical asymptotes

  11. gjhfdfg
    • 2 years ago
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    Ah, ok. Thank you.! Just out of curiosity, what happened to the x on top of the fraction?

  12. jim_thompson5910
    • 2 years ago
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    the numerator doesn't play any role in finding the vertical asymptotes unless you can make it cancel with something in the denominator

  13. jim_thompson5910
    • 2 years ago
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    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

  14. gjhfdfg
    • 2 years ago
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    So basically the numerator was pointless in this equation?

  15. jim_thompson5910
    • 2 years ago
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    pretty much

  16. jim_thompson5910
    • 2 years ago
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    the basic thing is to simplify as much as possible (which couldn't be done in this case) then look at the denominator only

  17. gjhfdfg
    • 2 years ago
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    Got it, thank you again.!

  18. jim_thompson5910
    • 2 years ago
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    sure thing

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