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anonymous
 3 years ago
Find the vertical asymptotes, if any, of the graph of the rational function.
anonymous
 3 years ago
Find the vertical asymptotes, if any, of the graph of the rational function.

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0f(x)= \[\frac{ x }{ x^2 +1 }\]

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1solve x^2 + 1 = 0 for x

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1tell me what you get

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Would it just be x^2 = 1 ?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1good so far, what's next?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there will be no vertical asymptote

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hmm, Im not sure? Do I replace the variable with 1?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1take 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
 3 years ago
Best ResponseYou've already chosen the best response.1so x^2 + 1 = 0 has no real solutions

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1leading to the fact that \[\Large \frac{ x }{ x^2 +1 }\] has no vertical asymptotes

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Ah, ok. Thank you.! Just out of curiosity, what happened to the x on top of the fraction?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1the numerator doesn't play any role in finding the vertical asymptotes unless you can make it cancel with something in the denominator

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1if 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
 3 years ago
Best ResponseYou've already chosen the best response.0So basically the numerator was pointless in this equation?

jim_thompson5910
 3 years ago
Best ResponseYou've already chosen the best response.1the basic thing is to simplify as much as possible (which couldn't be done in this case) then look at the denominator only

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Got it, thank you again.!
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