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

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gjhfdfgBest ResponseYou've already chosen the best response.0
f(x)= \[\frac{ x }{ x^2 +1 }\]
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
solve x^2 + 1 = 0 for x
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
tell me what you get
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Would it just be x^2 = 1 ?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
good so far, what's next?
 one year ago

k.rajabhishekBest ResponseYou've already chosen the best response.0
there will be no vertical asymptote
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Hmm, Im not sure? Do I replace the variable with 1?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
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
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
so x^2 + 1 = 0 has no real solutions
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
leading to the fact that \[\Large \frac{ x }{ x^2 +1 }\] has no vertical asymptotes
 one year ago

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

jim_thompson5910Best ResponseYou've already chosen the best response.1
the numerator doesn't play any role in finding the vertical asymptotes unless you can make it cancel with something in the denominator
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
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
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
So basically the numerator was pointless in this equation?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
the basic thing is to simplify as much as possible (which couldn't be done in this case) then look at the denominator only
 one year ago

gjhfdfgBest ResponseYou've already chosen the best response.0
Got it, thank you again.!
 one year ago
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