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Mathematics 9 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

f(x)= \[\frac{ x }{ x^2 +1 }\]

jimthompson5910 (jim_thompson5910):

solve x^2 + 1 = 0 for x

jimthompson5910 (jim_thompson5910):

tell me what you get

OpenStudy (anonymous):

Would it just be x^2 = -1 ?

jimthompson5910 (jim_thompson5910):

good so far, what's next?

OpenStudy (anonymous):

there will be no vertical asymptote

OpenStudy (anonymous):

Hmm, Im not sure? Do I replace the variable with -1?

jimthompson5910 (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

jimthompson5910 (jim_thompson5910):

so x^2 + 1 = 0 has no real solutions

jimthompson5910 (jim_thompson5910):

leading to the fact that \[\Large \frac{ x }{ x^2 +1 }\] has no vertical asymptotes

OpenStudy (anonymous):

Ah, ok. Thank you.! Just out of curiosity, what happened to the x on top of the fraction?

jimthompson5910 (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

jimthompson5910 (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

OpenStudy (anonymous):

So basically the numerator was pointless in this equation?

jimthompson5910 (jim_thompson5910):

pretty much

jimthompson5910 (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

OpenStudy (anonymous):

Got it, thank you again.!

jimthompson5910 (jim_thompson5910):

sure thing

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