How do you find all the asymptotes for function #f(x)= ((3x^16)+28)/ ((15x^9)+33)#?

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Answer 1 · 2015-10-17T13:27:58

#f(x)# has a vertical asymptote #x = root(9)(-11/5)#

It has no horizontal or oblique asymptotes.

#f(x) = (3x^16+28)/(15x^9+33)#

#=1/5(15x^16+140)/(15x^9+33)#

#=1/5((15x^16+33x^7)-33x^7+140)/(15x^9+33)#

#=x^7/5-(33x^7-140)/(15(5x^9+11))#

When #x = root(9)(-11/5)# then denominator is zero, but the numerator is non-zero. So there is a vertical asymptote #x = root(9)(-11/5)#.

As #x->+-oo#, #(33x^7-140)/(15(5x^9+11)) ->0#

since the #9#th power of #x# in the denominator will dominate the #7#th power of #x# in the numerator.

So #f(x)# is asymptotic to #x^7/5# as #x->+-oo#.

This is not a linear asymptote.