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.

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