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I just started learning about asymptotes in my Advanced Functions class, and as I was taking a look at all this stuff, a question came up. Do all rational functions that have vertical asymptotes also have an oblique asymptote? Or is an oblique asymptote only formed when the degree of the numerator is 1 higher than the degree of the denominator, and so only functions with a vertical asymptote with a degree of 1 can also have an oblique asymptote?

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    $\begingroup$ What do you think about $1/x$? $\endgroup$ – Rahul Oct 13 '18 at 4:01
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    $\begingroup$ Take $f(x) = \tan( x)$ as an example. It do not have oblique asymptotes. $\endgroup$ – xbh Oct 13 '18 at 4:25
  • $\begingroup$ @xbh Perhaps the question intended to consider algebraic functions? I’m not sure if that’s the right terminology. But yes—good example! $\endgroup$ – gen-z ready to perish Oct 13 '18 at 6:30
  • $\begingroup$ @ChaseRyanTaylor I assume you mean the rational functions, which is a quotient of two polynomial functions? In that case my example is not applicable here. Thank you for informing me! $\endgroup$ – xbh Oct 13 '18 at 8:39
  • $\begingroup$ Yes, I meant rational functions, not just functions. I added in the edit. $\endgroup$ – Korvexius Oct 13 '18 at 16:07
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You are dealing with dividing two functions and the bottom function approaches zero while the top function approaches a non zero value.

For example $$ f(x) = \frac {2x+1}{(x-5)(2x+3)}$$ where $x=5$ and $x=-3/2$ are vertical asymptotes.

Oblique asymptotes happen when your function behaves like a non-horizontal straight line as $x$ goes to $\infty$ or $-\infty$ We find slant asymptotes by dividing the top by the bottom and ignoring the remainder.

For example $$ f(x) = \frac {2x^2+1}{2x+3}$$ where your function behaves like $$ g(x)=x-3/2$$ which is a straight line.

A function may have both vertical and oblique asymptote but not both horizontal and slant.

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  • $\begingroup$ this might be a rookie question, but why isn't the equation of the oblique (gx) = x + 1/3 instead of x - 3/2 here? $\endgroup$ – Korvexius Oct 15 '18 at 1:30
  • $\begingroup$ @Korvexius Because $\frac{2x^2+1}{2x+3} = x - \frac {3x -1}{2x+3}.$ $\endgroup$ – David K Oct 15 '18 at 1:56

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