Do all functions with vertical asymptotes also have oblique asymptotes?

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?

• What do you think about $1/x$? – Rahul Oct 13 '18 at 4:01
• Take $f(x) = \tan( x)$ as an example. It do not have oblique asymptotes. – xbh Oct 13 '18 at 4:25
• @xbh Perhaps the question intended to consider algebraic functions? I’m not sure if that’s the right terminology. But yes—good example! – gen-z ready to perish Oct 13 '18 at 6:30
• @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! – xbh Oct 13 '18 at 8:39
• Yes, I meant rational functions, not just functions. I added in the edit. – Korvexius Oct 13 '18 at 16:07

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.
• @Korvexius Because $\frac{2x^2+1}{2x+3} = x - \frac {3x -1}{2x+3}.$ – David K Oct 15 '18 at 1:56