Graphing Rational Equations with Non-Linear Asymptotes

[waffles]

I'm currently studying asymptotes of rational functions, and how to obtain the oblique/slant, vertical and horizontal from them as well as graph/sketch them. I'm fine with this however no problems. My question is for equations where the power of the numerator exceeds the denominator by more then 1.
for example: x^3/x+5 or x^4+2x^2+1/x^2-x+1. I tried to find some answers on the internet but to no avail. If I follow the same method as you do to obtain an oblique and use long / synthetic division, so:

1. for x^3/x+5 I get (x+5)(x^2-5x+25) with -125 as the remainder which I ignore.

2. for x^4+2x^2+1/x^2-x+1 I get (x^2-x+1)(x^2+x+2) with x-1 as the remainder.

So If I assume that the asymptote is the function x^2-5x+25 in 1. and x^2+x+2 in 2. Both of them are quadratics with complex roots, and I found that no matter what functions I make up I always end up getting complex roots for the functions remaining after division. So what do I do now, If I needed to sketch this?. Is there ever a situation where the quadratic or whatever function remaining is factorable? And If so do I graph it, and say its a parabolic asymptote do I treat it as any other asymptote and see what x approaches from either end of it? Also do these functions have vertical asymptotes? Because I tried using an internet function grapher and the image was a parabola above the x-axis and it crossed -5 on the left side for 1. Unless it was wrong. Thanks in advance for anyone who can help me.

 

[galactus]

The rational expressions in your examples have no asymptote. Asymptotes are not assured.

If the power of the numerator is greater than the power of the denominator(by more than 1), then there is no horizontal asymptote. If it exceeds by exactly 1, then it has an oblique asymptote.

Each one of those you posted result in quadratics, which are parabolas. No asymptotes.

The latter one does not cross the x-axis, so it has complex roots.

For instance, try:

\(\displaystyle \frac{x^{4}+3x^{3}-2x^{2}+x-1}{x^{2}+4x+4}\)

Upon dividing we get \(\displaystyle \frac{13x+7}{(x+2)^{2}}+x^{2}-x-2\)

\(\displaystyle x^{2}-x-2=(x-2)(x+1)\)

This one factors. Is that what you were getting at?.

 

[waffles]

I'm confused. From \(\displaystyle \frac{x^{4}+3x^{3}-2x^{2}+x-1}{x^{2}+4x+2}\) I got \(\displaystyle (x^2+4x+2)(x^2-x) + 3x-1\) so I can factor the \(\displaystyle x^2-x\) to \(\displaystyle x(x-1)\). I don't know how to get to your solution. And how do I proceed to graph this, do I just use \(\displaystyle x^2-x\)?

 

[galactus]

I'm sorry, I had a typo. I fixed it.

 

[soroban]

Hello, waffles!

I don't understand why you're concerned with roots of asymptotes, real or otherwise.


\(\displaystyle \text{Example: }\;y \:=\:\frac{x^3 + x^2 - 4x - 2}{x+1}\)

\(\displaystyle \text{With long division, we get: }\:y \;=\;x^2 -4 + \frac{2}{x+1}\)

\(\displaystyle \text{As }x\to\pm\infty\text{, the fraction approaches 0}\)
. . \(\displaystyle \text{and the function approaches the parabola: } \:y \:=\:x^2-4\)

\(\displaystyle \text{The parabola happ\!ens to have }x\text{-intercepts }\pm2\quad \text{ . . . but who cares?}\)

 

[waffles]

Thank you guys, I think I understand it now. If the power is greater then 1 in the numerator then in the denominator there is no asymptotes period and just a normal function I assume. Oh and sorobon in your example with x+1 on the bottom does that mean theres a vertical asymptote at -1 and thats it?

 

[tkhunny]

Whoever said asymptotes had to be lines? This is a problem in the nomenclature, I think. All we usually see is "oblique" or "slant". These terms imply lines. Well, if the asymptote has an equation of degree 2 or more, it's still an asymptote, it's just not a line. You entitled this post "Non-Linear Asymptotes". Now, down at the bottom you decide they don't exist? Don't do that. Things should make sense.