Oblique Asymptotes

[joeyjon]

I can find oblique asymptotes.

My question is : Is it possible to have points on the oblique asymptote that a line can "pass through" similar to a horizontal asymptote?

Usually, after finding my horizontal asymptote, I check for any points on the asymptote before trying to find my line points on the graph.

Should i check for points on the oblique asymptote before proceeding to find the points of the function, and if so how?

(If it is the same as finding points on the horizontal - setting the equation equal to the numerical value of the asymptote, then I know how to do it.)

Thank you!!

 

[tkhunny]

Sure! What did you have in mind?

Something like this? \(\displaystyle y = \frac{x^{3}}{(x+1)\cdot(x-1)}\) with a little intersection at x = 0?

Or maybe a little farther out on the asymptote: \(\displaystyle y = \frac{2.30430\cdot x^{3}\cdot (x-4.96744)}{(x+1)\cdot(x-1)\cdot(x-4.95116)}\) with a bit of an encounter at x = 5.43161 and x = 55.9733?

 

[joeyjon]

Ok, Thanks.

My College Algebra book never even brings up the subject that a line can intersect an oblique asymptote, so...

For something simple like...

f (x) = x²-1 / x-2

I know the line doesn't intersect the asymptote because it's an example in the book, but if i didn't know that, how would I check for possible intersections?

and shouldn't it be standard practice to test for intersections just like I should test each horizontal asymptote I encounter?

I'm also wondering that maybe this falls beyond the scope of algebra, maybe calculus tools are needed. It just seems odd to me that the book doesn't tell us to check for intersections.

 

[tkhunny]

In order to draw the rational function completely properly, you will need to know which side of the asymptote you are on as the independent variable increases without bound. Generally, for linear asymptotes, (horizontal or oblique), the equation of the line is very simple and an algebraic solution is easy enough.

In your example, \(\displaystyle y = \frac{x^{2}-1}{x-2}\), we have the oblique asymptote \(\displaystyle y = x+2\). Then...


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

Produces any x-value where they intersect. In this case, it is seen rather quickly that there is no Real Number solution to this equation.

Excellent exploration. Keep asking questions!!

 

[joeyjon]

Thank you!

same way to find the intersections as a horizontal asymptote. That's what I figured intuitively, but I wasn't sure.

That's a load off my mind.