Function with two horizontal asymptotes and three vertical asymptotes

futoo

For the life of me I cannot figure out this problem: give an example of a function that has two horizontal asymptotes and three vertical asymptotes. Any help on this topic would be greatly appreciated.

berkeman

That's a hard one. What's an example of a function that has the two horizontal asymptotes? What about a function that meets the 2nd part of the question? Seems like maybe for the vertical ones, we could use x = zero and +/- infinity.... Then how to mix in the horizontal ones...?

We can't give you the answers, but we can try to ask some questions to help you figure it out.

Mark44

A graph could have a vertical asymptote at x = 0, but couldn't have vertical asymptotes at +/- infinity. They would have to occur for some finite numbers. It could have them at, say, x = 1 and x = -1. Vertical asymptotes are determined by factors in the denominator of a rational function that are zero at certain values of x. Horizontal asymptotes arise from the limit of the rational expression being different as x approaches infinity and as x approaches neg. infinity.

berkeman

Ah, thanks for the clarification -- my bad. I was just thinking in terms of infinities, but you're right, to be an asymptote, the value would have to go to infinity for some finite argument in the domain of the fuction.

futoo

since vertical asymptotes arise from 0 being in the denominator, would 1/(x^2-1)(x-3) satisfy having three vertical asymptotes? and wouldn't they be -1, 1, and 3

Mark44

Yes. Now you need to add something in the numerator so that you get two different hor. asymptotes. The function you have has a single hor. asymptote--the line y = 0. All you need to get a different horizontal asymptote is have a third degree polynomial in the numerator whose highest-degree term is x³. There's a bit of a snag, because you want the numerator and denominator to have the same sign for large, positive x, and different signs for very negative x. There's a way around this problem, though.

HallsofIvy

By the way, a "function" does not have to correspond to a single "formula".

futoo

I also know that tan^-1(x) has two horizontal asymptotes, do I then incorporate x^3 into the numerator of the equation? Also thanks a lot for all your help.

Mark44

Well, tan^-1(x) doesn't have anything to do with your problem other than as an example of a function with two hor. asymptotes.

If you have x³ in the numerator, with the denominator you mentioned earlier, you'll get only one hor. asymtote - the line y = 1. Your denominator is negative for very negative x, and positive for very positive x, which is the same for x³.

You need something so that the numerator (only) is positive for large, positive x and for very negative x. HallsOfIvy suggested having two formulas for your function. I'm thinking of something different that involves x³. The basic function I'm thinking of has a V shape.

futoo

since the numerator has to be positive for large positive x's and negative for negative x's, (x-4)^3/(x^2-1)(x-3) satisfies the requirements that you gave, but when I plug it into my calculator the graph does not show the necessary requirements

Bohrok

What about a piecewise-defined function?

AUMathTutor

Hint: adding the unit step function to one of the functions already given will give two different horizontal asymptotes and 3 different vertical asymptotes.

Mark44

You misread what I wrote. The numerator has to be positive for both x << 0 and for x >> 0 (x << 0 means x is very negative). Your function has only one hor. asymptote. If you can keep the numerator positive for large x and very negative x, you'll have two different hor. asymptotes. A simpler function in the numerator is x³. Is there anything you can do to it to make it always >= 0?

liquidsnak

an absolute value of a function is always a positive result

I just had this problem in calculus class at PSU

**Edit: realised your question was to teach**

Mark44

Bingo!
Are liquidsnak and futoo the same person?

liquidsnak

No, I was just eager to help out, since I just spent several hours this weekend trying to figure this out.