Question

How do you find the vertical, horizontal or slant asymptotes for `(x^3-x+1)/(2x^4+x^3-x^2-1)`?

Answer 1

This function has horizontal asymptote `y=0` and vertical asymptotes at approximately `x ~~ -1.2029` and `x ~~ 0.86170`

It has no slant asymptotes or holes.

Explanation:

Given:

`f(x) = (x^3-x+1)/(2x^4+x^3-x^2-1)`

Note that the numerator has degree `3` and the denominator degree `4`.

We can deduce that this rational function has horizontal asymptote `y=0`, since the denominator will dominate the numerator for large positive and negative values of `x`.

By Descartes' Rule of Signs, we can tell that the denominator has one positive real zero, since the pattern of signs of its coefficients is `+ + - -` with `1` change.

The pattern of signs of the coefficients of `f(-x)` is `+ - - -`, telling us that `f(x)` has exactly one negative real zero.

These two real zeros are potentially vertical asymptotes. They can only fail to be so if the numerator is zero at them (which could result in a hole instead). This happens if the numerator and denominator have a common factor - so let's find their GCF using Euclid's method...

`2x^4+x^3-x^2-1 = (x^3-x+1)(2x+1)+x^2-x-2`

`x^3-x+1 = (x^2-x-2)(x+1)+2x+3`

`x^2-x-2 = (2x+3)(1/2x-5/4)+7/4`

Since we found no exact division, the GCF is constant. There is no common linear or higher degree factor.

So `f(x)` does have vertical asymptotes at the zeros of the denominator.

Slant (a.k.a. oblique) asymptotes of a rational function can only arise if the degree of the numerator is one more than that of the denominator. Such is not the case in our example.

To actually find the zeros of `2x^4+x^3-x^2-1` algebraically is very messy. You would be better off using a numerical method such as Newton's method to find approximations:

`x_1 ~~ -1.2029`

`x_2 ~~ 0.86170`

The graph of `f(x)` looks like this:
graph{(x^3-x-1)/(2x^4+x^3-x^2-1) [-10, 10, -5, 5]}