Question

How do you find the asymptotes for `b(x)= (x^3-x+1)/(2x^4+x^3-x^2-1)`?

Answer 1

`b(x)` has two vertical asymptotes and one horizontal asymptote...

Explanation:

First let us find the zeros of the numerator:

Given:

`x^3-x+1 = 0`

Let `x = u+v`, to get:

`u^3+v^3+(3uv-1)(u+v)+1 = 0`

Add the constraint `v = 1/(3u)` to eliminate the expression in `(u+v)` and get:

`u^3+1/(27u^3)+1 = 0`

Multiply by `27u^3` and rearrange slightly to get:

`27(u^3)^2+27(u^3)+1 = 0`

Hence using the quadratic formula, we find:

`u^3 = (-27+-sqrt(27^2-4*27))/54`

`color(white)(u^3) = (-27+-3sqrt(81-12))/54`

`color(white)(u^3) = (-27+-3sqrt(69))/54`

Since this is real and the derivation is symmetric in `u` and `v`, we can deduce that the only real zero of `x^3-x+1` is:

`root(3)((-27+3sqrt(69))/54) + root(3)((-27-3sqrt(69))/54) ~~ -1.324717957`

The quartic is somewhat more tedious to solve algebraically, so suffice it to say that its two real zeros are approximately:

`-1.20291275`

`0.86169984`

These can be found numerically using a Newton-Raphson method, or a Durand-Kerner method.

Since these are distinct from the real zero of the cubic, these are points of vertical asymptotes and `b(x)` has no holes.

Since the numerator is of degree `3` which is less than that of the denominator, the given rational expression has a horizontal asymptote `y=0` and no oblique asymptotes.

graph{(x^3-x+1)/(2x^4+x^3-x^2-1) [-5, 5, -2.5, 2.5]}