Question

How do you find the asymptotes for `(x^4-3x^3-21x^2+43x+60)/(x^4-6x^3+x^2+24x-20)`?

Answer 1

Horizontal asymptote (left and right): `y = 1`

Vertical asymptotes at `x=1`, `x=2`, `x=-2`

Hole at `(5, 9/7)`

Explanation:

Let's attempt to factorise the numerator and denominator first.

`color(white)()`
Numerator

Substituting `x=-1` in the numerator we get:

`1+3-21-43+60 = 0`

So `x=-1` is a zero and `(x+1)` a factor:

`x^4-3x^3-21x^2+43x+60 = (x+1)(x^3-4x^2-17x+60)`

Substituting `x=3` in the remaining cubic, we get:

`27-36-51+60 = 0`

So `x=3` is a zero and `(x-3)` a factor:

`x^3-4x^2-17x+60=(x-3)(x^2-x-20) = (x-3)(x-5)(x+4)`

In summary:

`x^4-3x^3-21x^2+43x+60 = (x+1)(x-3)(x-5)(x+4)`

`color(white)()`
Denominator

Substituting `x=1` in the denominator we get:

`1-6+1+24-20 = 0`

So `x=1` is a zero and `(x-1)` a factor:

`x^4-6x^3+x^2+24x-20 = (x-1)(x^3-5x^2-4x+20)`

The remaining cubic factors by grouping:

`x^3-5x^2-4x+20`

`=(x^3-5x^2)-(4x-20)`

`=x^2(x-5)-4(x-5)`

`=(x^2-4)(x-5)`

`=(x-2)(x+2)(x-5)`

In summary:

`x^4-6x^3+x^2+24x-20 = (x-1)(x-2)(x+2)(x-5)`

`color(white)()`
Together

`(x^4-3x^3-21x^2+43x+60)/(x^4-6x^3+x^2+24x-20)`

`=((x+1)(x-3)color(red)(cancel(color(black)((x-5))))(x+4))/((x-1)(x-2)(x+2)color(red)(cancel(color(black)((x-5)))))`

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

with exclusion `x != 5`

Looking at the leading terms of the numerator and denominator, the degrees and coefficients are identical, so there is a horizontal asymptote (left and right): `y = 1`

When `x=1` or `+-2` the denominator is zero and the numerator non-zero, so these values of `x` correspond to vertical asymptotes.

Substituting the value `x=5` into the simplified rational function, we find the coordinate of the hole as: `(5, 9/7)`

graph{(x^4-3x^3-21x^2+43x+60)/(x^4-6x^3+x^2+24x-20) [-41.83, 38.17, -10.24, 29.76]}