Question

How do you find the asymptotes for `g(x) = (x^4 -81) /(x^3+3x^2 - x - 3)`?

Answer 1

Vertical asymptotes at `x=-1`and `x=1` and oblique or slanting asymptote at `y=x`. Wealso have a hole at `x=-3`.

Explanation:

Let us first factorize the numerator `x^4-81` and denominator `x^3+3x^2-x-3`

`x^4-81=(x^2)^2-9^2=(x^2+9)(x^2-9)=(x^2+9)(x+3)(x-3)`

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

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

Hence `g(x)=((x^2+9)(x+3)(x-3))/((x+3)(x-1)(x+1))`

Observe that when we have `x+3=0` or `x-1=0` or `x+1=0`, `g(x)` is not defined. However, as `x+3` appears both in numerator and denominator, this cancels out and we can arrive at limiting value of `f(x)` when `x+3=0` or `x=-3`.

Hence, we have a hole at `x=-3`.

Further, when `x->1` or `x->-1` `g(x)->+-oo` depending on whether we approach these values from left or right.

Hence, we have vertical asymptotes at `x=-1`and `x=1`.

Further as degree of numerator is just one more that of denominator, we should have an oblique or slanting asymptote.

As `y=g(x)=(x^4-1)/(x^3+3x^2-x-3)=(x-1/x^3)/(1+3/x-1/x^2-3/x^3)`

and as `x->+-oo`, `y->x`

Hence, we have an oblique or slanting asymptote at `y=x`

graph{(x^4-81)/(x^3+3x^2-x-3) [-84, 76, -33, 47]}