Question

How do you find vertical, horizontal and oblique asymptotes for `(x^4 -81) /(x^3+3x^2 - x - 3)`?

Answer 1

Invisible hole at (-3, -27/2). Slant asymptote : `x-y+3=0`. Vertical asymptotes : `uarr x = +-1 darr`. See graphs, The third is not to scale but gets graph for great `|y|` near.

Explanation:

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

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

Sans the hole at `H(-3, -27/2)`,

`y= ((x-3)(x^2+9))/((x-1)(x+1))`
graph{(x-3)(1+10/(x^2-1)) [-164, 164.1, -82.1, 82]}

graph{((x-3)(1+10/(x^2-1))-y)(y-x-3)(x-1+.01y)(x+1-.01y)=0 [-164, 164.1, -82.1, 82]}

graph{(x-3)(1+10/(x^2-1)) [-16, 16, -160, 160]}

Answer 2

We have vertical asymptote at `x=1` and `x=-1` and oblique asymptote at `y=x-3`

Explanation:

First factorize numerator and denominators

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

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

Hence `f(x)=(x^4-81)/(x^3+3x^2-x-3)=((x+3)(x-3)(x^2+9))/((x+1)(x-1)(x+3))`

= `((x-3)(x^2+9))/((x+1)(x-1))=(x^3-3x^2+9x-27)/(x^2-1)`

Observe that although we have cancelled out `x+3`, we have an undefined function `(x^4-81)/(x^3+3x^2-x-3)` at `x=-3`, though close to `x=-3`, on either side, `f(x)` is defined. As such we have a hole at `x=-3`

Further at `x->1` as also for `x_>-1`, `f(x)->+-oo` depending on whether we approach limits from left or right and hence

we have vertical asymptotes at `x=1` and `x=-1`

Here as degree of polynomial in numerator is just one more than that of denominator, we do not have a horizontal asymptote, but we do have a oblique or slant asymptote.

As `f(x)=(x^4-81)/(x^3+3x^2-x-3)=x-3+(10x^2-90)/(x^3+3x^2-x-3)`

= `x-3+(10/x-90/x^3)/(1+3/x-1/x^2-3/x^3)`

and as `x->+-oo`, `y->x-3`

we have an oblique or slant asymptote `y=x-3`
graph{(x^4-81)/(x^3+3x^2-x-3) [-20, 20, -40, 40]}