Question

How do you find the vertical, horizontal and slant asymptotes of: `f(x)=(x^4-x^2)/(x(x-1)(x+2))`?

Answer 1

See below.

Explanation:

Start by simplifying the expression:

`(x^4-x^2)/(x(x-1)(x+2))`

`(x^2(x^2-1))/(x(x-1)(x+2))`

`(x^2-1)=(x+1)(x-1)` ( Difference of two squares )

`(x^2(x+1)(x-1))/(x(x-1)(x+2))`

`(x^cancel(2)(x+1)cancel((x-1)))/(cancelxcancel((x-1))(x+2))`

`(x(x+1))/(x+2)=(x^2+x)/(x+2)`

Vertical asymptotes occur where the function is undefined. This can be seen to be when `x=-2` (division by zero)

So the line `x=-2` is a vertical asymptote.

We now divide:

`(x^2+x)/(x+2)`

The quotient of this will be the oblique asymptote:

`(x+2)|bar(x^2+x)`

`color(white)(888888888)x`
`(x+2)|bar(x^2+x)`

`color(white)(888888888)x-1`
`(x+2)|bar(x^2+x)`
`color(white)(88888888)x^2+2x`
`color(white)(88888888888)-x`
`color(white)(88888888888)-x-2`
`color(white)(88888888888.888888)2`

We do not need to be concerned with the remainder. So the oblique asymptote is the line:

`y=x-1`

This is the line `(x^2+x)/(x+2)` approaches as `x->+-oo`

Graph: