Question

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

Answer 1

The vertical asymptote is `x=-2`
A hole at `x=2`
The slant asymptote is `y=x-2`
No horizontal asymptote.

Explanation:

The denominator is `x^2-4=(x+2)(x-2)`

Let's do a long division

`color(white)(aaaa)` `x^3-2x^2-x+2` `color(white)(aaaa)` `∣` `x^2-4`

`color(white)(aaaa)` `x^3` `color(white)(aaaaaa)` `-4x` `color(white)(aaaaaaa)` `∣` `x-2`

`color(white)(aaaa)` `0-2x^2` `color(white)(aaa)` `3x+2`

`color(white)(aaaaaa)` `-2x^2` `color(white)(aaaaaa)` `+8`

`color(white)(aaaaaaaaaa)` `0` `color(white)(aaa)` `3x-6`

Therefore,

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

`=(x-2)+(3cancel((x-2)))/((x+2)cancel((x-2))`

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

Let `f(x)=(x-2)+3/(x+2)`

As we cannot divide by `0`, `x!=-2`

The vertical asymptote is `x=-2`

`lim_(x->-oo)f(x)-(x-2)= lim_(x->-oo)3/x=0^-`

`lim_(x->+oo)f(x)-(x-2)= lim_(x->+oo)3/x=0^+`

The slant asymptote is `y=x-2`

graph{(y-(x^3-2x^2-x+2)/(x^2-4))(y-x+2)(y-50x-100)=0 [-23.33, 22.3, -16.34, 6.48]}

Answer 2

Alternative format for long division + graph

Explanation:

Note that I have used place keepers where there is no value solely so that things line up in formatting. Example `0x^2`

`" "x^3-2x^2-x+2`
`color(red)(x)(x^2-4) ->ul( x^3+0x^2-4x) larr" subtract"`
`" "0color(white)(.) -2x^2+3x+2`
`color(red)(-2)(x^2-4)->ul(color(white)(.) -2x^2 +0x+8)larr" subtract"`
`color(red)("Remainder") ->" "0color(white)(.)+3x-6`

`y" "=" "(x^3-2x-x+2)/(x^2-4)" " =" " (x-2) + (3x-6)/(x^2-4)`