Question

How do you find the slant asymptote of `(2x^3+2x)/(x^2-1)`?

Answer 1

The slant asymptote is `y=2x`.

Explanation:

Let `f(x)=(2x^2+2x)/(x^2-1)`

First, perform a long division

`color(white)(aaaa)` `2x^3+0x^2+2x` `color(white)(aaaa)` `|` `x^2-1`

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

`color(white)(aaaa)` `0x^3+0x^2+4x`

Therefore,

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

The slant asymptote is `y=2x`

`lim_(x->+oo)f(x)-2x=lim_(x->+oo)(4x)/(x^2-1)=lim_(x->+oo)4/x=0^+`

`lim_(x->-oo)f(x)-2x=lim_(x->-oo)(4x)/(x^2-1)=lim_(x->-oo)4/x=0^-`

graph{(y-(2x^3+2x)/(x^2-1))(y-2x)=0 [-35.66, 37.4, -15.9, 20.63]}