Question

How do you find the vertical, horizontal or slant asymptotes for `f(x)=(x^3)/((x-1)^2)`?

Answer 1

The vertical asymptote is `x=1`
No horozontal asym`ptote.
The slant asymptote is `y=x+2`

Explanation:

The domain of `f(x)` is `D_f(x)=RR-{1}`

As you cannot divide by `0`, `x!=1`

The vertical asymptote is `x=1`

As the degree of the numerator is `>` than the degree of the denominator, we have a slant asymptote.

The denominator is

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

We do a long division

`color(white)(aaaa)` `x^3` `color(white)(aaaaaaaaaaaaaaa)` `∣` `x^2-2x+1`

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

`color(white)(aaaaa)` `0+2x^2-x`

`color(white)(aaaaaaa)` `+2x^2-4x+2`

`color(white)(aaaaaaaaa)` `+0+3x-2`

So,

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

Therefore,

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

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

So,

The slant asymptote is `y=x+2`

graph{(y-(x^3/(x-1)^2))(y-x-2)(y-100(x-1))=0 [-27.35, 30.37, -11.67, 17.2]}