Question

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

Answer 1

The vertical asymptote is `x=1`
The slant asymptote is `y=x+2`
No horizontal asymptote

Explanation:

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

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

The vertical asymptote is `x=1`

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

Let's do a long division

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

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

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

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

`color(white)(aaaaaaa)` `+0+5`

Therefore,

`f(x)=(x^2+x+3)/(x-1)=(x+2)+5/(x-1)`

`lim_(x->-oo)(f(x)-(x+2))=lim_(x->-oo)5/x=0^-`

`lim_(x->+oo)(f(x)-(x+2))=lim_(x->+oo)5/x=0^+`

The slant asymptote is `y=x+2`

graph{(y-(x^2+x+3)/(x-1))(y-x-2)=0 [-32.48, 32.44, -16.2, 16.32]}