Question

How do you find the vertical, horizontal or slant asymptotes for `(x + 1)/(x^2 + 5x − 6)`?

Answer 1

The vertical asymptotes are `x=1` and `x=6`
No slant asymptote
The horizontal asymptote is `y=0`

Explanation:

We start by factorising the denominator

`x^2+5x-6=(x-1)(x+6)`

Let `f(x)=(x+1)/((x-1)(x+6))`

The domain of `f(x)` is `D_f(x)=RR-{1,-6}`

As we cannot divide by `0`, `x!=1` and `x!=-6`

So,

The vertical asymptotes are `x=1` and `x=6`

As the degree of the numerator is `<` than the degree of the denominator, there is no slant asymptote.

For the horizontal asymptote, we calculate the `lim` of `f(x)` as `x->+-oo`

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

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

Therefore,

The horizontal asymptote is `y=0`

graph{(y-(x+1)/(x^2+5x-6))(y)=0 [-10, 10, -5, 5]}