Question

How do you find vertical, horizontal and oblique asymptotes for `(x+6)/(x^2-9x+18)`?

Answer 1

The vertical asymptotes are `x=3` and `x=6`
No oblique asymptote.
The horizontal asymptote is `y=0`

Explanation:

Let's factorise the denominator

`x^2-9x+18=(x-6)(x-3)`

Let

`f(x)=(x+6)/(x^2-9x+18)=(x+6)/((x-6)(x-3))`

The domain of `f(x)` is `D_f(x)=RR-{3,6}`

As we cannot divide by `0`, `x!=3` and `x!=6`

The vertical asymptotes are `x=3` and `x=6`

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

To calculate the limits as `x->+-oo`, we take the highest coefficients in the numerator and denominator

`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^(+)`

The horizontal asymptote is `y=0`

graph{(y-(x+6)/(x^2-9x+18))(y)=0 [-22.8, 22.8, -11.4, 11.4]}

Answer 2

Vertical asymototes x=6 and x=3

Horizontal asymptote y=0

Explanation:

`f(x) = (x+6)/(x^2-9x+18)`. Factorise the denominator:

`=(x+6)/((x-6)(x-3))`

f(x) has no oblique asymptotes. It has two vertical asymptotes given by x-6=0 and x-3=0.

Since the degree of numerator is less than that of denominator, divide both by x , giving `f(x)= (1+6/x)/(x-9+18/x)` . Now find the limit as x`-> oo` giving f(x)=y=0, which is a horizontal asymptotes