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

##### 2 Answers

#### Answer:

The vertical asymptotes are

No oblique asymptote.

The horizontal asymptote is

#### Explanation:

Let's factorise the denominator

Let

The domain of

As we cannot divide by

The vertical asymptotes are

As the degree of the numerator is

To calculate the limits as

The horizontal asymptote is

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

#### Answer:

Vertical asymototes x=6 and x=3

Horizontal asymptote y=0

#### Explanation:

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