# How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)= (x^2+4x+3)/(x^2 - 9)#?

##### 2 Answers

using lim calculus

#### Explanation:

vertical asymptote:

first find domain:

then calculate

but is not the same for

so you have vertical asymptote x=3

horizontal asymptote;

calculate

so you have horizontal asymptote for y=1

when you have horizontal asymptote you haven't oblique ones

vertical asymptote x = 3

horizontal asymptote y = 1

#### Explanation:

The first step is to factorise and simplify f(x).

#f(x)=(cancel((x+3))(x+1))/(cancel((x+3))(x-3))=(x+1)/(x-3)# Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation set the denominator equal to zero.

solve : x - 3 = 0 → x = 3 is the asymptote

Horizontal asymptotes occur as

#lim_(xto+-oo),f(x)toc" ( a constant)"# divide terms on numerator/denominator by x

#(x/x+1/x)/(x/x-3/x)=(1+1/x)/(1-3/x)# as

#xto+-oo,f(x)to(1+0)/(1-0)#

#rArry=1" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ). Hence there are no oblique asymptotes.

graph{(x^2+4x+3)/(x^2-9) [-10, 10, -5, 5]}