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

2 Answers
Jun 4, 2016

Answer:

using lim calculus

Explanation:

vertical asymptote:
first find domain: #x^2-9!=0# then #x!=+-3#
then calculate
#lim f(x) #when #x->+3# and it is #24/0 =oo#
but is not the same for #x->-3#
so you have vertical asymptote x=3

horizontal asymptote;
calculate
#lim f(x)# when #x->oo# that's 1
so you have horizontal asymptote for y=1

when you have horizontal asymptote you haven't oblique ones

Jun 4, 2016

Answer:

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]}