# How do you find vertical, horizontal and oblique asymptotes for #y= ((x+2)(x+1))/((3x-1)(x+2))#?

##### 1 Answer

vertical asymptote at

horizontal asymptote at

#### Explanation:

The first step here is to simplify the function y.

#y=(cancel((x+2))(x+1))/((3x-1)cancel((x+2)))=(x+1)/(3x-1)# The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve:

# 3x-1=0rArrx=1/3" is the asymptote"# Horizontal asymptotes occur as

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

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

#xto+-oo,yto(1+0)/(3-0)#

#rArry=1/3" is the asymptote"# Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both degree 1 ) Hence there are no oblique asymptotes.

graph{(x+1)/(3x-1) [-10, 10, -5, 5]}