How do you find the asymptotes for #(3x-2) / (x+1)#?

2 Answers
May 21, 2018

When #x=-1 =>-5/0# so there is an asymptote when #x=-1#

May 21, 2018

Answer:

#"vertical asymptote at "x=-1#
#"horizontal asymptote at "y=3#

Explanation:

#"let "f(x)=(3x-2)/(x+1)#

The denominator of f(x) cannot be zero as this would make f(x) 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 "x+1=0rArrx=-1" is the asymptote"#

#"horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" (a constant)"#

#"divide terms on numerator/denominator by "x#

#f(x)=((3x)/x-2/x)/(x/x+1/x)=(3-2/x)/(1+1/x)#

#"as "xto+-oo,f(x)to(3-0)/(1+0)#

#rArry=3" is the asymptote"#
graph{(3x-2)/(x+1) [-10, 10, -5, 5]}