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

1 Answer
Feb 22, 2017

Answer:

#"vertical asymptote at "x=-2#
#"horizontal asymptote at "y=1#

Explanation:

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+2=0rArrx=-2" is the asymptote"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

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

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

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