How do you identify all asymptotes for #f(x)=(x+1)/(x-3)#?

1 Answer
May 14, 2017

Answer:

#"vertical asymptote at " x=3#
#"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-3=0rArrx=3" 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-3/x)=(1+1/x)/(1-3/x)#

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

#rArry=1" is the asymptote"#
graph{(x+1)/(x-3) [-11.25, 11.25, -5.625, 5.625]}