How do you find the vertical, horizontal or slant asymptotes for #f(x)=(x+1)/(x-3)#?

1 Answer
Sep 16, 2016

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"#

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