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

1 Answer
Jan 13, 2016

Answer:

I found:
Vert. #x=1#
Horiz. #y=1#

Explanation:

When you see this kind of functions (division between two expressions both with #x#) we start checking the denominator, in this case, #x-1#;

The denominator CANNOT BO EQUAL TO ZERO! In this case there is only one possibility to be zero, i.e., when:
#x=1#
This expression represents a vertical line passing through #x=1# called VERTICAL asymptote.
The graph of your function will tend to get near to this vertical line as much as possible (never touching it though!).

Next we look at the behaviour of your function for big values of the variable #x#:
let use #x=1,000,000#:
using this value you can see that #+3# and #-1# become negligible and you have:
#y=f(1,000,000)~~(1,000,000)/(1,000,000)=1#
This happens when you increase further #x# so that the graph wil tend to get near to the horizontal line of equation:
#y=1#
your HORIZONTAL asymptote.

Graphically you can see them:
graph{(x+3)/(x-1) [-10, 10, -5, 5]}