How do you find the vertical, horizontal or slant asymptotes for #x / (x^2 -1)#?

1 Answer
Jul 28, 2018

Answer:

Below

Explanation:

Vertical Asymptotes
The denominator cannot equal to #0#. Hence, if we let the denominator equal to #0#, we can see at what points the graph cannot cross.

#x^2-1=0#
#x^2=1#
#x=+-1#

Horizontal Asymptotes
When you sub different values into #x#, we can see that the value of the numerator is always less than the denominator. Hence, when you divide a small number by a larger number, then your answer will be close to #0#. Thus #y=0#.

Hint:
If the degree of the numerator is LESS THAN the denominator, then the asymptote is #y=0#

If the degree of the numerator is EQUAL to the denominator, then the asymptote is the ratio of the leading coefficients

If the degree of the numerator is GREATER THAN the denominator, then you need to use synthetic division to find what the OBLIQUE asymptote is.

Below is what the graph #x/(x^2-1)# looks like
graph{x/(x^2-1) [-10, 10, -5, 5]}