How do you graph #f(x)=x/(x^2-1)# using holes, vertical and horizontal asymptotes, x and y intercepts?

#f(x)=x/(x^2-1)#

For horizontal asymptotes, #y=0#. This is because if you replace #x# with different numbers, you will notice that the numerator will always have a smaller value than the denominator. Hence, if you divide a smaller number by a larger number, the answer will be close to #0#.

Heads up: If the degree of the numerator is less than the degree of the denominator, then the horizontal asymptote is #y=0#

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

If the degree of the numerator is greater than the degree of the denominator, then you have to use synthetic division to find the oblique asymptote (in most cases, it is an oblique asymptote)

For vertical asymptotes, look at the denominator. It cannot equal to #0# as the graph will be undefined. Hence, we let #x^2-1=0# so we can find at what points is the graph undefined.

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

For intercepts,
When #y=0#, #x=0#
When #x=0#, #y=0#

Therefore, after plotting in your intercept and drawing in your vertical and horizontal asymptote, you can hopefully see the outline of your graph. Remember, the asymptotes only influence the end points of your graph and that's it. So the graph can actually cross the asymptotes anywhere else

graph{x/(x^2-1) [-10, 10, -5, 5]}
Above is the graph