How do you graph #y=x/(x^2-9)# using asymptotes, intercepts, end behavior?

1 Answer
Nov 11, 2016

Answer:

The vertical asymptotes are #x=3# and #x=-3#
The horizontal asymptote is #y=0#
No slant asymptotes
Only one intercept at #(0,0)#

Explanation:

Let's factorise the denominator #x^2-9=(x+3)(x-3)#
As we canot divide by #0#, the vertical asymptotes are #x=3# and #x=-3#

As the degree of the numerator is #<# the degree of the denominator, we do not have a slant asymptote.

#lim_(x->-oo)y=lim_(x->-oo)x/x^2=lim_(x->-oo)1/x=0^(-)#

#lim_(x->+oo)y=lim_(x->+oo)x/x^2=lim_(x->+oo)1/x=0^(+)#

So, #y=0# is a horizontal asymptote

The intercept is, when #x=0##=>##y=0#
graph{x/(x^2-9) [-11.25, 11.24, -5.63, 5.62]}