How do you graph #f(x)=(2x^2)/(x^2-9)# using holes, vertical and horizontal asymptotes, x and y intercepts?

To summarize issue on asymptotes and holes for such algebraic expressions

1. Factorize numerator and denominator
2. Those monomials, binomials or polynomials that cancels out provide us with holes
3. Values of #x#, the vaariable, that make denominatr provide us vertical asymptotes
4. If degree of numerator is equal to that of denominator, we may have a horiizontal asymptote
5. If degree of numerator is just one more than that of denominator, we may have a slanting or oblique asymptote

Here we have #f(x)=(2x^2)/((x^2-9))#, which can be factorized as

#f(x)=(2x^2)/((x+3)(x-3))#

As there is no common factor between numerator and denominator, there s no hole.

Further vertical asymptotes are #x=-3# and #x=3#

and as #f(x)=(2x^2)/((x^2-9))=2/(1-9/x^2)#, as #x->oo#, #f(x)->2#, hence horizontal asymptote is #y=2#.

Observe that #f(-x)=f(x)# and hence graph is symmetric w.r.t. #y#-axis. Further as #x=0#, #f(x)=0#. Using calculas we can find that at #(0,0)# there is a local maxima as #(dy)/(dx)=(-36x)/((x^2-9)^2# and at #x=0# it is #0#. Further while for #x<-3# and #x>3#, function is positive, for #-3 < x < 3# function is negative.

Now take a few values of #x# say #{-10,-7,-4,-2,-1,1,2,4,7,10}# and corresponding values of #f(x)# are #{2 18/91,2 9/20,4 4/7,-1 3/5,-1/4,-1/4,-1 3/5,4 4/7,2 9/20,2 18/91}#

The graph appears as shown below:

graph{(2x^2)/(x^2-9) [-20, 20, -10, 10]}