Question 143151
First, simplify each equation a little:

{{{f(x)=4/(x^2-1)=4/((x+1)(x-1))}}}
This tells us right away that x cannot equal +/- 1!!! (No dividing by zero.) Moreover, the lines x=1 and x=-1 will be vertical asymptotes. Now, for horizontal asymptotes, we must consider the degree of the x term in the numerator compared to the denominator. That is, 0 compared with 2. Since 0<2 we will have a horizontal asymptote at y=0.

Now, having already said that the denominator cannot be zero, we can say that there will be NO x-intercepts (because the denominator is the only part with x-terms). Y-intercept is easily found, however:
f(0)=4/(-1)=-4. So (0,-4) is the y-intercept. I now have enough information to plot, however you may want to write out some points to test.
{{{graph( 300, 300, -10, 10, -10, 10, 4/(x^2-1))}}}
Note that the method this site uses to graph inaccurately draws vertical lines where the vertical asymptotes are. Do not include this on your graph.


#2. {{{f(x)=3x/(x^2-3x)=3x/(x(x-3))=3/(x-3)}}}
Much of the same applies here process-wise as in#1. x=3 is the vertical asymptote, y=0 is the horizontal asymptote, (0,-1) is the y-intercept.
{{{graph( 300, 300, -10, 10, -10, 10, 3x/(x^2-3x))}}}