Question 975202
{{{"f(x)"=(x^2+x)/(3x^2-9x)}}}
<pre>
I'll just do the first one. 

Factor the numerator and denominator:

{{{"f(x)"=(x(x+1))/(3x(x-3))}}}

Setting the denominator = 0,  3x(x-3) = 0, tells us that
we have discontinuities at x=0, and at x=3

We must decide which type of discontinuity we have at 0 and 3.

1. If we can cancel a common factor in the numerator and denominator
and remove the discontinuity, then it is a "removable discontinuity" 
or "a hole in the graph".

2. If we can't cancel a factor, then the discontinuity is infinite and 
there is an asymptote there.

We have one of each type.

1. We can remove the discontinuity at x=0 by cancelling the x in the 
numerator and denominator.  To find out where the hole is, we cancel
the x and get a new function which is like the original function everywhere
except at the hole.  Let's call it g(x):

{{{"g(x)"=(x+1)/(3(x-3))}}}
{{{"g(x)"=(x+1)/(3x-9))}}}

g(x) doesn't have a hole at x=0, we substitute and find 
  
{{{"g(0)"=(0+1)/(3(0)-9)=1/(-9))=-1/9}}}

So f(x) has a hole at {{{(matrix(1,3,0,",",-1/9))}}}


2. We cannot remove the discontinuity at x=3 by cancelling, so there is
a vertical asymptote there, a "non-removable" discontinuity.

Since the degree of the numerator and denominator have the same degree, 1,
there is a horizontal asymptote at y = the ratio of the two leading
coefficients, so the horizontal asymptote has equation y=1/3,

So we plot the hole at {{{(matrix(1,3,0,",",-1/9))}}}, the vertical asymptote at x=3, and the horizontal 
asymptote at y=1/3:

{{{drawing(800,2400/7,-7,7,-3,3, graph(800,2400/7,-7,7,-3,3,(x^2+x)/(3x^2-9x)),
locate(-1.2,-.8,hole),
line(-0.8,-0.9,-0.2,-0.3), line(-0.22046602,-0.36694133,-0.2,-0.3), line(-0.26694133,-0.32046602,-0.2,-0.3),







circle(0,-1/9,.09), green(line(3,10,3,-10),line(-10,1/3,10,1/3)) )}}} 
 



Edwin</pre>