Question 815092
<pre>
f(x) = {{{(x^2+7x+10)/(x^2-7x-18)}}}

Since the degree of the numerator and denominator are both 2,
the horizontal asymptote has the equation

y = {{{(LEADING_COEFFICIENT_OF_NUMERATOR)/(LEADING_COEFFICIENT_OF_DENOMINATOR)}}}

y = {{{1/1}}}

y = 1

Let's draw the horizontal asymptote y = 1 (in green) 

{{{drawing(400,400,-25,25,-7,7,graph(400,400,-25,25,-7,7), green(line(-50,1,50,1)) )}}}

This function f(x) is not defined when the denominator = 0.
It either has a vertical asymptote there or else it
has a hole in the curve.

We set the denominator = 0.

  x²-7x-18 = 0
(x-9)(x+2) = 0
x-9=0;  x+2=0
  x=9;    x=-2

So the function is not defined at either x=9 or x=-2.

The domain of f(x) is (-oo,-2)U(-2,9)U(9,oo)

Next we find out if there is a vertical asymptote or
a "hole in the curve" (removable discontinuity)
at x=9 and x=-2

Factor the numerator x²+7x+10 as (x+5)(x+2)
We have already factored the denominator x²-7x-18 as (x-9)(x+2).


f(x) = {{{((x+5)(x+2)/(x-9)(x+2))}}}

Now we may ONLY cancel the (x+2)'s if we specify that x is not
equal to -2, for f(x) is not defined at x=-2 or at x=9.

But the fact that we have a factor (x-2) in the numerator and 
also an (x-2) factor in the denominator tells us there is a
removable discontinuity at x=-2.  And since there is no (x-9)
factor in the numerator tells us that there is a vertical
asymptote at x=9.

So we draw the vertical asymptote (also in green) which has 
the equation x=9

{{{drawing(400,400,-25,25,-7,7,graph(400,400,-25,25,-7,7), green(line(-50,1,50,1),line(9,-50,9,50)) )}}}
 

Now if we cancel the (x+2)'s we will get a function that we will
call g(x).  g(x) is exactly like f(x) except it will have a value at 
x=2 whereas f(x) does not have a value there.

So the graph that removes the dicontinuity (plugs up the hole) is

g(x) = {{{(x+5)/(x-9)}}} has an asymptote at x=9

Let's first draw g(x), which does not have a hole:

{{{drawing(400,400,-25,25,-7,7,graph(400,400,-25,25,-7,7,(x^2+7x+10)/(x^2-7x-18)), green(line(-50,1,50,1),line(9,-50,9,50)) )}}}

In fact when x=-2 we have g(-2)={{{(x+5)/(x-9)}}}={{{(-2+5)/(-2-9)}}}={{{-3/11)}}}


But we don't want g(x), we want f(x), so we must put a hole in
the curve at the point (-2,{{{-3/11}}}), for that is a removable
discontinuity.  

{{{drawing(400,400,-25,25,-7,7,graph(400,400,-25,25,-7,7,(x^2+7x+10)/(x^2-7x-18)), green(line(-50,1,50,1),line(9,-50,9,50)),circle(-2,-3/11,.7),
line(-7.7,-2,-4,-.9),line(-4.8,-.95,-4,-.9),line(-3.94,-1.2,-4,-.9),
locate(-10,-2,HOLE!)

 )}}}

Edwin</pre>