f(x) = 

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

y = 

y = 

y = 1

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



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) = 

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


 

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) =  has an asymptote at x=9

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



In fact when x=-2 we have g(-2)===


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



Edwin