# uc3rmw_0204e04

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Hi there.
Sometimes the graph of
a rational function may
have missing points or holes, with
or without vertical asymptotes.
In this example, we will be working
with rational functions whose
graphs have a hole.
We are asked to find all vertical asymptotes
of the graph of each rational function,
as seen in parts a and b.
In part a, we have h of x equal to,
in the numerator, x squared minus 9
over, in the denominator, x minus 3.
In part B, we have g of x equal to x plus
2 in the numerator over x squared minus 4
in the denominator.
The vertical asymptotes
of a rational function
are defined by the values of x that make
the reduced form of the original function
undefined.
The reduced form of a
rational function is obtained
by completely factoring the numerator
and denominator of the function
and canceling out factors common
to the numerator and denominator.
For the function in part A, we start by
factoring the numerator and denominator.
We have that h of x is equal to--
in the numerator, we can factor that
using the difference of squares formula.
That's factored as x plus 3 times x minus 3.
And in the denominator, that
factor is already prime,
so we leave it alone as x minus 3.
At this point, we see that the factors x
minus 3 in the numerator and denominator
can be cancelled, and so h of x gets
reduced to x plus 3 if x is not equal to 3.
We see that there is no value for x
that makes this reduced form undefined.
Therefore, there is no vertical
asymptote for the function h of x.
We do note that the domain of x excludes
the value x equals 3, referred to as a hole.
This is because the function is undefined
at that value in its original form
prior to reducing it.
However, x equals 3 is
not a vertical asymptote.
For the function in part B, we start by
factoring the numerator and denominator,
as we did for the function in part a.
We have that g of x is equal
to x plus 2 in the numerator.
That factor is already prime.
And then we have the difference
of squares formula coming in again
for x squared minus 4, which
factors as x plus 2 times x minus 2.
Again, we see that x plus 2 cancels
both in the numerator and denominator,
and so we have that g of x reduces
to the form 1 over x minus 2
if x is not equal to minus 2.
Thus, there is a vertical asymptote
for this function at x equal 2.
Again, we note, as we did in
part A, that the domain of x also
excludes the value x equal
minus 2, called a hole,
at x equal minus 2 because
the function is undefined
at that value in its original
form prior to reducing it.
However, x equal 2 is a vertical
asymptote for this function.