Chapter 7 Rational Functions - College of the Redwoods

642 Chapter 7 Rational Functions
So, what point should we remove from the graph of g? We should remove the point
that has an x-value equal to 2. Therefore, we evaluate the function g(x) = 1/(x + 2)
at x = 2 and find
g(2) = 1
2 + 2 = 1 4 .
Because g(2) = 1/4, we remove the point (2, 1/4) from the graph of g to produce the
graph of f. The result is shown in Figure 3.
y
5
y=0 (2,1/4)
x
5
x=−2
Figure 3. The graph of f(x) = (x −
2)/((x − 2)(x + 2)) exhibits a vertical
asymptote at its restriction x = −2 and
a hole at its second restriction x = 2.
We pause to make an important observation. In Example 3, we started with the
function
f(x) =
x − 2
(x − 2)(x + 2) ,
which had restrictions at x = 2 and x = −2. After reducing, the function
g(x) = 1
x + 2
no longer had a restriction at x = 2. The function g had a single restriction at x = −2.
The result, as seen in Figure 3, was a vertical asymptote at the remaining restriction,
and a hole at the restriction that “went away” due to cancellation. This leads us to the
following procedure.
Version: Fall 2007