Chapter 7 Rational Functions - College of the Redwoods

Section 7.7 Solving Rational Equations 719
Let’s look at another example.
◮ Example 10.
Solve the following equation for x, both graphically and analytically.
1
x + 2 − x
2 − x = x + 6
x 2 (11)
− 4
We start the graphical solution in the usual manner, loading the left- and right-hand
sides of equation (11) into Y1 and Y2, as shown in Figure 9(a). Note that in the
resulting plot, shown in Figure 9(b), it is very difficult to interpret where the graph
of the left-hand side intersects the graph of the right-hand side of equation (11).
(a)
(b)
Figure 9. Sketch the left- and
right-hand sides of equation (11).
In this situation, a better strategy is to make one side of equation (11) equal to zero.
1
x + 2 − x
2 − x − x + 6
x 2 − 4 = 0 (12)
Our approach will now change. We’ll plot the left-hand side of equation (12), then
find where the left-hand side is equal to zero; that is, we’ll find where the graph of the
left-hand side of equation (12) intercepts the x-axis.
With this thought in mind, load the left-hand side of equation (12) into Y1,
as shown in Figure 10(a). Note that the graph in Figure 10(b) appears to have
only one vertical asymptote at x = −2 (some cancellation must remove the factor
of x − 2 from the denominator when you combine the terms of the left-hand side of
equation (12) 20 ). Further, when you use the zero utility in the CALC menu of the
graphing calculator, there appears to be a zero at x = −4, as shown in Figure 10(b).
(a)
(b)
Figure 10. Finding the zero of
the left-hand side of equation (12).
20 Closer analysis might reveal a “hole” in the graph, but we push on because our check at the end of the
problem will reveal a false solution.
Version: Fall 2007