College Algebra, 2013a

4.2 Graphs of Rational Functions 325
3
y
(−) ” (+) 0 (−) ” (+)
−2 0 2
−3 −1 1 3
2
1
−5 −4 −3 −1 1 3 4 5
−1
−2
−3
x
A couple of notes are in order. First, the graph of y = f(x) certainly seems to possess symmetry
with respect to the origin. In fact, we can check f(−x) =−f(x) to see that f is an odd function.
In some textbooks, checking for symmetry is part of the standard procedure for graphing rational
functions; but since it happens comparatively rarely 9 we’ll just point it out when we see it. Also
note that while y = 0 is the horizontal asymptote, the graph of f actually crosses the x-axis at (0, 0).
The myth that graphs of rational functions can’t cross their horizontal asymptotes is completely
false, 10 as we shall see again in our next example.
Example 4.2.2. Sketch a detailed graph of g(x) = 2x2 − 3x − 5
x 2 − x − 6 .
Solution.
1. Setting x 2 − x − 6=0givesx = −2 andx = 3. Our domain is (−∞, −2) ∪ (−2, 3) ∪ (3, ∞).
2. Factoring g(x) givesg(x) = (2x−5)(x+1)
(x−3)(x+2)
. There is no cancellation, so g(x) isinlowestterms.
3. To find the x-intercept we set y = g(x) = 0. Using the factored form of g(x) above, we find
the zeros to be the solutions of (2x − 5)(x + 1) = 0. We obtain x = 5 2
and x = −1. Since
both of these numbers are in the domain of g, wehavetwox-intercepts, ( 5
2 , 0) and (−1, 0).
To find the y-intercept, we set x = 0 and find y = g(0) = 5 6 ,sooury-intercept is ( 0, 5 6)
.
4. Since g(x) was given to us in lowest terms, we have, once again by Theorem 4.1 vertical
asymptotes x = −2 andx = 3. Keeping in mind g(x) = (2x−5)(x+1)
(x−3)(x+2)
, we proceed to our
analysis near each of these values.
ˆ The behavior of y = g(x) as x →−2: As x →−2 − , we imagine substituting a number
a little bit less than −2. We have
g(x) ≈
9 And Jeff doesn’t think much of it to begin with...
10 That’s why we called it a MYTH!
(−9)(−1)
(−5)(very small (−)) ≈ 9
≈ very big (+)
very small (+)