College Algebra, 2013a

4.2 Graphs of Rational Functions 329
the x-axis at (−1, 0). It should make you uncomfortable plugging x = −1 intothereduced
formula for h(x), especially since we’ve made such a big deal concerning the stipulation about
not letting x = −1 for that formula. What we are really doing is taking a Calculus short-cut
to the more detailed kind of analysis near x = −1 which we will show below. Speaking of
which, for the discussion that follows, we will use the formula h(x) = (2x+1)(x+1)
x+2
, x ≠ −1.
ˆ The behavior of y = h(x) as x →−2: As x →−2 − , we imagine substituting a number
(−3)(−1)
a little bit less than −2. We have h(x) ≈
(very small (−)) ≈ 3
(very small (−))
≈ very big (−)
thus as x →−2 − , h(x) →−∞. On the other side of −2, as x →−2 + , we find that
3
h(x) ≈
very small (+)
≈ very big (+), so h(x) →∞.
ˆ The behavior of y = h(x) as x →−1. As x →−1 − , we imagine plugging in a number
(−1)(very small (−))
a bit less than x = −1. We have h(x) ≈
1
= very small (+) Hence, as
x →−1 − , h(x) → 0 + . This means that as x →−1 − , the graph is a bit above the point
(−1, 0). As x →−1 + (−1)(very small (+))
,wegeth(x) ≈
1
=verysmall(−). This gives us
that as x →−1 + , h(x) → 0 − , so the graph is a little bit lower than (−1, 0) here.
Graphically, we have
y
−3
x
5. For end behavior, we note that the degree of the numerator of h(x), 2x 3 +5x 2 +4x +1,
is 3 and the degree of the denominator, x 2 +3x + 2, is 2 so by Theorem 4.3, the graph of
y = h(x) has a slant asymptote. For x →±∞, we are far enough away from x = −1 tousethe
reduced formula, h(x) = (2x+1)(x+1)
x+2
, x ≠ −1. To perform long division, we multiply out the
numerator and get h(x) = 2x2 +3x+1
x+2
, x ≠ −1, and rewrite h(x) =2x − 1+ 3
x+2
, x ≠ −1. By
Theorem 4.3, the slant asymptote is y =2x − 1, and to better see how the graph approaches
3
the asymptote, we focus our attention on the term generated from the remainder,
x+2 .
ˆ The behavior of y = h(x) as x →−∞: Substituting x = −1 billion into 3
x+2 ,wegetthe
3
3
estimate
−1 billion
≈ very small (−). Hence, h(x) =2x−1+
x+2
≈ 2x−1+very small (−).
This means the graph of y = h(x) is a little bit below the line y =2x − 1asx →−∞.