Download CHAPTER 13: GRAPHS OF RATIONAL FUNCTIONS 1. Rational

CHAPTER 13: GRAPHS OF RATIONAL FUNCTIONS
1. Rational Functions and Asymptotes
Recall that a rational function is a ratio of two polynomials; that is, it is a
function, f , which can be expressed in the form
p(x)
where p(x) and q(x) are polynomials.
f (x) =
q(x)
The domain of such a function is the real line with the roots of q(x) removed.
What happens, then, at the roots of q(x)?
If a is a root of q(x), then the vertical line x = a is (virtually always) a
vertical asymptote of the function f .
1.1. Vertical Asymptotes. Recall (see the Chapter on Limits): If q(a) =
0 and p(a) 6= 0, then limx→a f (x) doesn’t exist. In this case the line x = a
is a vertical asymptote of the function. This means that as x → a, f (x) →
±∞; that is, as the inputs x get closer to a, the outputs get arbitrarily
large (either negative or positive). The graph of the function ‘explodes’ at
a vertical asymptote.
[It may (very rarely) happen for a number a that q(a) = 0 and p(a) = 0.
Recall (Chapter 5, section 2), that to find the limit in this case we cancel a
factor of x − a above and below and start again. If the limit doesn’t exist,
the line x = a is a vertical asymptote. (This is a special feature of rational
functions.)]
1.2. Horizontal Asymptotes.
Definition 1.1. The line y = a is a horizontal asymptote of the curve
y = f (x) if
lim f (x) = a
x→±∞
If f is a rational function, then it is very easy to calculate limx→∞ f (x).
A rational function will always be of the form
f (x) =
p(x)
axd + lower powers
= e
.
q(x)
bx + lower powers
(Thus d is the degree of p(x), e is the degree of q(x). a is the leading
coefficient of p(x) and b is the leading coefficient of q(x).)
The horizontal asymptote (if it exists) can be read off as follows:
(1) If d < e then the x-axis, y = 0, is a horizontal asymptote. (i.e.,
limx→∞ f (x) = 0 )
a
(2) If d = e then the line y = is a horizontal asymptote.
b
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(3) If d > e there is no horizontal asymptote.
Example 1.1.
100x5 + 4x3 + 27
2x7 + 1
y = 0 is a horizontal asymptote since 5 < 7. (Thus limx→∞ f (x) = 0.)
f (x) =
Example 1.2.
4x3 + 20x2 − 52x + 3
7x3 + 23x2 + 9x − 108
Here d = e = 3. So y = 4/7 is a horizontal asymptote. (Thus limx→∞ f (x) =
4/7.)
f (x) =
Example 1.3.
10x3 + 3x + 1
50x + 7
Since d = 3, e = 1, there is no horizontal asymptote. (Thus limx→∞ f (x)
does not exist.)
f (x) =
1.3. Method of Graphing rational functions. The asymptotes of a
rational function are often the dominant feature of its graph, and it is
always advisable to begin by determining the asymptotes:
(1) Find the asymptotes: To find the vertical asymptotes, find the roots
of the denominator.
To find the horizontal asymptotes, use the procedure described
above.
(2) Find the x- and y-intercepts:
To find the y-intercept, find f (0); that is, let x = 0.
To find the x-intercepts (the roots), solve f (x) = 0; i.e., solve
p(x) = 0 (where p(x) is the numerator).
(3) Find the critical points; i.e., solve f 0 (x) = 0.
(4) Determine whether f 0 > 0 or f 0 < 0 on each of the intervals between
the critical points and vertical asymptotes (the sign of the derivative
of a rational function can be different on both sides of a vertical
asymptote, even though it is not a critical point).
(5) Plot some points and sketch the graph.
2. Examples
Example 2.1. Consider the rational function
x
f (x) =
.
x−1
Find the asymptotes, the intercepts, intervals of increase and decrease, the
critical points and local extreme points (if any).
Solution:
(1) Vertical asymptote: Solve x − 1 = 0. Thus the line x = 1 is the only
vertical asymptote.
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(2) Horizontal asymptote. d = e = 1 here. The leading coefficients are
both equal to 1. So y = 1/1 = 1 is a horizontal asymptote.
(3) y-intercept: f (0) = 0: (0, 0) is the y-intercept.
(4) x-intercept: Solve x = 0: (0, 0) is the only x-intercept.
(5) Find the critical points:
(x − 1) · 1 − x · 1
f 0 (x) =
(x − 1)2
−1
=
(x − 1)2
There are no critical points (since −1 is never equal to 0). In particular, there are no local extreme points.
(6) The real line is cut in two by the unique vertical asymptote:
(1, ∞)
(−∞, 1)
0
0
f (0) = −1 f (2) = −1/12
f0 < 0
f0 < 0
&
&
So (−∞, 1) and (1, ∞) are both intervals of decrease. There are no
intervals of increase.
(7) Some points: (0, 0), (−1, 1/2), (2, 2).
Conclusions: The interval(s) of decrease are (−∞, 1) and (1, ∞).
There are no critical points or local extreme points.
Example 2.2.
f (x) =
Solution:
3x − 2
x3
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(1) To find the vertical asymptotes, solve x3 = 0: x = 0 (the y-axis) is
the only vertical asymptote.
(2) To find the horizontal asymptote: d = 1 and e = 3, so y = 0 (the
x-axis) is a horizontal asymptote.
(3) There is no y-intercept (the y-axis is a vertical asymptote). The
only x-intercept is x = 2/3.
(4)
f 0 (x) =
=
=
=
=
x3 · 3 − (3x − 2) · 3x2
(x3 )2
3x3 − 9x3 + 6x2
x6
2
6x − 6x3
x6
6x2 (1 − x)
x6
6(1 − x)
.
x4
Thus, to find the critical points we must solve 0 = 1 − x. x = 1 is
the only critical point.
(5) The vertical asymptote (x = 0) and the critical point (x = 1) divide
the x-axis into three parts:
(−∞, 0)
(0, 1)
(1, ∞)
6·2
6 · (1/2)
6 · (−1)
f 0 (1/2) =
f 0 (2) =
4
4
(−1)
(1/2)
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0
0
0
f >0
f >0
f <0
%
%
&
f 0 (−1) =
Thus, each of the intervals (−∞, 0) and (0, 1] is an interval of increase. The interval [1, ∞) is an interval of decrease.
(6) Some points: (−1, 5), (2/3, 0), (1, 1), (2, 1/2).
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Note that the unique critical point, 1, is a local maximum.
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