Calculus
Of One Real Variable – By Pheng Kim Ving 
5.6 
Return
To Contents
Go To Problems & Solutions
1. Asymptotes 
Fig. 1.1
Horizontal And Vertical Asymptotes.

Fig. 1.2
Horizontal Asymptotes. 
The horizontal line y = L is called a horizontal asymptote of the graph of y = f(x) if:

Fig. 1.3
Vertical Asymptotes.

The vertical line x = a is called a vertical asymptote of the graph of y = f(x) if one of the following is satisfied:

The graph of f(x) = (x^{2} + 5x – 4)/(2x – 2) is
sketched in Fig. 1.4. Long division of the numerator x^{2} + 5x – 4 by the
denominator 2x – 2 gives: quotient = (1/2)x + 3 and remainder = 2. So:
Fig. 1.4
Oblique asymptote of graph of y = f(x) = (x^{2} + 5x – 4)/(2x – 2) is


Remark 1.1
The graph of a function can intersect
a horizontal or oblique asymptote, but can never intersect a vertical asymptote
(why?
hint: definition of a function).
iii. The graph of a function f may have two onesided horizontal asymptotes. For example, see Fig. 1.5.
iv. The notions of one and twosided asymptotes also apply to vertical and oblique asymptotes.
Fig. 1.5
The graph of y = f(x) has two onesided
horizontal asymptotes.

Go To Problems & Solutions Return To Top Of Page
2. Asymptotes Of Rational Functions 
Let f(x) = P(x)/Q(x)
be a rational function, ie, a ratio or fraction of two polynomials P(x) and Q(x). Let's
denote the
degree of a polynomial p by deg(
p). Then:
The graph of f has a vertical asymptote at every point x where Q(x) = 0. 
If deg(P) < deg(Q), then the line y = 0 (the xaxis) is a horizontal asymptote of the graph of f. 
iii. Suppose deg(P) = deg(Q). Let a and b be the coefficients of the dominating terms of P and Q respectively; see
If deg(P) = deg(Q), then the
line y
= a/b,
where a and b
are the coefficients of the dominating terms of P
and 
iv. Suppose deg(P) = deg(Q) + 1. For an
example, see the discussion of the function f(x) = (x^{2} + 5x – 4)/(2x – 2)
above and its graph sketched in Fig. 1.4. Divide P by Q using long division to get:

Go To Problems & Solutions Return To Top Of Page
3. Sketching Graphs Of Functions 
When sketching the graph of a
function y = f(x), we have three sources of useful
information: f itself, f
', and f
''. We'll
follow these suggested steps:
i. Information From f:
Determine each of the following if any:
a.
The domain of f.
b. Intercepts: xintercepts
and yintercept.
c. Symmetry of the graph (is f even, odd, or neither?).
d. Limits at points of discontinuity
and at positive and negative infinity.
e. Asymptotes: determine
vertical and horizontal ones from the limits above, and oblique ones by long
division if
appropriate.
ii. Information From f ':
a. Find points x where f '(x) = 0 or f '(x) doesn't exist (critical points). Calculate the
value of f at each of them if
defined.
b. Determine the signs of f ';
if it's not obvious to do so by simply examining the expression of f '
, draw a chart for
f
'; see Section
5.3 Part 3 and Part
5.
iii. Information From f '':
a. Find points x
where f
''(x)
= 0 or f ''(x) doesn't exist. Calculate
the value of f at each of them if defined.
b. Determine the signs of f '';
if it's not obvious to do so by simply examining the expression of f ''
, draw a chart for
f
''; see Section
5.4 Part 4.
iv. Draw the chart for f, where the intervals of increase, decrease, and concavity are determined; see Section 5.4 Part 4. Indicate local maxima, local minima, and inflection points if any.
v. Determine additional points of the graph if helpful.
vi. Sketch the graph using all the information obtained above.
Example 3.1
Sketch the graph of the function:
Asymptotes:
long division yields:
vertical: lines x = –1 and x = 1,
horizontal: none,
oblique: line y = x.
First Derivative:
The chart for y'' is shown in Fig. 3.1. The chart for y is shown in Fig. 3.2. The graph of y is shown in Fig. 3.3.
Fig. 3.1 Chart For y''. 
Fig. 3.2
Chart For y.

Fig. 3.3 Graph of: 
EOS
Remark 3.1
In the charts, the double vertical
line below a point x and on a
row means that the function on
that row isn't defined or
doesn't exist at that point.
Problems & Solutions 
1. Sketch the graph of the function y = x^{2}(x^{2} – 1), making use of any suitable information you can
obtain from the
function and its first and second
derivatives.
Solution
Domain: R.
y = x^{2}(x^{2} – 1) = x^{2}(x + 1)(x – 1);
y = x^{2}(x^{2} – 1) = x^{4} – x^{2}.
Intercepts:
xintercepts:
x = 0, –1, and 1,
yintercept: y = 0.
Symmetry:
y(–x)
= (–x)^{4} – (–x)^{2} = x^{4} – x^{2} = y(x);
function is even; graph is symmetric about yaxis.
Limits:
Second Derivative:
from y' = 4x^{3} – 2x we get:
2. Sketch the graph of the function:
using information from the function and its first and second derivatives.
Domain: R – {–1, 1}.
Intercepts:
xIntercepts: x = 0,
yIntercept: y = 0.
Symmetry:
Second Derivative:
3. Sketch the graph of the function:
using information from the function and its first and second derivatives.
Solution
Domain: R – {1}.
Intercepts:
xintercepts:
x = – 2,
yintercept: y = – 4.
Symmetry:
y'
is never 0;
y'
doesn't exist at x = 1 where y doesn't exist either;
4. Sketch the graph of the function:
utilizing any useful information you can get from f , f ', and f ''.
Solution
Domain: R – {–1}.
Asymptotes:
long division yields:
vertical: line x = –1,
horizontal: none,
oblique: line y = x – 1.
First Derivative:
5. Sketch the graph of the function:
employing any useful information you can get from y , y ', and y ''.
Solution
Domain: R – {1}.
Intercepts:
xIntercepts:
x = 0,
yIntercept: y = 0.
Symmetry:
Asymptotes:
vertical: line x = 1,
horizontal: none,
oblique: none.
First Derivative:
y'' = 0 at x = 0,
y''
doesn't exist at x = 1 where y doesn't exist either;
if x = 0 then y
= 0, point (0, 0);
sign of y'' is same as
that of x/(x
– 1)^{3};
if x < 0 then x
< 1 and so (x – 1)^{3} < 0, thus y'' > 0,
if 0 < x < 1 then y''
< 0,
if 1 < x then y''
> 0.
Return To Top Of Page Return To Contents