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Chapter 1
Coordinates, Graphs, Lines
1.1 Real Numbers, Sets, and Inequalities (A Review)
1.2 Absolute Value
1.3 Coordinate Planes and Graphs
1.4 Lines
1.5 Distance; Circles; Equations of the Form y D ax2 C bx C c
1.1 REAL NUMBERS, SETS, AND INEQUALITIES (A REVIEW) : : : : : : : : : : : : : : :
Although this section covers material which you have probably already studied in ear-
lier mathematics courses, there is one topic which deserves very close and careful
attention: solving inequalities. As you will see, many calculus operations involve
solving inequalities. It is not surprising, therefore, that errors with inequalities are a
major source of mistakes in calculus!
1.1.1 Simple Inequalities and Common Errors : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
The five basic rules governing inequalities are given in Theorem 1.1.1. While all five
are important, you should concentrate on rule (c) because it causes by far the most
trouble:
If a < b then ac < bc when c is positive and ac > bc when c is negative.
Note this! Forgetting to reverse inequality signs when multiplying by a negative
number is one of the most common and costly mistakes in calculus. It occurs most
often when a person forgets that an expression containing a variable might take on
negative values, i.e., the negative values are “hidden.” Here are some typical exam-
ples:
98
 
Section 1.1 Real Numbers, Sets, and Inequalities (A Review) 99
Example A
Solve
x C 2
x C 1 > 0.
Incorrect “solution”
x C 2
x C 1 > 0
.x C 1/
µ
x C 2
x C 1
¶
> .x C 1/
"
⁄
¢ 0
* Oops! Error ! Not true when x C 1 < 0!
x C 2 > 0
x > ¡2
¾
⁄
* Since the previous step was incorrect, the inequalities which follow are incorrect.
The mistake in this solution lies in not realizing that x C 1 can be negative. We’ll
now give two correct solutions for this problem.
First solution
In order to multiply the inequality xC2
xC1 > 0 by x C 1 we need to consider two cases:
the case when x C 1 is positive and the case when x C 1 is negative:
Case 1:
x C 1 > 0; i.e.; x > ¡1:
In this case multiplying our inequality by x C 1 will not reverse the inequality
sign:
x C 2
x C 1 > 0
.x C 1/ ¢
µ
x C 2
x C 1
¶
> .x C 1/ ¢ 0
inequality in same direction since x C 1 > 0
x C 2 > 0
x > ¡2
There is one more step needed to obtain the set of solutions: we must discard
those x-values which are less than or equal to ¡1 since we are in Case 1 where
 
100 Chapter 1 Coordinates, Graphs, Lines
we only allow x > ¡1. Thus x > ¡2 is “cut back” to
x > ¡1
This is the set of solutions for our inequality in Case 1.
Case 2:
x C 1 < 0; i.e., x < ¡1:
In this case multiplying our inequality by x C 1 will reverse the inequality sign:
x C 2
x C 1 > 0
.x C 1/ ¢
µ
x C 2
x C 1
¶
< .x C 1/ ¢ 0
(reverse inequality since x C 1 < 0/
x C 2 < 0
x < ¡2
All these solutions are allowed in Case 2 since our only restriction is x < ¡1. (If
x is less than ¡2, then it is certainly less than ¡1.) Hence
x < ¡2
is the set of solutions for our inequality in Case 2. Putting the two cases together
gives the solution set
x > ¡1 or x < ¡2
When written in interval notation this is
.¡1;¡2/ [ .¡1;C1/
Second solution
As you can see, things can be a bit complicated when you are forced to deal with sep-
arate cases. Fortunately there is an alternate method of solution which avoids the use
of separate cases. The idea of the method is simple: the inequality
x C 2
x C 1 > 0
will be true only when the sign of x C 2 multiplied by the sign of x C 1 is positive.
To determine when this happens, draw two number lines, one for x C 2 and one for
 
Section 1.1 Real Numbers, Sets, and Inequalities (A Review) 101
x C 1, and indicate by C and ¡ signs where each term is positive or negative. Then
draw a third line for .x C 2/=.x C 1/; the signs along this line are determined by
multiplying the signs of the previous two lines.
For example, above ¡3 we find ¡ and ¡, the product of which is C. Hence we
assignC to¡3. On the other hand, above¡32 we findC and¡, the product of which
is¡. Hence we assign¡ to¡32. Determining the signs in this way can be done quickly
and the final result shows that .x C 2/=.x C 1/ is positive on the set
.¡1;¡2/ [ .¡1;C1/
This agrees with the answer in our previous solution.
Example B
Solve
.5¡ x/
.x C 2/ < 1.
Solution
We first move all the terms to one side of the inequality1:
0 < 1¡ 5¡ x
x C 2 D
.x C 2/¡ .5¡ x/
x C 2
D 2x ¡ 3
x C 2
Thus the inequality can be written
0 <
2x ¡ 3
x C 2
Since we now have a zero on one side of the inequality, we can use the solution meth-
ods of Example A. Lines for each of the terms 2x ¡ 3 and x C 2 are shown, and they
determine the signs for the quotient 2x¡3
xC2 as shown on the third line.
1See Companion Sections 0.2.2 and 0.2.3, if you find the fraction algebra confusing.
 
102 Chapter 1 Coordinates, Graphs, Lines
From this we can see that 2x¡3
xC2 is positive whenever x < ¡2 or x > 32. Hence the
solution set for our original inequality is
.¡1;¡2/ [
‡
3
2;C1
·
Example C
Solve
1
.x C 1/ <
2
.x ¡ 1/ .
Solution
As in Example B we first move all the terms to one side of the inequality:
0 <
2
x ¡ 1 ¡
1
x C 1 D
2.x C 1/¡ .x ¡ 1/
.x ¡ 1/.x C 1/
0 <
x C 3
.x ¡ 1/.x C 1/
We now have to consider the signs of three terms: x C 3, x ¡ 1 and x C 1. So we
draw lines for each of these terms, and then determine the signs for the quotient (by
multiplying three C or ¡ signs) as shown on the fourth line.
From this
x C 3
.x ¡ 1/.x C 1/
 
Section 1.1 Real Numbers, Sets, and Inequalities (A Review) 103
is positive whenever ¡3 < x < ¡1 or 1 < x . Hence the solution set for our original
inequality is
.¡3;¡1/ [ .1;C1/
1.1.2 Solving Complicated Inequalities: The Interval Method : : : : : : : : : : : : : : : : : : : : : : : : : : :
As you can see from Examples A, B and C, solving inequalities can be a tricky busi-
ness. This is especially true when dealing with the inequalities that so often arise
in calculus. For this reason we describe another easy-to-use technique that solves
nearly all inequalities encountered in applications: the interval method. Sometimes
it is called the method of test points, which is the name Anton uses.
Suppose we have an inequality of the form f .x/ > 0, where f .x/ is an expres-
sion (i.e., a function2) in the variable x . We first observe that a normal 3 f .x/ cannot
change sign on an interval unless it first becomes zero or undefined. Said another way,
any normal f .x/ cannot go from below the x-axis to above the x-axis without crossing
the x-axis ( f .x/ D 0) or jumping over it through a missing point ( f .x/ undefined).
To see this principle in picture form, consider the f .x/ whose graph is shown.
Notice that f .x/ changes sign at the points 1; 2 and 3. As predicted above, these
are the points at which f .x/ crosses the x-axis . f .1/ D f .3/ D 0/, or at which f .x/
is missing ( f .2/ is undefined). The crossing points (x D 1 and x D 3) and missing
points (x D 2) can be used to divide the real line into four open intervals:
.¡1; 1/; .1; 2/; .2; 3/; .3;C1/
2You may already be familiar with the “function” symbol f .x/. If not, do not let it confuse you. Here it is merely shorthand
for “an expression in x .” For example, in Example C we considered the inequality
x C 3
.x ¡ 1/ .x C 1/ > 0:
This inequality could also be expressed as f .x/ > 0 where
f .x/ D x C 3
.x ¡ 1/ .x C 1/ .
Functions will be studied in Chapter 1.
3 f .x/ is “normal” if it is “continuous on its domain of definition.” Continuity is defined and studied in Section 2.7.
 
104 Chapter 1 Coordinates, Graphs, Lines
Here’s the criticalobservation: on each of these open intervals the sign of f .x/ must
be always positive or always negative. To determine the solution set for the inequality
f .x/ > 0 we have only to determine on which of our four intervals f .x/ is positive.
Since we can check that f .x/ is positive only on .1; 2/ and .3;C1/, then our solution
set is .1; 2/ [ .3;C1/. This example motivates the Interval Method:
The Interval Method (Method of Test Points) for Solving f .x/ > 0
Step 1 Crossing points. Determine those x-values for which f .x/ D 0.
Step 2 Missing points. Determine those x-values for which f .x/ is undefined.
Step 3 Pick a point x⁄ from each of the open intervals formed by the crossing points
and the missing points. Plug x⁄ into f .x/ to get f .x⁄/. If f .x⁄/ > 0, then the
corresponding interval is in the solution set for f .x/ > 0. Don’t let the length
of this discussion mislead you: the interval method is an easy-to-use technique.
Here are two examples, the first of which is a reworking of Example C:
Example D
Solve
1
.x C 1/ <
2
.x ¡ 1/ by the Interval Method.
Solution
As in Example C we move all the terms to one side of the inequality and obtain
0 <
x C 3
.x ¡ 1/.x C 1/
Now we apply the three steps of the Interval Method to
f .x/ D x C 3
.x ¡ 1/.x C 1/
Step 1 Crossing points. We must determine those x-values for which
x C 3
.x ¡ 1/.x C 1/ D 0
 
Section 1.1 Real Numbers, Sets, and Inequalities (A Review) 105
That’s easy to do since a fraction can equal zero only when the numerator equals
zero. Thus x C 3 D 0, which gives x D ¡3 as the only crossing point.
Step 2 Missing points. We must determine these x-values for which xC3
.x¡1/.xC1/ is
undefined. The only thing that could “go wrong” in this fraction is that the de-
nominator would equal zero if x D 1 or ¡1. Thus x D 1 and ¡1 are the missing
points.
Step 3 Our points x D ¡3;¡1 and 1 divide the real line into four open intervals.
From each we choose a point. Then we plug that point into .x C 3/=.x ¡ 1/.x C
1/ and determine if it makes this fraction positive or negative. In this way we
determine if the interval is in our solution for f .x/ > 0.
For each interval we:
.1/ choose a point in each interval
.2/ plug the chosen point into f .x/ D x C 3
.x ¡ 1/ .x C 1/ :
.1/ .2/
# + Is interval in solution set
interval x⁄ f ¡x⁄¢ for f .x/ > 0?
.¡1;¡3/ ¡4 ¡ 115 NO, since f
¡
x⁄
¢
< 0
.¡3;¡1/ ¡2 13 YES, since f
¡
x⁄
¢
> 0
.¡1; 1/ 0 ¡3 NO, since f ¡x⁄¢ < 0
.¡1;C1/ 2 53 YES, since f
¡
x⁄
¢
> 0
(Notice how we chose the x⁄-values so that each f .x⁄/ is easy to compute.) Thus
the solution set is .¡3;¡1/ [ .1;C1/, which agrees with the answer arrived at in
Example C.
Example E
Solve 0 •
p
x ¡ 2
4¡px C 8
 
106 Chapter 1 Coordinates, Graphs, Lines
Solution
Since the expression
f .x/ D
p
x ¡ 2
4¡px C 8
contains the terms
p
x and
p
x C 8;
we must have x ‚ 0 and x C 8 ‚ 0 (since we cannot take the square root of negative
numbers). Hence we are restricted to considering only those x-values for which x ‚
0. (These are the x-values for which both x ‚ 0 and x ‚ ¡8 are true.)
Step 1 Crossing points. We can have
p
x ¡ 2
4¡px C 8 D 0
only when the numerator equals zero, i.e., when
p
x ¡ 2 D 0p
x D 2
x D 4;
the only crossing point.
Step 2 Missing points. Having already restricted our attention to x ‚ 0, we will
not have any negative numbers occurring under square root signs (which would
make the expression undefined). Thus our expression is undefined only when the
denominator is zero, i.e., when
4¡px C 8 D 0p
x C 8 D 4
x C 8 D 16
x D 8;
the only missing point.
Step 3 The points x D 4 and 8 divide the positive x-axis into three open intervals:
 
Section 1.1 Real Numbers, Sets, and Inequalities (A Review) 107
Is interval in solution set
interval x⁄ f ¡x⁄¢ for f .x/ ‚ 0?
.0; 4/ 1 ¡1 NO, since f ¡x⁄¢ < 0
.4; 8/ 4:25 2
‡p
4:25¡ 2
·
YES, since f ¡x⁄¢ > 0
.8;C1/ 17 2¡p17 NO, since f ¡x⁄¢ < 0
The x⁄-values were chosen so that
p
x C 8 is easy to compute:
p
1C 8 D
p
9 D 3;p
4:25C 8 D
p
12:25 D 3:5;p
17C 8 D
p
25 D 5
Thus the solution set to Example F contains the open interval .4; 8/. However,
notice that our inequality is of the form f .x/ ‚ 0, i.e., f .x/ D 0 is allowed. Thus we
must also include the crossing point x D 4, so our solution set is [4; 8/:
1.1.3 Reversable Steps (Optional) : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
In mathematical reasoning, a series of steps is reversible if it is true back-to-front as
well as front-to-back. For example, the steps
if (a) 2x ¡ 3 < 7
then (b) 2x < 10
and
then (c) x < 5
are reversible since
if (c) x < 5
then (b) 2x < 10
and
then (a) 2x ¡ 3 < 7
Here is an example of a non-reversible series of steps:
(a) x > 1
(b) x2 > 1
 
108 Chapter 1 Coordinates, Graphs, Lines
Notice that if (a) holds true, then (b) must be true. However, if (b) is true, then it
does not follow that (a) is true. For example, x D ¡2 is in the solution set of (b), but it
is not in the solution set of (a). Thus “(a) implies (b)” is not a reversible step because
it is not true that (b) implies (a).
Example F
Solve
p
3x ¡ 2 < x
Solution
p
3x ¡ 2 < x
3x ¡ 2 < x2
(if 0 • a < b , then a2 < b2)
0 < x2 ¡ 3x C 2
0 < .x ¡ 2/.x ¡ 1/
In order for this last inequality to be true, either x¡2 and x¡1 must both be positive
(i.e., x > 2 and x > 1) or must both be negative (i.e., x < 2 and x < 1). Thus x < 1
(both negative) or x > 2 (both positive) One might be tempted to conclude that the
solution set for our inequality is .¡1; 1/[.2;C1/. This is false. The very first step
is not reversible (e.g., x D 0 satisfies the second equation, but not the first). The step
is reversible only when 3x ¡ 2 ‚ 0, i.e., when x ‚ 23. For values of x less than 23
the term
p
3x ¡ 2 is not even defined. Thus only part of our initial “solution set” is
correct. The actual solution set ish
2
3; 1
·
[ .2;C1/
As this example shows, you must check your answers, especially when your compu-
tations involve possibly non-reversible steps. �
The lesson to be learned is this: when solving inequalities (or equalities) it is not
at all uncommon “accidentally” to introduce some incorrect solutions along the way.
This is especially true when squares of quantities are involved. Thus, when solving
an equation, do not claim that your answer is the solution set until:
i. you have checked that all your solution steps are reversible, or
ii. if even one step is non-reversible, you have checked all your solutions back in the
original equation.
 
Section 1.2 Absolute Value 109
If the steps used to show that a statement A implies a statement B are all reversible,
then we can say that A is true if and only if B is true. The term “if and only if” is
commonly abbreviated to“iff.”
1.2 ABSOLUTE VALUE : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
1.2.1 The Definition of Absolute Value : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
When asked to find the absolute value of a specific number—for example, 5, ¡23,¡…—the chance that you’ll make a mistake is very small:
j5j D 5;
flflflfl¡23
flflflfl D 23; j¡… j D …
Perhaps it is because the intuitive idea of absolute value is so simple that many people
never bother to learn the precise definition:
jaj D
‰
a if a ‚ 0
¡a if a < 0
Notice how the definition captures the intuitive idea that “jaj is a made positive.”
i. if a is already non-negative, then jaj is just a,
ii. if a is negative, then jaj is defined to be ¡a, and this is a positive quantity since“the negative of a negative is positive.”
The precise definition is important for uses in calculus; you must learn it carefully.
The following two examples are typical of what you will encounter in many calculus
applications:
Example A
Simplify the expression
2x ¡ jx j
x
.
Solution
In order to deal with jx j, we must consider two cases: x ‚ 0 and x < 0.
Case 1: x ‚ 0. If x D 0, then the expression is undefined since we can’t divide by
zero. If x > 0, then jx j D x by its definition. Thus
2x ¡ jx j
x
D 2x ¡ x
x
D x
x
D 1
 
110 Chapter 1 Coordinates, Graphs, Lines
Case 2: x < 0. If x < 0, then jx j D ¡x by its definition. Thus
2x ¡ jx j
x
D 2x ¡ .¡x/
x
D 3x
x
D 3
In summary, our answer becomes
2x ¡ jx j
x
D
8<: 1 when x > 0undefined when x D 03 when x < 0
�
Example B
Simplify the expression x ¡ jx C 1j.
Solution
The two cases here are x C 1 ‚ 0 and x C 1 < 0.
Case 1: x C 1 ‚ 0, i.e., x ‚ ¡1. Thus jx C 1j D x C 1, which gives
x ¡ jx C 1j D x ¡ .x C 1/ D ¡1
Case 2: x C 1 < 0, i.e., x < ¡1. Thus jx C 1j D ¡(x C 1), which gives
x ¡ jx C 1j D x ¡ .¡.x C 1// D 2x C 1
In summary, our answer becomes
x ¡ jx C 1j D
‰ ¡1 if x ‚ ¡1
2x C 1 if x < ¡1
�
1.2.2 Algebraic Properties of Absolute Value : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
The behavior of absolute values with products, quotients and powers is all quite pre-
dictable and easily memorized (Theorem B.3 and Equation (2)). Add to that list the
simple but useful equality
¡ jaj • a • jaj
Unfortunately the rules governing addition and subtraction are not so nice; in fact,
their misuse occurs very often in calculus:
 
Section 1.2 Absolute Value 111
Common Errors:
ja C bj is NOT equal to jaj C jbj
ja ¡ bj is NOT equal to jaj ¡ jbj
There are, in fact, no simple rules which equate ja C bj orja ¡ bjwith expressions
involving only jaj and jbj. The best we can do with either ja C bj orja ¡ bj is to bound
them by inequalities as follows:
jjaj ¡ jbjj • ja C bj • jaj C jbj
jjaj ¡ jbjj • ja ¡ bj • jaj C jbj
Of the four inequalities shown here, by far the most important is the upper bound
for ja C bj:
ja C bj • jaj C jbj
Known as the triangle inequality, it is proved in Anton’s Theorem B.5. The other
three inequalities can be derived from clever applications of the triangle inequality.
This is the subject of Anton’s Exercises 41–43 in Appendix B.
Here is another common (and oftentimes costly) mistake involving absolute val-
ues:
Common Error: p
a2 is NOT always equal to a.
The correct result uses absolute values (Theorem B:2):
p
a2 D jaj (1.1)
This occurs because the symbol
p
x denotes the non-negative square root of x , e.g.,q
.¡3/2 D
p
9 D 3 D j¡3j
Carefully read over Anton’s discussion of this preceding Theorem B.2.
Example C
Simplify the expression 1¡
p
x2 ¡ 2x C 1.
 
112 Chapter 1 Coordinates, Graphs, Lines
Solution
1¡
p
x2 ¡ 2x C 1 D 1¡
q
.x ¡ 1/2
D 1¡ jx ¡ 1j by .1.1/
D⁄
‰
1¡ .x ¡ 1/ D 2¡ x if x ‚ 1
1C .x ¡ 1/ D x if x < 1
* using the method of Example B.
Example D
Solve for x : jx ¡ 2j C 1 D 2x .
Solution
First we isolate the absolute value on one side of the equation: jx ¡ 2j D 2x ¡ 1.
Now, depending on whether x¡2 is positive or negative, this equation can be written
as
x ¡ 2 D 2x ¡ 1 or¡ .x ¡ 2/ D 2x ¡ 1
Solving each of these equations we obtain
x D ¡1 or x D 1
Don’t jump to conclusions, however! It is not uncommon “accidentally” to introduce
some incorrect solutions while solving an absolute value problem. Hence we must
check our two answers in the original equation. In this case x D 1 solves the original
equation but x D ¡1 does NOT. Thus x D 1 is the desired answer.
1.2.3 Absolute Value and Distance : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
The fundamental fact which relates the algebraic concept of absolute value with
the geometric notion of distance is the following:
the quantity ja ¡ bj equals the distance between a and
b when a and b are considered as points on the x-axis. (1.2)
In particular, jaj equals the distance from a to the origin since jaj D ja ¡ 0j.
You must become entirely comfortable with statement .1:2/. By choosing some
specific values of a and b, convince yourself that it is true regardless of whether a
 
Section 1.2 Absolute Value 113
and b are positive or negative, and regardless of whether a < b or b < a. Statement
.1:2/ allows us to interpret inequalities such as jx ¡ 2j < 3 in a geometric fashion:
The solution set for jx ¡ 2j < 3 consists of all those x values which are
less than 3 units away from the point 2, i.e.,
¡1 < x < 5
Thus the expression jx ¡ 2j < 3 can be read in two ways:
algebraic: the absolute value of x minus 2 is less than 3
geometric: the distance between x and 2 is less than 3
Inequalities such as jx ¡ 2j < 3 are important in calculus because of their
interpretation in terms of distance.
1.2.4 Common Absolute Value Inequalities : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
One general type of inequality which occurs quite often in calculus is
j f .x/j < k
where f .x/ is an expression involving x (i.e., a function of x). Using the
geometric interpretation of absolute value this inequality can be read as “the
distance between f .x/ and the origin is less than k units.”
Thus we see that¡k < f .x/ < k , which is equivalent to the two (simul-
taneous) inequalities ¡k < f .x/ and f .x/ < k . Summarizing:
The following expressions are equivalent:
1. j f .x/j < k
2. ¡k < f .x/ < k
3. ¡k < f .x/ and f .x/ < k
Given an inequality of the form j f .x/j < k your automatic reaction
should be to convert it to ¡k < f .x/ < k , and, if necessary, to convert
it further to the pair of simultaneous inequalities ¡k < f .x/ and f .x/ < k.
Here are some examples:
 
114 Chapter 1 Coordinates, Graphs, Lines
Example E
Solve for x :
2 jx ¡ 1j ¡ 3 < x
Solution
First we isolate the absolute value: j2x ¡ 2j < x C 3. Ah-ha! We now have
an inequality of the form j f .x/j < k , and our next two steps are automatic
reactions:
All the steps we used are reversible, and hence all the x-values just found
are indeed solutions of our original inequality. Our solution set is thus the
interval
‡
¡13; 5
·
. �
Example E can also be solved by using the Interval Method of Section
1:1:2 of The Companion. Try it!
Example F
Solve for x :
j2x ¡ 3j < jx C 4j :
First solution
Division by jx C 4j yields the inequality
flflfl2x¡3xC4 flflfl < 1 which is of the form
j f .x/j < k. Thus our next two steps are automatic reactions:
¡1 < 2x ¡ 3
x C 4 < 1
¡1 < 2x ¡ 3
x C 4 and
2x ¡ 3
x C 4 < 1
Now we solve each inequality as we did in Section 1.1:
First inequality:
0 <
2x ¡ 3
x C 4 C 1 D
3x C 1
x C 4
 
Section 1.2 Absolute Value 115
Thus .¡1;¡4/ [
‡
¡13;C1
·
is the solution set for the 1st inequality.
Second inequality:
0 < 1¡ 2x ¡ 3
x C 4 D
¡x C 7
x C 4
Thus .¡4; 7/ is the solution set for the 2nd inequality. The solution set
for the original inequality is the set of points which are in both sets, i.e., it is
the intersection of the solution sets just obtained:
.¡1;¡4/ [
‡
¡13;C1
·
intersected with .¡4; 7/
This intersection is the interval
‡
¡13; 7
·
Second solution
Using the Interval Method described in Section 1:1:2 of The Companion will
give a simpler solution for the inequality j2x ¡ 3j < jx C 4j. We move all
the terms to one side of the inequality to obtain
0 < jx C 4j ¡ j2x ¡ 3j
Now we can apply the three steps of the Interval Method to
f .x/ D jx C 4j ¡ j2x ¡ 3j
 
116 Chapter 1 Coordinates, Graphs, LinesStep 1 Crossing points.
jx C 4j ¡ j2x ¡ 3j D 0
jx C 4j D j2x ¡ 3j
x C 4 D 2x ¡ 3 or x C 4 D ¡.2x ¡ 3/
x D 7 or x D ¡1
3
(Here we have used solution techniques similar to those employed in An-
ton’s Example 2 in Appendix B.)
Step 2 Missing points. There are none since jx C 4j ¡ j2x ¡ 3j is defined
for all values of x .
Step 3 The points x D ¡13 and 7 give us three intervals:
Is interval in
solution set
interval x⁄ f ¡x⁄¢ for f .x/ > 0?‡
¡1;¡13
·
¡1 ¡2 NO‡
¡13; 7
·
0 1 YES
.7;C1/ 10 ¡9 NO
Thus the solution set is
‡
¡13; 7
·
, agreeing with the first solution.
1.2.5 Another Absolute Value Inequality : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
Inequalities of the form
k < j f .x/j ;
(although not quite as common as those studied in the previous subsection) do
occur in applications. In general they are best handled by the Interval Method
as in the following example:
Example G
Solve for x :
x • 2 jx ¡ 2j :
 
Section 1.2 Absolute Value 117
Solution
We rewrite the inequality as
0 • 2 jx ¡ 2j ¡ x
so that we can apply the Interval Method to
f .x/ D 2 jx ¡ 2j ¡ x
Step 1 Crossing points.
2 jx ¡ 2j ¡ x D 0
j2x ¡ 4j D x
2x ¡ 4 D x or ¡ 2x C 4 D x
x D 4 or x D 4
3
Step 2 Missing points. There are none.
Step 3 The points x D 43 and 4 give us three intervals:
Is interval in
solution set
interval x⁄ f ¡x⁄¢ for f .x/ ‚ 0?‡
¡1; 43
·
0 4 YES‡
4
3; 4
·
2 ¡2 NO
.4;C1/ 5 1 YES
Since our inequality allows f .x/ D 0, we must include the two crossing
points x D 43 and 4 in the solution set for our inequality. Our final answer is
therefore µ
¡1; 4
3
‚
[ [4;C1/
�
 
118 Chapter 1 Coordinates, Graphs, Lines
1.3 COORDINATE PLANES AND GRAPHS : : : : : : : : : : : : : : : : : : : : : : : : : : : :
1.3.1 Analytic Geometry: Geometry into Algebra : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
Most beginning calculus students are familiar and comfortable with the xy-
coordinate plane, so much so that they overlook the incredible power and im-
portance of the technique. Identifying points on the plane with ordered pairs
of real numbers converts very hard geometry problems into generally much
easier algebra problems: this is the power of “analytic geometry.” For ex-
ample, for the early Greeks the study of parabolas (Section 0.5) involved in-
tricate and tricky geometric techniques; however, through analytic geometry
we can simply work with the algebraic properties of equations of the form
y D fix2Cfl. This conversion of geometry into algebra is the cornerstone of
calculus.
1.3.2 Solution Sets : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
The solution set of an equation in two variables x and y is merely the col-
lection of all ordered pairs of real numbers .a; b/ such that the equation is
satisfied when we substitute x D a and y D b. The equation itself is an alge-
braic object; however, we can turn its solution set into a geometric object by
identifying each solution with the corresponding point in the xy-plane, thus
obtaining the graph of the equation.
The interplay between equations (algebraic objects) and their graphs (ge-
ometric objects) is important in real-world applications. For example, sup-
pose a chemist determines that two quantities, x and y, are related according
to some particular equation. To be specific, suppose the equation is
y D x4=3 ¡ x1=3 C 1
How then does y vary when x is changed from 0 to 1? What value of x gives
the lowest value of y? Is there a maximum value of y? All of these questions
can be answered by determining the graph of the equation; it is drawn very
accurately here.
 
Section 1.3 Coordinate Planes and Graphs 119
y
x
How did we get this graph? Black magic? No: : : CALCULUS (which
in some cases is more powerful than black magic!) The accurate sketching
of graphs is an important application of calculus, and will be developed in
Section 5.3.
1.3.3 The Uses of Symmetry : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
In the discussion surrounding the symmetry tests of Theorem 1:4:3, Anton
illustrates their uses in curve sketching:
† if a plane curve is symmetric about the y-axis (i.e., if replacing x by ¡x
leaves the equation unchanged) then we have only to graph that portion
of the curve for which x ‚ 0, and then reflect it about the y-axis
† if a plane curve is symmetric about the x-axis (i.e., if replacing y by ¡y
leaves the equation unchanged) then we have only to graph that portion
of the curve for which y ‚ 0, and then reflect it about the x-axis
 
120 Chapter 1 Coordinates, Graphs, Lines
† if a plane curve is symmetric about the origin (i.e., if replacing .x; y/with
.¡x;¡y/ leaves the equation unchanged) then given any line through
.0; 0/, we have only to graph that portion of the curve which lies on one
side of the line, and then reflect it through the origin.
Simply put, symmetry can cut our graphing work almost in half, as the
following example shows:
Example A
Sketch the graph of y D 4x
1C x2 .
Solution
We first look for symmetries: replacing x by¡x will change the equation, as
will replacing y with ¡y. However, replacing x and y with ¡x and ¡y will
not change the equation:
.¡y/ D 4.¡x/
1C .¡x/2 D ¡
4x
1C x2
so
y D 4x
1C x2 ;
the original equation.
Hence, by our symmetry tests, the graph will be symmetric about the ori-
gin (but not about the x-axis or the y-axis). We therefore have only to graph
 
Section 1.3 Coordinate Planes and Graphs 121
that portion of the curve which lies on one side of a line through .0; 0/, and
then reflect it through the origin. We will graph the curve for x ‚ 0 (i.e., on
the right side of the y-axis) and then reflect it through .0; 0/. Using a hand
calculator to compute y-values from given x-values we obtain the following:
x y D 4x1Cx2
0 0
0.5 1:6
1.0 2:0
1.5 1:85
2.0 1:6
3.0 1:2
4.0 0.94
Plotting these points leads us to fill in the curve shown in the figure; the
dotted portion of the curve (for x < 0) is obtained by reflection through the
origin.
�
WARNING: See Anton’s Remark following Section 1.3. Sketching
a graph by plotting points will only yield an approximation to the graph.
In Sections 5.3 we will use the techniques of differential calculus
to obtain exact graphs of curves.
 
122 Chapter 1 Coordinates, Graphs, Lines
1.4 LINES : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
Since trigonometry is used in this section, Anton recommends that
people who need a trigonometry review consult his Appendix E. A
somewhat shorter trigonometry refresher can also be found in The Com-
panion.
1.4.1 The Definition of Slope of a Line : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
As mentioned in Section 1.3, the power of analytic geometry lies in turning
geometry problems into algebra problems. The slope of a line is a good exam-
ple. To any non-vertical line (a geometric object) in the xy-coordinate plane
we can associate a number called the slope (an algebraic object). This num-
ber contains a vast amount of information about the line. You cannot learn
calculus if you do not understand slope. Indeed, the first major concept of
calculus, the derivative, is defined in terms of slopes. Anton provides two
basic ways to define the slope m of a line in Appendix C:
I. The slope is the difference in the y-values over the difference in the x-
values, i.e.,
m D riserun D y2¡y1x2¡x1 (1.3)
where .x1; y1/ and .x2; y2/ are any two distinct points on the line.
II. The slope is the tangentof the angle of inclination ` made between the
line and the x-axis, i.e.,
m D tan`; 0 • ` < …; ` 6D …=2 (1.4)
 
Section 1.4 Lines 123
Essentially the proof is nothing but applying Formula .1:3/ to the
picture below:
m D rise
run
D sin`
cos`
D tan`
by Formula .1:3/ :
Two remarks are in order concerning Formula .1:3/:
(a) Any two points on a line can be used to compute its slope (that’s estab-
lished by Anton’s Figure C.1.5). Thus, if you have a choice of points : : :
choose the two which are most convenient for the calculation at hand!
(b) In formula .1:3/ it is important that each y-value be placed over its cor-
responding x-value.
This is a common mistake, so be careful!
 
124 Chapter 1 Coordinates, Graphs, Lines
You must know equations .1:3/ and .1:4/ very well! Moreover you must
“see” and “feel” the slope of a line as a measure of the steepness of the line. In
this regard be sure you completely understand Anton’s Figure C:17: In par-
ticular:
† A horizontal line has slope 0.
† A vertical line has an undefined slope (speaking informally, we some-
times say an “infinite” slope).
† A line which “points upward” has a positive slope.
† A line which “points downward” has a negative slope.
1.4.2 Slopes of Pairs of Lines : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
Suppose L1 and L2 are two non-vertical lines with slopes m1 and m2. An-
ton’s Theorem 1.4.4 says:
L1 and L2 are parallel if and only if m1 D m2
L1 and L2 are perpendicular if and only if
m2 D ¡
1
m1
These results are perfect illustrations of our claim that the slopes (alge-
braic quantities) of lines contain geometric information. These are important
results and you must know them well. When someone says “Lines L1 and L2
are parallel,” you must automatically convert this information into “Lines L1
and L2 have the same slope.” When someone says “Lines L1 and L2 are per-
pendicular,” you must automatically convert this information into “Lines L1
and L2 have slopes which are the negative reciprocals of each other.”
Example A
Use slopes to determine if .¡2; 1/, .¡1; 3/ and .2;¡1/ are the vertices of a
right triangle.
Solution
We have a right triangle only if two of the three lines forming the sides of our
triangle are perpendicular to each other. By Theorem C.3 this is the case
only if two of the three slopes are the negative reciprocals of each other. Thus
we have translated a geometric problem into a simpler algebraic problem.
Now we compute our three slopes: The slope of the line connecting .¡2; 1/
 
Section 1.4 Lines 125
to .¡1; 3/ is
3¡ 1
¡1¡ .¡2/ D
2
1
D 2
The slope of the line connecting .¡1; 3/ to .2;¡1/ is
¡1¡ 3
2¡ .¡1/ D
¡4
3
D ¡43
The slope of the line connecting .2;¡1/ to .¡2; 1/ is
1¡ .¡1/
¡2¡ 2 D
2
¡4 D ¡
1
2
Ah-ha! Our first and third lines have slopes 2 and ¡12 respectively. Since
these numbers are the negative reciprocals of each other, the corresponding
lines are perpendicular. Hence we do have a right triangle, with the right an-
gle at the vertex .¡2; 1/ (the vertex which lies on both of the perpendicular
sides). Note: The previous example will be solved by different methods in
Example A of Section 1:5:1.
1.4.3 Why is Theorem C.3 True? : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
The part of Theorem C.3 on parallel lines is easily remembered and intu-
itively clear; its proof is quite straightforward. However, none of this is true
for the part on perpendicular lines.
The reason is not that mysterious. Suppose L1 and L2 are two perpendic-
ular lines. As shown in the figure, things can be arranged so that the rise and
run of L1 become the run and negative rise of L2, respectively.
 
126 Chapter 1 Coordinates, Graphs, Lines
When written in terms of slope this yields:
m1 D fslope of L1g D
‰
rise of L1
run of L1
¾
D
‰
run of L2
¡ rise of L2
¾
D ¡
‰
1
slope of L2
¾
D ¡ 1
m2
This is the formula of the second part of Theorem C:3! (This informal
justification can be turned into a rigorous proof by using elementary geometry.)
See paragraphs following Theorem C.3.
1.4.4 Equations of Straight Line: Standard Forms : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
Straight lines and their equations play a central role in the development and
application of calculus, and for that reason you must understand them well .
When trying to find the equation of a line, you need only determine cer-
tain geometric information. In fact, the basic concept which you should learn
from this section is:
To determine the equation of a line you need
i. a point on the line and its slope, or
ii. the y-intercept of the line and its slope, or
iii. two distinct points on the line.
The information in any one of these three groups is enough to determine a
line, i.e., given values for the quantities in i, ii or iii, there is one and only one
line in the coordinate plane which has those given specifications. For exam-
ple, we learn in geometry that two distinct points determine a line (i.e., that iii
determines a line). The same principle holds true for i and ii. Anton further
shows how the equation of the line can be derived from the information given
in i and ii, and the derivation of the equation of the line from the information
given in iii is easy once you know how to do it for i.
 
Section 1.4 Lines 127
These methods (for non-vertical lines) are summarized as follows:
Necessary Equation of
Name Quantities the line
Point- The slope m
Slope The coordinates y ¡ y1 D
form .x1; y1/ of one m .x ¡ x1/
point on the line
y-int. The slope m
form The y-int., b y D mx C b
Two The coordinates y ¡ y1 D
point .x1; y1/ and m .x ¡ x1/
form .x2; y2/ of two where
points on the line m D y2¡y1x2¡x1
Memorization of these forms is easy:
i. The point-slope form is just a rewrite of the slope definition formula
m D y ¡ y1
x ¡ x1
ii. The y-intercept form is a well-known analytic geometry formula.
iii. The the two point form is just the point-slope form combined with the
slope definition formula.
Remember: When determining the equation of a line, always focus on finding the quan-
tities needed for one of the three standard forms.
Example B
Determine the equation of the line which passes through the point .2;¡1/
and has slope ¡23.
Solution
We are given the exact specifications required for the point-slope form of a
line:
m D y ¡ y1
x ¡ x1
 
128 Chapter 1 Coordinates, Graphs, Lines
(point-slope form of a line),
¡2
3
D y C 1
x ¡ 2
(plugging in given information),
y C 1 D
µ
¡2
3
¶
.x ¡ 2/
y D ¡
µ
2
3
¶
x C 1
3
Example C
Determine the equation of the line L which contains all the points .x; y/ equidis-
tant from .4;¡2/ and .3; 1/.
Solution
There is more work to do here than in the previous example because we are
not given the exact specifications for any one of our three standard forms.
Nonetheless we can compute the quantities needed for the point-slope form
from our given information.
A point on L is easily found: the midpoint .x1; y1/ of the line segment
from .4;¡2/ to .3; 1/:
x1 D
4C 3
2
D 7
2
y1 D
¡2C 1
2
D ¡1
2
Thus
.x1; y1/ D
µ
7
2
;¡1
2
¶
To determine the slope of L , notice, in the figure, that L is perpendicular
to the line segment from .3; 1/ to .4;¡2/;
 
Section 1.4 Lines 129
since the slope of this line segment is
m0 D
¡2¡ 1
4¡ 3 D ¡
3
1
D ¡3
then the slope of L is given by
m D ¡1=m0 D ¡1=.¡3/ D
1
3
Using the point-slope form of a line we thus have
m D y ¡ y1
x ¡ x1
(point-slope form of a line),
1
3
D y C
1
2
x ¡ 72
(plugging in information calculatedabove),
y C 1
2
D
µ
1
3
¶
x ¡ 7
6
y D
‡
1
3
·
x ¡ 53
Vertical lines Since vertical lines have an undefined slope, the point-slope
and y-intercept forms are irrelevant to such lines. When given two points,
however, it is always easy to see if they determine a vertical line: they will
have the same x-coordinate, say x0.
 
130 Chapter 1 Coordinates, Graphs, Lines
The equation of the line is then
x D x0
1.4.5 The General Form of a Line : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
Every equation for a line, including that of a vertical line, can be put in the
form of a first degree equation in x and y,
Ax C By C C D 0 (1.5)
where at least one of the constants A or B is not zero. This result is part of
Anton’s Theorem C.7 and is easy to prove. However, the converse of this
result is even more interesting: any first degree equation in x and y is an
equation for a line, i.e., has a line as its graph. (This is also part of Theorem
C.7.) Thus equations such as
3x ¡ 4y C 2 D 0;
…x C
p
2y ¡ 2 D 0;
or
4x ¡
p
3 D 0
are all equations for lines. To determine the important quantities associated
with the corresponding line (e.g., slope, y-intercept, etc.) we simply convert
our first degree equation into an appropriate standard form and then “read
off” the desired quantities. The y-intercept form is particularly useful in this
regard.
Example D
Determine the slope, the y-intercept, and at least two points of the line given
by the first degree equation
…x C
p
2y ¡ 2 D 0:
 
Section 1.4 Lines 131
Solution
We convert our equation to y-intercept form simply by solving for y in terms
of x :
…x C
p
2y ¡ 2 D 0p
2y D ¡…x C 2
y D
µ
¡ …p
2
¶
x C
µ
2p
2
¶
Compare this result with the point-slope formula:
y D
‡
¡…
p
2
2
·
x Cp2
l l
y D mxC b
From the comparison we see that the slope is
m D ¡…
p
2
2
and the y-intercept is
b D
p
2:
To find specific points on the line we have only to choose convenient x-values
and compute the corresponding y-values:
x D 0 gives y D
p
2
x D 1 gives y D ¡…
p
2=2C
p
2
Thus .0;
p
2/ and .1;¡…p2=2Cp2/ are two points on the line. �
1.4.6 Problems Involving Lines : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
We’ll end our section with two examples of analytic geometry problems which
require the material developed in this section about lines. Concentrate, in par-
ticular, on the problem solving techniques used in Example F.
Example E
Determine whether the following two lines are parallel, perpendicular, or nei-
ther:
2x C 3y ¡ 6 D 0 and 3x ¡ 2y ¡ 4 D 0:
 
132 Chapter 1 Coordinates, Graphs, Lines
Solution
Determining whether two lines are parallel or perpendicular requires compar-
ing their slopes (Section 1.4). We’ll determine these slopes as in Example D
by converting each equation to the y-intercept form:
2x C 3y ¡ 6 D 0 becomes y D ¡
µ
2
3
¶
x C 2
3x ¡ 2y ¡ 4 D 0 becomes y D
µ
3
2
¶
x ¡ 2
As we now see, the slope of one line is the negative reciprocal of the other
(i.e., 32 D ¡ 1¡¡ 23¢). Thus the lines are perpendicular by Theorem C:3: �
Example F
Find the (shortest) distance from the point P.¡2; 9/ to the line L given by
2x ¡ 3y C 5 D 0.
Solution
To determine the plan of attack in such a problem it is often useful to
think backwards from where you want to end.
In this problem our ultimate goal (where we want to end) is to
1. Compute the shortest distance from P to L , which is the length of the
shortest possible line segment joining P to L .
A little reflection should convince you that, to compute this length we
must
2. Determine the point Q on L such that the line segment P Q is perpendic-
ular to L .
 
Section 1.4 Lines 133
We see, however, that Q is the intersection of L with the line L 0 that passes
through P and is perpendicular to L . If we can find an equation for L 0,
then Q will be found by solving the equations for L and L 0 simultane-
ously. Thus, to find Q it is sufficient to
3. Find an equation for the line L 0 which contains P and is perpendicular to
L .
If we knew the slope of L 0 we could use the point-slope formula to com-
pute the equation for L 0. Since L and L 0 are perpendicular, Theorem C.3
says that to compute the slope of L 0 it is sufficient to
4. Compute the slope of L . Ahh: : : Our “backwards reasoning” has finally
bumped into a computation which we know how to do.
Our game plan for the solution of Example F is thus to perform all the
computations we laid out above: : : but in the order 4 ¡ 3 ¡ 2 ¡ 1 of
course! Here’s the execution of our plan:
4. L is given by 2x ¡ 3y C 5 D 0. So we solve for y in terms of x to obtain
the slope of L:
3y D 2x C 5
y D
µ
2
3
¶
x C 5
3
Thus the slope of L is m D 23.
3. The slope of L 0 is
m0 D ¡1=m
(Theorem C:3),
D ¡3
2
 
134 Chapter 1 Coordinates, Graphs, Lines
A point on L 0 is P.¡2; 9/. Thus the point-slope equation for L 0 is
¡3
2
D y ¡ 9
x C 2;
or
y D ¡
µ
3
2
¶
x C 6:
2. The intersection point Q of L and L 0 is found by solving the simultaneous
equations 8<: y D
‡
2
3
·
x C 53
y D ¡
‡
3
2
·
x C 6
Equating these two different expressions for y yieldsµ
2
3
¶
x C 5
3
D
µ
¡3
2
¶
x C 6
.13=6/x D 13=3
x D 2;
which gives y D 3: Thus Q D .2; 3/.
1. The distance between P and Q is
d D
q
.¡2¡ 2/2 C .9¡ 3/2
D
p
52 D 2
p
13
�
 
Section 1.5 Distance; Circles; Equations of the Form y D ax2 C bx C c 135
1.5 DISTANCE; CIRCLES; EQUATIONS OF THE FORM
y D ax2 C bx C c : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
1.5.1 The Distance Formula : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
Perhaps the most basic result of analytic geometry is the distance formula:
the distance between two points P1.x1; y1/ and P2.x2; y2/ is given by
d D
p
.x2 ¡ x1/2 C .y2 ¡ y1/2 (1.6)
(Anton’s Theorem D:1). Unfortunately it also seems to be one of the
first formulas that people forget! This is unfortunate for two reasons:
i. The distance formula is vital for calculus for, as you will soon learn, cal-
culus is built on the concept of distance.
ii. The distance formula is easy to remember: it is really nothing but the
Pythagorean Theorem (Companion Section 0.1.3) applied to the figure.
One Small Remark: Since .x1 ¡ x2/2 D .x2 ¡ x1/2, the order in which
x1 and x2 appear in (1.6) is unimportant. The same is true of y1 and y2.
However, be sure to group x variables with x variables and y variables
with y variables.
Example A
Is the triangle with vertices .¡2; 1/, .¡1; 3/, .2;¡1/ an equilateral triangle?
An isosceles triangle? A right triangle?
 
136 Chapter 1 Coordinates, Graphs, Lines
Solution
From the figure we strongly suspect that it is a right triangle. However, a
rough sketch proves nothing (are you sure that the angle at .¡2; 1/ is a full
90–? Might it not be 88–?) To prove that we have a right triangle let us com-
pute the length of each side, i.e., compute the distances between the vertices:
The distance between .¡2; 1/ and .¡1; 3/ isq
.¡1¡ .¡2//2 C .3¡ 1/2 D
p
12 C 22
D
p
5
The distance between .¡1; 3/ and .2;¡1/ isq
.2¡ .¡1//2 C .¡1¡ 3/2 D
q
32 C .¡4/2
D
p
25 D 5
The distance between .2;¡1/ and .¡2; 1/ isq
.¡2¡ 2/2 C .1¡ .¡1//2 D
q
.¡4/2 C 22
D
p
20
Since the side lengths
p
5, 5 and
p
20 are all different, then we do not have
an equilateral or an isosceles triangle. However,‡p
5
·2 C ‡p20·2 D 5C 20 D 25 D .5/2
Thus thePythagorean Theorem is valid for this triangle, proving it to be
a right triangle with right angle at the vertex .¡2; 1/ (the vertex between the
two sides of lengths
p
5 and
p
20:)
 
Section 1.5 Distance; Circles; Equations of the Form y D ax2 C bx C c 137
1.5.2 The General Pattern for Lines : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
In its logical framework, Anton’s development of equations for circles and
parabolas is very similar to his development of equations for lines in App-
endix C. From that section, given certain geometric data about a line (e.g.,
the slope m and the coordinates .x1; y1/ of one point on the line) we can
write down a standard form for the line (e.g., the point-slope form y ¡ y1 D
m.x ¡ x1/) and then a general equation for the line (Ax C By C C D 0).
The reverse procedure was also possible: given a general equation for a line
we can convert to a standard form and then read off geometric data about
the line (see Example D in Section 1:4:5 of The Companion). This can be
summarized in a diagram:‰
geometric information
for a line
¾
‰
standard forms
for a line
¾
‰
general equation
for a line
¾
Remember that there were three standard forms for a line studied in Appendix
C (See the table in Section 1 . 4 . 4 of The Companion).
1.5.3 Circles : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
The above pattern continues to hold for circles: we can go back and forth
between geometric data for a circle, a standard equation for a circle, and a
general equation for a circle.
Geometric data: To determine the equation of a circle you need the coordi-
nates .x0; y0/ of the center and the value r of the radius.
Given values for the center and the radius, there is one and only one circle
which has these specifications:
Standard equation: for a circle with center .x0; y0/ and radius r :
.x ¡ x0/2 C .y ¡ y0/2 D r2
General equation: for a circle:
Ax2 C Ay2 C Dx C Ey C F D 0; A 6D 0
 
138 Chapter 1 Coordinates, Graphs, Lines
Note that the coefficients of x2 and y2 are equal.
Here are examples of how to go back and forth between these objects:
Example B
Find a general equation for the circle with center .2;¡1/ and radius 3.
Solution
We are given geometric data and wish to find the general equation; the proce-
dure is to find the standard equation as an intermediate step. Since the stan-
dard equation is
.x ¡ x0/2 C .y ¡ y0/2 D r2
we merely plug in the values .x0; y0/ D .2;¡1/ and r D 3 to obtain
.x ¡ 2/2 C .y C 1/2 D 9
To go from this standard equation to a general equation only requires expand-
ing the squared terms and combining the constants:‡
x2 ¡ 4x C 4
·
C
‡
y2 C 2y C 1
·
D 9
x2 C y2 ¡ 4x C 2y ¡ 4 D 0
�
Example C
Determine the center and radius of the circle given by the general equation
2x2 C 2y2 C 4x ¡ 2y C 2 D 0:
Solution
As in Example B, finding the standard equation is our intermediate step. To
go from a general to a standard equation requires completing the square in
both x and y (Section 0.4.7):‡
2x2 C 4x
·
C
‡
2y2 ¡ 2y
·
C 2 D 0‡
x2 C 2x
·
C
‡
y2 ¡ y
·
C 1 D 0
.x2 C 2x C 1/C
µ
y2 ¡ y C 1
4
¶
C 1¡ 1¡ 1
4
D 0
 
Section 1.5 Distance; Circles; Equations of the Form y D ax2 C bx C c 139
.x C 1/2 C
µ
y ¡ 1
2
¶2
¡ 1
4
D 0
.x C 1/2 C
µ
y ¡ 1
2
¶2
D 1
4
This is the standard equation for a circle with center
‡
¡1; 12
·
and radius
1
2. �
WARNING: As Anton points out in Theorem D:3, an equation of the gen-
eral form
Ax2 C Ay2 C Dx C Ey C F D 0
does NOT necessarily have a circle for a graph; in some instances a single
point or no graph at all will result. For example, in Example C, if the
constant term 2 is replaced by 52, then the graph will be the single point‡
¡1; 12
·
; if 2 is replaced by any number higher than 52, then there will be
no graph at all.
Example D
Find the standard equation for the circle with center .¡3;¡4/ which passes
through the origin.
Solution
The geometric data needed in order to determine a circle are its center and
radius; when given any other type of data, immediately try to determine the
center and radius. We know that the radius must be
r D distance from .¡3;¡4/ to .0; 0/
D
q
.¡3¡ 0/2 C .¡4¡ 0/2
D
p
25 D 5
Thus, using .x0; y0/ D .¡3;¡4/ and r D 5, we obtain the standard equa-
tion
.x C 3/2 C .y C 4/2 D 25
 
140 Chapter 1 Coordinates, Graphs, Lines
1.5.4 Parabolas : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : : :
Anton’s discussion of parabolas can be organized in the way we did for cir-
cles. A quadratic equation in x ,
y D ax2 C bx C c .a 6D 0/ (1.7)
is the analogue for parabolas of the general equation for a circle discussed in
the previous section. To get a standard equation for the parabola we complete
the square (Companion Section 0.4.7):
y D a
‡
x2 C ba x
·
C c
y D a
ˆ
x2 C b
a
x C b
2
4a2
!
¡ b
2
4a2
C c
y D a
µ
x C b
2a
¶2
C
ˆ
c ¡ b
2
4a2
!
Let x0 D ¡
b
2a
and y0 D c ¡
b2
4a2
:
Then
y D a.x ¡ x0/2 C y0 .a 6D 0/ (1.8)
An equation of type .1:8/ is called a standard equation for a parabola. It is
not hard to show that the vertex of such a parabola is .x0; y0/ (for a justifica-
tion of this statement, see Companion Section 0:5:1); since
x0 D ¡
b
2a
from above, this corresponds to Anton’s equation (8) in Appendix D.
The geometric data which determine a parabola are the vertex .x0; y0/
and the number a, which we will call the parabolic steepness. As shown in
Anton’s Figure D:7 , if a is positive, then the parabola turns upward; if a is
negative then the parabola turns downward. Moreover, as jaj gets large, the
steepness of the parabola increases (this is shown in Section 0:4:1).
Here’s a summary of our approach to parabolas:
Geometric data: To determine the equation of a parabola you need the coor-
dinates .x0; y0/ of the vertex and the value a of the parabolic steepness.
Standard equation: for a parabola with vertex .x0; y0/ and parabolic steep-
ness a:
y D a.x ¡ x0/2 C y0
 
Section 1.5 Distance; Circles; Equations of the Form y D ax2 C bx C c 141
General equation: for a parabola (quadratic equation):
y D ax2 C bx C c .a 6D 0/
Here are examples showing how these are used:
Example E
Find an equation for the parabola passing through .4; 1/with vertex at .2;¡1/.
Sketch the graph.
Solution
We are given the vertex .x0; y0/ D .2;¡1/, and thus the standard equation
becomes
y D a.x ¡ 2/2 ¡ 1
We have only to determine the value of the parabolic steepness a; to do this
we note that .x; y/ D .4; 1/ must satisfy our standard equation:
1 D a.4¡ 2/2 ¡ 1
1 D 4a ¡ 1
a D 1
2
Thus our equation is
y D 1
2
.x ¡ 2/2 ¡ 1
Expanding the squared term leads to the quadratic equation
y D 1
2
x2 ¡ 2x C 1
To graph the equation we first plot the vertex .2;¡1/ and then find other
points on the graph using either of our above equations:
x ¡1 0 2 4 5
y 72 1 ¡1 1 72
 
142 Chapter 1 Coordinates, Graphs, Lines
These points enable us to make the sketch of our parabola.
Example F
Find the vertex and the intersections with the x-axis of the parabola given by
y D ¡2x2 C 2x C 1
Solution
Anton’s method of solution would be to use the equation x0 D ¡b=2a to find
the x-coordinate of the vertex (i.e., x0 D ¡b=2a D ¡.2/=2.¡2/ D 12), and
then to use the quadratic formula (Companion Section 0.4.5) on 0 D ¡2x2C
2x C 1 to find the intersections with the x-axis (i.e.,
x D ¡b §
p
b2 ¡ 4ac
2a
D ¡2§
p
4C 8
¡4D 1§
p
3
2
:
We’ll give an alternate method: first determine the standard equation for
the parabola. This is done by completing the square (Section 0.4.7 of The
 
Section 1.5 Distance; Circles; Equations of the Form y D ax2 C bx C c 143
Companion):
y D ¡2x2 C 2x C 1
D ¡2
‡
x2 ¡ x
·
C 1
D ¡2
‡
x2 ¡ x C 14 ¡ 14
·
C 1
D ¡2
‡
x ¡ 12
·2 C 12 C 1
D ¡2
0BB@x ¡ 12"
x0
1CCA
2
C 32
"
y0
Thus the vertex is .x0; y0/ D
‡
1
2;
3
2
·
. To obtain the intersections with the
x-axis we merely set y D 0:
0 D ¡2
µ
x ¡ 1
2
¶2
C 3
2µ
x ¡ 1
2
¶2
D 3
4
x ¡ 1
2
D §
p
3
2
x D 1§
p
3
2
… ¡:366 and 1:366
�
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