Table of Contents

Chapter 1 Sets 1

Section 1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Section 1.2 Union and Intersection . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Section 1.3 Complements and Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Chapter 2 The Real Number System 2

Section 2.1 The Real Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Section 2.2 Absolute Value . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Section 2.3 Exponents and Radicals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

Chapter 3 Expressions and Equations 9

Section 3.1 Algebraic Expressions and Polynomials . . . . . . . . . . . . . . . . . . . . . 9

Section 3.2 Factoring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Section 3.3 Rational Expressions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

Section 3.4 Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Section 3.5 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

Chapter 4 Coordinate Geometry 17

Section 4.1 The Cartesian Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Section 4.2 The Distance and Midpoint Formulas . . . . . . . . . . . . . . . . . . . . . . 18

Section 4.3 Graphs and Intercepts . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Section 4.4 Circles . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Section 4.5 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

Chapter 5 Lines 23

Section 5.1 Slope . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Section 5.2 Equations of Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Section 5.3 Parallel Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

i
 Section 5.4 Perpendicular Lines . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

Chapter 6 Complex Numbers 25

Section 6.1 Arithmetic with Complex Numbers . . . . . . . . . . . . . . . . . . . . . . . 25

Section 6.2 The Complex Conjugate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Chapter 7 Functions 26

Section 7.1 The Definition of Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

Section 7.2 Graphs of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

Section 7.3 Even and Odd Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Section 7.4 Combinations of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Section 7.5 Transformations of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

Section 7.6 One-to-One and Onto Functions . . . . . . . . . . . . . . . . . . . . . . . . . 36

Section 7.7 Inverses of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

Chapter 8 Polynomial Functions 42

Section 8.1 Linear Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Section 8.2 Quadratic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Section 8.3 Polynomial Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

Section 8.4 Roots of Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

Chapter 9 Rational Functions and Asymptotes 50

Section 9.1 Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

Section 9.2 Infinite Limits and Vertical Asymptotes . . . . . . . . . . . . . . . . . . . . . 50

Section 9.3 Limits at Infinity and Horizontal Asymptotes . . . . . . . . . . . . . . . . . . 52

Section 9.4 Oblique Asymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Chapter 10 Conic Sections 57

Section 10.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Section 10.2 Parabolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

Section 10.3 Ellipses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58

Section 10.4 Hyperbolas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

Chapter 11 Exponential and Logarithmic Functions 61

Section 11.1 Exponential Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

Section 11.2 Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

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 Section 11.3 Exponential Growth and Decay . . . . . . . . . . . . . . . . . . . . . . . . . 64

Section 11.4 Logarithmic Scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Subsection 11.4.1 The pH Scale . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Subsection 11.4.2 The Richter Scale . . . . . . . . . . . . . . . . . . . . . . . . . . . 65

Subsection 11.4.3 The Decibel Scale . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Section 11.5 Logistic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Section 11.6 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

Chapter 12 Sequences and Sigma Notation 69

Section 12.1 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

Section 12.2 Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

Chapter 13 The Mathematics of Finance 71

Section 13.1 Interest . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

Section 13.2 Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

Section 13.3 Installment Buying . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

Chapter 14 Induction and the Binomial Theorem 74

Section 14.1 Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

Section 14.2 The Binomial Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

iii
 Chapter 1: Sets

Section 1.1: Introduction

Definition 1: A set is an unordered collection of objects.

Example 2: The following are examples of sets.

(a) {1, 2, 7};

(b) {a, b, c};

(c) {a, b, 9, 10, δ};

(d) {2, 4, 6, 8, 10, . . .}.

Notation 3: If an object x is contained in a set X, we say x is an element of X and write x ∈ X.

Otherwise, we say x is not an element of X and write x ∈

/ X.

Example 4: Consider the following.

(a) 2 ∈ {1, 2, 3};

(b) a ∈ {a, b, c};

(c) 3 ∈
/ {1, 2, 4};

(d) d ∈
/ {a, b, c}.

Definition 5: The empty set, written either ∅ or {}, is the set with no elements.

Note 6: The set {∅} is not empty; it is the set containing the empty set.

Definition 7: A singleton is a set containing one element.

Definition 8: A set X is a subset of a set Y if every element of X is contained in Y and write

X ⊆Y.

Section 1.2: Union and Intersection

Definition 9: Let X and Y be sets.

(i) The union of X and Y is the set X ∪ Y = {x | x ∈ X or x ∈ Y }.

(ii) The intersection of X and Y is the set X ∩ Y = {x | x ∈ X and x ∈ Y }.

Example 10: Set X = {1, 2, 7, a, c} and Y = {2, 3, 4, a, b}.

(a) X ∪ Y = {1, 2, 3, 4, 6, a, b, c}

(b) X ∩ Y = {2, a}.

Definition 11: Two sets are disjoint if their intersection is the empty set.

Example 12: The sets {1, 2, 3} and {4, 5, 6} are disjoint.

1
 Section 1.3: Complements and Products
Definition 13: Let X and Y be sets. The complement of X relative to Y is the set X \ Y = {x ∈

X|x∈
/ Y }.

Example 14: Set X = {1, 2, 7, a, c} and Y = {2, 3, 4, a, b}.

(a) X \ Y = {1, 7, c}

(b) Y \ X = {3, 4, b}.

Definition 15: Let X and Y be sets. The Cartesian product of X and Y is the set X × Y =

{(x, y) | x ∈ X and y ∈ Y }.

Example 16: Set X = {1, a} and Y = {2, b}.

(a) X × Y = {(1, 2), (1, b), (a, 2), (a, b)}

(b) Y × X = {(2, 1), (2, a), (b, 1), (b, a)}.

Chapter 2: The Real Number System

Section 2.1: The Real Numbers

Notation 17: The following are common sets of numbers we will use throughout this text.

(i) (Natural Numbers) N = {1, 2, 3, 4, 5, . . .};

(ii) (Whole Numbers) ω = N ∪ {0};

(iii) (Integers) Z = {±n | n ∈ ω};

 
p
(iv) (Rational Numbers) Q = | p, q ∈ Z and q ̸= 0 ;
q
(v) (Real Numbers) R (definition below);

(vi) (Irrational Numbers) R \ Q.

Definition 18: The set of real numbers is the set

R = {(−1)b (x0 .x1 x2 x3 x4 x5 . . .) | b ∈ {0, 1}, x0 ∈ ω, and xi ∈ {0, 1, 2, . . . , 9} for every i ∈ N}.

Note 19: Consider R with the usual operations of addition and multiplication.

A1. (Closure and Commutativity of Addition) For every a, b ∈ R, a + b ∈ R and a + b = b + a

A2. (Associativity of Addition) For every a, b, c ∈ R, a + (b + c) = (a + b) + c

A3. (Additive Identity) There is a 0 ∈ R such that for all a ∈ R, a + 0 = 0 + a = a

A4. (Additive Inverse) For all a ∈ R, there is a −a ∈ R such that a + (−a) = −a + a = 0

2
M1. (Closure and Commutativity of Multiplication) For every a, b ∈ R, ab ∈ R and ab = ba

M2. (Associativity of Multiplication) For every a, b, c ∈ R, a(bc) = (ab)c

M3. (Multiplicative Identity) There is a 1 ∈ R such that for all a ∈ R, 1a = a1 = a

M4. (Multiplicative Inverse) For all a ∈ R \ {0}, there is an a−1 ∈ R such that aa−1 = a−1 a = 1

AM1. (Distributivity) For all a, b, c ∈ R, (a + b)c = ac + bc

Proposition 20 (Uniqueness of Identity and Inverse):

(i) If a, b ∈ R and a + b = a, then b = 0.

(ii) If a, b ∈ R and ab = a, then b = 1.

(iii) If a, b ∈ R and a + b = 0, then b = −a.

(iv) If a, b ∈ R and ab = 1, then b = a−1 .

Proof: For (i), suppose a, b ∈ R and a + b = a. Then,

b
=b+0
= b + (a + (−a))
= (b + a) + (−a)
= (a + b) + (−a)
= a + (−a)
= 0.

Thus, b = 0.

The proofs of parts (ii) and (iii) are left as an exercise for the reader. ■

For (iv), suppose a, b ∈ R, a ̸= 0, and ab = 1. Then,

b
= b1
= b(aa−1 )
= (ba)a−1
= (ab)a−1
= 1a−1
= a−1 .

Thus, b = a−1 . ■

Proposition 21:

(i) If a ∈ R, then a · 0 = 0.

(ii) If a, b ∈ R and ab = 0, then one of a and b is 0.

(iii) If a ∈ R, then (−1)a = −a.

Proof: For (i), note that a0 = a(0 + 0) = a0 + a0, and so, a0 = 0.

For (ii), suppose a, b ∈ R and ab = 0. If a = 0, there is nothing to show. So suppose a ̸= 0 and

consider the following.

b
= b1

3
= b(aa−1 )
= (ba)a−1
= (ab)a−1
= 0a−1
= 0.

Therefore, b = 0.

For (iii), consider the following.

(−1)a + a
= (−1)a + 1a
= (−1 + 1)a
= 0a
= 0.

Thus, (−1)a + a = 0, and so, (−1)a = −a. ■

Note 22: By part (iii) of the theorem above and the uniqueness of additive inverses, (−1)2 =

(−1)(−1) = 1.

Notation 23: Suppose a, b ∈ R.

(i) a + (−b) = a − b;
1 a
(ii) If b ̸= 0, then a ·
= .
b b
Theorem 24 (Order Properties of R): The set R+ has the following properties.

(i) For all a, b ∈ R+ , a + b ∈ R+ ;

(ii) For all a, b ∈ R+ , ab ∈ R+ ;

(iii) (Trichotomy Property) For all a ∈ R, either a ∈ R+ , −a ∈ R+ , or a = 0.

Theorem 25:

(i) If a ∈ R \ {0}, then a2 ∈ R+ ;

(ii) 1 ∈ R+ ;

(iii) N ⊆ R+ ;
1
(iv) If a ∈ R+ , then ∈ R+ .
a
Proof: For (i), suppose a ∈ R \ {0}. By the trichotomy property, a ∈ R+ or −a ∈ R+ . If a ∈ R+ ,

then a · a = a2 ∈ R+ by part (ii) of the order properties of R+ . If −a ∈ R+ , then by an argument

similar to the one above, (−a)(−a) ∈ R+ . Now,

(−a)(−a)
= [(−1)a][(−1)a]
= (−1)(−1)aa
= 1aa
= a2 .

For (ii), since 1 = 12 , 1 ∈ R+ by part (i) of this theorem.

For (iii), use induction.

4
 1 1
For (iv), toward a contradiction, suppose ∈ / R+ . By the trichotomy property, − ∈ R+ . Then,
  a a
1 1 1
−1 = a − = −1 · a · ∈ R+ . This is a contradiction since 1 ∈ R+ Therefore, ∈ R+ . ■
a a a
Definition 26: Let a, b ∈ R. Then,

(i) a is less than b, written a < b, if b − a ∈ R+ ;

(ii) a is less than or equal to b, written a ≤ b, if b − a ∈ R+ or b − a = 0.

Theorem 27: Let a, b, c ∈ R.

(i) If a < b and b < c, then a < c.

(ii) If a < b, then a + c < b + c

(iii) If a < b and c ∈ R+ , then ac < bc.

(iv) If a < b and −c ∈ R+ , then bc < ac.

1
(v) If a ∈ R+ such that 1 < a, then < 1.
a
Proof: For (i), since b − a ∈ R+ and c − b ∈ R+ , (c − b) + (b − a) = c − a ∈ R+ .

For (ii), since a < b, b − a ∈ R+ . Also, since c ∈ R+ , c(b − a) = cb − ca ∈ R+ .

For (iii), since a < b, b − a ∈ R+ . Also, since c ∈ R+ , c(b − a) = cb − ca ∈ R+ .

For (iv), since b − a ∈ R+ and −c ∈ R+ , −c(b − a) = ca − cb ∈ R+ .

For (v), consider the following.

1<a
1 1
⇐⇒ 1 · < a ·
a a
1
⇐⇒ < 1. ■
a
Proposition 28: If a ∈ R with 0 ≤ a < ε for every ε > 0, then a = 0.
a 1 1 a
Proof: We prove the contrapositive. If a > 0, set ε = . Then, 0 = · 0 < a = = ε.
2 2 2 2
a
Also, ε = < 1a = a. This implies that a > ε, as desired. ■
2
Theorem 29: For every a, b ∈ R, ab < 0 if and only if one of a and b is positive, and the other is

negative.

Proof: (=⇒) We will prove the contrapositive. If a, b ∈ R+ , then ab > 0. Also, if a, b < 0, then

−a, −b ∈ R+ , and so ab = (−a)(−b) > 0.

(⇐=) Suppose one of a and b is positive and the other is negative. Without loss of generality,

suppose a > 0 and b < 0. Then, by the trichotomy property, −b > 0, and so, a(−b) = −ab > 0. So,

ab < 0. ■

5
 Section 2.2: Absolute Value


x
 if x ≥ 0;
Definition 30: Let x ∈ R. The absolute value of x is |x| =

−x
 if x < 0.
Proposition 31: Let x, y ∈ R. Then

(i) |x|2 = x2 ;
√
(ii) |x| = x2 ;

(iii) |xy| = |x| · |y|;

(iv) xy ≤ |xy|.

Proof: For (i), if x ≥ 0, then |x|2 = x2 . If x < 0, then |x|2 = (−x)2 = x2 . So, in any case,

|x|2 = x2 .
p √
For (ii), by the definition of square root and part (i), |x| = |x|2 = x2 .
p p √ p
For (iii), by part (ii), |xy| = (xy)2 = x2 y 2 = x2 · y 2 = |x| · |y|.

For (iv), if x, y ≥ 0 or x, y ≤ 0, then xy ≥ 0 and |xy| = xy. If one of x and y is positive and the

other is negative, then xy < 0 and xy < |xy|. ■

Theorem 32 (The Triangle Inequality): For all x, y ∈ R, |x + y| ≤ |x| + |y|.

Proof: By part (i) of the previous proposition, |x + y|2 = (x + y)2 . So,

|x + y|2

= (x + y)2 = x2 + 2|x| · |y| + y 2

= x2 + 2xy + y 2 = |x|2 + 2|x| · |y| + |y|2

≤ x2 + 2|xy| + y 2 = (|x| + |y|)2 .

Since |x + y|2 ≤ (|x| + |y|)2 and |x| + |y| ≥ 0,

|x + y|
p 2
= |x + y| = |(|x| + |y|)|
p
≤ (|x| + |y|)2 = |x| + |y|. ■

Corollary 33: For all x, y, z ∈ R, |x − y| ≤ |x − z| + |y − z|.

Proof: Exercise. ■

Proposition 34: For all x, y ∈ R, 2|xy| ≤ x2 + y 2 .

Proof: Exercise. ■

Proposition 35: Suppose that a, b ∈ R with a ≤ b. The distance between a and b in R is

6
|a − b| = |b − a|.

Proof: Note that a + |a − b| = a + (b − a) = b and b − |a − b| = b − (b − a) = a. So |a − b| is the

distance between a and b. ■

Notation 36: For x ∈ R and ε > 0, B(x, ε) = (x − ε, x + ε).

 
1
Example 37: B 2, = (1.75, 2.25) = {x ∈ R | 1.75 < x < 2.25}.
4
Proposition 38: For x, y ∈ R and ε > 0, the following are equivalent.

(i) |x − y| < ε;

(ii) y ∈ B(x, ε);

(iii) x ∈ B(y, ε).

Proof: ((i) =⇒ (ii)) Suppose |x − y| < ε and consider two cases.

Case 1: y ≤ x Case 2: x ≤ y

Note the following. Note the following.

|x − y| < ε |x − y| < ε

⇐⇒ x − y < ε ⇐⇒ y − x < ε

⇐⇒ x − ε < y ⇐⇒ y < x + ε

⇐⇒ x − ε < y < x < x + ε ⇐⇒ x − ε < x < y < x + ε

⇐⇒ y ∈ (x − ε, x + ε) = B(x, ε). ⇐⇒ y ∈ (x − ε, x + ε) = B(x, ε).

((ii) =⇒ (iii)) Suppose that y ∈ B(x, ε). Then x − ε < y < x + ε. Since x − ε < y, x < y + ε. Also,

since y < x+ε, y −ε < x. Therefore, y −ε < x < y +ε, which means that x ∈ (y −ε, y +ε) = B(y, ε).

((iii) =⇒ (i)) Depending on the relative values of x and y, |x − y| is equal to either x − y or y − x.

We will show that both x − y and y − x are less than ε. Since x ∈ B(y, ε), y − ε < x < y + ε. Since

x < y + ε, x − y < ε and since y − ε < x, y − x < ε. Therefore, |x − y| < ε. ■

Corollary 39: For any a ∈ R and c ≥ 0, |a| ≤ c if and only if −c ≤ a ≤ c.

Proof: Consider the following.

|a| ≤ c ⇐⇒ a ∈ B(0, c)
⇐⇒ |a − 0| ≤ c ⇐⇒ −c ≤ a ≤ c. ■

Proposition 40: Suppose x ∈ R. If |x| < ε for all ε > 0, then x = 0.

|x|
Proof: We will prove the contrapositve. Suppose x ̸= 0. Then |x| > 0. Set ε = and note that
2
|x|
x> = ε. ■
2
Corollary 41: For all x, y ∈ R, the following are equivalent.

(i) x = y;

7
(ii) |x − y| = 0;

(iii) |x − y| < ε for all ε > 0.

Proof: Exercise. ■

Section 2.3: Exponents and Radicals

Definition 42: Let a ∈ R and n ∈ N. Then an = a · a · . . . · a, where a is called the base and n is

called the exponent.

Theorem 43: Let a, b, x, y ∈ R.

(i) ax+y = ax · ay ; 1
ax (vi) a−x = x ;
(ii) ax−y = y ; a
a
 a −n bn
(vii) = n;
(iii) (a ) = axy ;
x y b a
a−n bm
(viii) −m = n ;
(iv) (ab)x = ax bx ; b a
 a x ax (ix) If a ̸= 0, then a0 = 1.
(v) = x;
b b

Example 44: Simplify the following as much as possible.

 2 4 −2
28 x
(c)
49x−2
3 2
49x−2

x3 y −2 z 4

(b) =
(a) (2x3 y 2 )(3xy 4 )3 x−2 y 3 z 3 282 x4
x9 y −6 z 12 492 x−4
= 2x3 y 2 (27x3 y 12 ) = −6 9 9 =
x y z 284 x8
= 54x6 y 14 . x15 z 3 72 · 72 · x−4
= 15 . = 2 2 2 2 8
y 7 ·7 ·4 ·4 ·x
1
= .
256x12

Definition 45: A real number x ∈ R+ is in scientific notation if x = a × 10n where a ∈ [1, 10)

and n ∈ Z.

Example 46:

(a) The distance from the Sun to the star Proxima Centauri is 4 × 1013 kilometers.

(b) The mass of a hydrogen atom is 1.66 × 10−24 grams.

√
Definition 47: Let a, b ∈ R and n ∈ N. The principal nth root of a is defined by n
a = b if and

only if bn = a.

Note 48: In the definition above, if n is even, then a ≥ 0 and b ≥ 0.

8
 √ √
n √
Theorem 49: Let a, b ∈ R and n ∈ N such that n
a, b, and
√
m
a exist.
√ √ √
(i) a1/n = n a; (iv) n ab = n a n b;
r √
√ n a n
a
(ii) ( n a) =a; (v) n = √ n
;
b b √
p √

a if n is odd (vi) m n a = mn a;
√

n
(iii) an = √
 (vii) am/n = n am .
|a| if n is even;


Example 50: Simplify the following as much

√ as√possible. √ √
p (b) 32 + 200 (c) (2 x)(3 3 x)
(a) 4 81x8 y 4 √ √
√4
√
4
p = 16 · 2 + 100 · 2 = 6x1/2 x1/3
81 x8 y 4
4
√ √
= 4 2 + 10 2 = 6x5/6 .
= 3x2 |y|. √
= 14 2.
Definition 51: The procedure of removing a root from a denominator is called rationalizing the

denominator.

Example 52: For each of the following, rationalize the denominator and simplify.
s
2
5 x
r
2 (c)
(a) 1 y3
√ 3 (b) √ √
2
4
x√ 5
x 2
=√ 1
4
x 3 = p
√3 √ = √ · √ √
5
y3 p
4
2 3
4
x x 3 5
x2 5 y 2
=√ ·√ √ 3 = · p
4
x
√3 3
p
= . 5
y3 5 y2
6 |x| p
= .
5
x2 y 2
3 = .
y

Chapter 3: Expressions and Equations

Section 3.1: Algebraic Expressions and Polynomials

Definition 53:

(i) A variable is a letter that can represent any number from a given set of numbers.

(ii) Combinations of variables and real numbers with addition, subtraction, multiplication, division,

powers, and roots are called algebraic expressions.

Definition 54:

(i) A monomial is an expression of the form axk where a ∈ R, k ∈ ω, and x is a variable.

(ii) A binomial is a sum of two monomials.

(iii) A trinomial is a sum of three monomials.

9
(iv) A polynomial is a sum of monomials.

Example 55: Perform the indicated operation.

(a) (x3 − 6x2 + 2x + 4) + (x3 + 5x2 − 7x)

= (x3 + x3 ) + (−6x2 + 5x2 ) + (2x − 7x) + 4

= 2x3 − x2 − 5x + 4.

(b) (x3 − 6x2 + 2x + 4) − (x3 + 5x2 − 7x)

= (x3 − x3 ) + (−6x2 − 5x2 ) + (2x − (−7x)) + 4

= −11x2 + 9x + 4.

(c) (2x + 1)(3x − 5)

= 2x(3x − 5) + 1(3x − 5)

= 6x2 − 10x + 3x − 5

= 6x2 − 7x − 5.

Proposition 56: Let A and B be algebraic expressions.

(i) (A − B)(A + B) = A2 − B 2 ;

(ii) (A + B)2 = A2 + 2AB + B 2 ;

(iii) (A − B)2 = A2 − 2AB + B 2 ;

(iv) (A + B)3 = A3 + 3A2 B + 3AB 2 + B 3 ;

(v) (A − B)3 = A3 − 3A2 B + 3AB 2 − B 3 .

Example 57: Multiply and simplify.

(a) (3x + 5)2 √ √
(c) (2x + y)(2x − y)
= (3x)2 + 2(3x)(5) + 52 √
= (2x)2 − ( y)2
= 9x2 + 30x + 25.
= 4x2 − y.
(b) (x2 − 2)3
(d) (x + y − 1)(x + y + 1)
= (x2 )3 − 3(x2 )2 (2) + 3(x2 )(2)2 − (2)3
= (x + y)2 − (1)2
= x6 − 6x4 + 12x2 − 8.
= x2 + 2xy + y 2 − 1.

Section 3.2: Factoring

Lemma 58: For any algebraic √ expressions
! with a ̸= 0,!
a, b, c √
b − b2 − 4ac b + b2 − 4ac
ax2 + bx + c = a x + x+ .
2a 2a
Proof: Consider
√ the!following. √ !
b − b2 − 4ac b + b2 − 4ac
x+ x+
2a 2a
√ ! √ ! √ ! √ !
b + b2 − 4ac b − b 2 − 4ac b − b2 − 4ac b+ b2 − 4ac
= x2 + ·x+ ·x+ ·
2a 2a 2a 2a

10
 " √ √ # √ √
2 b+ b2 − 4ac b − b2 − 4ac (b − b2 − 4ac)(b + b2 − 4ac)
=x + + ·x+
2a 2a 4a2
2 2
2b b − (b − 4ac)
= x2 + x +
2a 4a2
2 b 4ac
=x + x+ 2
a 4a
2 b c
=x + x+
a a √ ! √ !
2 b − b2 − 4ac b + b2 − 4ac
⇐⇒ ax + bx + c = a x + x+ . ■
2a 2a
Theorem 59 (Factor Theorem for Quadratics): For any algebraic expressions a, b, c with a ̸= 0,
1 p p
ax2 + bx + c = (2ax + b − b2 − 4ac)(2ax + b + b2 − 4ac).
4a
Proof: By the lemma above, √ ! √ !
2 b − b2 − 4ac b + b2 − 4ac
ax + bx + c = a x + x+
2a 2a
1 1 p p
= · · a(2ax + b − b2 − 4ac)(2ax + b + b2 − 4ac)
2a 2a
1 p p
= (2ax + b − b2 − 4ac)(2ax + b + b2 − 4ac). ■
4a
Example 60: Factor the following expressions.

(a) 6x2 + 7x − 5
1 h p ih p i
= 2(6)x + (7) − (7)2 − 4(6)(−5) 2(6)x + (7) + (7)2 − 4(6)(−5)
4(6)
1 √ √
= (12x + 7 − 49 + 120)(12x + 7 + 49 + 120)
24
1
= (12x + 7 − 13)(12x + 7 + 13)
24
1
= (12x − 6)(12x + 20)
24
1
= · 6(2x − 1) · 4(3x + 5)
24
= (2x − 1)(3x + 5).

(b) x2 − 2x − 3
1 h p ih p i
= 2(1)x + (−2) − (−2)2 − 4(1)(−3) 2(1)x + (−2) + (−2)2 − 4(1)(−3)
4(1)
1 √  √ 
= 2x − 2 − 4 + 12 2x − 2 + 4 + 12
4
1
= (2x − 2 − 4)(2x − 2 + 4)
4
1
= (2x − 6)(2x + 2)
4
1
= · 2(x − 3) · 2(x + 1)
4
= (x − 3)(x + 1).

11
(c) (5x + 1)2 − 2(5x + 1) − 3

Set u = 5x + 1. Then, (e) x3 + x2 − 4x − 4

(5x + 1)2 − 2(5x + 1) − 3 = x2 (x + 1) − 4(x + 1)
= u2 − 2u − 3 = (x2 − 4)(x + 1)
= (u − 3)(u + 1) = (x + 2)(x − 2)(x + 1).
= [(5x + 1) − 3][(5x + 1) + 1] (f ) x3 − 2x2 − 3x + 6
= (5x − 2)(5x + 2). = x2 (x − 2) − 3(x − 2)
(d) x5 y 2 − xy 6 = (x2 − 3)(x − 2)
√
√

= xy 2 (x4 − y 4 ) = x + 3 x − 3 (x − 2).
= xy 2 (x2 − y 2 )(x2 + y 2 )

= xy 2 (x + y)(x − y)(x2 + y 2 ).

Note 61: If a and b are algebraic expressions, then

(i) a3 − b3 = (a + b)(a2 − ab + b2 );

(ii) a3 + b3 = (a + b)(a2 + ab + b2 ).

Section 3.3: Rational Expressions

Definition 62:

(i) A fractional expression is a quotient of two algebraic expressions.

(ii) A rational expression is a quotient of two polynomials.

Proposition 63: Let A, B, C, and D be algebraic expressions.

AC A
(i) If C ̸= 0, then = .
BC B
A C AC
(ii) If B, D ̸= 0, then · = .
B D BD
A C AD
(iii) If B, C, D ̸= 0, then ÷ = .
B D BC
A+B A B
(iv) If C ̸= 0, then = + .
C C C
A A AC + AB
(v) If B, C ̸= 0, then + = .
B C BC
A C
(vi) If B, D ̸= 0, then = if and only if AD = BC.
B D
A
(vii) If B ̸= 0, then = 0 if and only if A = 0.
B
Definition 64: A complex fraction is a fractional expression where the numerator or denominator

consists of one or more fractions.

Example 65: Simplify the following.

2 1 2x + x − 2
+ 3x − 2 x−5 x−5
x(x − 2)
(a) x − 2 x = = · = .
3x 2 3x − 2 x(x − 2) 3x − 2 x(x − 2)
−
x−5 x−5 x−5

12
 2x 2x 2x2 4(x + 2) − 2x2 −2x2 + 4x + 8
(b) 4 − =4− =4− = = .
x−2 2x − (x − 2) x+2 x+2 x+2
2−
x x
1 1
c−1 c ab
(c) −1 = = c = .
a + b−1 1 1 a+b c(a + b)
+
a b ab

Section 3.4: Equations

Definition 66:

(i) An equation is a statement that two expressions are equivalent.

(ii) The values of a variable that make an equation true are the solutions (roots) of the equation.

(iii) The process of finding the solutions of an equation is called solving the equation.

(iv) Two equations with the same solutions are equivalent equations.

Proposition 67: Let A, B, and C be expressions and n ∈ N. Then the following are equivalent.

(i) A = B;

(ii) A + C = B + C;

(iii) A − C = B − C;

(iv) If C ̸= 0, then AC = BC;

(v) An = B n .

Definition 68: A linear equation is an equation equivalent to one of the form ax + b = 0, where

a, b, c ∈ R and x is a variable.

Example 69: Solve the equation 7x − 4 = 3x + 8.

Consider the following.

7x − 4 = 3x + 8 ⇐⇒ 7x − 3x = (3x + 12) − 3x
⇐⇒ (7x − 4) + 4 = (3x + 8) + 4 ⇐⇒ 4x = 12
⇐⇒ 7x = 3x + 12 ⇐⇒ x = 3.

Definition 70: A quadratic equation is an equation equivalent to one of the form ax2 +bx+c = 0,

where a, b, c ∈ R and a ̸= 0.
√
Proposition 71: Let c ∈ R+ . Then x2 = c if and only if x = ± c.

Example 72: Solve each equation.

(a) x2 = 5
√
By the proposition above, x = ± 5.

(b) (x + 5)2 = 4

By the proposition above,

13
 √
x + 5 = ± 4 = ±2

⇐⇒ x = −5 ± 2

⇐⇒ x = −3 or x = −7.

Lemma 73 (Completing the Square Lemma): For any a, b, c ∈ R with a ̸= 0,

2
b2 − 4ac

2 b
ax + bx + c = a x + − .
2a 4a
Proof: Consider the following.
2
b2 − 4ac

b
a x+ −
 2a 4a
2 b b2 b2 − 4ac
=a x + x+ 2 −
a 4a 4a
2 2
b b − 4ac
= ax2 + bx + −
4a 4a
2 4ac
= ax + bx +
4a
2
= ax + bx + c. ■

Example 74: Use the completing the square lemma to solve the equation x2 − 5x + 6 = 0.

Consider the following.

= x2 − 5x + 6
2
(−5)2 − 4(1)(6)

−5
=1 x+ −
2 4(1)
 2
5 25 − 24
= x− −
2 4
 2
5 1
= x− −
2 4
 2
5 1
⇐⇒ x − =
2 4
5 1
⇐⇒ x − = ±
2 2
5 1
⇐⇒ x = ±
2 2
⇐⇒ x = 3 or x = 2.

Definition 75: Let a, b, c ∈ R with a ̸= 0 and x be a variable. The discriminant of the expression

ax2 + bx + c is the quantity b2 − 4ac.

Note 76: Let a, b, c ∈ R with a ̸= 0, ax2 + bx + c = 0 be a quadratic equation, and D denote the

discriminant of the expression ax2 + bx + c.

(i) If D > 0, then the equation ax2 + bx + c = 0 has two real solutions;

(ii) If D = 0, then the equation ax2 + bx + c = 0 has one real solution;

(iii) If D < 0, then the equation ax2 + bx + c = 0 has two complex solutions.

Theorem 77 (Quadratic
√ Formula): Let a, b, c ∈ R with a ̸= 0. Then ax2 + bx + c = 0 if and
2
−b ± b − 4ac
only if x = .
2a

14
Proof: By the completing the square lemma,

= ax2 + bx + c
2
b2 − 4ac

b
=a x+ −
2a 4a
2
b2 − 4ac

b
⇐⇒ =a x+
4a 2a
2
 2
b − 4ac b
⇐⇒ = x +
√4a2 2a
b2 − 4ac b
⇐⇒ ± =x+
2a √ a √
b b2 − 4ac −b ± b2 − 4ac
⇐⇒ x = − ± = . ■
2a 2a 2a
Example 78: Solve the following equations.
3 5
(b) + =2
x x+2
(a) 3x2 − x − 7 = 0p ⇐⇒ 3(x + 2) + 5x = 2x(x + 2)
−(−1) ± (−1)2 − 4(3)(−7) ⇐⇒ 3x + 6 + 5x = 2x2 + 4x
⇐⇒ x =
√ 2(3)
1 ± 1 + 84 ⇐⇒ 8x + 6 = 2x2 + 4x
=
√6 ⇐⇒ 2x2 − 4x − 6 = 0
1 ± 85
= .
6 ⇐⇒ x2 − 2x − 3 = 0

⇐⇒ (x − 3)(x + 1) = 0

√ ⇐⇒ x = 3 or x = −1.
(c) 2x = 1 − 2−x
√
⇐⇒ 2x − 1 = − 2 − x (d) x1/3 + x1/6 − 2 = 0
√
⇐⇒ 1 − 2x = 2 − x Set u = x1/6 . Then,
⇐⇒ (1 − 2x)2 = 2 − x x1/3 + x1/6 − 2 = 0
⇐⇒ 1 − 4x + 4x2 = 2 − x ⇐⇒ u2 + u − 2 = 0
⇐⇒ 4x2 − 3x − 1 = 0 ⇐⇒ (u + 2)(u − 1) = 0
⇐⇒ (4x + 1)(x − 1) = 0 ⇐⇒ u = −2 or u = 1
1
⇐⇒ x = − or x = 1. ⇐⇒ x1/6 = −2 or x1/6 = 1
4
Note that x = 1 does not satisfy the equation ⇐⇒ x = 64 or x = 1.
(is an extraneous solution). Note that x = 64 is an extraneous solution.

15
(e) 4x4 − 25x2 + 36 = 0 ⇐⇒ (4u − 9)(u − 4) = 0
9
Set u = x2 . Then, ⇐⇒ u = or u = 4
4
9
4x4 − 25x2 + 36 = 0 ⇐⇒ x2 = or x2 = 4
4
3
⇐⇒ 4u2 − 25u + 36 = 0 ⇐⇒ x = ± or x = ±2.
2

Section 3.5: Inequalities

Theorem 79: Let A, B, C, D be expressions and c ∈ R.

(i) A ≤ B if and only if A + C ≤ B + C;

(ii) A ≤ B if and only if A − C ≤ B − C;

(iii) If C > 0, then A ≤ B if and only if AC ≤ BC;

(iv) If C < 0, then A ≤ B if and only if AC ≥ BC;

1 1
(v) If A > 0 and B > 0, then A ≤ B if and only if ≥ ;
A B
(vi) If A ≤ B and C ≤ D, then A + C ≤ B + D;

(vii) |A| ≤ c if and only if −c ≤ A ≤ c;

(viii) |A| ≥ c if and only if A ≤ −c or A ≥ c.

Theorem 80: Let p be a polynomial. If p(x1 ) = 0 = p(x2 ), where x1 , x2 are consecutive zeros of p,

then for all x ∈ (x1 , x2 ), either p(x) > 0 or p(x) < 0.

Definition 81: A critical value (number) of a rational expression is a number that makes the

numerator or the denominator of the expression equal to zero.

Example 82: Solve the following inequalities.

(a) 3x + 1 < 4x − 2 ⇐⇒ 3x + 3 − 3x < 4x − 3x

⇐⇒ 3x + 1 + 2 < 4x − 2 + 2 ⇐⇒ x > 3.

So the solution set is (3, ∞).

(b) x3 + 3x2 − 4x − 12 ≥ 0

⇐⇒ (x + 3)(x + 2)(x − 2) ≥ 0.

If (x + 3)(x + 2)(x − 2) = 0, then x = −3, −2, 2. Consider four cases.

Case 1: x < −3

In this case, x + 3 < 0, x + 2 < 0, and x − 2 < 0. This means that (x + 3)(x + 2)(x − 2) < 0.

Case 2: x ∈ (−3, −2)

In this case, x + 3 > 0, x + 2 < 0, and x − 2 < 0. This means that (x + 3)(x + 2)(x − 2) > 0.

Case 3: x ∈ (−2, 2)

In this case, x + 3 > 0, x + 2 > 0, and x − 2 < 0. This means that (x + 3)(x + 2)(x − 2) < 0.

16
Case 4: x > 2

In this case, x + 3 > 0, x + 2 > 0, and x − 2 > 0. This means that (x + 3)(x + 2)(x − 2) > 0.

Therefore, the solution set is [−3, −2] ∪ [2, ∞).

3x + 4
(c) ≤2
x+1
3x + 4 3x + 4 − 2x − 2
⇐⇒ −2≤0 ⇐⇒ ≤0
x+1 x+1
3x + 4 − 2(x + 1) x+2
⇐⇒ ≤0 ⇐⇒ ≤ 0.
x+1 x+1

The above implies that x = −2 and x = −1 are critical values for the expression. Consider three

cases.

Case 1: x < −2 Case 2: x > −1 Case 3: x ∈ (−2, −1)

x+2 x+2 x+2
In this case, > 0. In this case, > 0. In this case, < 0.
x+1 x+1 x+1

Thus, the solution set is [−2, −1).

Chapter 4: Coordinate Geometry

Section 4.1: The Cartesian Plane

Definition 83: The Cartesian Plane is the set R × R = R2 = {(x, y) | x, y ∈ R}.

Definition 84: Let (x, y) ∈ R2 .

(i) x is called the abcissa;

(ii) y is called the ordinate;

(iii) x and y are called the coordinates of (x, y);

(iv) The pair (x, y) is called either a coordinate (pair), an ordered pair, or simply a point;

(v) The origin is the point (0, 0), and it is usually denoted O.

Example 85:

17
 Section 4.2: The Distance and Midpoint Formulas
Theorem 86 (Pythagorean Theorem): Suppose the legs of a right triangle have lengths a, b and

the hypotenuse has length c. Then a2 + b2 = c2 .

Proof: Suppose that the triangle described in the theorem is a right triangle and consider the figure

below.

1
Recall the area of the triangle is ab. Now consider the figure below.
2

18
Since α + β = 90, the figure contains a small square of side length c, a large square of side length
1
a + b, and four copies of the triangle. Now, the area of the small square is c2 = (a + b)2 − 4 · ab =
2
(a + b)2 − 2ab = a2 + 2ab + b2 − 2ab = a2 + b2 . Therefore, a2 + b2 = c2 . ■

Corollary 87 (Distance Formula): Let (x1 , y1 ), (x2 , y2 ) ∈ R2 . The distance d between (x1 , y1 )
p
and (x2 , y2 ) is d = (x2 − x1 )2 + (y2 − y1 )2 .

Proof: Consider the figure below.

Let d denote the distance between (x1 , y1 ) and (x2 , y2 ). By the Pythagorean Theorem and a previous

proposition,

d2

= |x2 − x1 |2 + |y2 − y1 |2

= (x2 − x1 )2 + (y2 − y1 )2
p
⇐⇒ d = (x2 − x1 )2 + (y2 − y1 )2 . ■

19
 p
Example 88: The distance between the points (−3, 1) and (5, −2) is (−3 − 5)2 + (1 − (−2))2 =
√
73.

Theorem 89 (Midpoint Formula): The midpoint between the points (x1 , y1 ) and (x2 , y2 ) is the
 
x1 + x2 y1 + y2
points , .
2 2  
−3 + 5 1 − 2
Example 90: The midpoint between the points (−3, 1) and (5, −2) is the point , =
  2 2
−1
1, .
2

Section 4.3: Graphs and Intercepts

Definition 91: Let (x, y) ∈ R2 . Then (x, y) satisfies an equation if the equation is a true statement

when x and y are substituted in it.

Example 92:

(a) The point (3, 10) satisfies the equation y = x2 + 1.

(b) The point (1, 3) does not satisfy the equation y = x2 + 1.

Definition 93: The graph of an equation is the set {(x, y) ∈ R2 | the equation is satisfied}.

Theorem 94 (Fundamental Principle of Analytic Geometry): A point in R2 lies on the

graph of an equation if and only if its coordinates satisfy the equation.

Proof: (=⇒) Suppose that the point (x, y) ∈ R2 lies on the graph of an equation. Since the point

(x, y) is on the graph of the equation, the equation is a true statement, and so, the equation is

satisfied.

(⇐=) Suppose the point (x, y) satisfies an equation. Then, by the definition of graph, (x, y) is on

the graph of the equation. ■

Definition 95:

(i) The x-coordinates where a graph of an equation intersects the x-axis are called the x-intercepts.

(ii) The y-coordinates where a graph of an equation intersects the y-axis are called the y-intercepts.

Note 96:

(i) To find the x-intercepts, set y = 0 in the equation and solve for x.

(ii) To find the y-intercepts, set x = 0 in the equation and solve for y.

Example 97: Find the x- and y-intercepts of the equation y = x2 − 2.

To find the x-intercepts, we set y = 0. Doing this, we obtain that 0 = x2 − 2, or 2 = x2 , which

√ √ √
means that x = ± 2. So, the x-intercepts are ( 2, 0) and (− 2, 0). Now, to find the y-intercepts,

we set x = 0. Doing this, we obtain that y = 02 − 2 = −2. So, the y-intercept is (−2, 0).

20
 Section 4.4: Circles
Definition 98:

(i) A circle is the set of points in R2 which are equidistant from a given point.

(ii) r is called the radius of the circle;

(iii) The point (h, k) is called the center of the circle.

Theorem 99: The equation of the circle with center (h, k) and radius r is (x − h)2 + (y − k)2 = r2 .

Proof: Exercise. ■

Theorem 100: The general equation of a circle is given by x2 + y 2 − 2hx − 2ky + h2 + k 2 − r2 = 0.

Proof: Consider the following.

= x2 + y 2 − 2hx − 2ky + h2 + k 2
2 2 2
r = (x − h) + (y − k)
⇐⇒ x2 + y 2 − 2hx − 2ky + h2 + k 2 − r2 = 0. ■
2 2 2 2
= x − 2hx + h + y − 2ky + y
Example 101: For each of the following, find the center and radius of the given circle.

(a) (x − 1)2 + (y + 2)2 = 4

The circle is centered at (1, −2) and has radius 2;

(b) x2 + 4x + y 2 + 10y + 26 = 0

Note that 0 = x2 + 4x + y 2 + 10y + 26 = (x + 2)2 + (y + 5)2 − 3, and so (x + 2)2 + (y + 5)2 = 3. So,

√
the circle is centered at (−2, −5) and has radius 3;

(c) x2 + y 2 − 4x + 6y − 23 = 0

Note that −2h = −4, −2k = 6, and h2 + k 2 − r2 = −23. So, h = 2, k = −3, and r2 = 4 + 9 + 23 = 36,

and so, r = 6. Thus, the circle is centered at (2, −3) and has radius 6.

Section 4.5: Symmetry

Definition 102: A graph of an equation

(i) is symmetric (has symmetry) with respect to the x-axis if whenever the point (x, y) is

on the graph, then the point (x, −y) is on the graph.

(ii) is symmetric (has symmetry) with respect to the y-axis if whenever the point (x, y) is

on the graph, then the point (−x, y) is on the graph.

(iii) is symmetric (has symmetry) with respect to the origin if whenever the point (x, y) is

on the graph, then the point (−x, −y) is on the graph.

Example 103: Determine whether the graph of the given equation has symmetry.

(a) x = y 2 .

Note that x = y 2 = (−y)2 . So, the graph of the equation has symmetry with respect to the x-axis.

Also, if x = y 2 , then −x ̸= y 2 , and so, the graph of the equation does not have symmetry with

21
respect to the y-axis. By a similar argument, the graph of the equation is not symmetric with respect

to the origin.

(b) y = x3 − 9x.

Note that −y = (−x)3 − 9(−x) = −x3 + 9x, or y = x3 − 9x. So, the graph of the equation has

symmetry with respect to the origin.

22
 Chapter 5: Lines

Section 5.1: Slope

Definition 104: The slope of a non-vertical line passing through the points (x1 , y1 ) and (x2 , y2 )
y2 − y1 y1 − y2
is m = = .
x2 − x1 x1 − x2
Note 105:

(i) There is no definition of the slope of a vertical line.

(ii) Since every point on a horizontal line has the same y-coordinate, the slope of a horizontal line

is 0.

Example 106: For each of the following, calculate the slope of the line passing through the given

pair of points.

(a) (2, −3) and (0, 4)

4 − (−3) 7
m= =− .
0−2 2
(b) (2, 6) and (2, −4)
−4 − 6 −10
m= = , which is undefined.
2−2 0
(c) (3, 1) and (−2, 1)
1−1
m= = 0.
3 − (−2)

Section 5.2: Equations of Lines

Theorem 107 (Point-Slope Formula): The equation of a line passing through the point (x1 , y1 )

with slope m is y − y1 = m(x − x1 ).

Proof: Suppose the point (x, y) is on the line passing through the point (x1 , y1 ) with slope m.
y − y1
Then, m = , which means y − y1 = m(x − x1 ). ■
x − x1
Corollary 108: The equation of a horizontal line passing through the point (x1 , y1 ) is y = y1 .

Proof: Exercise. ■

Corollary 109 (Slope-Intercept Formula) The equation of a line with slope m passing through

y-intercept (0, b) is y = mx + b.

Proof: By the point-slope formula, the equation of the line with slope m passing through the point

(0, b) is y − b = m(x − 0), which means y = mx + b. ■

Example 110: Find the equation of the line with slope 3 passing through the point (0, −7).

By the point-slope formula, the desired line is given by the equation y − (−7) = 3(x − 0) = 3x, or

y = 3x − 7.

23
 Section 5.3: Parallel Lines
Theorem 111: Two distinct non-vertical lines are parallel if and only if they have the same slope.

Proof: (=⇒) We will prove the contrapositive. Suppose two lines have equations y1 = m1 x + b1

and y2 = m2 x + b2 with m1 ̸= m2 . Then, since m1 ̸= m2 , the equation m1 x + b1 = m2 x + b2 has

 
b2 − b1 b2 − b1 m1 b2 − m2 b1
solution . So, the lines intersect at the point , .
m1 − m2 m1 − m2 m1 − m2
(⇐=) Suppose two lines have the same slope. If they share a point, the point-slope formula implies

the lines passing through the point in common are the same line. Thus, two different lines with the

same slope are parallel. ■

Note 112: Two distinct vertical lines are parallel.

Section 5.4: Perpendicular Lines

Theorem 113: Two lines with slope m1 , m2 ∈ R are perpendicular if and only if m1 m2 = −1.

Proof: Suppose that two lines with slope m1 and m2 respectively intersect at the point (a, b). Then,

the equations of the lines are y = m1 (x − a) + b (call this line l1 ) and y = m2 (x − a) + b (call this

line l2 ). Note that the point (a + 1, b + m1 ) is on l1 and the point (a + 1, b + m2 ) is on l2 . By the

p
distance formula, the distance between (a, b) and (a + 1, b + m1 ) is 1 + m21 , the distance between
p
(a, b) and (a + 1, b + m2 ) is 1 + m22 , and the distance between (a + 1, b + m1 ) and (a + 1, b + m2 )
p
is (m1 − m2 )2 . Consider the triangle with vertices (a, b), (a + 1, b + m1 ), and (a + 1, b + m2 ) and

note the following.

l1 and l2 are perpendicular

⇐⇒ the triangle is a right triangle

⇐⇒ (1 + m1 )2 + (1 + m2 )2 = (m1 − m2 )2

⇐⇒ m21 + m22 + 2 = m21 + m22 − 2m1 m2

⇐⇒ 2 = −2m1 m2

⇐⇒ m1 m2 = −1 ■

Note 114: A horizontal and a vertical line are perpendicular.

Example 115: For each of the following, give the equation of the line (if possible) with the given

properties. If it is not possible, explain why it is not.

(a) Perpendicular to the line 2x + 4y = 8 and passing through the point (2, 1).

Note the following.

2x + 4y = 8

⇐⇒ x + 2y = 4

24
⇐⇒ 2y = 4 − x
1
⇐⇒ y = 2 − x.
2
So the desired slope is 2. Now, by the point-slope formula,

y − 1 = 2(x − 2) = 2x − 4, or y = 2x − 3.

(b) Parallel to the line 2x + 4y = 8 and passing through the point (2, 1).

This is not possible since the point (2, 1) is on the line 2x + 4y = 8.

(c) Passing through the point (0, 3) with no x-intercept.

The line is given by the equation y = 3.

Chapter 6: Complex Numbers

Section 6.1: Arithmetic with Complex Numbers

√
Definition 116: The number i, called the imaginary unit, is defined to be −1.
√
Notation 117: C = {a + bi | a, b ∈ R, and i = −1}.

Definition 118: Let z = a + bi, w = c + di ∈ C for some a, b, c, d ∈ R.

(i) a is the real part of z and is denoted Re(z);

(ii) b is the imaginary (complex) part of z and is denoted Im(z);

(iii) If Re(z) = 0 and Im(z) ̸= 0, then z is called pure imaginary;

(iv) z and w are equal if Re(z) = Re(w) and Im(z) = Im(w).

Definition 119: Let z = a + bi, w = c + di ∈ C.

(i) z + w = (a + c) + (b + d)i;

(ii) zw = (a + bi)(c + di) = ac + adi + bci + bdi2 = ac − bd + (ad + bc)i;

z a + bi a + bi c − di (ac + bd) + (bc − ad)i
(iii) = = · = .
w c + di c + di c − di a2 + b2
Example 120: Let z = 3 + 4i and w = −2 + 5i.

(a) z + w = 1 + 9i;

(b) z − w = 5 − i;

(c) w − z = −5 + i;

(d) zw = (3 + 4i)(−2 + 5i) = −26 + 7i.

25
 Section 6.2: The Complex Conjugate
Definition 121: Let z = a + bi ∈ C. The conjugate of z is z = a − bi.

Example 122: Let z = 3 + 4i and w = −2 + 5i.

(a) z = 3 − 4i;

(b) w = −2 − 5i.

Note 123: If z = a + bi ∈ C, then

(i) z + z = (a + bi) + (a − bi) = 2a;

(ii) zz = (a + bi)(a − bi) = a2 − abi + abi − bi2 = a2 + b2 .

Proposition 124: Let z, w ∈ C.

(i) z + w = z + w;

(ii) zw = z · w.

Proof: Let a, b, c, d ∈ R such that z = a + bi and w = c + di. Then,

z+w zw
= a + c + (b + d)i = (a + bi)(c + di)
= a + c − (b + d)i = ac − bd + (ad + bc)i
= a + c − bi − di = (ac − bd) − (ad + bc)i
= a − bi + c − di = (ac − bd) + (−bc − ad)i
= a + bi + c + di = (a − bi)(c − di)
= z + w, and = z · w. ■

Chapter 7: Functions

Section 7.1: The Definition of Function

Definition 125: Let X and Y be sets.

(i) A relation from X to Y is a rule that assigns one or more elements of Y to each element of X.

(ii) A function from X to Y is a relation that assigns a unique element of Y to each element of X.

(iii) The set X is the domain of the function and will be denoted dom(f ).

(iv) The set Y is the codomain of the function and will be denoted cod(f ).

Notation 126: Let X and Y be sets.

(i) If f is a function from X to Y , we write f : X → Y .

(ii) Y X = {f | f : X → Y }.

Example 127: The following are examples of relations. Which are functions?

26
(a)

(b)

(c)

(d)

27
Each of (a), (b), and (c) are functions, while (d) is not a function.

Definition 128: Suppose that f : X → Y and x ∈ X.

(i) The unique element of Y that f assigns to x is the function value of x or the image of x under

f and is usually denoted f (x).

(ii) The range of f is the set f (X) = ran(f ) = {f (x) | x ∈ X}.

(iii) The natural domain of f is the set {x ∈ X | f (x) is defined}.

Example 129: Define f : R → R by f (x) = x2 .

(a) f (3) = 9;

(b) f (−3) = 9;

(c) dom(f ) = R;

(d) cod(f ) = R;

(e) ran(f ) = [0, ∞).

√
 
1
Example 130: Define f : − , ∞ → R by f (x) = 2x + 1.
  2
1
(a) dom(f ) = − , ∞ ;
2
(b) ran(f ) = [0, ∞).

Definition 131: Suppose that f : [a, b] → R for some a, b ∈ R with a < b. The average rate of
f (b) − f (a)
change of f over [a, b] (or from a to b) is the quantity .
b−a
Example 132: For each of the following, find the average rate of change of the function on the

given interval.

(a) f (x) = 3x + 8; [0, 3]

f (3) − f (0) 17 − 8
By the definition above, the average rate of change is = = 3.
3−0 3
(b) g(x) = x2 − 2x + 8; [1, 5]
g(5) − g(1) 23 − 7
By the definition above, the average rate of change is = = 4.
5−1 4
√
(c) h(x) = − x + 1 + 3; [3, 8]

28
 h(8) − h(3) 0−1 1
By the definition above, the average rate of change is = =− .
8−3 5 5

Section 7.2: Graphs of Functions

Definition 133: The graph of a function f : X → Y is the set G (f ) = {(x, y) ∈ X × Y | y =

f (x)} = {(x, f (x)) ∈ X × Y | x ∈ X}.

Example 134: Set X = {−1, 0, 1, 2, 3} and define f : X → Z by f (x) = x2 + 1.

(a) dom(f ) = X = {−1, 0, 1, 2, 3}

(b) cod(f ) = Z

(c) ran(f ) = {2, 1, 2, 5, 10} = {1, 2, 5, 10}

(d) The graph of f is the set G (f ) = {(−1, 2), (0, 1), (1, 2), (2, 5), (3, 10)}.
x2 + 3x + 2
Example 135: Consider the functions f (x) = and g(x) = x + 2. Since g(−1) = 1 and
x+1
f (−1) is undefined, the functions are not the same. Note that dom(f ) = R \ {−1} and dom(g) = R.

Since ran(g) = R and g(−1) = 1, ran(f ) = R \ {1}. The sketches of f and g are below.

29
Example 136: For each of the following functions, the domain, range, and sketch of the graph is

given.

(a) f (x) = |x|

dom(f ) = R

ran(f ) = [0, ∞)

√
(b) g(x) = 3 − 2x
 
3
dom(g) = −∞,
2
ran(f ) = (−∞, 0]

30
 √
(c) h(x) = 16 − x2

dom(h) = [−4, 4]

ran(h) = [0, 4]

|x|
(d) j(x) =
x
dom(j) = R \ {0}

ran(j) = {−1} ∪ {1} = {−1, 1}

31
Theorem 137 (Vertical Line Test): A curve in R2 is the graph of a function if and only if every

vertical line intersects the curve in at most one point.

Proof: (=⇒) Suppose that X ⊆ R, f : X → R, and G (f ) ⊆ R2 is the graph of f . To see that

every vertical line in R2 intersects G (f ) in at most one point, let a ∈ R and consider the vertical

line x = a.

Case 1: a ∈
/ X.

In this case, f (a) does not exist. So, there is no y ∈ R such that (a, y) ∈ G (f ), which means the

vertical line does not intersect G (f ).

Case 2: a ∈ X.

Since G (f ) is the graph of f and f (a) is unique, the vertical line x = a intersects G (f ) only at the

point (a, f (a)).

(⇐=) Suppose G (f ) ⊆ R2 such that every vertical line intersects G (f ) in at most one point. Set

X = {x ∈ R | there is a y ∈ R such that (x, y) ∈ G (f )}. Since every vertical line intersects G (f )

in at most one point, for each x ∈ X, there is a unique y ∈ R such that (x, y) ∈ G . So, the rule

f : X → R defined by f (x) = y if and only if (x, y) ∈ G (f ) is a function. By design, G (f ) is the

graph of f . ■

Example 138: The following is not the graph of a function since there is a vertical line that

intersects the curve in more than one point.

32
 Section 7.3: Even and Odd Functions
Definition 139: A function f is

(i) even if f (−x) = f (x) for every x ∈ dom(f );

(ii) odd if f (−x) = −f (x) for every x ∈ dom(f ).

Theorem 140:

(i) A function is even if and only if its graph is symmetric about the y-axis.

(ii) A function is odd if and only if its graph is symmetric about the origin.

Question 141: Does a function have to be even or odd? No.

Example 142: Determine whether the following functions are even, odd, or neither.

(a) f (x) = x2 + x

Note that f (2) = 6 ̸= 2 = f (−2). So f is neither even nor odd.

(b) g(x) = −|x|

Note that g(−x) = −| − x| = −|x| = g(x). Thus, g is even.

x
(c) h(x) =
|x|
−x x
Note that h(−x) = =− = −h(x). Thus, h is odd.
| − x| |x|
(d) f (x) = x3 + x2 − 1

Note that f (2) = 11 ̸= −5 = f (−2). Thus, f is neither even nor odd.

Section 7.4: Combinations of Functions

Definition 143: Let f and g be functions.

(i) The sum of f and g is the function (f + g)(x) = f (x) + g(x).

(ii) The difference of f and g is the function (f − g)(x) = f (x) − g(x).

33
(iii) The product of f and g is the function (f g)(x) = f (x)g(x).
 
f f (x)
(iv) The quotient of f and g is the function (x) = .
g g(x)
Note 144: If f and g are functions, then dom(f + g) = dom(f − g) = dom(f g) = dom(f ) ∩ dom(g)
 
f
and dom = {x ∈ dom(f ) ∩ dom(g) | g(x) ̸= 0}.
g
√
Example 145: Set f (x) = x and g(x) = x + 1. Calculate each of the following.

(a) (f + g)(9) = f (9) + g(9) = 3 + 10 = 13

(b) (f g)(9) = f (9) · g(9) = 3 · 10 = 30.

Definition 146: Let f : A → B and g : B → C.

(i) The function (g ◦ f ) : A → C defined by (g ◦ f )(x) = g(f (x)) is the composition of g with f .

(ii) dom(g ◦ f ) = {x ∈ dom(f ) | f (x) ∈ dom(g)}.

√
Example 147: Let f (x) = x and g(x) = x + 1. Calculate each of the following.
√
(a) (f ◦ g)(x) = f (x + 1) = x + 1
√ √
(b) (f ◦ g)(9) = 9 + 1 = 10
√
(c) (g ◦ f )(x) = x + 1

(d) (g ◦ f )(9) = 4
√
Example 148: Let f (x) = (x2 − 1)3 + x2 − 2. Find functions h(x) and g(u) such that f (x) =

g[h(x)].

There are infinitely many solutions to this. Listed below are three solutions.
√
Set h(x) = x2 − 1 and g(u) = u3 + u − 1
√
Set h(x) = x2 and g(u) = (u − 1)3 + u − 2
√
Set h(x) = x2 − 2 and g(u) = (u + 1)3 + u.

Definition 149: Let f be a function and h ∈ R\{0}. The difference quotient of f is the quantity
f (x + h) − f (x)
.
h

34
Example 150: Calculate the difference quotient of the function f (x) = x2 + 7.
f (x + h) − f (x)
h
(x + h)2 + 7 − (x2 + 7)
=
h
x2 + 2xh + h2 + 7 − x2 − 7
=
h
2xh + h2
=
h
h(2x + h)
=
h
= 2x + h.

Section 7.5: Transformations of Functions

Theorem 151: The graph of y = f (x) + c is the graph of y = f (x) shifted

(i) up c units if c ≥ 0;

(ii) down |c| units if c < 0.

Theorem 152: The graph of y = f (x + c) is the graph of y = f (x) shifted

(i) left c units if c ≤ 0;

(ii) right |c| units if c < 0.

Theorem 153: The graph of the function y = −f (x) is the graph of the function y = f (x) reflected

about the x-axis.

Theorem 154: The graph of the function y = f (−x) is the graph of the function y = f (x) reflected

about the y-axis.

Theorem 155: The graph of y = a · f (x) is the graph of y = f (x)

(i) stretched vertically if a > 1;

(ii) compressed vertically if a ∈ (0, 1).

Theorem 156: The graph of y = f (ax) is the graph of the function y = f (x)

(i) stretched horizontally if a ∈ (0, 1);

(ii) compressed horizontally if a > 1.

Example 157: Given the graph of f (x) = |x|, use transformations to sketch the graph of the

function g(x) = −2|x − 1| + 3.

35
The transformations are as follows:

(i) reflection about the x-axis;

(ii) vertical stretch of factor 2;

(iii) horizontal shift of 1 unit right;

(iv) vertical shift of 3 units up.

The graph of the function g is below.

Section 7.6: One-to-One and Onto Functions

Theorem 158: The following are equivalent for any f : X → Y and any x1 , x2 ∈ X.

(i) f (x1 ) = f (x2 ) if and only if x1 = x2 ;

(ii) If x1 ̸= x2 , then f (x1 ) ̸= f (x2 );

36
(iii) If f (x1 ) = f (x2 ), then x1 = x2 .

Proof: Let f : X → Y .

((i) =⇒ (ii)) Suppose that (i) is satisfied and let x1 , x2 ∈ X with x1 ̸= x2 . Since x1 ̸= x2 , it is

clear that f (x1 ) ̸= f (x2 ).

((ii) =⇒ (iii)) Suppose (ii) is satisfied. To see that f (x1 ) = f (x2 ) implies that x1 = x2 , suppose

x1 , x2 ∈ X such that f (x1 ) = f (x2 ). If x1 ̸= x2 , then f (x1 ) ̸= f (x2 ) by the assumption. Since

f (x1 ) = f (x2 ), it must be that x1 = x2 .

((iii) =⇒ (i)) Suppose that (iii) is satisfied. To see that f (x1 ) = f (x2 ) if and only if x1 = x2 ,

suppose x1 , x2 ∈ X. If x1 = x2 , then f (x1 ) = f (x2 ) since f is a function. Also, if f (x1 ) = f (x2 ),

then by the assumption, we obtain x1 = x2 .

Therefore, (i) is satisfied, and the theorem is proved. ■

Definition 159: A function f : X → Y is one-to-one if it satisfies any of the conditions in the

theorem above.

Notation 160: We sometimes write “1-1” instead of “one-to-one.”

Example 161: Determine whether the following functions are one-to-one.

(a) f (x) = x

To see that f is one-to-one, note that for x1 ̸= x2 , then f (x1 ) = x1 ̸= x2 = f (x2 ). Thus, f is

one-to-one.

(b) g(x) = x2

To see that f is not one-to-one, note that g(−3) = 9 = g(3). Thus, g is not one-to-one. ■

Theorem 162 (Horizontal Line Test): Let X ⊆ R. Then a function f : X → R is one-to-one if

and only if every horizontal line intersects the graph of the function in at most one point.

Proof: (=⇒) Suppose that f is one-to-one and y = c is a horizontal line. Attempting a contradiction,

suppose the line y = c intersects the graph of f at two points, (x1 , c) and (x2 , c). Since (x1 , c) lies

on the graph of f , f (x1 ) = c. Likewise, f (x2 ) = c. This is a contradiction since f is one-to-one.

(⇐=) Suppose that every horizontal line intersects the graph of f in at most one point. To see that

f is one-to-one, suppose x1 , x2 ∈ X with x1 ̸= x2 . Let c = f (x1 ) and consider the horizontal line

y = c. By the assumption, the line y = c intersects the graph of f only at the point (x1 , c). This

implies the point (x2 , f (x2 )) is not on the line y = c which means f (x2 ) ̸= c = f (x1 ). Since x1 and

x2 were chosen arbitrarily, f is one-to-one. ■

Definition 163: Suppose I ⊆ R is an interval and f : I → R.

(i) f is increasing (nondecreasing) on I if f (x1 ) ≤ f (x2 ) whenever x1 ≤ x2

(ii) f is decreasing (nonincreasing) on I if f (x1 ) ≥ f (x2 ) whenever x1 ≤ x2

37
(iii) f is strictly increasing (increasing) on I if f (x1 ) < f (x2 ) whenever x1 < x2

(iv) f is strictly decreasing (decreasing) on I if f (x1 ) > f (x2 ) whenever x1 < x2

(v) f is monotonic (monotone) if it is either increasing or decreasing.

Proposition 164: A strictly increasing or strictly decreasing function is one-to-one.

Proof: Let f : X → R be either strictly increasing or strictly decreasing.

Case 1: f is strictly increasing

To see that f is one-to-one, suppose x1 , x2 ∈ X with x1 ̸= x2 . Without loss of generality, suppose

x1 < x2 . Since f is strictly increasing, f (x1 ) < f (x2 ), which implies that f (x1 ) ̸= f (x2 ). Therefore,

f is one-to-one.

Case 2: f is strictly decreasing

To see that f is one-to-one, use an argument similar to the one above. ■

Definition 165: A function f has a

(i) local (relative) maximum at c ∈ R if there is an open interval U containing c such that

f (x) ≤ f (c) for all x ∈ U .

(ii) local (relative) minimum at c ∈ R if there is an open interval U containing c such that

f (x) ≥ f (c) for all x ∈ U .

(iii) local (relative) extremum at c ∈ R if c is a local maximum or a local minimum.

Definition 166: A function f : X → Y is onto if for each y ∈ Y , there is an x ∈ X such that

f (x) = y.

Example 167:

(a) The function f : R → R defined by f (x) = x2 is not onto since ran(f ) = [0, ∞) ̸= R = cod(f ).

(b) The function f : R → [0, ∞) defined by f (x) = x2 is onto since ran(f ) = [0, ∞) = cod(f ).

Theorem 168: Let f : X → Y , g : Y → Z, and h : Z → W . Then

(i) f ◦ (g ◦ h) = (f ◦ g) ◦ h;

(ii) If f and g are one-to-one, then g ◦ f is one-to-one;

(iii) If f and g are onto, then g ◦ f is onto.

Proof: Exercise. ■

Section 7.7: Inverses of Functions

Definition 169: A function is a bijection if it is one-to-one and onto.

Definition 170: Let X be a set and f : X → X. Then f is an identity function if f (x) = x for

all x ∈ X.

Proposition 171: The identity function is unique.

38
Proof: Let f, g : X → X be distinct identity functions. Then, for all x ∈ X, f (x) = x = g(x), a

contradiction. Thus, the identity function is unique. ■

Definition 172: Let f : X → Y be a function. Then a function g : ran(f ) → dom(f ) is called the

inverse of f if (f ◦ g)(x) = (g ◦ f )(x) = x.

Definition 173: A function which has an inverse is called invertible.

Notation 174: If f is an invertible function, then the inverse of f is usually denoted f −1 .

Note 175: To find the inverse of an invertible function y = f (x), we proceed as following.

(i) Interchange x and y;

(ii) Solve for y.

Example 176: For each of the following, find the inverse of the given function.

(a) f : [0, ∞) → [−4, ∞); f (x) = x2 − 4

Consider y = x2 − 4. Interchanging x and y, we obtain the following.

x = y2 − 4 √
⇐⇒ y = x + 4.
⇐⇒ x + 4 = y 2

(b) g : [−3, ∞) → [−4, ∞); g(x) = x2 + 6x + 5

Consider y = x2 + 6x + 5. Interchanging x and y, we obtain the following.

x = y 2 + 6y + 5 ⇐⇒ x + 4 = (y + 3)2
√
⇐⇒ x = (y + 3)2 − 4 ⇐⇒ y = −3 + x + 4

x+1
(c) h : R \ {1} → R \ {3}; h(x) =
x−3
x+1
Consider y = . Interchanging x and y, we obtain the following.
x−3
y+1 ⇐⇒ xy − y = 3x + 1
x=
y−3
⇐⇒ x(y − 3) = y + 1 ⇐⇒ y(x − 1) = 3x + 1
3x + 1
⇐⇒ xy − 3x = y + 1 ⇐⇒ y = .
x−1

Theorem 177: Let f : X → Y be invertible.

(i) f −1 is unique;

(ii) (f −1 )−1 = f .

Proof: Exercise. ■

Theorem 178: Let f : X → Y and g : Y → Z. Then

(i) f is invertible if and only if it is a bijection;

(ii) If f and g are invertible, then g ◦ f is invertible and (g ◦ f )−1 = f −1 ◦ g −1 .

Proof: For (i),

39
(=⇒) Suppose f is invertible. Then for any x, y ∈ X, f (x) = f (y) implies that x = [f −1 ◦ f ](x) =

[f −1 ◦ f ](y) = y. So f is one-to-one. Also, if u ∈ Y , then [f ◦ f −1 ](u) = u. Set x = [f −1 ](u). Note

that f (x) = u. Since u was chosen arbitrarily, we obtain f is onto.

(⇐=) Suppose f is a bijection. Define g : Y → X as follows. For any y ∈ Y , let g(y) = x ∈ X such

that f (x) = y. Note that this x exists since f is onto. Also, since f is one-to-one, the function g

will clearly be well-defined. Furthermore, by definition of g, f (g(x)) = x. Hence, g = f −1 , and so,

f is invertible.

The proof of (ii) is left as an exercise to the reader. ■

Theorem 179: Let X, Y ⊆ R and suppose f : X → Y is invertible. Then the graph of f −1 is

obtained by reflecting the graph of f about the line y = x.

Proof: Note that (u, v) ∈ R2 lies on the graph of f if and only if the point (v, u) lies on the graph

of f −1 . If u = v, then the point (u, v) = (v, u) lies on the line y = x. So suppose u ̸= v. Recall that
 
u+v u+v
the midpoint of the line segment connecting the points (u, v) and (v, u) is , , which
2 2
is on the line y = x. Also, recall that the slope of the line containing the points (u, v) and (v, u) is

−1. So, the line containing the points (u, v) and (v, u) is perpendicular to the line y = x. Therefore,

(u, v) is the reflection of (v, u) through the line y = x. ■

40
Example 180: Below is the graph of an invertible function f . Sketch the graph of the function

f −1 .

The sketch of f −1 is below.

Proposition 181: If X, Y ⊆ R and f : X → Y is invertible and odd, then f −1 is odd.

Proof: Exercise. ■

41
 Chapter 8: Polynomial Functions

Section 8.1: Linear Functions

Definition 182: A linear function is a function of the form f (x) = mx + b, where m, b ∈ R.

Note 183:

(i) A line is the graph of a linear function if and only if it is not vertical.

(ii) Two intersecting lines intersect in a single point.

Section 8.2: Quadratic Functions

Definition 184: A quadratic function is a function of the form f (x) = ax2 + bx + c, where a ̸= 0.

Theorem 185: The graph of a quadratic function f (x) = ax2 + bx + c

(i) opens upward if a > 0;

(ii) opens downward if a < 0;

  
−b −b
(iii) has vertex ,f .
2a 2a
Definition 186: The axis (of symmetry) of the graph of of a quadratic function is the vertical

line that passes through the vertex.

Example 187: Find the vertex and x-intercept(s) (if any) of the function f (x) = x2 + x − 6. Use

your answers to sketch the graph.

 
1 25
By the theorem above, the vertex is given by the point − ,− . Now, setting f (x) = 0, we
2 4
obtain 0 = x2 + x − 6 = (x + 3)(x − 2), or x = 2, −3. Thus, the x-intercepts are (2, 0) and (−3, 0).

The graph of f (x) is given below.

42
 Section 8.3: Polynomial Functions
Definition 188: A polynomial is a function of the form f (x) = an xn + an−1 xn−1 + . . . + a1 x + a0 ,

where n ∈ ω, ai ∈ R for all 0 ≤ i ≤ n with an ̸= 0 (unless f (x) = 0).

Notation 189: R[x] = {p(x) | p(x) is a polynomial with coefficients in R}.

Definition 190: A polynomial in R[x] is monic if its leading term is 1.

Definition 191: The degree of a nonconstant polynomial in R[x] is a function deg : R[x] → N

defined by deg(an xn + an−1 xn−1 + . . . + a2 x2 + a1 x + a0 ) = n.

Note 192:

(i) The degree of the zero polynomial is not defined.

(ii) The degree of a constant polynomial is zero.

(iii) A linear function is a polynomial of degree at most 1.

(iv) A quadratic function is a polynomial of degree 2.

Example 193: Which of the following are polynomials?

√
(a) f (x) = x2 + 3x − x

(b) g(x) = x5 − 6

(c) h(x) = 3x
x3 − x2 + x
(d) j(x) =
x
Theorem 194 (Leading Coefficient Test): The end behavior of a polynomial is determined by

its leading term.

Section 8.4: Roots of Polynomials

Definition 195: Let f ∈ R[x] and c ∈ R. Then

(i) c is a root (zero) if f (c) = 0.

(i) c is algebraic if there is a polynomial in Q[x] such that c is a root.

(ii) c is transcendental if it is not algebraic.

Theorem 196: Suppose that f ∈ R[x] and c ∈ R. Then the following are equivalent.

(i) c is a root of f ;

(ii) c is a solution to the equation f (x) = 0;

(iii) x − c is a factor of f (x);

(iv) The point (c, 0) is an x-intercept of the graph of f .

Theorem 197 (Division Algorithm in R[x]): Let f (x), g(x) ∈ R[x] such that g(x) ̸= 0. Then,

there are unique q(x), r(x) ∈ R[x] such that f (x) = g(x)q(x) + r(x) and either r(x) = 0 or

43
deg(r) < deg(q).

Proof: Induction. ■

Example 198: Set p(x) = 2x3 + x2 − x + 1.

(a) Divide p(x) by x − 1.

Long division:

2x2 + 3x + 2


x−1 2x3 + x2 − x + 1

− 2x3 + 2x2

3x2 − x

− 3x2 + 3x

2x + 1

− 2x + 2

3
Synthetic division (Horner’s Scheme):

2 1 −1 1

1 2 3 2

2 3 2 3
(b) Calculate p(1).

p(1) = 3.

Theorem 199 (Remainder Theorem): Let f (x) ∈ R[x] and a ∈ R. Then f (a) is the remainder

when dividing f (x) by x − a.

Proof: By the division algorithm, there are q(x), r(x) ∈ R[x] such that f (x) = q(x)(x − a) + r(x)

and either r(x) = 0 or deg(r) < deg(x − a) = 1. So, in either case, r(x) = c, where c ∈ R. Thus,

f (x) = q(x)(x − a) + c. Note that f (a) = c = r(x). ■

Definition 200: Let f (x) ∈ R[x], and a ∈ R be a zero of f . Then

(i) the multiplicity of a is k ∈ N if (x − a)k is a factor of f and (x − a)k+1 is not a factor of f ;

(ii) a is a simple zero if it is a zero of multiplicity 1.

Theorem 201: A polynomial in R[x] with degree n has at most n zeros.

Proof: Induction. ■

Corollary 202: A polynomial in R[x] of degree n has at most n − 1 local extrema.

Proof: Let p(x) = an xn + an−1 xn−1 + . . . + a1 x + a0 ∈ R[x]. Define the derivative of p by

p′ (x) = nan xn−1 + (n − 1)xn−2 + . . . + 2a2 x + a1 ∈ R[x]. It is shown in Calculus that the zeros of

44
p′ represent the local extrema of p. Hence, p has at most n − 1 local extrema. ■

Theorem 203 (Intermediate Value Theorem for Polynomials): Let f (x) ∈ R[x] and suppose

there are a, b ∈ R with f (a) < 0 and f (b) > 0. Then there is a c ∈ R such that f (c) = 0.

Corollary 204: A polynomial in R[x] with odd degree has at least one root in R.

Proof: Let f (x) = xn + an−1 xn−1 + . . . a1 x + a0 ∈ R[x], where n is odd, and let

u = 1 + |an−1 | + . . . + |a1 | + |a0 |, and consider f (u) = un + an−1 un−1 + . . . + a1 u + a0 .

Then |an−1 un−1 + . . . + a1 u + a0 | ≤ |an−1 |un−1 + . . . + |a1 |u + |a0 |.

Since |ai | ≤ |an−1 | + . . . + |a1 | + |a0 | = u − 1,

we obtain that |an−1 un−1 + . . . + a1 u + a0 | ≤ (u − 1)(un−1 + . . . + u + 1) = un − 1 < un .

Therefore, f (u) ≥ un − |an−1 un−1 + . . . a1 u + a0 | > 0.

Since n is odd, (−u)n = −un and

f (−u) ≤ −un + |an−1 un−1 + . . . a1 u + a0 | < 0.

Thus, by the Intermediate Value Theorem for Polynomials, there is a c ∈ R such that f (c) = 0, as

desired. ■

Example 205: Show that the polynomial p(x) = x5 − 3x4 + 2x + 2 has a root between −1 and 0.

Proof: By the corollary above, p has a real root. To see that a real root is between −1 and 0, note

that p(−1) = −4 < 0 and p(0) = 2 > 0. So by the Intermediate Value Theorem for Polynomials,

there is a c ∈ (−1, 0) such that p(c) = 0. ■

Theorem 206: Suppose that f ∈ R[x] and a ∈ R is a root of f with multiplicity k.

(i) If k is odd, then the graph of f crosses the x-axis at x = a;

(ii) If k is even, then the graph of f touches but does not cross the x-axis at x = a;
p
Theorem 207 (Rational Roots Theorem): Let f (x) = a0 + a1 x + . . . + an xn ∈ Z[x]. If ∈Q
q
is a zero of f (x), where gcd(p, q) = 1, then p divides a0 and q divides an .

Proof: Note the following.

 
p
0=f
q
p pn
= a0 + a1 + . . . + an n
q q
n n−1
= a0 q + a1 pq + . . . + an pn

⇐⇒ −a0 q n = a1 pq n−1 + . . . + an pn .

This means that p divides −a0 q n , and so, p divides a0 .

Also, −an pn = a0 q n + a1 pq n−1 + . . . + an−1 pn−1 q, which means that q divides an pn , and so, q divides

an . ■

45
Example 208: Find all the roots of the polynomial p(x) = 2x3 + x2 − 13x + 6.

The factors of 6 are ±1, ±2, ±3, and ±6. By the Rational Roots Theorem, the potential roots are
1 2 3 6 1 2 3 6
± , ± , ± , ± , ± , ± , ± , and ± . Eliminating duplicates, the potential roots are ±1, ±2,
1 1 1 1 2 2 2 2
1 3
±3, ±6, ± , and ± . If x = 1, then p(1) = −4 ̸= 0. If x = −1, then p(−1) = 18 ̸= 0. If x = 2,
2 2
then p(2) = 0. So 2 is a root of p. Dividing p by x − 2, we obtain p(x) = (x − 2)(2x2 + 5x − 3) =
1
(x − 2)(2x − 1)(x + 3). Therefore, the roots of p are x = 2, x = , and x = −3.
2
Theorem 209 (Descartes’ Rule of Signs): Let p(x) ∈ R[x] and suppose that j is the total

multiplicities of the positive roots and k is the total multiplicities of the negative roots.

(i) If s is the number of sign changes of p(x), then s − j ≥ 0 and s − j is even.

(ii) If t is the number of sign changes of p(−x), then t − k ≥ 0 and t − k is even.

Example 210: Determine the number of possible positive and negative real zeros of the function

f (x) = x4 − 5x3 + 5x2 + 5x − 6.

Note that f (x) has three sign changes. So, by Descartes’ Rule of Signs, f (x) either has one or three

positive real zeros. Also, note that f (−x) = x4 + 5x3 + 5x2 − 5x − 6 has one sign change. So, by

Descartes’ Rule of Signs, f (x) has one negative real zero.

Theorem 211 (Upper and Lower Bound Rules): Suppose that p(x) ∈ R[x] with a positive

leading coefficient and p is divided by x − c for some c ∈ R using synthetic division.

(i) If c > 0 and each number of the last row is either positive or zero, then c is an upper bound for

the real zeros of p.

(ii) If c < 0 and the numbers of the last row are alternately positive and negative (with zero con-

sidered both positive and negative), then c is a lower bound for the real zeros of p.

Proof: Exercise. ■

Example 212: What is the smallest positive integer which is an upper bound and the largest neg-

ative integer which is a lower bound for the real zeros of the function f (x) = 2x3 + 7x2 − 4x − 14?

Performing synthetic division with c = 2,

2 7 −4 − 14

2 4 22 36

2 11 18 22
we obtain that c = 2 is the smallest positive integer upper bound by the upper and lower bound

rules theorem. Also, performing synthetic division with c = −4,

46
 2 7 −4 − 14

−4 −8 4 0

2 −1 0 − 14
we obtain that c = −4 is the largest negative integer lower bound by the upper and lower bound

rules theorem.

Definition 213: A field F is algebraically closed if every polynomial in F [x] has a root.

Theorem 214 (Fundamental Theorem of Algebra): The field C is algebraically closed.

Corollary 215 (Linear Factor Theorem): If p ∈ C[x] is of positive degree, then it can be ex-

pressed as a product of linear factors.

Proof: Induction. ■

Example 216: Find all zeros of the polynomial p(x) = x4 − 4x3 + 8x2 − 16x + 16 and factor.

By the linear factor theorem, p(x) has four zeros. By the rational root theorem, the potential zeros

of p are ±1, ±2, ±4, ±8, and ±16. If x = 1, then p(1) = 5 ̸= 0. If x = −1, then p(−1) = 45 ̸= 0. If

x = 2, then p(2) = 0. Applying synthetic division with divisor x − 2,

1 −4 8 − 16 16

2 2 −4 8 − 16

1 −2 4 −8 0
we obtain that p(x) = (x − 2)(x3 − 2x2 + 4x − 8). Set q(x) = x3 − 2x2 + 4x − 8. By the rational

roots theorem, the potential zeros of q are ±1, ±2, ±4, and ±8. By a similar argument to the one

above, we obtain that x = 2 is a root of q. Applying synthetic division with x − 2,

1 −2 4 −8

2 2 0 8

1 0 4 0
we obtain that q(x) = (x − 2)(x2 + 4), which means that p(x) = (x − 2)2 (x2 + 4). Thus, the four

zeros are given by x = 2 and x = ±2i and p(x) = (x − 2)2 (x + 2i)(x − 2i).

Theorem 217 (Conjugate Pair Theorem): Let p(x) ∈ R[x]. If z ∈ C is a root of p(x) with

multiplicity n for some n ∈ N, then z is a root of p(x) with multiplicity n.

Proof: Let p(x) = an xn + an−1 xn−1 + . . . + a2 x2 + a1 x + a0 ∈ C[x] and suppose z ∈ C is a root of

p(x) with multiplicity n. We prove the theorem with two claims.

Claim 1: The conjugate z is a root of p.

Proof (Claim 1): Consider the following.

47
p(z) = 0

⇐⇒ p(z) = 0 = 0.

Since p ∈ R[x],

p(z) = an z n + an−1 z n−1 + . . . + a2 z 2 + a1 z + a0

= an z n + an−1 z n−1 + . . . + a2 z 2 + a1 z + a0

= p(z).

Thus, since p(z) = 0 and p(z) = p(z), we obtain that z is a root of p. ■Claim 1

Claim 2: The conjugate z is a root with multiplicity n.

Proof (Claim 2): Consider the following.

p(x) = (x − z)n q(x)

⇐⇒ p(x) = (x − z)n q(x)

= (x − z)n · q(x)

= (x − z)n · q(x)

⇐⇒ p(x) = (x − z)n · q(x).

Therefore, z is a root with multiplicity n. ■Claim 2

The proof of claim 2 completes the proof of the theorem. ■

Corollary 218: A non-constant irreducible polynomial in R[x] has degree 1 or 2.

Proof: Let p ∈ R[x] with deg(p) = n > 0. By the Linear Factor Theorem, there are c1 , c2 , . . . , cn ∈ C

and a d ∈ R such that p(x) = d(x − c1 )(x − c2 ) . . . (x − cn ). If ci ∈ R for every i, 1 ≤ i ≤ n, then

there is nothing to show. So suppose there is a cj ∈ C for some j, 1 ≤ j ≤ n. Then there are a, b ∈ R

such that cj = a + bi. Since p ∈ R[x], the conjugate cj = a − bi is a root by the Conjugate Pair

Theorem. So,

(x − cj )(x − cj )

= [x − (a + bi)][x − (a − bi)]

= x2 − 2ax + (a2 + b2 ),

which means that p(x) = d(x − c1 )(x − c2 ) . . . (x − cn )[x2 − 2ax + (a2 + b2 )]. ■

Example 219: If 3 − 5i is a zero, find all the zeros of the polynomial p(x) = x5 − 10x4 + 65x3 −

184x2 + 274x − 204.

By a previous theorem and the linear factor theorem, the polynomial p has 5 zeros. By the conjugate

pair theorem, since 3−5i is a zero, 3+5i is a zero. Dividing p by (x−3+5i)(x−3−5i) = x2 −6x+34,

48
 x3 − 4x2 + 7x −6


x2 − 6x + 34 x5 − 10x4 + 65x3 − 184x2 + 274x − 204

− x5 + 6x4 − 34x3

− 4x4 + 31x3 − 184x2

4x4 − 24x3 + 136x2

7x3 − 48x2 + 274x

− 7x3 + 42x2 − 238x

− 6x2 + 36x − 204

6x2 − 36x + 204

0
we obtain p(x) = (x − 3 + 5i)(x − 3 − 5i)(x3 − 4x2 + 7x − 6). Set q(x) = x3 − 4x2 + 7x − 6. By

Descartes’ Rule of Signs, q has either one or three positive real zeros and no negative zeros. By the

rational root theorem, 1, 2, 3, and 6 are potential zeros. If x = 1, then q(1) = −2 ̸= 0. If x = 2, then

q(2) = 0. Applying synthetic division with divisor x − 2,

1 −4 7 −6

2 2 −4 6

1 −2 3 0
we obtain that q(x) = (x − 2)(x2 − 2x − 3), which means that p(x) = (x − 3 + 5i)(x − 3 − 5i)(x −
√
2)(x2 − 2x + 3). By the quadratic formula, the remaining two zeros are x = 1 ± i 2.

49
 Chapter 9: Rational Functions and Asymptotes

Section 9.1: Rational Functions

Definition 220:
p(x)
(i) A rational function is a function of the form r(x) = , where p(x), q(x) ∈ R[x] are relatively
q(x)
prime polynomials and q(x) ̸= 0 for every x ∈ R.
ax + b
(ii) A rational function of the form r(x) = , where a, b, c, d ∈ R, is called a linear fractional
cx + d
transformation.
p(x)
Note 221: The domain of a rational function r(x) = is the set dom(r) = {x ∈ R | q(x) ̸= 0}.
q(x)

Section 9.2: Infinite Limits and Vertical Asymptotes

Concept 222: Let f be a function and c ∈ R be not necessarily in the domain of f . Then

(i) lim f (x) = ∞ if f (x) can be made arbitrarily large by making x sufficiently close to c but not
x→c

equal to c.

(ii) lim f (x) = −∞ if f (x) can be made arbitrarily small by making x sufficiently close to c but
x→c

not equal to c.

(iii) lim− f (x) = ∞ if f (x) can be made arbitrarily large by making x sufficiently close to, and less
x→c
than, c.

(iv) lim f (x) = −∞ if f (x) can be made arbitrarily large by making x sufficiently close to, and
x→c−
less than, c.

(v) Then lim+ f (x) = ∞ if f (x) can be made arbitrarily large by making x sufficiently close to, and
x→c
greater than, c.

(vi) Then lim+ f (x) = −∞ if f (x) can be made arbitrarily small by making x sufficiently close to,
x→c
and greater than, c.

Example 223: Analyze each limit.

1
(a) lim 2
x→0 x
1 1 1
Note that lim 2 = ∞ = lim 2 . Thus, lim 2 = ∞.
x→0 x
− x→0 x+ x→0 x
1
(b) lim
x→5 (x − 5)3
1
Note that x − 5 > 0 as x > 5. So, lim+ = ∞. Likewise, x − 5 < 0 as x < 5. So,
x→5 (x − 5)3
1 1
lim− 3
= −∞. Thus, lim does not exist.
x→5 (x − 5) x→5 (x − 5)3
x+1
(c) lim
x→−2 (x − 3)(x + 2)

50
 x+1 x+1 x+1
Note that lim = ∞ and lim − = −∞. Thus, lim
x→−2+ (x − 3)(x + 2) x→−2 (x − 3)(x + 2) x→−2 (x − 3)(x + 2)
does not exist.

Definition 224: The line x = c is a vertical asymptote of the graph y = f (x) if at least one of

the following is true.

(i) lim f (x) = ∞; (iii) lim− f (x) = ∞; (v) lim+ f (x) = ∞;

x→c x→c x→c
(ii) lim f (x) = −∞; (iv) lim f (x) = −∞; (vi) lim f (x) = −∞.
x→c x→c− x→c+
p(x)
Theorem 225: Let r(x) = be a rational function.
q(x)
(i) If q(a) ̸= 0, then the vertical line x = a is not a vertical asymptote of r.

(ii) If q(a) = 0 and p(a) ̸= 0, then the vertical line x = a is a vertical asymptote of r.

(iii) If p(a) = 0 = q(a), then r may or may not be a vertical asymptote of r.

Example 226: For each of the following, find the vertical asymptotes, if any, and sketch a graph.
x2 + x + 1
(a) f (x) =
x
2
If x = 0, then x + x + 1 = 1. So, by the theorem above, there is a vertical asymptote at x = 0, and

a sketch of the graph is below.

x2 + 3x + 2
(b) f (x) =
x+1
x2 + 3x + 2 (x + 2)(x + 1)
Suppose that x = −1. Then x2 +3x+2 = 0 and x+1 = 0. So f (x) = = =
x+1 x+1
x+2
. By the theorem above, there are no vertical asymptotes. A sketch of the graph is below.
1

51
Section 9.3: Limits at Infinity and Horizontal Asymptotes
Concept 227: Let f be a function.

(i) The limit as x tends to infinity of f is L if f (x) can be made arbitrarily close to L by making x

sufficiently large.

(ii) The limit as x tends to negative infinity of f is L if f (x) can be made arbitrarily close to L by

making x sufficiently small.

Notation 228:

(i) If the limit as x tends to infinity of f is L, then we write lim f (x) = L.

x→∞

(ii) If the limit as x tends to negative infinity of f is L, then we write lim f (x) = L.
x→−∞

Example 229: Evaluate the following limits.

3x2 + 3x + 1
3x3 + 4x2 − 172 3x2 + x − 1 (c) lim
(a) lim (b) lim
x→−∞ x2 + 1
x→∞ x4 + 1 x→∞ x+1 Consider the following.
Consider the following. Consider the following. 3x2 + 3x + 1
3x3 + 4x2 − 172 3x2 + x − 1 lim
lim lim
x→−∞ x2+ 1 
x→∞ x4 + 1 x→∞ x+ 1 2 3 1
4 172

1
 x 3+ + 2
x3 3 + − 3 x 3x + 1 − x x
x x x = lim  
= lim = lim
x→−∞
2
1
x→∞

1

x→∞

1
 x 1+ 2
x3 x + 3 x 1+ x
x x 3 1
4 172 1 3+ + 2
3+ − 3 3x + 1 − = lim x x
= lim x x x x→−∞ 1
1 = lim 1+ 2
x→∞ x→∞ 1 x
x+ 3 1+ = 3.
x x
= 0. = ∞.

52
 p(x)
Theorem 230: Let r(x) = be a rational function, where p(x) = am xm + am−1 xm−1 + . . . +
q(x)
a2 x2 + a1 x + a0 and q(x) = bn xn + bn−1 xn−1 + . . . + b2 x2 + b1 x + b0 .

(i) If m < n, then lim r(x) = 0.

x→∞
am
(ii) If m = n, then lim r(x) = .
x→∞ b
n




 ∞ if am > 0 and bn > 0;




∞
 if am < 0 and bn < 0;
(iii) If m > n, then lim r(x) =
x→∞ 



 −∞ if am > 0 and bn < 0;




−∞
 if am < 0 and bn < 0.
Proof: Apply the Leading Coefficient Test. ■

Example 231: Evaluate each of the following.

3x2 + x + 1
(a) lim =∞
x→∞ 100x + 10
30x + 1
(b) lim =0
x→∞ x2
2x + 5x3 + 10x
4
2
(c) lim = .
x→−∞ 3x4 + 38x 3
Definition 232: The horizontal line y = L is a horizontal asymptote of the function y = f (x) if

at least one of the following is true.

(i) lim f (x) = L;

x→∞

(ii) lim f (x) = L.

x→−∞
p(x)
Theorem 233: Let r(x) = be a rational function, where p(x) = am xm + am−1 xm−1 + . . . +
q(x)
a2 x2 + a1 x + a0 and q(x) = bn xn + bn−1 xn−1 + . . . + b2 x2 + b1 x + b0 .

(i) If m < n, then the horizontal line y = 0 is a horizontal asymptote of r.

am
(ii) If m = n, then the horizontal line y = is a horizontal asymptote of r.
bn
(iii) If m > n, then the function r has no horizontal asymptotes.

Proof: Apply the theorem above and the definition of horizontal asymptote. ■

Example 234: For each of the following, find the horizontal and vertical asymptotes, if any, of the

graph of the given function. Then sketch a graph.

x2 − 1
(a) f (x) = 2
x − 3x + 2
Since deg(x2 − 1) = 2 = deg(x2 − 3x + 2), the line y = 1 is a horizontal asymptote of the graph of

f . Also, if x = 2, then x2 − 1 = 3 ̸= 0 and x2 − 3x + 2 = 0, and so, the line x = 2 is a vertical

asymptote of the graph of f . A sketch of the graph is below.

53
 4x2
(b) f (x) =
x2
+1
Since deg(4x2 ) = 2 = deg(x2 + 1), there is a horizontal asymptote at y = 4. Also, since x2 + 1 ̸= 0

for all x ∈ R, there are no vertical asymptotes on the graph of f . A sketch of the graph is below.

Note 235: A function can have at most two horizontal asymptotes. There is no limit to the number

of vertical asymptotes a function can have, however.

Section 9.4: Oblique Asymptotes

Definition 236: Let m, b ∈ R with m ̸= 0. Then the line y = mx + b is an oblique asymptote of

the graph of a function y = f (x) if either lim f (x) − (mx + b) = 0 or lim f (x) − (mx + b) = 0.
x→∞ x→−∞
p(x)
Theorem 237: Let r(x) = be a rational function with deg(p) = m and deg(q) = n. Then the
q(x)
graph of r has an oblique asymptote if and only if m = n + 1.

Corollary 238: If the graph of a rational function has an oblique asymptote, then it does not have

a horizontal asymptote.

54
 p(x)
Proof: Let r(x) = be a rational function. Since the graph of r has an oblique asymptote,
q(x)
deg(p) = deg(q) + 1, and so, deg(p) > deg(q). By a previous theorem, the graph of r does not have

a horizontal asymptote. ■

Note 239: If a rational function has an oblique asymptote, apply the division algorithm. The

quotient will be the oblique asymptote.

Example 240: For each of the following functions, find all asymptotes and sketch a graph.
x2 + x + 1
(a) f (x) =
x−1
Note that there are no horizontal asymptotes, the line x = 1 is a vertical asymptote, and f (x) =
3
x+2+ , which means that the line y = x + 2 is an oblique asymptote. A sketch of the graph
x−1
is below.

x3 + 6x2 + 8x
(b) g(x) =
x2 + 3x + 2
Note that there are no horizontal asymptotes, the line x = −1 is a vertical asymptote, and g(x) =
−3x + 6
x+3+ 2 , which means that the line y = x + 3 is an oblique asymptote. A sketch of the
x + 3x + 2
graph is below.

55
56
 Chapter 10: Conic Sections

Section 10.1: Definitions

Definition 241: A conic section is the intersection of a plane and a double-napped cone.

Definition 242: A conic section is

(i) degenerate if the intersecting plane passes through the vertex of the cone;

(ii) non-degenerate if the intersecting plane does not pass through the vertex of the cone.

Note 243: The graph of a degenerate conic is either a point, a line, or two intersecting lines.

Example 244: Pictures.

Section 10.2: Parabolas

Definition 245: A parabola is the set of points in R2 that are equidistant from a fixed point F

(called the focus) and a fixed line l (called the directrix).

Definition 246:

(i) The midpoint between the focus and the directrix of a parabola is called the vertex of the

parabola;

(ii) The line that passes through the vertex and the focus of a parabola is called the axis (of

symmetry) of the parabola.

Theorem 247: The graph of the equation (x − h)2 = 4p(y − k) with p ̸= 0 is a parabola with vertex

(h, k), a vertical axis, focus (h, k + p), and directrix y = k − p.

Proof: If a point (x, y) ∈ R2 lies on the parabola, then it is equidistant to the focus (h, k + p) and

the directrix y = k − p. So,

p
(x − h)2 + [y − (k + p)]2 = y − (k − p)

⇐⇒ (x − h)2 + [y − (k + p)]2 = [y − (k − p)]2

⇐⇒ (x − h)2 + y 2 − 2y(k + p) + (k + p)2 = y 2 − 2y(k − p) + (k − p)2

⇐⇒ (x − h)2 + y 2 − 2ky − 2py + k 2 + 2pk + p2 = y 2 − 2ky + 2py + k 2 − 2pk + p2

⇐⇒ (x − h)2 − 2py + 2pk = 2py − 2pk

⇐⇒ (x − h)2 = 4py − 4pk = 4p(y − k). ■

Corollary 248: The graph of the equation x2 = 4py is a parabola with vertex (0, 0), focus (0, p),

and directrix y = −p.

Theorem 249: The graph of the equation (y − k)2 = 4p(x − h) with p ̸= 0 is a parabola with vertex

(h, k), a horizontal axis, focus (h + p, k), and directrix x = h − p.

57
Proof: If a point (x, y) ∈ R2 lies on the parabola, then it is equidistant to the point (h + p, k) and

the directrix x = h − p. So,

p
[x − (h + p)]2 + (y − k)2 = x − (h − p)

⇐⇒ [x − (h + p)]2 + (y − k)2 = [x − (h − p)]2

⇐⇒ x2 − 2x(h + p) + (h + p)2 + (y − k)2 = x2 − 2x(h − p) + (h − p)2

⇐⇒ x2 − 2x(h + p) + h2 + 2ph + p2 + (y − k)2 = x2 − 2x(h − p) + h2 − 2hp + p2

⇐⇒ −2px + 2ph + (y − k)2 = 2px − 2ph

⇐⇒ (y − k)2 = 4px − 4ph = 4p(x − h). ■

Corollary 250: The graph of the equation y 2 = 4px is a parabola with vertex (0, 0), focus (p, 0),

and directrix x = −p.

Example 251: Determine the equation of the parabola with vertex (0, 0) and focus (2, 0).

Note that the axis is horizontal. So the equation of the parabola is y 2 = 4px with h = 0, k = 0, and

p = 2. In other words, the equation of the parabola is y 2 = 8x.

1 1
Example 252: Find the focus of the parabola y = − x2 − x + .
2 2
Consider the following.
1 1
y = − x2 − x +
2 2
2
⇐⇒ −2y = x + 2x − 1

⇐⇒ 1 − 2y = x2 + 2x = (x + 1)2 − 1

⇐⇒ 2 − 2y = (x + 1)2

⇐⇒ −2(y − 1) = (x + 1)2 .
 
1 1
Hence, h = −1, k = 1, and p = − . Therefore, the focus is the point (h, k + p) = −1, .
2 2

Section 10.3: Ellipses

Definition 253: An ellipse is the set of all points P such that the sum of the distances to two

fixed points F1 and F2 , called the foci is a constant K > 0. In other words, P F1 + P F2 = K.

Definition 254:

(i) The intersection of an ellipse and the line passing through its foci is called the set of vertices;

(ii) The chord joining the vertices of an ellipse is called the major axis;

(iii) The midpoint of the major axis of an ellipse is called the center;

(iv) The chord perpendicular to the major axis of an ellipse at the center is called the minor axis.
(x − h)2 (y − k)2
Theorem 255: The graph of the equation 2
+ = 1 is an ellipse centered at the
a b2 √
point (h, k) with vertices (h ± a, k), foci (h ± c, k) with c = a2 − b2 , a horizontal major axis of

length 2a, and a vertical minor axis of length 2b.

58
Example 256: Find the foci of the ellipse 16x2 + 9y 2 = 144.
x2 y2
Dividing by 144, we obtain + = 1. Since 16 > 9, this is an ellipse with its foci on the y-axis
9 16 √ √
and with a = 4 and b = 3. Hence, c = 7, and so the foci are the points (0, ± 7).
(x − h)2 (y − k)2
Theorem 257: The graph of the equation + = 1 is an ellipse centered at the
b2 √a
2

point (h, k) with vertices (h, k ± a), foci (h, k ± c) with c = a2 − b2 , a vertical major axis of length

2a, and a horizontal minor axis of length 2b.

Example 258: Find the center, vertices, and foci of the ellipse 4x2 + y 2 − 8x + 4y − 8 = 0.

Consider the following.

0 = 4x2 + y 2 − 8x + 4y − 8

= 4x2 − 8x + y 2 + 4y − 8

= 4(x − 1)2 + (y + 2)2 − 16

⇐⇒ 16 = 4(x − 1)2 + (y + 2)2

(x − 1)2 (y + 2)2
⇐⇒ + = 1.
4 16 √
√
Since 16 ¿ 4, the major axis is vertical, h = 1, k = −2, a = 4, b = 2, and c = 16 − 4 = 2 3.

Therefore, the center is the point (1, −2), the vertices are the points (1, −6) and (1, 2), and the foci
√ √
are the points (1, −2 − 2 3) and (1, −2 + 2 3).

Section 10.4: Hyperbolas

Definition 259: A hyperbola is the set of all points P in a plane, the difference of whose distances

from two fixed points F1 and F2 , called the foci, is a constant K > 0. In other words, P F1 − P F2 =

±K.

Definition 260:

(i) The intercepts of a hyperbola are called the vertices of the hyperbola.

(ii) The line segment containing the vertices of a hyperbola is called the transverse axis.

(iii) The line segment perpendicular to the transverse axis passing through the center of a hyperbola

is called the conjugate axis.

(x − h)2 (y − k)2
Theorem 261: The graph of the equation − = 1 is a hyperbola centered at the
a2 b2
point (h, k) with vertices (h ± a, k), a horizontal transverse axis of length 2a, a vertical conjugate
√
axis of length 2b, and foci (h ± c, k), where a, b > 0 and c = a2 + b2 .
(y − k)2 (x − h)2
Theorem 262: The graph of the equation − = 1 is a hyperbola centered at the
a2 b2
point (h, k) with vertices (h, k ± a), a vertical transverse axis of length 2a, a horizontal conjugate
√
axis of length 2b, and foci (h, k ± c), where a, b > 0 and c = a2 + b2 .

Example 263: Find the equation of the hyperbola with foci (−1, 2) and (5, 2) and vertices (0, 2)

59
and (4, 2).

By the Midpoint Formula, the center of the hyperbola is the point (2, 2). Furthermore, c = 5 − 2 = 3
√ √
and a = 4 − 2 = 2, and so b = c2 − a2 = 5. So the transverse axis is horizontal and the equation
(x − 2)2 (y − 2)2
of the hyperbola is − = 1.
4 5
(x − h)2 (y − k)2
Theorem 264: The asymptotes of the hyperbola − = 1 with horizontal transverse
a2 b2
b
axis are the lines y = k ± (x − h).
a
(y − k)2 (x − h)2
Theorem 265: The asymptotes of the hyperbola − = 1 with vertical transverse
a2 b2
a
axis are the lines y = k ± (x − h).
b
Example 266: Determine the foci and asymptotes of the hyperbola 4x2 − 3y 2 + 8x + 16 = 0.

Consider the following.

0 = 4x2 − 3y 2 + 8x + 16

= 4x2 + 8x − 3y 2 + 16

= 4(x + 1)2 − 3y 2 + 12

⇐⇒ −12 = 4(x + 1)2 − 3y 2

y2 (x + 1)2
⇐⇒ − = 1.
4 3 √ √ √
Hence, k = 0, a = 2, h = −1, b = 3, and c = a2 + b2 = 7. So, the asymptotes are the lines
2 √ √
y = ± √ (x + 1) and the foci are the points (−1, −2 − 7) and (−1, −2 + 7).
3

60
 Chapter 11: Exponential and Logarithmic Functions

Section 11.1: Exponential Functions

Definition 267: Let a ∈ R+ \ {1}. The function f : R → R+ defined by f (x) = ax is an exponen-

tial function.

Example 268: The sketch of the function f (x) = 2x is given below.

Note 269: Define f : R → R+ by f (x) = bx for some b ∈ R+ \ {1}.

(i) If b > 1, then f is strictly increasing;

(ii) If b ∈ (0, 1), then f is strictly decreasing.

(iii) The graph of f has a horizontal asymptote at y = 0.

(iv) f is invertible.

Example 270: Solve the following equations.

 2x
1
(b) 7x+1 =
(a) 23x+1 = 25 7
Consider the following.
Consider the following.  2x
1
7x+1 = = 7−2x
23x+1 = 25 7
⇐⇒ 7x+1 = 7−2x
⇐⇒ 3x + 1 = 5
⇐⇒ x + 1 = −2x
⇐⇒ 3x = 4
4 ⇐⇒ 3x = −1
⇐⇒ x = .
3 −1
⇐⇒ x = .
3

61
  n
1
Definition 271: The number e = lim 1+ = lim (1 + x)1/x is called Euler’s number.
n→∞ n x→0

Section 11.2: Logarithmic Functions

Definition 272: Let a ∈ R+ \ {1}. The function loga : R+ → R defined by loga x = y if and only

if x = ay is called a logarithmic function of base a.

Proposition 273:

(i) Logarithmic functions are invertible.

(ii) Exponential and logarithmic functions are inverses.

Proof: Exercise. ■

Example 274: Calculate the following.

(a) log2 (8) = 3;

(g) loga (a5 ) = 5;
(b) log5 (25) = 2;
1 (h) loga (ax ) = x;
(c) log9 (3) = ;
2 (i) loga (a) = 1;
(d) log3 (9) = 2;
 
1 (j) loga (1) = 0;
(e) log3 = −2;
9 (k) loga (0) is undefined.
(f ) loga (a2 ) = 2;

Notation 275:

(i) The function log10 is denoted log;

(ii) The function loge is denoted ln.

Theorem 276 (Properties of Logarithms): Let a ∈ R+ \ {1}.

(i) loga (ax ) = x for all x ∈ R; (iv) loga (xy) = loga (x) + loga (y) for all x, y > 0;

(ii) aloga (x) = x for all x > 0; (v) loga (xr ) = r loga (x) for all x > 0 and all r ∈ R;
 
x
(iii) loga (1) = 0; (vi) loga = loga (x) − loga (y) for all x, y > 0.
y

Proof:

For (iii), consider the following. For (iv), consider the following.

loga (1) = x xy = aloga x · aloga y

⇐⇒ ax = 1 = aloga x+loga y

⇐⇒ x = 0. ⇐⇒ loga (xy) = loga x + loga y.

62
For (v), consider the following. For (vi), consider the following.
x aloga x
xr = aloga x

r = log y = aloga x · a− loga y
y a a
= ar loga x = aloga x−loga y
 
x
⇐⇒ loga (xr ) = r loga x. ⇐⇒ loga = loga x − loga y. ■
y

Example 277: Simplify the following.

(a) log6 (9) + log6 (4) = log6 (36) = 2.

√
 
1 1 1
(b) ln √ = ln(1) − ln( e) = − ln(e) = − .
e √ 2 2
(c) 10 logb (b3 ) − 4 logb ( b) = 30 − 2 = 28.

Theorem 278 (Change of Base for Logarithms): Let a, b ∈ R+ \ {1}. Then for any x > 0,
logb (x)
loga (x) = .
logb (a)
Proof: Consider the following.
aloga (x) = x ⇐⇒ (loga x) (logb a) = logb x
logb x


⇐⇒ logb aloga (x) = logb x ⇐⇒ loga x = . ■
log a b

Example 279: Calculate each of the following.

log27 (27) 1 ∼
(a) log4 (27) = = = 2.377443751.
log27 (4) log27 (4)
log18 (18) 1 ∼
(b) log3 (18) = = = 2.630929754.
log18 (3) log18 (3)
Example 280: Calculate each of the following without a calculator.
3
(a) e3 ln(x) = eln(x )
= x3 . (b) ln(e1000 ) = 1000.

Example 281: Solve the following equations for x.

(b) 5x+1 = 12
(a) 4x = 6
Consider the following.
Consider the following.
5x+1 = 12
4x = 6
⇐⇒ log5 (5x+1 ) = log5 (12)
⇐⇒ log4 (4x ) = log4 (6)
⇐⇒ x + 1 = log5 (12)
⇐⇒ x = log4 (6).
⇐⇒ x = log5 (12) − 1.
(c) e2x = 8x
⇐⇒ 2x − x ln(8) = 0
Consider the following.
⇐⇒ x(2 − ln(8)) = 0
e2x = 8x
⇐⇒ x = 0.
⇐⇒ ln(e2x ) = ln(8x )

⇐⇒ 2x = x ln(8)
Example 282: The bacteria population in a bottle at time t (in hours) has size P (t) = 1000e0.35t .

After how many hours will there be 5000 bacteria in the bottle?

63
 ln(5) ∼
Solving 1000e0.35t = 5000 for t, we obtain 0.35t = ln(5), or t = = 4.6 hours.
0.35

Section 11.3: Exponential Growth and Decay

Definition 283:

(i) A quantity Q grows exponentially if Q = Q0 ekt , where Q0 is the quantity at time t = 0 and

k > 0 is the growth constant.

(ii) A quantity Q decays exponentially if Q = Q0 e−kt , where Q0 is the quantity at time t = 0

and k > 0 is the decay constant.

Example 284: A bacteria culture containing 200 bacteria is growing at a rate proportional to its

size. In an hour the culture will contain 300 bacteria.

(a) Express the number of bacteria in the culture as a function of time.

Consider the following. 3

⇐⇒ = ek
2
Q(t) = Q0 ekt
 
3
⇐⇒ k = ln
2
⇐⇒ 300 = 200ek  t
3
⇐⇒ Q(t) = 200eln(3/2)t = 200 · .
2
(b) How many bacteria will the culture contain in three hours?
 3
3
Q(3) = 200 · = 675.
2
Theorem 285: Suppose the half-life of a substance is λ. If Q is the amount present at time t and
 t/λ
1
Q0 is the initial amount present at time t = 0, then Q = Q0 · = Q0 e−[ln(2)t]/λ .
2
Corollary 286: Suppose the half-life of a substance is λ and k is a decay constant. Then kλ = ln(2).
 t/λ
1
Proof: By the theorem above and the definition of exponential decay, Q(t) = Q0 = Q0 e−kt .
2
So,
 t/λ
1
Q0 = Q0 e−kt t
2 ⇐⇒ − ln(2) = −kt
 t/λ λ
1 ⇐⇒ ln(2) = kλ. ■
⇐⇒ = e−kt
2

Example 287: Find the half-life of a radioactive substance that decays from 100 g to 40 g in three

hours.

Consider the following.

 3/λ  
 t/λ 2 1 3 1
1 ⇐⇒ = = ln
Q(t) = Q(0) · 5 2  λ 2
2 1
 3/λ 3 ln
1 2 3 ln(2)
⇐⇒ 40 = 100 · ⇐⇒ λ =   = hours.
2 2 ln(5) − ln(2)
ln
5

64
 Section 11.4: Logarithmic Scales

Subsection 11.4.1: The pH Scale

Definition 288: The pH of a solution is defined by pH= − log[H + ], where [H + ] is the concentration

of hydrogen ions measured in moles per liter (M ).

Definition 289: A solution is

(i) basic if pH > 7; (ii) acidic if pH < 7; (iii) neutral if pH = 7.

Note 290: If the pH of a solution increases by 1, then the concentration of hydrogen ions decreases

by a factor of 10.

Example 291: The hydrogen ion concentration of a sample of human blood was measured to be

[H + ] = 3.16 × 10−8 M . Find the pH value and classify the blood as acidic or basic.

Note that pH = − log[H + ] = − log(3.16 × 10−8 ) ∼

= 7.5. Since 7.5 > 7, the blood is basic.
Example 292: The most acidic rainfall ever measured occurred in Scotland in 1974. The pH of

the rain was 2.4. Find the hydrogen ion concentration of the rain.

To determine the hydrogen ion concentration, we solve the equation log[H + ] = −pH.

The equation implies [H + ] = 10−pH = 10−2.4 ∼

= 4 × 10−3 M .

Subsection 11.4.2: The Richter Scale

Definition 293: An earthquake is standard if its intensity is 1 micron.
 
I
Definition 294: The magnitude of an earthquake is defined by log , where I is the intensity
S
of the earthquake, which is measured by the amplitude of a seismograph reading taken 100 km from

the epicenter of the earthquake, and S is the intensity of a standard earthquake.

Note 295: The magnitude of a standard earthquake is zero.

Example 296: The 1906 earthquake in San Francisco had an estimated magnitude of 8.3 on the

Richter scale. In the same year, a powerful earthquake occurred on the Colombia-Ecuador border

that was four times as intense. What was the magnitude of the Colombia-Ecuador earthquake on

the Richter scale?

I
If I is the intensity of the San Francisco earthquake, then M = log = 8.3. The intensity of the
S
4I I
Columbia-Ecuador earthquake is 4I, so its magnitude is M = log = log 4 + log = log 4 + 8.3 ∼
=
S S
8.9.

Example 297: The 1989 Loma Prieta earthquake that shook San Francisco had a magnitude of 7.1

on the Richter scale. How many times more intense was the 1906 earthquake than the 1989 event?

65
If I1 and I2 are the intensities of the 1906 and the 1989 earthquakes, respectively, then we must find
I1
. Note the following.
I2
I1 I1 /S
log = log
I2 I2 /S
I1 I2
= log − log
S S
= 8.3 − 7.1 = 1.2.
I1
Therefore, = 101.2 ∼
= 16, and so, the 1906 earthquake was about 16 times stronger than the 1989
I2
earthquake.

Subsection 11.4.3: The Decibel Scale

I
Definition 298: The intensity level of a sound, measured in decibels (dB), is B = 10 log ,
I0
where I is the intensity of the sound and I0 ∼
= 10−12 watts per square meter is the threshold of

hearing.

Note 299: The intensity level of a sound with intensity the threshold of hearing is zero.

Example 300: Find the decibel intensity level of a jet engine during takeoff if the intensity was

measured at 100 W/m2 .

102
By the definition of intensity level, B = 10 log = 10 log 1014 = 10 · 14 = 140 dB.
10−12

Section 11.5: Logistic Functions

Definition 301: Let a, b, c, k ∈ R.
c
(i) A logistic growth function in x is a function of the form f (x) = or f (x) =
1 + a · bx
c
, where c is the growth constant.
1 + a · e−kx
(ii) If b > 1 or k < 0, the function in part (i) is called a logistic decay function.
1
(iii) The function f (x) = is called the logistic function.
1 + e−x
Proposition 302: Let f (x) denote a logistic growth function.

(i) lim f (x) = 0; (ii) lim f (x) = c, where c is the limit of growth.
x→−∞ x→∞

Proof: Exercise. ■

Section 11.6: Hyperbolic Functions

Definition 303: The hyperbolic functions are defined as follows.
ex − e−x 1 2
(i) sinh x = ; (iv) csch(x) = = x ;
2 sinh(x) e − e−x
ex + e−x 1 2
(ii) cosh x = ; (v) sech(x) = = x ;
2 cosh(x) e + e−x
x −x −x
sinh(x) e −e 1 x
e +e
(iii) tanh x = = x ; (vi) coth(x) = = x .
cosh(x) e + e−x tanh(x) e − e−x

66
Proposition 304:

(i) The function sinh(x) is odd.

(ii) The function cosh(x) is even.

(iii) The function f (x) = 1 + tanh(x) is a logistic function.

Proof: Exercise. ■

Proposition 305:

(i) For any x ∈ R, sinh(x) + cosh(x) = ex .

(ii) For any x ∈ R cosh(x) − sinh(x) = e−x .

Proof: Exercise. ■

Corollary 306: cosh2 (x) − sinh2 (x) = 1 for all x ∈ R.

Proof: By the proposition above, cosh(x) + sinh(x) = ex and cosh(x) − sinh(x) = e−x . Hence,

[cosh(x) + sinh(x)][cosh(x) − sinh(x)] = ex · e−x

⇐⇒ cosh2 (x) − sinh2 (x) = 1,

as desired. ■

Theorem 307: Let x, y ∈ R.

(i) sinh(x + y) = sinh(x) cosh(y) + cosh(x) sinh(y);

(ii) cosh(x + y) = cosh(x) cosh(y) + sinh(x) sinh(y);

tanh(x) + tanh(y)
(iii) tanh(x + y) = .
1 + tanh(x) tanh(y)
Proof: Exercise. ■

Corollary 308: Let x ∈ R.

(i) sinh(2x) = 2 sinh(x) cosh(x);

(ii) cosh(2x) = cosh2 (x) + sinh2 (x).

Proof: Set x = x and y = x and apply the theorem above. ■

Theorem 309: The functions below are bijections, and hence invertible.

(i) f : R → R defined by f (y) = x if and only if sinh(y) = x;

(ii) g : [1, ∞) → [0, ∞) defined by g(y) = x if and only if cosh(y) = x;

(iii) h : (−1, 1) → R defined by h(y) = x if and only if tanh(y) = x.

Theorem 310:
√
(i) For each x ∈ R, sinh−1 (x) = ln(x + x2 + 1);
√
(ii) For each x ∈ [1, ∞), cosh−1 (x) = ln(x + x2 + 1);
 
1 1+x
(iii) For each x ∈ (−1, 1), tanh−1 (x) = ln .
2 1−x
Proof: We prove (i) and leave the remaining proofs as exercises.

(i) Set y = sinh−1 (x). Then

67
 ey − e−y
x = sinh(y) =
2
⇐⇒ e2y − 2x − e−y = 0

⇐⇒ e2y − 2xey − 1 = 0.

Set u = ey . Then

u2 − 2xu − 1 = √
0
2x ± 4x2 + 4 √
⇐⇒ ey = = x ± x2 + 1.
√ 2 √ √
Since x < x2 + 1 and ey > 0 for all y ∈ R, ey = x + x2 + 1, and so y = ln(x + x2 + 1). ■

68
 Chapter 12: Sequences and Sigma Notation

Section 12.1: Sequences

Definition 311: A sequence in R is a function ϕ : N → R.

Notation 312:

(i) We usually write xn instead of ϕ(n) = xn .

(ii) A sequence is usually denoted {xn }∞

n=1 instead of ϕ : N → R.

Example 313: {1, −2, 3, −4, 5, . . .} = {(−1)n+1 · n}∞

n=1 .

Example 314 (Fibonacci Sequence): {1, 1, 2, 3, 5, 8, 13, 21, 34, 55, . . .}

Set x1 = 1 and x2 = 1. For n ≥ 2, define xn+1 = xn + xn−1 .

Example 315: {1, 2, 4, 8, 16, . . .} − {2n }∞

n=0

Section 12.2: Sigma Notation

n
X
Notation 316: If a1 , a2 , . . . , an ∈ R, then ai = a1 + a2 + . . . + an .
i=1
Example 317: Calculate each of the following.
4
X
(a) 5i = 5 + 10 + 15 + 20 = 50
i=1
X6
(b) i2 = 9 + 16 + 25 + 36 = 86
i=3
5
X
(c) 2n = 2 + 4 + 8 + 16 + 32 = 62
i=1
Note 318: The following properties follow from the usual properties of arithmetic.
n
X n
X n
X
(i) ai + bi = ai + bi .
i=1 i=1 i=1
Xn Xn Xn
(ii) ai − bi = ai − bi .
i=1 i=1 i=1
Xn n
X
(iii) cai = c ai .
i=1 i=1
Xn
(iv) c = nc for every c ∈ R.
i=1
Theorem 319: For any n ∈ N,

n 2  2
X n(n + 1) X n(n + 1)
(i) i= ; (iii) i3 = ;
i=1
2 2
i=1
2 2
X n(n + 1)(2n + 1) X n(n + 1)(6n3 + 9n2 + n − 1)
(ii) i2 = ; (iv) i4 = .
i=1
6 30
i=1

69
Example 320: Calculate each of the following.

20
X
(a) 4i + 2 20
X
i=1
20 20 (b) 2i2 + 3i − 1
X X
i=1
= 4i + 2 20 20
X X
i=1 i=1 2
20 =2 i +3 i − 20
X
i=1 i=1
=4 i + 40
20(21)(41) 20(21)
i=1 =2· +3· − 20
20(21) 6 2
=4· + 40 = 20(7)(41) + 3(10)(21) − 20
2
= 2(20)(21) + 40 = 140(41) + 30(21) − 20
= 40(21) + 40 = 6740 + 610
= 22(40) = 7350
= 880
10
X
50
X
3 2
(d) 2i4 + i2
(c) 4i + i + 3i + 1 i=1
i=1 10
X 10
X
50
X
3
50
X
2
50
X 50
X =2 i4 + i2
= 4i + i + 3i + 1 i=1 i=1
i=1 i=1 i=1 i=1 10(11)(6000 + 900 + 9) 10(11)(21)
50 50 =2· +
X 50(51)(101) X 30 6
=4 i3 + +3 i + 50 20(11)(6909) + 50(11)(21)
i=1
6 i=1 =
30
= 4 · [25(51)]2 + 25(17)(101) + 75(51) + 50 1519980 + 11550
=
30
= 6502500 + 42925 + 3875 1531530
=
= 6549300 30
= 51051
10
X
(e) i(3i − 1)
i=1
10
X
= 3i2 − i
i=1 100
10 10 1
X 1
=
X
2
3i −
X
i (f ) −
i=1
k+2 k+4
i=1 i=1
1 1 1 1
10(11)(21) 10(11) = + − −
=3· − 3 4 103 104
6 2
= 5(11)(21) − 5(11)

= 20(55)

= 1100

Example 321: Write each of the following in sigma notation.

21
X
(a) 5 + 10 + 15 + . . . + 105 = 5i
i=1
14
X
(b) 5 + 9 + 13 + 17 + . . . + 57 = 4i + 1
i=1

70
 Chapter 13: The Mathematics of Finance

Section 13.1: Interest

Definition 322: Simple interest is calculated by the formula I = P rt, where

(i) I is the interest (measured in dollars);

(ii) P is the principal amount (the amount before interest is applied; measured in dollars);

(iii) r is the (nominal annual) interest rate (expressed as a decimal);

(iv) t is the time (measured in years).

Example 323: Suppose that $12,000 is invested at 6% interest for four years. How much money

will be in the account if interest is simple?

12000 + 12000 · 0.06 · 4 = 14880.

 r nt
Definition 324: Compound interest is calculated by the formula A = P 1 + , where
n
(i) A is the total amount after interest is applied;

(ii) P is the principal amount (measured in dollars);

(iii) r is the nominal annual interest rate (expressed as a decimal);

(iv) n is the number of times per year the interest is applied;

(v) t is the time (measured in years).

Example 325: Suppose that $12,000 is invested at 6% interest for four years. How much money

will be in the account if interest is compounded as indicated?

(a) Quarterly?
 4·4
0.06 ∼
12000 1 + = 15227.83.
4
(b) Annually?
 1·4
0.06 ∼
12000 1 + = 15149.72.
1
(c) Daily?
 365·4
0.06 ∼
12000 1 + = 15254.69.
365
Question 326: Suppose that $1 is invested at 100% interest for one year. Can the investment grow

to any amount if the interest is compounded often enough?

 1·k
1
A=1· 1+
k
 k
1
= 1+ .
k
Make k larger and larger.
 k
1
lim 1 + = e.
k→∞ k

71
Hence, the answer to the question is no.

Definition 327: Continuously compounded interest is given by the formula A = P ert , where

(i) A is the total amount after t years;

(ii) P is the principal amount (measured in dollars);

(iii) r is the nominal annual interest rate (expressed as a decimal);

(iv) t is the time (measured in years).

Example 328: Suppose that $12,000 is invested at 6% interest for four years. How much money

will be in the account if interest is compounded continuously?

12e0.06·4 ∼
= 12454.99
 r k
Example 329: The effective interest rate for compounded interest is rE = 1 + − 1,
k
where

(i) r is the nominal annual interest rate (expressed as a decimal);

(ii) k is the number of times per year that the interest is applied.

Definition 330: The effective interest rate for continuously compounded interest is given

by rE = er − 1, where r is the nominal annual interest rate expressed as a decimal.

Example 331: For each of the following, calculate the effective interest rate.

(a) 5% compounded semiannually.

 2
0.05
rE = 1 + − 1 = 0.050625.
2
(b) 5% compounded continuously.

rE = e0.05 − 1 ∼
= 0.0512710964.
 r −tk
Definition 332: The present value for compounded interest is given by P = A 1 + ,
k
where

(i) P is the present value (measured in dollars);

(ii) A is the future value (measured in dollars);

(iii) r is the nominal annual interest rate (measured as a decimal);

(iv) t is the time (measured in years);

(v) k is the number of times per year that interest is compounded.

Definition 333: The present value for continuous interest is given by P = Ae−rt , where

(i) P is the present value (measured in dollars);

(ii) A is the future value (measured in dollars);

(iii) r is the nominal annual interest rate (measured as a decimal);

(iv) t is the time (measured in years).

Example 334: What amount of money should be invested in an account that pays 6% interest

72
compounded quarterly in order to have $10000 in five years?

Consider the following.

 4·5
0.06
10000 = P · 1 + = P · (1.015)20
4
10000
⇐⇒ P = = 7424.704182.
(1.015)20
Hence, $7424.70 should be invested in the account.

Section 13.2: Annuities

Definition 335:

(i) An annuity is a sum of money that is paid in regular equal payments.

(ii) The amount of an annuity is the sum of all the individual payments from the time of the first

payment until the last payment is made, together with all the interest.

Theorem 336: The amount of an annuity consisting of n equal payments of size R with interest
(1 + i)n − 1
rate i per time period is given by Af = R · .
i
Example 337: How much money should be invested every month at 12% interest per year, com-

pounded monthly, in order to have $4000 in 18 months?

0.12
Set i = = 0.01, Af = 4000, and n = 18. By the annuity formula,
12
(1 + 0.01)18 − 1
4000 = R ·
0.01 
0.01
⇐⇒ R = 4000 ·
(1 + 0.01)18 − 1
∼
= 203.928.
Therefore, the monthly investment should be $203.93.

Definition 338: The present value of an annuity consisting of n regular equal payments of size
1 − (1 + i)−n
R and interest rate i per time period is given by Ap = R · .
i
Example 339: A person wins $10,000,000 in the California lottery, and the amount is paid in yearly

installments of half a million dollars each for 20 years. What is the present value of their winnings?

Assume that they can earn 10% interest, compounded annually.

Since the amount won is paid as an annuity, we need to find its present value. Set i = 0.1,

R = $500, 000, and n = 20. Then

1 − (1 + 0.1)−20 ∼
Ap = 500000 · = 4256781.859.
0.1
Therefore, the winner really won only $4,256,781.86 if it were paid immediately.

Section 13.3: Installment Buying

Theorem 340: If a loan is to be repaid in n equal payments with interest rate i per time period,
iAp
then the size R of each payment is given by R = .
1 − (1 + i)−n

73
Example 341: A couple borrows $100,000 at 9% interest as a mortage loan on a house. They

expect to make monthly payments for 30 years to repay the loan. What is the size of each payment?

The mortgage payments form an annuity whose present value is $100,000. Set i = 0.09/12 = 0.0075,

and n = 12 · 30 = 360. By the theorem above,

(0.0075)(100, 000) ∼
R= = 804.623.
1 − (1 + 0.0075)−360
Thus the monthly payments are $804.62.

Chapter 14: Induction and the Binomial Theorem

Section 14.1: Induction

Note 342: Let P (n) denote a statement about n ∈ N.

(i) (Basis Step) Show that P (1) is true.

(ii) (Induction Hypothesis) Suppose that P (k) is true and use P (k) to show that P (k + 1) is

true.
n
X n(n + 1)
Example 343: Show that i= for every n ∈ N.
i=1
2
n
X n(n + 1)
Proof: We will proceed by induction on n. Let P (n) denote the statement “ i = .”
i=1
2
1
X 1+1
Since i=1= , P (1) is true. Now suppose that k ∈ N such that P (k) is true. To see that
i=1
2
P (k + 1) is true, consider the following.
k+1
X
i
i=1
k
X
=k+1+ i
i=1
k(k + 1)
=k+1+
2
2(k + 1) k(k + 1)
= +
2 2
2(k + 1) + k(k + 1)
=
2
(k + 2)(k + 1)
= .
2
Therefore, P (k+1) is true, and so, the example is proved by the Principle of Mathematical Induction.

Definition 344: Let n ∈ N. Then the factorial of n is n! = n(n − 1)(n − 2) . . . (3)(2)(1).

74
Note 345: It is possible to define a factorial for any real number.

Example 346: Prove that for any n ∈ N with n > 4, 2n < n!.

Proof: We will proceed by induction on n. For each n ∈ N, let P (n) be the statement “2n < n!.”

Note that P (4) is true since 24 = 16 < 24 = 4!. Now suppose that P (k) is true for some k ∈ N. To

see that P (k + 1) is true, consider the following.

2k < k!

⇐⇒ 2 · 2k < (k + 1)k!

⇐⇒ 2k+1 < (k + 1)!.

Therefore, P (k+1) is true, and so, the example is proved by the Principle of Mathematical Induction.

Section 14.2: The Binomial Theorem

n
X n!
Theorem 347 (Binomial Theorem): For every a, b ∈ R and every n ∈ N, (a+b)n = an−k bk .
k!(n − k)!
k=0
n
X n!
Proof: For each n ∈ N, let P (n) denote the statement “(a + b)n = an−k bk .” Note
k!(n − k)!
k=0
that

(a + b)1

=a+b
1! 1−0 0 1! 0 1
= a b + a b
0!1! 1!0!
1
X 1!
= a1−k bk .
k!(1 − k)!
k=0
Hence, P (1) is true. So suppose P (ℓ) is true for some ℓ ∈ N. To see that P (ℓ + 1) is true, consider

the following.

(a + b)ℓ+1

= (a + b)(a + b)ℓ
ℓ
X ℓ!
= (a + b) aℓ−k bk
k![(ℓ − k)!]
k=0
ℓ ℓ
X ℓ! X ℓ!
=a aℓ−k bk + b aℓ−k bk
k!(ℓ − k)! k!(ℓ − k)!
k=0 k=0
ℓ ℓ
X ℓ! X ℓ!
= aℓ+1−k bk + aℓ−k bk+1
k!(ℓ − k)! k!(ℓ − k)!
k=0 k=0
ℓ ℓ+1
X ℓ! X ℓ!
= aℓ+1−k bk + aℓ−(k−1) bk
k!(ℓ − k)! (k − 1)![ℓ − (k − 1)]!
k=0 k=1
ℓ ℓ
X ℓ! X ℓ!
= aℓ+1 + aℓ+1−k bk + aℓ−(k−1) bk + bℓ+1
k!(ℓ − k)! (k − 1)![ℓ − (k − 1)]!
k=1 k=1

75
 ℓ ℓ
X ℓ! X ℓ!
= aℓ+1 + aℓ+1−k bk + aℓ−(k−1) bk + bℓ+1
k!(ℓ − k)! (k − 1)!(ℓ − k + 1)!
k=1 k=1
ℓ  
X ℓ! ℓ!
= aℓ+1 + aℓ+1−k bk + aℓ−(k−1) bk + bℓ+1
k!(ℓ − k)! (k − 1)!(ℓ − k + 1)!
k=1
ℓ  
X ℓ! ℓ!
= aℓ+1 + aℓ+1−k bk + + bℓ+1
k!(ℓ − k)! (k − 1)!(ℓ − k + 1)!
k=1
ℓ  
X ℓ!(ℓ − k + 1) kℓ!
= aℓ+1 + aℓ+1−k bk + + bℓ+1
k!(ℓ − k)!(ℓ − k + 1) k(k − 1)!(ℓ − k + 1)!
k=1
ℓ  
X ℓ!(ℓ − k + 1) kℓ!
= aℓ+1 + aℓ+1−k bk + + bℓ+1
k!(ℓ − k + 1)! k!(ℓ − k + 1)!
k=1
ℓ  
X ℓ!(ℓ + 1 − k) + kℓ!
= aℓ+1 + aℓ+1−k bk + bℓ+1
k!(ℓ + 1 − k)!
k=1
ℓ  
X ℓ!(ℓ + 1) − kℓ! + kℓ!
= aℓ+1 + aℓ+1−k bk + bℓ+1
k!(ℓ + 1 − k)!
k=1
ℓ  
ℓ+1
X
ℓ+1−k k ℓ!(ℓ + 1)
=a + a b + bℓ+1
k!(ℓ + 1 − k)!
k=1
ℓ  
ℓ+1
X
ℓ+1−k k (ℓ + 1)!
=a + a b + bℓ+1
k!(ℓ + 1 − k)!
k=1
ℓ  
(ℓ + 1)! ℓ+1−0 0
X
ℓ+1−k k (ℓ + 1)! (ℓ + 1)!
= a b + a b aℓ+1−k bk + aℓ+1−(ℓ+1) bℓ+1
0!(ℓ + 1 − 0)! k!(ℓ + 1 − k)! (ℓ + 1)![ℓ + 1 − (ℓ + 1)]!
k=1
ℓ+1
X (ℓ + 1)!
= aℓ+1−k bk .
k!(ℓ + 1 − k)!
k=0
Therefore, P (ℓ+1) is true, and so, the theorem is proved by the Principle of Mathematical Induction.

76