PRECALCULUS - Kkuniyuk.com

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Algebra: Blitzer, Lial, Tussy and Gustafson. Trigonometry: Lial, Smith. Precalculus : Axler, Cohen, Larson, Stewart, Sullivan; REA Problem Solvers. Calculus: ...
PRECALCULUS KEN KUNIYUKI SAN DIEGO MESA COLLEGE

TABLE OF CONTENTS Chapter 0: Preliminary Topics Chapter 1: Functions Chapter 2: Polynomial and Rational Functions Chapter 3: Exponential and Logarithmic Functions Chapter 4: Trigonometric Functions Chapter 5: Analytic Trigonometry Chapter 6: Topics in Trigonometry Chapter 7: Systems Chapter 8: Matrices and Determinants Chapter 9: Discrete Mathematics Chapter 10: Conic Sections, Polar Coordinates, and Plane Curves COLOR CODING Warnings are in red. Tips are in purple. Calculus comments are (sometimes) in green (clover). TECHNOLOGY USED This work was produced on Macs with Microsoft Word, MathType, Adobe Illustrator, Adobe Acrobat, and Mathematica and Calculus WIZ. CONTACT INFORMATION Ken Kuniyuki: Email address: [email protected] or [email protected] Website: http://www.kkuniyuk.com • You may download these and other course notes, exercises, and exams. Feel free to send emails with suggestions, improvements, tricks, etc. LICENSING This work may be freely copied and distributed without permission under the specifications of the Creative Commons License at: http://www.kkuniyuk.com/Math141 PARTIAL BIBLIOGRAPHY / SOURCES Algebra: Blitzer, Lial, Tussy and Gustafson Trigonometry: Lial, Smith Precalculus: Axler, Cohen, Larson, Stewart, Sullivan; REA Problem Solvers Calculus: Hamming, Larson, Stewart, Swokowski, Tan, [Peter D.] Taylor Complex Variables: Churchill and Brown, Schaum’s Outlines Discrete Mathematics: Rosen Online: Britannica Online Encyclopedia: http://www.britannica.com Wikipedia: http://www.wikipedia.org Wolfram MathWorld: http://mathworld.wolfram.com/ Other: Harper Collins Dictionary of Mathematics People: Larry Foster, Laleh Howard, Terrie Teegarden, Tom Teegarden (especially for the Frame Method for graphing trigonometric functions), and many more.

(Assumptions and Notation) A.1

ASSUMPTIONS and NOTATION Unless otherwise specified, we assume that: • f , g, and s denote functions. (See Section 1.1.) •• g sometimes denotes Earth’s gravitational constant. •• h may denote a function, or it may denote the “run” in some difference quotients in Sections 1.10 and 1.11. •• s often denotes a position or height function. • a, b, c, k, m, and n denote real numbers. (See Section 0.1.) •• “Let a ∈ ” means “let a be an arbitrary real number.” By “arbitrary,” we mean that any one will do. •• c sometimes denotes the speed of light in a vacuum. •• d sometimes denotes a distance function. •• e denotes a mathematical constant defined in Chapter 3. e ≈ 2.718 .

(

)

•• i sometimes denotes the imaginary unit defined in Chapter 2 i = −1 . In Chapter 9, we will use i as a generic subscript (as in ai ) and as an index n ⎛ ⎞ of summation ⎜ as in ∑ ai ⎟ ; we also use j and k for these purposes. ⎝ ⎠ i =1

•• n might be restricted to be an integer ( n ∈ ) . (See Section 0.1.) • x, y, t, and θ typically denote variables that take on real numbers as values. •• z can denote a variable or a complex number (see Chapter 2). • The domain of a function f , which we will denote by Dom ( f ) , is the implied (or mathematical) domain of f . (See Section 1.1.) •• This might not be the case in applied “word problems.” •• We assume that the domain and the range of a function only consist of real numbers, as opposed to imaginary numbers. That is, Dom ( f ) ⊆  , and Range ( f ) ⊆  . ( ⊆ means “is a subset of”; see Section 0.1.) • Graphs extend beyond the scope of a figure as expected, unless endpoints are clearly shown. Arrowheads help to make this clearer. (See Section 1.2.)

(Assumptions and Notation) A.2

MORE NOTATION Sets of Numbers (Section 0.1) Notation + , Z +

, Z

, Q

, R , C

Meaning

Comments This is the set (collection) {1, 2, 3, ...} .

“Zahlen” is a related German word.  is in blackboard bold typeface; it is more commonly used than Z. This set consists of the positive integers, the the set of integers negative integers ( −1, − 2, − 3, …), and 0. This set includes the integers and numbers 1 9 the set of rational numbers such as , − , 7.13, and 14.3587 . 3 4  comes from “Quotient.” This set includes the rational numbers and irrational numbers such as 2 , π , e, and the set of real numbers 0.1010010001…. Think: “all decimal numbers.” This set includes the real numbers and the set of complex numbers imaginary numbers such as i and 2 + 3i . the set of positive integers

The Venn diagram below indicates the (proper) subset relations:  ⊂  ⊂  ⊂  . For example, every integer is a rational number, so  ⊂  . ( ⊆ permits equality.) Each disk is contained within each larger disk.

(Assumptions and Notation) A.3

Set Notation (Section 0.1) Notation

Meaning



in, is in

∉ ∍

∀ ∃

not in, is not in such that such that (in set-builder form) for all, for any there is, there exists

∃!

there exists a unique, there is one and only one

∀x ∈

for every real number (denoted by x)

∀x, y ∈

for every pair of real numbers (denoted by x and y)

| or :

∅ or {}

empty set (or null set)



set union



set intersection

\ or −

set difference, set complement

Comments This denotes set membership. Example: 7 ∈ . Example: 1.7 ∉ . Example: { x ∈ x > 3} , or { x ∈: x > 3} , is the set of all real numbers greater than 3. This is called the universal quantifier. This is called the existential quantifier. This is called the unique quantifier. Example: ∃!x ∈ ∍ x = 3 , which states that there exists a unique real number equal to 3. More precisely: for any arbitrary member of the set of real numbers; this member will be denoted by x. Example: ∀x ∈, x < x + 1 ; that is, every real number is less than one added to itself. More precise notation: ∀ ( x, y ) ∈ 2 . This is the set consisting of no members. Example: The solution set of the equation x = x + 1 is ∅ . The symbol ∅ is not to be confused with the Greek letter phi ( φ ). Example: If f ( x ) = csc x , then

Dom ( f ) = ( − ∞, −1] ∪ [1, ∞ ) .

∪ is used to indicate that one or more real number(s) is/are being skipped over. Think: “all members are invited.” Example: [ 4, 6 ] ∩ [ 5, 7 ] = [ 5, 6 ] . Think: “overlap.”

1 , then Dom ( f ) is x  \ {0} , or  − {0} . We exclude 0 from  .

Example: If f ( x ) =

(Assumptions and Notation) A.4

Logical Operators (Sections 0.1, 0.2) Notation

Meaning



or, disjunction

Comments Example: If f ( x ) = csc x , then

Dom ( f ) = { x ∈ x ≤ −1 ∨ x ≥ 1} .



and, conjunction

 or ¬

not, negation

⇒ ⇔

implies if and only if (iff)

x−3 , then x−4 Dom ( f ) = { x ∈ x ≥ 3 ∧ x ≠ 4 } .

Example: If f ( x ) =

Example: The statement  ( x = 3) is equivalent to the statement x ≠ 3 . Example: x = 2 ⇒ x 2 = 4 . Example: x + 1 = 3 ⇔ x = 2 .

Greek Letters The lowercase Greek letters below (especially θ ) often denote angle measures. Notation α

Name alpha

β

beta

γ

gamma

θ

theta

φ or ϕ

phi

Comments This is the first letter of the Greek alphabet. This is the second letter of the Greek alphabet. This is the third letter of the Greek alphabet. This is frequently used to denote angle measures. This is not to be confused with ∅ , which denotes the empty set (or null set).

1+ 5 , 2 which is about 1.618. Tau (τ ) is also used.

φ also denotes the golden ratio,

The lowercase Greek letters below often denote (perhaps infinitesimally) small positive quantities in calculus, particularly when defining limits. Notation

Name

δ

delta

ε

epsilon

Comments This is the fourth letter of the Greek alphabet. This is the fifth letter of the Greek alphabet. This is not be confused with ∈, which denotes set membership.

(Assumptions and Notation) A.5 Some other Greek letters of interest: Notation

Name

Δ

(uppercase) delta

κ

(lowercase) kappa

λ

(lowercase) lambda

π

(lowercase) pi

Π

(uppercase) pi

ρ

(lowercase) rho

Σ

(uppercase) sigma

τ

(lowercase) tau

ω

(lowercase) omega

Ω

(uppercase) omega

Comments This denotes “change in” or increment. Δy Example: slope is often written as . Δx It also denotes the discriminant, b 2 − 4ac , from the Quadratic Formula. This denotes the curvature of a curve. This denotes an eigenvalue (in linear algebra), a Lagrange multiplier (in multivariable optimization), and a wavelength (in physics). This is a famous mathematical constant. It is the ratio of a circle’s circumference to its diameter. π ≈ 3.14159 . It is irrational. This is the product operator. This denotes mass density and also the distance between a point in 3-space and the origin ( ρ is a spherical coordinate). This is the summation operator. See Chapter 9. This denotes the golden ratio, though phi (φ ) is more commonly used. This is the last letter of the Greek alphabet. It denotes angular velocity. This denotes ohm, a unit of electrical resistance.

More lowercase Greek letters: zeta (ζ ) , eta (η ) , iota (ι ) , mu ( µ ) , nu (ν ) , xi (ξ ) , omicron (ο ) , sigma (σ ) , upsilon (υ ) , chi ( χ ) , psi (ψ )

(Assumptions and Notation) A.6

Geometry Notation ∠ 



Meaning angle is parallel to is perpendicular to, is orthogonal to, is normal to

Comments See Section 4.1. See Section 0.14 and Chapter 6. See Section 0.14 and Chapter 6.

Vector Operators Notation

• ×

Meaning dot product, Euclidean inner product cross product, vector product

Comments See Chapter 6. See Chapter 8.

Other Notations Notation

Meaning



therefore

Q.E.D., or 

end of proof

≈, ≅

is approximately, is about

⎢⎣ ⎥⎦ or



floor, greatest integer

∞ min max Dom ( f ) deg ( f ( x ))

infinity minimum maximum domain of a function f degree of a polynomial f ( x)



composition of functions

Comments This is placed before a concluding statement. Q.E.D. stands for “quod erat demonstrandum,” which is Latin for “which was to be demonstrated / proven / shown.” Think: “round down.” Examples: ⎢⎣ 2.9 ⎥⎦ = 2 , ⎢⎣ − 2.9 ⎥⎦ = − 3 See Section 0.1. The least of … The greatest of … The set of legal (real) input values for f See Section 0.6. Example: ( f  g ) ( x ) = f ( g ( x )) . See Section 1.6.

CHAPTER 0: Preliminary Topics 0.1: Sets of Numbers 0.2: Logic 0.3: Rounding 0.4: Absolute Value and Distance 0.5: Exponents and Radicals: Laws and Forms 0.6: Polynomial, Rational, and Algebraic Expressions 0.7: Factoring Polynomials 0.8: Factoring Rational and Algebraic Expressions 0.9: Simplifying Algebraic Expressions 0.10: More Algebraic Manipulations 0.11: Solving Equations 0.12: Solving Inequalities 0.13: The Cartesian Plane and Circles 0.14: Lines 0.15: Plane and Solid Geometry 0.16: Variation • This chapter will review notation, concepts, skills, techniques, and formulas needed in precalculus and calculus. Sections 0.2, 0.8, 0.9, 0.10, and 0.16 may be largely unfamiliar to incoming precalculus students.

(Section 0.1: Sets of Numbers) 0.1.1

SECTION 0.1: SETS OF NUMBERS LEARNING OBJECTIVES • Be able to identify different sets of numbers. • Know how to write sets of real numbers using set-difference, set-builder, graphical, and interval forms. • Be able to take the union and intersection of intervals of real numbers. PART A: DISCUSSION • Theorems and formulas require constants (denoted by c, a1 , a2 , etc.) to be from a particular set of numbers, usually the set of real numbers (denoted by  ). • Sets of real numbers can correspond to solutions of equations (in Section 0.11), solutions of inequalities (in Section 0.12), and domains and ranges of functions (in Section 1.1). There are several ways to describe these sets. PART B: SETS OF NUMBERS A set is a collection of objects, called the members (or elements) of the set. • Two sets A and B are equal (that is, A = B ) when they consist of the same members. Typically, order is irrelevant, and members are not repeated.

∅ denotes the empty set (or null set), the set consisting of no members. Let A and B be sets. A is a subset of B, denoted by A ⊆ B , when every member of A is also a member of B. If A ⊆ B , but A ≠ B , then A is a proper subset of B, denoted by A ⊂ B . This means that B contains all of the members of A, as well as at least one other member not in A. Some important sets of numbers are: + , or Z + , the set of positive integers.

• This set consists of 1, 2, 3, etc. •  comes from the German word “Zahlen.” •  is in blackboard bold typeface. It is more commonly used than Z. • WARNING 1: Sources differ as to whether counting numbers, whole numbers, and natural numbers also include 0.

(Section 0.1: Sets of Numbers) 0.1.2

 , or Z, the set of integers. • This set consists of 1, 2, 3, etc.; their opposites, −1 , − 2 , − 3 , etc. (making up − , the set of negative integers); and 0.  , or Q, the set of rational numbers.

• This is the set of all numbers that can be written in the form:

integer nonzero integer • It is the set of numbers that can be written as finite (or terminating) decimals or repeating decimals. • Examples include: Fraction form 1 2 1 3 823 − 9900 7 , or 7 1

Decimal form 0.5

0.3, or 0.3333…

− 0.831, or − 0.8313131… 7

• As demonstrated by the last example, every integer is a rational number. That is,  ⊆  . •  comes from “Quotient.”

 , or R, the set of real numbers. • This is the set of all numbers that can be written as decimals. • This set can be represented by the real number line below. There should be tick marks at 0 and at least one other number to indicate scale.

• This set consists of rational numbers and irrational numbers, such as π and

2.

(Section 0.1: Sets of Numbers) 0.1.3

 , or C, the set of complex numbers. • This set consists of real numbers and imaginary numbers, such as i and 2 + 3i. (See Chapter 2.) The Venn diagram below indicates the (proper) subset relations:  ⊂  ⊂  ⊂  . Each disk is contained within each larger disk.

PART C: SET NOTATION Set Notation ∈ denotes “in,” or “is in.” It denotes set membership. ∉ denotes “is not in.” ∍ denotes “such that.” ∀ (the universal quantifier) denotes “for all” or “for every.” ∃ (the existential quantifier) denotes “there exists” or “there is.” Example Set 1 (Using Set Notation)

7 ∈ 1 ∉ 2 ∀x ∈ , … ∀a, b ∈ , … ∀x ∈, x < x + 1 ∀x ∈, ∃y ∈ ∍ x = 2y §

Plain English translation 7 is an integer. 1 is not an integer. 2 For all real numbers x, … For all integers a and b, … Every real number is less than one added to itself. Every real number is twice some real number.

(Section 0.1: Sets of Numbers) 0.1.4 PART D: INFINITY Infinity, denoted by the lemniscate ∞ , is a quantity that is greater than any real number. Negative infinity, denoted by − ∞ , is a quantity that is lesser than any real number. WARNING 2: ∞ and − ∞ are not real numbers, though they may be handled differently in higher math. PART E: REPRESENTING SETS OF REAL NUMBERS There are various ways to represent a set of real numbers. We sometimes list the members of a set and surround them with braces. For example, the set 1, π consists of the two members 1 and π .

{ }

An interval corresponds to a connected (“unbroken”) piece of the real number line. A bounded interval has finite length on the real number line. An unbounded interval has infinite length. An open interval excludes its endpoints. A closed interval includes its endpoints. Example 2 (A Bounded, Open Interval) The set of all real numbers x such that 3 < x < 5 … … in set-builder form is:

{x ∈

}

3 < x < 5 , or

{ x ∈ : 3 < x < 5 }

… in graphical form is: … in interval form is:

(3, 5)

The set is an interval with 3 and 5 as its endpoints. The set is an open interval, because it excludes its endpoints. The exclusion of 3 and 5 is indicated by the use of: • strict inequality signs ( 3 } {x ∈ x ≥ 3 } {x ∈ x < 3 } {x ∈ x ≤ 3 } {x x ∈ } , which is 

Graphical Form

Interval Form

)

⎡⎣3, 5

(3, 5⎤⎦ (3, ∞ ) ⎡⎣3, ∞ ) ( − ∞, 3) ( − ∞, 3⎤⎦ ( − ∞, ∞ )

Type of Interval Bounded; Half-Open Bounded; Half-Open Unbounded Unbounded Unbounded Unbounded Unbounded

§ WARNING 3: In interval form, parentheses are always placed next to ∞ and − ∞ , because they are not real numbers and are therefore excluded from the set. However,  may be considered to be both an open interval and a closed interval.

(Section 0.1: Sets of Numbers) 0.1.6.

A ∪ B , the union of set A and set B, consists of all elements in one or both sets. Think: “All members are invited.”

{

}

• A ∪ B = x x ∈ A or x ∈B , where “or” is taken to be “inclusive or,” or “and/or.” The disjunctive symbol ∨ denotes “or.”

A ∩ B , the intersection (“overlap”) of set A and set B, consists of all elements in both sets.

{

}

• A ∩ B = x x ∈ A and x ∈B . The conjunctive symbol ∧ denotes “and.” Example 5 (A Union of Intervals)

{ }

The set  \ 1, π , which is written in set-difference form, consists of all real numbers except 1 and π . The set … … in set-builder form is:

{x ∈

}

x ≠ 1 and x ≠ π , or

{ x ∈ : x ≠ 1 and

x≠π }

… in graphical form is: … in interval form is:

( − ∞, 1) ∪ (1, π ) ∪ (π , ∞ )

TIP 1: The graphical form can help us write the interval form. TIP 2: The union symbol ∪ separates intervals in interval form whenever a real number must be “skipped over.” § Example 6 (Union and Intersection of Intervals)

( ) ( )

) )

a) Simplify 1, 5 ∪ ⎡⎣3, ∞ . (Think: “All members are invited.”) b) Simplify 1, 5 ∩ ⎡⎣3, ∞ . (Think: “Overlap.”) § Solution Interval Form

(1, 5) ⎡⎣3, ∞ )

( ) ) ( ) b) (1, 5) ∩ ⎡⎣3, ∞ ) = ⎡⎣3, 5)

a) 1, 5 ∪ ⎡⎣3, ∞ = 1, ∞

§

Graphical Form

(Section 0.2: Logic) 0.2.1

SECTION 0.2: LOGIC LEARNING OBJECTIVES • Be able to identify and use logical notation and terminology. • Understand the structure of “if-then” and “if and only if” statements. • Understand counterexamples and logical equivalence. • Know how to find the converse, inverse, and contrapositive of an “if-then” statement. • Understand necessary conditions and sufficient conditions. PART A: DISCUSSION • Although logic is a subject that is often relegated to courses in discrete mathematics, computer science, and electrical engineering, its fundamentals are essential for clear and precise mathematical thought. • Many theorems are “if-then” or “if and only if” statements. PART B: PROPOSITIONS AND “IF-THEN” STATEMENTS A proposition is a statement that is either true or false. “If-Then” Statements The statement “If p, then q” can be written as “ p → q .” • The proposition p is called the hypothesis, assumption, or condition. • The proposition q is called the conclusion. • If there are no cases where p is true and q is false, we say that the statement is true. • Otherwise, the statement is false, and any case where p is true and q is false is called a counterexample. If the statement is known to be true, we can write “ p ⇒ q .” “ ⇒ ” may be read as “implies.” • WARNING 1: “ → ” denotes “approaches” when we discuss limits in calculus.

(Section 0.2: Logic) 0.2.2 Example 1 (An “If-Then” Statement) Consider the statement: “If I get an A, then I pass.” This statement is of the form “If p, then q,” where: p is the hypothesis “I get an A,” and q is the conclusion “I pass.” The statement is true, because there is no case where a student gets an A but does not pass. § The converse of p → q is q → p . Example 2 (An “If-Then” Statement and Its Converse) Consider the statement: “If a number is an integer, then it is a rational number.” • That is, ( x ∈ ) → ( x ∈) . The statement is true, because every integer is also a rational number. We can write: ( x ∈ ) ⇒ ( x ∈) . Now, consider the converse of the above statement: “If a number is a rational number, then it is an integer.” • That is, ( x ∈) → ( x ∈ ) . This second statement is false, because a counterexample exists. 1 Observe that is a rational number, yet it is not an integer. 2 WARNING 2: The discovery of even one counterexample can be used to disprove a statement (that is, to prove that a statement is false). However, a single example is usually not enough to prove that a statement is true. To prove that a statement is true, one often needs to present a rigorous and general argument that applies to all cases where the hypotheses hold. Because the converse is false, we can write: ( x ∈) ⇒ ( x ∈ ) , where “ ⇒ ” denotes “does not imply.” §

(Section 0.2: Logic) 0.2.3 PART C: “IF AND ONLY IF” (or “IFF”) STATEMENTS “If and Only If ” (or “Iff”) Statements The statement “p if and only if q,” or “p iff q,” can be written as “ p ↔ q .” If the statement is known to be true, we can write “ p ⇔ q ,” or “ p ≡ q .” • If p ⇒ q , and if q ⇒ p , then p ⇔ q . • Then, the propositions p and q are logically equivalent; either both are true, or both are false.

(

) (

)

• For example, 2x = 6 ⇔ x = 3 . (See Footnote 1.) • Definitions are essentially “iff” statements. For example, a number is a rational integer number iff it can be written in the form . nonzero integer • True “iff” statements arise when an “if-then” statement and its converse are true. PART D: CONVERSE, INVERSE, AND CONTRAPOSITIVE The converse of p → q is: The inverse of p → q is:

q→ p ~ p → ~q

• “~” and “ ¬ ” are used to denote “not.” They are negation operators. The contrapositive of p → q is: ~ q → ~ p • TIP 1: Take the original statement, switch the propositions, and negate them. Contrapositive Theorem If an “if-then” statement is true, then its contrapositive must be true, and vice-versa. In other words, they are logically equivalent.

(

) (

)

That is, p → q ⇔ ~ q → ~ p . • This can be proven using truth tables in a discrete mathematics class.

As a result, any “if-then” statement you know to be true has a contrapositive associated with it that will automatically be true. WARNING 3: An “if-then” statement may or may not be logically equivalent to its converse or its inverse. (However, the converse and the inverse must be logically equivalent to each other. Why? How are they related?)

(Section 0.2: Logic) 0.2.4 Example 3 (Converse, Inverse, and Contrapositive of an “If-Then” Statement; Revisiting Example 1) Consider the true “if-then” statement: If I get an A , then I pass .   p q What are its converse, inverse, and contrapositive? Which of these are true? § Solution Converse:

If I pass , then I get an A .   q p

This statement is false, because we can find a counterexample. A student can pass the class with a grade of B or C. Inverse:

If I do not get an A , then I do not pass .       ~p ~q

This statement is false, because we can find a counterexample. A student can have a grade of B or C, and the student can pass. Contrapositive: If I do not pass , then I do not get an A .       ~q ~p By the Contrapositive Theorem, this must be true, since the original statement is. The Venn diagram below may clarify matters. (We ignore “+” and “ − ” modifiers, as well as pass/fail grading.)

§

(Section 0.2: Logic) 0.2.5. Example 4 (An “Iff” Statement) Consider the (true) statement:

If I get an A, B, or C , then I pass .    p q

Its converse is true:

If I pass , then I get an A, B, or C .    q p

Therefore, this is true:

I get an A, B, or C iff I pass . §    p q

PART E: NECESSARY CONDITIONS and SUFFICIENT CONDITIONS

(

p is a necessary condition for q ⇔ q ⇒ p

)

• That is, for q to be true, it is required that p must be true.

(

p is a sufficient condition for q ⇔ p ⇒ q

)

• That is, if p is true, then it is guaranteed that q must be true. Example 5 (Necessary Conditions and Sufficient Conditions; Revisiting Example 1) Consider the “if-then” statement: If I get an A , then I pass .   p q Is p sufficient for q? Is p necessary for q? § Solution p is sufficient for q, because p ⇒ q . (Equivalently, q is necessary for p.) Think: If I get an A, then I am guaranteed to pass. p is not necessary for q, because q ⇒ p . Think: I don’t need to get an A to pass. § Example 6 (“Necessary and Sufficient” Conditions; Revisiting Example 4) Consider the “iff ” statement: I get an A, B, or C iff I pass .    p q This is true, so p ⇔ q , and we say that p is necessary and sufficient for q. Therefore, p and q are logically equivalent. § FOOTNOTES 1. Propositions. Unlike some sources such as Rosen’s Discrete Mathematics and Its Applications, we call “ 2x = 6 ” a proposition, even though its truth value (“true” or “false”) depends on the value of x.

(Section 0.3: Rounding) 0.3.1

SECTION 0.3: ROUNDING LEARNING OBJECTIVES • Be able to round off decimals to a specified decimal place, a specified number of decimal places, or a specified number of significant digits (or figures). • Understand scientific notation. PART A: DISCUSSION (WARNINGS and TIPS) • Using Calculators Memory buttons on calculators can help preserve decimal accuracy. Taking too few decimal places may ruin the accuracy of final answers. A calculator might approximate π as 3.14159265. Though seemingly impressive, this is not an exact representation of π . Calculators cannot give exact decimal representations of such numbers as 2 , ln5 , and sin 37 . • Exact vs. Approximate Answers Math instructors typically expect exact answers, though decimal approximations may be required for “word problems.” For example, if an answer is π , you should write π . You may also be asked to give a decimal approximation such as 3.14. • “Word Problems” Depending on context, decimal answers might need to be rounded down, rounded up, or rounded off. Write any appropriate units such as feet, pounds, etc. as part of your answer. PART B: ROUNDING “ ≈ ” means “is approximately” or “is about.” Example 1 (Rounding Off to a Specified Decimal Place) π ≈ 3.14159 . π rounded off to the nearest … … integer is: 3 … tenth (that is, to one decimal place) is: 3.1 … hundredth (that is, to two decimal places) is: 3.14 … thousandth (that is, to three decimal places) is: 3.142 • In the U.S., the “1” in the third decimal place of π is rounded up to “2.” This is because the digit in the next decimal place is 5 or higher. (Different countries have different rules if that digit is a “5.”) §

(Section 0.3: Rounding) 0.3.2. Consider a number written in decimal form. To count decimal places, count all digits to the right of the decimal point. To count significant digits (or figures), locate the leftmost nonzero digit, and count it and all digits to the right of it. • We may want numbers written out to the same number of decimal places when adding and subtracting them, while we may focus on significant digits when multiplying and dividing them. We can also use significant digits to roughly retain the same level of accuracy when handling different units such as inches, feet, and miles. Example 2 (Counting Decimal Places and Significant Digits) Consider: 70.1230 • This is written out to four decimal places. We count the four digits after the decimal point. • This is written out to six significant digits. We include the two digits to the left of the decimal point. • The “0” at the end indicates that we claim accuracy to four decimal places. Writing “70.123” might not have had that effect. • We call 70 the integer part of this decimal. § Example 3 (Counting Decimal Places and Significant Digits; Scientific Notation) Consider: 0.001020, or .001020 (the integer part is 0; writing it is optional). • These are written out to six decimal places. We count all six digits after the decimal point, including the two zeros after the decimal point and before the “1.” • These are written out to four significant digits. We do not include the aforementioned zeros, but we include the “1” and all digits after it. • These can be written as 1.020 × 10− 3 using scientific notation. The “ − 3 ” exponent tells us to move the decimal point three places to the left to obtain .001020 in standard form. Note that: 1.020 × 103 = 1020. = 1020 , since the “3” exponent tells us to move the decimal point three places to the right. §

(Section 0.4: Absolute Value and Distance) 0.4.1

SECTION 0.4: ABSOLUTE VALUE AND DISTANCE LEARNING OBJECTIVES • Be able to formally define absolute value. • Know how to take the absolute value of a real number. • Know properties of the absolute value operation. • Relate absolute value and distances on the real number line. PART A: DISCUSSION • Understanding the definition of absolute value is crucial to understanding the absolute value function and its graph. (See Section 1.3.) PART B: ABSOLUTE VALUE Definition of Absolute Value The absolute value of a is denoted by a , where

⎧a, if a ≥ 0 a =⎨ ⎩− a, if a < 0 • That is, the absolute value of a nonnegative real number is itself, while the absolute value of a negative real number is its opposite. • This is an example of a piecewise definition because of the different rules for different values of a. (See Section 1.5.) TIP 1: The absolute value of a real number is never negative. WARNING 1: − a does not necessarily represent a negative value. In fact, if a < 0 , then − a > 0 . That is, the opposite of a negative real number is positive. Example 1 (Finding Absolute Value) a) Find 7 . b) Find 0 . c) Find − 3 . § Solution a) 7 ≥ 0 , so we use the top rule in the definition. 7 = 7 . b) 0 ≥ 0 , so we again use the top rule in the definition. 0 = 0 . c) − 3 < 0 , so we use the bottom rule in the definition: − 3 = − ( − 3) = 3 . §

(Section 0.4: Absolute Value and Distance) 0.4.2 PART C: PROPERTIES OF ABSOLUTE VALUE Properties of Absolute Value 1) a ≥ 0

Absolute values are never negative. (TIP 1)

2) − a = a

Opposites have the same absolute value. The absolute value of a product equals the product of the absolute values. The absolute value of a quotient equals the quotient of the absolute values, provided the denominators are nonzero.

3) ab = a b 4)

a a = , if b ≠ 0 b b

PART D: ABSOLUTE VALUE, DISTANCE, and OPPOSITES Interpreting Absolute Value as a Distance

a = the distance between a and 0 on the real number line. TIP 2: Just as for absolute value, a distance cannot be negative. However, we do use “signed distances” and “signed lengths” in trigonometry and motion problems. Example 2 (Interpreting Absolute Value as a Distance)

− 3 = 3 and 3 = 3 , because both − 3 and 3 are at a distance of 3 units away from 0 on the real number line.

§ The Distance Between a and b

b − a = a − b = the distance between a and b on the real number line. • In particular, a − 0 = a = the distance between a and 0 on the real number line. • Absolute value notation allows us to denote such distances without worrying about whether a is greater than b or vice versa. • In calculus, this notation helps us rigorously define limits and analyze sequences and series. The following (important!) statement explains why b − a = a − b .

(Section 0.4: Absolute Value and Distance) 0.4.3. “Opposite Differences” Rule

( b − a ) = − ( a − b) . That is, ( b − a ) and ( a − b) are opposites. • Prove this by rewriting − ( a − b) . • For example, ( 7 − 3) = 4 , while ( 3− 7 ) = − 4 , its opposite. • Because they are opposites, ( b − a ) and ( a − b) must have the same absolute value by Property 2) in Part C. That is, b − a = a − b .

Example 3 (The Distance Between a and b)

x − 3 and 3 − x both represent the distance between x and 3 on the real number line, regardless of whether x is greater than, less than, or equal to 3.

• For example, if x = 7 , its distance from 3 is given by: 7 − 3 = 4 = 4 , or by: 3− 7 = − 4 = 4 .

• For example, if x = 1, its distance from 3 is given by: 1− 3 = − 2 = 2 , or by: 3− 1 = 2 = 2 . §

(Section 0.5: Exponents and Radicals: Laws and Forms) 0.5.1

SECTION 0.5: EXPONENTS AND RADICALS: LAWS AND FORMS LEARNING OBJECTIVES • Understand radicals and how they are related to powers. • Know laws of exponents and radicals and how to apply them, particularly when simplifying expressions. • Be able to rewrite expressions in different forms. • Recognize how expressions are restricted. PART A: DISCUSSION • Rewriting expressions can help us solve problems. PART B: RADICALS is a radical symbol. In

n

x , n is the index, and x is the radicand.

• We assume n is an integer such that n ≥ 2 .

x , also written as x1/ 2 , is the principal square root of x. • Here, the index n = 2 . • If x ≥ 0 ,

x is the unique nonnegative real number whose square is x.

• If x < 0 ,

x cannot be defined as a real value.

(We will define the imaginary unit i as

()

−1 in Chapter 2.)

2

• For example, 9 = 3 and 91/ 2 = 3 , because 3 = 9 , and 3 is nonnegative. Although 9 has two square roots, 3 and − 3 , we take the nonnegative square root (3) as our principal square root. • TIP 1: Remember that n

0 = 0.

x , also written as x1/ n , is the principal n th root of x. • If n is even and x ≥ 0 ,

n

x is the unique nonnegative real number whose

n th power is x. (If x < 0 , then • If n is odd,

n

n

x cannot be defined as a real value.)

x is the unique real number whose n th power is x.

()

4

• For example, 4 16 = 2 and 161/4 = 2 , because 2 = 16 , and 2 is nonnegative. • TIP 2: A principal odd root of a negative real number is negative. For example,

3

( )

−8 = − 2 and −8

1/3

= − 2 , because ( − 2 ) = −8 . In Chapter 6, we 3

( )

will see that −8 has three complex cube roots, but only one of them − 2 is real.

(Section 0.5: Exponents and Radicals: Laws and Forms) 0.5.2

x m/ n can be rewritten as

n

x m or as

( ). n

x

m

• We assume that m and n are positive integers (not both even; see Footnote 1). • Think: take the mth power and the nth root, in either order. • For example, x 3/5 can be written as

5

x 3 or as

( ) 5

3

x .

PART C: LAWS OF EXPONENTS AND RADICALS The term power usually refers to an exponent. However, we often refer to x n as a power of x. Laws of Exponents If the expressions involved are real-valued, then the following laws apply. #

Law

1

x m x n = x m+ n

2

xm = x m− n n x

3

(x ) m

n

( )( )

To divide powers of x, subtract the exponents.

= x mn

To raise a power to a power, multiply the exponents.

= xn yn

The power of a product is the product of the powers.

4

( xy )

5

⎛ x⎞ xn ⎜⎝ y ⎟⎠ = y n

6

x0 = 1

7

1 = x− n n x

n

In Plain English / Comments To multiply powers of x, add the exponents. For example, x 2 x 3 = xx xxx = x 5 .

n

⎛ x⎞ 8 ⎜ ⎟ ⎝ y⎠

−n

⎛ y⎞ =⎜ ⎟ ⎝ x⎠

The power of a quotient is the quotient of the powers. This is consistent with Law 2, with m = n . (We define 00 to be 1 here. See Section 1.3, Part H.) This is consistent with Laws 2 and 6. n

⎛ x⎞ ⎜⎝ y ⎟⎠

−1

=

1 y = ⎛ x⎞ x ⎜⎝ y ⎟⎠

( y ≠ 0) . Use Law 3 with m = −1 .

WARNING 1: These laws may or may not apply when the expressions

( )

involved are not real-valued. − 2

1/ 2

, for example, cannot be defined as a

real value. Also, we require nonzero denominators in Laws 2, 5, 7, and 8.

(Section 0.5: Exponents and Radicals: Laws and Forms) 0.5.3 Laws of Radicals Assume that m and n are positive integers. If the expressions involved are real-valued, then the following laws apply. • Law 13 fundamentally distinguishes between even and odd roots. • For Laws 9 through 12, the square root laws extend to even roots. They also extend to odd roots, but we allow cases where x < 0 or y < 0 . For example,

3

xy =

( x )( y ) for all real values of x and y.

#

Law

9

xy = x y , if x ≥ 0 and y ≥ 0

10

x = y

x

,

y if x ≥ 0 and y > 0 n m

11

x = mn x (if m or n is even, we require x ≥ 0 )

( x) = x, 2

12

3

3

In Plain English / Comments The root of a product equals the product of 1 1 the roots. See Law 4 with n = , , etc. 2 3 The root of a quotient equals the quotient of 1 1 the roots. See Law 5 with n = , , etc. 2 3 For example, 3 x = 6 x , if x ≥ 0 . This is related to Law 3, with m and n there being the reciprocals of m and n here. More generally,

if x ≥ 0

13

3

( x) n

n

= x , if n = 2, 3, 4, etc.

If n is even, we require x ≥ 0 .

x2 = x

n

x n = x , if n is even; see Warning 4.

x3 = x

n

x n = x, if n is odd.

• We simplify 18 , for example, using Law 9: 18 = 9 ⋅ 2 = 9 2 = 3 2 . (The greatest perfect square that divides 18 is 9. Then, the 9 “comes out” as a 3.) • WARNING 2: Although Laws 9 and 10 cover the square root of a product or a quotient, there is no similar law for the square root of a sum or a difference. In particular,

x + y is not equivalent to

x + y . (See the Exercises.)

• WARNING 3: Do not apply Laws 9 through 12 to even roots if x < 0 or y < 0 . As we will see in Chapter 2,

−2 −3 ≠ 6 .

• WARNING 4: See Law 13. The statement For example, if x = − 3 ,

( − 3)2

x 2 = x is incorrect if x < 0.

= 9 = 3 , not − 3 .

(Section 0.5: Exponents and Radicals: Laws and Forms) 0.5.4 PART D: FRACTIONAL, RADICAL, AND EXPONENTIAL FORMS Example Set 1 (Equivalent Expressions) The following tables list pairs of equivalent expressions, together with relevant restrictions (we will discuss domains in Section 1.1). The restrictions guarantee that we never (do the equivalent of): 1) divide by zero, or 2) take the even root of a negative value. WARNING 5: Do not confuse exponential form with exponential functions (see Chapter 3). “Power form” may be more appropriate. Fractional Form 1 x 1 x2

Restrictions

Exponential Form

x≠0

x −1

x≠0

x− 2

x≠0

1 4/3 x ( x ≠ 0) 7 (See Warnings below.)

1 7x − 4 /3

WARNING 6: In the last example, we need to state the “hidden” 1 restriction ( x ≠ 0 ) , because the expression implies it, yet the 7x − 4 /3 1 expression x 4 /3 does not. (See Warnings 9 and 10.) 7 1 WARNING 7: It is incorrect to re-express as 7x 4 /3 , a common − 4 /3 7x 1 1 1 ⋅ error. Think of as . The negative exponent only applies to x. 7x − 4 /3 7 x − 4 /3 Radical Form x 3 x

3

( x ) or 3( x ) 7

5

7

( )(

Exponential Form x1/2 x1/3

None

3x 5 /7

5

TIP 3: The parentheses in 3 expression as 37

Restrictions x≥0 None

)

( x ) prevent us from misreading the 7

5

x5 .

WARNING 8: The taking of odd roots (such as cube roots) does not impose any new restrictions. See Tip 2 in Part B.

(Section 0.5: Exponents and Radicals: Laws and Forms) 0.5.5 Combined Form 1

Restrictions

Exponential Form

x>0

x −1/2

x≠0

x −1/3

x>0

7x − 3/4

x 1 3

4

x 7

x

3

PART E: SIMPLIFYING EXPRESSIONS Example 2 (Simplifying an Expression)

⎛ 3 ⎞ Simplify ⎜ ⎝ − 2x 2 ⎟⎠

−4

. Nonpositive exponents (that is, exponents that are

negative or zero) are not allowed in the final expression. § Solution

⎛ 3 ⎞ ⎜⎝ − 2x 2 ⎟⎠

−4

⎛ − 2x 2 ⎞ =⎜ ⎝ 3 ⎟⎠

4

( x ≠ 0)

by Law 8

WARNING 9: The given expression should be examined for restrictions. Here, since division by 0 is forbidden, we require that ( x ≠ 0 ) . WARNING 10: Whenever a step “hides” a restriction, we should write the restriction, although we sometimes wait until the final expression. (See Warning 6 and Example 3.) Restrictions can be hidden when we apply laws of exponents and radicals, when we take reciprocals (see the Comment for Law 8), and when we cancel (see Section 0.9).

( − 2x ) =

2 4

34

=

§

( − 2 )4 ( x 2 )

16x 8 = 81

81

( x ≠ 0)

by Law 5

( x ≠ 0)

by Law 4

( x ≠ 0)

by Law 3

4

(Section 0.5: Exponents and Radicals: Laws and Forms) 0.5.6. Example 3 (Simplifying an Expression)

x x . Nonpositive exponents are not allowed in the final x 7 /2 x 0 expression. Simplify

§ Solution

(

)

The given expression has the restriction x > 0 , because

(

)

(

)

x and x 7/2

require x ≥ 0 , and the powers of x x 7/2 and x 0 that are factors of the

(

)

denominator require x ≠ 0 .

x x x1 x1/2 = x 7 /2 x 0 x 7 /2 ⋅1

by Law 6

1 2

1+

x x 7 /2 x 3/2 = 7 /2 x =

=x

by Law 1

3 7 − 2 2

= x− 2 =

1 x2

( x > 0) ( x > 0) ( x > 0)

by Law 2

by Law 7

(

)

It might be unclear which step “hides” the restriction x > 0 , so we might

(

)

wait until the end to write x > 0 . § FOOTNOTES 1. Interpreting x m / n . x m/ n , where m and n are positive, even integers, is interpreted differently by different sources. • For example, if x 2 / 6 is interpreted as 6

x2 =

3

x2 =

3

x = x

1/3

6

x 2 , there are no restrictions, and

.

• On the other hand, if x 2 / 6 is interpreted as

( x ) , we need the restriction x ≥ 0 . 6

2

2. Irrational exponents. See Section 3.1 on how to interpret something like 2π .

(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.1

SECTION 0.6: POLYNOMIAL, RATIONAL, AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES • Be able to identify polynomial, rational, and algebraic expressions. • Understand terminology and notation for polynomials. PART A: DISCUSSION • In Chapters 1 and 2, we will discuss polynomial, rational, and algebraic functions, as well as their graphs. PART B: POLYNOMIALS Let n be a nonnegative integer. An n th -degree polynomial in x , written in descending powers of x, has the following general form:

an x n + an 1 x n

1

+ ... + a1 x + a0 ,

( an

0)

The coefficients, denoted by a1 , a2 ,…, an , are typically assumed to be real numbers, though some theorems will require integers or rational numbers.

an , the leading coefficient, must be nonzero, although any of the other coefficients could be zero (i.e., their corresponding terms could be “missing”).

an x n is the leading term. a0 is the constant term. It can be thought of as a0 x 0 , where x 0 = 1 . • Because n is a nonnegative integer, all of the exponents on x indicated above must be nonnegative integers, as well. Each exponent is the degree of its corresponding term.

(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.2 Example 1 (A Polynomial)

5 2 x + 1 is a 3rd-degree polynomial in x with leading coefficient 4, 2 leading term 4x 3 , and constant term 1. The same would be true even if the terms were reordered: 5 2 1 x + 4x 3 . 2 4x 3

5 2 x + 1 fits the form 2 an x n + an 1 x n 1 + ... + a1 x + a0 , with degree n = 3. It can be rewritten as: The polynomial 4x 3

4x 3

5 2 x + 0x + 1 , which fits the form 2

a3 x 3 + a2 x 2 + a1 x + a0 , where the coefficients are:

( leading coefficient )

a3 = 4 a2 = a1 = 0 a0 = 1

5 2

(constant term )

§ Example Set 2 (Constant Polynomials) 7 is a 0th-degree polynomial. It can be thought of as 7x 0 . 0 is a polynomial with no degree. §

(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.3 PART C: CATEGORIZING POLYNOMIALS BY DEGREE Degree 0 1 2 3 4 5

Type [Nonzero] Constant Linear Quadratic Cubic Quartic Quintic

Examples 7 3x + 4 5x 2 x + 1 x 3 + 4x x4 x5

PART D: CATEGORIZING POLYNOMIALS BY NUMBER OF TERMS Number of Terms 1 2 3

Type

Examples

Monomial Binomial Trinomial

x5 x 3 + 4x 5x 2 x + 1

PART E: SQUARING BINOMIALS Formulas for Squaring Binomials

( a + b) ( a b)

2

= a 2 + 2ab + b2

2

= a2

2ab + b2

WARNING 1: When squaring binomials, don’t forget the “middle term” of the resulting Perfect Square Trinomial (PST). For example, ( x + 3) = x 2 + 6x + 9 . Observe that 6x is twice the product of the 2

( )( )

terms x and 3: 6x = 2 x 3 .

(

The figure below implies that x + y

)

2

= x 2 + 2xy + y 2 for x > 0 and y > 0 .

(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.4 PART F: RATIONAL AND ALGEBRAIC EXPRESSIONS A rational expression in x can be expressed in the form:

polynomial in x nonzero polynomial in x Example Set 3 (Rational Expressions) Examples of rational expressions include: a)

1 . x

5x 3 1 b) 2 . Irrational coefficients such as x + 7x 2 as coefficients of either polynomial.

2 are permissible

x7 + x c) x + x which equals . In fact, all polynomials are rational 1 expressions. § 7

An algebraic expression in x is also permitted to contain non-integer rational powers of variable expressions (and their equivalents in radical form). Example Set 4 (Algebraic Expressions) Examples of algebraic expressions include: a) x1/ 2 , or

x.

x 3 + 7x 5 /7 b) x 3 x+5+

.

• (See Footnote 1.)

All rational expressions are algebraic. §

(Section 0.6: Polynomial, Rational, and Algebraic Expressions) 0.6.5 The Venn diagram below is for expressions in x that correspond to functions (see Chapter 1):

FOOTNOTES 1. Algebraic expressions. Some sources forbid the presence of in an algebraic expression, since is a transcendental (i.e., non-algebraic) number. That means that is not a zero of any polynomial with integer coefficients, as, say,

2 is.

(Section 0.7: Factoring Polynomials) 0.7.1

SECTION 0.7: FACTORING POLYNOMIALS LEARNING OBJECTIVES • Know techniques and formulas for factoring polynomials. • Know the Test for Factorability for factoring quadratic trinomials. • Recognize polynomials in quadratic form and be able to factor them. PART A: DISCUSSION • Factoring is a very commonly used technique in precalculus and calculus. Factoring helps us simplify expressions, find zeros, solve equations and inequalities, and find partial fraction decompositions (see Section 7.3). • Rewriting a sum of terms as a product of factors helps us perform sign analyses, as we will see in Sections 2.4 and 2.10. PART B: FACTORING OUT GCFs For now, when we factor a polynomial, we factor it completely over the integers ( ), meaning that the factors cannot be broken down further using only integer coefficients. That is, the factors must be prime (or irreducible) over the integers. • In Chapter 2, we will factor over other sets, such as

,

, or

.

TIP 1: The Greatest Common Factor (GCF), if it is not 1, should typically be factored out first, although it can be factored out piece-by-piece for more complicated expressions. (Unfortunately, there is no simple, standard definition for the GCF.) Example 1 (Factoring out a GCF) We factor 8x + 6 as 2 ( 4x + 3) , because 2 is the GCF. 2 is the greatest common divisor of 8 and 6. § Example 2 (Factoring out a GCF)

(

)

We factor x 5 + x 3 as x 3 x 2 + 1 , because x 3 is the GCF.

x 3 is the power of x with the least exponent. §

(Section 0.7: Factoring Polynomials) 0.7.2 TIP 2: Sometimes, it is helpful to factor out 1 , particularly when a polynomial has a negative leading coefficient. Example 3 (Factoring out -1 First)

8x 5

Factor

6x 3 .

§ Solution

6x 3 =

8x 5

=

( 8x + 6x ) 2x ( 4x + 3) 5

3

3

2

WARNING 1: Sometimes, people confuse signs if they try to factor out 2x 3 immediately. § WARNING 2: Be careful when factoring the base of a power. Make sure to apply the exponent to all factors. Example 4 (Factoring out of a Power)

(x

+x

)

(

5

)

5

is not equivalent to x x 2 + 1 . The following is correct: 3

(x

3

+x

)

5

( (x

) + 1)

5

= x x2 + 1 = x5

2

5

Each factor of the base must be raised to the exponent, 5.

( )

See Section 0.5, Law 4: xy

n

= xn yn . §

(Section 0.7: Factoring Polynomials) 0.7.3 PART C: FACTORING FORMULAS Factoring Formulas Factoring a …

Formula

Perfect Square Trinomial (PST) Sum of Two Squares Difference of Two Squares Sum of Two Cubes Difference of Two Cubes

( ) = ( a b)

a 2 + 2ab + b2 = a + b a2

2ab + b2

2 2

(A rule will be provided for a 2 + b2 when we discuss imaginary numbers in Section 2.1. As is, it is prime for now.) a 2 b2 = a + b a b

(

( )( = ( a b) ( a

)(

a 3 + b3 = a + b a 2 a3

b3

2

)

) + ab + b ) ab + b2 2

WARNING 3: Many math students forgot or never learned the last two formulas. WARNING 4: In the last two formulas, there is no “2” or “ 2 ” coefficient on the ab term of the trinomial factor. If a and b have no common factors (aside from 1 and 1 ), the trinomial factors are typically prime. Sometimes, people confuse these trinomials with Perfect Square Trinomials (PSTs), which we introduced in Section 0.6, Part E. TIP 3: In the last two formulas, observe that the binomial factor is “as expected”: a + b for a 3 + b3 , and a b for a 3 b3 . The visible signs on the right-hand

(

)

(

)

sides follow the pattern: “same,” “different,” and “+.”

(Section 0.7: Factoring Polynomials) 0.7.4 PART D: TEST FOR FACTORABILITY and PRACTICE EXAMPLES Test for Factorability This test applies to any quadratic trinomial of the form ax 2 + bx + c , where a, b, and c are nonzero, integer coefficients. (Assume the GCF is 1 or 1 ; if it is not, factor it out.) The discriminant of the trinomial is b2

4ac .

• If the discriminant is a perfect square (such as 0, 1, 4, 9, etc.; these are squares of integers), then the trinomial can be factored over the integers. For example, x 2 + 3x + 2 has discriminant 1 and can be factored as x + 2 x +1 .

(

)(

)

•• In fact, if the discriminant is 0, then the trinomial is a perfect square trinomial (PST) and can be factored as the square of a binomial with integer coefficients. For example, x 2 6x + 9 has

(

)

2

discriminant 0 and can be factored as x 3 . • If the discriminant is not a perfect square, then the trinomial is prime over the integers. This test may be applied in Example Set 5, a) through g), which serve as review exercises for the reader. • The discriminant is denoted by (uppercase delta), though that symbol is also used for other purposes. It is seen in the Quadratic Formula in Section 0.11. We will discuss a method for factoring quadratic trinomials using the Quadratic Formula in Chapter 2.

(Section 0.7: Factoring Polynomials) 0.7.5 Example Set 5 (Factoring Polynomials) Factor the following polynomials over the integers. a)

x 2 + 9x + 20

b)

x2

20x + 100

c)

x2

4x 12

d)

3x 2

e)

4x 2 + 11x + 6

f)

2x 2 + 10x + 5

g)

3x 2 + 6x 3

(Hint: This is a Perfect Square Trinomial (PST).)

20x 7

h)

x 4 16

i)

a 3 3a + 2a 2 b 6b (Hint: Use Factoring by Grouping. This is when we group terms and factor each group “locally” before we factor the entire expression “globally” by factoring out the GCF.)

j)

4x 2 + 9 y 2

k)

x 3 + 125y 3

l)

x 3 125y 3

(Section 0.7: Factoring Polynomials) 0.7.6 § Solution a)

ax + bx + c = x + 9x + 20 = ( x + 5 ) ( x + 4 ) 2

2

We want 5 and 4, because they have product = c = 20 and (since this is the a = 1 case) sum = b = 9. We can rearrange the factors: x + 4 x + 5 .

(

b)

x

2

20x + 100 = ( x 10 ) , or 2

(10)2

( x )2

x2

10 ) ( x 10 ) = ( x 10 )

)

2

Check: 2( x )( 10 ) = 20 x

Guess that this is a PST for now.

c)

(x

)(

(

)(

4x 12 = x 6 x + 2

)

How do we know we need

6 and + 2?

The constant term, c, is negative, so use opposite signs: one "+" and one " ." The middle coefficient, b, is negative, so the negative number must be higher in absolute value than the positive number; it “carries more weight.” 2-factorizations of –12 (which is c) Think: What? • What?? = –12. 12, +1 6, +2 4, +3 d)

F + (O + I ) + L = 3x

2

20x

7 = ( 3x + 1) ( x

Sum = b = 4? ( a = 1 case ) No Yes Can stop No 7)

F = First product (product of the First terms) O = Outer product (product of the Outer terms) I = Inner product (product of the Inner terms) L = Last product (product of the Last terms)

( 3x

)( x

)

2

F = 3x ; factors must be 3x and x Need L = 7

+7

1

1

+7

Makes O + I = 20x. We need O + I to be

20x,

which is the middle term of the trinomial. We're only off by a sign, so we change both signs. +1

7

7

+1

Makes O + I = 20x. This works.

Also, b = 20 , a "very negative" coefficient, so we are inclined to pair up the 3x and the 7 to form the outer product, since they form 21x .

(Section 0.7: Factoring Polynomials) 0.7.7 e)

(

)(

4x 2 + 11x + 6 = 4x + 3 x + 2

)

Method 1: Trial-and-Error ("Guess") Method

(

)(

)

4x

x

2x

2x

F = 4x

+1

+6

+6

+1

+2

+3

+3

+2

2

L = 6; need both " + " because of + 11x

Method 2: Factoring by Grouping 4 and 6 are neither prime nor “1,” so we may prefer this method. We want two integers whose product is ac = (4)(6) = 24 and whose sum is b = 11. We want 8 and 3; split the middle term accordingly. 4 x + 11x + 6 = 4 x + 8x + 3x + 6 2

2

OK to switch

(

f)

)

= 4 x + 8x + ( 3x + 6 )

Group terms

= 4 x ( x + 2 ) + 3( x + 2 )

"Local factoring"

= ( 4 x + 3) ( x + 2 )

"Global factoring"

2

2x 2 + 10x + 5 is prime or irreducible over the integers (i.e., it cannot be broken down further using integer coefficients). None of these combinations work:

( 2x

)( x

)

2

F = 2x ; factors must be 2x and x Need L = 5; need both " + " because of + 10x

+1

+5

+5

+1

We could also apply the Test for Factorability. The discriminant

b 4ac = (10 ) 4 ( 2 ) ( 5 ) = 100 40 = 60 , which is not a perfect square, and the GCF = 1, so the polynomial is prime. 2

2

g)

3x + 6x 2

3=

(

3 x GCF

)

2x + 1

2

a PST

You should usually factor out the GCF first. =

3 ( x 1)

2

(Section 0.7: Factoring Polynomials) 0.7.8 h)

[

Apply the Difference of Two Squares formula a

(

16 = x 2 + 4

x4

( x2 )

2

( 4)

2

)

prime

(

)(

x2

4

( x)

( 2)

2

]

b ) twice:

2

2

)(

= x2 + 4 x + 2 x 2 i)

b = (a + b)(a

2

)

Use Factoring by Grouping:

(

) ( = a ( a 3) + 2b ( a = ( a + 2b) ( a 3)

) 3)

a 3 3a + 2a 2 b 6b = a 3 3a + 2a 2 b 6b 2

2

2

j)

4x 2 + 9 y 2 is prime. The GCF = 1, and we have no formula for the Sum of Two Squares (for now…; this will change when we discuss imaginary numbers in Section 2.1).

k)

Apply the Sum of Two Cubes formula a + b =

3

(a + b)

3

"Expected factor"

a

ab + b

2

2

:

NOT 2ab

The visible signs follow the pattern: same, different, "+"

(

x + 125y = ( x + 5y ) x 3

( x)

l)

3

3

(5 y)

2

5xy + 25y

2

)

3

Apply the Difference of Two Cubes formula a

3

3

b =

(a

b)

"Expected factor"

a + ab + b 2

2

NOT + 2ab

The visible signs follow the pattern: same, different, "+"

x

3

( x )3

125y = ( x 3

(5 y)

3

(

5y ) x + 5xy + 25y 2

2

). §

:

(Section 0.7: Factoring Polynomials) 0.7.9 PART E: FACTORING EXPRESSIONS IN QUADRATIC FORM An expression is in quadratic form It can be expressed as au 2 + bu + c after performing a u substitution, where a 0 , and a, b, and c are real coefficients. • The term “quadratic form” is defined differently in higher math; that definition requires each term to have degree 2.

Example 6 (Factoring an Expression in Quadratic Form) Factor 2x 6

x 3 1 over the integers.

§ Solution The trinomial 2x 6 x 3 1 is in quadratic form, because the exponent on x in the first term is twice that in the second term (6 is twice 3), and the third term is a constant. We will use the substitution u = x 3 , the power of x in the “middle” term.

( )

Then, u 2 = x 3

2

= x6 . 2x 6

x 3 1 = 2u 2 u 1 Now, factor as usual. = ( 2u + 1) ( u 1)

(

Substitute back. Replace u with x 3 .

)(

)

= 2x 3 + 1 x 3 1

With practice, the substitution process can be avoided. Either way, we are not done yet! It is true that 2x 3 + 1 is prime over the integers; Chapter 2

(

(

)

)

will help us verify that. However, x 3 1 is not prime, because we can apply the Difference of Two Cubes formula.

( 2x

3

)(

) (

)

(

)

+ 1 x 3 1 = 2x 3 + 1 ( x 1) x 2 + x + 1

This is factored completely over the integers. The Test for Factorability can be used to show that the trinomial factor x 2 + x + 1 is prime, as

(

expected. §

)

(Section 0.8: Factoring Rational and Algebraic Expressions) 0.8.1

SECTION 0.8: FACTORING RATIONAL AND ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES • Know techniques for factoring rational and algebraic expressions. PART A: DISCUSSION • In this section, we will extend techniques for factoring polynomials to other rational and algebraic expressions, including those with negative and fractional exponents. Prior to a precalculus course, most students have no experience factoring such expressions. PART B: FACTORING OUT GCFs

(

)

When we factor x 5 + x 3 as x 3 x 2 + 1 , we factor out x 3 , the power of x with the least exponent; x 3 is the GCF. We then divide each term of x 5 + x 3 by x 3 to obtain the other factor, x 2 + 1 . When we divide x 5 by x 3 , we subtract the

(

)

exponents in that order and get x 2 . TIP 1: Think: “We’re factoring x 3 out of x 5 . 5 takeaway 3 is 2.” These techniques apply even when the exponents involved are negative and/or fractional.

(Section 0.8: Factoring Rational and Algebraic Expressions) 0.8.2 Example 1 (Factoring with Negative Exponents) Factor x

7

+x

4

2x

1

over the integers.

§ Solution Observe that 7 is the least exponent on x. Our GCF is x 7 , so we will factor it out and subtract 7 from each of the exponents.

x

7

+x

4

2x

1

=x

7

=x

7

=x

7

(1 + x (1 + x (1 + x

4

( 7)

4+7 3

2x 2x

2x 6

)

1

1+ 7

)

( 7)

)

TIP 2: Observe that this last trinomial has no negative exponents on x. This is a sign that we have factored out the GCF correctly. WARNING 1: We usually try to avoid negative exponents in final answers, so we will rewrite the expression as a fraction.

=

1 + x 3 2x 6 x7 We are not done yet! We can factor the numerator further over the integers. TIP 3: We will first factor out 1 so that the new leading coefficient is positive. This tends to make factoring easier.

=

(

1 x 3 + 2x 6 x

)

7

The indicated 1 factor can be moved in front of the fraction with no other sign changes. WARNING 2: This is because it is a factor of the entire numerator.

=

1 x 3 + 2x 6 x7

(Section 0.8: Factoring Rational and Algebraic Expressions) 0.8.3 TIP 4: We will now rewrite the numerator in descending powers of x. This tends to make factoring easier.

2x 6

=

x3 1 x7

Fortunately, we have already factored the numerator in Section 0.7, Example 6.

= =

( 2x (

3

)(

)

+ 1 x3 1

x7 2x 3 + 1 ( x 1) x 2 + x + 1

)

(

)

x7

§

Example 2 (Factoring with Negative and Fractional Exponents)

(

Factor x 3 + 2

)

1/3

(

+ x3 + 2

)

5 /3

over the integers.

WARNING 3: Exponents do not typically distribute over sums. WARNING 4: Likewise, the root of a sum is not typically equal to the sum of the roots. § Solution WARNING 5: All negative exponents are less than all positive exponents.

5 is the least exponent on x 3 + 2 . Our GCF is x 3 + 2 3 5 so we will factor it out and subtract from each of the exponents on 3 x3 + 2 .

Observe that

(

)

(

)

(

)

5 /3

,

(Section 0.8: Factoring Rational and Algebraic Expressions) 0.8.4

(x

3

+2

)

1/3

(

+ x +2 3

)

5 /3

(

3

)

(

3

= x +2

)

= x +2

( = (x = (x

) + 2) + 2)

5 /3

5 /3

(x

3

+2

)

1 3

(x

3

+2

)

1 5 + 3 3

(x

3

+2

)

+1

+1

= x3 + 2

5 /3

3

5 /3

x 6 + 4x 3 + 4 + 1

5 /3

x 6 + 4x 3 + 5

=

3

2

5 3

+1

x 6 + 4x 3 + 5

(x

3

+2

)

5 /3

Although x 6 + 4x 3 + 5 is in quadratic form, it is prime over the integers. §

(Section 0.9: Simplifying Algebraic Expressions) 0.9.1

SECTION 0.9: SIMPLIFYING ALGEBRAIC EXPRESSIONS LEARNING OBJECTIVES • Be able to use factoring and canceling (i.e., division) to simplify algebraic expressions. • Be able to rationalize the numerator or the denominator of a fraction containing radical(s). • Be aware of common errors when rewriting and simplifying expressions.

PART A: DISCUSSION • The factoring techniques from Sections 0.7 and 0.8 will help us simplify fractions by canceling (i.e., dividing out) common factors. • We can also re-express or simplify a fraction by rationalizing the numerator or the denominator. PART B: CANCELING COMMON FACTORS IN A FRACTION We are typically required to provide answers in simplified form. While there are different opinions as to what that means, there is common agreement that factors (other than 1 and 1 ) that are common to both the numerator and the denominator of a fraction must be canceled (i.e., divided out). WARNING 1: Some instructors object to the term “canceling,” because students often abuse the idea by inappropriately and carelessly deleting matching expressions. Remember to rely on mathematical rules, not merely wishful thinking! Canceling Rule for Fractions As we simplify a fraction, we can “cancel”: a nonzero factor of the entire numerator with an equivalent factor of the entire denominator. • That is, we can divide the entire numerator and the entire denominator by equivalent nonzero factors.

(Section 0.9: Simplifying Algebraic Expressions) 0.9.2 For example, we can divide the numerator and the denominator of by x and obtain

(

1

x x+4

)

(

x2 x + 4

)

. We have canceled a pair of x factors:

(1)

(

x

x2 x + 4

( x)

)

=

1

(

x x+4

)

WARNING 2: We cannot cancel a pair of x factors in not shown as a factor of the entire denominator.

x , because x is x2 + 4

Canceling and Restrictions When canceling, if a factor is eliminated from the denominator of a fraction, check to see if any restrictions have been “hidden.” Such restrictions must be written out separately. • It is usually acceptable to wait until the end to write such restrictions. For example,

( x)

(

x

x2

x x+4

)

=

x x+4

(1) =

(

x2

)

x x x+4

(

( x 0) 0

)

(Section 0.9: Simplifying Algebraic Expressions) 0.9.3 PART C: CANCELING OPPOSITE FACTORS

( a b) and ( b a ) are opposites. Canceling Opposite Factors The quotient of nonzero, opposite factors is 1 . Example 1 (Canceling Opposite Factors)

9 x2 Simplify . x 3 § Solution Method 1

(

)(

3+ x 3 x 9 x2 = x 3 x 3

(

)

)

WARNING 3: It is incorrect to factor 9 which is the factorization of x 2

(

)(

9.

(3 x ) and ( x 3) are opposites, so their quotient in either order is

1.

( 1)

(3 + x ) (3 x ) = ( x 3) x 3 ( ) (1)

TIP 1: The parentheses around the “ 1 ” remind us that it is a factor of the numerator, not a term.

=

(3 + x ) ( x 3) , or 3 x ( x 3)

The choice between

(3 + x ) and

3

)

x 2 as x + 3 x 3 ,

x may depend on context. §

(Section 0.9: Simplifying Algebraic Expressions) 0.9.4 § Solution Method 2

9 x2 = x 3 =

=

(x

2

9

)

x 3 x2 9 x 3

(1)

( x + 3) ( x 3) ( x 3) ( x 3) (1)

= §

( x + 3) ( x 3) , or x 3 ( x 3)

PART D: SIMPLIFYING ALGEBRAIC EXPRESSIONS Example 2 (Simplifying an Algebraic Expression)

( 4x + 7 )1/3 ( 2x ) Simplify

( x ) 13 ( 4x + 7 ) ( 4 ) 2 /3

2

( 4x + 7 )2 /3

.

Do not leave nonpositive exponents in the final expression. • In calculus, the given expression is obtained by applying the Quotient Rule for x2 Differentiation to 3 . 4x + 7

§ Solution We begin by “cleaning up” the numerator.

( 4x + 7 )1/3 ( 2x )

( x ) 13 ( 4x + 7 ) ( 4 ) 2 /3

2

( 4x + 7 )2 /3

4 2 x ( 4x + 7 ) 3 ( 4x + 7 )2 /3

2x ( 4x + 7 )

1/3

=

2 /3

(Section 0.9: Simplifying Algebraic Expressions) 0.9.5 Method 1 (Factor the Numerator) We will factor out the GCF of the numerator, which is x ( 4x + 7 )

2 /3

.

TIP 2: It may be easier to factor out x in one step and then ( 4x + 7 ) 2 /3 in another step. The GCF does not have to be factored out in one step. TIP 3: It may be easier to substitute u = 4x + 7 and obtain 4 2 2 /3 2xu1/3 x u 3 . Make sure to substitute back later. u 2 /3

4 2 x ( 4x + 7 ) 3 ( 4x + 7 )2 /3

2x ( 4x + 7 )

1/3

=

=

x ( 4x + 7 )

2 ( 4x + 7 )

2 /3

2 /3

or 2 3

1 3

( 4x + 7 )2 /3 x ( 4x + 7 )

2 /3

2 ( 4x + 7 )

( 4x + 7 )2 /3

( 4x + 7 ) 2 /3 ( 4x + 7 )2 /3

2 /3

=

( 4x + 7 )4 /3 x 8x + 14

=

4 x 3

4 x 3

( 4x + 7 )4 /3

2 /3

u 2 /3 2 /3

xu

2u

or

2 3

1 3

4 x 3

u 2 /3 xu

2 /3

or

2u

4 x 3

u 2 /3

by ( 4x + 7 ) . 2 /3

u 2 /3 = 2 /3 = u u

The division yields ( 4x + 7 )

x 2 ( 4x + 7 )

4 x 3

4 x 3

We now divide ( 4x + 7 ) If we let u = 4x + 7 :

4 2 x u 3

2xu1/3

4 /3

2 3

2 3

=u

4 3

=

1 u

4 /3

=

in the denominator.

1

( 4x + 7 )4 /3

(Section 0.9: Simplifying Algebraic Expressions) 0.9.6 WARNING 4: Compound fractions are typically unacceptable in a simplified expression. We will multiply the numerator and the denominator by 3.

4 x 3

3x 8x + 14 = = =

3 ( 4x + 7 ) 24x 2 + 42x

3 ( 4x + 7 )

4 /3

4x 2 4 /3

20x 2 + 42x

3 ( 4x + 7 )

, or 4 /3

2x (10x + 21) 3 ( 4x + 7 )

4 /3

WARNING 5: We should factor the numerator and see if there are any common factors with the denominator, aside from 1 and 1 . Such factors must be canceled (divided out). WARNING 6: Do not distribute the 3 in the denominator. The 4/3 exponent forbids that. • No restrictions need to be written. The only restriction on the given 7 expression was x , and it is implied by the final expression. 4 Method 2 (Rewrite as a Compound Fraction First)

2 , the expressions in blue below can be 3 rewritten as fractions containing positive exponents. Because of the negative exponent

Again, the substitution u = 4x + 7 may help.

4 2 x ( 4x + 7 ) 3 ( 4x + 7 )2 /3

2x ( 4x + 7 )

1/3

2x ( 4x + 7 )

1/3

=

4x 2 2 /3 3 ( 4x + 7 )

( 4x + 7 )

2 /3

2 /3

4 2 x u 3

2xu1/3 or

u 2 /3 2xu1/3

or

u 2 /3

4x 2 3u 2 /3

2 /3

(Section 0.9: Simplifying Algebraic Expressions) 0.9.7 We will multiply the numerator and the denominator by the Least 2 /3 Common Denominator (LCD), 3 ( 4x + 7 ) , or 3u 2 /3 .

3 ( 4x + 7 ) =

1/3

3 ( 4x + 7 ) 3u

2 /3

or

4x 2 2 /3 3 ( 4x + 7 )

2x ( 4x + 7 )

2 /3

2xu 3u 2 /3

2 /3

1/3

( 4x + 7 )2 /3 4x 2 3u 2 /3

u 2 /3

WARNING 7: Do not cancel the expressions above in red. We must apply the Distributive Property. If it helps, write the following step; with experience, you can skip it.

3 ( 4x + 7 )

2x ( 4x + 7 )

2 /3

=

3 ( 4x + 7 ) 3u

2 /3

3 ( 4x + 7 )

1/3

2xu

or

1/3

3u 2 /3

3u

2 /3

2 /3

6xu 4x 2 = 3u 4 /3

= = = §

6x ( 4x + 7 )

3 ( 4x + 7 )

24x 2 + 42x

3 ( 4x + 7 )

20x 2 + 42x

3 ( 4x + 7 )

4x 2 4 /3

4x 2 4 /3

, or 4 /3

2x (10x + 21) 3 ( 4x + 7 )

( 4x + 7 )2 /3 4x 2 3u 2 /3

u 2 /3

4 /3

2 /3

4x 2 3 ( 4x + 7 )

2 /3

(Section 0.9: Simplifying Algebraic Expressions) 0.9.8 PART E: RATIONALIZING DENOMINATORS AND NUMERATORS When we rationalize the denominator of a fraction, we eliminate radicals and non-integer exponents in the denominator but possibly introduce them in the numerator. For example,

1

1 2

is unacceptable in a simplified expression, so we

2 . For more complicated expressions, there 2 2 2 2 are different opinions as to when denominators need to be rationalized. rewrite it:

=

1

2

=

(See Footnote 1)

When we rationalize the numerator of a fraction, we eliminate radicals and non-integer exponents in the numerator but possibly introduce them in the denominator. Example 3 (Rationalizing a Numerator) Re-express

x+h h

x

by rationalizing the numerator.

• This expression is an example of a difference quotient (see Section 1.10). Rationalizing the numerator will help us find a derivative (see Section 1.11).

§ Solution We will multiply the numerator and the denominator by the conjugate of

x+h h

x+h

x.

( =

x+h

x

x

h

)( (

x+h+ x,

) x)

x+h+ x x+h+

To multiply out the numerator, we use the rule: a b a + b = a 2 b2 .

(

)(

)

(Section 0.9: Simplifying Algebraic Expressions) 0.9.9

=

(

) ( ) h( x + h + x) x+h

2

x

2

WARNING 8: Do not forget the radical expression in the denominator.

=

= =

h

h

h

(

( x + h)

( (

x

x+h+ x x +h

x

x+h+ x h x+h+ x

) ) )

We now cancel the h factors and note the restriction h 0 .

(

)

(1) =

h

(1) = §

(

h x+h+ x 1

x+h+ x

)

( h 0)

( h 0)

(Section 0.9: Simplifying Algebraic Expressions) 0.9.10 PART F: COMMON ERRORS Beware of the following errors when rewriting or simplifying an expression. Errors involving Radicals and Exponents Error (crossed out)

Related Correct Formulas

Comments

a2 = a 3

a = a, if a < 0 2

a =a

( ) a

a+b = a + b a b= a

( a + b) ( a b)

The formula

3

2

= a, if a

1/ 2

= a1/ 2 + b1/ 2

1/ 2

= a1/ 2

b1/ 2

(

ab + ac = a b + c

if a

1/ 2

(

= a b+ c

)

There is no general formula for the square root of a sum (or a difference), as there is for a product or a quotient.

(

= a b+ c

(

1/ 2

)

= a b + c, 0 and b + c 0.

( ab + ac) 1/ 2

Do not apply radicals or exponents term-by-term.

a a , = b b if a 0 and b > 0 .

ab + ac = a b + c

( ab + ac)

often forgotten.

0

ab = a b , if a 0 and b 0 .

b

)

= a1/ 2 b + c if a

0 and b + c

a 2 = a is

1/ 2

)

1/ 2

0.

,

When factoring a radicand or the base of a power, make sure to apply the appropriate radical or exponent to all factors.

(Section 0.9: Simplifying Algebraic Expressions) 0.9.11 Errors involving Fractions Error (crossed out)

(1) x x +4 2

( x)

=

Related Correct Formulas

Comments

( x)

(

x2

x x+4

1 x+4

)

=

x x+4

(1) =

(

x2

( x 0)

)

x x x+4

(

0

We can cancel (nonzero, equivalent) factors of the entire numerator and the entire denominator.

)

If we add like fractions, we add the numerators and keep the common denominator.

a b a+b + = . 1 1 1

1 1 1 + = a b a+b

a d

1 1 1 b 1 + = + a b a b b b+ a = , or ab

b+c a b+c = d d 1 =x 1+x 2 x+x

a d

a b+c = d

a a a+b . ab

d

1 1 4 /3 = x (x 7x 4 /3 7

1 = 7x 4 /3 4 /3 7x

A fraction groups together its numerator (and groups together its denominator).

(b + c )

1 1 + 2 =x 1+x x x

2

If we add unlike fractions, we do not add the denominators.

Reciprocals cannot be taken term-by-term.

2

The negative exponent only applies to x.

0)

FOOTNOTES 1. Rationalizing denominators. The expression

2x (10x + 21)

from Example 2 is often 4/3 3( 4x + 7 ) considered simplified. If we were to rationalize the denominator, we would have (nontrivial) powers of 4x + 7 as factors of the numerator and the denominator:

(

)

2x (10x + 21) 3( 4x + 7 )

4/3

=

2x (10x + 21) 3( 4x + 7 )

4/3

( 4x + 7 )2 / 3 = 2x (10x + 21) ( 4x + 7 )2 / 3 2 3( 4x + 7 ) ( 4x + 7 )2 / 3

We may be inclined to divide those powers and go back to

2x (10x + 21) 3( 4x + 7 )

4/3

.

(Section 0.10: More Algebraic Manipulations) 0.10.1

SECTION 0.10: MORE ALGEBRAIC MANIPULATIONS LEARNING OBJECTIVES • Learn additional ways to write equivalent expressions.

PART A: DISCUSSION • The techniques of this section will help us rewrite expressions so that they are easier to manipulate. • They will also help us fit expressions to particular forms and templates. PART B: SPLITTING A FRACTION THROUGH ITS NUMERATOR Example 1 (Splitting a Fraction Through its Numerator) Fill in the boxes below with appropriate, simplified numbers:

7x + 3 = 4 x

x

+

x

§ Solution We begin by rewriting 4 x as x1/ 4 , because the template on the right-hand side suggests a preference for exponential form over radical form.

7x + 3 7x + 3 = 1/4 4 x x We will rewrite this as a sum by splitting it through its numerator.

7x1 3 = 1/4 + 1/4 x x 3/4 = 7x + 3x

1/4

(Section 0.10: More Algebraic Manipulations) 0.10.2

7x + 3 Answer: 4 = x

7 x

3 4

+

3 x

1 4

• In calculus, the right-hand side is much easier to differentiate and integrate. §

WARNING 1: Do not split a fraction in this way through its denominator. 1 1 1 For example, is not equivalent to + . x+y x y PART C: MAKING COMPOUND FRACTIONS When we divide by a nonzero number, we multiply by its reciprocal. When we multiply by a nonzero number, we divide by its reciprocal. Example 2 (Making Compound Fractions) Fill in the boxes below with appropriate, simplified numbers:

9x 2 +

3y 2 = 4

x2

+

y2

§ Solution

3y 2 9 2 3 2 9x + = x + y 1 4 4 2

WARNING 2:

3 2 3 y . is not equivalent to 2 4 4y

x2 y2 = + 1 4 9 3

3y 2 Answer: 9x + = 4 2

x2 1 9

+

y2 4 3

§ • This technique will be used in Chapter 10 to set up standard forms for equations of conics.

(Section 0.10: More Algebraic Manipulations) 0.10.3 PART D: COMPENSATION Sometimes, we can alter an expression if we compensate for the change and maintain equivalence. Example 3 (Compensation Using Addition and Subtraction) Fill in the box below with an appropriate, simplified expression:

x = 1 x +1 § Solution

x +1 =1 x 1 , so we would like to add 1 to the x +1 x numerator of . To compensate for that, we must also subtract 1 from x +1 the numerator. We know that

(

)

WARNING 3: If we add 1 to the numerator of a fraction, we cannot 1 2 . compensate by adding 1 to the denominator. For example, 2 3

x x +1 1 = x +1 x +1 x +1 1 = x +1 x +1 1 =1 x +1 Answer:

x = 1 x +1

1 x +1

§ • A similar compensation technique is used when Completing the Square (CTS) in Sections 0.11, 0.13, and 2.2 and Chapter 10.

(Section 0.10: More Algebraic Manipulations) 0.10.4 Example 4 (Compensation Using Multiplication and Division) Fill in the box below with an appropriate, simplified number:

( 4x + 1)5 =

( 4x + 1)5 ( 4 )

§ Solution The expression on the left-hand side has been multiplied by 4. We compensate by dividing by 4, or by multiplying by its reciprocal,

Answer:

( 4x + 1)5 =

1 4

( 4x + 1)5 ( 4 )

• In calculus, this technique helps us integrate using the u-substitution method.

1 . 4

(Section 0.11: Solving Equations) 0.11.1

SECTION 0.11: SOLVING EQUATIONS LEARNING OBJECTIVES • Know how to solve linear, quadratic, rational, radical, and absolute value equations. PART A: DISCUSSION • Much of precalculus is devoted to solving equations of various types. In this section, we will solve basic algebraic equations. • We will solve polynomial equations more generally in Chapter 2, exponential and logarithmic equations in Chapter 3, and trigonometric equations in Chapter 4. We will solve systems of equations in Chapters 7 and 8. PART B: SOLVING EQUATIONS A solution to an equation in x is a number that makes the equation true when the number is substituted for x. For now, we only consider real solutions. We solve an equation by finding its solution set, the set of all solutions. When solving an equation, we often write a sequence of equivalent equations, which have the same solution set. • Adding, subtracting, multiplying by, and dividing by the same nonzero number on both sides of an equation maintains equivalence. Example Set 1 (Solving Equations) • When we solve the linear equation 2x = 6 , we divide both sides by 2 and obtain the equivalent equation x = 3 . The solution set of both equations is 3 . We can check a solution by verifying that it satisfies the original

{}

()

equation: 2 3 = 6 , so 3 checks out. • The equation x = x + 1 has no real solutions. Its solution set is ∅ , the empty set (or null set). • The equation x + 1 = x + 1 is solved by all real numbers. Its solution set is  , so the equation is automatically called an identity.

1 1 = is solved by all nonzero real numbers. Its solution set x x is  \ {0} , also written as ( − ∞, 0 ) ∪ ( 0, ∞ ) . The equation is an identity, in part because the only excluded real number (0) corresponds to a restriction. § • The equation

(Section 0.11: Solving Equations) 0.11.2 WARNING 1: There is a difference between simplifying an expression and solving an equation. For example, when we simplify the expression 2x + x , we write “ 2x + x = 3x ,” and 3x is our answer. On the other hand, when we solve the equation 2x + x = 3x , we state that the solution set is  . PART C: SOLVING QUADRATIC EQUATIONS The general form of a quadratic equation in x is given by: ax 2 + bx + c = 0 , where a ≠ 0 . Its solutions are given by the Quadratic Formula:

− b ± b2 − 4ac x= 2a • Sometimes, the solutions are not real, but imaginary (see Chapter 2). The discriminant of ax 2 + bx + c is b2 − 4ac , the radicand in the formula. • It may be denoted by D or Δ (uppercase delta). • It helps us classify (or “discriminate between”) types of solutions. WARNING 2: Make sure the fraction bar in the formula goes all the way

b2 − 4ac across. The formula is not: x = − b ± . 2a WARNING 3: Here, the plus-minus sign ± indicates that we take both the result from the “+” case and the result from the “ − ” case. (The results are equal ⇔ the discriminant b2 − 4ac is 0.) Sometimes in precalculus, the ± sign indicates that we do not yet know which sign to take. Example 2 (Using the Quadratic Formula) Solve the equation 2x 2 − 7x = 15 using the Quadratic Formula. § Solution WARNING 4: We must rewrite the equation in general form before we apply the Quadratic Formula. We must isolate 0 on one side of the equation. (Sometimes, it is easier to isolate 0 on the left-hand side.)

2x 2 − 7x = 15 2x 2 − 7x − 15 = 0 TIP 1: If we had been asked to solve 4x 2 − 14x − 30 = 0 , we could have easily divided both sides by 2.

(Section 0.11: Solving Equations) 0.11.3 TIP 2: To avoid sign errors, we will identify a, b, and c before we apply the Quadratic Formula. Make sure to get them from the general form! Here, a = 2 , b = − 7 , and c = −15 .

− b ± b2 − 4ac x= 2a =

− (− 7) ±

( − 7 ) − 4 ( 2)( −15) = 7 ± 2 ( 2) 2

49 + 120 7 ± 169 = 4 4

WARNING 5: We must simplify the radical. (If we had obtained, say, to simplify.)

=

7± 2 , there would be no need 4

7 ± 13 4 “+” case : x =

7 + 13 20 = =5 4 4

“ − ” case: x =

7 − 13 − 6 3 = =− 4 4 2

⎧ 3 ⎫ The solution set is: ⎨− , 5⎬ . ⎩ 2 ⎭ (Some instructors list solutions in increasing order, although solution sets are technically unordered, meaning order doesn’t matter.) § We also use the Factoring Method, the Square Root Method, and the Completing the Square (CTS) Method to solve quadratic equations. The Factoring Method for solving equations relies on the following Zero Factor Property. Zero Factor Property (or Zero Product Property) If a and b represent real quantities, then: ab = 0 ⇔ a = 0 or b = 0 .

(

)

• Essentially, a product is 0 ⇔ a factor is 0, provided all factors are defined.

(Section 0.11: Solving Equations) 0.11.4 Example 3 (Using the Factoring Method; Revisiting Example 2) Solve 2x 2 − 7x = 15 using the Factoring Method. § Solution WARNING 6: Again, we must first isolate 0 on one side.

2x 2 − 7x = 15 2x 2 − 7x − 15 = 0

( 2x + 3) ( x − 5) = 0 By the Zero Factor Property,

2x + 3 = 0 x=−

or 3 2

x −5= 0 x=5

⎧ 3 ⎫ Again, the solution set is: ⎨− , 5⎬ . § ⎩ 2 ⎭ In Example 2, the discriminant of 2x 2 − 7x − 15 was 169, a perfect square, which led to the elimination of the radical sign. As a result, 2x 2 − 7x − 15 = 0 had rational numbers as solutions. Also, by the Test for Factorability from Section 0.7, 2x 2 − 7x − 15 can be factored over the integers,  . In Example 3, we used the Factoring Method as a quicker alternative to the Quadratic Formula when solving 2x 2 − 7x − 15 = 0 , and we obtained the same rational numbers as solutions. Test for Factorability and Types of Solutions The Test for Factorability applies to ax 2 + bx + c , where a, b, and c are nonzero integers. (Assume the GCF is 1 or −1 ; otherwise, factor it out.) If the discriminant b2 − 4ac is … … a perfect square … in fact, 0 … not a perfect square

Then, ax 2 + bx + c …

(Distinct) solutions to ax 2 + bx + c = 0

… can be factored over  … and is a PST

two rational numbers one rational number

(Perfect Square Trinomial)

(a “double root”)

… is prime over 

… and positive

two irrational numbers

… and negative

two imaginary numbers (see Chapter 2)

(Section 0.11: Solving Equations) 0.11.5 Square Root Method If d > 0 , then x 2 = d ⇔ x = ± d . If d = 0 , then x 2 = d ⇔ x 2 = 0 ⇔ x = 0 . If d < 0 , then x 2 = d has no real solutions. • For example, x 2 = 3 ⇔ x = ± 3 , while x 2 = − 3 has no real solutions. WARNING 7: Remember that squares of real numbers are never negative. • This method can be extended to u 2 = d , where u is an expression in x or some other variable. • This is often the quickest method for solving ax 2 + bx + c = 0 when b = 0 . Example 4 (Using the Square Root Method) Solve 3x 2 − 4 = 0 using the Square Root Method. § Solution We begin by isolating x 2 on one side of the equation.

3x 2 − 4 = 0 3x 2 = 4 4 x2 = 3 We now apply the Square Root Method.

x=±

4 3

WARNING 8: Do not forget the “ ± ” sign. WARNING 9: If we have a numerical fraction as a radicand, we usually have to simplify. Here, we will rationalize the denominator.

x=± x=±

2 3 2 3 3

⎧⎪ 2 3 2 3 ⎫⎪ , Technically, the solution set is: ⎨− ⎬. § 3 3 ⎪⎩ ⎪⎭

(Section 0.11: Solving Equations) 0.11.6 Completing the Square (“CTS”) Method This method creates a perfect square trinomial (PST), which can be factored as the square of a binomial. That square is then isolated, and the Square Root Method is applied. • This method can be easy to apply if a = 1 and b is even. (Other cases will be discussed in Chapters 2 and 10.) • The Quadratic Formula can be derived using this method. • CTS will be used to set up standard forms for equations of conics in Sections 0.13 and Chapters 2 and 10. Example 5 (Using the “CTS” Method) Solve x 2 + 8x + 5 = 0 using the “CTS” Method. § Solution We begin by isolating the x 2 and x terms on one side of the equation and isolating a constant term on the other side.

x 2 + 8x + 5 = 0 x 2 + 8x = − 5 The coefficient of x 2 is 1, so we may now complete the square. We accomplish this by adding 16 to both sides of the equation. Why 16? We take the coefficient of x (here, 8), halve it (resulting in 4), and then square the result (the square of 4 is 16).

x 2 + 8x + 16 = −5 + 16 WARNING 10: Remember to add 16 to the right-hand side, also. We now have a PST on the left-hand side.

( x + 4)

2

= 11

x + 4 = ± 11

( by factoring the PST ) ( by the Square Root Method )

x = − 4 ± 11

{

}

Technically, the solution set is: − 4 − 11, − 4 + 11 . §

(Section 0.11: Solving Equations) 0.11.7 PART D: SOLVING RATIONAL EQUATIONS We often solve a rational equation by first multiplying both sides by the LCD. WARNING 11: Indicate restrictions that are “hidden” by this step. Example 6 (Solving a Rational Equation) Solve

9 1 = . x 3 4x

§ Solution We multiply both sides by the LCD, 4x 3 .

9 1 = 3 4x x 9 1 4x 3 = 4x 3 3 4x x 36 = x 2 ( x ≠ 0)

( )

( )

x 2 = 36

( x ≠ 0)

x = ±6

{

}

The solution set is: − 6, 6 . WARNING 12: If we had obtained x = 0 , we would have had to reject it. § PART E: SOLVING RADICAL EQUATIONS We often solve a radical equation by isolating radicals on one or both sides of the equation and then raising both sides to the appropriate positive integer power. WARNING 13: If we raise both sides of an equation to an even power at any step, we must check any tentative solutions at the end and reject extraneous solutions. (Raising to an odd power does not require such a check.) • Observe that, if we square both sides of x = 2 , we obtain x 2 = 4 , which has both 2 and − 2 as solutions. However, − 2 is an extraneous solution that must be rejected. • Although x = 2 ⇒ x 2 = 4 , the equations x = 2 and x 2 = 4 are not equivalent. That is, x = 2 ⇔ x 2 = 4 . We lose “reversibility” here.

(Section 0.11: Solving Equations) 0.11.8 Example 7 (Solving a Radical Equation) Solve

x + 4 = x 2 + x − 21 .

§ Solution We square both sides of the equation. Since it is cumbersome to write restrictions that are “hidden” by this step, namely x + 4 ≥ 0 and x 2 + x − 21 ≥ 0 , we will instead check our tentative solutions at the end.

x + 4 = x 2 + x − 21 ⇒

(

x+4

) =( 2

x + x − 21 2

)

2

x + 4 = x 2 + x − 21 25 = x 2 x 2 = 25 x = ±5 We must check our tentative solutions.

(5) + 4 = (5) + (5) − 21 2

x = 5 checks out:

TIP 3: Beyond mechanical errors, we need to check that radicands of even roots are nonnegative.

9= 9 3= 3

x = − 5 does not check out:

( − 5) + 4 = ( − 5) + ( − 5) − 21 2

−1 = −1 WARNING 14: We must reject − 5 , because it yields a non-real expression,

{}

The solution set is: 5 . §

−1 , in the check.

(Section 0.11: Solving Equations) 0.11.9. PART F: SOLVING ABSOLUTE VALUE EQUATIONS Solving Absolute Value Equations If d > 0 , then x = d ⇔ x = ± d . If d = 0 , then x = d ⇔ x = 0 ⇔ x = 0 . If d < 0 , then x = d has no solutions. • For example, x = 3 ⇔ x = ± 3 , while x = − 3 has no solutions. WARNING 15: Remember that absolute values are never negative. • This method can be extended to u = d , where u is an expression in x or some other variable. Example 8 (Solving an Absolute Value Equation) Solve x − 1 = 2 . § Solution

x −1 = 2 x −1= ±2 “ − ” case:

“+” case :

x −1= 2 x=3

{

x −1= −2 x = −1

}

The solution set is: −1, 3 . • Observe that −1 and 3 lie at a distance of two units away from 1 on the real number line. This is consistent with our discussion of absolute value and distance in Section 0.4.

§

(Section 0.12: Solving Inequalities) 0.12.1

SECTION 0.12: SOLVING INEQUALITIES LEARNING OBJECTIVES • Know how to solve linear inequalities and absolute value inequalities. PART A: DISCUSSION • We will solve inequalities when we perform sign analyses and find domains of radical functions (see Section 1.1 and Chapter 2). • Absolute value inequalities allow us to write compound inequalities more efficiently. They also help us describe an interval on the real number line with respect to its center. • We will solve nonlinear inequalities in Chapter 2. PART B: SOLVING LINEAR INEQUALITIES Strict inequalities involve the “ < ” (is less than) or the “ > ” (is greater than) signs. Weak inequalities involve the “ ≤ ” (is less than or equal to) or the “ ≥ ” (is greater than or equal to) signs. Inequations involve the “ ≠ ” (is not equal to) sign. Solving linear inequalities is similar to solving linear equations, except that: • An inequality typically has infinitely many solutions, and the solution set is often written in interval form. • WARNING 1: We must reverse the direction of the inequality sign if we switch the sides of an inequality, or if we multiply or divide both sides by a negative number. Solving inequations is also similar to solving equations, although “ ≠ ” never needs to be reversed.

(Section 0.12: Solving Inequalities) 0.12.2 Example 1 (Solving a Linear Inequality) Solve − 3x > x + 8 . § Solution Method 1

− 3x > x + 8

Now subtract x from both sides.

− 4x > 8

Now divide both sides by − 4 . We must reverse the direction of the inequality sign.

x < −2 The solution set … … in set-builder form is:

{x ∈ x < − 2} , or { x ∈ : x < − 2}

… in graphical form is: … in interval form is:

( − ∞, − 2)

§ § Solution Method 2

− 3x > x + 8 − 8 > 4x

Now add 3x to, and subtract 8 from, both sides. Now switch sides. We must reverse the direction of the inequality sign.

4x < − 8 x < −2 See Method 1 for the solution set. §

(Section 0.12: Solving Inequalities) 0.12.3 PART C: SOLVING ABSOLUTE VALUE INEQUALITIES Solving Absolute Value Inequalities If d > 0 , then:

x < d ⇔ − d < x < d , and x ≤ d ⇔ −d ≤ x ≤ d . Also,

( ⇔ (x ≥ d

) or x ≤ − d ) .

x > d ⇔ x > d or x < − d , and x ≥d

• TIP 1: Think of x as the distance between x and 0 on the real number line. • For example, x < 1 ⇔ −1 < x < 1 . This is a compound inequality that means:

(

)

x > −1 and x < 1. The solution set is the interval −1, 1 . It is the set of numbers

that lie strictly within one unit of 0 on the real number line.

(

)

• Also, x > 1 ⇔ x > 1 or x < −1 . This is a different kind of compound

(

) ( )

inequality. The solution set is − ∞, −1 ∪ 1, ∞ . It is the set of numbers that are further than one unit from 0 on the real number line.

WARNING 2: Students incorrectly write 1 < x < −1 here.

(

)

This actually means x > 1 and x < −1 , which corresponds to ∅ , the empty (or null) set. • These methods can be extended to u , where u is an expression in x or some other variable.

(Section 0.12: Solving Inequalities) 0.12.4. Example 2 (Solving an Absolute Value Inequality; Related to Section 0.11, Example 8) Solve x − 1 < 2 . § Solution

x −1 < 2 −2 < x −1< 2 We can add 1 to all three parts of this compound inequality.

−1 < x < 3 The solution set … … in set-builder form is:

{x ∈ −1 < x < 3} , or { x ∈ : −1 < x < 3}

… in graphical form is: … in interval form is:

( −1, 3)

• The solution set is the set of numbers that lie strictly within two units of 1 on the real number line. This is consistent with our discussion of absolute value and distance in Section 0.4.

§

(Section 0.13: The Cartesian Plane and Circles) 0.13.1

SECTION 0.13: THE CARTESIAN PLANE and CIRCLES LEARNING OBJECTIVES • Understand the Cartesian plane and associated terminology. • Know how to plot points and graph equations in the Cartesian plane. • Know the Distance and Midpoint Formulas. • Be able to recognize, write, standardize, and graph equations of circles. PART A: DISCUSSION • We typically graph an equation in x and y in the Cartesian (or rectangular) plane, named after René Descartes. We plot points using their Cartesian coordinates. • There are alternate coordinate systems. We will discuss polar coordinates in Chapter 6. In three dimensions, we use Cartesian, cylindrical, and spherical coordinates.

PART B: THE CARTESIAN (OR RECTANGULAR) PLANE The Cartesian (or rectangular) plane is a plane with the Cartesian coordinate system imposed on it. We usually graph in the Cartesian xy-plane, though other variables could be used. We locate a horizontal line called the x-axis and a vertical line called the y-axis. These coordinate axes are real number lines; at least one nonzero tick mark should be placed on each axis to indicate scale. The axes intersect at the origin, O. A point in the plane corresponds to an ordered pair of the form x-coordinate, y-coordinate . For example, the origin O corresponds to the ordered

(

( )

)

( )

pair 0, 0 , and the red point below corresponds to 2, 3 . If we name the red point

( )

P, we can write P 2, 3 .

(Section 0.13: The Cartesian Plane and Circles) 0.13.2 PART C: DISTANCE AND MIDPOINT FORMULAS Distance Formula

(

)

(

)

The distance between points P x1 , y1 and Q x2 , y2 in the Cartesian plane is given by:

d=

( x2

x1 ) + ( y2 2

y1 )

2

or, equivalently,

( x1

x2 ) + ( y1

y2 )

2

2

• This is proven using the Pythagorean Theorem, which we will discuss in Chapter 4 on trigonometry.

Midpoint Formula The midpoint of PQ , the line segment with endpoints P and Q, is given by: x1 + x2 y1 + y2 , 2 2 • Observe that the x-coordinate is the average of the x-coordinates of the endpoints, and the y-coordinate is the average of the y-coordinates. • For example, in the figure below, the distance between the points

( )

and 3, 3 is: d =

( 3 ( 2 )) + ( 3 1) 2

2

(

)

2, 1

= 29 . If the coordinate axes are

scaled in (say) meters, then the distance is

29 meters.

• The midpoint M of the red line segment is:

1 2 + 3 1+ 3 = , ,2 2 2 2

(Section 0.13: The Cartesian Plane and Circles) 0.13.3 PART D: THE GRAPH OF AN EQUATION and CIRCLES The Graph of an Equation; the “Basic Principle of Graphing” The graph of an equation consists of all points whose coordinates satisfy the equation. The points correspond to solutions of the equation. Circles A circle is the set of all points in a plane that are a fixed distance (r, the radius) away from a fixed point (the center). The diameter d is twice the radius. Distance and Squared Distance from the Origin

(

)

( )

x 2 + y 2 is the distance between a point x, y and the origin 0, 0 .

x 2 + y 2 is the squared distance between them.

( )

Equation of a Circle with Center 0, 0

The standard form of the equation of a circle (in the xy-plane) with center 0, 0 and radius r, where r > 0 , is given by:

( )

x 2 + y2 = r 2

(

)

• This is because such a circle consists of all points x, y whose squared

( )

distance from 0, 0 is r 2 .

(Section 0.13: The Cartesian Plane and Circles) 0.13.4 Example 1 (The Graph of an Equation; A Circle Centered at the Origin)

( )

The graph of x 2 + y 2 = 9 is the circle below with center 0, 0 and radius 3.

( ) (

The ordered pairs 3, 0 ,

) ( )

(

)

3, 0 , 0, 3 , and 0, 3 are all solutions of

the equation x 2 + y 2 = 9 , and their corresponding points lie on the circle. Other points such as

(

)

7, 2 also lie on the circle.

§

( )

Equation of a Circle with Center h, k

( )

More generally, if the circle has center h, k , the standard form is given by:

(x

h) + ( y k ) = r 2 2

2

(

)

• The left-hand side is the squared distance between the points x, y and

( h, k ) . The circle consists of all points ( x, y ) whose squared distance from ( h, k ) is r . 2

(Section 0.13: The Cartesian Plane and Circles) 0.13.5 Example 2 (Finding the Equation of a Circle) Find an equation of the circle in the xy-plane with center

(

)

2, 1 and

radius 3. § Solution

( ( )) + ( y 1) = (3) , which we usually

We obtain the equation x

(

) ( 2

2

)

2

2

2

2

rewrite as: x + 2 + y 1 = 9 . WARNING 1: The center of the circle is

(

)

(

)

2, 1 , not 2, 1 . If we are

given the equation of a circle in standard form, we can find the center by asking, “What makes the left-hand side equal to 0?” (The center has a squared distance of 0 from itself.) § The circles from Examples 1 and 2 are graphed below:

The circles are translations of one another. (See Section 1.4.)

(Section 0.13: The Cartesian Plane and Circles) 0.13.6 Example 3 (Finding the Standard Form of the Equation of a Circle) A circle has as its equation:

4x 2 + 4 y 2 16x + 4 y 11 = 0 Find the standard form of this equation, and identify the center and the radius of the circle. § Solution The common coefficient of x 2 and y 2 is 4, so we will divide both sides of the equation by 4.

4x 2 + 4 y 2 16x + 4 y 11 = 0 x2 + y2

4x + y

11 =0 4

Now, we group together the x 2 and x terms, and we group together the y 2 and y terms. We isolate constant terms on the right-hand side.

(x

) (

)

4x + y 2 + y =

2

11 4

We now Complete the Square (CTS) in both groups.

(x

2

)

4x + 4 + y 2 + y +

11 1 1 = +4+ 4 4 4

WARNING 2: Do not forget to add 4 and the right-hand side, also.

1 to 4

We now factor both of the resulting Perfect Square Trinomials (PSTs).

(

1 x 2 + y+ 2

)

2

2

=7

We now have the desired form, although the equation could be rewritten as:

( x 2) The circle has center 2,

2

1 2

+ y

1 2

and radius

2

=7

7.§

(Section 0.14: Lines) 0.14.1

SECTION 0.14: LINES LEARNING OBJECTIVES • Understand and compute the slope of a line. • Distinguish between equations of horizontal lines and those of vertical lines. • Know how to write equations of lines in various forms, including Point-Slope Form, Slope-Intercept Form, and Two-Intercept Form. • Understand parallel and perpendicular lines and relate their slopes. • Know how to find the intercepts of a line.

PART A: DISCUSSION • We frequently graph lines in precalculus and calculus. • In this section, we will graph lines in the xy-plane, though we can work with different variables. • There are many ways to write an equation for a line. The form we select may depend on the information we have about the line, or on the information we want to find or display. • In Section 1.11, we will discuss linear approximations of functions and graphs by tangent lines, a crucial idea in calculus.

(Section 0.14: Lines) 0.14.2 PART B: NOTATION Assume that m, a, b, c, x1 , x2 , y1 , y2 , A, B, and C represent real numbers. Let ( x1 , y1 ) and ( x2 , y2 ) be two distinct points on a line. m is the slope of the line (if defined).

( ) b, or the point ( 0, b) , is the y-intercept of the line (if there is exactly one).

a, or the point a, 0 , is the x-intercept of the line (if there is exactly one), and

(Section 0.14: Lines) 0.14.3 PART C: m, THE SLOPE OF A LINE Formulas for m, the Slope of a Line

m=

rise y y2 = = run x x2

y1 x1

or, equivalently,

y1 x1

y2 x2

• If the line is vertical, then m is undefined. • A negative rise can be interpreted as a drop. •

(uppercase delta) denotes “change in.”

Interpretation of Slope m as Marginal Change For every unit increase in x along the line, y changes by m. • This idea will be developed in Section 1.11.

(Section 0.14: Lines) 0.14.4 Interpreting the Sign of Slope m m>0 m=0 m