Precalculus Review I Precalculus Review II The Cartesian Coordinate System Straight Lines

11

Precalculus Review IPrecalculus Review I Precalculus Review IIPrecalculus Review II The Cartesian Coordinate SystemThe Cartesian Coordinate System Straight LinesStraight Lines

PreliminariesPreliminaries

1.11.1Precalculus Review IPrecalculus Review I

– 4 – 3 – 2 – 1 0 1 2 3 4

Origin

Positive DirectionNegative Direction

2 –

The Real Number LineThe Real Number Line

We can represent real numbers We can represent real numbers geometricallygeometrically by points on by points on a a real numberreal number, or, or coordinate coordinate,, line line::

This line includes This line includes allall real numbersreal numbers.. Exactly one point on the lineExactly one point on the line is associated with is associated with each real each real

numbernumber, and vice-versa., and vice-versa.

– 4 – 3 – 2 – 1 0 1 2 3 4

Origin


2

Finite IntervalsFinite Intervals

Open IntervalsOpen Intervals The The set of real numbersset of real numbers that lie that lie strictlystrictly between two fixed between two fixed

numbersnumbers aa and and bb is called an is called an open intervalopen interval ((aa, , bb)).. It consists of all the real numbers that It consists of all the real numbers that satisfy the satisfy the

inequalitiesinequalities a < x < ba < x < b.. It is called It is called “open”“open” because because neither of its endpoints is neither of its endpoints is

includedincluded in the interval. in the interval.


Closed IntervalsClosed Intervals The The set of real numbersset of real numbers that lie that lie between two fixed numbersbetween two fixed numbers

aa and and bb, , that includesthat includes aa and and bb, is called a , is called a closed intervalclosed interval [[aa, , bb]].. It consists of all the real numbers that It consists of all the real numbers that satisfy the inequalitiessatisfy the inequalities

a a x x b b.. It is called It is called “closed”“closed” because because both of its endpoints are includedboth of its endpoints are included

in the interval.in the interval.


Half-Open Intervals Half-Open Intervals The The set of real numbersset of real numbers that that between two fixed numbersbetween two fixed numbers aa

and and bb, , that contains that contains only oneonly one of its endpoints of its endpoints aa or or bb, is , is called a called a half-open intervalhalf-open interval ((aa, , bb]] or or [[aa, , bb))..

It consists of all the real numbers that It consists of all the real numbers that satisfy the satisfy the inequalitiesinequalities a a << x x b b or or a a x x << b b..

Infinite IntervalsInfinite Intervals

Examples of infinite intervals include:Examples of infinite intervals include: ((aa, , )),, [ [aa, , )),, (– (–, , aa)), and, and (– (–, , aa]].. The above are defined, respectively, by the set of real The above are defined, respectively, by the set of real

numbers that satisfy numbers that satisfy xx > > aa, , x x a a, , x x << a a, , x x a a..

Exponents and RadicalsExponents and Radicals

If If b b is any is any real numberreal number and and n n is a is a positive integerpositive integer, then the , then the expression expression bbnn is defined as the number is defined as the number

bbnn = b = b · · b b b b ·· · … · · … · bb

The number The number bb is called the is called the basebase, and the superscript , and the superscript nn is is called the called the powerpower of the of the exponential expressionexponential expression bbnn..

For example: For example:

If If bb ≠ 0≠ 0, we define , we define bb00 = 1 = 1.. For example:For example:

2200 = 1 = 1 and and (–(–))00 = 1 = 1, but , but 0000 is undefined. is undefined.

n factors

52 2 2 2 2 2 32

32 2 2 2 8

3 3 3 3 27

Exponents and RadicalsExponents and Radicals

If If nn is a is a positive integerpositive integer, then the expression , then the expression bb1/1/nn is defined is defined to be the number that, when raised to the to be the number that, when raised to the nnthth power, is power, is equal to equal to bb, thus, thus

Such a number is called the Such a number is called the nnthth root ofroot of bb, also written as, also written as

Similarly, the expression Similarly, the expression bbpp//qq is defined as the number is defined as the number

((bb1/1/qq))pp or or Examples: Examples:

n b

q pb

3 33/2 1/22 2 1.41412 2.8283

5/2

55/2 51/2

1 1 1 14

4 2 324

((bb1/1/nn))nn = = bb

Laws of ExponentsLaws of Exponents

LawLaw ExampleExample

1.1. aamm · a· ann = a = am + nm + n xx22 · x· x33 = x = x22 + + 33 = = xx55

2.2.

3.3. ((aamm))nn = a = am · nm · n ((xx44))33 = x = x44 · · 33 = = xx1212

4.4. ((abab))nn = a = an n · b· bnn (2(2xx))44 = = 224 4 ·· xx 44 = =

1616xx44

5.5.

( 0) m

m nn

aa a

a ( 0)

mm n

n

aa a

a

77 4 3

4

xx x

x

77 4 3

4

xx x

x

n n

n

a a

b b

n n

n

a a

b b

3 3 3

32 2 8

x x x

3 3 3

32 2 8

x x x

ExamplesExamples SimplifySimplify the expressions the expressions

2 3

5/ 4

1/ 2

32/3

23 2

23/ 2

1/ 4

3 4

16

16

6

x x

x y

y

x

a.

b.

c.

d.

e.

2 3 512 12x x

33/4 3416 16 2 8

(2/3)(3) 6/3 26 6 6 36

4

2 23 2 (3)( 2) ( 2)( 2) 6 46

yx y x y x y

x

(3/2)( 2) 3 1/2

(1/4)( 2) 1/2 3

y y x

x x y

Example 1, page 6

ExamplesExamples SimplifySimplify the expressions the expressions

4 84

3 5

3 6

33

16

12 3

27

8

x y

m n m n

x

y

a.

b.

c.

1/44 8 1/4 4/4 8/4 216 16 2x y x y xy

1/28 2 8 2 1/2 8/2 2/2 436 36 36 6m n m n m n m n

1/36 1/3 6/3 2

1/3 1/3 3/33

27 27 3

8 28

x x x

y yy

(assume (assume xx, , yy, , mm, and , and nn are are positivepositive))

Example 2, page 7

ExamplesExamples RationalizeRationalize the the denominatordenominator of the expression of the expression

3

2

x

x

2

3

2

3

2

x x

x x

x x

x

3

23

2

x x

x

x

Example 3, page 7

ExamplesExamples RationalizeRationalize the the numeratornumerator of the expression of the expression

3

2

x

x2

3

2

3

2

x x

x x

x

x x

3

23

2

x

x x

x

Example 4, page 7

Operations With Algebraic ExpressionsOperations With Algebraic Expressions

An algebraic expression of the form An algebraic expression of the form axaxnn, where the , where the coefficient coefficient aa is a real number and is a real number and nn is a nonnegative is a nonnegative integer, is called a integer, is called a monomialmonomial, meaning it consists of , meaning it consists of one term.one term.

Examples:Examples:

77xx22 22xyxy 1212xx33yy44

A A polynomialpolynomial is a monomial or the sum of two or is a monomial or the sum of two or more monomials.more monomials.

Examples:Examples:

xx22 + 4 + 4xx + 4 + 4 xx44 + 3 + 3xx22 – 3 – 3 xx22yy – – xyxy + + yy

Operations With Algebraic ExpressionsOperations With Algebraic Expressions

Constant terms, or terms containing the same Constant terms, or terms containing the same variable factors are called variable factors are called likelike, or, or similar similar,, terms terms..

Like terms may be combined by adding or Like terms may be combined by adding or subtracting their numerical coefficients.subtracting their numerical coefficients.

Examples:Examples:

33x x + 7+ 7xx = 10 = 10x x 1212xyxy – 7 – 7xyxy = 5 = 5xyxy

ExamplesExamples SimplifySimplify the expression the expression

4 3 4 3 2(2 3 4 6) (3 9 3 )x x x x x x

4 3 4 3 22 3 4 6 3 9 3x x x x x x

4 4 3 3 22 3 3 9 3 4 6x x x x x x

4 3 26 3 4 6x x x x

Example 5, page 8

ExamplesExamples SimplifySimplify the expression the expression

3 22 { [ (2 1)] 4}t t t t

3 22 { [ 2 1] 4}t t t t

3 22 { [ 1] 4}t t t

3 22 { 1 4}t t t

3 22 1 4t t t

3 22 3t t t

Example 5, page 8

ExamplesExamples Perform Perform the operationthe operation andand simplify simplify the expression the expression

2 2( 1)(3 10 3)x x x

2 2 2(3 10 3) 1(3 10 3)x x x x x

4 3 2 23 10 3 3 10 3x x x x x

4 3 23 10 6 10 3x x x x

Example 6, page 8

ExamplesExamples Perform Perform the operationthe operation andand simplify simplify the expression the expression

( ) ( )t t t t t te e e e e e 2 0 2 0t te e e e

1 1

2

Example 6, page 9

FactoringFactoring

Factoring is the process of Factoring is the process of expressingexpressing an an algebraic algebraic expressionexpression as a as a productproduct of other of other algebraic expressions algebraic expressions..

Example:Example:23 (3 1)x x x x

FactoringFactoring

To factor an algebraic expression, first To factor an algebraic expression, first checkcheck to see if it to see if it contains any contains any common termscommon terms..

If so, If so, factor outfactor out the greatest common term the greatest common term.. For example, the greatest common factor for the For example, the greatest common factor for the

expressionexpression

is is 22aa, because, because

22 4 6a x ax a

22 4 6 2 32

(

2 2

2 2 3)

a x ax a axa

a ax

axa

x

ExamplesExamples Factor out Factor out thethe greatest common factorgreatest common factor in each expressionin each expression

2 2

2

3/ 2 1/ 2

3

0.3 3

2 3

2 2xy xy

t t

x x

ye xy e

a.

b.

c.

0.3 ( 10)t t

1/2(2 3)x x

2 22 (1 )xyye xy

Example 7, page 9

ExamplesExamples Factor out Factor out thethe greatest common factorgreatest common factor in each expressionin each expression

2 2

3 4 2 6

ax ay bx by

x y y x

a.

b.

2 ( ) ( )

(2 )( )

a x y b x y

a b x y

3 2 6 4

(3 2) 2(3 2)

(3 2)( 2)

x y y x

y x x

x y

Example 8, page 10

Factoring Second Degree PolynomialsFactoring Second Degree Polynomials

The The factorsfactors of the of the second-degree polynomialsecond-degree polynomial with integral with integral coefficientscoefficients

pxpx22 + qx + r + qx + r

are are ((ax + bax + b)()(cx + dcx + d)), where , where ac = pac = p, , ad + bc = qad + bc = q, and , and bd = rbd = r.. Since Since only a limited number of choices are possibleonly a limited number of choices are possible, we use , we use

a a trial-and-error methodtrial-and-error method to factor polynomials having this to factor polynomials having this form.form.

ExamplesExamples

Find the correct factorization for Find the correct factorization for xx22 – 2 – 2xx – 3 – 3

SolutionSolution The The xx22 coefficient coefficient is is 11, so the only possible first degree , so the only possible first degree

terms areterms are((xx )( )(xx ) )

The product of the The product of the constant termconstant term is is –3–3, which gives us the , which gives us the following possible factorsfollowing possible factors

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

We check to see which set of factors yields We check to see which set of factors yields –2–2 for the for the xx coefficientcoefficient: :

(–1)(–1)(1) + (1)(1) + (1)(3)(3) = 2 = 2 or or (1)(1)(1) + (1)(1) + (1)(–3)(–3) = –2 = –2

and conclude that the and conclude that the correct factorizationcorrect factorization is is

xx22 – 2 – 2xx – 3 = ( – 3 = (x x ++ 1)(1)(x x –3)–3)

ExamplesExamples

Find the correct factorization for the expressionsFind the correct factorization for the expressions

2

2

3 4 4

3 6 24

x x

x x

(3 2)( 2)x x

23( 2 8)

3( 4)( 2)

x x

x x

Example 9, page 11

Roots of Polynomial ExpressionsRoots of Polynomial Expressions

AA polynomial equation polynomial equation of degree of degree nn in the variable in the variable x x is an is an equation of the form equation of the form

where where nn is a is a nonnegative integernonnegative integer and and aa00, , aa11, … , , … , aann are are real real

numbersnumbers with with aann ≠ 0≠ 0..

For exampleFor example, the equation, the equation

is a polynomial equation of is a polynomial equation of degree 5degree 5..

11 0 0n n

n na x a x a

5 3 22 8 6 3 1 0x x x x

Roots of Polynomial ExpressionsRoots of Polynomial Expressions

TheThe roots roots of a polynomial equation are the values of of a polynomial equation are the values of xx that that satisfy the equation.satisfy the equation.

One way to factor the roots of a polynomial equation is to One way to factor the roots of a polynomial equation is to factor the polynomial and then solve the equation.factor the polynomial and then solve the equation.

For exampleFor example, the polynomial equation , the polynomial equation

may be written in the formmay be written in the form

For the product to be zero, For the product to be zero, at least one of the factors must at least one of the factors must be zerobe zero, therefore, we have, therefore, we have

xx = 0 = 0 xx – 1 = 0 – 1 = 0 xx – 2 = 0 – 2 = 0 So, So, the roots of the equationthe roots of the equation are are xx = 0 = 0, , 11, and , and 22..

2( 3 2) 0 ( 1)( 2) 0 or x x x x x x

3 23 2 0x x x

The Quadratic FormulaThe Quadratic Formula

The solutions of the equation The solutions of the equation

axax22 + + bxbx + + cc = 0 ( = 0 (aa ≠ 0)≠ 0)

are given byare given by

2 4

2

b b acx

a

2 4

2

b b acx

a

ExamplesExamples

Solve the equation using the Solve the equation using the quadratic formulaquadratic formula::

SolutionSolution For this equation, For this equation, aa = 2 = 2, , bb = 5 = 5, and , and cc = –12 = –12, so, so

22 5 12 0x x 22 5 12 0x x

22 5 5 4(2)( 12)4

2 2(2)

b b acx

a

22 5 5 4(2)( 12)4

2 2(2)

b b acx

a

5 121 5 11

4 43

42

or

5 121 5 11

4 43

42

or

Example 10, page 12

ExamplesExamples

Solve the equation using the Solve the equation using the quadratic formulaquadratic formula::

SolutionSolution First, rewrite the equation in the standard formFirst, rewrite the equation in the standard form

For this equation, For this equation, aa = 1 = 1, , bb = 3 = 3, and , and cc = – = – 88, so, so

2 3 8x x 2 3 8x x

22 3 3 4(1)( 8)4

2 2(1)

b b acx

a

22 3 3 4(1)( 8)4

2 2(1)

b b acx

a

2 3 8 0x x 2 3 8 0x x

3 41

2

3 41 3 411.7 4.7

2 2x x

so, or

3 41

2

3 41 3 411.7 4.7

2 2x x

so, or

Example 10, page 12

1.21.2Precalculus Review IIPrecalculus Review II

1 1h

h

2 21 11 1 1 1

1 1 1 1

hh h

h h h h

2 21 11 1 1 1

1 1 1 1

hh h

h h h h

1 1

1 1 1 1

1

1 1

h h

h h h h

h

1 1

1 1 1 1

1

1 1

h h

h h h h

h

Rational ExpressionsRational Expressions

Quotients of polynomials are called rational Quotients of polynomials are called rational expressions.expressions.

For exampleFor example

3 8

2 3

x

x

3 8

2 3

x

x

2 35 2

4

x y xy

x

2 35 2

4

x y xy

x

2

5ab

2

5ab

Rational ExpressionsRational Expressions

The The properties of real numbersproperties of real numbers apply to rational apply to rational expressions.expressions.

ExamplesExamples Using the properties of number we may writeUsing the properties of number we may write

where where aa, , bb, and , and c c are any real numbers and are any real numbers and bb and and c c are not zero.are not zero.

Similarly, we may writeSimilarly, we may write

1ac a c a a

bc b c b b 1

ac a c a a

bc b c b b

( 2)( 3) ( 2)( 2,3)

( 2)( 3) ( 2)

x x xx

x x x

( 2)( 3) ( 2)

( 2,3)( 2)( 3) ( 2)

x x x

xx x x

ExamplesExamples Simplify the expressionSimplify the expression

2

2

2 3

4 3

x x

x x

2

2

2 3

4 3

x x

x x

( 3)( 1)

( 3)( 1)

1

1

x x

x x

x

x

( 3)( 1)

( 3)( 1)

1

1

x x

x x

x

x

Example 1, page 16

ExamplesExamples Simplify the expressionSimplify the expression

2 2 2

2 4

( 1) ( 2) (2 )(2)( 1)(2 )

( 1)

x x x x

x

2 2 2

2 4

( 1) ( 2) (2 )(2)( 1)(2 )

( 1)

x x x x

x

2 2

2 4

2 2 2

2 4

( 1) ( 1)( 2) (2 )(2)(2 )

( 1)

( 1)( 2 2 8 )

( 1)

x x x x

x

x x x

x

2 2

2 4

2 2 2

2 4

( 1) ( 1)( 2) (2 )(2)(2 )

( 1)

( 1)( 2 2 8 )

( 1)

x x x x

x

x x x

x

2 2

2 4

2

2 3

2

2 3

( 1)(6 2)

( 1)

(6 2)

( 1)

2(3 1)

( 1)

x x

x

x

x

x

x

2 2

2 4

2

2 3

2

2 3

( 1)(6 2)

( 1)

(6 2)

( 1)

2(3 1)

( 1)

x x

x

x

x

x

x

Example 1, page 16

Rules of Multiplication and DivisionRules of Multiplication and Division

If If PP, , QQ, , RR, and , and SS are are polynomialspolynomials, then, then MultiplicationMultiplication

DivisionDivision

( , 0) P R PR

Q SQ S QS ( , 0)

P R PRQ S

Q S QS

( , , 0) P R P S PS

Q R SQ S Q R QR ( , , 0)

P R P S PSQ R S

Q S Q R QR

ExampleExample Perform the indicated operation and simplifyPerform the indicated operation and simplify

2

2

2 8 4 4

2 16

x x x

x x

2

2

2 8 4 4

2 16

x x x

x x

22( 4) ( 2)

2 ( 4)( 4)

2( 4)( 2)( 2)

( 2)( 4)( 4)

2( 2)

4

x x

x x x

x x x

x x x

x

x

22( 4) ( 2)

2 ( 4)( 4)

2( 4)( 2)( 2)

( 2)( 4)( 4)

2( 2)

4

x x

x x x

x x x

x x x

x

x

Example 2, page 16

Rules of Addition and SubtractionRules of Addition and Subtraction

If If PP, , QQ, , RR, and , and SS are are polynomialspolynomials, then, then AdditionAddition

SubtractionSubtraction

( 0)P Q P Q

RR S R

( 0)

P Q P QR

R S R

( 0)P Q P Q

RR S R

( 0)

P Q P QR

R S R

ExampleExample Perform the indicated operation and simplifyPerform the indicated operation and simplify

1 1

x h x

( )

( )

( )

( )

x x h

x x h

x x h

x x h

h

x x h

Example 3b, page 17

Other Algebraic FractionsOther Algebraic Fractions

The techniques used to simplify The techniques used to simplify rational expressionsrational expressions may may also be used to simplify algebraic fractions in which the also be used to simplify algebraic fractions in which the numeratornumerator and and denominatordenominator are are not polynomialsnot polynomials..

ExamplesExamples SimplifySimplify

11

14

x

xx

2 2

1 121

4 1 4

xx xx

x x xx

2

1 ( 2)( 2)

( 1)( 2)

x x

x x x

x

x x

Example 4a, page 18

ExamplesExamples SimplifySimplify

1 1

2 2

x y

x y

2 2

2 2 2 2

1 1

1 1

y xx y xy

y xx y x y

2 2 2

2 2

( )

( )( )

y x x y y x xy

xy y x xy y x y x

xy

y x

Example 4b, page 18

Rationalizing Algebraic FractionsRationalizing Algebraic Fractions

When the denominator of an algebraic fraction contains When the denominator of an algebraic fraction contains sums or differences involving radicals, we may rationalize sums or differences involving radicals, we may rationalize the denominator.the denominator.

To do so we make use of the fact thatTo do so we make use of the fact that

2 2

a b a b a b

a b

ExamplesExamples Rationalize the Rationalize the denominatordenominator

22

1

1

1

1

x

x

x

x

1

1 x1 1

1 1

x

x x

Example 6, page 19

ExamplesExamples Rationalize the Rationalize the numeratornumerator

1 1h

h

2 21 11 1 1 1

1 1 1 1

hh h

h h h h

2 21 11 1 1 1

1 1 1 1

hh h

h h h h

1 1

1 1 1 1

1

1 1

h h

h h h h

h

1 1

1 1 1 1

1

1 1

h h

h h h h

h

Example 7, page 19

Properties of InequalitiesProperties of Inequalities

If If aa, , bb, and , and cc, are any , are any real numbersreal numbers, then, then

Property 1Property 1 If If aa << bb and and b < cb < c, then , then a < ca < c..

Property 2Property 2 If If aa << bb, then , then a + c < b + ca + c < b + c..

Property 3Property 3 If If aa << bb and and c > c > 00, then , then ac < bcac < bc..

Property 4Property 4 If If aa << bb and and c < c < 00, then , then ac > bcac > bc..

ExamplesExamples

Find the set of real numbers that satisfy Find the set of real numbers that satisfy

––1 1 2 2xx – 5 < 7 – 5 < 7SolutionSolution Add Add 55 to each member of the given double inequality to each member of the given double inequality

4 4 2 2xx < 12 < 12 Multiply each member of the inequality by Multiply each member of the inequality by ½½

2 2 xx < 6 < 6 So, the solution is the set of all values ofSo, the solution is the set of all values of x x lying in the lying in the

interval interval [2, 6)[2, 6)..

Example 8, page 20

ExamplesExamples

Solve the inequality Solve the inequality xx22 + 2 + 2xx – 8 < 0 – 8 < 0..SolutionSolution FactorizingFactorizing we get we get ((xx + 4)( + 4)(xx – 2) < 0 – 2) < 0.. For the product to be For the product to be negativenegative, the , the factors must have factors must have

opposite signsopposite signs, so we have , so we have two possibilitiestwo possibilities to consider: to consider:

1.1. The inequality holds if The inequality holds if ((xx + 4) < 0 + 4) < 0 andand ((xx – 2) > 0 – 2) > 0, , which meanswhich means xx < – < – 44, and, and xx > 2 > 2, but , but this is impossiblethis is impossible: : xx cannot meet these cannot meet these two conditionstwo conditions simultaneouslysimultaneously..

2.2. The inequality also holds if The inequality also holds if ((xx + 4) > 0 + 4) > 0 andand ((xx – 2) < 0 – 2) < 0, , which means which means xx > – > – 44, and, and xx < 2 < 2, or, or – – 4 < 4 < xx < 2 < 2..

So, the So, the solutionsolution is the set of all values of is the set of all values of x x lying in the lying in the interval interval (–(– 4, 2)4, 2)..

Example 9, page 20

ExamplesExamples

Solve the inequalitySolve the inequality

SolutionSolution For the quotient to be For the quotient to be positivepositive, the , the numerator numerator andand

denominator denominator must havemust have thethe same sign same sign, so we have , so we have two two possibilitiespossibilities to consider: to consider:

1.1. The inequality holds if The inequality holds if ((xx + 1) + 1) 0 0 andand ((xx – 1) < 0 – 1) < 0, , which meanswhich means xx – – 11, and, and xx < 1 < 1, both of these , both of these conditions are met only when conditions are met only when xx – – 11..

2.2. The inequality also holds if The inequality also holds if ((xx + 1) + 1) 0 0 andand ((xx – 1) > 0 – 1) > 0, , which means which means xx – – 11, and, and xx > 1 > 1, both of these , both of these conditions are met only when conditions are met only when xx > > 11..

So, the So, the solutionsolution is the set of all values of is the set of all values of x x lying in the lying in the intervals intervals (–(– , –, – 1] 1] andand (1, (1, ))..

10

1

x

x

1

01

x

x

Example 10, page 21

Absolute ValueAbsolute Value

The The absolute valueabsolute value of a number of a number a a is is denoted denoted |a||a| and is and is defineddefined by by

0

0

if

if

a aa

a a

0

0

if

if

a aa

a a

Absolute Value PropertiesAbsolute Value Properties

If If aa, , bb, and , and cc, are any , are any real numbersreal numbers, then, then

Property 5Property 5 |–|– a|a| == |a||a|

Property 6Property 6 |ab||ab| == |a| |b||a| |b|

Property 7Property 7 ((bb ≠ 0)≠ 0)

Property 8Property 8 |a + b| ≤ |a||a + b| ≤ |a| + + |b||b|

aa

b b

aa

b b

ExamplesExamples

Evaluate the expressionEvaluate the expression

||– 5| + 3– 5| + 3SolutionSolution Since Since – 5 < 0– 5 < 0, we see that , we see that

||– 5|– 5| = –(= –(– 5)– 5). . ThereforeTherefore

||– 5| + 3– 5| + 3 = –= – ((– 5) +3 – 5) +3

= 8 – = 8 – ≈ ≈ 4.85844.8584

Example 12a, page 22

ExamplesExamples

Evaluate the expressionEvaluate the expression

SolutionSolution Since Since , we see that , we see that

Similarly, Similarly, , so, so

Therefore,Therefore,

3 2 2 3

3 2 0 3 2 3 2

2 3 0 2 3 2 3

3 2 2 3 3 2 2 3

4 2 3

2 2 3

0.5359

Example 12b, page 22

ExamplesExamples

Evaluate the inequality Evaluate the inequality ||xx| | 5 5..

SolutionSolution If If xx 0 0, then , then ||xx| =| = x x, so , so ||xx| | 5 5 implies implies xx 5 5.. If If xx < 0 < 0, then , then ||xx| =| = – – x x , so , so ||xx| | 5 5 implies implies –– xx 5 5 or or xx –– 55.. So, So, ||xx| | 5 5 means means –– 55 xx ≤ 5 ≤ 5, and the , and the solutionsolution is is [[–– 5, 5]5, 5]..

Example 13, page 22

ExamplesExamples

Evaluate the inequality Evaluate the inequality |2|2x x – 3| – 3| 1 1..

SolutionSolution From our From our last examplelast example, we know that , we know that |2|2x x – 3| – 3| 1 1 is is

equivalent to equivalent to –1 –1 2 2x x – 3 – 3 1 1.. Adding Adding 33 throughout we get throughout we get 2 2 2 2x x 4 4.. Dividing by Dividing by 22 throughout we get throughout we get 1 1 x x 2 2, so the , so the

solutionsolution is is [1, 2][1, 2]..

Example 14, page 22

1.31.3The Cartesian Coordinate SystemThe Cartesian Coordinate System

C(h, k)

h

k r

P(x, y)

The Cartesian Coordinate SystemThe Cartesian Coordinate System

At the beginning of the chapter we saw a one-to-one At the beginning of the chapter we saw a one-to-one correspondence between the set of real numbers and the points correspondence between the set of real numbers and the points on a straight line on a straight line (one dimensional space)(one dimensional space)..

– 4 – 3 – 2 – 1 0 1 2 3 4


The The Cartesian coordinate systemCartesian coordinate system extends this concept to a extends this concept to a plane plane (two dimensional space)(two dimensional space) by adding a by adding a vertical axisvertical axis..

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4


The The horizontalhorizontal line is called the line is called the xx-axis-axis, and the , and the verticalvertical line is line is called the called the yy-axis-axis..

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4

xx

yy


The point where these two lines intersect is called the The point where these two lines intersect is called the originorigin..

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4

xx

yy

Origin


In theIn the xx-axis-axis, , positive numberspositive numbers are to the are to the rightright and negative and negative numbersnumbers are to the are to the leftleft of the origin. of the origin.

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4

xx

yy



In theIn the yy-axis-axis, , positive numberspositive numbers are are aboveabove and negative and negative numbersnumbers are are belowbelow the origin. the origin.

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4

xx

yy

Pos

itive

Dir

ecti

onN

egat

ive

Dir

ecti

on

(– 2, 4)

(– 1, – 2)

(4, 3)


A A pointpoint in the plane can now be represented uniquely in this in the plane can now be represented uniquely in this coordinate system by coordinate system by an ordered pair of numbersan ordered pair of numbers ((xx, , yy))..

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4

xx

yy

(3,–1)


The axes divide the plane into The axes divide the plane into four quadrantsfour quadrants as shown below. as shown below.

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4

xx

yy

Quadrant I(+, +)

Quadrant II(–, +)

Quadrant IV(+, –)

Quadrant III(–, –)

The Distance FormulaThe Distance Formula

The The distancedistance between any two pointsbetween any two points in the plane may be in the plane may be expressed in terms of their coordinatesexpressed in terms of their coordinates..

Distance formulaDistance formula The distance The distance dd between two points between two points PP11((xx11, , yy11))

and and PP22((xx22, , yy22)) in the plane is given by in the plane is given by

2 2

2 1 2 1d x x y y 2 2

2 1 2 1d x x y y

ExamplesExamples

Find the Find the distancedistance between the points between the points (–(– 4, 3)4, 3) and and (2, 6)(2, 6)..

SolutionSolution Let Let PP11(–(– 4, 3)4, 3) and and PP22(2, 6)(2, 6) be points in the plane. be points in the plane.

We have We have

xx11 = = –– 44 yy11 = = 33 xx22 = = 22 yy22 = = 66

Using the Using the distance formuladistance formula, we have, we have

2 2

2 1 2 1

22

2 2

2 ( 4) 6 3

6 3 45 3 5

d x x y y

2 2

2 1 2 1

22

2 2

2 ( 4) 6 3

6 3 45 3 5

d x x y y

Example 1, page 26

ExamplesExamples

Let Let PP((xx, , yy)) denote a point lying on the denote a point lying on the circlecircle with with radiusradius rr and and centercenter CC((hh, , kk)). Find a relationship between . Find a relationship between xx and and yy..

SolutionSolution By definition in a circle, the distance between By definition in a circle, the distance between PP((xx, , yy)) and and

CC((hh, , kk)) is is rr.. With distance formula we getWith distance formula we get

Squaring both sides givesSquaring both sides gives

2 2x h y k r

2 2 2x h y k r

CC((hh, , kk))

hh

kkrr

PP((xx, , yy))

y

x

Example 2, page 27

Equation of a CircleEquation of a Circle

An equation of a circle with An equation of a circle with centercenter CC((hh, , kk)) and and radiusradius rr is given by is given by

2 2 2x h y k r 2 2 2x h y k r

ExamplesExamples

Find an Find an equation of the circle equation of the circle withwith radiusradius 22 and and centercenter (–1, 3)(–1, 3). .

SolutionSolution We use the circle formula with We use the circle formula with rr = 2 = 2, , hh = = –1–1, and, and kk = 3 = 3::

2 2 2

22 2

2 2

( 1) 3 2

1 3 4

x h y k r

x y

x y

(–1, 3)(–1, 3)

––11

33

22

y

x

Example 3, page 27

ExamplesExamples

Find an Find an equation of the circle equation of the circle withwith radiusradius 33 and and centercenter located at the origin. located at the origin.

SolutionSolution We use the circle formula with We use the circle formula with rr = 3 = 3, , hh = = 00, and, and kk = 0 = 0::

2 2 2

2 2 2

2 2

0 0 3

9

x h y k r

x y

x y

33

y

x

Example 3, page 27

1.41.4Straight LinesStraight Lines

11 22 33 44 55 66

(2, 5)(2, 5)

yy

xx

LL

y y = = ––44

x x = 3= 366

55

44

33

22

11 (5, 1)(5, 1)

Slope of a Vertical LineSlope of a Vertical Line

Let Let LL denote the unique straight line that passes through denote the unique straight line that passes through the two distinct points the two distinct points ((xx11, , yy11)) and and ((xx22, , yy22))..

If If xx11 = = xx22, then , then LL is a is a vertical linevertical line, and , and the slope is the slope is

undefinedundefined..

((xx11, , yy11))

((xx22, , yy22))

y

x

L

Slope of a Nonvertical LineSlope of a Nonvertical Line

If If ((xx11, , yy11)) and and ((xx22, , yy22)) are two distinct points on a are two distinct points on a

nonvertical linenonvertical line LL, then the , then the slopeslope mm of of LL is given by is given by

((xx11, , yy11))

((xx22, , yy22))

y

x

2 1

2 1

y y ym

x x x

L

yy22 – – yy11 = = yy

xx22 – – xx11 = = xx


If If mm > 0 > 0, the line , the line slants slants upwardupward fromfrom left to rightleft to right..

y

x

L

y y = 2= 2

x x = 1= 1

m = 2

m = –1


If If mm < 0 < 0, the line , the line slants slants downwarddownward fromfrom left to rightleft to right..

y

x

L

y y = = –1

x x = 1= 1

ExamplesExamples

Sketch the straight line that passes through the point Sketch the straight line that passes through the point (2, 5)(2, 5) and has slope and has slope –– 4/34/3..

SolutionSolution

1.1. Plot the Plot the pointpoint (2, 5)(2, 5)..

2.2. A A slopeslope of of –– 4/34/3 means means that if that if xx increasesincreases by by 33, , yy decreasesdecreases by by 44..

3.3. Plot the Plot the pointpoint (5, 1)(5, 1)..

4.4. Draw a Draw a lineline across the across the two points.two points.

11 22 33 44 55 66

(2, 5)(2, 5)

yy

xx

LL

y y = –= – 44

x x = 3= 366

55

44

33

22

11 (5, 1)(5, 1)

Example 1, page 34

ExamplesExamples

Find the Find the slopeslope m m of the line that goes through the of the line that goes through the pointspoints (–1, 1)(–1, 1) and and (5, 3)(5, 3). .

SolutionSolution Choose Choose ((xx11, , yy11)) to be to be (–1, 1)(–1, 1) and and ((xx22, , yy22)) to be to be (5, 3)(5, 3). .

With With xx11 = = –1–1, , yy11 = 1 = 1, , xx22 = = 55, , yy22 = = 33, we find, we find

2 1

2 1

3 1 2 1

5 ( 1) 6 3

y ym

x x

2 1

2 1

3 1 2 1

5 ( 1) 6 3

y ym

x x

Example 2, page 35

Equations of LinesEquations of Lines

Let Let LL be a be a straight linestraight line parallelparallel to the to the yy-axis-axis..

Then Then LL crossescrosses the the xx-axis-axis at at some some pointpoint ((aa, 0), 0) , with the , with the xx-coordinate-coordinate given by given by x = ax = a, , where where a a is a real number.is a real number.

Any other point on Any other point on LL has has the form the form ((aa, , )), where , where is is an appropriate number.an appropriate number.

The The vertical linevertical line LL can can therefore be described astherefore be described as

x = ax = a

((aa, , ))

y

x

L

((aa, 0), 0)

y y

y

Equations of LinesEquations of Lines

Let Let LL be a be a nonvertical linenonvertical line with a slope with a slope mm.. Let Let ((xx11, , yy11)) be a be a fixed pointfixed point lying on lying on LL and and ((xx, , yy)) be be

variable point on variable point on LL distinct from distinct from ((xx11, , yy11))..

Using the slope formula by letting Using the slope formula by letting ((xx, , yy) =) = ((xx11, , yy11)) we get we get

Multiplying both sides by Multiplying both sides by xx – – x x22 we get we get

1

1

y ym

x x

1 1( )y y m x x

Point-Slope FormPoint-Slope Form

1 1( )y y m x x 1 1( )y y m x x

An equation of the line that has slope An equation of the line that has slope mm and and passes through point passes through point ((xx11, , yy11)) is given by is given by

ExamplesExamples

Find an equation of the line that passes through the point Find an equation of the line that passes through the point (1, 3)(1, 3) and has slope and has slope 22..

SolutionSolution Use the Use the point-slope formpoint-slope form

Substituting for Substituting for pointpoint (1, 3)(1, 3) and and slopeslope mm = 2 = 2, we obtain, we obtain

SimplifyingSimplifying we get we get

1 1( )y y m x x

3 2( 1)y x

2 1 0x y

Example 5, page 36

ExamplesExamples

Find an equation of the line that passes through the points Find an equation of the line that passes through the points (–3, 2)(–3, 2) and and (4, –1)(4, –1)..

SolutionSolution The slope is given byThe slope is given by

Substituting in the point-slope form for point Substituting in the point-slope form for point (4, –1)(4, –1) and and slope slope mm = – 3/7 = – 3/7, we obtain, we obtain

31 ( 4)

7y x

3 7 5 0x y

1

1

1 2 3

4 ( 3) 7

y ym

x x

7 7 3 12y x

Example 6, page 36

Perpendicular LinesPerpendicular Lines

If If LL11 and and LL22 are two distinct nonvertical lines that are two distinct nonvertical lines that

have slopes have slopes mm11 and and mm22, respectively, then , respectively, then LL11 is is

perpendicularperpendicular to to LL22 (written (written LL1 1 ┴┴ LL22) if and only if ) if and only if

12

1m

m1

2

1m

m

ExampleExample

Find the equation of the line Find the equation of the line LL11 that passes through the that passes through the point point (3, 1)(3, 1) and is perpendicular to the line and is perpendicular to the line LL22 described by described by

SolutionSolution LL22 is described in is described in point-slope formpoint-slope form, so its , so its slopeslope is is mm22 = 2 = 2.. Since the lines are Since the lines are perpendicularperpendicular, the , the slopeslope of of LL11 must be must be

mm11 = –1/2 = –1/2

Using the Using the point-slope formpoint-slope form of the equation for of the equation for LL11 we obtain we obtain

3 2( 1)y x

11 ( 3)

22 2 3

2 5 0

y x

y x

x y

Example 7, page 37

((aa, 0), 0)

(0, (0, bb))

Crossing the AxisCrossing the Axis

A straight line A straight line LL that is that is neither horizontal nor verticalneither horizontal nor vertical cuts the cuts the x-x-axisaxis and the and the y-y-axisaxis at , say, points at , say, points ((aa, 0), 0) and and (0, (0, bb)), respectively., respectively.

The numbers The numbers aa and and b b are called the are called the xx--intercept intercept andand y y--interceptintercept, respectively, of , respectively, of LL..

y

x

L

y-intercept

x-intercept

Slope Intercept FormSlope Intercept Form

An equation of the line that has An equation of the line that has slopeslope mm and and intersectsintersects the the yy-axis-axis at the at the pointpoint (0, (0, bb)) is given by is given by

y y = = mxmx + + bb

ExamplesExamples

Find the equation of the line that has Find the equation of the line that has slopeslope 33 and and yy--interceptintercept of of –– 44..

SolutionSolution We substitute We substitute mm = 3 = 3 and and bb = – = – 44 into into y y = = mxmx + + bb, and get, and get

y y = 3= 3xx – 4 – 4

Example 8, page 38

ExamplesExamples

Determine the Determine the slopeslope and and yy-intercept -intercept of the line whose of the line whose equation is equation is 33xx – 4 – 4y y = 8= 8..

SolutionSolution Rewrite the given equation in the Rewrite the given equation in the slope-intercept formslope-intercept form.. Thus,Thus,

Comparing to Comparing to y y = = mxmx + + bb we find that we find that mm = ¾ = ¾ , and , and bb = – = – 22.. So, the So, the slopeslope is is ¾¾ and the and the yy-intercept-intercept is is –– 22..

3 4 8

4 8 3

32

4

x y

y x

y x

Example 9, page 38

Applied ExampleApplied Example

An art object An art object purchasedpurchased for for $50,000$50,000 is expected to is expected to appreciate in valueappreciate in value at a at a constant rateconstant rate of of $5000$5000 per year for per year for the next the next 55 years. years.

Write an Write an equationequation predicting the value of the art object for predicting the value of the art object for any given year.any given year.

What will be its What will be its valuevalue 33 years after the purchase? years after the purchase?SolutionSolution Let Let xx == time (in years) since the object was purchasedtime (in years) since the object was purchased

yy == value of object (in dollars)value of object (in dollars)

Then, Then, yy = 50,000 = 50,000 when when xx = 0 = 0, so the , so the yy-intercept-intercept is is bb = = 50,00050,000.. Every year the value rises by Every year the value rises by 50005000, so the , so the slopeslope is is mm = 5000 = 5000.. Thus, the equation must be Thus, the equation must be yy = 5000 = 5000xx + 50,000 + 50,000.. After After 33 years the years the value of the objectvalue of the object will be will be $65,000$65,000::

yy = 5000(3) + 50,000 = 65,000 = 5000(3) + 50,000 = 65,000Applied Example 11, page 39

General Form of an Linear EquationGeneral Form of an Linear Equation

The equationThe equation

AxAx + + ByBy + + CC = 0 = 0

where where AA, , BB and and CC are are constantsconstants and and AA and and BB are not both zero, is called the are not both zero, is called the general form general form of a linear equationof a linear equation in the in the variablesvariables xx and and yy..

TheoremTheorem 1 1

An equation of a An equation of a straight linestraight line is a is a linear linear equationequation; conversely, every ; conversely, every linear equationlinear equation represents a represents a straight linestraight line..

ExampleExample

Sketch the straight line represented by the equationSketch the straight line represented by the equation

33xx – 4 – 4yy – 12 = 0 – 12 = 0

SolutionSolution Since every straight line is Since every straight line is uniquely determineduniquely determined by by two two

distinct pointsdistinct points, we need find only two such points through , we need find only two such points through which the line passes in order to sketch it.which the line passes in order to sketch it.

For convenience, let’s compute the For convenience, let’s compute the xx-- and and yy-intercepts-intercepts::✦ Setting Setting yy = 0= 0, we find , we find xx = 4= 4; so the ; so the xx-intercept-intercept is is 44..

✦ Setting Setting xx = 0= 0, we find , we find yy = –3= –3; so the ; so the yy-intercept-intercept is is –3–3.. Thus, Thus, the line goes through the pointsthe line goes through the points (4, 0)(4, 0) and and (0, –3)(0, –3)..

Example 12, page 40

ExampleExample

Sketch the straight line represented by the equationSketch the straight line represented by the equation

33xx – 4 – 4yy – 12 = 0 – 12 = 0

SolutionSolution Graph the line going through the points Graph the line going through the points (4, 0)(4, 0) and and (0, –3) (0, –3)..

1 2 3 4 5 6

(0, –(0, – 3)3)

y

x

L1

– 1

– 2

– 3

– 4

(4, 0)(4, 0)

Example 12, page 40

End of End of Chapter Chapter

Precalculus Review I Precalculus Review II The Cartesian Coordinate System Straight Lines

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