Bccalcet01 App A

1061

The goal of this appendix is to establish the essential notation, terminology, and algebraicskills that are used throughout the book.

AlgebraEXAMPLE 1 Algebra review

a. Evaluate

b. Simplify

c. Solve the equation

SOLUTION

a. Recall that Because we have

Another option is to write

b. Finding a common denominator and simplifying leads to

c. Notice that cannot be a solution of the equation because the left side of theequation is undefined at Because both sides of the equation can be multiplied by to produce After factoring, this equation becomes which implies or

The roots of are and the roots of are Excluding the roots of the original equation are

and Related Exercises 15–26

Sets of Real NumbersFigure A.1 shows the notation for open intervals, closed intervals, and various boundedand unbounded intervals. Notice that either interval notation or set notation may be used.

➤

x = ;2.x = -1x = 1,x = ;1.x2

- 1 = 0x = ;2x2

- 4 = 0x2- 1 = 1x - 121x + 12 = 0.

x2- 4 = 1x - 221x + 22 = 01x2

- 421x2- 12 = 0,

x4- 5x2

+ 4 = 0.x - 1x Z 1,x = 1.

x = 1

1

x - 2-

1

x + 2=

1x + 22 - 1x - 22

1x - 221x + 22=

4

x2- 4

.

1-3222>5 = 31-322241>5 = 10241>5= 4.

1-3222>5 = 1-222 = 4.1-3221>5 = 25 -32 = -2,1-3222>5 = 31-3221>542.

x4- 5x2

+ 4

x - 1= 0.

1

x - 2-

1

x + 2 .

1-3222>5.

A

Appendix

Z01_BRIG0561_01_SE_APP A pp4.qxd 12/5/09 3:19 PM Page 1061

Copyright © 2011 Pearson Education, Inc. All rights reserved.

1062 APPENDIX A

a

a

ba

ba

ba

ba

b

b

FIGURE A.1

EXAMPLE 2 Solving inequalities Solve the following inequalities.

a. b.

SOLUTION

a. We multiply by reverse the inequality, and then factor:

Multiply by

Factor.

The roots of the corresponding equation are and These roots partition the number line (Figure A.2) into three intervals: ,and On each interval, the product does not change sign. To determine the sign of the product on a given interval, a test value is selected and the sign of is determined at .x1x - 221x - 32

x1x - 221x - 3213, q2.

12, 321- q , 22,x = 3.x = 21x - 221x - 32 = 0

1x - 221x - 32 7 0

-1. x2- 5x + 6 7 0

-1,

x2- x - 2

x - 3… 0-x2

+ 5x - 6 6 0

4 53210

� � �Sign of

(x � 2)(x � 3)

FIGURE A.2

A convenient choice for in is At this test value,

Using a test value of in the interval (2, 3), we have

A test value of in gives

Therefore, on and We conclude that theinequality is satisfied for all in either or (Figure A.2).

13, q21- q , 22x-x2+ 5x - 6 6 0

13, q2.1- q , 221x - 221x - 32 7 0

1x - 221x - 32 = 122112 7 0.

13, q2x = 4

1x - 221x - 32 = 10.521-0.52 6 0.

x = 2.5

1x - 221x - 32 = 1-221-32 7 0.

x = 0.1- q , 22x

➤ The set of numbers {x: x is in or } may alsobe expressed using the union symbol:

1- q , 22´ 13, q2

13, q21- q , 22

(-q, q)

(-q, b) = {x: x 6 b}

(-q, b] = {x: x … b}

(a, q) = {x: x 7 a}

[a, q) = {x: x Ú a}

(a, b) = {x: a 6 x 6 b}

[a, b) = {x: a … x 6 b}

(a, b] = {x: a 6 x … b}

[a, b] = {x: a … x … b} Closed, bounded interval

Bounded interval

Bounded interval

Open, bounded interval

Unbounded interval

Unbounded interval

Unbounded interval

Unbounded interval

Unbounded interval



APPENDIX A 1063

b. The expression changes sign only at points where the numerator or

denominator of equals 0. Because

the numerator is 0 when and and the denominator is 0 at

Therefore, we examine the sign of on the intervals

, and

Using test values on these intervals, we see that on

and (2, 3). Furthermore, the expression is 0 when and

Therefore, for all values of in either or

(Figure A.3).

32, 321- q , -14xx2

- x - 2

x - 3… 0

x = 2.x = -11- q , -12

1x + 121x - 22

x - 36 0

13, q2.1-1, 22, 12, 32

1- q , -12,1x + 121x - 22

x - 3

x = 3.x = 2,x = -1

x2- x - 2

x - 3=

1x + 121x - 22

x - 3,

x2- x - 2

x - 3

x2- x - 2

x - 3

43210

� ��

�1�2

(x � 1)(x � 2)x � 3

Sign of

FIGURE A.3 Related Exercises 27–30

Absolute ValueThe absolute value of a real number , denoted is the distance between and the originon the number line. (Figure A.4) More generally, is the distance between the points

and on the number line. The absolute value has the following definition and properties:yxƒ x - y ƒ

xƒ x ƒ ,x➤

4 units 5 units

�4 50��4� � 4 �5� � 5

x units x units

�x x0�x� � x��x� � x

For x � 0

FIGURE A.4

➤ The absolute value is useful insimplifying square roots. Because

is nonnegative, we have

For example, and

Note that thesolutions of are orx = ;3.

ƒ x ƒ = 3x2= 9

21-322 = 19 = 3.

232= 3

2a2= ƒ a ƒ .1a

Definition and Properties of the Absolute Value

The absolute value of a real number is defined as

Let be a positive real number.

1. 2.

3. or 4.

5. or 6. ƒ x + y ƒ … ƒ x ƒ + ƒ y ƒx … -aƒ x ƒ Ú a 3 x Ú a

ƒ x ƒ … a 3 -a … x … ax 6 -aƒ x ƒ 7 a 3 x 7 a

ƒ x ƒ 6 a 3 -a 6 x 6 aƒ x ƒ = a 3 x = ;a

a

ƒ x ƒ = ex if x Ú 0

-x if x 6 0

x

➤ Property 6 is called the triangleinequality.

Test Value x � 1 x � 2 x � 3 Result

-2 - - - -

0 + - - +

2.5 + + - -

4 + + + +



EXAMPLE 3 Inequalities with absolute values Solve the following inequalities.Then sketch the solution on the number line and express it in interval notation.

a. b.

SOLUTION

a. Using Property 2 of the absolute value, is written as

Adding 2 to each term of these inequalities results in (Figure A.5). This set of numbers is written as in interval notation.

b. Using Property 5, the inequality implies that

We add 6 to both sides of the first inequality to obtain which implies Similarly, the second inequality yields (Figure A.6). In interval notation, thesolution is . Related Exercises 31–34

➤

1- q ,-24 or 38, q2x … -2

x Ú 8.2x Ú 16,

2x - 6 Ú 10 or 2x - 6 … -10.

ƒ 2x - 6 ƒ Ú 10

1-1, 52-1 6 x 6 5

-3 6 x - 2 6 3.

ƒ x - 2 ƒ 6 3

ƒ 2x - 6 ƒ Ú 10ƒ x - 2 ƒ 6 3

1064 APPENDIX A

Cartesian Coordinate SystemThe conventions of the Cartesian coordinate system or xy-coordinate system are illus-trated in Figure A.7.

Distance Formula and CirclesBy the Pythagorean theorem (Figure A.8), we have the following formula for the distancebetween two points and P21x2, y22.P11x1, y12

�1 520

3 3

�x: �x � 2� � 3�

FIGURE A.5

�2 80

�x: �2x � 6� � 10�

FIGURE A.6

y

xO

Quadrant IVx � 0, y � 0

Quadrant Ix � 0, y � 0

Quadrant IIIx � 0, y � 0

Quadrant IIx � 0, y � 0

(x, y)

x

y

FIGURE A.7

➤ The familiar coordinate system is named after René Descartes(1596–1650). However, it was introducedindependently and simultaneously byPierre de Fermat (1601–1665).

1x, y2

�(x2 � x1)2 � (y2 � y1)2

For any right triangle,a2 � b2 � c2.

P2(x2, y2)

P1(x1, y1)�x2 � x1�

�y2 � y1�

(x2, y1)

bc

a

FIGURE A.8

Distance Formula

The distance between the points and is

ƒ P1P2 ƒ = 21x2 - x122

+ 1y2 - y122

.

P21x2, y22P11x1, y12

A circle is the set of points in the plane whose distance from a fixed point (the center)is a constant (the radius). This definition leads to the following equations that describea circle.



APPENDIX A 1065

Equation of a Circle

The equation of a circle centered at with radius is

Solving for , the equations of the upper and lower halves of the circle (Figure A.9) are

upper half of the circle

lower half of the circle.y = b - 2r2- 1x - a22

y = b + 2r2- 1x - a22

y

1x - a22 + 1y - b22 = r2.

r1a, b2

a

y

b

Upper half: y � b � �r2 � (x � a)2

Lower half: y � b � �r2 � (x � a)2

center: (a, b)

radius: r

x

FIGURE A.9

EXAMPLE 4 Sets involving circles

a. Find the equation of the circle with center passing through

b. Describe the set of points satisfying

SOLUTION

a. The radius of the circle is the length of the line segment between the center andthe point on the circle which is

Therefore, the equation of the circle is

b. To put this inequality in a recognizable form, we complete the square on the left sideof the inequality:

Therefore, the original inequality becomes

This inequality describes those points that lie within the circle centered at with radius 5 (Figure A.10). Note that a dashed curve is used to indicate that the circle itself is not part of the solution.

12, 32

1x - 222 + 1y - 322 - 13 6 12, or 1x - 222 + 1y - 322 6 25.

= 1x - 222 + 1y - 322 - 13

= x2- 4x + 4 + y2

- 6y + 9 - 4 - 9

x2+ y2

- 4x - 6y = x2- 4x + 4 - 4 + y2

- 6y + 9 - 9

1x - 222 + 1y - 422 = 25.

212 - (-22)2+ 14 - 122 = 5.

1-2, 12,12, 42

x2+ y2

- 4x - 6y 6 12.

1-2, 12.12, 42

➤ A circle is the set of all points whosedistance from a fixed point is a constant.A disk is the set of all points within andpossibly on a circle.

Add and subtract the squareof half the coefficient of x.

Add and subtract the squareof half the coefficient of y.

∂ ∂➤ Recall that the procedure shown here forcompleting the square works when thecoefficient on the quadratic term is 1.When the coefficient is not 1, it must befactored out before completing the square.

1

1

The solution to(x � 2)2 � (y � 3)2 � 25is the interior of a circle.

y

x

r � 5

(2, 3)

FIGURE A.10

t

1x - 222

t

1y - 322

Related Exercises 35–36

➤



1066 APPENDIX A

Equations of LinesThe slope of the line passing through the points and is the rise overrun (Figure A.11), computed as

m =

change in vertical coordinate

change in horizontal coordinate=

y2 - y1

x2 - x1 .

P21x2, y22P11x1, y12m

FIGURE A.11

y

xO

y

xO

(x2, y2)

(x2, y2)

(x1, y1)

(x1, y1)

x2 � x1, y2 � y1, m � 0

x2 � x1, y2 � y1, m � 0

Equations of a Line

Point-slope form The equation of the line with slope passing through the pointis

Slope-intercept form The equation of the line with slope and -intercept (0, )is (Figure A.12a).

General linear equation The equation describes a line in theplane, provided and are not both zero.

Vertical and horizontal lines The vertical line that passes through ( , 0) has anequation its slope is undefined. The horizontal line through (0, ) has anequation with slope equal to 0 (Figure A.12b).y = b,

bx = a;a

BAAx + By + C = 0

y = mx + bbym

y - y1 = m1x - x12.1x1, y12m

O x

(a)

y � mx � b(m � 0)

y � mx � b(m � 0)

(0, b)

y

O

(b)

x � a(m undefined)

y � b(m � 0)

b

a

y

x

FIGURE A.12

➤ Given a particular line, we often talkabout the equation of a line. But theequation of a specific line is not unique.Having found one equation, we canmultiply it by any nonzero constant to produce another equation of the same line.

EXAMPLE 5 Working with linear equations Find an equation of the line passingthrough the points and

SOLUTION The slope of the line through the points and is

Using the point the point-slope form of the equation is

Solving for yields the slope-intercept form of the equation:


➤

y = -

7

5 x -

3

5

y

y - 1-22 = -

7

5 1x - 12.

11, -22,

m =

5 - 1-22

-4 - 1=

7

-5= -

7

5 .

1-4, 5211, -22

1-4, 52.11, -22

➤ Because both points and lie on the line and must satisfy theequation of the line, either point can beused to determine an equation of the line.

1-4, 5211, -22



APPENDIX A 1067

Parallel and Perpendicular LinesTwo lines in the plane may have either of two special relationships to each other: they maybe parallel or perpendicular.

Parallel Lines

Two distinct nonvertical lines are parallel if they have the same slope; that is, thelines with equations and are parallel if and only if

Two distinct vertical lines are parallel.m = M.y = Mx + By = mx + b

EXAMPLE 6 Parallel lines Find an equation of the line parallel tothat intersects the -axis at

SOLUTION Solving the equation for , we have

This line has a slope of and any line parallel to it has a slope of Therefore, the line

that passes through with slope has the point-slope equation

After simplifying, an equation of the line is

Notice that the slopes of the two lines are the same; only the -intercepts differ.Related Exercises 41–42

➤

y

y =

1

2 x - 2.

y - 0 =12

1x - 42.1214, 02

12

.12

y =

1

2 x + 2.

y3x - 6y + 12 = 0

14, 02.x3x - 6y + 12 = 0

Perpendicular Lines

Two lines with slopes and are perpendicular if and only ifor equivalently, m = -1>M.mM = -1,

M Z 0m Z 0➤ The slopes of perpendicular lines arenegative reciprocals of each other.

y

xO

slope m

slope M

mM � �1

Perpendicularlines

y

xO

slope m

slope M

m � M

Parallellines

EXAMPLE 7 Perpendicular lines Find an equation of the line passing through thepoint perpendicular to the line

SOLUTION The equation of can be written which reveals that its slope is 2.

Therefore, the slope of any line perpendicular to is The line with slope passing

through the point is


APPENDIX A EXERCISES

➤

y - 5 = -

1

2 1x + 22, or y = -

x

2+ 4.

1-2, 52

-12-

12

./

y = 2x +72

,/

/: 4x - 2y + 7 = 0.1-2, 52

Review Questions1. State the meaning of in words. Express the

set using interval notation and draw it on anumber line.

2. Write the interval in set notation and draw it on a number line.

3. Give the definition of ƒ x ƒ .

1- q , 22

5x: -4 6 x … 1065x: -4 6 x … 106

4. Write the inequality without absolute value symbols.

5. Write the inequality without absolute value symbols.

6. Write an equation of the set of all points that are a distance 5 units from the point (2, 3).

ƒ 2x - 4 ƒ Ú 3

ƒ x - 2 ƒ … 3



1068 APPENDIX A

7. Explain how to find the distance between two points whose coordinates are known.

8. Sketch the set of points

9. What is the equation of the upper half of the circle centered at the origin with radius 6?

10. What are the possible solution sets of the equation

11. Give an equation of the line with slope that passes through thepoint

12. Give an equation of the line with slope and -intercept (0, 6).

13. What is the relationship between the slopes of two parallel lines?

14. What is the relationship between the slopes of two perpendicularlines?

Basic Skills15–20. Algebra review Simplify or evaluate the following expressionswithout a calculator.

15. 16.

17. 18.

19. 20.

21–26. Algebra review

21. Factor 22. Solve

23. Solve 24. Solve

25. Simplify for

26. Rewrite where without square roots in

the numerator.

27–30. Solving inequalities Solve the following inequalities and drawthe solution on a number line.

27. 28.

29. 30.

31–34. Inequalities with absolute values Solve the following inequal-ities. Then draw the solution on a number line and express it usinginterval notation.

31. 32.

33. 34. 2 6 ƒ

x2 - 5 ƒ 6 63 6 ƒ 2x - 1 ƒ 6 5

1 … ƒ x ƒ … 10ƒ 3x - 4 ƒ 7 8

x1x - 1 7 0x2

- 9x + 20

x - 6… 0

x + 1

x + 26 6x2

- 6x + 5 6 0

h Z 0,1x + h - 1x

h,

h Z 0.1x + h23 - x3

h

4x- 612x2 = -8.u4

- 11u2+ 18 = 0.

x3- 9x = 0.y2

- y-2.

2

x + 3-

2

x - 3

1

x + h-

1

x

1a + h22 - a2

h1u + v22 - 1u - v22

23 -125 + 21>2511>82-2>3

ym

14, -22.m

x2+ y2

+ Cx + Dy + E = 0?

51x, y2: x2+ 1y - 222 7 166.

35–36. Circle calculations Solve the following problems using the distance formula.

35. Find the equation of the lower half of the circle with centerand radius 3.

36. Describe the set of points that satisfy

37–40. Working with linear equations Find an equation of the line that satisfies the given condition. Then draw the graph of

37. has slope and -intercept

38. has undefined slope and passes through

39. has -intercept and -intercept

40. is parallel to the -axis and passes through the point

41–42. Parallel lines Find an equation of the following lines and drawtheir graphs.

41. the line with -intercept parallel to the line

42. the line with -intercept parallel to the line

43–44. Perpendicular lines Find an equation of the following lines.

43. the line passing through perpendicular to the line

44. the perpendicular bisector of the line joining the points and

Further Explorations45. Explain why or why not State whether the following statements

are true and give an explanation or counterexample.

a.

b.c. There are two real numbers that satisfy the condition

d.e. The point is inside the circle of radius 1 centered at the

origin.

f. for all real numbers .

g. implies for all real numbers and .

46–48. Intervals to sets Express the following intervals in setnotation. Use absolute value notation when possible.

46. 47. or

48. or

49–50. Sets in the plane Graph each set in the -plane.

49. 50. 51x, y2: ƒ x ƒ = ƒ y ƒ651x, y2: ƒ x - y ƒ = 06

xy

34, 5212, 34

34, q21- q , -241- q , 122

baa 6 b2a26 2b2

x2x4= x2

11, 12ƒp2

- 9 ƒ 6 0.ƒ x ƒ = -2.

242= 21-422

216 = ; 4

13, -521-9, 22

y = -3x + 213, -62

2x - 5 = 0.1-6, 02x

x + 2y = 810, 122y

12, 32.x/

15, 02.x10, -42y/

10, 52./

10, 42.y5>3/

/./

x2+ y2

+ 6x + 8y Ú 25.

1-1, 22



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Bccalcet01 App A