Linear Equations and Inequalities ch02 section2

2-2 Linear Equations and Inequalities 105

92. Petroleum Consumption.Analyzing data from theUnited States Energy Department for the period between1920 and 1960 reveals that petroleum consumption as apercentage of all energy consumed (wood, coal, petro-leum, natural gas, hydro, and nuclear) increased almostlinearly. Percentages for this period are given in the table.

Year

1920

1930

1940

1950

1960

Consumption (%)

11

22

29

37

44

(A) Use regression analysis to find a linear regressionfunction f (x) for this data, where x is the number ofyears since 1900.

(B) Use f (x) to estimate (to the nearest one percent) thepercent of petroleum consumption in 1932. In 1956.

(C) If we assume that f (x) continues to provide a gooddescription of the percentage of petroleumconsumption after 1960, when would this percentagereach 100%? Is this likely to happen? Explain.

Section 2-2 Linear Equations and Inequalities

Solving Linear EquationsSolving Linear InequalitiesSolving Equations and Inequalities Involving Absolute ValueApplication

In this section we discuss methods for solving equations and inequalities thatinvolve linear functions. Some problems are best solved using algebraic tech-niques, while others benefit from a graphical approach. Since graphs often giveadditional insight into relationships, especially in applications, we will usu-ally emphasize graphical techniques over algebraic methods. But you must be cer-tain to master both. There are problems in this section that can only be solvedalgebraically. Later we will also encounter problems that can only be solvedgraphically.

Solving Linear Equations

In the preceding section we found the x intercept of a linear function f (x) 5 mx1 b by solving the equation f (x) 5 0. Now we want to apply the same ideas tosome more complicated equations.

Solving an Equation Algebraically

Solve 5x 2 8 5 2x 1 1 and check.

S o l u t i o n We use the familiar properties of equality to transform the given equation intoan equivalent equation that has an obvious solution (see Section A-8).

Original equation

Add 8 to both sides. 5x 2 8 1 8 5 2x 1 1 1 8

5x 2 8 5 2x 1 1

E X A M P L E

1

106 2 LINEAR AND QUADRATIC FUNCTIONS

Combine like terms.

Subtract 2x from both sides.

Combine like terms.

Divide both sides by 3.

Simplify.

The solution set for this last equation is obvious:

Solution set: {3}

It follows from the properties of equality that {3} is also the solution set of allthe preceding equations in our solution. [Note: If an equation has only one ele-ment in its solution set, we generally use the last equation (in this case, x 5 3)rather than set notation to represent the solution.]

C h e c k Original equation

Substitute x 5 3.

Simplify each side.

A true statement

Solve and check: 2x 1 1 5 4x 1 5

We can also use a graphing utility to solve equations of this type. From agraphical viewpoint, a solution to an equation of the form f (x) 5 g(x) is an inter-section point of the graphs of f and g. Figure 1 shows a graphical solution toExample 1 using a built-in intersection routine. Most graphing utilities have sucha routine (consult your owner’s manual or the graphing utility manual for thistext; see the Preface). If yours does not, then use zoom and trace to approximateintersection points.

210

210

10

10

y2 5 2x 1 1

y1 5 5x 2 8

FIGURE 1Graphical solution of 5x 2 8 5 2x 1 1.

M A T C H E D P R O B L E M

1

7 ⁄ 7

15 2 8 0 6 1 1

5(3) 2 8 0 2(3) 1 1

5x 2 8 5 2x 1 1

x 5 3

3x

35

9

3

3x 5 9

5x 2 2x 5 2x 1 9 2 2x

5x 5 2x 1 9

106 2 LINEAR AND QUADRATIC FUNCTIONS106 2 LINEAR AND QUADRATIC FUNCTIONS


Explore/Discuss

1An equation that is true for all permissible values of the variable iscalled an identity. An equation that is true for some values of the vari-able and false for others is called a conditional equation. Use algebraicand/or graphical techniques to solve each of the following and identifyany identities.

(A) 2(x 2 4) 5 2x 2 8

(B) 2(x 2 4) 5 3x 2 12

(C) 2(x 2 4) 5 2x 2 12

Solving an Equation with a Variable in the Denominator

Solve algebraically and graphically:

S o l u t i o n We begin with an algebraic solution. Note that 0 must be excluded from the per-missible values of x because division by 0 is not permitted. To clear the fractions,we multiply both sides of the equation by 3(2x) 5 6x, the least common denom-inator (LCD) of all fractions in the equation. (For a discussion of LCDs and howto find them, see Section A-4.)

x Þ 0

The solution set is { }.

Figure 2 shows the graphical solution. Note that to seven dec-imal places.

Solve algebraically and graphically:

Remark Which solution method should you use—algebraic or graphical? In Example 1,both the algebraic solution and the graphical solution produced the exact solution,

7

3x1 2 5

1

x2

3

5M A T C H E D P R O B L E M

2

11134 < 3.2647059,

11134 x 5

111

34

234x 5 2111

The equation is now free offractions.

21 2 18x 5 16x 2 90

6x ?7

2x2 6x ? 3 5 6x ?

8

32 6x ?

15

x

Multiply by 6x, the LCD. This and the next step usually canbe done mentally.

6x1 7

2x2 32 5 6x18

32

15

x 2

7

2x2 3 5

8

32

15

x

7

2x2 3 5

8

32

15

x

E X A M P L E

2

210

210

10

10

y2 515x

83

2

y1 5 72x

2 3

FIGURE 2Graphical solution of

8

32

15

x.

7

2x2 3 5


x 5 3. In Example 2, the algebraic solution again produced the exact solution, x 5 111/34, while the graphical solution produced x 5 3.2647059, a seven-decimal-place approximation to the solution. Some like to argue that this makesthe algebraic method superior to the graphical method. But exact solutions havelittle relevance to most applications of mathematics and decimal approximationsare usually quite satisfactory.

We encourage you to choose the method that seems best to you, and when pos-sible, use the other method to confirm your answer. In a simple problem, likeExample 1, choose either method. In Example 2, we would recommend the alge-braic method over the graphical method because of the complexity of the graphs.We have not yet studied graphs of functions involving fractions with x in thedenominator. It was a fortunate accident that the intersection point was visible ina standard viewing window.

We frequently encounter equations involving more than one variable. Forexample, if L and W are the length and width of a rectangle, respectively, the areaof the rectangle is given by (see Fig. 3):

A 5 LW

Depending on the situation, we may want to solve this equation for L or W. Tosolve for W, we simply consider A and L to be constants and W to be a variable.Then the equation A 5 LW becomes a linear equation in W, which can be solvedeasily by dividing both sides by L:

Solving an Equation with More than One Variable

Solve for P in terms of the other variables: A 5 P 1 Prt

S o l u t i o n Think of A, r, and t as constants.

Factor to isolate P.

Divide both sides by 1 1 rt.

Restriction: 1 1 rt Þ 0

Solve for r in terms of the other variables: A 5 P 1 Prt

Solving Linear Inequalities

Now we want to turn our attention to inequalities. Any inequality that can be reducedto one of the four forms in (1) is called a linear inequality in one variable.


3

P 5A

1 1 rt

A

1 1 rt5 P

A 5 P(1 1 rt)

A 5 P 1 Prt

E X A M P L E

3

W 5A

L L Þ 0

A 5 LW W

L

FIGURE 3Area of a rectangle.

Explore/Discuss

2


mx 1 b . 0

mx 1 b $ 0

mx 1 b , 0

mx 1 b # 0

As was the case with equations, the solution set of an inequalityis the set ofall values of the variable that make the inequality a true statement. Each elementof the solution set is called a solution. Two inequalities are said to be equivalentif they have the same solution set.

Associated with the linear equation and inequalities

3x 2 12 5 0 3x 2 12 , 0 3x 2 12 . 0

is the linear function

f (x) 5 3x 2 12

(A) Graph the function f.

(B) From the graph of f determine the values of x for which

f (x) 5 0 f (x) , 0 f (x) . 0

(C) How are the answers to part B related to the solutions of

3x 2 12 5 0 3x 2 12 , 0 3x 2 12 . 0

As you discovered in Explore/Discuss 2, solving inequalities graphically is bothintuitive and efficient. On the other hand, algebraic methods can become quite com-plicated. So we will emphasize the graphical approach when solving inequalities.

Solving a Linear Inequality

Solve and graph on a number line: 0.5x 1 1 # 0

S o l u t i o n The graph of f (x) 5 0.5x 1 1 is shown in Figure 4. It is clear from the graphthat f (x) is negative to the left of x 5 22 and positive to the right. Thus, the solu-tion set of the inequality

0.5x 1 1 # 0

is

x # 22 or (2`, 22]

E X A M P L E

4

Linear inequalities (1)

210

210

10

10

FIGURE 4f(x) 5 0.5x 1 1.

Explore/Discuss

3


Figure 5 shows a graph of the solution set on a number line. A similar graph canbe produced on most graphing utilities by entering y1 5 0.5x 1 1 # 0 (Fig. 6).The expression 0.5x 1 1 # 0 is assigned the value 1 for those values of x thatmake it a true statement and the value 0 for those values of x that make it a falsestatement.

Solve and graph on a number line: 2x 2 6 $ 0

Associated with the following equations and inequalities

1 4 5 x 2 2

1 4 , x 2 2

1 4 . x 2 2

are the two linear functions

f (x) 5 1 4 and g(x) 5 x 2 2

(A) Graph both f and g in the same viewing window.

(B) From the graph in part A determine the value(s) of x for which

f (x) 5 g(x) f (x) , g(x) f (x) . g(x)

(C) How are the answers to part B related to the solutions of

1 4 5 x 2 2

1 4 , x 2 2

1 4 . x 2 2

Most inequalities can be solved graphically. If you need to algebraicallymanipulate an inequality, Theorem 1 lists the properties that govern operations oninequalities.

212x

212x

212x

212x

212x

212x

212x


4

FIGURE 6FIGURE 5

210

210

10

10

025 522x]

T H E O R E M

1


INEQUALITY PROPERTIES

An equivalent inequality will result and the sense will remain the same,if each side of the original inequality

1. Has the same real number added to or subtracted from it.

2. Is multiplied or divided by the same positive number.

An equivalent inequality will result and the sense will reverse,if eachside of the original inequality

3. Is multiplied or divided by the same negative number.

Note: Multiplication by 0 and division by 0 are not permitted.

To gain some experience with these properties, we will solve the next exam-ple two ways, algebraically and graphically.

Solving a Double Inequality

Solve and graph on a number line: 23 # 4 2 7x , 18

S o l u t i o n To solve algebraically, we perform operations on the double inequality until wehave isolated x in the middle with a coefficient of 1.

Subtract 4 from each member.

or or (2)

To solve graphically, enter y1 5 23, y2 5 4 2 7x, y3 5 18, graph [Fig. 7(a)], andfind the intersection points [Figs. 7(b) and 7(c)]. It is clear from the graph that y2

is between y1 and y3 for x between 22 and 1. Since y2 5 23 at x 5 1, we include1 in the solution set, obtaining the same solution, as shown in equation (2).

(a) (b) (c)

215

25

30

5

215

25

30

5

215

25

30

5

y3 5 18

y1 5 23

y2 5 4 2 7xFIGURE 7

xx025 1 522

( ]

(22, 1]22 , x # 1 1 $ x . 22

Divide each member by 27 andreverse each inequality.

27

27$

27x

27 .

14

27

27 # 27x , 14

23 2 4 # 4 2 7x 2 4 , 18 2 4

23 # 4 2 7x , 18

E X A M P L E

5

Explore/Discuss


Solve algebraically and graphically and graph on a number line: 23 , 7 2 2x # 7

Solving Equations and Inequalities Involving Absolute Value

Recall the definition of the absolute value function (see Section 1-4):

f (x) 5 _x_ 5

(A) Graph the absolute value function f (x) 5 _x_ and the constant func-tion g(x) 5 3 in the same viewing window.

(B) From the graph in part A, determine the values of x for which

_x_ , 3 _x_ 5 3 _x_ . 3

(C) Discuss methods for using the definition of _x_ and algebraic tech-niques to solve part B.

The algebraic solution of an equation or inequality involving the absolutevalue function usually must be broken down into two or more cases. For exam-ple, to solve the equation

_x 2 4_ 5 2 (3)

we consider two cases:

x 2 4 5 2 or x 2 4 5 22

x 5 6 x 5 2

We can also solve equation (3) graphically by graphing y1 5 _x 2 4_ and y2 5 2,and finding their intersection points, as shown in Figure 8.

(a) (b)

Algebraic solutions for inequalities involving absolute values can becomequite involved. However, as the next example illustrates, even problems thatappear to be complicated are easily solved with a graphing utility.

210

210

10

10

210

210

10

10

y2 5 2

y1 5 ux 2 4uFIGURE 8Graphical solution of _x 2 4_ 5 2.

52x

x

if x , 0

if x $ 0


5

4


Solving Absolute Value Problems Graphically

Solve graphically. Write solutions in both inequality and interval notation andgraph on a number line.

(A) _2x 2 5_ . 4 (B) _0.5x 1 2_ $ 3x 2 5

S o l u t i o n s (A) Graph y1 5 _2x 2 5_ and y2 5 4 in the same viewing window and find the intersection points (Fig. 9). Examining the graphs in Figure 9, we see ifx , 0.5 or x . 4.5, then the graph of y1 is above the graph of y2. Thus, thesolution is

x , 0.5 or x . 4.5 Inequality notation

(2`, 0.5) < (4.5, ̀ )* Interval notation

(a) (b)

(B) Figure 10 shows the appropriate graphs for the inequality _0.5x 1 2_ $3x 2 5. The graph in Figure 10 shows that y1 . y2 for x , 2.8. Since y1 5y2 for x 5 2.8, we must include this value of x in the solution set:

x # 2.8 or (2`, 2.8]

Solve graphically and write solutions in both inequality and interval notation.

(A) _ _ $ 2 (B) _2x 2 5_ , 0.4x 1 2

Application

Break-Even, Profit, and Loss

A recording company produces compact discs (CDs). One-time fixed costs fora particular CD are $24,000, which include costs such as recording, album

E X A M P L E

7

23x 1 1


6

x025 2.8 5

]

210

210

10

10

y2 5 3x 2 5

y1 5 u0.5x 1 2u210

210

10

10

210

210

10

10

y1 5 u2x 2 5u

y2 5 4

FIGURE 10FIGURE 9

25 0 5

4.50.5() x

E X A M P L E

6

*The symbol < denotes the union operation for sets. See Section A-8 for a discussion of interval notationand set operations.


design, and promotion. Variable costs amount to $5.50 per CD and includethe manufacturing, distribution, and royalty costs for each disc actually manufactured and sold to a retailer. The CD is sold to retail outlets for $8.00each.

(A) Find the level of sales for which the company will break even. Describeverbally and graphically the sales levels that result in a profit and thosethat result in a loss.

(B) Find the sales level that will produce a profit of $20,000.

S o l u t i o n s (A) Let

x 5 Number of CDs sold

C 5 Total cost for producing x CDs

R 5 Revenue (return) on sales of x CDs

Now form the cost and revenue functions.

C(x) 5 Fixed costs 1 Variable costs

5 24,000 1 5.5x Cost function

R(x) 5 8x Revenue function

The company will break even when revenue 5 cost; that is, when R(x) 5C(x). The solution to this equation is often referred to as the break-evenpoint. Graphs of both functions and their intersection point are shown in Fig-ure 11. Examining this graph, we see that the company will break even ifthey sell 9,600 CDs. If they sell more than 9,600 CDs, then revenue is greaterthan cost, and the company will make a profit. If they sell fewer than 9,600CDs, then cost is greater than revenue and the company will lose money.These sales levels are illustrated in Figure 12.

(B) The profit function for this manufacturer is

P(x) 5 R(x) 2 C(x)

5 8x 2 (24,000 1 5.5x)

5 2.5x 2 24,000

FIGURE 12FIGURE 11

x

y

20,000

200,000

y 5 R(x) 5 8x

y 5 C(x) 5 24,000 1 5.5x

Loss Profit

Break-even point

9,600250,000

0

200,000

20,000

y2 5 8x

y1 5 24,000 1 5.5x

Explore/Discuss

5


The sales level x that will produce a profit of $20,000 is the solution of theequation P(x) 5 20,000. Figure 13 shows a graphical solution of this linearequation. Thus, we see that the company will make a profit of $20,000 whenthey sell 17,600 CDs.

Repeat Example 7 if fixed costs are $28,000, variable costs are $6.60 per CD,and the CDs are sold for $9.80 each.

(A) Find the x intercept of the profit function in Example 7 (see Fig. 13).

(B) Discuss the relationship between the x intercept of the profit func-tion and the sales levels for which the company incurs a loss, breakseven, or makes a profit.

(C) In general, compare the graphical solutions of the inequalities

f (x) . g(x) and f(x) 2 g(x) . 0

A n s w e r s t o M a t c h e d P r o b l e m s

1. 22 2. ' 20.5128205 3. r 5 4. x $ 3 or [3, ̀ ) 5. 0 # x , 5 or [0, 5)

6. (A) x # 24.5 or x $ 1.5; (2`, 24.5] < [1.5, `)

(B) 1.25 , x , 4.375; (1.25, 4.375)

7. (A) The company breaks even if they sell 8,750 CDs, makes a profit if they sell more than 8,750 CDs,and loses money if they sell less than 8,750 CDs.

(B) The company must sell 15,000 CDs to make a profit of $20,000.

xx0 525

1.25 4.375( )

0

1.5

5

24.5] [ xx

25

xx025 5

)[x025 3 5

[ x

A 2 P

Pt2

20

39


7

250,000

0

50,000

20,000

y3 5 2.5x 2 24,000

y4 5 20,000FIGURE 13

x

y

20,0008,750

200,000

Loss Profit

Break-even point

y 5 C(x) 5 28,000 1 6.6x

y 5 R(x) 5 9.8x


EXERCISE 2-2

A

Use the graphs of functions u and v in the figure to solve theequations and inequalities in Problems 1–8. (Assume thegraphs continue as indicated beyond the portions shown here.)Express solutions to inequalities in interval notation.

1. u(x) 5 0 2. v(x) 5 0

3. u(x) 5 v(x) 4. u(x) 2 v(x) 5 0

5. u(x) . 0 6. v(x) $ 0

7. v(x) $ u(x) 8. v(x) , 0

Solve Problems 9–14 algebraically and check graphically.

9. 3(x 1 2) 5 5(x 2 6) 10. 5x 1 10(x 2 2) 5 40

11. 5 1 4(t 2 2) 5 2(t 1 7) 1 1

12. 5w 2 (7w 2 4) 2 2 5 5 2 (3w 1 2)

13. 5 2 14.

Solve Problems 15–20 algebraically and check graphically.Represent each solution using inequality notation, intervalnotation, and a graph on a real number line.

15. 7x 2 8 , 4x 1 7 16. 4x 1 8 $ x 2 1

17. 25t , 210 18. 27n $ 21

19. 24 , 5t 1 6 # 21 20. 2 # 3m 2 7 , 14

B

In Problems 21–36, solve each equation or inequality. Whenapplicable, write answers using both inequality notation andinterval notation.

21. _y 2 5_ 5 3 22. _x 1 1_ 5 5

23. _5t 1 3_ # 7 24. _2w 2 9_ , 6

25. 26.

27. 212 , (2 2 x) # 24 28. 24 # (x 2 5) , 362

3

3

4

2

3x1

1

25

4

x1

4

3

1

m2

1

95

4

92

2

3m

x 1 3

42

x 2 4

25

3

8

2x 2 1

45

x 1 2

3

x

y

y 5 u(x)

y 5 v(x)

e fdc

b

a

29. _3t 2 7_ 5 30. _2s 1 3_ 5 6 2 0.5s

31. _1.5x 1 6_ . 0.3x 1 7.5 32. _7 2 2x_ $ x 2 0.8

33. 34.

35. 6 , _x 2 2_ 1 _x 1 1_ , 12

36. _x 1 1_ 2 _x 2 2_ , 0.4x

In Problems 37–44, solve for the indicated variable in terms ofthe other variables.

37. an 5 a1 1 (n 2 1)d for d (arithmetic progressions)

38. F 5 1 32 for C (temperature scale)

39. for f (simple lens formula)

40. for R1 (electric circuit)

41. A 5 2ab 1 2ac for a (surface area of a rectangular solid)

42. A 5 2ab 1 2ac 1 2bc for c

43. 44.

45. Discuss the relationship between the graphs of y1 5 x andy2 5

46. Discuss the relationship between the graphs of y1 5 _x_ andy2 5

47. Discuss the possible signs of the numbers a and b giventhat

(A) ab . 0 (B) ab , 0

(C) (D)

48. Discuss the possible signs of the numbers a, b,and c giventhat

(A) abc. 0 (B)

(C) (D)

In Problems 49–52, replace each question mark with , or .and explain why your choice makes the statement true.

49. If a 2 b 5 1, then a ? b.

50. If u 2 v 5 22, then u ? v.

51. If a , 0, b , 0, and . 1, then a ? b.

52. If a . 0, b . 0, and . 1, then a ? b.b

a

b

a

a2

bc , 0

a

bc . 0

ab

c , 0

a

b , 0

a

b . 0

Ïx2.

Ïx2.

x 53y 1 2

y 2 3 for yy 5

2x 2 3

3x 1 5 for x

1

R5

1

R1

11

R2

1

f5

1

d1

11

d2

95 C

2x

x 1 45 7 2

6

x 1 4

2x

x 2 35 7 1

4

x 2 3

4

3t 1

1

2



C

Problems 53–56 are calculus-related. Solve and graph. Writeeach solution using interval notation.

53. 0 , _x 2 3_ , 0.1 54. 0 , _x 2 5_ , 0.01

55. 0 , _x 2 c_ , d 56. 0 , _x 2 4_ , d

57. What are the possible values of ?

58. What are the possible values of ?

APPLICATIONS

59. Sales Commissions.One employee of a computer store ispaid a base salary of $2,150 a month plus an 8% commis-sion on all sales over $7,000 during the month. How muchmust the employee sell in 1 month to earn a total of$3,170 for the month?

60. Sales Commissions.A second employee of the computerstore in Problem 59 is paid a base salary of $1,175 a monthplus a 5% commission on all sales during the month.

(A) How much must this employee sell in 1 month to earna total of $3,170 for the month?

(B) Determine the sales level where both employeesreceive the same monthly income. If employees canselect either of these payment methods, how wouldyou advise an employee to make this selection?

61. Approximation. The area A of a region is approximatelyequal to 12.436. The error in this approximation is lessthan 0.001. Describe the possible values of this area bothwith an absolute value inequality and with intervalnotation.

62. Approximation. The volume V of a solid is approxi-mately equal to 6.94. The error in this approximation isless than 0.02. Describe the possible values of this volumeboth with an absolute value inequality and with intervalnotation.

63. Break-Even Analysis.An electronics firm is planning tomarket a new graphing calculator. The fixed costs are$650,000 and the variable costs are $47 per calculator.The wholesale price of the calculator will be $63. For thecompany to make a profit, it is clear that revenues must begreater than costs.

(A) How many calculators must be sold for the companyto make a profit?

(B) How many calculators must be sold for the companyto break even?

(C) Discuss the relationship between the results in parts Aand B.

|x 2 1|x 2 1

x

|x|

64. Break-Even Analysis.A video game manufacturer isplanning to market a 64-bit version of its game machine.The fixed costs are $550,000 and the variable costs are$120 per machine. The wholesale price of the machinewill be $140.

(A) How many game machines must be sold for thecompany to make a profit?

(B) How many game machines must be sold for thecompany to break even?

(C) Discuss the relationship between the results in parts Aand B.

65. Break-Even Analysis.The electronics firm in Problem 63finds that rising prices for parts increases the variablecosts to $50.5 per calculator.

(A) Discuss possible strategies the company might use todeal with this increase in costs.

(B) If the company continues to sell the calculators for$63, how many must they sell now to make a profit?

(C) If the company wants to start making a profit at thesame production level as before the cost increase,how much should they increase the wholesale price?

66. Break-Even Analysis.The video game manufacturer inProblem 64 finds that unexpected programming problemsincreases the fixed costs to $660,000.

(A) Discuss possible strategies the company might use todeal with this increase in costs.

(B) If the company continues to sell the game machinesfor $140, how many must they sell now to make aprofit?

(C) If the company wants to start making a profit at thesame production level as before the cost increase,how much should they increase the wholesale price?

★★ 67. Significant Digits.If N 5 2.37 represents a measurement,then we assume an accuracy of 2.37 6 0.005. Express theaccuracy assumption using an absolute value inequality.

★ 68. Significant Digits.If N 5 3.65 3 1023 is a number from ameasurement, then we assume an accuracy of 3.65 3 1023

6 5 3 1026. Express the accuracy assumption using anabsolute value inequality.

★ 69. Finance.If an individual aged 65–69 continues to workafter Social Security benefits start, benefits will be reducedwhen earnings exceed an earnings limitation. In 1989,benefits were reduced by $1 for every $2 earned in excessof $8,880. Find the range of benefit reductions for individ-uals earning between $13,000 and $16,000.

★ 70. Finance.Refer to Problem 69. In 1990 the law waschanged so that benefits were reduced by $1 for every $3earned in excess of $8,880. Find the range of benefit re-ductions for individuals earning between $13,000 and$16,000.


71. Celsius/Fahrenheit.A formula for converting Celsius de-grees to Fahrenheit degrees is given by the linear function

F 5 1 32

Determine to the nearest degree the Celsius range in tem-perature that corresponds to the Fahrenheit range of 60°Fto 80°F.

72. Celsius/Fahrenheit.A formula for converting Fahrenheitdegrees to Celsius degrees is given by the linear function

C 5 (F 2 32)

Determine to the nearest degree the Fahrenheit range intemperature that corresponds to a Celsius range of 20°C to30°C.

★ 73. Earth Science.In 1984, the Soviets led the world indrilling the deepest hole in the Earth’s crust—more than12 kilometers deep. They found that below 3 kilometersthe temperature T increased 2.5°C for each additional 100meters of depth.

(A) If the temperature at 3 kilometers is 30°C and x is thedepth of the hole in kilometers, write an equationusing x that will give the temperature T in the hole atany depth beyond 3 kilometers.

(B) What would the temperature be at 15 kilometers?[The temperature limit for their drilling equipmentwas about 300°C.]

5

9

9

5C

(C) At what interval of depths will the temperature bebetween 200°C and 300°C, inclusive?

★ 74. Aeronautics.Because air is not as dense at high altitudes,planes require a higher ground speed to become airborne.A rule of thumb is 3% more ground speed per 1,000 feet ofelevation, assuming no wind and no change in air tempera-ture. (Compute numerical answers to 3 significant digits.)

(A) Let

Vs 5 Takeoff ground speed at sea level for a particularplane (in miles per hour)

A 5 Altitude above sea level (in thousands of feet)

V 5 Takeoff ground speed at altitude A for the sameplane (in miles per hour)

Write a formula relating these three quantities.

(B) What takeoff ground speed would be required at LakeTahoe airport (6,400 feet), if takeoff ground speed atSan Francisco airport (sea level) is 120 miles perhour?

(C) If a landing strip at a Colorado Rockies hunting lodge(8,500 feet) requires a takeoff ground speed of 125miles per hour, what would be the takeoff groundspeed in Los Angeles (sea level)?

(D) If the takeoff ground speed at sea level is 135 milesper hour and the takeoff ground speed at a mountainresort is 155 miles per hour, what is the altitude of themountain resort in thousands of feet?

Section 2-3 Quadratic Functions

Quadratic FunctionsCompleting the SquareProperties of Quadratic Functions and Their GraphsApplications

Quadratic Functions

The graph of the square function, h(x) 5 x2, is shown in Figure 1. Notice that thegraph is symmetric with respect to the y axis and that (0, 0) is the lowest pointon the graph. Let’s explore the effect of applying a sequence of basic transfor-mations to the graph of h. (A brief review of Section 1-5 might prove helpful atthis point.)

h(x)

5

25 5x

FIGURE 1Square function h(x) 5 x2.

Linear Equations and Inequalities ch02 section2

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