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4 Quadratic Models; More on Rates of Change

# 4 Quadratic Models; More on Rates of Change

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College Algebra G&M—

3–49

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Section 3.4 Quadratic Models; More on Rates of Change

EXAMPLE 1

Growth of Online Sales of Pet Supplies

Since the year 2000, there has been a tremendous increase in the

online sales of pet food, pet medications, and other pet supplies.

That data shown in the table shows the amount spent (in billions

of dollars) for the years indicated.

Source: 2009 Statistical Abstract of the United States, Page 646, Table 1016.

a. Input the data into a graphing calculator, set an appropriate

window and view the scatterplot, then use the context and the

scatterplot to decide on an appropriate form of regression.

b. Determine the regression equation, and use it to

Year

find the amount of online sales projected for the (2001 S 1)

year 2012.

1

c. If the predicted trends continue, in what

5

year will online sales of pet supplies surpass

\$25 billion dollars?

6

Solution

Sales

(billions)

0.8

4.1

5.6

7

7.4

a. Begin by entering the years in L1 and the dollar

amounts in L2. From the data given, a viewing

8

9.1

window of Xmin ϭ 0, Xmax ϭ 15, Ymin ϭ Ϫ2,

and Ymax ϭ 15 seems appropriate. The

Figure 3.53

scatterplot in Figure 3.53 shows a definite

15

increasing, non-linear pattern and we opt for

b. Using STAT

0

15

places this option on the home screen.

Pressing

at this point will give us the

quadratic coefficients, but we can also have

the calculator paste the equation itself

Ϫ2

directly into Y1 on the Y= screen, simply

by appending Y1 to the QuadReg command (Figure 3.54). The resulting

equation is y Ϸ 0.118x2 ϩ 0.136x ϩ 0.537 (to three decimal places), and the

graph is shown in Figure 3.55. Using the home screen, we find that

Y1 1122 Ϸ 19.2 (Figure 3.56), indicating that in 2012, about \$19.2 billion is

projected to be spent online for pet supplies.

c. To find the year when spending will surpass \$25 billion, we set Y2 ϭ 25 and

use the Intersection-of-Graphs method (be sure to increase Ymax so this graph

can be seen: Ymax ϭ 30). There result is shown in Figure 3.57 and indicates

that spending will surpass \$25 billion late in the year 2013.

ENTER

Figure 3.54, 3.55

Figure 3.57

Figure 3.56

15

0

15

Ϫ2

30

0

15

Ϫ2

Now try Exercises 7 through 10

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CHAPTER 3 Quadratic Functions and Operations on Functions

Applications of quadratic regression tend to focus on the defining characteristics of

a quadratic graph, even if only a part of the graph is used. The data may simply indicate

an increasing rate of growth beyond an initial or minimum point as in Example 1, or

show values that decrease to a minimum and increase afterward. These would be data

sets modeled by an equation where a 7 0. Applications where a 6 0 are also common.

In actual practice, data sets are often very large and need not be integer-valued.

In addition, as we explore additional functions and forms of regression, the decision

of which form to use must be carefully evaluated as many functions exhibit similar

characteristics.

EXAMPLE 2

Apollo 13, Retro-Rocket Option

About 56 hr into the mission and 65,400 km from the Moon, the

oxygen tanks exploded on Apollo 13. At this point, mission control

explored two options: (1) firing the retro-rockets for a direct return

to Earth, or (2) the so called, “sling-shot around the Moon,” the

option actually chosen. The data shown in the table explore a

scenario where option 1 was taken, and give the distance d from

the Moon, t seconds after the retro-rockets have fired.

a. Input the data into a graphing calculator, set an

appropriate window and view the scatterplot,

then use the context and the scatterplot to decide

Time

on an appropriate form of regression.

(sec)

b. Determine the regression equation, and use it to

0

find how close Apollo 13 would have come to the

Moon (the minimum distance). How many

1

seconds after the retro-rockets fired would this

2

have occurred?

3

c. How many seconds would it have taken for

4

Apollo 13 to “turn around,” and get back to

5

the original distance of 65,400 km (where the

10

accident occurred)?

Solution

a. After entering the data (seconds in L1 and

distance in L2), we set a viewing window

of Xmin ϭ 0, Xmax ϭ 15, and after some

trial and error, Ymin ϭ 65,300, and

Ymax ϭ 65,450. The scatterplot shows

(be sure that PLOT1 is activated on the

Y=

nonlinear pattern and a quadratic regression

seems appropriate (Figure 3.58).

b. We find the regression equation

using STAT

Y1 (pasting the equation to Y1):

Y1 Ϸ 0.464X2 Ϫ 11.519X ϩ 65,400.013.

Next, press GRAPH to obtain the graph and

scatterplot shown in Figure 3.59. The

result indicates that we should increase

the value of Xmax to see the full graph and

to help locate the minimum value. Using

Xmax ϭ 30 and the 2nd TRACE (CALC)

3:minimum option shows that Apollo 13

would have come within 65,328.5 km of the

Moon, after the retro-rockets had fired for

12.4 sec (Figure 3.60).

Distance from

Moon (km)

65,400

65,389.0

65,378.8

65,369.6

65,361.4

65,354.0

65,331.2

Figure 3.58

65,450

0

15

65,300

Figure 3.59

65,450

0

15

65,300

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Section 3.4 Quadratic Models; More on Rates of Change

Figure 3.60

Figure 3.61

65,450

65,450

0

30

0

30

65,300

65,300

c. For the time required to return to a distance of 65,400 km, we enter

Y2 ϭ 65,400 and use the 2nd TRACE (CALC) 5:intersect option. The result

shown in Figure 3.61 indicates a required time of about 25 sec. Actually, we

could have reasoned that if it took 12.5 sec to reach the minimum distance, we

A. You’ve just seen how

function models from a set

of data

Now try Exercises 11 through 14 and 45 through 48

B. Nonlinear Functions and Rates of Change

As noted in Section 1.2, one of the defining characteristics of linear functions is that

¢y

y2 Ϫ y1

ϭ

ϭ m. For nonlinear functions the rate

their rate of change is constant:

x2 Ϫ x1

¢x

of change is not constant, but to aid in their study and application, we use a related concept called the average rate of change, given by the slope of a secant line through

points (x1, y1) and (x2, y2) on the graph.

EXAMPLE 3

Calculating Average Rates of Change

The graph shown displays the number of units shipped of vinyl records, cassette

tapes, and CDs for the period 1980 to 2005.

Units shipped in millions

1000

CDs

900

Units shipped (millions)

800

700

600

500

400

300

Cassettes

200

Vinyl

100

0

2

4

6

8

10

12

14

Year (1980 → 0)

Source: Swivel.com

16

18

20

22

24

26

Year

(1980 S 0)

Vinyl

Cassette

0

323

110

0

2

244

182

0

4

205

332

6

6

125

345

53

8

72

450

150

10

12

442

287

12

2

366

408

14

2

345

662

16

3

225

779

18

3

159

847

20

2

76

942

24

1

5

767

25

1

3

705

CDs

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CHAPTER 3 Quadratic Functions and Operations on Functions

a. Find the average rate of change in CDs shipped and in cassettes shipped from

1994 to 1998. What do you notice?

b. Does it appear that the rate of increase in CDs shipped was greater from 1986

to 1992, or from 1992 to 1996? Compute the average rate of change for each

period and comment on what you find.

Solution

Using 1980 as year zero (1980 S 0), we have the following:

a.

CDs

1994: 114, 6622, 1998: 118, 8472

¢y

847 Ϫ 662

ϭ

¢x

18 Ϫ 14

185

ϭ

4

ϭ 46.25

Cassettes

1994: 114, 3452, 1998: 118, 1592

¢y

159 Ϫ 345

ϭ

¢x

18 Ϫ 14

186

ϭϪ

4

ϭ Ϫ46.5

The decrease in the number of cassettes shipped was roughly equal to the

increase in the number of CDs shipped (about 46,000,000 per year).

b. From the graph, the secant line for 1992 to 1996 appears to have a greater

slope.

1986–1992 CDs

1986: 16, 532, 1992: 112, 4082

¢y

408 Ϫ 53

ϭ

¢x

12 Ϫ 6

355

ϭ

6

ϭ 59.16

B. You’ve just seen how

we can calculate the average

rate of change for nonlinear

functions using points on a

graph

1992–1996 CDs

1992: 112, 4082, 1996: 116, 7792

¢y

779 Ϫ 16

ϭ

¢x

16 Ϫ 12

371

ϭ

4

ϭ 92.75

For the years 1986 to 1992, the average rate of change for CD sales was about

59.2 million per year. For the years 1992 to 1996, the average rate of change

was almost 93 million per year, a significantly higher rate of growth.

Now try Exercises 15 through 26, 49 and 50

C. The Average Rate of Change Formula

The importance of the rate of change concept would be hard to overstate. In many business, scientific, and economic applications, it is this attribute of a function that draws the

most attention. In Example 3 we computed average rates of change by selecting two

¢y

y2 Ϫ y1

ϭ

points from a graph, and computing the slope of the secant line: m ϭ

.

x2 Ϫ x1

¢x

With a simple change of notation, we can use the function’s equation rather than relying

on a graph. Note that y2 corresponds to the function evaluated at x2: y2 ϭ f 1x2 2 . Likewise,

f 1x2 2 Ϫ f 1x1 2

¢y

, giving

y1 ϭ f 1x1 2 . Substituting these into the slope formula yields

ϭ

x 2 Ϫ x1

¢x

the average rate of change between x1 and x2 for any function f.

Average Rate of Change

For a function f and [x1, x2] a subset of the domain,

the average rate of change between x1 and x2 is

f 1x2 2 Ϫ f 1x1 2

¢y

ϭ

, x1

x2 Ϫ x1

¢x

x2

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