4 Quadratic Models; More on Rates of Change
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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
a quadratic regression.
b. Using STAT
(CALC) 5:QuadReg
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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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=
screen) a gradually decreasing,
nonlinear pattern and a quadratic regression
seems appropriate (Figure 3.58).
b. We find the regression equation
using STAT
(CALC) 5:QuadReg
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
could double this time to return to the original distance.
A. You’ve just seen how
we can develop quadratic
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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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