课程： 10.12 功能模型的应用

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功能模型的应用

Applications of Function Models
::功能模型的应用

Here you’ll learn how to perform exponential and quadratic regression to find equations for curves that fit non-linear data sets.
::这里您将学习如何进行指数回归和二次回归, 以找到适合非线性数据集的曲线方程式。

The following table shows how many miles per gallon a car gets at different speeds.
::下表显示一辆汽车以不同速度每加仑能达到多少英里。

Speed (mph)	Miles per gallon
30	18
35	20
40	23
45	25
50	28
55	30
60	29
65	25
70	25

Using a graphing calculator, draw the scatterplot of the data, find the quadratic function of best fit, draw the quadratic function of the best fit on the scatterplot, find the speed that maximizes the miles per gallon, and predict the miles per gallon for the car if you drive at a speed of 48 mph:
::使用图形计算器,绘制数据散射图,发现最佳的二次函数,绘制散射图上最佳的二次函数,找到最大每加仑英里的速度,并预测汽车每加仑的英里,如果驾驶速度为48英里:

Step 1: Input the data.
::第1步:输入数据。

Press [STAT] and choose the [EDIT] option.
::按[STAT]键,并选择[EDIT]选项。

Input the values of $\begin{aligned} x \end{aligned}$ in the first column $\begin{aligned} (L_{1}) \end{aligned}$ and the values of $\begin{aligned} y \end{aligned}$ in the second column $\begin{aligned} (L_{2}) \end{aligned}$ . ( Note: in order to clear a list, move the cursor to the top so that $\begin{aligned} L_{1} \end{aligned}$ or $\begin{aligned} L_{2} \end{aligned}$ is highlighted. Then press [CLEAR] and then [ENTER] .)
::在第一列(L1)中输入 x 的值,在第二列(L2)中输入 y 的值。 (注:为了清除列表,将光标移动到顶部,以便突出显示 L1 或 L2 。然后按 [CLEAR] 键,然后[ENTEER]。 )

Step 2: Draw the scatterplot.
::步骤 2: 绘制散射图。

First press [Y=] and clear any function on the screen by pressing [CLEAR] when the old function is highlighted.
::在突出显示旧函数时按 [CLEAR] 并清除屏幕上的任何功能。

Press [STATPLOT] [STAT] and [Y=] and choose option 1.
::请按[统计组 [统计组和[Y=]并选择备选案文1。

Choose the ON option; after TYPE, choose the first graph type (scatterplot) and make sure that the Xlist and Ylist names match the names on top of the columns in the input table.
::选择 ON 选项; 在 TYPE 之后, 请选择第一个图形类型( 缓存) , 并确保 X 列表和 Y 列表中的名称符合输入表格列上方的名称。

Press [GRAPH] and make sure that the window is set so you see all the points in the scatterplot. In this case, the settings should be $\begin{aligned} 30 \leq x \leq 80 \end{aligned}$ and $\begin{aligned} 0 \leq y \leq 40 \end{aligned}$ . You can set the window size by pressing the [WINDOW] key at the top.
::按 [GRAPH] 键并确保窗口设置, 以便您看到散射图中的所有点。在此情况下, 设置应该是 30 x80 和 0y40。您可以按上方 [WINDOW] 键来设置窗口大小。

Step 3: Perform quadratic regression.
::第三步:进行二次回归。

Press [STAT] and use the right arrow to choose [CALC] .
::按[STAT]键,使用右箭头选择 [CALC]。

Choose Option 5 (QuadReg) and press [ENTER] . You will see “QuadReg” on the screen.
::选择选项 5 (QuadReg) 并按 [ENTER] 键。您将在屏幕上看到“ QuadReg ” 。

Type in $\begin{aligned} L_{1}, L_{2} \end{aligned}$ after ‘QuadReg’ and press [ENTER] . The calculator shows the quadratic function: $\begin{aligned} y = - 0.017 x^{2} + 1.9 x - 25 \end{aligned}$
::L1,L2 类型为 L1, L2, 后 adReg 和按 [ENTER] 键。计算器显示二次函数 : y 0.017x2+1. 9x- 25

Step 4: Graph the function.
::第4步:绘制函数图。

Press [Y=] and input the function you just found.
::按 [Y=] 键并输入您刚刚找到的函数。

Press [GRAPH] and you will see the curve of best fit drawn over the data points.
::按[GRAPH]键,你会看到数据点上最合适的曲线。

To find the speed that maximizes the miles per gallon, use [TRACE] and move the cursor to the top of the parabola . You can also use [CALC] [2nd] [TRACE] and option 4:Maximum, for a more accurate answer. The speed that maximizes miles per gallon is 56 mph.
::要找到最大每加仑里程数的速度, 请使用 [TRACE] , 并将光标移动到抛物线顶部。您也可以使用 [CALC][ 2(nd) [TRACE] 和选项 4: Meximum, 以获得更准确的答案。每加仑里程数的最大速度为56 英里。

Finally, plug $\begin{aligned} x = 48 \end{aligned}$ into the equation you found: $\begin{aligned} y = - 0.017 (48)^{2} + 1.9 (48) - 25 = 27.032 m i l e s p e r g a l l o n . \end{aligned}$
::最后,在您找到的方程式中插入 x=48: y 0.017( 48)( 482)+1. 9( 48)- 25=27. 032英里/加仑。

Note: The image above shows our function plotted on the same graph as the data points from the table. One thing that is clear from this graph is that predictions made with this function won’t make sense for all values of $\begin{aligned} x \end{aligned}$ . For example, if $\begin{aligned} x < 15 \end{aligned}$ , this graph predicts that we will get negative mileage, which is impossible. Part of the skill of using regression on your calculator is being aware of the strengths and limitations of this method of fitting functions to data.
::注意 : 上面的图像显示我们与表格中的数据点相同的图表所绘制的函数。从这个图表中可以清楚地看到一件事, 使用这个函数所作的预测对于 x 的所有值来说是没有意义的。例如, 如果 x < 15 , 这个图预测我们将获得负里程, 而这是不可能的。在您的计算器上使用回归法的技巧的一部分, 正在意识到这个对数据进行适当函数的方法的优点和局限性。

Real-World Application: Investing Money
::现实世界应用:投资投资

The following table shows the amount of money an investor has in an account each year for 10 years.
::下表显示投资者在10年内每年在一个账户中存入的金额。

Year	Value of account
1996	$5000
1997	$5400
1998	$5800
1999	$6300
2000	$6800
2001	$7300
2002	$7900
2003	$8600
2004	$9300
2005	$10000
2006	$11000

Using a graphing calculator, draw a scatterplot of the value of the account as the dependent variable , and the number of years since 1996 as the independent variable, find the exponential function that fits the data, draw the exponential function on the scatterplot, and determine what the value of the account will be in 2020:
::使用图形计算器, 绘制账户作为附属变量的值的散射图, 以及自1996年以来作为独立变量的年数, 找到符合数据的指数函数, 在散射图上绘制指数函数, 并确定2020年账户的值 :

Step 1: Input the data.
::第1步:输入数据。

Press [STAT] and choose the [EDIT] option.
::按[STAT]键,并选择[EDIT]选项。

Step 2: Draw the scatterplot.
::步骤 2: 绘制散射图。

First press [Y=] and clear any function on the screen.
::第一按 [Y=] 并清除屏幕上的任何功能。

Press [GRAPH] and choose Option 1.
::按[GRAPH]键,并选择备选案文1。

Choose the ON option and make sure that the Xlist and Ylist names match the names on top of the columns in the input table.
::选择 ON 选项,并确保 Xlist 和 Y list 名称符合输入表格列顶部的名称。

Press [GRAPH] , and make sure that the window is set so you see all the points in the scatterplot. In this case the settings should be $\begin{aligned} 0 \leq x \leq 10 \end{aligned}$ and $\begin{aligned} 0 \leq y \leq 11000 \end{aligned}$ .
::按 [GRAPH] 键, 并确保窗口被设置, 这样您就可以看到撒布点中的所有点。在此情况下, 设置应该为 0x10 和 0y11000 。

Step 3: Perform exponential regression.
::步骤3: 进行指数回归。

Press [STAT] and use the right arrow to choose [CALC] .
::按[STAT]键,使用右箭头选择 [CALC]。

Choose Option 0 and press [ENTER] . You will see “ExpReg” on the screen.
::选择选项 0 并按 [ENTER] 键。您可以在屏幕上看到“ Expreg ” 。

Press [ENTER] . The calculator shows the exponential function: $\begin{aligned} y = 4975.7 (1.08)^{x} \end{aligned}$
::按 [ENTER] 按下。计算器显示指数函数: y=4975.7(1.08)x

Step 4: Graph the function.
::第4步:绘制函数图。

Press [Y=] and input the function you just found. Press [GRAPH].
::按 [Y=] 键并输入您刚刚找到的函数。按 [GRAPH] 键。

Finally, plug $\begin{aligned} x = 2020 - 1996 = 24 \end{aligned}$ into the function: $\begin{aligned} y = 4975.7 (1.08)^{24} = \underline{$ 31551.81} \end{aligned}$
::最后,在函数中插入 x=2020 - 1996=24:y=4975.7(1.0824)=31551.81_

In 2020, the account will have a value of $31551.81.
::2020年,该账户的价值将为31551.81美元。

Note: The function above is the curve that comes closest to all the data points. It won’t return $\begin{aligned} y - \end{aligned}$ values that are exactly the same as in the data table, but they will be close. It is actually more accurate to use the curve fit values than the data points.
::注意 : 上面的函数是最接近所有数据点的曲线。它不会返回与数据表格完全相同的 y - 值, 但将会接近。使用曲线匹配值比数据点更准确。

Solve Applications by Comparing Function Models
::通过比较函数模型解决应用程序

Real-World Application: School Enrollment
::现实世界应用:学校入学

The following table shows the number of students enrolled in public elementary schools in the US (source: US Census Bureau). Make a scatterplot with the number of students as the dependent variable, and the number of years since 1990 as the independent variable. Find which curve fits this data the best and predict the school enrollment in the year 2007.
::下表显示了美国公立小学入学学生人数(资料来源:美国人口普查局)。用学生人数作为依附变量,用1990年以来作为独立变量的年数做一个散射点。找出哪个曲线最适合这一数据,并预测2007年的入学率。

Year	Number of students (millions)
1990	26.6
1991	26.6
1992	27.1
1993	27.7
1994	28.1
1995	28.4
1996	28.1
1997	29.1
1998	29.3
2003	32.5

We need to perform linear, quadratic and exponential regression on this data set to see which function represents the values in the table the best.
::我们需要在这个数据集上进行线性、二次和指数回归,以确定哪个函数代表表中的数值是最好的。

Step 1: Input the data.
::第1步:输入数据。

Step 2: Draw the scatterplot.
::步骤 2: 绘制散射图。

Set the window size: $\begin{aligned} 0 \leq x \leq 10 \end{aligned}$ and $\begin{aligned} 20 \leq y \leq 40. \end{aligned}$
::设定窗口大小: 0x10 和 20y40 。

Here is the scatterplot:
::以下是散射图 :

Step 3: Perform Regression.
::第3步:执行倒退。

Linear Regression
::线回归

The function of the line of best fit is $\begin{aligned} y = 0.51 x + 26.1 \end{aligned}$ . Here is the graph of the function on the scatterplot:
::最合适行的函数是 y=0.51x+26.1。这是分布图上的函数图 :

Quadratic Regression
::二次倒退

The quadratic function of best fit is $\begin{aligned} y = 0.064 x^{2} - .067 x + 26.84 \end{aligned}$ . Here is the graph of the function on the scatterplot:
::最佳的二次函数为 y= 0.064x2 -. 067x+26.84。这是散射点上的函数图 :

Exponential Regression
::指数回归

The exponential function of best fit is $\begin{aligned} y = 26.2 (1.018)^{x} \end{aligned}$ . Here is the graph of the function on the scatterplot:
::最适合的指数函数为 Y=26.2(1.018x) 。这是散射图上的函数图示 :

From the graphs, it looks like the quadratic function is the best fit for this data set. We’ll use this function to predict school enrollment in 2007.
::从图表看,四边函数似乎是最适合这一数据集的。我们将利用这一功能预测2007年的入学率。

$\begin{aligned} x = 2007 - 1990 = 17 \end{aligned}$ so $\begin{aligned} y = 0.064 (17)^{2} - .067 (17) + 26.84 \end{aligned}$ which is 44.2 million students.
::x=2007-1990=17,y=0.064(17)2-.067(17)+26.84,共4 420万名学生。

Example
::示例示例示例示例

Example 1
::例1

The profits in dollars of a company are given in the table below. Find the model that describes the relationship to profit as a function of time in years:
::下表列出公司以美元计的利润。

\begin{aligned} Time in Years & 0 & 1 & 2 & 3 & 4 \\ Profit in Dollars & 0 & 5000 & 20000 & 45000 & 80000 \end{aligned}

::01234 以05 000200004500080000美元计的利润

Start by graphing the points, to get a sense of the shape.
::以绘制点形图开始, 以获得形状感知。

This curve looks like it could be quadratic or exponential. If you check the ratios, you would see they are not the same. Checking the differences of differences yields:
::这个曲线看起来可能是二次曲线或指数曲线。如果您检查了比率, 你会看到它们不一样。检查差异的产量差异 :

\begin{aligned} Time in Years & 0 & 1 & 2 & 3 & 4 \\ Profit in Dollars & 0 & 5000 & 20000 & 45000 & 80000 \\ Differences & 5000 - 0 = 5000 & 20000 - 5000 = 15000 & 45000 - 20000 = 25000 & 80000 - 45000 = 35000 \\ Differences of Differences & 15000 - 5000 = 10000 & 25000 - 15000 = 10000 & 35000 - 25000 = 10000 \end{aligned}

::时间 01234 以 05 000 000200004 5000080000美元以 05 000 000200004 5000080000美元以5 0000-0 = 50002000000 - 50000= 15 00045000 - 20000= 2500080000 - 45000= 35000 以差异 15 000 - 50000= 100002 50000 - 1 500000=100035000 - 2 50000=1000

Since the differences of differences are the same, this is a quadratic model.
::由于差异的差别相同,这是一个二次模型。

Review
::回顾

For 1-5, suppose as a ball bounces up and down, the maximum height that the ball reaches continually decreases from one bounce to the next. For a given bounce, this table shows the height of the ball with respect to time:
::对于1-5,假设一个球向上和向下跳动,球达到的最大高度从一个向下不断下降。对于一个特定弹跳,本表显示球相对于时间的高度:

Time (seconds)	Height (inches)
2	2
2.2	16
2.4	24
2.6	33
2.8	38
3.0	42
3.2	36
3.4	30
3.6	28
3.8	14
4.0	6

Using a graphing calculator, answer the following questions:
::使用图形计算计算器,回答下列问题:

Draw the scatterplot of the data
::绘制数据分布图

Find the quadratic function of best fit
::找到最适合的二次函数

Draw the quadratic function of best fit on the scatterplot
::在散点图上绘制最适合的二次函数

Find the maximum height the ball reaches on the bounce
::找到弹跳时球达到的最大高度

Predict how high the ball is at time $\begin{aligned} t = 2.5 s e c o n d s \end{aligned}$
::预测球在时间t=2.5秒时有多高

For 6-9, a chemist has a 250 gram sample of a radioactive material. She records the amount of radioactive material remaining in the sample every day for a week and obtains the data in the following table:
::就6-9而言,化学家拥有250克放射性材料样本,她每周每天记录在样品中的放射性物质数量,并在下表中获得数据:

Day	Weight (grams)
0	250
1	208
2	158
3	130
4	102
5	80
6	65
7	50

Use a graphing calculator to answer the following questions:
::使用图形计算计算器回答下列问题:

Draw a scatterplot of the data
::绘制数据散射图

Find the exponential function of best fit
::查找最适合的指数函数

Draw the exponential function of best fit on the scatterplot
::在散点绘图上绘制最适合的指数函数

Predict the amount of material after 10 days
::10天后预测材料数量

For 10-12 use the following table, which shows the rate of pregnancies (per 1000) for US women aged 15 to 19. (source: US Census Bureau).
::10-12使用下表显示美国15至19岁妇女的怀孕率(每1000人)(资料来源:美国人口普查局)。

Make a scatterplot with the rate of pregnancies as the dependent variable and the number of years since 1990 as the independent variable.
::将怀孕率作为依附变量进行撒布,将1990年以来作为独立变量的年数作为独立变量。

Find which type of curve fits this data best.
::查找哪种曲线最适合此数据。

Predict the rate of teen pregnancies in the year 2010.
::预测2010年青少年怀孕率。

Year	Rate of pregnancy (per 1000)
1990	116.9
1991	115.3
1992	111.0
1993	108.0
1994	104.6
1995	99.6
1996	95.6
1997	91.4
1998	88.7
1999	85.7
2000	83.6
2001	79.5
2002	75.4

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

章节大纲

常规

功能模型的应用

Applications of Function Models ::功能模型的应用

Real-World Application: Investing Money ::现实世界应用:投资投资

Real-World Application: School Enrollment ::现实世界应用:学校入学

Example ::示例示例示例示例

Example 1 ::例1

Review ::回顾

Review (Answers) ::回顾(答复)

Applications of Function Models
::功能模型的应用

Real-World Application: Investing Money
::现实世界应用:投资投资

Real-World Application: School Enrollment
::现实世界应用:学校入学

Example
::示例示例示例示例

Example 1
::例1

Review
::回顾

Review (Answers)
::回顾(答复)