8.12: 职务职能的应用

Section outline

• Select section General

  Collapse Expand

  General

  Collapse all Expand all

  • Select activity 新闻通告

    

    新闻通告 Forum
• Select section A. 职务应用

  Collapse Expand

  A. 职务应用

  播放段落

  Applications of Exponential Functions 
  ::A. 职务应用

  播放段落

  For her eighth birthday, Shelley’s grandmother gave her a full bag of candy. Shelley counted her candy and found out that there were 160 pieces in the bag. As you might suspect, Shelley loves candy, so she ate half the candy on the first day. Then her mother told her that if she eats it at that rate , the candy will only last one more day—so Shelley devised a clever plan. She will always eat half of the candy that is left in the bag each day. She thinks that this way she can eat candy every day and never run out.
  ::她八岁生日时,雪莱的祖母给了她一整包糖果。雪莱点数了她的糖果,发现袋子里有160块。 正如你可能怀疑的那样,雪莱喜欢糖果,所以她在第一天吃了一半糖果。 接着,她妈妈告诉她,如果她以这个速度吃了糖果,糖果只能多吃一天 — — 所以雪莱设计了一个聪明的计划。 她总是吃袋子里剩下的一半糖果。 她认为这样她可以每天吃糖果,永远不跑。

  播放段落

  How much candy does Shelley have at the end of the week? Will the candy really last forever?
  ::雪莉周末有多少糖果?

  播放段落

  Let’s make a table of values for this problem.
  ::让我们为这个问题绘制一个价值表。

  播放段落

  $$\begin{array}{r}\begin{array}{|cc|}\hline \text{Day}& \text{\# of candies}\\ 0& 160\\ 1& 80\\ 3& 20\\ 4& 10\\ 5& 5\\ 6& 2.5\\ 7& 1.25\\ \hline\end{array}\end{array}$$

  
  ::农庄节01601803204105562.571.25

  播放段落

  You can see that if Shelley eats half the candies each day, then by the end of the week she only has 1.25 candies left in her bag.
  ::你可以看到,如果雪莱每天吃一半的糖果, 那么在周末 她只有1.25只糖果留在她的包里。

  播放段落

  Let’s write an equation for this exponential function . Using the formula $\begin{array}{r}y=A\cdot b^x\end{array}$ , we can see that $\begin{array}{r}A\end{array}$ is 160. The number of candies she starts out with and $\begin{array}{r}b\end{array}$ is $\begin{array}{r}\frac{1}{2}\end{array}$ , so our equation is $\begin{array}{r}y=160\cdot {\left(\frac{1}{2}\right)}^x\end{array}$
  ::让我们为这个指数函数写一个方程式。 使用公式 y= Abx, 我们可以看到 A 是 160。 她开头的糖果数量是 12, 因此我们的方程式是 y=160( 12) x

  播放段落

  Now let’s graph this function . The resulting graph is shown below.
  ::现在让我们来图解这个函数。 由此产生的图表显示在下面。

  

  播放段落

  So, will Shelley’s candy last forever? We saw that by the end of the week she has 1.25 candies left, so there doesn’t seem to be much hope for that. But if you look at the graph, you’ll see that the graph never really gets to zero. Theoretically there will always be some candy left, but Shelley will be eating very tiny fractions of a candy every day after the first week!
  ::因此,雪莱的糖果会永远存在吗?我们看到,在她只剩下1.25只糖果的周末,所以似乎没有多少希望。 但是,如果你看看图表,你就会看到图表永远不会真正达到零。 从理论上讲,总是有一些糖果,但雪莱在第一周之后每天都会吃很少的糖果。

  播放段落

  This is a fundamental feature of an exponential decay function . Its values get smaller and smaller but never quite reach zero. In mathematics, we say that the function has an asymptote at $\begin{array}{r}y=0\end{array}$ ; in other words, it gets closer and closer to the line $\begin{array}{r}y=0\end{array}$ but never quite meets it.
  ::这是指数衰变函数的一个基本特征。 它的值越来越小, 越来越小, 却从未达到零。 在数学中, 我们说函数在 y=0 时是零; 换句话说, 它越来越接近 y=0 线, 但从未达到 y=0 线 。

  播放段落

  Problem-Solving Strategies
  ::解决问题战略

  播放段落

  Remember our problem-solving plan from earlier?
  ::记得我们以前的解决问题计划吗?

  播放段落
  1. Understand the problem.
     ::了解问题所在。
  播放段落
  2. Devise a plan – Translate.
     ::制定计划—翻译。
  播放段落
  3. Carry out the plan – Solve.
     ::执行计划——解决。
  播放段落
  4. Look – Check and Interpret.
     ::- 检查和解释。

  播放段落

  We can use this plan to solve application problems involving . Compound interest, loudness of sound, population increase, population decrease or radioactive decay are all applications of exponential functions. In these problems, we’ll use the methods of constructing a table and identifying a pattern to help us devise a plan for solving the problems.
  ::我们可以用这个计划来解决应用问题,其中涉及:复合兴趣、声音的响亮、人口增长、人口减少或放射性衰减都是指数函数的应用。 在这些问题上,我们将使用方法构建一张桌子并确定一种模式来帮助我们制定解决问题的计划。

  播放段落

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

  播放段落

  Suppose $4000 is invested at 6% interest compounded annually. How much money will there be in the bank at the end of 5 years? At the end of 20 years?
  ::假设每年投资4000美元,利率为6%,加上年利率。 5年后银行将有多少资金? 20年后?

  播放段落

  Step 1: Read the problem and summarize the information.
  ::第1步:阅读问题并总结信息。

  播放段落

  $4000 is invested at 6% interest compounded annually; we want to know how much money we have in five years.
  ::每年投资4000美元,利率为6%;我们想知道五年来我们有多少资金。

  播放段落

  Assign variables:
  ::指派变量 :

  播放段落

  Let $\begin{array}{r}x=\end{array}$ time in years
  ::Let x= 年时间

  播放段落

  Let $\begin{array}{r}y=\end{array}$ amount of money in investment account
  ::Let y= 投资账户中的金额

  播放段落

  Step 2: Look for a pattern.
  ::第2步:寻找一个模式。

  播放段落

  We start with $4000 and each year we add 6% interest to the amount in the bank.
  ::我们从4000美元开始,每年在银行金额中增加6%的利息。

  播放段落

  $$\begin{array}{rlrl}& \text{Start:}& & \mathrm{\$}4000\\ & 1^{st}\text{year:}& & \text{Interest}=4000\times (0.06)=\mathrm{\$}240\\ & & & \text{This is added to the previous amount:}\mathrm{\$}4000+\mathrm{\$}4000\times (0.06)\\ & & & =\mathrm{\$}4000(1+0.06)\\ & & & =\mathrm{\$}4000(1.06)\\ & & & =\mathrm{\$}4240\\ & 2^{nd}\text{year}& & \text{Previous amount}+\text{interest on the previous amount}\\ & & & =\mathrm{\$}4240(1+0.06)\\ & & & =\mathrm{\$}4240(1.06)\\ & & & =\mathrm{\$}4494.40\end{array}$$

  
  ::起始年:400001美元: Interest=40000x(0.06)=240美元 240美元 上一个数额中加:400美元+400x(0.06)=400美元(1+0.06)=400美元(1+0.06)=40000(11.06)=4240美元=4240美元=4240美元(1+0.06)=4240(1.06)=4494美元。

  播放段落

  The pattern is that each year we multiply the previous amount by the factor of 1.06.
  ::模式是,每年我们将前一数额乘以1.06系数。

  播放段落

  Let’s fill in a table of values:
  ::让我们填写一个数值表:

  播放段落

  $$\begin{array}{r}\begin{array}{|cc|}\hline \text{Time (years)}& \text{Investments amount}(\mathrm{\$})\\ 0& 4000\\ 1& 4240\\ 2& 4494.40\\ 3& 4764.06\\ 4& 5049.90\\ 5& 5352.90\\ \hline\end{array}\end{array}$$

  
  ::投资金额(美元)

  播放段落

  We see that at the end of five years we have $5352.90 in the investment account .
  ::我们看到,在五年结束时,我们在投资账户中有5352.90美元。

  播放段落

  Step 3: Find a formula.
  ::步骤3:找到一个公式。

  播放段落

  We were able to find the amount after 5 years just by following the pattern, but rather than follow that pattern for another 15 years, it’s easier to use it to find a general formula. Since the original investment is multiplied by 1.06 each year, we can use exponential notation. Our formula is $\begin{array}{r}y=4000\cdot (1.06{)}^x\end{array}$ , where $\begin{array}{r}x\end{array}$ is the number of years since the investment.
  ::5年后,我们通过遵循模式找到了这笔金额,但与其再遵循15年,不如用它寻找一个通用公式。 由于最初的投资每年乘以1.06,我们可以使用指数符号。 我们的公式是y=4000(1.06)x,其中x是投资以来的年数。

  播放段落

  To find the amount after 5 years we plug $\begin{array}{r}x=5\end{array}$ into the equation:
  ::为了在5年后找到这个数量,我们把X=5插入方程式:

  播放段落

  $$\begin{array}{r}y=4000\cdot (1.06{)}^5=\mathrm{\$}5352.90\end{array}$$

  
  ::y=4,000_______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

  播放段落

  To find the amount after 20 years we plug $\begin{array}{r}x=20\end{array}$ into the equation:
  ::为了在20年后找到这个数量,我们把x=20插入方程式:

  播放段落

  $$\begin{array}{r}y=4000\cdot (1.06{)}^{20}=\mathrm{\$}12828.54\end{array}$$

  
  ::y=4,000_____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

  播放段落

  Step 4: Check.
  ::第四步:检查。

  播放段落

  Looking back over the solution, we see that we obtained the answers to the questions we were asked and the answers make sense.
  ::回顾解决办法,我们看到,我们得到了对我们所提问题的答案,答案是有道理的。

  播放段落

  To check our answers, we can plug some low values of $\begin{array}{r}x\end{array}$ into the formula to see if they match the values in the table:
  ::为了检查我们的答案, 我们可以在公式中插入一些低的 x 值, 看看它们是否与表格中的值匹配 :

  播放段落

  $\begin{array}{r}x=0:y=4000\cdot (1.06{)}^0=4000\end{array}$
  ::x=0: Y=4000( 1.060)=4000

  播放段落

  $\begin{array}{r}x=1:y=4000\cdot (1.06{)}^1=4240\end{array}$
  ::x=1: y= 4000( 1. 061) = 4240

  播放段落

  $\begin{array}{r}x=2:y=4000\cdot (1.06{)}^2=4494.4\end{array}$
  ::x= 2: y= 4000( 1. 062) = 4494.4

  播放段落

  The answers match the values we found earlier. The amount of increase gets larger each year, and that makes sense because the interest is 6% of an amount that is larger every year.
  ::答案与我们早些时候发现的值相符。 增加的金额逐年增加, 这很有意义, 因为利息是每年6%, 而每年的利息都比较大。

  播放段落

  Real-World Application: Population 
  ::现实世界应用:人口

  播放段落

  In 2002 the population of schoolchildren in a city was 90,000. This population decreases at a rate of 5% each year. What will be the population of school children in year 2010?
  ::2002年,一个城市的学龄儿童人口为90 000人,每年下降5%,2010年在校儿童人口是多少?

  播放段落

  Step 1: Read the problem and summarize the information.
  ::第1步:阅读问题并总结信息。

  播放段落

  The population is 90,000; the rate of decrease is 5% each year; we want the population after 8 years.
  ::人口为90 000人;每年下降5%;8年后我们希望人口。

  播放段落

  Assign variables:
  ::指派变量 :

  播放段落

  Let $\begin{array}{r}x=\end{array}$ time since 2002 (in years)
  ::Let x= 2002年以来的时间(年)

  播放段落

  Let $\begin{array}{r}y=\end{array}$ population of school children
  ::让y=在校儿童人口

  播放段落

  Step 2: Look for a pattern.
  ::第2步:寻找一个模式。

  播放段落

  Let’s start in 2002, when the population is 90,000.
  ::让我们从2002年开始, 当时人口为90,000人。

  播放段落

  The rate of decrease is 5% each year, so the amount in 2003 is 90,000 minus 5% of 90,000, or 95% of 90,000.
  ::下降率为每年5%,因此2003年的下降率为90,000减去90,000的5%,即90,000的95%。

  播放段落

  $$\begin{array}{rl}& \text{In}2003:\text{Population}=90,000\times 0.95\\ & \text{In}2004:\text{Population}=90,000\times 0.95\times 0.95\end{array}$$

  
  ::2003年:人口=90 000x0.95 2004年:人口=90 000x0.95x0.95

  播放段落

  The pattern is that for each year we multiply by a factor of 0.95
  ::模式是每年我们乘以0.95乘以0.95

  播放段落

  Let’s fill in a table of values:
  ::让我们填写一个数值表:

  播放段落

  $$\begin{array}{r}\begin{array}{|cc|}\hline \text{Year}& \text{Population}\\ 2002& 90,000\\ 2003& 85,500\\ 2004& 81,225\\ 2005& 77,164\\ 2006& 73,306\\ 2007& 69,640\\ \hline\end{array}\end{array}$$

  
  ::人口 2002年 人口 2002年 人口 2002年 人口

  播放段落

  Step 3: Find a formula.
  ::步骤3:找到一个公式。

  播放段落

  Since we multiply by 0.95 every year, our exponential formula is $\begin{array}{r}y=90000\cdot (0.95{)}^x\end{array}$ , where $\begin{array}{r}x\end{array}$ is the number of years since 2002. To find the population in 2010 (8 years after 2002), we plug in $\begin{array}{r}x=8\end{array}$ :
  ::由于我们每年乘以0.95,我们的指数公式为y=90000(0.95x)x,x是2002年以来的年数。 为了在2010年找到人口(2002年之后8年),我们插入x=8:

  播放段落

  $\begin{array}{r}y=90000\cdot (0.95{)}^8=59,708\end{array}$ schoolchildren.
  ::y=90000(0.958)8=59,708名学童。

  播放段落

  Step 4: Check.
  ::第四步:检查。

  播放段落

  Looking back over the solution, we see that we answered the question we were asked and that it makes sense. The answer makes sense because the numbers decrease each year as we expected. We can check that the formula is correct by plugging in the values of $\begin{array}{r}x\end{array}$ from the table to see if the values match those given by the formula.
  ::回顾解决方案,我们看到我们回答了我们被问到的问题,这是有道理的。答案是有道理的,因为数字每年都会像我们预期的那样下降。我们可以通过从表格中插入x的值来检查公式是否正确,看看这些值是否与公式给出的值相符。

  播放段落

  $$\begin{array}{rl}& \text{Year}2002,x=0:\text{Population}=y=90000\cdot (0.95{)}^0=90,000\\ & \text{Year}2003,x=1:\text{Population}=y=90000\cdot (0.95{)}^1=85,500\\ & \text{Year}2004,x=2:\text{Population}=y=90000\cdot (0.95{)}^2=81,225\end{array}$$

  
  ::2002年,x=0:人口=y=y=90000__(0.950=900000=900000__(0.950)=900000×2003年,x=1:人口=y=90000__(0.95)=85 500 2004年,x=2:人口=y=90000__(0.952=81 225)

  播放段落

  Solve Real-World Problems Involving Exponential Growth
  ::解决涉及指数增长的现实世界问题

  播放段落

  Now we’ll look at some more real-world problems involving exponential functions. We’ll start with situations involving .
  ::现在,我们将审视一些涉及指数函数的更真实的世界问题。 我们将从涉及的情况开始。

  播放段落

  The population of a town is estimated to increase by 15% per year. The population today is 20 thousand. Make a graph of the population function and find out what the population will be ten years from now.
  ::一个城镇的人口估计每年增加15%,今天的人口为20 000人,绘制人口功能图,并找出十年后的人口。

  播放段落

  First, we need to write a function that describes the population of the town.
  ::首先,我们需要写一个描述该镇居民的功能。

  播放段落

  The general form of an exponential function is $\begin{array}{r}y=A\cdot b^x\end{array}$ .
  ::指数函数的一般形式为 y= Abx。

  播放段落

  Define $\begin{array}{r}y\end{array}$ as the population of the town.
  ::将y定义为该镇的人口。

  播放段落

  Define $\begin{array}{r}x\end{array}$ as the number of years from now.
  ::将x定义为从现在起的年数。

  播放段落

  $\begin{array}{r}A\end{array}$ is the initial population, so $\begin{array}{r}A=20\end{array}$ (thousand).
  ::A是初始人口,因此A=20(千)。

  播放段落

  Finally we must find what $\begin{array}{r}b\end{array}$ is. We are told that the population increases by 15% each year. To calculate percents we have to change them into decimals: 15% is equivalent to 0.15. So each year, the population increases by 15% of $\begin{array}{r}A\end{array}$ , or $\begin{array}{r}0.15A\end{array}$ .
  ::最后,我们必须找到什么是b。我们被告知,人口每年增加15 % 。为了计算百分比,我们必须将其转换为小数点:15 % 等于 0.15。因此,人口每年增加15 % A, 即0.15A。

  播放段落

  To find the total population for the following year, we must add the current population to the increase in population. In other words, $\begin{array}{r}A+0.15A=1.15A\end{array}$ . So the population must be multiplied by a factor of 1.15 each year. This means that the base of the exponential is $\begin{array}{r}b=1.15\end{array}$ .
  ::为了找到下一年的总人口,我们必须在人口增加的同时加上目前的人口。换句话说,A+0.15A=1.15A。所以人口每年必须乘以1.15倍。这意味着指数的基数是b=1.15。

  播放段落

  The formula that describes this problem is $\begin{array}{r}y=20\cdot (1.15{)}^x\end{array}$ .
  ::描述这个问题的公式为 Y=20( 1. 15x) 。

  播放段落

  Now let’s make a table of values.
  ::现在让我们绘制一个数值表。

  $\begin{array}{r}x\end{array}$	$\begin{array}{r}y=20\cdot (1.15{)}^x\end{array}$
  -10	4.9
  -5	9.9
  0	20
  5	40.2
  10	80.9

  播放段落

  Now we can graph the function.
  ::现在我们可以绘制函数图。

  

  播放段落

  Notice that we used negative values of $\begin{array}{r}x\end{array}$ in our table of values. Does it make sense to think of negative time? Yes; negative time can represent time in the past. For example, $\begin{array}{r}x=-5\end{array}$ in this problem represents the population from five years ago.
  ::注意我们在数值表中使用了 x 的负值。 想象负时间是否合理 ? 是的; 负时间可以代表过去的时间 。 例如, 这一问题中的 x+% 5 代表了五年前的人口 。

  播放段落

  The question asked in the problem was: what will be the population of the town ten years from now? To find that number, we plug $\begin{array}{r}x=10\end{array}$ into the equation we found: $\begin{array}{r}y=20\cdot (1.15{)}^{10}=80,911\end{array}$ .
  ::问题在于:从现在起十年之后,该镇的人口将是什么?为了找到这个数字,我们将x=10插入我们发现的方程式:y=20(1.15)10=80,911。

  播放段落

  The town will have 80,911 people ten years from now.
  ::10年后 镇上将有80,911人

  播放段落

  Real-World Application: Compounding Interest 
  ::现实世界应用:增加利息

  播放段落

  Peter earned $1500 last summer. If he deposited the money in a bank account that earns 5% interest compounded yearly, how much money will he have after five years?
  ::彼得去年夏天赚了1500美元。 如果他把钱存入一个银行帐户,每年赚取5%的利息,那么五年后他有多少钱?

  播放段落

  This problem deals with interest which is compounded yearly. This means that each year the interest is calculated on the amount of money you have in the bank. That interest is added to the original amount and next year the interest is calculated on this new amount, so you get paid interest on the interest.
  ::这个问题涉及每年复杂的利息问题。 这意味着每年的利息是根据银行的金额计算的。 利息加到原来的金额, 明年的利息加到新的金额上, 这样您就可以得到利息的利息。

  播放段落

  Let’s write a function that describes the amount of money in the bank.
  ::让我们写一个描述银行金额的函数。

  播放段落

  The general form of an exponential function is $\begin{array}{r}y=A\cdot b^x\end{array}$ .
  ::指数函数的一般形式为 y= Abx。

  播放段落

  Define $\begin{array}{r}y\end{array}$ as the amount of money in the bank.
  ::将y定义为银行中的资金数额。

  播放段落

  Define $\begin{array}{r}x\end{array}$ as the number of years from now.
  ::将x定义为从现在起的年数。

  播放段落

  $\begin{array}{r}A\end{array}$ is the initial amount, so $\begin{array}{r}A=1500\end{array}$ .
  ::A 是初始金额, 所以 A= 1500 。

  播放段落

  Now we have to find what $\begin{array}{r}b\end{array}$ is.
  ::现在我们必须找到b是什么。

  播放段落

  We’re told that the interest is 5% each year, which is 0.05 in decimal form. When we add $\begin{array}{r}0.05A\end{array}$ to $\begin{array}{r}A\end{array}$ , we get $\begin{array}{r}1.05A\end{array}$ , so that is the factor we multiply by each year. The base of the exponential is $\begin{array}{r}b=1.05\end{array}$ .
  ::我们被告知,每年的利息为5%,小数点为0.05。 当我们把0.05A添加到A时,我们得到1.05A,这就是每年我们乘以的系数。指数的基数是b=1.05。

  播放段落

  The formula that describes this problem is $\begin{array}{r}y=1500\cdot {1.05}^x\end{array}$ . To find the total amount of money in the bank at the end of five years, we simply plug in $\begin{array}{r}x=5\end{array}$ .
  ::描述这个问题的公式是y=15001.05x。为了在5年结束时找到银行中的资金总额,我们只需插入 x=5。

  播放段落

  $$\begin{array}{r}y=1500\cdot (1.05{)}^5=\mathrm{\$}1914.42\end{array}$$

  
  ::y=1500__________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

  播放段落

  Solve Real-World Problems Involving
  ::解决涉及的现实世界问题

  播放段落

  Exponential decay problems appear in several application problems. Some examples of these are half-life problems and depreciation problems . Let’s solve an example of each of these problems.
  ::某些应用问题中出现了指数衰变问题,其中有一些是半衰期问题和折旧问题。 让我们解决其中每一个问题的例子。

  播放段落

  A radioactive substance has a half-life of one week. In other words, at the end of every week the level of radioactivity is half of its value at the beginning of the week. The initial level of radioactivity is 20 counts per second.
  ::放射性物质的半衰期为一周,换言之,在每周结束时,放射性水平是本周初值的一半,最初的放射性水平是每秒20次。

  播放段落

  Draw the graph of the amount of radioactivity against time in weeks.
  ::绘制周内按时间计的放射量图。

  播放段落

  Find the formula that gives the radioactivity in terms of time.
  ::找到给放射性时间的公式。

  播放段落

  Find the radioactivity left after three weeks.
  ::3周后找到剩下的放射物

  播放段落

  Let’s start by making a table of values and then draw the graph.
  ::让我们先绘制一个数值表,然后绘制图表。

  Time	Radioactivity
  0	20
  1	10
  2	5
  3	2.5
  4	1.25
  5	0.625

  

  播放段落

  Exponential decay fits the general formula $\begin{array}{r}y=A\cdot b^x\end{array}$ . In this case:
  ::指数衰变符合一般公式y=Abx。在此情况下:

  播放段落

  $\begin{array}{r}y\end{array}$ is the amount of radioactivity
  ::y 是放射量

  播放段落

  $\begin{array}{r}x\end{array}$ is the time in weeks
  ::x 周中的时间

  播放段落

  $\begin{array}{r}A=20\end{array}$ is the starting amount
  ::A=20 是起始金额

  播放段落

  $\begin{array}{r}b=\frac{1}{2}\end{array}$ since the substance losses half its value each week
  ::b=12,因为物质每星期损失一半价值

  播放段落

  The formula for this problem is $\begin{array}{r}y=20\cdot {\left(\frac{1}{2}\right)}^x\end{array}$ or $\begin{array}{r}y=20\cdot 2^{-x}\end{array}$ . To find out how much radioactivity is left after three weeks, we plug $\begin{array}{r}x=3\end{array}$ into this formula.
  ::这个问题的公式是 y= 20( 12x) 或 y= 202- x。 要找出三周后还剩下多少放射性, 我们在此公式中插入 x= 3 。

  播放段落

  $$\begin{array}{r}y=20\cdot {\left(\frac{1}{2}\right)}^3=20\cdot \left(\frac{1}{8}\right)=2.5\end{array}$$

  
  ::y=20(12)3=20(18)=2.5

  播放段落

  Example
  ::示例示例示例示例

  播放段落

  Example 1
  ::例1

  播放段落

  The cost of a new car is $32,000. It depreciates at a rate of 15% per year. This means that it loses 15% of each value each year.
  ::新车成本为32,000美元,以每年15%的速度贬值,每年损失15%。

  播放段落

  Draw the graph of the car’s value against time in year.
  ::绘制汽车每年按时间计值的图形 。

  播放段落

  Find the formula that gives the value of the car in terms of time.
  ::找到给出汽车在时间上价值的公式 。

  播放段落

  Find the value of the car when it is four years old.
  ::在汽车四岁时找到车的价值。

  播放段落

  Let’s start by making a table of values. To fill in the values we start with 32,000 at time $\begin{array}{r}t=0\end{array}$ . Then we multiply the value of the car by 85% for each passing year. (Since the car loses 15% of its value, that means it keeps 85% of its value). Remember that 85% means that we multiply by the decimal .85.
  ::让我们从建立一个数值表开始。 要填充我们从时间t=0开始的32,000个数值。 然后我们每年将汽车价值乘以85 % 。 (由于汽车损失了其价值的15%,这意味着其价值的85%。 记住,85%意味着我们乘以小数点.85。 )

  Time	Value (thousands)
  0	32
  1	27.2
  2	23.1
  3	19.7
  4	16.7
  5	14.2

  播放段落

  Now draw the graph:
  ::现在绘制图形 :

  

  播放段落

  Let’s start with the general formula $\begin{array}{r}y=A\cdot b^x\end{array}$
  ::让我们从一般公式y=Abx开始

  播放段落

  In this case:
  ::在这种情况下:

  播放段落

  $\begin{array}{r}y\end{array}$ is the value of the car,
  ::y 是汽车的价值,

  播放段落

  $\begin{array}{r}x\end{array}$ is the time in years,
  ::x 是年中的时间,

  播放段落

  $\begin{array}{r}A=32\end{array}$ is the starting amount in thousands,
  ::A=32是起始数 以千计,

  播放段落

  $\begin{array}{r}b=0.85\end{array}$ since we multiply the amount by this factor to get the value of the car next year
  ::b=0.85,因为我们乘以乘以这个系数得出下一年汽车的价值

  播放段落

  The formula for this problem is $\begin{array}{r}y=32\cdot (0.85{)}^x\end{array}$ .
  ::这个问题的公式是 Y= 32( 0. 85x) 。

  播放段落

  Finally, to find the value of the car when it is four years old, we plug $\begin{array}{r}x=4\end{array}$ into that formula: $\begin{array}{r}y=32\cdot (0.85{)}^4=16.7\end{array}$ thousand dollars, or $16,704 if we don’t round.
  ::最后,为了在汽车四岁时找到车的价值,我们将x=4插入该公式:y=32(0.854)=1.67亿美元,或者如果我们不圆的话,16 704美元。

  播放段落

  Review 
  ::回顾

  播放段落

  Solve the following problems involving exponential growth.
  ::解决下列涉及指数增长的问题。

  播放段落
  1. Nadia received $200 for her $\begin{array}{r}{10}^{th}\end{array}$ birthday. If she saves it in a bank with a 7.5% interest rate compounded yearly, how much money will she have in the bank by her $\begin{array}{r}{21}^{st}\end{array}$ birthday?
     ::Nadia10岁生日时收到200美元。 如果她在银行储蓄,年利率为7.5%,加上每年的利率,那么她到21岁生日时在银行会有多少钱?
  播放段落
  2. Suppose again that Nadia received $200 for her $\begin{array}{r}{10}^{th}\end{array}$ birthday. But what if she saves it in a bank, also with a 7.5% interest rate, but this bank compounds compounds quarterly - how much money will she have in the bank by her $\begin{array}{r}{21}^{st}\end{array}$ birthday?
     ::假设Nadia10岁生日时又收到200美元。 但如果她用7.5%的利率存到银行呢? 但是这个银行的化合物每季度增加一次 — — 21岁生日前她会在银行存多少钱?
  播放段落
  3. The population of a city grows 15% each year. If the town started with 105 people, how many people will there be in 10 years?
     ::一个城市的人口每年增长15%,如果该镇开始时有105人,十年内将有多少人?

  播放段落
  4. Half-life:  Suppose a radioactive substance decays at a rate of 3.5% per hour.
     播放段落
     1. What percent of the substance is left after 6 hours?
        ::6小时后剩下物质的百分比是多少?
     播放段落
     2. What percent is left after 12 hours?
        ::12小时后还剩下多少百分比?
     播放段落
     3. The substance is safe to handle when at least 50% of it has decayed. Make a guess as to how many hours this will take.
        ::物质在至少50%的衰变后可以安全处理。 猜猜这需要多少小时 。
     播放段落
     4. Test your guess. How close were you?
        ::测试你的猜测 你离那有多近?
     
     ::半衰期: 假设放射性物质以每小时3.5%的速度衰减。 6小时后剩下多少百分比的物质? 12小时后还剩下什么百分比? 12小时后剩下什么百分比? 当至少50%的放射性物质衰减后, 处理该物质是安全的。 猜猜这需要多少小时。 测试一下你的猜测。 您离这有多近 ?
  播放段落
  5. Population decrease:  In 1990 a rural area has 1200 bird species.
     播放段落
     1. If species of birds are becoming extinct at the rate of 1.5% per decade (ten years), how many bird species will be left in the year 2020?
        ::如果鸟类物种以每十年(十年)1.5%的速度灭绝,2020年将剩下多少鸟类物种?
     播放段落
     2. At that same rate, how many were there in 1980?
        ::以同样的速度,1980年有多少人?
     
     ::人口减少:1990年,农村地区有1,200个鸟类物种;如果鸟类物种以每十年(10年)1.5%的速度灭绝,2020年还剩下多少鸟类物种?以同样的速度,1980年有多少鸟类物种?
  播放段落
  6. Growth: Janine owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007?
     ::增长:Janine拥有一连串快餐餐馆,1999年经营200家商店,如果年增长率为8%,该餐馆2007年经营多少家商店?
  播放段落
  7. Investment:  Paul invests $360 in an account that pays 7.25% compounded annually.
     播放段落
     1. What is the total amount in the account after 12 years?
        ::12年后账户中的总金额是多少?
     播放段落
     2. If Paul invests an equal amount in an account that pays 5% compounded quarterly (four times a year), what will be the amount in that account after 12 years?
        ::如果保罗在一个每季度支付5%(每年四次)的账户中投资同等数额,12年后该账户中的数额会是多少?
     播放段落
     3. Which is the better investment?
        ::哪种投资更好?
     
     ::投资: Paul每年投资360美元, 在一个每年支付7. 25 % 的账户中。 12年后账户的总金额是多少? 如果保罗在一个每季度支付5%(每年4次)的账户中投资同等金额,那么12年后该账户的金额会是多少? 哪种投资更好?
  播放段落
  8. The cost of a new ATV (all-terrain vehicle) is $7200. It depreciates at 18% per year.
     播放段落
     1. Draw the graph of the vehicle’s value against time in years.
        ::绘制车辆用年计时值的图形 。
     播放段落
     2. Find the formula that gives the value of the ATV in terms of time.
        ::找到给出ATV在时间上值的公式 。
     播放段落
     3. Find the value of the ATV when it is ten years old.
        ::找到ATV的价值 当它已经10岁了。
     
     ::新建ATV(全地形车辆)的费用为7200美元,每年贬值18 % 。 绘制车辆年年价值图。 找到以时间表示ATV价值的公式。 10年前ATV的值。
  播放段落
  9. The percentage of light visible at $\begin{array}{r}d\end{array}$ meters is given by the function $\begin{array}{r}V(d)={0.70}^d\end{array}$ .
     播放段落
     1. What is the multiplication factor?
        ::乘数因素是什么?
     播放段落
     2. What is the initial value?
        ::初始值是什么 ?
     播放段落
     3. Find the percentage of light visible at 25 meters.
        ::发现25米处可见的光的百分率
     
     ::函数 V(d) = 0. 70d 给出了 d米可见光的百分比。 乘法是多少? 初始值是多少? 查找25米可见光的百分比 。
  播放段落
  10. A person is infected by a certain bacterial infection. When he goes to the doctor the population of bacteria is 2 million. The doctor prescribes an antibiotic that reduces the bacteria population to $\begin{array}{r}\frac{1}{4}\end{array}$ of its size each day.
      播放段落
      1. Draw the graph of the size of the bacteria population against time in days.
         ::绘制细菌群规模的图形,以天数计时间。
      播放段落
      2. Find the formula that gives the size of the bacteria population in terms of time.
         ::找到一个公式 以时间计算细菌人口的规模。
      播放段落
      3. Find the size of the bacteria population ten days after the drug was first taken.
         ::在首次服用该药物10天后,发现细菌人口的规模。
      播放段落
      4. Find the size of the bacteria population after 2 weeks (14 days).
         ::在2周(14天)后找到细菌群的规模。
      
      ::一个人被某种细菌感染。当他去看医生时,细菌人口为200万。医生规定一种抗生素,每天将细菌人口减少到14个其体积。 绘制细菌人口规模的图表,按天数逐天绘制。 从时间上找到计算细菌人口规模的公式。 在首次服用药物10天之后,找出细菌人口规模。 在2周之后(14天)找到细菌人口规模。

  播放段落

  Review (Answers)
  ::回顾(答复)

  播放段落

  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 。

  播放段落

  Texas Instruments Resources
  ::得克萨斯州工具资源

  播放段落

  In the CK-12 Texas Instruments Algebra I FlexBook® resource, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See .
  ::在CK-12得克萨斯州仪器代数I FlexBook资源中,有图表计算活动,旨在补充本章某些经验教训的目标。