Course: 4.13 合理职能和等同-interactive

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理性函数和公式
The Purpose of this Lesson
::本课程的目的

In this lesson, you will model scenarios with rational functions . You will explore the characteristics of rational functions and their relationship to polynomial functions.
::在此教训中, 您将模拟具有理性功能的情景。您将探索理性功能的特点及其与多元函数的关系。

Activity 1: Modeling with Rational Functions
::活动1:用合理功能建模

Example 1-1
::例1-1

The force acting on an object is given by the equation :
::方程式给出了在物体上行动的力量:

$\begin{aligned} F = m a \end{aligned}$

::F=马

The units for mass are kilograms, and the units for acceleration are meters per second per second. The resulting units for force are defined as Newtons:
::质量单位为公斤,加速单位为每秒每秒米。

$\begin{aligned} 1 N = 1 \frac{k g \cdot m}{s^{2}} \end{aligned}$

::1 N=1千克+平方米2

Ten Newtons is equivalent to the amount of force required to carry 1 kg (about 2.2 pounds) on the surface of planet E arth. One Newton is about equivalent to the amount of force required to carry a large apple on the surface of the earth. How far you are from the center of the earth changes the amount of force required to carry an object because the acceleration due to gravity decreases as two objects get further from each other.
::十牛顿相当于在地球表面携带1公斤(约2.2磅)的强度。一牛顿相当于在地球表面携带大苹果所需的强度。你离地球中心多远就改变了运载物体所需的强度,因为重力加速度随着两个物体的距离越来越远而下降。

This idea is expressed in the equation:
::这种想法在等式中表现为:

$\begin{aligned} F = \frac{G \cdot m_{1} \cdot m_{2}}{r^{2}} \end{aligned}$

::F=Gm1m2r2

There are two masses in this equation because any two masses have gravity. E arth exerts a gravitational pull on the apple, but the apple also exerts a gravitational pull on E arth! The strength of that pull is decreased as two objects get further from each other. That's why, in the equation, you divide by $\begin{aligned} r^{2} . \end{aligned}$ The variable $\begin{aligned} r \end{aligned}$ is the distance between the two objects. $\begin{aligned} G \end{aligned}$ is the universal gravitation constant - it's the same everywhere in the universe! The above equation is called the universal gravitation equation.
::这个方程式有两个质量, 因为任何两个质量都有引力。地球对苹果有引力拉力, 但苹果也对地球有引力拉力拉力! 当两个对象相互距离更远时, 拖力就会减弱。这就是为什么, 在方程式中, 您除以 r2 。变量 r 是两个对象之间的距离。 G 是通用引力常数。 G 是通用引力常数, 宇宙的每个角落都是相同的。上述方程式被称为通用引力方程。

Now use the equation to check and see if an apple on earth is subject to a force of 1 Newton. Here are some important values and their explanations:
::现在使用方程式来检查和查看地球上的苹果是否受 1 牛顿力的影响。以下是一些重要的值及其解释 :

$\begin{aligned} \begin{array}{ll} Value & Explanation \\ 5.972 \cdot 10^{24} kg & The mass of the earth. \\ \frac{1}{10} kg & The mass of an apple. \\ 6.674 \cdot 10^{- 11} & G \\ 6.371 \cdot 10^{6} m & The distance from the center of the apple to the center of the earth. \\ That's the distance "between" the apple and the earth. \end{array} \end{aligned}$

::5.9721024公斤地球质量 110公斤苹果质量 6.67410-11G6.371106 m 苹果中心到地球中心的距离。这就是苹果和地球之间的距离。

Solution:
::解决方案 :

$\begin{aligned} \begin{array}{l} F = \frac{(6.674 \cdot 10^{- 11}) (5.972 \cdot 10^{24}) (\frac{1}{10})}{{(6.371 \cdot 10^{6})}^{2}} \\ F \approx 1 N \end{array} \end{aligned}$

::F=6.674-10-11 (5.972-1024)(110)(6.371-106)2F_1 N

Interactive
::交互式互动

Use the interactive to visualize orbits from these locations relative to the size of the earth. What do you think would happen to the force applied to the above apple from these various locations?
::使用交互方式将这些位置的轨道与地球大小相对可视化。您认为从这些位置对上面苹果应用的力会如何?

$\begin{aligned} \begin{array}{ll} Location & Distance above surface. (km) \\ The beginning of space. & 100 \\ International Space Station. & 408 \\ Geo-stationary satellites, orbiting "with" earth's rotation. & 36, 000 \\ The moon. & 384, 400 \end{array} \end{aligned}$

:公里) 空间的开端。 100个国际空间站408地球静止卫星,在“与”地球旋转的轨道上运行。

INTERACTIVE

Orbiting Earth

Move the red point to zoom out through the locations listed in the table above.
::将红色点移到上面表格所列的位置上缩放。

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Example 1-2
::例1-2

Use the universal gravitation equation to create a function for the force applied to the above apple (0.1 kilograms) as a function of its distance above the surface of the earth. What kind of function is this? What is the practical domain ? Approximate the forces applied if the apple is at the locations indicated in the table below. Use graphing technology to approximate when the force applied to the apple is a hundredth of a Newton.
::使用通用重力方程式来创建上述苹果(0. 1公斤)所应用的强制力函数, 以该方程式在地球表面的距离为函数。这是哪一类函数? 什么是实际域? 如果苹果位于下表所示位置, 则该方程式几乎可以应用。使用图形技术来估计苹果所应用的强制力是牛顿的千分之一。

$\begin{aligned} \begin{array}{ll} Location & Distance above surface. (km) \\ The beginning of space. & 100 \\ International Space Station. & 408 \\ Geo-stationary satellites, orbiting "with" earth's rotation. & 36, 000 \\ The moon. & 384, 400 \end{array} \end{aligned}$

:公里) 空间的开端。 100个国际空间站408地球静止卫星,在“与”地球旋转的轨道上运行。

Solution:  Simplifying the top of the equation and keeping the variable $\begin{aligned} r : \end{aligned}$
::解决方案: 简化方程顶部并保留变量 r:

$\begin{aligned} F (r) = \frac{3.986 \cdot 10^{13}}{r^{2}} \end{aligned}$

::F(r)=3.986_1013r2

The variable $\begin{aligned} r \end{aligned}$ represents the distance from the center of the earth to the center of the apple, so we need to add the distance above the earth to the radius of the earth. Also, the given values for distance above the earth are in kilometers. So to substitute correctly, each value needs to be converted to meters, then added to the radius of the earth. For example:
::变量 r 代表从地球中心到苹果中心之间的距离, 所以我们需要将地球的距离加到地球的半径上。另外, 给定的距离值以公里计。因此, 要正确替换, 每个值需要转换为米, 然后添加到地球的半径上。例如 :

$\begin{aligned} \begin{array}{ll} Value & Explanation \\ 100 km & The beginning of space. \\ 100, 000 m & Multiplying by 1000 to convert to meters. \\ 6.371 \cdot 10^{6} + 100, 000 & Adding this to the radius of the earth. \\ 6.471 \cdot 10^{6} & Simplifying. \end{array} \end{aligned}$

::100000 mm moltip 以 1000 开始转换为 6.371106+100000 将它添加到地球半径。 6.471106 简化。

In general, you want to take an orbit $\begin{aligned} x \end{aligned}$ given in kilometers, multiply it by 1000, then add the earth's radius. (This step ensures that all length measurements use the same units.) That value goes in place of $\begin{aligned} r \end{aligned}$ in the function:
::一般而言,您想要使用以公里表示的轨道x,乘以1000,然后添加地球的半径。 (这一步骤确保所有长度测量都使用相同的单位。 ) 该值取代函数中的 r:

$\begin{aligned} F (x) = \frac{3.986 \cdot 10^{13}}{{(1000 x + 6, 371, 000)}^{2}} \end{aligned}$

::F(x)=3.986-1013(1000x+6,371,000)2

Interactive
::交互式互动

The previous example wa s a rational function. Recall that a rational function is the ratio of two . The numerator was a constant and the denominator was linear — those are both polynomials. The practical domain was $\begin{aligned} [0, \infty) . \end{aligned}$ U se the interactive to approximate the force applied to the apple at different distances from the earth . You can also use it to approximate when the force applied to the apple is about a hundredth of a Newton.
::上一个示例是一个理性函数。回顾一个理性函数是两个比。分子是一个常数, 分母是线性的, 两者都是多元的。实际域是 [ 0\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

INTERACTIVE

Force and Distance Relationship

Use the slider to explore the coordinates along the curve representing the relationship between distance and force, $F(x).$
::使用滑动器沿代表距离与力( F( x)) 之间的关系的曲线, 探索坐标。

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Rational Functions
::理性函数

Rational functions are the ratio of two polynomials.
::合理函数是两个多数值之比。

Here are some examples:
::以下是一些例子:

$\begin{aligned} f (x) = \frac{1}{x} \end{aligned}$ is the ratio of a constant to a linear expression .
::f(x)=1x 是常数对线性表达式的比例。

$\begin{aligned} f (x) = \frac{3 x + 5}{4 x + 1} \end{aligned}$ is the ratio of a linear expression to a linear expression .
::f(x)=3x+54x+1 是线性表达式与线性表达式之比。

$\begin{aligned} f (x) = \frac{x^{2} + 5 x + 4}{4 x + 1} \end{aligned}$ is the ratio of a quadratic expression to a linear expression.
::f(x) =x2+5x+44x+1 是二次表达式与线性表达式之比。

$\begin{aligned} f (x) = \frac{(x + 2) (x - 1) (x + 5)}{x^{2}} \end{aligned}$ is the ratio of a cubic expression to a quadratic expression.
::f(x) = (x+2)(x-1)(x+5)x2 是立方表达式与二次表达式之比。

Work it Out
::工作出来

If a car is traveling at 50 miles per hour, for 3 hours, its distance traveled is 150 miles. The relationship between distance, rate , and time is expressed by the equation $\begin{aligned} d = r \cdot t . \end{aligned}$ Solve this equation for $\begin{aligned} r . \end{aligned}$ If the distance is fixed at 320 miles, create a function for the rate as a function of time . Graph this function. Give the rates for various times of travel. What is the practical domain for this scenario? What happens to the values returned by the function as $\begin{aligned} t \end{aligned}$ progresses towards positive infinity? Why? Are there values for $\begin{aligned} t \end{aligned}$ that the function cannot accept as input? Why or why not?
::如果一辆汽车以每小时50英里的速度行驶,3小时,其距离是150英里。距离、速度和时间之间的关系由等式 d=rt 表示。为 r. 解决此等式。如果距离固定在320英里,则按时间的函数来设定速率的函数。绘制此函数。给出不同旅行时间的速率。这个假想的实用域是什么? 函数返回的正无限度的值会怎样? 为什么? 该函数无法被接受为输入的值是多少? 为什么或为什么不接受?

Activity 2: Modeling Rates with Rational Functions
::活动2:有合理功能的建模率

Example 2-1
::例2-1

Micah makes 5 pizzas per hour and works 7 hours. So he makes 35 pizzas. The relationship between the work Micah accomplishes (number of pizzas made), his rate (in pizzas per hour), and his time (in hours) is given by the equation:
::Micah每小时做5个比萨饼,工作7小时。所以他做了35个比萨饼。Micah完成的工作(披萨数量)、工资(每小时比萨)和时间(小时)之间的关系由等式来决定:

$\begin{aligned} w = r \cdot t \end{aligned}$

::w=rt

S ometimes, you may not be given the rates in the form of pizzas per hour. Suppose the following represents the hours per pizza for several pizza chefs. Convert each to pizzas per hour:
::有时,你可能得不到比萨饼每小时的价格。假设以下是几个比萨饼厨师每比萨饼的时数。将每个比萨饼的时数转换为每小时比萨饼:

$\begin{aligned} \begin{array}{ll} Chef & Hours per Pizza \\ Isaiah & \frac{1}{3} \\ Jayla & \frac{2}{5} \\ Kiran & \frac{4}{7} \\ Laura & 2 \end{array} \end{aligned}$

::每支比萨厨师时速13Jayla25Kiran47Laura2

Solution:    Pizzas per hour is the reciprocal of the number of hours per pizza:
::答案:每小时披萨是每比萨时数的对等乘数:

$\begin{aligned} \begin{array}{lll} Chef & Hours per Pizza & Pizzas per Hour \\ Isaiah & \frac{1}{3} & 3 \\ Jayla & \frac{2}{5} & \frac{5}{2} = 2.5 \\ Kiran & \frac{4}{7} & \frac{7}{4} = 1.75 \\ Laura & 2 & \frac{1}{2} \end{array} \end{aligned}$

::每小时每比萨匹萨披萨厨师车位133Jayla2552=2.5Kiran4774=1.75Laura212

Example 2-2
::例2-2

If it takes a landscaper 4 hours to landscape a lawn , and they work for 9 hours, how many lawns can they complete?
::如果一个园艺员需要4小时才能在草坪上露面他们工作9小时他们能完成多少草坪?

Solution:    4 hours per lawn is $\begin{aligned} \frac{1}{4} \end{aligned}$ lawns per hour.
::解决办法:每片草坪4小时为每小时14个草坪。

Work is rate times time:
::工作是率乘以时间:

$\begin{aligned} \begin{array}{l} w = \frac{1}{4} \frac{lawns}{hr} \cdot 9 hr \\ w = \frac{9}{4} lawns \\ w = 2.25 lawns \end{array} \end{aligned}$

::w=14 草原9 小时=94 lawnsw=2.25 草地

Example 2-3
::例2-3

Ayla is a landscaper. It takes her 3 hours to landscape a lawn in a particular neighborhood, and she works 9 hour days. She is interviewing potential partners and wants to know how many lawns they will be able to complete in a day, working together. Each individual reports the hours it takes for them to complete a lawn . Create a function for the number of lawns the team can complete as a function of the time it takes the second individual to complete a lawn . Evaluate the function for different values, and interpret the results.
::Ayla是一个园艺人。她需要3小时才能在某个街区的草坪上露面,她工作了9小时。她正在采访潜在的合伙人,想知道他们每天能够一起工作完成多少草坪。每个人报告他们完成草坪所需的时间。为团队完成草坪的数量设定一个函数,作为第二个人完成草坪所需时间的函数。评估不同价值的功能,并解释结果。

Solution:    If Ayla takes 3 hours to landscape 1 lawn, her rate is $\begin{aligned} \frac{1}{3} \end{aligned}$ lawns per hour.
::解决方案:如果艾拉花3小时到1片草坪,她的比率是每小时13片草。

If the prospective partner takes $\begin{aligned} x \end{aligned}$ hours to landscape 1 lawn, her rate is $\begin{aligned} \frac{1}{x} \end{aligned}$ lawns per hour.
::如果未来的伴侣需要x小时才能在1片草坪上露面,她的比率是每小时1x草坪。

Their combined rate is $\begin{aligned} \frac{1}{3} + \frac{1}{x} \end{aligned}$ lawns per hour.
::其合并比率为每小时13+1草坪。

Work equals rate times time. They will work 9 hour days. So the function for the number of lawns they will complete is:
::工作率等于时间。他们工作9小时。所以他们完成的草坪数的函数是:

$\begin{aligned} \begin{array}{ll} Equation & Explanation \\ w (x) = (\frac{1}{3} + \frac{1}{x}) \cdot 9 & w = r \cdot t \\ w (x) = (\frac{x}{x} \cdot \frac{1}{3} + \frac{1}{x} \cdot \frac{3}{3}) \cdot 9 & Common denominator is 3 x . \\ w (x) = (\frac{x}{3 x} + \frac{3}{3 x}) \cdot 9 & Common denominator is 3 x . \\ w (x) = (\frac{x + 3}{3 x}) \cdot 9 & Adding fractions. \\ w (x) = \frac{3 (x + 3)}{x} & Simplifying. \end{array} \end{aligned}$

::EquationExplanationw(x) = (13+1x)%9w=r}tw(x) = (xxx) = (xx}13+1xx) 33=9公分母为 3x.w(x) = (x3x+33x) =9公分母为 3x.w(x) = (x+33x) =9 添加分数.w(x)= 3(x+3)x简化。

As the number of hours increases, the number of lawns decreases. If the second worker can complete a lawn in less than 3 hours, small decreases in their time result in substantial increases in the number of lawns they can complete. At 3 hours, the team completes 6 lawns in a 9 hour day. But at 2 hours, the team completes 7.5 lawns, and at 1 hour, the team completes 12 lawns.
::随着时数的增加,草坪数量减少;如果第二名工人能够在不到3小时的时间内完成草坪,他们的时间略有减少,导致他们能够完成的草坪数量大幅度增加;在3小时的时间内,小组在9小时的时间内完成6次草坪;但在2小时的时间内,小组完成了7.5次草坪,在1小时的时间内,小组完成了12次草坪。

Interactive
::交互式互动

Use the interactive of the graph of the function f rom the previous example to see the number of l awns Ayla and her partner can landscape each day, depending on her partner's rate.
::使用上一个示例中函数图的互动图显示艾拉草坪及其伴侣每天的景观数量,视其伴侣的速率而定。

INTERACTIVE

Lawn Rates

Use the slider to explore how many lawns can be landscaped each day.
::使用滑块来探索每天有多少草坪可以覆盖景观。

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Reciprocals of Rates
::利率对等

Given a rate of work per unit time, then $\begin{aligned} w = r \cdot t . \end{aligned}$
::按单位时间的劳动率计算,然后是按单位时间的劳动率计算。

Given a time $\begin{aligned} T \end{aligned}$ per unit work for an individual, then $\begin{aligned} r = \frac{1}{T} . \end{aligned}$ For example, if Jane takes 2 hours per pizza, her rate of work is $\begin{aligned} r = \frac{1}{2} \end{aligned}$ pizzas per hour.
::例如,如果Jane每批披萨需要2小时,她的工作时间是每小时12个比萨。

Given multiple times per unit work for various workers, $\begin{aligned} T_{1}, T_{2}, T_{3}, . . . \end{aligned}$ , then if they work together, their combined rate is $\begin{aligned} \frac{1}{T_{1}} + \frac{1}{T_{2}} + \frac{1}{T_{3}} + . . . \end{aligned}$
::根据不同工人的单位工作数倍,T1,T2,T3,.,如果他们一起工作,他们的合并费率为1,T1+1,T2+1,T3+......。

This sum of rates can be substituted in place of $\begin{aligned} r \end{aligned}$ in $\begin{aligned} w = r \cdot t . \end{aligned}$
::可以用这一总和的汇率取代w=rt中的r。

Work it Out
::工作出来

It takes Fernando 2 hours to paint a room, and he works 10 hour days. He wants to hire an assistant to work with him. Create an equation for the number of rooms they can complete in a 10 hour day, working together, as a function of the time it takes the assistant to paint a room. Use graphing technology to explore the function. Evaluate the function for different values and interpret the results.
::Fernando需要2小时来油漆一个房间,他工作10小时。他想聘请一名助手来与他一起工作。他想为每天10小时可以完成的房间创建一个方程,他们可以一起工作,这取决于助理需要多少时间来油漆一个房间。他使用图形技术来探索这个功能。他想用不同的值来评估函数,并解释结果。

It takes Elon 4 years to build a rocket. Dania isn't sure how long it will take her to build one. Greg is sure it will take him twice as long as Dania. Create a function for the number of rockets the three can complete in one year, working together, as a function of the time it takes Dania to build one. Use graphing technology to explore this function. Evaluate the function for different values and interpret the results. What value for Dania's time allows the team to complete a rocket in 1 year?
::建造一枚火箭需要Elon 4年时间。 Dania不确定她要多久才能造出一枚火箭。 Greg肯定他需要时间是Dania的两倍。创建一个三枚能在一年内完成的火箭数量的函数, 一起工作, 取决于Dania建造一枚火箭需要的时间。使用图形化技术来探索这个功能。评估不同值的函数, 并解释结果。对于 Dania的时间有什么价值可以让团队在1年内完成一枚火箭?

Add fractions to write each of the following as the ratio of two polynomials.
::添加分数, 将以下各写成为两个多数值之比。

$\begin{aligned} \begin{array}{ll} a. & \frac{1}{5} + \frac{1}{6} \\ b. & \frac{1}{5} + \frac{1}{x} \\ c. & \frac{1}{7} + \frac{1}{14} \\ d. & \frac{1}{x} + \frac{1}{2 x} \\ e. & \frac{1}{5} + \frac{1}{x} + \frac{1}{3 x} \\ f. & \frac{1}{7} + \frac{3}{x} + \frac{4}{5 x} \end{array} \end{aligned}$
::a.15+16b.15+1xc.17+114d.1x+12xe15+1x+13xf.17+3x+45x

Activity 3: Transformations of Rational Functions
::活动3:合理职能的转变

Example 3-1
::例3-1

Graph the function $\begin{aligned} f (x) = \frac{1}{x} . \end{aligned}$ Give its domain and range . Explain. As $\begin{aligned} x \end{aligned}$ approaches positive infinity, what happens to the values returned by the function? As $\begin{aligned} x \end{aligned}$ approaches negative infinity, what happens to the values returned by the function? As the $\begin{aligned} x \end{aligned}$ -values become small positive or negative numbers, what happens to the values returned from the function? Explain.
::函数 f( x) = 1x 的图形。给出它的域和范围。解释。当 x 接近正无穷时, 函数返回的值会怎样? 当 x 接近负无穷时, 函数返回的值会怎样? 当 x 值变成小正负数时, 从函数返回的值会怎样? 解释。

Solution:    The domain is all real numbers except 0. The function does not accept 0 as input because 1 divided by 0 is undefined .
::解析 : 域名是除 0 外的所有真实数字。函数不接受 0 作为输入, 因为 1除以 0 未定义。

The range is all real numbers except 0. The equation $\begin{aligned} \frac{1}{x} = 0 \end{aligned}$ has no solution:
::除 0 外, 区域都是真实数字。方程式 1x=0 无法解析 :

$\begin{aligned} \begin{array}{ll} Equation & Explanation \\ \frac{1}{x} = 0 & Equation to determine an x -value that returns a y -value of 0. \\ 1 = x \cdot 0 & Cross multiplying. \\ 1 = 0 & This statement is false. Therefore, the equation has no solution. \end{array} \end{aligned}$

::EquationExplation1x=0 Equation 以确定返回 y 值 0.1=x%0Cross 乘以1=0的 x 值。此语句是假的。因此, 方程式没有解析法。

A s $\begin{aligned} x \end{aligned}$ approaches positive infinity, the values returned get smaller and smaller, that is, closer to 0. The same is true as $\begin{aligned} x \end{aligned}$ approaches negative infinity.
::随着x接近正无穷度,返回的值越来越小,也就是接近0。与x接近负无穷一样。

A s the $\begin{aligned} x \end{aligned}$ values become very small, the values returned get very large. As $\begin{aligned} x \end{aligned}$ approaches 0 from below, the values returned are increasingly large negative numbers. You can say they approach negative infinity. As $\begin{aligned} x \end{aligned}$ approaches 0 from above, the values returned are increasingly large positive numbers. You can say they approach positive infinity.
::当 x 值变得非常小时, 返回的值会变得非常大。当 x 从下面接近 0 时, 返回的值会越来越大的负数。您可以说它们接近负无穷。当 x 从上面接近 0 时, 返回的值会越来越大。您可以说它们接近正无穷。

Interactive
::交互式互动

Use the interactive below to further explore how t he graph of the rational function $\begin{aligned} f (x) = \frac{1}{x} \end{aligned}$ behaves .
::使用下面的交互功能来进一步探索合理函数 f( x) = 1x 的图形行为方式。

INTERACTIVE

Rational Functions

Move the red point on the slider to explore the coordinates on the curve.
::移动滑块上的红色点以探索曲线上的坐标。

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Vertical and Horizontal Asymptotes
::和水平垂直单数

A vertical asymptote for a function is a vertical line representing an $\begin{aligned} x \end{aligned}$ -value that doesn't return a value for the function.
::函数的垂直同量点是代表不返回函数值的 x 值的垂直线条。

As $\begin{aligned} x \end{aligned}$ approaches this $\begin{aligned} x \end{aligned}$ -value, the function returns values that approach positive or negative infinity.
::x 接近此 x 值时, 函数返回正或负无穷的值。

A horizontal asymptote for a function is a horizontal line representing the value that the function approaches as $\begin{aligned} x \end{aligned}$ approaches positive or negative infinity.
::函数的水平同量点为水平线,表示函数当 x 向正向或负向无穷度方向时所处理的值。

Interactive
::交互式互动

Recall that given a function $\begin{aligned} y = f (x) : \end{aligned}$
::回顾给定的函数 y=f( x) :

$\begin{aligned} y = f (x) + k \end{aligned}$ represents a vertical shift of $\begin{aligned} f \end{aligned}$ by $\begin{aligned} k \end{aligned}$ units.
::y=f(x)+k 表示 f 乘 k 单位的垂直移动。

$\begin{aligned} y = f (x - h) \end{aligned}$ represents a horizontal shift of $\begin{aligned} f \end{aligned}$ by $\begin{aligned} h \end{aligned}$ units.
::y=f(x-h) 表示 f 乘 h 单位的水平移动。

$\begin{aligned} y = a f (x) \end{aligned}$ represents a vertical stretch of $\begin{aligned} f \end{aligned}$ by a factor of $\begin{aligned} a \end{aligned}$ units.
::y=af(x) 表示以单位因数为单位的f的垂直伸展。

In the interactive below, you are given $\begin{aligned} f (x) = \frac{1}{x} . \end{aligned}$ Explore changing the parameters to transform the function. The interactive also shows dotted lines that represent the graphs of the vertical and horizontal asymptotes. Observe how transforming the function impacts the intersection point of the two asymptotes. Explain the impact on the domain and range. Summarize the results. Suggest methods for graphing transformations of $\begin{aligned} f (x) = \frac{1}{x} . \end{aligned}$
::在下面的互动中,给您给 f( x) =1x 。探索修改参数以转换函数。互动还显示代表垂直和水平静态图图的虚线。观察函数如何转换影响两个静态的交叉点。解释对域和范围的影响。总结结果。建议的 f( x) =1x 的图形转换方法。

INTERACTIVE

Transformations of Rational Functions

Move the colored slider to transform the function.
::移动彩色滑动器以转换函数。

You can move the points to find values on the graph or shift it on the grid.
::您可以移动点以在图形中找到值,或者在网格中移动。

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Summary
::摘要

Rational functions are the ratio of two polynomials.
::合理函数是两个多数值之比。

They often feature vertical and horizontal asymptotes.
::它们往往具有纵向和横向的零星特征。

4.13 合理职能和等同-interactive

Section outline

General

理性函数和公式

The Purpose of this Lesson
::本课程的目的

Activity 1: Modeling with Rational Functions
::活动1:用合理功能建模

Example 1-1
::例1-1

$\begin{aligned} \begin{array}{l} F = \frac{(6.674 \cdot 10^{- 11}) (5.972 \cdot 10^{24}) (\frac{1}{10})}{{(6.371 \cdot 10^{6})}^{2}} \\ F \approx 1 N \end{array} \end{aligned}$

::F=6.674-10-11 (5.972-1024)(110)(6.371-106)2F_1 N

Interactive
::交互式互动

Example 1-2
::例1-2

Interactive
::交互式互动

Rational Functions
::理性函数

Work it Out
::工作出来

Activity 2: Modeling Rates with Rational Functions
::活动2:有合理功能的建模率

Example 2-1
::例2-1

Example 2-2
::例2-2

Example 2-3
::例2-3

Interactive
::交互式互动

Reciprocals of Rates
::利率对等

Work it Out
::工作出来

Activity 3: Transformations of Rational Functions
::活动3:合理职能的转变

Example 3-1
::例3-1

Interactive
::交互式互动

Vertical and Horizontal Asymptotes
::和水平垂直单数

Interactive
::交互式互动

Summary
::摘要

Section outline

General

理性函数和公式

The Purpose of this Lesson ::本课程的目的

Activity 1: Modeling with Rational Functions ::活动1:用合理功能建模

Example 1-1 ::例1-1

F = ( 6.674 ⋅ 10 − 11 ) ( 5.972 ⋅ 10 24 ) ( 1 10 ) ( 6.371 ⋅ 10 6 ) 2 F ≈ 1 N ::F=6.674-10-11 (5.972-1024)(110)(6.371-106)2F_1 N

Interactive ::交互式互动

Example 1-2 ::例1-2

Interactive ::交互式互动

Rational Functions ::理性函数

Work it Out ::工作出来

Activity 2: Modeling Rates with Rational Functions ::活动2:有合理功能的建模率

Example 2-1 ::例2-1

Example 2-2 ::例2-2

Example 2-3 ::例2-3

Interactive ::交互式互动

Reciprocals of Rates ::利率对等

Work it Out ::工作出来

Activity 3: Transformations of Rational Functions ::活动3:合理职能的转变

Example 3-1 ::例3-1

Interactive ::交互式互动

Vertical and Horizontal Asymptotes ::和水平垂直单数

Interactive ::交互式互动

Summary ::摘要

The Purpose of this Lesson
::本课程的目的

Activity 1: Modeling with Rational Functions
::活动1:用合理功能建模

Example 1-1
::例1-1

$\begin{aligned} \begin{array}{l} F = \frac{(6.674 \cdot 10^{- 11}) (5.972 \cdot 10^{24}) (\frac{1}{10})}{{(6.371 \cdot 10^{6})}^{2}} \\ F \approx 1 N \end{array} \end{aligned}$

::F=6.674-10-11 (5.972-1024)(110)(6.371-106)2F_1 N

Interactive
::交互式互动

Example 1-2
::例1-2

Interactive
::交互式互动

Rational Functions
::理性函数

Work it Out
::工作出来

Activity 2: Modeling Rates with Rational Functions
::活动2:有合理功能的建模率

Example 2-1
::例2-1

Example 2-2
::例2-2

Example 2-3
::例2-3

Interactive
::交互式互动

Reciprocals of Rates
::利率对等

Work it Out
::工作出来

Activity 3: Transformations of Rational Functions
::活动3:合理职能的转变

Example 3-1
::例3-1

Interactive
::交互式互动

Vertical and Horizontal Asymptotes
::和水平垂直单数

Interactive
::交互式互动

Summary
::摘要