课程： 4.1 绝对值-interactive

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绝对绝对值

The Purpose of This Lesson
::本课程的目的

In this lesson, you will develop and solve a wide variety of equations to answer questions about scenarios and functions. You'll use both inverse operations and properties of equality in your efforts. You'll also solve equations related to absolute value functions.
::在此课中, 您将开发并解答各种各样的方程式, 以解答关于情景和函数的问题。您将同时使用反向操作和平等特性来努力。您还将解答与绝对值函数相关的方程式。

Activity 1 : Equations From Physics and Geometry
::活动1:物理和几何等等同

Interactive
::交互式互动

Equations are used to describe relationships between quantities in physics, geometry and more. Use the interactive questions below to review some of the most common formulas by matching each formula with its description:
::方程式用于描述物理、几何及更多数量之间的关系。以下交互式问题用来审查一些最常见的公式,将每种公式与其描述相匹配:

INTERACTIVE

Matching Equations

Match the equations with their given descriptions by dragging the blue boxes into the corresponding gray box.
::将蓝箱拖入相应的灰盒,使方程式与给定的描述匹配。

When you are done, drag the slider to see your work.
::完成后,拖动滑动器查看您的工作。

Progress

0 / 4

If an object at rest accelerates at 4 meters per second per second, what is its velocity at 10 seconds? Enter the value only, no units.
::如果休息对象每秒加速4米/秒,10秒的速度是多少?只输入数值,不输入单位。

Work it Out
::工作出来

Depending on what is known or unknown, you may want to solve for one or another variable in an equation . Use properties of equality and inverse operations to isolate the indicated variable for each equation:
::取决于已知或未知的情况,您可能需要在方程式中解析一个或另一个变量。使用等同属性和反向操作来分离每个方程式的指定变量:

Solve $\begin{aligned} C = π d \end{aligned}$ for $\begin{aligned} d . \end{aligned}$
::给D. Solve Cád for D. 解决了D.

Solve $\begin{aligned} a^{2} + b^{2} = c^{2} \end{aligned}$ for $\begin{aligned} a . \end{aligned}$
::为 a 解决 a2+b2=c2。

Solve $\begin{aligned} P = \frac{W}{t} \end{aligned}$ for $\begin{aligned} t . \end{aligned}$
::解决 P=wt for t.

Solve $\begin{aligned} E = m c^{2} \end{aligned}$ for $\begin{aligned} c . \end{aligned}$
::解决 E=mc2 c.

Solve $\begin{aligned} v_{f} = v_{i} + a t \end{aligned}$ for $\begin{aligned} a . \end{aligned}$
::a 的 Solve vf=vi+at 解决 vf=vi+at 。

Solve $\begin{aligned} F = \frac{G m_{1} m_{2}}{r^{2}} \end{aligned}$ for $\begin{aligned} r^{2} . \end{aligned}$
::F=Gm1m2r2为 r2 溶解 F=Gm1m2r2。

Solve $\begin{aligned} F = \frac{G m_{1} m_{2}}{r^{2}} \end{aligned}$ for $\begin{aligned} r . \end{aligned}$
::F=Gm1m2r2为r.

Solve $\begin{aligned} D = \frac{m}{V} \end{aligned}$ for $\begin{aligned} V . \end{aligned}$
::为 V 解决 D=mV 。

Solving Equations Algebraically
::溶解等量代数

Frequently , you can solve equations by using the properties of equality and inverse operations.

Interactive
::交互式互动

Use the interactive below to practice solving for variables using properties of equality and inverse operations.
::使用以下互动方式,使用平等和反向操作的特性解决变量。

INTERACTIVE

Isolating a Variable

Isolate the stated variable in the given equation
::分离给定方程中指定变量

Use the drop down lists to correctly isolate the varible
::使用向下下下载列表来正确分隔可动

Press the next equation button to move on to the next problem
::按下一个方程式按钮来继续到下一个问题

Activity 2 : Absolute Value Functions and Equations
::活动2:绝对值函数和等式

So far, when you've solved equations, you've done so using the properties of equality and inverse operations. Depending on the type of equation, there may be other tools or methods you can use. Absolute value equations provide an opportunity to explore these other tools. With absolute value equations, use the definition of absolute value to create two equations that are then solved.
::到目前为止,当您解析了方程式时, 您已经使用平等和反向操作的属性。取决于方程式的类型, 可能有其它工具或方法可供您使用。绝对值方程式为探索这些其他工具提供了一个机会。在绝对值方程式中, 使用绝对值定义来创建两个公式, 然后解答这两个方程式。

Example 2-1
::例2-1

Recall the scenario about Rafael shining a flashlight at a mirror from the lesson on Simple Composite Functions. The diagram represents a top view of the scenario. The units on the axes are feet. Rafael is standing at point A with a flashlight. A mirror is placed along the x-axis at point B. Rays of light travel in straight lines.
::回顾拉斐尔在简单复合函数课的一面镜子上照亮手电灯的情景。图表代表了情景的顶端视图。斧头上的单位是脚。拉斐尔站在A点, 手电筒是手电筒。在B点, 光线穿梭的Rays 沿X轴放置了一道镜子。

Find the equation for the line representing the path of light from Rafael to the mirror.
::从拉斐尔到镜子寻找代表光线路径的线的方程

The line graphed shows the path the light would take if it passed through the mirror. Is this what actually happens?
::图表的线条显示光穿过镜子时会走的路。这是真的发生吗 ?

How is the path of the light after the mirror different from the line graphed? Modify the graph accordingly.
::镜子后面的光线路径与图解的线条有何不同? 相应修改图形。

Do you think the reflected light will be seen by Rafael's friend Venus along the opposite wall at point C? Why or why not?
::你认为拉斐尔的朋友维纳斯会在C点的对面墙上看到反光光吗?

Is there a way to modify the equation so that the path of the light is accurately reflected?
::有没有办法修改方程,以便准确反映光线路径?

Solution: The equation for the line is $\begin{aligned} y_{1} = - \frac{5}{9} x + 10. \end{aligned}$ The path of the reflected light is shown below. As you can see, Venus will not be able to see the reflected light.
::解析度 : 线条的方程式是 y1\\\\ 59x+ 10。反射光的路径显示在下面。您可以看到, 金星将无法看到反射光。

The path of the reflected light
::反光光的路径

One option would be to make a piecewise function consisting of two lines, one we use for the interval $\begin{aligned} 0 \leq x \leq 18, \end{aligned}$ and one for the interval $\begin{aligned} 18 < x \leq 24. \end{aligned}$ The more efficient option here is to use the absolute value :
::一种选项是使一个小字函数由两行组成,一行用于间距0x18,一行用于间距18<x24。更有效率的选项是使用绝对值:

$\begin{aligned} y_{2} = | - \frac{5}{9} x + 10 | \end{aligned}$

::y259x+10

The absolute value has no effect when the values returned for the expression $\begin{aligned} - \frac{5}{9} x + 10 \end{aligned}$ are positive or 0. But when the expression $\begin{aligned} - \frac{5}{9} x + 10 \end{aligned}$ returns negative values, the absolute value makes them positive. Complete a short table of values to see that this is the case:
::当为表达式 - 59x+10 返回的值为正或0时,绝对值没有效果。但当表达式 - 59x+10 返回负值时,绝对值使这些值为正。填写一个简短的数值表,以了解这种情况:

\begin{aligned} \begin{array}{cc} x & y_{1} \\ 0 & 10 \\ 9 & 5 \\ 27 & - 5 \end{array} \end{aligned}

\begin{aligned} \begin{array}{cc} x & y_{2} \\ 0 & 10 \\ 9 & 5 \\ 27 & 5 \end{array} \end{aligned}

Absolute Value Functions
::绝对值函数

An absolute value function has the effect of taking all the negative values returned by the expression inside the absolute value and making them positive.
::绝对值函数的效果是,将绝对值内表达式返回的所有负值都计算在内,使其为正值。

This means that the graph of absolute value functions of the form $\begin{aligned} y = | f (x) | \end{aligned}$ is always on or above the $\begin{aligned} x \end{aligned}$ -axis, regardless of the expression represented by $\begin{aligned} f (x) . \end{aligned}$
::这意味着表yf(x) 绝对值函数的图形总是在x轴上方或以上,而不管 f(x) 表示的表达式如何。

The range for an absolute value function of the above form never includes negative values.
::上述窗体的绝对值函数范围绝不包括负值。

Example 2-2
::例2-2

The light in the scenario above doesn't hit Venus. Where should Venus stand along the opposite wall to be hit by the light?
::维纳斯应该站在对面的墙上被灯光击中?

Solution: To find where Venus should stand, you substitute $\begin{aligned} x = 24 \end{aligned}$ into the absolute value function:
::解决方案: 要找到维纳斯应该站的位置, 您将 x=24 替换为绝对值函数 :

\begin{aligned} \begin{array}{l} y_{2} = | - \frac{5}{9} x + 10 | \\ y_{2} = | - \frac{5}{9} (24) + 10 | \\ y_{2} = | - \frac{40}{3} + 10 | \\ y_{2} = | - \frac{10}{3} | \\ y_{2} = \frac{10}{3} \\ y_{2} = 3 \frac{1}{3} \end{array} \end{aligned}

::y2\\ {Y2\\\ {Y2( 24) +10\\\\\\\\ _403+10\\\\\\\\\\\\\\\\\ 103\\\\\\\\\\\\\2=103y2=313

She should stand at $\begin{aligned} (24, 3 \frac{1}{3}) . \end{aligned}$
::她应该站着(24 313人)

Example 2-3
::例2-3

In the above scenario, a projectile is thrown along the horizontal line $\begin{aligned} y = 2. \end{aligned}$ At how many points does the projectile intersect the path of the light? Write and solve an equation to find these points.
::在上述假设情景中, 沿着水平线 y=2. 抛出一个投射体。投射体在多少个点上交叉光线路径? 写入并解析一个方程式以找到这些点。

Solution: You have to solve:
::解答: 你必须解答:

\begin{aligned} 2 = | - \frac{5}{9} x + 10 | \end{aligned}

::259x+10{_______________________________________________________________________________________________________________________________________________________________________________

How do you remove the absolute value? There isn't a clear inverse operation for absolute value, but the following is known:
::您如何去掉绝对值? 绝对值没有明显的反向操作, 但以下是已知的 :

Any value of $\begin{aligned} x \end{aligned}$ that returns a value of $\begin{aligned} 2 \end{aligned}$ in the expression $\begin{aligned} - \frac{5}{9} x + 10 \end{aligned}$ will make the equation true. That's because the absolute value of $\begin{aligned} 2 \end{aligned}$ is $\begin{aligned} 2. \end{aligned}$
::x 返回表达式 - 59x+10 中的 2 值为 2 的任何值将使方程成为真实。因为 2 的绝对值是 2 。

Any value of $\begin{aligned} x \end{aligned}$ that returns a value of $\begin{aligned} - 2 \end{aligned}$ in the expression $\begin{aligned} - \frac{5}{9} x + 10 \end{aligned}$ will also make the equation true. That's because the absolute value of $\begin{aligned} - 2 \end{aligned}$ is $\begin{aligned} 2. \end{aligned}$
::返回 ~ 59x+10 表达式中的 2- 2 值的 x 任何值也将使方程成为真实。因为 ~ 2 的绝对值是 2 。

So you need to find the 2 $\begin{aligned} x \end{aligned}$ -values such that:
::所以您需要找到 2 x 值, 以便 :

\begin{aligned} - \frac{5}{9} x + 10 = 2 o r - \frac{5}{9} x + 10 = - 2 \end{aligned}

::- 59x+10=2or-59x+10=2or-59x+10=2

Solving each gives you:
::解决每一个问题给你带来:

\begin{aligned} \begin{array}{l} x = 14.4 \\ x = 21.6 \end{array} \end{aligned}

::x=14.4x=21.6

The projectile intersects the path of light at (14.4, 2) and (21.6, 2).
::抛射体在(14.4,2)和(21.6,2)处交叉光线路径。

Solving Absolute Value Equations
::解决绝对值

Given an absolute value function of the form $\begin{aligned} | f (x) | = a, \end{aligned}$ and $\begin{aligned} a \geq 0, \end{aligned}$ you solve by writing and solving:
::给定窗体 \\ f( x) a 和 a0 的绝对值函数, 您通过书写和解析解决 :

$\begin{aligned} f (x) = a \end{aligned}$
:fx)=a)

$\begin{aligned} f (x) = - a \end{aligned}$
:fx) a

Work it Out
::工作出来

A top view of two walls of a room is represented by the $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ -axis, with units in meters. A ball is rolled from the point (0,15). It hits the adjacent wall at (20,0). Find the absolute value function that models the path of the ball. Determine when the ball passes within 3 meters of the wall represented by the $\begin{aligned} x \end{aligned}$ -axis.
::房间两面墙的顶部视图由 X 和 y 轴代表, 单位以米计。球从点( 0, 15) 滚动, 击中相邻的墙壁( 20,0) 。找到模拟球路径的绝对值函数。确定球何时在X 轴代表的墙3米以内通过。

Solve: $\begin{aligned} | - \frac{4}{5} x + 7 | = 15 \end{aligned}$
::解决:+45x+7+15

Solve: $\begin{aligned} | 8 x + \frac{6}{7} | = 34 \end{aligned}$
::解决:% 8x+67 @34

Solve: $\begin{aligned} | \frac{2}{x} | = 3 \end{aligned}$
::解决:% 2x% 3

Solve: $\begin{aligned} | 3 x | = - 2 \end{aligned}$
::解决:% 3x% 2

Be sure and check your answer for # 5 by substituting the values into the original equation. What do you notice? Why is this the case? Explain by referring to the original equation, as well as the graph of $\begin{aligned} y = | 3 x | \end{aligned}$ .
::请确定并检查您对 # 5 的答案, 将值替换为原始方程。您注意到什么? 为什么是这个案例? 引用原始方程和 y 3x 的图形解释。

Solve: $\begin{aligned} | 4 x + 7 | = 0 \end{aligned}$
::解决: @% 4x+7 @%0

How many solutions are there? Why? Explain by referring to the original equation, as well as the graph of $\begin{aligned} y = | 4 x + 7 | . \end{aligned}$
::有多少解决方案?为什么?通过提及原始方程以及y4x+7的图表来解释。

Summary
::摘要

Solving equations algebraically involves using properties of equality, inverse operations, and more techniques that you'll add to your toolbox as you go along.
::解析方程式代数涉及使用平等的属性,反向操作, 以及更多的技术, 来添加到工具箱中。

Absolute value functions of the form $\begin{aligned} y = | f (x) | \end{aligned}$ never return negative values.
::yf(x) 窗体的绝对值函数从不返回负值。

Absolute value equations of the form $\begin{aligned} | f (x) | = a \end{aligned}$ with $\begin{aligned} a \geq 0 \end{aligned}$ can be solved by solving $\begin{aligned} f (x) = a \end{aligned}$ and $\begin{aligned} f (x) = - a . \end{aligned}$
::窗体 f(x) a 的绝对值方程式加上 a0 可以通过 f(x) =a 和 f(x) a 解析。

章节大纲

常规

绝对绝对值

The Purpose of This Lesson ::本课程的目的

Activity 1 : Equations From Physics and Geometry ::活动1:物理和几何等等同

Interactive ::交互式互动

Work it Out ::工作出来

Solving Equations Algebraically ::溶解等量代数

Interactive ::交互式互动

Activity 2 : Absolute Value Functions and Equations ::活动2:绝对值函数和等式

Example 2-1 ::例2-1

Absolute Value Functions ::绝对值函数

Example 2-2 ::例2-2

Example 2-3 ::例2-3

Solving Absolute Value Equations ::解决绝对值

Work it Out ::工作出来

Summary ::摘要

The Purpose of This Lesson
::本课程的目的

Activity 1 : Equations From Physics and Geometry
::活动1:物理和几何等等同

Interactive
::交互式互动

Work it Out
::工作出来

Solving Equations Algebraically
::溶解等量代数

Interactive
::交互式互动

Activity 2 : Absolute Value Functions and Equations
::活动2:绝对值函数和等式

Example 2-1
::例2-1

Absolute Value Functions
::绝对值函数

Example 2-2
::例2-2

Example 2-3
::例2-3

Solving Absolute Value Equations
::解决绝对值

Work it Out
::工作出来

Summary
::摘要