4.2 方形图为“解析集”。-interactive

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  方形图形是解析集

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  The Purpose of This Lesson
  ::本课程的目的

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  In this lesson, you will explore the relationship between equations and graphs. You'll develop equations from relationships between given quantities, and from relationships between objects in the coordinate plane .
  ::在此教训中, 您将探索方程式和图形之间的关系。 您将从给定数量之间的关系和坐标平面上的天体之间的关系中开发方程式 。

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  Activity 1: Equations and Graphs Represent Relationships Between Quantities
  ::活动1:数量之间的关系的方形和图表

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  Example 1-1
  ::例1-1

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  Zachery is employed as a painter. He buys two colors of pigment in powder form. The blue pigment is $5 per pound, and the green pigment is $7 per pound. He spends $52.50 on the pigments. After using all the pigment, Zachery needs to be reimbursed by the company for his expenses, but he's not sure how much of each pigment he purchased. What are the possible quantities of blue and green pigment that would be consistent with the prices of each and the amount he spent? Write an equation to express the relationship between the two quantities of pigment. Graph this equation. What is the practical domain and range for this scenario? Find a plausible point on the graph that might represent the quantities of each pigment that Zachery bought. Is it possible that Zachery bought the same amount of each pigment? How do you know? Find this quantity.
  ::Zachery是一名画家,他以粉色形式购买两种颜色的色素。蓝色色素是每磅5美元,绿色色素是每磅7美元。他在颜料上花费了52.5美元。在使用所有的颜料后,Zachery需要公司偿还他的开支,但他不确定他购买的每张颜料的多少。他购买的蓝色和绿色色素可能与每张色素的价格和花费的金额相符多少?写一个方程式来表达两批色素之间的关系。绘制这个方程式。这个方程式是什么实际领域和范围?在图表上找到一个可能代表Zachehry购买的每一批色料数量的合理点。Zachery能否购买每批色料的同样数量?你怎么知道?找到这个数量。

   

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  Solution:  Let  $\begin{align*}b\end{align*}$  represent the amount of blue pigment, and  $\begin{align*}g\end{align*}$  represent the amount of green pigment:
  ::溶液: Letb 代表蓝色色素的数量, g 代表绿色色素的数量 :

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  $$\begin{align*}5b+7g=52.5\end{align*}$$
  ::5b+7g=52.5

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  This equation is a line in standard form . Let the  $\begin{align*}x\end{align*}$ -axis represent  $\begin{align*}b,\end{align*}$  and the  $\begin{align*}y\end{align*}$ -axis represent  $\begin{align*}g.\end{align*}$
  ::此方程式是标准格式的直线。 让 x 轴代表 b, y 轴代表 g 。

  

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  Blue and green pigment amounts that total a cost of $52.50
  ::蓝色和绿色色素合计费用52.50美元

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    Finding the  $\begin{align*}x\end{align*}$  and  $\begin{align*}y\end{align*}$   intercepts :
  ::查找 x 和 Y 拦截 :

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  $$\begin{align*}\begin{array}{l|l} \text{x-intercept}&\text{y-intercept}\\ \hline 5b+7(0)=52.5 & 5(0)+7g=52.5\\ b=\frac{52.5}{5} & g=\frac{52.5}{7}\\ b=10.5 & b=7.5\\ (10.5,0) & (0,7.5) \end{array}\end{align*}$$
  ::x- 截取- 截取5b+7(0)=52.55(0)+7g=52.5b=52.55g=52.57b=10.5b=7.5(10.5,0)(0,7.5)

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  The practical domain and range:
  ::实际领域和范围:

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  $$\begin{align*}\begin{array}{ll} \text{Domain:} & [0,10.5]\\ \text{Range:} & [7.5,0]\\ \end{array}\end{align*}$$
  ::域:[0,10.5],Range:[7.5,0]

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  To find points on the graph, substitute any value in the domain for  $\begin{align*}b,\end{align*}$  or any value in the range for  $\begin{align*}g,\end{align*}$  and calculate the value of the remaining variable :
  ::要在图形上找到点,请将域中的任何值替换为 b 或 g 区域中的任何值,并计算剩余变量的值:

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  $$\begin{align*}\begin{array}{c|c} b & g\\ \hline 3.5 & 5\\ 2.1 & 6\\ 2.8 & 5.5\\ \end{array} \end{align*}$$
  ::bg3.552.62.85.5

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  The second line represents all the points such that  $\begin{align*}b=g.\end{align*}$
  ::第二行代表b=g的所有点。

  

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  The red line indicates amounts of blue and green pigment that total a cost of $52.50. The blue line indicates when the amounts of each are equal. Where do the two functions intersect?
  ::红线表示蓝色和绿色色素的总价值为52.5美元。 蓝色线表示每件数量相等的时间。 这两种功能在哪里交叉?

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  The original equation for the relationship between  $\begin{align*}b\end{align*}$  and  $\begin{align*}g\end{align*}$  is:
  

  $$\begin{align*}5b+7g=52.5\end{align*}$$
  ::b和g之间的关系的原始方程式是:5b+7g=52.5。

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  If  $\begin{align*}b\end{align*}$  and  $\begin{align*}g\end{align*}$  are equal, we can substitute  $\begin{align*}b\end{align*}$  for  $\begin{align*}g\end{align*}$  in the equation:
  ::如果b和g相等,我们可以在等式中用b代替g:

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  $$\begin{align*}\begin{array}{l} 5b+7b=52.5\\ 12b=52.5\\ b=4.375~\text{pounds} \end{array}\end{align*}$$
  ::5b+7b=52.512b=52.5b=4.375磅

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  If $\begin{align*}b\end{align*}$  and  $\begin{align*}g\end{align*}$  are equal then  $\begin{align*}g=4.375~\text{pounds}.\end{align*}$
  ::如果b和g等于g=4.375磅,则g=4.375磅。

  roblem-solving-with-linear-graphs" quiz-url="https://www.ck12.org/assessment/ui/embed.html?test/view/5d237df181a341201221013b&collectionHandle=algebra&collectionCreatorID=3&conceptCollectionHandle=algebra-:roblem-solving-with-linear-graphs&mode=lite" test-id="5d237df181a341201221013b">

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  Activity 2:  Graphs Represent the Set of Points That Make an Equation True
  ::活动2:代表一组点的图示,使方形真实

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  Example 2-1
  ::例2-1

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  Acceleration due to gravity on earth is about 9.8 meters per second per second. The force applied to an object is the mass of the object times its acceleration:
  ::地球重力加速度每秒约9.8米。

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  $$\begin{align*}F=ma\end{align*}$$
  ::F=马

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  Praveen isn't a weightlifter, but he can comfortably carry two tires on Earth. T ogether, they have a mass of 20 kg. How much force must Praveen withstand to carry the   tires on earth? Praveen aspires to be an astronaut and visit all the planets and the moon. He wants to know, how much mass will he be able to comfortably carry on each world? First, what  is the relationship between the amount he can carry and acceleration due to gravity on each world? Write an equation relating the amount he can carry in kilograms to the acceleration due to gravity on each world. What type of function is this? Graph the points corresponding to the acceleration due to gravity on each world shown in the table below. Interpret the trends in the graph.
  ::普拉文不是重量提升器, 但他可以舒适地在地球上携带两个轮胎。 它们一起有20公斤的重量。 普拉文必须承受多少力才能在地球上携带轮胎? 普拉文渴望成为一名宇航员, 并访问所有行星和月球。 他想知道, 他能将多少力来舒适地携带到每个世界? 首先, 他能够携带的数量和由于每个世界的重力而加速度之间的关系是什么? 写一个方程式, 将他能够携带的重量与每个世界的重力加速度联系起来。 这是哪种功能? 绘制与下表显示的每个世界的重力加速度相对应的点。 解释图表中的趋势 。

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  $$\begin{align*}\begin{array}{l|l} \text{World} & \text{Gravity}~\left(\frac{\text{m}}{\text{s}^2}\right)\\ \hline \text{Mercury} & 3.7\\ \text{Venus} & 8.9\\ \text{Moon} & 1.6\\ \text{Mars} & 3.7\\ \text{Jupiter} & 25\\ \text{Pluto} & 0.62\\ \end{array} \end{align*}$$
  ::世界重力(毫升2) 3.7Venus8.9Moon1.6Mars3.7Jupiter25Pluto0.62

   

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  Solution:  The equation is  $\begin{align*}F=ma.\end{align*}$  The force Praveen can comfortably support is the mass of the two tires, times acceleration due to gravity on Earth:
  ::解答: 方程式是 F=ma 。 Praveen 的力量可以令人放心地支持的是两个轮胎的质量, 由于地球的重力, 乘以加速度 :

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  $$\begin{align*}\begin{array}{l} F=(20)(9.8)\\ F=196~\text{Newtons}\\ \end{array}\end{align*}$$
  ::F=20(9.8)F=196牛顿

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  He can use this same amount of force on any world. The relationship between the mass Praveen can carry and the acceleration due to gravity on the world is given by the equation:
  ::他可以在任何世界上使用同样的力气。 质子Praveen可以携带与因世界重力而加速之间的关系由等式来决定:

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  $$\begin{align*}196=ma \end{align*}$$
  ::196=ma

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  Solving for  $\begin{align*}m\end{align*}$  gives us  $\begin{align*}m\end{align*}$  as a function of  $\begin{align*}a:\end{align*}$
  ::m的溶解给我们m 的函数是:

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  $$\begin{align*}m=\frac{196}{a}\end{align*}$$
  ::m=196a m=196a

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  This is a rational function . For worlds with a lower acceleration due to gravity, Praveen can carry a greater mass. As acceleration due to gravity increases, he carries less:
  ::这是一个合理的函数。 对于重力加速度较低的世界来说, Praveen 能够携带更大的质量。 当重力加速度增加时, 他可以携带更少的:

  

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     Graphs Represent the Set of Points That Make an Equation True
  ::表示一组点的图表,这些点使方程式真实

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  Every point on the graph of an equation makes the equation true.
  ::方程式图上的每一点 都使方程式成为真实

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  Every point that makes an equation true is on the graph of the equation.
  ::每一个使方程式真实的点 都在方程式的图上

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  Work it Out
  ::工作出来

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  1. Daria and Martin custom paint shirts to order . Daria makes 12 shirts  per hour , and Martin makes 10 per hour. Together they want to make 539 shirts. How can they divide the labor? Write an equation that expresses the number of hours they each work. Graph the equation. What is the practical domain and range for this scenario? Find a plausible point on the graph that might represent the hours they each work. Is it possible for them to work the same number of hours? How do you know? Find the single number of hours they can each work so the task is completed. Explain your solution by graphing the line  $\begin{align*}y=x.\end{align*}$
     ::Daria 和 Martin 定制的油漆衬衣。 Daria 每小时12件衬衫, Martin 每小时10件。 他们想一起做539件衬衫。 他们想一起做539件衬衫。 他们如何分工? 写一个表达他们每个工作小时数的方程式 。 绘制方程式 。 这个方程式的实用域和范围是什么? 在图表中找到一个可能代表他们每个工作小时数的合理的点 。 他们能否工作相同的小时数? 你如何知道? 找到他们每件工作的单一小时数, 从而完成任务 。 通过绘制线 y=x 来解释你的解决办法 。
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  2. A graph is given, and several possible equations are provided. The graph represents the solution set for only one of the equations. Which one? Explain your reasoning .
     ::给出了一个图形, 并提供了几个可能的方程式。 图形代表只为其中一个方程式设定的解决方案。 哪一个? 请解释您的理由 。

  

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  $\begin{align*}\quad\begin{array}{l} \text{a.} x^2+y^2=36\\ \text{b.} (x-4)^2+y^2=36\\ \text{c.} 2(x-3)^2+(y-2)^2=36\\ \text{d.} (x-3)^2+(y-2)^2=36\\ \text{e.} y=2(x-3)^2-36\\ \end{array} \end{align*}$
  ::a.x2+y2=36b.(x-4)2+y2=36c.2(x-33)2+(y-2)2+(y-2)2=36d.(x-3)2+(y-2)2+(y-2)2=36e.y=2(x-33)2-36

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  3. Match each equation with its graph:
     ::匹配每个方程式的图形 :

  

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  $\begin{align*}\quad\begin{array}{ll} \text{a.} & (x-5)^2+3(y-4)^2=4\\ \text{b.} & (x-1)^3+(y-5)^2=4\\ \text{c.} & (x-10)^2+(y-3)^2=4\\ \end{array}\end{align*}$
  :x-5)2+3(y-4)2=4b.(x-1)3+(y-5)2=4c.(x-10)2+(y-3)2=4

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  Activity 3 :  The Equation f or a Circle
  ::活动3:圆形的方程式

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  Example 3-1
  ::例3-1

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  As you've explored previously, a circle is the set of all points equidistant from a point called the center.  Consider the graph below.   $\begin{align*}x\end{align*}$  and  $\begin{align*}y\end{align*}$  are the sides of the right triangle shown. The equation  $\begin{align*}x^2+y^2=5^2\end{align*}$  relates the coordinates to the length of the hypotenuse . All the values for  $\begin{align*}x\end{align*}$  and  $\begin{align*}y\end{align*}$   that satisfy this equation are the coordinates of points 5 units from the origin. That means all values for $\begin{align*}x\end{align*}$  and $\begin{align*}y\end{align*}$  that satisfy the equation are points on a circle. If the center is at the origin, the Pythagorean Theorem can be used to relate the coordinates of a point on the circle to the radius.  Create the equation for a circle with a radius of 13. Create the equation for a circle with a radius of 10. Create the equation for a circle with a radius of 7. For each, give at least 4  points that you know are on the graph of the circle. Sketch the graphs.
  ::正如您以前所探索的, 一个圆形是所有点的等距, 从一个名为中点的点开始。 考虑下方的图。 x 和 y 是显示的右三角的侧面。 方程式 x2+y2=52 将坐标与下方的长度联系起来。 x 和 y 满足此方程的所有值都是点 5 单位的坐标。 这意味着 x 和 y 满足方程的所有值都是圆的点。 如果中心位于圆的起源点, 可以使用 Pythagorean Theorem 将圆圆上点的坐标与半径联系起来。 为圆13 的半径创建方程式。 为圆10 创建圆的方程式。 为圆10 创建圆的方程式, 圆的半径为 7 。 对于每一个方形, 请在圆的图中至少给出您知道的 4 点 。

  

   

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  Solution:  A circle with radius 13:
  ::解决方案:半径13的圆圈:

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  $$\begin{align*}\begin{array}{l} x^2+y^2=13^2\\ x^2+y^2=169\\ \end {array}\end{align*}$$
  ::x2+y2=132x2+y2=169

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  Because the equation also represents the relationship between 2 sides of a right triangle and a hypotenuse of 13, it is possible to recognize values for  $\begin{align*}x\end{align*}$  and  $\begin{align*}y\end{align*}$  that complete a Pythagorean Triple. Negative values also make the equation true:
  ::因为方程还代表右三角两侧与13的下方之间的关系, 所以可以识别完成 Pytagoren Triple 的 x 和 y 的值。 负值也使方程成为真实 :

  $$\begin{align*}\begin{array}{l} (5,12)\\ (-5,12)\\ (-5,-12)\\ (5,-12)\\ (12,5)\\ (-12,5)\\ (-12,-5)\\ (12,-5)\\ \end{array}\end{align*}$$

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  A similar process leads to the equation for a circle with radius 10, and points on the circle:
  ::类似进程导致圆的方程,圆半径为10,圆的点数如下:

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  $$\begin{align*}x^2+y^2=100\end{align*}$$   $$\begin{align*}\begin{array}{l} (3,4)\\ \text{etc.} \end{array}\end{align*}$$ The equation for a circle with radius 7:
  ::x2+y2=100 (3,4)etc. 半径为 7 的圆形方程式 :

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  $$\begin{align*}x^2+y^2=49\end{align*}$$
  ::x2+y2=49

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  This isn't a Pythagorean Triple, but substitute values for  $\begin{align*}x\end{align*}$  or  $\begin{align*}y\end{align*}$  and solve:
  ::这不是毕达哥伦三重力, 而是 x 或 y 的替代值, 然后解答 :

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  $$\begin{align*}\begin{array}{l} 3^2+y^2=49\\ 9+y^2=49\\ y^2=40\\ y=\pm \sqrt{40}\\ y=\pm 2\sqrt{10}\\ y\approx \pm 6.32\\ (3,2\sqrt{10})\\ (3,-2\sqrt{10}) \end{array} \end{align*}$$
  ::32+y2=499+y2=49y2=40y40y210y6.32(3,210)(3,210)(3,2-10)

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   The Equation for a Circle
  ::圆形的方程式

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  A circle is the set of points equidistant from a point called the center.
  ::a 圆是指从一个称为中点的点起的等距点的一组点。

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  The equation for a circle with center at the origin is  $\begin{align*}x^2+y^2=r^2,\end{align*}$  where  $\begin{align*}r\end{align*}$  is the radius.
  ::圆圆的方程式是 x2+y2=r2, 圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆圆

  roblem-solving-with-linear-graphs" quiz-url="https://www.ck12.org/assessment/ui/embed.html?test/view/5d237eabf590e33784c6d80b&collectionHandle=algebra&collectionCreatorID=3&conceptCollectionHandle=algebra-:roblem-solving-with-linear-graphs&mode=lite" test-id="5d237eabf590e33784c6d80b">

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  Work it Out
  ::工作出来

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  1. Determine if the given points are on the graph of the given equation. When the equation describes a function, or a circle, state the type of function or give the radius of the circle.
     ::确定给定点是否在给定方程式的图形中。当方程式描述函数或圆时,请说明函数类型或圆半径。

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  $\begin{align*}\qquad \quad\begin{array}{ll} \text{a.} & (24,-7) & x^2+y^2=625\\ \text{b.} & (7\sqrt2,-7\sqrt2) & x^2+y^2=14^2\\ \text{c.} & (4,0) & y=x^3+x^2-20x\\ \text{d.} & (1.5,-11.75) & y-7=-3(x-4)^2\\ \text{e.} & (1,-1) & y=x\\ \text{f.} & (-\frac35, \frac35) & x^2=y^2\\ \text{g.} & (\frac13,\frac23) & x^2+y^2=2\\ \text{h.} & (1,\sqrt3) & x^2+y^2=4\\ \text{i.} & (1,3) & x^2+y^2=9\\ \text{j.} & (-\frac59,0) & y=-\frac59\\ \end{array}\end{align*}$
  ::a. (24,-7)x2+y2+y2=625b. (7_2,-7_2)x2+y2+y2=142c. (4,0)y=x3+x2-20xd. (1.5,-11.75)y -7*3(x-4)2e.(1,-1)y=xf.(-35,35)x2=y2g.(13,23)x2=y2=2h.(1,_3)x2+y2+y2=4i.(1,3)x2+y2=9j.(59,0)y_59

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  2. Find any point on the graph of each of the following. When the equation describes a function, or a circle, state the type of function or give the radius of the circle.
     ::查找以下每个图形中的任意点。当公式描述函数或圆时,请说明函数类型或圆半径。

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  $\begin{align*}\qquad \quad\begin{array}{ll} \text{a.} & x^2+y^2=12\\ \text{b.} & y=x^4+2x^2-15\\ \text{c.} & \frac34(y-9)=(x+1)^2\\ \text{d.} & y=-x\\ \text{e.} & x^2=y^2+15\\ \text{f.} & y=\frac{6}{x}\\ \text{g.} & \frac{3}{y+4}=\frac{x+5}{2}\\ \text{h.} & y=4.5(\frac23)^x\\ \end{array}\end{align*}$
  ::a.x2+y2=12b.y=x4+2x2x2-15c.34(y-9)=(x+1)2d.yx2=y2+15f.y=6xg3y+4=x52hy=4.5(23)x

  ---

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  Activity 4 : A Parabola is a Set of Points Meeting Certain Geometric Conditions
  ::活动4:帕拉波拉是符合某些几何条件的一组点

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  The equation of a circle originates from the application of the Pythagorean Theorem to the coordinates of a point  $\begin{align*}(x,y).\end{align*}$  In this course so far, the equation for a quadratic has been developed from calculating the area of squares and rectangles, and from modeling an object subject to constant acceleration. The graph of a quadratic function is a parabola . Like a circle, a parabola can be defined geometrically, and the equation can be developed from this geometric definition.
  ::圆形的方程式来源于Pythagorean Theorem应用到一个点的坐标(x,y) 。在这个过程中,到目前为止,二次方程的方程式是从计算方形和矩形区域,以及从模拟一个受恒定加速度制约的物体中开发出来的。二次函数的图形是抛光线。像圆形一样,可按几何方式定义抛光线,而方程也可以从这个几何定义中开发。

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  Interactive
  ::交互式互动

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  The geometric definition of a parabola is the set of points that are equidistant from a given point  called the focus  and a given line called the  directrix . Use the interactive below to visualize this.
  ::抛物线的几何定义是一组从一个特定点(即焦点和直线)到一个特定点(即焦点和直线)相等的点。使用下面的交互式点来直观这个点。

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  Example 4-1
  ::例4-1

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  To create the equation for a simple parabola, consider the focus $\begin{align*}(0,1)\end{align*}$  and the  directrix   $\begin{align*}y=-1\end{align*}$  below. A parabola is the set of all points that are equidistant from the line and the point. That is to say, the distance from a point $\begin{align*}(x,y)\end{align*}$  to the focus equals the distance from  $\begin{align*}(x,y)\end{align*}$  to the directrix. An expression for the distance to the focus can be found with the , $\begin{align*}d=\sqrt{\left(\Delta x\right)^2+(\Delta y)^2}.\end{align*}$   The distance to the directrix can be found by subtraction . Set these distances equal to each other to find the equation for the resulting parabola.
  ::要为简单的抛物线创建方程式, 请考虑下面的焦点( 0, 1) 和直线 y 1。 抛物线是所有点的集合, 这些点与直线和点的距离相等。 也就是说, 从点( x,y) 到焦点的距离等于从点( x,y) 到直线的距离。 从 d 2+ 2 中可以找到到焦点的距离的表达方式。 与直线的距离可以通过减法找到。 设置这些距离, 以对等的方式找到由此产生的parbola 的方程 。

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  $$\begin{align*}d=\sqrt{\left(\Delta x\right)^2+(\Delta y)^2}\end{align*}$$
  :x)2+2

  

   

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  Solution:  The distance from the point to the focus:
  ::解决方案: 从点到焦点的距离 :

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  $$\begin{align*}\begin{array}{l} d_1=\sqrt{(x-0)^2+(y-1)^2}\\ d_1=\sqrt{x^2+(y-1)^2}\\ \end{array}\end{align*}$$
  ::d1(x-0)2+(y- 1)2d12+(y- 1)2+(y- 1)2

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  The distance from the point to the line:
  ::从点到线的距离 :

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  $$\begin{align*}\begin{array}{l} d_2=y-(-1)\\ d_2=y+1\\ \end{array}\end{align*}$$
  ::d2=y- (- 1) d2=y+1

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  For all points on the parabola, these distances are equal:
  ::对于抛物线上的所有点,这些距离相等:

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  $$\begin{align*}y+1=\sqrt{x^2+(y-1)^2}\end{align*}$$
  ::y+1x2+(y- 1)2

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  While that is not a very friendly-looking equation, it is the equation for all the points equidistant from the focus and directrix above. It's not necessary to simplify the equation or solve it for  $\begin{align*}y,\end{align*}$  but doing so makes the equation much more useful. It is more conventional to write quadratic functions in either vertex or standard form:
  ::虽然这不是一个非常友好的方程, 但它是所有点的方程, 从上面的焦点和直线直径相等。 不需要为y 简化方程或解决方程, 但这样做会使方程更加有用 。 以顶点或标准格式写入二次函数比较传统 :

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  $$\begin{align*}\begin{array}{l|l} \text{Equation} & \text{Explanation}\\ \hline y+1=\sqrt{x^2+(y-1)^2} & \text{Equation for a parabola with given focus and directrix.}\\ (y+1)^2=x^2+(y-1)^2 & \text{Squaring both sides.}\\ y^2+2y+1=x^2+y^2-2y+1 & \text{Squaring the binomials.}\\ 2y=x^2-2y & \text{Subtracting}~y^2~\text{and 1 from both sides.}\\ 4y=x^2 & \text{Adding}~2y~\text{to both sides.}\\ y=\frac14 x^2 & \text{Dividing by}~4.\\ \end{array}\end{align*}$$
  ::等式Explationy+1x2+(y- 1)2 具有给定焦点和直线的抛物线的等式 (y+1) 2=x2+(y- 1)2S 将两侧均分。 y2+2y+1=x2+y2- 2y+1S 将两侧的二元体2y=x2- 双向的 y2=x2=x2 将2y与两侧相相对。 y=14x2Diving 4

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  A Geometric Definition of a Parabola
  ::抛物线的几何定义

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  A parabola is the set of all points equidistant from the point called the focus and a line called the directrix.
  ::抛物线是所有点的一组等距, 从点称为焦点, 直线称为直线。

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  By writing expressions for those distances and setting them equal to each other, the equation for a parabola is created.
  ::通过书写这些距离的表达方式,并把它们等同起来,创造了抛物线的方程。

  roblem-solving-with-linear-graphs" quiz-url="https://www.ck12.org/assessment/ui/embed.html?test/view/5d237ed5f590e3361305bcf4&collectionHandle=algebra&collectionCreatorID=3&conceptCollectionHandle=algebra-:roblem-solving-with-linear-graphs&mode=lite" test-id="5d237ed5f590e3361305bcf4">

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  Work it Out
  ::工作出来

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  1. Use the above method to find the equation for a parabola with a focus of 2 and a directrix of $\begin{align*}y=-2.\end{align*}$  Support your work with a diagram.
     ::使用上述方法查找焦点为 2 的抛物线方程式和 y2 的直线方程式。 用图表支持您的工作 。
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  2. Find the equation for a circle with a radius of 20. Find the coordinates of at least four points on the circle.
     ::查找圆的方程,半径为20。 查找圆上至少四个点的坐标。
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  3. Create example equations for each of the following function types. At least two parameters must be neither 1 nor 0. For example, $\begin{align*}y=3x^2\end{align*}$  only has one parameter not equal to 1 or 0, while  $\begin{align*}y=(x-2)^2+3\end{align*}$  has two parameters that are neither 1 nor 0. Find at least 2  points on each.
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     1. constant
        ::常数
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     2. linear
        ::线
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     3. absolute value
        ::绝对绝对值
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     4. quadratic
        ::二次
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     5. square root
        ::平方根
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     6. exponential
        ::指数指数
     
     ::为以下函数类型中的每一类型创建示例方程式。 至少两个参数必须是 1 或 0 。 例如, y= 3x2 只有一个参数不等于 1 或 0, 而 y= (x- 2) 2+3 有两个参数, 两者不等于 1 或 0 。 在每个函数类型上至少要找到两个点。 恒定线绝对绝对值二次方根指数
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  4. Create 3 equations of your choice that are not functions of  the types represented in the last problem. Find at least two points on each.
     ::创建您选择的 3 个方程式, 这些方程式不是上一个问题中所代表类型中的函数。 在每个方程式上至少找到两个点 。

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    Summary
  ::摘要

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  • Equations represent a relationship between two quantities.
    ::方程式代表两个数量之间的关系。
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  • The graph of an equation is the set of points that make the equation true.
    ::方程图是使方程真实的一组点。