LESSON 4.1 Interpreting Graphs - KendallHuntmath.kendallhunt.com/documents/daa2/CL/DAA2CL010_04.pdf · LESSON 4.1 CONDENSED Interpreting Graphs Discovering Advanced Algebra Condensed

L E S S O N

4.1CONDENSED

Interpreting Graphs

Discovering Advanced Algebra Condensed Lessons CHAPTER 4 41©2010 Key Curriculum Press

(continued)

In this lesson you will

● interpret graphs that show information about real-world situations

● make a graph that reflects the information in a story

● invent a story that conveys the information in a graph

A graph can communicate a lot of information in a concise way. This graph shows the relationship between the number of people in a baseball stadium and the time since the game ended.

x

y

Time since game ended (min)

Nu

mb

er o

f p

eop

le in

sta

diu

m

The number of people depends on the time since the game ended. So the time in minutes is the independent variable and the number of people is the dependent variable.

As the time increases, the number of people decreases linearly, indicating that people are leaving the stadium at a constant rate. The slope is the number of people who leave each minute. The x-intercept represents the time when the stadium is empty. The y-intercept represents the number of people in the stadium at the time the game ends.

The example in your book shows a more complicated graph that indicates how the number of cans of juice in a vending machine changes over the course of a typical school day. Study the graph carefully for a few minutes, thinking about what might cause each increase, decrease, and change in slope. Then, work through the example, answering the questions on your own before reading the solutions.

The text after the example poses several more questions about the graph. Try to answer each question yourself, and then read the answers below.

● When do students arrive at school? The machine is full until 7 A.M., and then the number of cans decreases fairly quickly. This indicates that students probably begin arriving at around 7 A.M.

● What time do classes begin? The number of cans begins to decrease at a slower rate beginning at 8 A.M. This could be the time classes begin.

● When is lunch? Lunch most likely begins at around 11:15 A.M. Beginning at this time, the number of cans decreases very rapidly. This is probably because students and teachers are purchasing drinks to go with their lunches.

● When do classes let out for the day? The number of cans in the machine decreases rapidly beginning at around 3 P.M. This is most likely the time when classes let out.

DAA2CL_010_04.indd 41DAA2CL_010_04.indd 41 1/13/09 2:35:37 PM1/13/09 2:35:37 PM

42 CHAPTER 4 Discovering Advanced Algebra Condensed Lessons

©2010 Kendall Hunt Publishing

Notice that both the baseball-stadium graph and the vending-machine graph are continuous. In reality, the number of people in the stadium and the number of cans of juice in the machine must be discrete values. However, a continuous “graph sketch” makes it easier to see trends and patterns.

Investigation: Graph a StoryRead the story in Part 1 of the investigation in your book, and then sketch a graph that reflects all the information in the story. When you are finished, compare your graph with the graph below, which has been labeled to show how the various sections relate to the story.

64212108642A.M.

Empty

Full

1210P.M. Time

Leve

l of

wat

er in

bu

cket

Some water evaporates

Bucket is nearly empty

Rain slowsand stops

Heavyrain

Beforerain

starts

Trey drinksfrom bucket

Water evaporates in bright sun

Billy pulls plugs

Now, look at the graph in Part 2 of the investigation. Think carefully about the information shown in the graph. Then, write a story that conveys all the information, including when and how the rates of change increase or decrease. Below is one possible story, but write your own story before you read it.

“Luis and Loretta have a small outdoor swimming pool. The children want to use the pool, but the water level is very low, so Luis turns on the hose and begins filling the pool at a constant rate. The children are restless and persuade Luis to turn up the water flow. It fills at a faster constant rate than before. When the pool is completely full, Luis turns off the hose and the children are careful not to splash more water. After the children get out of the pool, he empties it. The water pours out rapidly at first, then more slowly as there is less and less water left. Luis leaves just a little water in the bottom, which will slowly evaporate.”

Lesson 4.1 • Interpreting Graphs (continued)


CONDENSED

Discovering Advanced Algebra Condensed Lessons CHAPTER 4 43©2010 Kendall Hunt Publishing


● review function notation

● determine whether a relation is a function based on its graph

● find values of functions using graphs and equations

A relation is any relationship between two variables. A function is a special type of relation in which, for every value of the independent variable, there is at most one value of the dependent variable. If x is the independent variable and y is the dependent variable, then a function pairs at most one y-value with each x-value.

The graph on the left below shows how Rachel’s height changed as she got older. The graph on the right shows the number of pages of an assigned novel Tom read each day for 9 days. The graph on the left is continuous, whereas the graph on the right is made up of discrete points.

Hei

ght (

in.)

25

30

35

40

45

50

55

60

65

70

202 4 6 8 10 12 14

Age (yr)160

40302010

0

50

6 7 8 9

6070

2 3 4 51

Day

Pag

es r

ead

Both of the relations shown above are functions: Rachel had only one height at any particular time in her life, and Tom read only one number of pages on any given day. You can say that Rachel’s height is a function of her age and that the number of pages Tom read is a function of the day.

In some cases there is a mathematical formula connecting the two variables of a function, but in other cases there is not. For example, there is no formula connecting the day to the number of pages Tom read.

You are probably familiar with the vertical line test, which helps you determine whether or not a graph represents a function. If no vertical line crosses the graph more than once, then the relation is a function.

Function

No vertical linecrosses the graphmore than once.

Not a function

Because a verticalline crosses the graphmore than once, thisis not a function.

(continued)

Function NotationL E S S O N

4.2




Use the vertical line test to verify that the graphs of Rachel’s height and the number of pages Tom read are functions.

Function notation emphasizes the dependent relationship between the variables used in a function. The notation y � f (x) indicates that the values of the dependent variable, y, are explicitly defined in terms of the independent variable, x, by the function f.

The example in your book gives you practice using function notation and finding the values of functions by using equations and graphs. Work through the example, solving the problems on your own before you read the solution.

Investigation: To Be or Not to Be (a Function)Complete the investigation in your book, and then read the solutions below.

Step 1

a. The graph passes the vertical line test, indicating that there is at most one y-value for every x-value. So the relation is a function.

b. You can draw a vertical line that intersects the graph in two points, so the relation is not a function.

c. The graph passes the vertical line test, so the relation is a function.

d. This relation pairs each x-value with exactly one y-value, so it is a function.

e. This relation pairs each x-value with only one y-value, so it is a function. (Note that, as in this example, a function can pair the same y-value with more than one x-value. Think about the horizontal line y � 2, for example. It is a function (it passes the vertical line test), yet it pairs the y-value 2 with every x-value).

f. This relation pairs the x-value 1 with two y-values, namely 1 and 3. It also pairs the x-value 3 with two y-values, namely 2 and 4. It is not a function.

g. There are students of the same age who have different heights. This means that one age value is paired with more that one height value. Therefore, the relation is not a function.

h. Because each automobile must have a unique license plate number, the relation is a function.

i. The answer depends on how you think about the situation. If you consider only the days in one year at one location, then the relation is a function because the sun sets at a particular time each day. If you consider the same day of any year, or if you consider different locations, then on the same day the sun sets at different times, so the relation is not a function.

Step 2 Answers are given only for the relations that are functions.

a. When x � 2, y � 2, so a (2) � 2. When y � 3, x � 0 or x � 1.5, so a (0) � 3 or a (1.5) � 3.

c. The graph has no point corresponding to x � 2, so c (2) is undefined; c (1) � 3, or c (3) � 3.

d. d(2) � 3

e. e (2) � 2; no x-value results in y � 3

Lesson 4.2 • Function Notation (continued)


CONDENSED


(continued)

Lines in MotionL E S S O N

4.3In this lesson you will

● explore what happens to the equation of a line when you translate the line

● learn to write an equation that translates a function horizontally h units and vertically k units

● describe the graph of an equation in the form y � k � f (x � h) by relating it to the graph of y � f (x)

With the exception of vertical lines, lines represent functions. Below the intercept and point-slope forms of a line are written in function notation.

x-y form Function notation

Intercept form y � a � bx f (x) � a � bx

Point-slope form y � y1 � b �x � x 1 � f (x) � f � x 1 � � b �x � x 1 �

In the investigation you will see how translating a line—that is, moving it horizontally or vertically—affects its equation.

Investigation: Movin’ AroundComplete Steps 1–5 on your own, and then read the results below.

Steps 1 and 2 The lines are shown at right. The equation for the

x

y

64–6 –4

6

4

–4

y = 3 � 2x y = 2x

y = �4 � 2x

line 3 units above y � 2x is y � 3 � 2x. If f (x) � 2x , you can write the equation of the new line as y � f (x) � 3. The equation for the line 4 units below y � 2x is y � �4 � 2x, or y � f (x) � 4.

Steps 3 and 4 The graphs of the new lines are added below. The line for Step 3 passes through the point (3, 0). It has slope 2, so its equation in point-slope form is y � 2(x � 3). If f (x) � 2x, the equation for the new line is y � f (x � 3).

The line for Step 4 passes through the point (�1, 2). It has slope 2, so its equation in point-slope form is y � 2 � 2(x � 1). If f (x) � 2x, the equation for the new line is y � 2 � f (x � 1).

x

y

64–6 –4

6

4

–4

y = 2(x�3)

y = 2 � 2(x �1)




Lesson 4.3 • Lines in Motion (continued)

Step 5 If you translate the function y � f (x) to a new point up k units and right h units, then the equation of the translated function is y � k � f (x � h), which can also be written y � k � f (x � h) or y � f (x � h) � k. (Note: If the graph is translated left h units, then the h-value will be negative. Likewise, if the graph is translated down k units, then the k-value will be negative.)

Read the rest of the investigation, up to Step 9, in your book. You probably will not have enough people or the equipment to conduct the experiment on your own, but try to imagine what would happen. See if you can answer the questions in Step 9, and then read the results below.

a. B’s graph is translated left 2 units and down slightly (depending on how far A is behind B) from A’s graph.

b. If you assume that A is 1 unit behind B, then the equation for B’s graph is y � f (x � 2) � 1. This indicates that the graph is the graph of y � f (x) shifted left 2 units (to account for the 2-second delay) and down 1 unit (because the distances B records are 1 unit less than those A records).

c. y � f (x � h) � k

You can generalize your findings from the investigation to any function. For example, no matter what the shape of the function y � f (x) is, the graph of y � f (x � 3) � 2 will look the same as the graph of y � f (x), but it will be translated up 2 units and right 3 units. Read the “Translation of a Function” box on page 200 of your book. You can use your calculator to verify the text in the box. For example, graph y � x 3, and then graph y � (x � 4)3 � 1. The graph of the second function should look like the graph of the first function shifted right 4 units and up 1 unit.

If a function is translated h units horizontally and k units vertically, then the point (x, y) is mapped onto a new point, (x � h, y � k). This new point is the image of the original point.

In your book, read Example A and the paragraph that follows it. In the investigation and Example A, you learned how to write the equations for translated lines. You can use the same principle to describe the translation of a function even when you don’t know the equation of the original function. Work through Example B in your book.



L E S S O N

4.4CONDENSED


● examine the graph of y � x 2, which is a parabola

● find equations for translations of the graph of y � x 2

● look at an application of a quadratic function

In the previous lesson you explored translations of linear functions. Read the text on page 205 of your book, which gives examples of how translations occur in other settings.

A translation is a type of transformation. A transformation is a change in the size or position of a figure. You may have learned about other transformations, such as reflections, dilations, and rotations, in earlier math courses.

The shape of the graph of the function y � x 2 is called a parabola. Parabolas always have a line of symmetry that passes through the vertex.

x

y

6

2 4 6–6–2

–4 –2

8

2

4

y � x2 The line of symmetry divides thegraph into mirror-image halves. Theline of symmetry of y � x2 is x � 0.

The vertex is the point where thegraph changes direction. The vertexof y � x2 is (0, 0).

The function y � x 2 is a parent function. By transforming the graph of this function, you can create a family of functions, consisting of an infinite number of related functions. The function y � x 2 and all the functions created by transforming it are called quadratic functions. As you will see throughout this course, quadratic functions are very useful. Below are three examples of quadratic functions.

y � x2 � 3 y � (x � 12)2 y � 5 � (x � 4)2

Investigation: Make My GraphStep 1 Each lettered part of Step 1 shows the graph of y � x 2 and a graph created by translating the graph of y � x 2. Your job is to find the equation that produces the translated graph. To help you find the equation, think about what you learned about translated linear functions in Lesson 4.3.

When you think you know the equation for a graph, check it by graphing both it and y � x 2 in the same calculator window. (Be sure to read about window settings in Calculator Note 4D.) Complete Step 1 by yourself before reading the results on the next page.

(continued)

Translations and theQuadratic Family




a. The red parabola is the graph of y � x2 translated down 4 units, so its equation is y � x2 � 4.

b. The red parabola is the graph of y � x2 translated up 1 unit, so its equation is y � x 2 � 1.

c. The red parabola is the graph of y � x2 translated right 2 units, so its equation is y � (x � 2)2.

d. The red parabola is the graph of y � x2 translated left 4 units, so its equation is y � (x � 4)2.

e. The red parabola is the graph of y � x 2 translated left 2 units and up 2 units, so its equation is y � (x � 2)2 � 2.

f. The red parabola is the graph of y � x 2 translated right 4 units and down 2 units, so its equation is y � (x � 4)2 � 2.

Step 2 For a translation right, you subtract the number of units from x. For a translation left, you add the number of units to x. For a translation up, you add the number of units to the right side of the equation (or subtract it from y). For a translation down, you subtract the number of units from the right side of the equation (or add it to y). The coordinates of the vertex of the translated parabola are (value of horizontal translation, value of vertical translation).

Step 3 In general, if the graph of y � x2 is translated horizontally h units and vertically k units, then the equation of the translated graph is y � (x � h)2 � k, or y � k � (x � h)2.

The example in your book shows an application involving translations of parabolas. Read the example and solution carefully, following along with a pencil and paper. To test your understanding, solve the same problem, but have the diver dive from a board 15 ft long that is 20 ft above the water. (The function for the new graph would be y � f(x � 10) � 5 and the diver’s maximum height would be 25 ft.)

You have seen that translations of quadratic functions work much the same as translations of linear functions. If you translate the graph of y � x 2 horizontally h units and vertically k units, then the equation of the translated parabola is y � (x � h)2 � k. You can also write this equation as y � k � (x � h)2 or y � k � (x � h)2. When you translate horizontally by h units, you can think of it as replacing x in the equation with (x � h). Likewise, a vertical translation by k units replaces y with (y � k).

Notice that the vertex of the translated parabola is (h, k). This is why finding the vertex is fundamental in determining translations of parabolas.

x

y

(0, 0)

y � x2

y � (x � h)2 � k

When the graph of y � x2 istranslated horizontally h unitsand vertically k units, the vertexof the translated parabola is (h, k).

(h, k)

Lesson 4.4 • Translations and the Quadratic Family (continued)



CONDENSED


● learn about the square root function and its graph

● reflect functions across the x- and y-axes

● learn about piecewise functions

At right is the graph of the square root function, y � � __

x . The

x

y

–10 –5 105

–5

5domain of the function is x � 0, and the range is y � 0.

Graph the square root function on your calculator and trace the graph to find �

__ 3 , �

__ 8 , and �

___ 31 . You should get approximately 1.732,

approximately 2.828, and approximately 5.568, respectively. If you try to trace for values of x � 0, the calculator gives no value for y. This is because the function is not defined for negative x-values (that is, you cannot take the square root of a negative number).

Investigation: Take a Moment to ReflectIn this investigation you explore a type of transformation called a reflection. You will apply reflections to linear, quadratic, and square root functions.

Step 1 Work through Step 1 in your book on your own. When you are finished, compare your results with those below.

a. f2(x) � �(0.5x � 2) � �0.5x � 2. The graph of f2(x) looks like the graph of f1(x) flipped over the x-axis. In other words, the graph of f2(x) is the reflection of the graph of f1(x) across the x-axis.

b. f2(x) � �(�2x � 4) � 2x � 4. The graph of f2(x) is the reflection of the graph of f1(x) across the x-axis.

c. f2(x) � ��x 2 � 1� � �x 2 � 1. The graph of f2(x) is the reflection of the graph of f1(x) across the x-axis.

d. In general, the graph of y � �f (x) is the reflection of the graph of y � f (x) across the x-axis.

Step 2 Work through Step 2 on your own, and then compare your results with those below.

a. f2(x) � 0.5(�x) � 2 � �0.5x � 2. The graph of f2(x) is the reflection of the graph of f1(x) across the y-axis.

(continued)

Reflections and the Square Root Family

L E S S O N

4.5




b. f2(x) � �2(�x) � 4 � 2x � 4. The graph of f2(x) is the reflection of the graph of f1(x) across the y-axis.

c. f2(x) � (�x)2 � 1 � x2 � 1. The graph of f2(x) is the same as the graph of f1(x) because the original graph has the y-axis as a line of symmetry, so a reflection across the y-axis maps the graph onto itself.

d. f2(x) � (�x � 3)2 � 2. The graph of f2(x) is the reflection of the graph of f1(x) across the y-axis.

e. The graph of y � f (�x) is the reflection of the graph of y � f (x) across the y-axis.

Step 3 Work through Step 3 on your own, and then compare your results with those below.

a. The graph of f2(x) � �f1(x) � � � __

x is a reflection of the square root function across the x-axis. The graph of f3(x) � f1(�x) � �

___ �x is a

reflection of the square root function across the y-axis.

b. The graph of f4(x) � �f1(�x) � � � ___

�x is a reflection of the square root function across both axes.

c. The graph of y � � __

x is not an entire horizontal parabola because it has a range of y � 0. Graphing y � � �

__ x would

complete the bottom half of the parabola.

The “Reflection of a Function” box summarizes what you discovered in the investigation. The example shows how to compose a piecewise function by limiting the domains of two or more different functions. Graphing and writing equations for piecewise functions gives you practice with different families of functions. Work through the example carefully and then read the rest of the lesson.

Lesson 4.5 • Reflections and the Square Root Family (continued)



CONDENSED


● graph the absolute-value function and transformations of the absolute-value function

● graph dilations of a parent function and observe how these transformations are related to the equation of the parent function

● use what you know about transformations to fit a function to data

Read page 221 in your book to review the idea of absolute value. Look at the graph of the parent absolute-value function, y � ⏐x⏐.

x

y

–8 –4 84

–8

–4

8

4

Notice that the right half of the graph looks like y � x, whereas the left half looks like y � �x. This makes sense because, when x � 0, ⏐x⏐ � x, and when x � 0, ⏐x⏐ � �x.

Example A shows three examples of transformations of the function y � ⏐x⏐. The transformations are all dilations. Read the example very carefully. Note that you can get the function in part a by substituting

y _ 2 for y in the parent function,

y � ⏐x⏐.

Substituting y _

b for y in the parent function dilates the graph of the parent

function vertically by a factor of b.

Similarly, as you can observe in part b, substituting x _ a for x in the parent function dilates the graph of the parent function horizontally by a factor of a.

Translations and reflections are rigid transformations, meaning that they produce images that are congruent to the original figure. By contrast, dilations are nonrigid transformations, which means they produce images that are not congruent to the original. Note that if you dilate a figure by the same factor both vertically and horizontally, the image will be similar to the original.

Example B in your book shows how you can sometimes use what you have learned about translations, reflections, and dilations to fit functions to data. Read Example B carefully, following along with a pencil and paper. Notice that we assume the point (1.14, 0.18) is the image of the point (1, 1) on the parent parabola. We could have assumed that (1.14, 0.18) was the image of any point on the parent parabola except (0, 0). We chose the point (1, 1) because it leads to fairly easy scale-factor calculations.

(continued)

L E S S O N

4.6Dilations and the Absolute-Value Family




Lesson 4.6 • Dilations and the Absolute-Value Family (continued)

Investigation: The PendulumStep 1 If you have a string, a small weight, and a stopwatch or a watch with a second hand, follow the Procedure Note in your book. Repeat the experiment for several different string lengths. If you do not collect your own data, complete the investigation using the sample data below. The results given here are based on these sample data.

Length (cm) 100 85 75 30 15 5 43 60 89 140 180 195x

Period (s) 2.0 1.9 1.8 1.4 0.9 0.6 1.4 1.6 2.0 2.4 2.6 2.9y

Step 2 The graph looks like a transformation of the square root function, y � �

__ x . The vertex is at the origin, so the graph has not been translated, and

it has clearly not been reflected across the x- or y-axis. So, it must be a dilation of y � �

__ x .

Step 3 The origin is (0, 0) because a pendulum of length 0 cm would have no period.

Step 4 Choose the point (100, 2.0) in the data set. Let’s assume it is the image of the point (100, 10) on the graph of the parent function, y � �

__ x . (We could

have chosen any point on the parent curve, but (100, 10) will lead to a simple calculation.) The point (100, 10) is 100 horizontal units and 10 vertical units away from the vertex of y � �

__ x , while (100, 2.0) is 100 horizontal units and

2 vertical units away from the vertex of the transformed graph. Therefore, there is no horizontal dilation, but there is a vertical dilation by a factor of 2 __ 10 , or 0.2. The equation is therefore

y __ 0.2 � �

__ x , or y � 0.2 �

__ x .

This equation seems to fit the data quite well, but you might want to start with a different data point to see if you can get a better fit.

Step 5 Points that are not too close to the vertex and that seem to follow the trend of the majority of points work best. The point (30, 1.4) is probably not a good choice, for example, because it is close to the origin and it does not seem to follow the general pattern of the rest of the points.



CONDENSED


● learn about the unit circle and its equation

● write equations for ellipses by transforming the equation of the unit circle

● graph circles and ellipses, which are not functions, on your calculator by entering equations for two functions

The unit circle is a circle with radius 1 unit and center at the origin.

1

(0, 0) (x , 0)

P(x , y)

x

y

x

y

Choose any point P (x, y) on the circle and draw a slope triangle.

The slope triangle has legs of lengths x and y and a hypotenuse of length 1. By the Pythagorean Theorem, x 2 � y 2 � 1. This is the equation of the unit circle.

The domain of the unit circle is �1 � x � 1, and the range is �1 � y � 1. The unit circle is not a function because most input values correspond to two output values. For example, the x-value 0.8 corresponds to both y � 0.6 and y � �0.6.

To graph the unit circle on your calculator, you must solve x 2 � y 2 � 1 for y. This gives two equations, y � � �

______ 1 � x 2 and y � � �

______ 1 � x 2 . These equations

are functions. You need to graph both of them to get a complete circle.

You can apply what you learned about transformations of lines, parabolas, absolute-value functions, and other functions to transform the unit circle. Read Example A to find out how to apply your previous knowledge to write the equations of different circles.

An ellipse is a circle that has been dilated vertically and/or horizontally. Example B in your book shows how you can use what you have learned about transformations to find an equation of an ellipse based on the equation of a unit circle. Read that example, and then read the example below.

EXAMPLE What is the equation of this ellipse?

� Solution The original unit circle has been dilated vertically by a factor of 3 and translated horizontally �2 units and vertically 2 units. The equation changes like this:

x2 � y2 � 1 Original unit circle.

x2 � � y __ 3 � 2 � 1 Dilate vertically by a factor of 3. �Replace y with

y _ 3 .�

(x � 2)2 � � y � 2 _____ 3 �

2 � 1 Translate center to (�2, 2). (Replace x with x � 2, and

replace y with y � 2.)

x

y

2–4

4

–2

x

y

2–4

4

–2

(continued)

L E S S O N

4.7Transformations and the Circle Family




To graph this equation on your calculator, you need to solve for y.

� y � 2 _____ 3 �

2 � 1 � (x � 2)2 Subtract (x � 2 ) 2 from both sides.

y � 2

_____ 3 � � � __________

1 � (x + 2 ) 2 Take the square root of both sides.

y � 2 � 3 � __________

1 � (x + 2 ) 2 Multiply both sides by 3, then add 2.

Graph both functions, y � 2 � 3 � __________

1 � (x + 2 ) 2 and y � 2 � 3 � __________

1 � (x + 2 ) 2 , to get the complete ellipse.

Investigation: When Is a Circle Not a Circle?Choose one of the ellipses below. Carefully use your ruler to draw axes inside the ellipse. Make sure they are perpendicular. Mark your axes in centimeters, with 0 at the center of the ellipse. Use your markings to write an equation for the ellipse.

Using the ellipse at right, the right edge is 2 units from the center

1–1

–2

2

x

y

and the top is 3 units from the center, so the equation is

� x __ 2 � 2 � � y __ 3 � 2 � 1

To check your equation, solve it for y. For this ellipse,

y � �3 � ________

1 � � x __ 2 � 2

Graph both equations on your calculator, and see if the ellipsehas the same dimensions as your original ellipse.

The equations for transformations of circles are easier to work with before you solve them for y. If you start with a function for the top or bottom half of the circle, you can transform it, but the result is often messy to deal with. This is illustrated in Example C in your book. Read this example, following along with a pencil and paper.

The “Transformations of Functions and Other Relations” box in your book summarizes all you have learned about transformations. Read the text in this box carefully, and refer to it if you need to when you do your homework.

Lesson 4.7 • Transformations and the Circle Family (continued)


CONDENSED



● compose two functions and find the equation and graph for the composite function

● explore real-world situations involving compositions of functions

● interpret expressions involving compositions of functions in the context of real-world situations

Read the discussion at the beginning of Lesson 4.8 very carefully. It presents a situation in which the output of one function is the input of another function. This situation illustrates a composition of functions.

Work through Example A in your book. Note that you can write a single equation for y � g ( f (x)) as follows:

g ( f (x)) � g � 3 __ 4 x � 3 � � | 3 __ 4 x � 3 | Investigation: Looking Up

Steps 1–3 In the investigation in your book, read the list of materials, the Procedure Note, and Steps 1–3. Because it is difficult to collect the data on your own, you can use the data below. These ordered pairs are in the form (d, h), where d is the distance from the observer’s toe to the center of the mirror and h is the height mark reflected in the mirror.

(50, 148), (70, 106), (100, 73.5), (130, 57), (160, 45)

Step 4 You can find a function in the form h � f (d ) � a _ d to fit the

data. If you rewrite this equation as a � d � f (d ), you can see that a is the distance times the height. Calculating this product for each of the five data points gives 7400, 7420, 7350, 7410, 7200. Using the median, 7400, for a gives the equation h � f (d ) � 7400

____ d . As you can

see, this equation fits the sample data very well.

Step 5 Now, imagine you are walking toward the mirror and that your position at 1-second intervals is given by this table:

Time (s) 0 1 2 3 4 5 6 7t

Distance to mirror (cm) 163 112 74 47 33 31 40 62d

The graph of these (time, distance) data is shaped like a parabola.

To find a function to fit these data, first note that the vertex appears to be about (4.75, 30). So, the function is a translation of the parent graph, y � x 2, horizontally 4.75 units and vertically 30 units.

Next, you need to determine the horizontal and vertical scale factors. Choose the data point (7, 62) and assume it is the image of the point (1, 1) on the parent graph. In the graph of y � x2, the point (1, 1) is 1 unit from the vertex (0, 0) both horizontally and vertically.

(continued)

Compositions of FunctionsL E S S O N

4.8




Lesson 4.8 • Compositions of Functions (continued)

The data point is 7 � 4.75 � 2.25 units horizontally from the vertex of the transformed graph and 62 � 30 � 32 units vertically from the vertex. So, the horizontal scale factor is 2.25, and the vertical scale factor is 32. Use the translations and scale factors to get the final equation.

d � 30 ______ 32 � � t � 4.75 _______ 2.25 � 2 or d � g (t) � 32 � t � 4.75 _______ 2.25 � 2 � 30

The function appears to be a good fit.

Step 6 Use the functions for h and d to answer the questions in Step 6 in your book. The answers are given below. Your answers will vary if your functions for h and d are different.

a. h � f (47) � 7400 ____ 47 � 157. You can see 157 cm up the wall when you are

47 cm from the mirror.

b. d � g (1.3) � 32 � 1.3 � 4.75 _______ 2.25 � 2 � 30 � 105. You are 105 cm from

the center of the mirror at 1.3 seconds.

c. You can use g to find your distance from the mirror at 3.4 seconds:

g (3.4) � 32 � 3.4 � 4.75 _______ 2.25 � 2 � 30 � 41.52. Then, use f to find out how

high you can see up the wall when you are 41.25 cm from the mirror:

f (41.52) � 7400 ____ 41.52 � 178 cm. Note that because you are using the output

of g as the input of f, this is equivalent to f (g (3.4)).

Step 7 Complete Step 7 in your book. Here are the results:

a. f (60) is the distance you can see up the wall when you are 60 cm

from the mirror. f (60) � 7400 ____ 60 � 123 cm.

b. g (5.1) is your distance from the mirror at 5.1 seconds.

g (5.1) � 32 � 5.1 � 4.75 _________ 2.25 � 2 � 30 � 31 cm

c. f (g (2.8)) is the distance you can see up the wall after 2.8 seconds. g (2.8) � 54, and f (54) � 137. So, f (g (2.8)) � 137 cm.

Step 8

H(t) � f (g (t)) � f � 32 � t � 4.75 _______ 2.25 � 2 � 30 � � 7400 ________________ 32 � t � 4.75 _______ 2.25 � 2 � 30

H(2.8) � 7400 __________________ 32 � 2.8 � 4.75 _________ 2.25 � 2 � 30

� 137 cm

Carefully read the text after the investigation. Then, work through Example B, which shows how to find the domain and range of a composite function.


Documents

LESSON 4.1 Interpreting Graphs - KendallHuntmath.kendallhunt.com/documents/daa2/CL/DAA2CL010_04.pdf · LESSON 4.1 CONDENSED Interpreting Graphs Discovering Advanced Algebra Condensed