lesson 4.1 interpreting graphs - high school...

L E S S O N

4.1CONDENSED

In this lesson, you

● 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 graphshows the relationship between the number of people in a baseball stadiumand the time since the game ended.

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

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

The example in your book shows a more complicated graph that indicates howthe number of cans in a soda machine changes over the course of a typical schoolday. Study the graph carefully for a few minutes, thinking about what might causeeach 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 toanswer each question yourself, and then read the answers below.

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

● What time do classes begin? The number of cans begins to decrease at aslower 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. Beginningat this time, the number of cans decreases very rapidly. This is probablybecause students and teachers are purchasing sodas to go with their lunches.

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

x

y

Time since game ended (min)

Nu

mb

er o

f p

eop

le in

sta

diu

mInterpreting Graphs

Discovering Advanced Algebra Condensed Lessons CHAPTER 4 41©2004 Key Curriculum Press

(continued)

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Lesson 4.1 • Interpreting Graphs (continued)

Notice that both the baseball-stadium graph and the soda-machine graph arecontinuous. In reality, the number of people in the stadium and the number ofsodas in the machine must be discrete values. However, a continuous “graphsketch” 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 agraph that reflects all the information in the story. When you are finished,compare your graph with the graph below, which has been labeled to showhow the various sections relate to the story.

Now, look at the graph in Part 2 of the investigation. Think carefully aboutthe information shown in the graph. Then, write a story that conveys all theinformation, 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 usethe pool, but the water level is very low, so Luis turns on the hose and beginsfilling the pool at a constant rate. The children are restless and get in the pooland begin splashing water out of the pool. It fills at a slower constant rate thanbefore. When the pool is completely full, Luis turns off the hose and the childrenare careful not to splash more water. After the children get out of the pool, heempties it. The water pours out rapidly at first, then more slowly as there is lessand less water left. Luis leaves just a little water in the bottom, which will slowlyevaporate.”

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

42 CHAPTER 4 Discovering Advanced Algebra Condensed Lessons

©2004 Key Curriculum Press

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L E S S O N

4.2CONDENSED

In this lesson, you

● 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 typeof relation in which, for every value of the independent variable, there is at mostone value of the dependent variable. If x is the independent variable and y is thedependent 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 Tomread each day for 9 days. The graph on the left is continuous, whereas the graphon the right is made up of discrete points.

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

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

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

Not a function

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

Function

No vertical linecrosses the graphmore than once.

40302010

0

50

6 7 8 9

6070

2 3 4 51

Day

Pag

es r

ead

Hei

ght (

in.)

25

30

35

40

45

50

55

60

65

70

202 4 6 8 10 12 14

Age (yr)160

Function Notation


(continued)

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Lesson 4.2 • Function Notation (continued)

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

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

The example in your book gives you practice using function notation and findingthe values of functions by using equations and graphs. Work through theexample, 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 oney-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 therelation 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 afunction.

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 withmore than one x-value. Think about the horizontal line y � 2, for example.It is clearly a function (it passes the vertical line test), yet it pairs they-value 2 with every x-value).

f. This relation pairs the x-value 1 with two y-values, namely 1 and 3. It alsopairs 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 meansthat 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, therelation is a function.

i. The answer depends on how you think about the situation. If you consideronly the days in one year at one location, then the relation is a functionbecause the sun sets at a particular time each day. If you consider the sameday of any year, or if you consider different locations, then on the sameday 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, so a(0) � 3, ora(1.5) � 3.

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

d. d(2) � 3

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



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L E S S O N

4.3CONDENSED

In this lesson, you

● 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 andvertically k units

● Describe the graph of an equation in the form y � k � f(x � h) by relatingit to the graph of y � f(x)

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

In the investigation, you will see how translating a line—that is, moving ithorizontally 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 line 3 units above y � 2x is y � 3 � 2x. The equation for the line 4 units below y � 2x is y � �4 � 2x.

Step 3 The graphs of both lines are shown below. The new linehas slope �

12� and passes through the point (3, 4), so its equation

in point-slope form is y � 4 � �12�(x � 3). You can rewrite this in

intercept form as y � 2.5 � �12�x.

Step 4 If you move every point on f(x) � �12�x up 1 unit and right 2 units,

the origin moves to (2, 1). The slope of the line does not change. Therefore,the equation of the new line in point-slope form would be y � 1 � �

12�(x � 2).

If you rewrite this equation in point-slope form, you get y � �12�x. This is the

same as the original equation.

x

y

64–4 2

6

–6

–4

–2

y � x1_2

y � 2.5 � x or1_2

y � 4 � (x � 3)1_2

x

y

64–6 –4

6

4

–4

y = 3 � 2x y = 2x

y = �4 � 2x

x-y form Function notation

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

Point-slope form y � y1 � b�x � x1� f(x) � f�x1� � b�x � x1�

Lines in Motion


(continued)

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Lesson 4.3 • Lines in Motion (continued)

Step 5 When you translate a line with an equation in the form y � ax, thenumber of units you move horizontally is subtracted from x and the number ofunits you move vertically is added to the right side of the equation. (Note: If youmove to the left, the number subtracted from x is negative; if you move down,the number added to the right side of the equation is negative.)

Read the rest of the investigation, up to Step 9, in your book. You probably willnot have enough people or the equipment to conduct the experiment on yourown, but try to imagine what would happen. See if you can answer the questionsin Step 9, and then read the results below.

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

b. If you assume that B is 1 unit behind A, then the equation for B’s graph isy � 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 up 1 unit(because the distances B records are 1 unit more than those A records).

c. y � f(x � h) � k

You can generalize your findings from the investigation to any function. Forexample, no matter what the shape of the function y � f(x) is, the graph ofy � f(x � 3) � 2 will look the same as the graph of y � f(x), but it will betranslated up 2 units and right 3 units. Read the “Translation of a Function” boxon page 188 of your book. You can use your calculator to verify the text in thebox. For example, graph y � x3, and then y � (x � 4)3 � 1. The graph of thesecond function should look like the graph of the first function shifted right4 units and up 1 unit.

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

In your book, read the example and the paragraph that follows it, and then readthe example below.

EXAMPLE Describe how the graph of f(x) � �3 � (x � 2) is a translation of the graph off(x) � x.

� Solution The graph of f(x) � �3 � (x � 2) passes through (�2, �3). Consider this point to be the translatedimage of (0, 0) on f(x) � x. It is translated left2 units and down 3 units from its original location,so the graph of f(x) � �3 � (x � 2) is simplythe graph of f(x) � x translated left 2 units anddown 3 units.

Notice that the graph of f(x) � �3 � (x � 2) in the example aboveis also the graph of f(x) � x � 1. So, translating f(x) left 2 units and down3 units is equivalent to translating it down 1 unit. Read the remainder of thelesson very carefully. It explains that there are infinitely many translations thatwill map a line onto another, parallel line.

x

y

4

4

–4

Down 3 units

Left2 units

y � x

y � �3 � (x � 2)



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L E S S O N

4.4CONDENSED

Translations and theQuadratic Family


In this lesson, you

● Examine the graph of y � x2, which is a parabola

● Find equations for translations of the graph of y � x2

● Look at a simple application of a quadratic function

In the previous lesson, you explored translations of linear functions. Read the texton page 193 of your book, which gives examples of how translations occur inother settings.

A translation is a type of transformation. A transformation is a change in thesize 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 � x2 is called a parabola. Parabolasalways have a line of symmetry that passes through the vertex.

The function y � x2 is a parent function. By transforming the graph of thisfunction, you can create a family of functions, consisting of an infinite numberof related functions. The function y � x2 and all the functions created bytransforming it are called quadratic functions. As you will see throughout thiscourse, quadratic functions are very useful. Below are three examples of quadraticfunctions.

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 � x2 and a graphcreated by translating the graph of y � x2. Your job is to find the equation thatproduces the translated graph. To help you find the equation, think about whatyou learned about translated linear functions in the previous lesson.

When you think you know the equation for a graph, check it by graphing bothit and y � x2 in the same calculator window. (Be sure to read about friendlywindows in Calculator Note 4C.) Complete Step 1 by yourself before readingthe results on the next page.

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).

(continued)

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Lesson 4.4 • Translations and the Quadratic Family (continued)

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

b. The red parabola is the graph of y � x2 translated up 1 unit, so itsequation is y � x2 � 1.

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

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

e. The red parabola is the graph of y � x2 translated left 2 units and up2 units, so its equation is y � (x � 2)2 � 2.

f. The red parabola is the graph of y � x2 translated right 4 units and down2 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 atranslation left, you add the number of units to x. For a translation up, you addthe number of units to the right side of the equation (or subtract it from y). Fora translation down, you subtract the number of units from the right side of theequation (or add it to y). The coordinates of the vertex of the translated parabolaare (value of horizontal translation, value of vertical translation).

Step 3 In general, if the graph of y � x2 is translated horizontally h units andvertically 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 ofparabolas. Read the example and solution carefully, following along with apencil and paper. To test your understanding, solve the same problem, buthave 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 thediver’s maximum height would be 20 ft.)

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

Notice that the vertex of the translated parabola is (h, k). This is why finding thevertex 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).



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L E S S O N

4.5CONDENSED

In this lesson, you

● Learn about the square root function and its graph

● Reflect functions across the x- and y-axes

At right is the graph of the square root function, y � �x�. The domain of the function is x � 0, and the range is y � 0.

Graph the square root function on your calculator and trace the graphto find �3�, �8�, and �31�. You should get approximately 1.732,approximately 2.828, and approximately 5.568, respectively. If you tryto trace for values of x � 0, the calculator gives no value for y. This isbecause the function is not defined for negative x-values (that is, youcannot take the square root of a negative number).

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

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

a. Y2 � �(0.5x � 2) � �0.5x � 2. The graph of Y2 looks like thegraph of Y1 flipped over the x-axis. In other words, the graphof Y2 is the reflection of the graph of Y1 across the x-axis.

b. Y2 � �(�2x � 4) � 2x � 4. The graph of Y2 is the reflection of the graph of Y1 across the x-axis.

[�9.4, 9.4, 1, �6.2, 6.2, 1]

c. Y2 � ��x 2 � 1� � �x 2 � 1. The graph of Y2 is the reflection of the graph of Y1 across the x-axis.

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

[�9.4, 9.4, 1, �6.2, 6.2, 1]

a. Y2 � 0.5(�x) � 2 � �0.5x � 2. The graph of Y2 is the reflection of the graph of Y1 across the y-axis.

x

y

–10 –5 105

–5

5

Reflections and the Square Root Family


(continued)

[�9.4, 9.4, 1, �6.2, 6.2, 1]

[�9.4, 9.4, 1, �6.2, 6.2, 1]

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

b. Y2 � �2(�x) � 4 � 2x � 4. The graph of Y2 is the reflection of thegraph of Y1 across the y-axis.

[�9.4, 9.4, 1, �6.2, 6.2, 1]

c. Y2 � (�x)2 � 1 � x 2 � 1. The graph of Y2 is the same as the graph of Y1 because the original graph has the y-axis as a line of symmetry, so a reflection across the y-axis maps the graphonto itself.

d. Y2 � (�x � 3)2 � 2. The graph of Y2 is the reflection of thegraph of Y1 across the y-axis.

[�9.4, 9.4, 1, �6.2, 6.2, 1]

[�9.4, 9.4, 1, �6.2, 6.2, 1]

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

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

a. The graph of Y2 � �Y1(x) � ��x� is a reflection of the square rootfunction across the x-axis. The graph of Y2 � Y1(�x) � ��x� is areflection of the square root function across the y-axis.

[�9.4, 9.4, 1, �6.2, 6.2, 1]

b. The graph of Y2 � �Y1(�x) � ��x� is a reflection of the square root function across both axes.

c. The graph of y � �x� is not an entire horizontal parabolabecause it has a range of y � 0. Graphing y � ��x� would complete the bottom half of the parabola.

[�9.4, 9.4, 1, �6.2, 6.2, 1]

The “Reflection of a Function” box summarizes what you discovered in theinvestigation. Read the text in this box, and then read the rest of the lesson.

Y2 � Y1(�x)

Y2 � �Y1(x)



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L E S S O N

4.6CONDENSED

In this lesson, you

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

● Graph stretches or shrinks of a parent function and observe how thesetransformations are related to the equation of the parent function

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

Read page 209 in your book to review the idea of absolute value. Look at thegraph of the parent absolute-value function, y � x.

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

Example A shows three examples of transformations of the function y �x.The transformations are all stretches. Read the example very carefully. Note thatyou can get the function in part a by substituting �2

y� for y in the parent function,

y �x.

Substituting �ay

� for y in the parent function stretches (if a � 1) or shrinks(if 0 � a � 1) the graph of the parent function vertically by a factor of a.

Similarly, as you can observe in part b, substituting �bx

� for x in the parent functionstretches (if b � 1) or shrinks (if 0 � b � 1) the graph of the parent functionhorizontally by a factor of b.

Translations and reflections are rigid transformations, meaning that they produceimages that are congruent to the original figure. By contrast, stretches and shrinksare nonrigid transformations, which means they produce images that are notcongruent to the original. Note that if you stretch or shrink a figure by the samefactor 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 havelearned about translations, reflections, and stretches to fit functions to data.Read Example B carefully, following along with a pencil and paper. Notice thatwe assume the point (1.14, 0.18) is the image of the point (1, 1) on the parentparabola. We could have assumed that (1.14, 0.18) was the image of any pointon the parent parabola except (0, 0). We chose the point (1, 1) because it leadsto fairly easy scale-factor calculations.

x

y

–8 –4 84

–8

–4

8

4

Stretches and Shrinks andthe Absolute-Value Family


(continued)

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

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

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, andit has clearly not been reflected across the x- or y-axis. So, it must simply be astretch or shrink of y � �x�.

Step 3 The origin is (0, 0) because a pendulum of length 0 cm would haveno period.

[0, 200, 50, 0, 3, 1]

Step 4 Choose the point (100, 2.0) in the data set. Let’s assume it is the imageof the point (100, 10) on the graph of the parent function, y � �x�. (We couldhave chosen any point on the parent curve, but (100, 10) will lead to a simplecalculation.) The point (100, 10) is 100 horizontal units and 10 vertical unitsaway from the vertex of y � �x�, while (100, 2.0) is 100 horizontal units and2 vertical units away from the vertex of the transformed graph. Therefore, thereis no horizontal stretch or shrink, but there is a vertical shrink by a factor of �1

20�,

or 0.2. The equation is therefore y � 0.2�x�.

[0, 200, 50, 0, 3, 1]

This equation seems to fit the data quite well, but you might want to start with adifferent 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 thetrend of the majority of points work best. The point (30, 1.4) is probably not agood choice, for example, because it is close to the origin and it does not seemto follow the general pattern of the rest of the points.

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



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L E S S O N

4.7CONDENSED

In this lesson, you

● Are introduced to the unit circle and its equation

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

● Graph ellipses, which are not functions, on your calculator by enteringequations for two functions

Read the first paragraph on page 217 of your book, which reviews thetransformations you have learned in this chapter.

The unit circle is a circle with radius 1 unit and center at the origin.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 theunit 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 totwo output values. For example, the x-value 0.8 corresponds to both y � 0.6and 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 equationsare functions. You need to graph both of them to get a complete circle.

An ellipse is a stretched or shrunken circle. Example A in your book shows howyou can use what you have learned about transformations to find an equation ofan ellipse based on the equation of a unit circle. Read that example, and thenread the example below.

EXAMPLE What is the equation of this ellipse?

� Solution The original unit circle has been stretched vertically by a factor of 3 andtranslated horizontally �2 units and vertically 2 units. The equation changeslike this:

x2 � y 2 � 1 Original unit circle.

x2 � ��3y

��2

� 1 Stretch vertically by a factor of 3. �Replace y with �3y

�.�

(x � 2)2 � ��y �3

2��

2

� 1 Translate center to (�2, 2).

x

y

2–4

4

–2

1

(0, 0) (x , 0)

P(x , y)

x

y

x

y

Transformations andthe Circle Family


(continued)

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Lesson 4.7 • Transformations and the Circle Family (continued)

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

��y �3

2��

2

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

�y �

32

� � �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 theellipse. Make sure they are perpendicular. Mark your axes in centimeters, with 0at the center of the ellipse.

Use your markings to write an equation for the ellipse.The right side of this ellipse is 2 cm from the center, andthe top is 3 cm from the center, so the equation is

��2x

��2

� ��2y��

2

� 1

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

y � 2��1 � ��2x

��2

Graph both equations on your calculator, and see if theellipse has the same dimensions as your original ellipse.Be sure to use a friendly window for your graph.

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

The “Transformations of Functions and Relations” box in your book summarizesall 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.

x

y

1–1

–2

2



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L E S S O N

4.8CONDENSED

In this lesson, you

● Compose two functions and find the equation and graph for the compositefunction

● Explore real-world situations involving compositions of functions

● Interpret expressions involving compositions of functions in the context ofreal-world situations

Read the discussion at the beginning of Lesson 4.8 very carefully. It presents asituation 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 equationfor y � g( f(x)) as follows:

g( f(x)) � g��34�x � 3� � �

34�x � 3

Investigation: Looking UpSteps 1–3 In the investigation in your book, read the list of materials, theProcedure Note, and Steps 1–3. Because it is difficult to collect the data on yourown, 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 andh 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) � �da

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

74d00�. As you can see, this

equation fits the data very well.

Step 5 Now, imagine you are walking toward the mirror and that your[0, 200, 10, 0, 200, 10]

position at 1-second intervals is given by this table:

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 tobe 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 stretch or shrink factors. Choose thedata point (7, 62) and assume it is the image of the point (1, 1) on theparent graph. In the graph of y � x 2, the point (1, 1) is 1 unit fromthe vertex (0, 0) both horizontally and vertically. The data point is

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

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

Compositions of Functions


(continued)

[0, 10, 1, 0, 200, 10]

DAA_CL_614_04.qxd 5/28/03 11:17 AM Page 55

Lesson 4.8 • Compositions of Functions (continued)

7 � 4.75 � 2.25 units horizontally from the vertex of the transformed graph and62 � 30 � 32 units vertically from the vertex. So, the horizontal scale factor is2.25, and the vertical scale factor is 32. Use the translations and scale factors toget the final equation.

�d �

3230

� � ��t �2.2

45.75

��2

or d � g(t) � 32��t �2.2

45.75

��2

� 30

The function appears to be a good fit.

[0, 10, 1, 0, 200, 10]

Step 6 Use the functions for h and d to answer the questions in Step 6 in yourbook. The answers are given below.

a. h � f(47) � �74

407

0� � 157. You can see 157 cm up the wall when you are

47 cm from the mirror.

b. d � g(1.3) � 32��1.32�.25

4.75��2

� 30 � 105. You are 105 cm fromthe 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��3.42�.25

4.75��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) � �4714.0502� � 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 cmfrom the mirror. f(60) � �

746

00

0� � 123 cm.

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

g(5.1) � 32��5.12�.25

4.75��

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 �2.2

45.75

��2

� 30� �

H(2.8) � � 137 cm

Read the very important text after the investigation. Then, work throughExample B, in which a function is composed with itself.

7400��

32��2.82�.25

4.75��

2

� 30

7400��

32��t �2.2

45.75

��2

� 30



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lesson 4.1 interpreting graphs - high school...

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