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194 5.1 Chapter Five TRANSFORMATIONSOFFUNCTIONSANDTHEIRGRAPHS VERTICALANDHORIZONTAL SHIFTS -- '1M _n"rr,""---- Suppose we shift the graph of some function vertically or horizontally, giving the graph of a new function. In this section we investigate the relationship between the formulas for the original func- tion and the new function. VerticalandHorizontal Shift:TheHeatingScheduleForanOfficeBuilding We start with an example of a vertical shift in the context of the heating schedule for a building. Example1 H, temperature (OF) 70 60 50 To save money, an office building is kept warm only during business hours. Figure 5.1 shows the temperature, H, in of, as a function of time, t, in hours after midnight. At midnight (t = 0), the building's temperature is 50°F. This temperature is maintained until 4 am. Then the building begins to warm up so that by 8 am the temperature is 70°F. At 4 pm the building begins to cool. By 8 pm, the temperature is again 50°F. Suppose that the building's superintendent decides to keep the building 5°F warmer than before. Sketch a graph of the resulting function. H, temperatureCF) 75 r 70 65 60 55 - 50 H = p(t): New, verticallyshifted schedule / H = f(t) '" H = f(t): Original heating schedule . I I I I t,time(hours 4 8 12 16 20 24 aftermidnight) Figure 5.2: Graph of new heating schedule, H = p( t), obtained by shifting original graph, H = f(t), upward by 5 units 4 L t, time(hours 24 after midnight) 8 12 16 20 Figure5.1: The heating schedule at an office building Solution The graph of f, the heating schedule function of Figure 5.1, is shifted upward by 5 units. The new heating schedule, H = p(t), is graphed in Figure 5.2. The building's overnight temperature is now 55°F instead of 50°F and its daytime temperature is 75°F instead of 70°F. The 5°F increase in temperature corresponds to the 5-unit vertical shift in the graph. The next example involves shifting a graph horizontally. Example 2 The superintendent then changes the original heating schedule to start two hours earlier. The build- ing now begins to warm at 2 am instead of 4 am, reaches 70°F at 6 am instead of 8 am, begins cooling off at 2 pm instead of 4 pm, and returns to 50°F at 6 pm instead of 8 pm. How are these changes reflected in the graph of the heating schedule?

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Page 1: DocumentEC

194

5.1Chapter Five TRANSFORMATIONSOF FUNCTIONSAND THEIRGRAPHS

VERTICALANDHORIZONTALSHIFTS-- '1M _n"rr,""----Suppose we shift the graph of some function vertically or horizontally, giving the graph of a newfunction. In this section we investigate the relationship between the formulas for the original func-tion and the new function.

VerticalandHorizontalShift:TheHeatingScheduleForanOfficeBuildingWe start with an example of a vertical shift in the context of the heating schedule for a building.

Example1

H, temperature(OF)

70

60

50

To save money, an office building is kept warm only during business hours. Figure 5.1 shows thetemperature, H, in of, as a function of time, t, in hours after midnight. At midnight (t = 0), thebuilding's temperature is 50°F. This temperature is maintained until 4 am. Then the building beginsto warm up so that by 8 am the temperature is 70°F. At 4 pm the building begins to cool. By 8 pm,the temperature is again 50°F.

Suppose that the building's superintendentdecides to keep the building 5°F warmer than before.Sketch a graph of the resulting function.

H, temperatureCF)

75

r

70

65

6055 -

50

H = p(t): New,verticallyshiftedschedule

/

H = f(t)

'"H = f(t):Originalheatingschedule .I I I I t, time(hours

4 8 12 16 20 24 aftermidnight)

Figure 5.2: Graph of new heating schedule, H = p( t),

obtained by shifting original graph, H = f(t), upward by 5units

4L t, time(hours

24 after midnight)8 12 16 20

Figure5.1: The heating schedule at an officebuilding

Solution The graph of f, the heating schedule function of Figure 5.1, is shifted upward by 5 units. The newheating schedule, H = p(t), is graphed in Figure 5.2. The building's overnight temperature is now55°F instead of 50°F and its daytime temperature is 75°F instead of 70°F. The 5°F increase intemperature corresponds to the 5-unit vertical shift in the graph.

The next example involves shifting a graph horizontally.

Example2 The superintendent then changes the original heating schedule to start two hours earlier. The build-ing now begins to warm at 2 am instead of 4 am, reaches 70°F at 6 am instead of 8 am, beginscooling off at 2 pm instead of 4 pm, and returns to 50°F at 6 pm instead of 8 pm. How are thesechanges reflected in the graph of the heating schedule?

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Solution

5.1 VERTICALAND HORIZONTALSHIFTS 195

Figure 5.3 gives a graph of H = q(t), the new heating schedule, which is obtained by shifting thegraph of the original heating schedule, H = f (t), two units to the left.

70 '"

60

50/

'I I I I I I I I I I I I t, time(hours4 8 12 16 20 24 aftermidnight)

Figure 5.3: Graph of new heating schedule, H = q(t),found by shifting, j, the original graph 2 units to the left

Notice that the upward shift in Example 1 results in a warmer temperature, whereas the leftwardshift in Example 2 results in an earlier schedule.

Formulasfora VerticalorHorizontalShift

How does a horizontal or vertical shift of a function's graph affect its formula?

Example3

Solution

In Example I, the graph of the original heating schedule, H = f (t), was shifted upward by 5 units;the result was the warmer schedule H = p(t). How are the formulas for f(t) and p(t) related?

The temperature under the new schedule, p(t), is always 5°F warmer than the temperature underthe old schedule, f (t). Thus,

New temperatureat time t

Old temperatureat time t

+5.

Writing this algebraically:p(t) f(t) +~ ~

New temperature Old temperatureat time t at time t

5.

The relationship between the formulas for p and f is given by the equation p(t) = f(t) + 5.

We can get information from the relationship p(t) = f(t) + 5, although we do not have anexplicit formula for f or p.

Suppose we need to know the temperature at 6 am under the schedule p(t). The graph of f (t)shows that under the old schedule f(6) = 60. Substituting t = 6 into the equation relating f and pgives p(6):

.. p(6) = f(6) + 5 = 60+5 = 65.

Thus,at 6 amthe temperatureunderthe newscheduleis 65°F.

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196 Chapter Five TRANSFORMATIONSOF FUNCTIONSAND THEIRGRAPHS

Example4 In Example 2 the heating schedule was changed to 2 hours earlier, shifting the graph horizontally 2units to the left. Find a formula for q, this new schedule, in terms of f, the original schedule.

Solution The old schedule always reaches a given temperature 2 hours after the new schedule. For example, at4 am the temperature under the new schedule reaches 60°. The temperature under the old schedulereaches 60° at 6 am, 2 hours later. The temperature reaches 65° at 5 am under the new schedul\.:, butnot until 7 am, under the old schedule. In general, we see that

Temperature under new schedule

at time t

Temperature under old schedule

at time (t + 2), two hours later.

Algebraically, we haveq(t) = f(t + 2).

This is a formula for q in terms of f.

Let's check the formula from Example 4 by using it to calculate q(14), the temperature underthe new schedule at 2 pm. The formula gives

q(14) = f(14 + 2) = f(16).

Figure 5.1 shows that f(16) = 70. Thus, q(14) = 70. This agrees with Figure 5.3.

Translationsofa FunctionandItsGraph

In the heating schedule example, the function representing a warmer schedule,

p(t) = f(t) + 5,

has a graph which is a vertically shifted version of the graph of f. On the other hand, the earlierschedule is represented by

q(t) = f(t + 2)

and its graph is a horizontally shifted version of the graph of f. Adding 5 to the temperature, oroutput value, f (t), shifted its graph up fiveunits. Adding 2 to the time, or input value, t, shifted itsgraph to the left two units. Generalizing these observations to any function g:

If y = 9(x) is a function and k is a constant, then the graph of. y = g(x) + k is the graph of y = g(x) shifted vertically Ikl units. If k is positive, the

shift is up; if k is negative, the shift is down.

. y = g(x + k) is the graph of y = g(x) shifted horizontally Ikl units. If k is positive, theshift is to the left; if k is negative, the shift is to the right.

A vertical or horizontal shift of the graph of a function is called a translation because it doesnot change the shape of the graph, but simply translates it to another position in the plane. Shifts ortranslations are the simplest examples of transformations of a function. We will see others in latersections of Chapter 5.

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5.1 VERTICALAND HORIZONTALSHIFTS 197

InsideandOutsideChangesSince y = g(x + k) involves a change to the input value, x, it is called an inside change to g.Similarly, since y = g(x) + k involves a change to the output value, g(x), it is called an outsidechange. In general, an inside change in a function results in a horizontal change in its graph, whereasan outside change results in a vertical change.

In this section, we consider changes to the input and output of a function. For the function

Q = f(t),

a change inside the function's parentheses can be called an "inside change" and a change outsidethe function's parentheses can be called an "outside change."

Example5 If n = f (A) gives the number of gallons of paint needed to cover a house of area A ft2, explain themeaning of the expressions f(A + 10) and f(A) + 10 in the context of painting.

These two expressions are similar in that they both involve adding 10. However, for f(A + 10), the10 is added on the inside, so 10 is added to the area, A. Thus,

Solution

n = f(A + 10) ='--v--'Area

The expressionf(A) + 10 representsan outsidechange.We are adding 10 to f(A), whichrepresents an amount of paint, not an area. We have

Amount of paint needed

to cover an area of (A + 10) ft2

Amount of paint needed to cover

an area 10 ft2 larger than A.

n = f(A) + 10 =~

Amount

of paint

In f(A + 10), we added 10 square feet on the inside of the function, which means that the area tobe painted is now 10 ft2 larger. In f(A) + 10, we added 10 gallons to the outside, which means thatwe have 10 more gallons of paint than we need.

Amount of paint needed

to cover region of area A+ 10gals= IO gallons more paint than

amount needed to cover area A.

Example6 Let 8(t) be the average weight (in pounds) of a baby at age t months. The weight, V, of a particularbaby named Jonah is related to the average weight function 8(t) by the equation

V = 8(t) + 2.

Find Jonah's weight at ages t = 3 and t = 6 months. What can you say about Jonah's weight ingeneral?

Solution At t = 3 months, Jonah's weight isV = 8(3)+ 2.

Since 8(3) is the average weight of a 3-month old boy, we see that at 3 months, Jonah weighs 2pounds more than average. Similarly, at t = 6 months we have

V = 8(6)+ 2,

which means that, at 6 months, Jonah weighs 2 pounds more than average. In general, Jonah weighs. 2 pounds more than average for babies of his age. ~

.,

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198 Chapter Five TRANSFORMATIONSOF FUNCTIONSAND THEIRGRAPHS

Example7 The weight, W, of another baby named Ben is related to s(t) by the equation

W = s(t + 4).

What can you say about Ben's weight at age t = 3 months? At t = 6 months? Assuming that babiesincrease in weight over the first year of life, decide if Ben is of average weight for his age, aboveaverage, or below average.

Solution Since W = s(t + 4), at age t = 3 months Ben's weight is given by

W = s(3 + 4) = s(7).

We defined s(7) to be the average weight of a 7-month old baby. At age 3 months, Ben's weight isthe same as the average weight of7-month old babies. Since, on average, a baby's weight increasesas the baby grows, this means that Ben is heavier than the average for a 3-month old. Similarly, atage t = 6, Ben's weight is given by

W = s(6 + 4) = s(lO).

Thus, at 6 months, Ben's weight is the same as the average weight of lO-month old babies. Inboth cases, we see that Ben is above average in weight.

Notice that in Example 7, the equation

W=s(t+4)

involves an inside change, or a change in months. This equation tells us that Ben weighs as much asbabies who are 4 months older than he is. However in Example 6, the equation

v = s(t) + 2

involves an outside change, or a change in weight. This equation tells us that Jonah is 2 poundsheavier than the average weight of babies his age. Although both equations tell us that the babiesare heavier than average for their age, they vary from the average in different ways.

CombiningHorizontalandVerticalShifts

We have seen what happens when we shift a function's graph either horizontally or vertically. Whathappens if we shift it both horizontally and vertically?

Example8 Let r be the transformation of the heating schedule function, H = f (t), defined by the equation

r(t) = f(t - 2) - 5.

(a) Sketch the graph of H = r(t).(b) Describe in words the heating schedule determined by r.

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Solution

5.1 VERTICALAND HORIZONTALSHIFTS 199

(a) To graph r, we break this transfonnation into two steps. First, we sketch a graph of H =f(t - 2). This is an inside change to the function f and it results in the graph of f being shifted2 unitsto theright.Next,we sketcha graphof H = f(t - 2) ~ 5. This graph can be found byshifting our sketch of H = f(t - 2) down 5 units. The resulting graph is shown in Figure 5.4.The graph of r is the graph of f shifted 2 units to the right and 5 units down.

(b) The function r represents a schedule that is both 2 hours later and 5 degrees cooler than theoriginal schedule.

H n) H = f(t)

~~

I

//, \60 /

/

55 /50 .45);"

/LJ ~ t, time(hours20 24 aftermidnight)

'--'--l

4 8 12 16

Figure5.4: Graph of r(t) = f(t - 2) - 5 is graph of H = f(t) shifted right by 2 and down by 5

We can use transformations to understand an unfamiliar function by relating it to a function wealready know.

Example9

Solution

A graph of f (x) = X2 is in Figure 5.5. Define 9 by shifting the graph of f to the right 2 units anddown 1unit; see Figure 5.6. Find a formula for 9 in terms of f. Find a formula for 9 in terms of x.

y

4f(x) = X2

g

x x-2 4 -2 4

-1 -1

Figure5.5: The graph of f (x) = x2 Figure5.6: The graph of g, a transformation of f

The graph of 9 is the graph of f shifted to the right 2 units and down 1 unit, so a formula for 9 isg(x) = f(x - 2) - 1.Sincef(x) = x2, wehavef(x - 2) = (x - 2)2. Therefore,

g(x) = (x - 2)2 - 1.

.. It is a good idea to check by graphing g(x) = (x - 2)2 -1 and comparing the graph with Figure 5.6.

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200 Chapter Five tRANSFORMAtiONS OF FUNC1\ONS AND tHEIR GRAPHS

ExercisesandProblemsforSection5.1

Exercises

8. Match the graphs in (a)-(f) with the formulas in (i)-(vi).1. Using Table 5.1, complete the tables for g, h, k, m,where:

Ca) g(x) = 1(x -1)

(c) k(x) = f(x) + 3

CD) h(x) = 1(x + 1)

(d) m(x)=f(x-1)+3

Explain how the graph of each function relates to thegraph of f(x).

Table5.1

~~

~~

~~~~~~

In Exercises 2-5, graph the transformations of f (x) in Fig-ure 5.7.

~~"02 3 64 5

Figure 5.7

2. y = f(x + 2)

4. y = f(x - 1) - 5

3. y = f(x) + 2

5. y = f(x + 6) - 4

(i) y = \x\(iii) y = \x - 1.2\(v) y = Ix + 3.41

(ii) y = \x\ - 1.2(iv) y = \x\ + 2.5'(vi) y=lx-31+2.7

(a) y (b) y

xx

~ y ~ y

-~x v IX\~\ '\J \\\ '\J

x 'X

9. The graph of f (x) contains the point (3, - 4). What pointmust be on the graph of

(a) f(x)+5?

(c) f(x - 3) - 2?

10. The domain of the function g(x) is -2 < x < 7. What\is'tb..~dQill'd.\\\Q{ <b~T,- '2.\1

11. The range of the function R( s) is 100 :; R( s) :; 200What is the range of R( s) - 150?

(b) f(x + 5)?

Write a formula and graph the transformations of m(n) =~n2 in Exercises 12-19.

6. Let f(x) = 4x, g(x) = 4x + 2, and h(x) = 4x - 3.What is the relationship between the graph of f(x) and 12. y = m(n) + 1thegraphsofh(x)andg(x)? 14 -

( ). y - m n - 3.7

(1)

x

(1

)X+4

(1

)X-2

~. Letf(x)= :3 ,g(x)=:3 ,andh(x)=:3 16. y=m(n)+yff3. How do the graphs of g(x) and h(x) compare to the

graph of f(x)? 18. y = m(n + 3) + 7

13. y = m(n + 1)

15. y = m(n - 3.7)

17. y = m(n + 2J2)

19. y=m(n-17)-159

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Write a fonnula and graph the transformations of k( w) = 3win Exercises 20-25.

20. y = k(w) - 3 21. y=k(w-3)

Problems

5.1 VERTICALAND HORIZONTALSHIFTS 201

22. y=k(w)+1.8

24. y = k(w + 2.1)- 1.3

23. y=k(w+V5)

25. y = k(w - 1.5) - 0.9

26. (a) yUsing Table 5.2, evaluate

(i) f(x) for x = 6.

(ii) f(5) - 3.

(iii) f(5 - 3).

(iv) g(x) + 6 for x = 2.

(v) g(x + 6) for x = 2.

(vi) 3g(x) for x =0.

(vii) f(3x) for x = 2.

(viii) f(x) - f(2) forx = 8.(ix) g(x + 1) - g(x) for x = 1.

(b) Solve

(i) g(x) =6.(iii) g(x) = 281.

(c) The values in the table were obtained using the for-mulas f(x) = X3 + X2 + x - 10 and g(x) =7X2 - 8x - 6. Use the table to find two solutionsto theequationX3+ X2+ x - 10 = 7X2 - 8x - 6.

(ii) f(x) = 574.

Table5.2

27. The graph of g(x) contains the point (-2,5). Write aformula for a translation of 9 whose graph contains thepoint

(a) (-2,8) (b) (0,5)

28. (a) Letf(x)= (~r +2.Calculatef(-6).(b) Solvef(x) = -6.

(c) Find points that correspond to parts (a) and (b) onthe graph of f(x) in Figure 5.8.

(d) Calculatef(4) - f(2). Drawa verticallinesegmenton the y-axis that illustrates this calculation.

(e) It a = -2, computef(a + 4) and f(a) + 4.(f);;In part (e), what x-value corresponds to f(a + 4)?

To f(a) + 4?

x

Figure5.8

29. The function P( t) gives the number of people in a certainpopulation in year t. Interpret in terms of population:

(a) P(t) + 100 (b) P(t + 100)

30. Describe a series of shifts which translates the graph ofy = (x + 3)3- 1 onto the graph of y = x3.

31. Graphf(x) = In(lx - 31)andg(x) = In(lxl). Findthevertical asymptotes of both functions.

32. Graph y = log x, y = log(10x), and y = log(100x).How do the graphs compare? Use a property of logs toshow that the graphs are vertical shifts of one another.

Explain in words the effects of the transformations in Exer-cises 33-38 on the graph of q(z). Assume a, b are positiveconstants.

33. q(z) + 3

35. q(z + 4)

34. q(z) - a

36. q(z - a)

37. q(z+b)-a 38. q(z - 2b) + ab

39. Suppose S(d) gives the height of high tide in Seattle ona specific day, d, of the year. Use shifts of the functionS(d) to findformulas for each of the following functions:

(a) T( d), the height of high tide in Tacoma on day d,given that high tide in Tacoma is always one foothigher than high tide in Seattle.

(b) P(d), the height of high tide in Portland on day d,given that high tide in Portland is the same height asthe previous day's high tide in Seattle.

x 0 1 2 3 4 5 6 7 8 9

f(x) -10 -7 4 29 74 145 248 389 574 809

g(x) -6 -7 6 33 74 129 198 281 378 489

0 f(x:

0 /

0 V./

-8 -6 (

rn-1Ojo-10

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202 Chapter Five TRANSFORMATIONSOF FUNCTIONSANDTHEIR GRAPHS

40. Table 5.3 contains values of f(x). Each function in parts(a)-(c) is a translation of f(x). Find a possible for-mula for each of these functions in terms of f. For ex-ample, given the data in Table 5.4, you could say thatk(x) = f(x) + 1.

Table5.3

x 7

24.5j(x)

Table5.4

x 7

25.5k(x)

41. For t :2:0, let H(t) = 68 + 93(0.91)t give the temper-ature of a cup of coffee in degrees Fahrenheit t minutesafter it is brought to class.

(a) Find formulas for H(t + 15) and H(t) + 15.(b) Graph H(t), H(t + 15), and H(t) + 15.(c) Describe in practical terms a situation modeled by

the function H(t + 15). What about H(t) + 15?(d) Which function, H(t+15) or H(t)+15, approaches

the same final temperature as the function H(t)?What is that temperature?

5.2 REFLECTIONSANDSYMMETRY_CC"L" ..~ --~-""""""~",gW_""'-__"""l"'l"*~-"~_~rm"'-_W£"""i'@-C~

42. At ajazz club, the cost of an evening is based on a covercharge of $20 plus a beverage charge of $7 per drink.

(a) Find a formula for t(x), the total cost for an eveningin which x drinks are consumed.

(b) If the price of the cover charge is raised by $5, ex-press the new total cost function, n(x ),as atransfor-mati on of t( x).

(c) The management increases the cover charge to $30,leaves the price of a drink at $7, but includes the first

two drinks for free. For x :2: 2, expressp( x), the newtotal cost, as a transformation oft(x).

7

22.5

43. A hot brick is removed from a kiln and set on the floorto cool. Let t be time in minutes after the brick was re-moved. The difference, D(t), between the brick's tem-perature, initially 350°F, and room temperature, 70°F,decays exponentially over time at a rate of 3% perminute. The brick's temperature, H(t), is a transforma-tion of D(t). Find a formula for H(t). Compare thegraphs of D(t) and H(t), paying attention to the asymp-totes.

7

3244. Suppose T( d) gives the average temperature in your

hometown on the dth day of last year (where d = 1 isJanuary 1st, and so on).7

30 (a) Graph T( d) for 1 <e:::d <e:::365.

(b) Give a possible value for each of the following:T(6); T(100); T(215); T(371).

(c) What is the relationship between T(d) and T(d +365)? Explain.

(d) If you were to graph wed) = T(d + 365) on thesame axes as T(d), how would the two graphs com-pare?

(e) Do you think the function T( d) + 365 has any prac-tical significance? Explain.

45. Let f(x) = eX and g(x) = 5ex. If g(x) = f(x - h),find h. ~

In Section 5.1 we saw that a horizontal shift of the graph of a function results from a change to theinput of the function. (Specifically, adding or subtracting a constant inside the function's parenthe-ses.) A vertical shift corresponds to an outside change.

In this section we consider the effect of reflecting a function's graph about the x or y-axis. Areflection about the x-axis corresponds to an outside change to the function's formula; a reflectionabout the y-axis and corresponds to an inside change.

-

(a) x-hex)

(b) - x-g(x)

(c)-x-i(x)

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A Formulafora Reflection

5.2 REFLECTIONSAND SYMMETRY 203

Figure 5.9 shows the graph of a function y = f(x) and Table 5.5 gives a corresponding table ofvalues. Note that we do not need an explicit formula for f.

Y

64y = f(x)

Table5.5 Valuesof thefunction y = f(x)

32

Sx

-3 -2 -1 1 2 3

-32

-64

x Y

1

2

4

8

16

32

64

-3

-2

-1

0

1

2

3

Figure5.9: A graph of the function y = f (x)

Figure 5.10 shows a graph of a function y = 9(x), resulting from a vertical reflection of thegraph of f about the x-axis. Figure 5.11 is a graph of a function y = h(x), resulting from a hori-zontal reflection of the graph of f about the y-axis. Figure 5.12 is a graph of a function y = k(x),resulting from a horizontal reflection of the graph of f about the y-axis followed by a vertical re-flection about the x-axis.

y y

h(x) 64 /8 f(x)/

I"....

_./

f (x) /8/

x x-3 -2 -1

-32

1 2 3

-64

Figure5.10: Graph reflectedabout x-axis

-64

Figure5.11: Graph reflected abouty-axis

Figure 5.12: Graph reflected abouty- and x-axes

Example1 Find a formula in terms of f for (a) y = g(x) (b) y = h(x) (c) y = k(x)

Solution (a) The graph of y = g(x) is obtained by reflecting the graph of f vertically about the x-axis. Forexample, the point (3,64) on the graph Qf f reflects to become the point (3, -64) on the graphof g. The point (2,32) on the graph of f becomes (2, -32) on the graph of g. See Table 5.6.

Table5.6 Valuesof thefunctions g(x) and f(x) graphed in Figure 5.10

~

y

64+ 8

f(x) /I32+ "

....

_./F=+- x2 3

x -3 -2 -1 0 1 2 3

g(x) -1 -2 -4 -8 -16 -32 -64

f(x) 1 2 4 8 16 32 64

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204 Chapter Five TRANSFORMATIONSOF FUNCTIONSANDTHEIR GRAPHS

Notice that when a point is reflected vertically about the x-axis, the x-value stays fixed,while the y-value changes sign. That is, for a given x-value,

y-value of 9 is the negative of y-value of f.

Algebraically, this meansg(x) = - f(x).

(b) The graph of y = h(x) is obtained by reflecting the graph of y = f(x) horizontally aboutthe y-axis. In part (a), a vertical reflection corresponded to an outside change in the formula,specifically,multiplying by -1. Thus, you might guess that a horizontal reflection of the graphcorresponds to an inside change in the formula. This is correct. To see why, consider Table 5.7.

Table5.7 Valuesof thefunctions hex) and f(x) graphed in Figure 5.11

Notice that when a point is reflected horizontally about the y-axis, the y-value remainsfixed, while the x-value changes sign. For example, since f( -3) = 1 and h(3) = 1, we haveh(3) = f( -3). Since f( -1) = 4 and h(l) = 4, we have h(l) = f( -1). In general,

h(x) = f( -x).

(c) The graph of the function y = k (x) results from a horizontal reflection of the graph of fab~ut the y-axis, followed by a vertical reflection about the x-axis. Since a horizontal reflectioncorresponds to multiplying the inputs by -1 and a vertical reflection corresponds to multiplyingthe outputs by -1, we have

Vertical reflectionacrossthe x-axist

k(x) = -fe-x).tHorizontalreflectionacrossthe y-axis

Let's check a point. If x = 1, then theformula k(x) = - f( -x) gives:

k(l) = -f(-l) =-4 since f (-1) = 4.

This result is consistent with the graph, since (1, -4) is on the graph of k (x).

0-

For a function f:

. The graph of y = - f (x) is a reflection of the graph of y = f (x) about the x-axis.

. Thegraphof y = f (- x) is a reflectionofthe graphof y = f (x) aboutthe y-axis.

x -3 -2 -1 0 1 2 3

hex) 64 32 16 8 4 2 1

f(x) 1 2 4 8 16 32 64

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5.2 REFLECTIONSANDSYMMETRY 205

SymmetryAboutthey-Axis

The graph of p(x) = X2 in Figure 5.13 is symmetric about the y-axis. In other words, the part of the

graph to the left of the y-axis is the mirror image of the part to the right of the y-axis. Reflecting thegraph of p( x) about the y-axis gives the graph of p( x) again.

yp(x) = X2

x

Figure5.13: Reflecting the graph of p( x) = X2 about the y-axis does not change its appearance

Symmetry about the y-axis is called even symmetry, because power functions with even expo-nents, such as y = x2, Y = X4, Y = x6, . . . have this property. Since y = p( -x) is a reflection ofthe graph of p about the y-axis and p( x) has even symmetry, we have

p(-x) = p(x).

To check this relationship, let x = 2. Then p(2) = 22 = 4, and p( -2) = (-2? = 4, so p( -2) =p(2). This means that the point (2,4) and its reflection about the y-axis, (-2,4), are both on thegraph of p(x).

Example2 For the function p(x) = X2, check algebraically that p( -x) = p(x) for all x.

Solution Substitute -x into the formula for p(x) giving

p(-x) = (-x? = (-x). (-x)= X2

= p(x).

Thus,p(-x) =p(x).

In general,

If f is a function, then f is called an even function if, for aUvalues of x in the domain off,

f(-x) = f(x).

- The graph of f is symmetric about the y-axis.

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206 Chapter Five TRANSFORMATIONSOFFUNCTIONSAND THEIRGRAPHS

SymmetryAbouttheOriginFigures 5.14 and 5.15 show the graph of q(x) = x3. Reflecting the graph of q first about the y-axisand then about the x-axis (or vice-versa) gives the graph of q again. This kind of symmetry is calledsymmetry about the origin, or odd symmetry.

In Example 1,we saw that y = - f (-x) is a reflection of the graph of y = f (x) about both they-axis and the x-axis. Since q(x) = x3 is symmetric about the origin, q is the same function as thisdouble reflection. That is,

q(x) = -q( -x) which means that q(-x) = -q(x).

To check this relationship, let x = 2. Then q(2) = 23 = 8, and q(-2) = (-2)3 = -8, soq( -2) = -q(2). This means the point (2,8) and its reflection about the origin, (-2, -8), are bothon the graph of q.

y

\ y = (_X)3\\\\ -+--, ,

',~

y

~e/~C7,~tx-a'j/f§

q(X)=X3/1q(x) = X3

/

// ~/

/ /,

,;;;:: (2,8)ft;/

x x

y = _(-x)3

Figure5.14: If the graph is reflected about the y-axis andthen about the x-axis, it does not change

/ /'?'

(-2 -8)if

';:<:/, ,///

/ // /

/ //

//

Figure 5.15: If every point on this graph isreflected about the origin, the graph is unchanged

Example3 For the function q(x) = x3, check algebraically that q(-x) = -q(x) for all x.

Solution We evaluate q(-x) giving

q(-x) = (-x)3 = (-x). (-x). (-x)= -x3

= -q(x).

Thus,q(-x) = -q(x).

In general,

If f is a function, then f is called an odd function if, for all values of x in the domain of f,

f( -x) = - f(x).

The graph of f is symmetric about the origin.

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5.2 REFLECTIONSAND SYMMETRY 207

Example4 Determine whether the following functions are symmetric about the y-axis, the origin, or neither.(a) f(x) = Ixl (b) g(x) = l/x (c) h(x) = _X3 - 3x2 + 2

Solution The graphs of the functions in Figures 5.16, 5.17, and 5.18 can be helpful in identifying symmetry.y y y

5-# \ 5

~(~)-5 I 5

-5t

Figure 5.16: The graph off(x) = Ixlappearsto be

symmetric about the y-axis

-5x x

g(x)

5

-5

Figure 5.17,:The graph ofg(x) = l/x appearsto besymmetric about the origin

-5

Figure 5.18:The graph ofhex) = _X3 - 3X2 + 2 is

symmetric neither about they-axis nor about the origin

From the graphs it appears that f is symmetric about the y-axis (even symmetry),9 is symmetricabout the origin (odd symmetry), and h has neither type of symmetry.However,how can we be surethat f(x) and g(x) are really symmetric? We check algebraically.

If f( -x) = f(x), then f has even symmetry.We check by substituting -x in for x:

f( -x) = I-xl= Ixl

= f(x).

Thus, f does have even symmetry.If g( -x) = -g(x), then 9 is symmetric about the origin. We check by substituting -x for x:

1g(-x) =- -x

1" x

= -g(x).

Thus, 9 is symmetric about the origin.The graph of h does not exhibit odd or even symmetry. To confirm, look at an example, say

x = 1:

h(l) = -13 - 3 .12 + 2 = -2.

Now substitute x = -1, giving

h(-l) = -(-1)3 - 3. (-1)2 + 2 = O.

Thus h(l) =1=h( -1), so the function is not symmetric about the y-axis. Also, h( -1) =1=-h(l),so the function is not symmetric about the origin.

Combil)ingShiftsandReflectionsWe can combine the horizontal and vertical shifts from Section 5.1 with the horizontal and verticalreflections of this section to make more complex transformations of functions

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208 ChapterFive TRANSFORMATIONSOF FUNCTIONSANDTHEIRGRAPHS

Example5 A cold yam is placed in a hot oven. Newton's Law of Heating tells us that the difference betweenthe oven's temperature and the yam's temperature decays exponentially with time. The yam's tem-perature is initially OaF,the oven's temperature is 300°F, and the temperature difference decreasesby 3% per minute. Find a formula for Y(t), the yam's temperature at time t.

Solution Let D(t) be the difference between the oven's temperature and the yam's temperature, which isgiven by an exponential function D(t) = abt. The initial temperature difference is 300°F - OaF =300°F, so a = 300.The temperature difference decreases by 3%per minute, so b = 1-0.03 = 0.97.Thus,

D(t) = 300(0.97)t.

lfthe yam's temperature is represented by Y(t), then the temperature difference is given by

D(t) = 300 - Y(t),

so, solving for Y(t), we haveY(t) = 300- D(t),

givingY(t) = 300 - 300(0.97)t.

Writing Y(t) in the formY(t) = -D(t) + 300

'-v-" '-v-"Reflect Shift

shows that the graph of Y is obtained by reflecting the graph of D about the t-axis and then shiftingit vertically up 300 units. Notice that the horizontal asymptote of D, which is on the t-axis, is alsoshifted upward, resulting in a horizontal asymptote at 300°F for Y.

Figures 5.19 and 5.20 give the graphs of D and Y. Figure 5.20 shows that the yam heats uprapidly at first and then its temperature levels off toward 3000F, the oven temperature.

temperaturedifference(OF)

300

temperature (OF)

300

iHorizontalasymptoteat 300°F

t, time (minutes)60 120 180 240

FigureS.~9:Graph of D(t) = 300(0.97)t, thetemperature difference between the )laID and the

oven

0t, time (minutes)

60 120 180 240

Figure5.20:The transformationY(t) = -D(t) + 300, where

D(t) = 300(0.97)t

Note that the temperature difference, D, is a decreasing function, so its average rate of chanis negative. However, Y, the yam's temperature, is an increasing function, so its average ratechange is positive. Reflecting the graph of D about the t-axis to obtain the graph of Y changed 1sign of the average rate of change.

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ExercisesandProblemsforSection5.2

Exercises

5.2 REFLECTIONSAND SYMMETRY 209

1. Thegraphof y = f (x) containsthepoint (2, - 3). Whatpoint must lie on the reflected graph if the graph is re-flected

(a) About the y-axis? (b) About the x-axis?

2. The graph of P = g(t) contains the point (-1, -5).

(a) If the graph has even symmetry, which other pointmust lie on the graph?

(b) What point must lie on the graph of -g(t)?

3. The graph of H (x) is symmetric about the origin. IfH( -3) = 7, what is H(3)?

4. The range of Q(x) is -2 :S Q(x) :S 12. What is therange of -Q(x)?

5. If the graph of y = eX is reflected about the x-axis, what

is the formula for the resulting graph? Check by graphingboth functions together.

6. If the graph of y = eX is reflected about the y-axis, what

is the formula for the resulting graph? Check by graphingboth functions together.

7. Completethefollowingtablesusingf (p) = p2+ 2p- 3,and g(p) = f( -p), and h(p) = - f(p). Graph the threefunctions. Explain how the graphs of g and h are relatedto the graph of f.

p

f(p)

3

3

p

g(p)

3

~h(p)

Problems

8. Graph y = f(x) = 4x and y = f( -x) on the same setof axes. How are these graphs related? Give an explicitformula for y = f (- x ) .

9. Graph y = g(x) = orand y = -g(x) on the sameset of axes. How are these graphs related? Give an ex-plicit formula for y = -g(x).

Give a formula and graph for each of the transformations ofm(n) = n2 - 4n + 5 in Exercises 10-13.

10. y = m(-n)

12. y = -m( -n)

11. y = -m(n)

13. y = -m(-n) + 3

Give a formula and graph for each of the transformations ofk(w) = 3w in Exercises 14-19.

14. y = k( -w)

16. y = -k( -w)

15. y = -k(w)

17. y=-k(w-2)

18. y = k(-w) +4 19. y = -k(-w)-l

In Exercises 20-23, show that the function is even, odd, orneither.

20~ f(x) = 7X2 - 2x + 1

22. f(x) = 8x6 + 12x2

21. f(x) = 4X7 - 3x5

23. f(x)=x5+3x3_2

24. (a) Graph the function obtained from f(x) = X3 byfirst reflecting about the x-axis, then translating uptwo units. Write a formula for the resulting function.

(b) Graph the function obtained from f by first trans-lating up two units, then reflecting about the x-axis.Write a formula for the resulting function.

(c) Are the functions in parts (a) and (b) the same?

25. (a) Graph the function obtained from g(x) = 2x by firstreflecting about the y-axis, then translating downthre'?, units. Write a formula for the resulting func-tion.

(b) Graph the function obtained from g by first trans-lating down three units, then reflecting about the y-axis. Write a formula for the resulting function.

(c) Are the functions in parts (a) and (b) the same?

26. If the graph of a line y = b + mx is reflected about they-axis, what are the slope and intercepts of the resultingline?

27. Graph y = 10g(1/x) and y = log x on the same axes.How are the two graphs related? Use the properties oflogarithms to explain the relationship algebraically.

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210 ChapterFive TRANSFORMATIONSOF FUNCTIONSANDTHEIRGRAPHS

28. The function d(t) graphed in Figure 5.21 gives the wintertemperature in of at a high school, t hours after midnight.

(a) Describe in words the heating schedule for thisbuilding during the winter months.

(b) Graph c(t) = 142 - d(t).(c) Explain why c might describe the cooling schedule

for summer months.

temperature(OF) d(t)68°

[60°

I

1

/ \t (hours)

4 8 12 16 20 24

Figure5.21

29. Using Figure 5.22, match the formulas (i)-(vi) with agraph from (a)-(f).

(i) y=f(-x)

(iii) y = f( -x) + 3(v) y=-f(-x)

(ii) y=-f(x)

(iv) y = -f(x -1)(vi) y = -2 - f(x)

-¥ f:X)

Figure5.22

(a) Y (b)

-*-x ~xY

~x

Y

(e) Y (d)

(e)

~_xY (I)

f-x ~x

Y

30. In Table 5.8, fill in as many y-values as you can if youknow that f is

(a) An even function (b) An odd function.

Table5.8

x 3

y

31. Figure 5.23 shows the graph of a function f in the sec-ond quadrant. In each of the following cases, sketchy = f (x), given that f is symmetric about

(a) They-axis. (b) The origin. (c) Theliney=x.

Y

f(x) ~Figure5.23

.J

32. For each table, decide whether the function could besymmetric about the y-axis, about the origin, or neither.

(a)

(b)3

8.1

x

g{x)

(c) x :.

I(x) + g(x)

(d) 3

13

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33. A function is called symmetric about the line y = x if in-terchanging x and y gives the same graph. The simplestexample is the function y = x. Graph another straightline that is symmetric about the line y = x and giveitsequation.

34. Show that the graph of the function h is symmetric aboutthe origin, given that

h(x)=I+x2x-x.3'

35. Comment on the following justification that the func-tion f (x) = .1:.3 - X2 + 1 is an even function: BecausefrO) = 11= -frO), we know that f(x) is not odd. If afunction is not odd, it must be even.

36. Is it possible for an odd function whose domain is all realnumbers to be strictly concave up?

5.3

5.3 VERTICAL STRETCHES AND COMPRESSIONS 211

37. If f is an odd function and defined at x = 0, what isthe value of frO)? Explain how you can use this result toshow that c(:r) = x + 1 and d(x) = 2x are not odd.

38. In the first quadrant an even function is increasing andconcave down. What can you say about the function'sbehavior in the second quadrant?

39. Show that the power function .f (x) = Xl/.3 is odd. Give acounterexample to the statement that all power functions

of the form f (x) = xP are odd.

40. Graph sex) = 2x + (~)x, c(x) = 2x - (~)x, andn(x) = 2x - (~)X-l. State whether you think thesefunctions are even, odd or neither. Show that your state-ments are true using algebra. That is, prove or disprovestatements such as s( -x) = s(x).

41. There are functions which are neither even nor odd. Isthere a function that is both even and odd?

42. Some functions are symmetric about the y-axis. Is it pos-sible for a function to be symmetric about the x-axis?

"""'" --VERTICALSTRETCHESANDCOMPRESSIONSOf""""""""""""""""'1JiA§""",,,,'"'.n "...-----......We have studied translations and reflections of graphs. In this section, we consider vertical stretchesand compressions of graphs. As with a vertical translation, a vertical stretch or compression of afunction is represented by an outside change to its formula.

VerticalStretch:A StereoAmplifierA stereo amplifier takes a weak signal from a cassette-tape deck, compact disc player, or radio tuner,and transforms it into a stronger signal to power a set of speakers.

Figure 5.24 shows a graph of a typical radio signal (in volts) as a function oftime, t, both beforeand after amplification. In this illustration, the amplifier has boosted the strength of the signal by afactor of 3. (The amount of amplification, or gain, of most stereos is considerably greater than this.)

Notice that the wave crests of the amplified signal are 3 times as high as those of the originalsignal; similarly, the amplified wave troughs are 3 times deeper than the original wave troughs. If Iis the original signal function and V is the amplified signal function, then

Amplified signal strength at time t = 3 . Original signal strength at time t,, / , /

J(t)Vet)

so we have

V(t) = 3. I(t).

This formula tells us that values of the amplified signal function are 3 times the values of the originalsignal. The graph of V is the graph of I stretched vertically by a factor of 3. As expected, a verticalstretch of the graph of I (t) corresponds to an outside change in the formula.

Notice that the t-intercepts remain fixed under a vertical stretch, because the I-value of thesepoints is 0, which is unchanged when multiplied by 3.

4

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216Chapter Five TRANSFORMATIONSOF FUNCTIONSAND THEIRGRAPHS

SolutionTo combine several transformations, always work from inside the parentheses outward as in Fig-ure 5.29. The graphs corresponding to each step are shown in Figure 5.30. Note that we did notneed a formula for f to graph g.

Step3: -~.y ~ -~f(X +» --:.' ~

Step4:1

g(x) = -- f(x + 3)-12

Figure5.30:The graph of y = f(x) transformed in four steps intog(x) = -(1/2)f(x + 3) - 1

Step 1: horizontalshift left 3 units

Step 2: v~rticalcompression

j~

g(x) = -~f(x + 3) - 1

St"3"".~1refl- \across the x-axis

Step 4: verticalshift down 1 unit

Figure5.29

ExercisesandProblemsforSection5.3

Exercises

Step1:y = f(x + 3)~\

\\\

Y'I-5

Step 2:1

Y = 2f(x + 3) --..

4

III

/'"I Original

/ y=f(x)/

//

x

-4

9. Using Table 5.14, create a table of values for1. Let y = f(x). Write a formula for the transformationwhich both increases the y-value by a factor of 10 andshifts the graph to the right by 2 units.

2. The graph of the function g(x) contains the point (5, ~).What point must be on the graph of y = 3g(x) + I?

3. The range of the function C(x) is -1 ::; C(x) ::; 1.What is the range of O.25C(x)?

In Exercises 4-7, graph and label f(x), 4f(x), -~ f(x), and-5f(x) on the same axes.

4. f(x) = yfX

6. f(x) = eX

5. f(x) = _X2 + 7x

7. f(x) = lnx

8. Using Table 5.13, make tables for the following transfor-mations of f on an appropriate domain.

(a) ~f(x)(d) f(x - 2)

(b) -2f(x + 1) (c) f(x) + 5

(e) f( -x) (0 - f(x)

Table5.13

x

I(x)

(a) f( -x) (b) - f(x) (c) 3f(x)

(d) Which of these tables from parts (a), (b), and (c) rep-resents an even function?

Table5.14

x

f{x)

10. Figure 5.31 is a graph of y = X3/2. Match the followingfunctions with the graphs in Figure 5.32.

(a) y = x3/2 - 1 (b) y = (x - 1)3/2

(c) Y = 1 - x3/2 (d) y = ~X3/2

y

1/I i

y

2

(I) (II) (III)

1

t- x3

1-2

t- x3

Figure 5.31 Figure 5.32

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Without a calculator, graph the transformations in Exer- (a)cises 11-16. Label at least three points.

11. y = f(x + 3) if f(x) = Ixl12. y = f(x) + 3 if f(x) = Ixl

13. y = -g(x) if g(x) = X2

14. y = g(-x) ifg(x) = X2

15. y = 3h(x) if hex) = 2x

16. y = 0.5h(x) if hex) = 2x

17. Using Figure 5.33, match the functions (i)-(v) with agraph (a)-(i).

(i) y = 2f(x)

(iii) y=-f(x)+l

(v) y=f(-x)

(ii) y = ~f(x)(iv) y = f(x + 2) + 1

y

~y = I(x)

x

Figure5.33

Problems

5.3 VERTICALSTRETCHESANDCOMPRESSIONS 217

y (b) y

LxU/ x

(c) y

-~x1

(d) y

T"(e) y (I) y

(9)

~Y (h) Y

:\L1x ~ x

+-1)-+ "x

(i) y

¥"

18. Describe the effect of the transformation 2f (x + 1) - 3on the graph ofy = f(x).

19. The function set) gives the distance (miles) in terms oftime (hours). If the average rate of change of set) on0 ::; t ::; 4 is 70 mph, what is the average rate of changeof ~s(t) on this interval?

In Problems 20-24, let f(t) = 1/(1 +X2). Graph thefunctiongiven, labeling intercepts and asymptotes.

20. y = f(t)

22. y = 0.5f(t)

21. y = f(t - 3)

23. y = - f(t)

24. y = f(t + 5) - 5

25. The number of gallons of paint, n = f(A), needed tocover a house is a function of the surface area, in ft2.

Match each story to one expression.

(a) I figured out how many gallons I needed and thenbought two extra gallons just in case.

(b) I bought enough paint to cover my house twice.(c) I bought enough paint to cover my house and my

}Velcomesign, which measures 2 square feet.

(i)~ 2f(A) (ii) f(A + 2) (iii) f(A) + 2

26. The US population in millions is pet) today and t is inyears. Match each statement (I)-(IV) with one of thefor-mulas (a)-(h).

I. The population 10years before today.

II. Today's population plus 10million immigrants.

III. Ten percent of the population we have today.

IV. The population after 100,000 people have emigrated.

(a) pet) - 10 (b) pet - 10)

(d) pet) + 10 (e) pet + 10)

(g) pet) + 0.1 (h) pet) - 0.1

(c) O.lP(t)

(f) P(t)/O.l

27. Let R = pet) be the number of rabbits living in the na-tional park in month t. (See Example 5 on page 5.) Whatdo the following expressions represent?

(a) pet + 1) (b) 2P(t)

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218 Chapter Five TRANSFORMATIONSOF FUNCTIONSAND THEIRGRAPHS

28. Without a calculator, match each formula (a)-(e) with agraph in Figure 5.34. There may be no answer or severalanswers.

(a) y = 3 . 2x (b) y = 5-x (c) y = -5x

(d) y = 2-Tx (e) y = l-(~r

y y

~x

x

y y

7F xhex)

x

Figure5.34

Graph the transformations of f in Problems 29-33 using Fig-ure 5.35. Label the points corresponding to A and B.

y10

B6

10

2x

-4

Figure5.35

29. y = f(x - 3)

31. y = f( -x)/3

30. y = f(x) - 3

32. y = -2f(x)

33. y = 5 - f(x + 5)

34. Using Figure 5.36, find formulas, in terms of f, for thehorizontal and vertical shifts of the graph of f in parts(a)-(c). What is the equation of each asymptote?

y

5

3

1

-3 -11 1 3 5 7

(a) y

Figure5.36

(b) y

2

-1

(c) y

-3

35. Using Figure 5.37, find formulas, in terms of f, for thetransformations of f in parts (a)-(c).

Y B3

~ (x)

A x2

Figure5.37

(a) Y

_6~-X(b}:~

Y

(c)y

l~x

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'2.'2.'2. ~X\o.\I\~~H\l~ \R~\\c;:.t~R~~\~\\c;:. ~t t~\\~\\~"c;:. ~"\) \'t\E\R G.R~\I't\c;:.

Exam1'\le 1 shows the effect of an inside multi1'\le of 1/'2. The next exam1'\le shows the effect onthe graph of an inside multiple of 2.

Exam\)\e2

So\u\io\'\

Let Hx) be the function in Example L Make a table and a gra1'\hfor the function h(x) = H'2x).

We use Table 5.15 and the formula h(x) = H'2x) to evaluate h(x) at several values of x. F01'Ch'O."ffi"\)\'C,\.\ T, = \, tn'C\:\

h(l) = H'2 .1) = H'2).

Table 5.15 shows that H'2) = -1, so h(l) = -1. These values are recorded in Table 5.Similarly, substituting x = 1.5, gives

h(1.5) = f(2. 1.5) = f(3) = 1.

Since h(O) = f(2 . 0) = f (0), the y-intercept remains fixed (at -1). In Figure 5.45 we see thatgraph of h is the graph of f compressed by a factor of 2 horizontally toward the y-axis.

Table5.17 Valuesofh(x) = f(2x) y

/8\ ~ 2/ \/ \I \

! \i \1I IIII

-1.5

-1.0

-0.5

0.0

0.5

1.0

1.5

x h(x)

0

2

0

-1

0

-1

1Figure5.45:The graph of h( x) = f (2x) is the

graph of y = f (x) compressed horizontally by afactor of 2

\\\

f(2x)

rf(x)IIII

x

In Chapter 3, we used the function P = 263eO.OO9tto model the US population in milli<This function is a transformation of the exponential function f (t) = et, since we can write

P = 263eOOO9t= 263f(0.009t).

The US population is f (t) = et stretched vertically by a factor of 263 and stretched horizontall~a factor of 1/0.009 ~ 111.

Example3 Match thefunctions f( t) = et, g(t) = eO.5t,h(t) = eO.8t,j (t) = e2twith the graphs in Figure 5.ABC D

100

t

50

5 10

Solution

Figure5.46

Since the function j(t) = e2t climbs fastest of the four and g(t) = eO.5t climbs slowest, grapmust be j and graph D must be g. Similarly, graph B is f and graph C is h.

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ExercisesandProblemsforSection5.4

Exercises

5.4 HORIZONTALSTRETCHESAND COMPRESSIONS 223

1. The point (2,3) lies on the graph of g(x). What pointmust lie on the graph of 9 (2x )?

2. Describe the effect of the transformation 10f ( fax ) onthe graph of f(x).

3. Using Table 5.18, make a table of values for f(~x) foran appropriate domain.

Table5.18

x

f(x)

4. Fill in all the blanks in Table 5.19 for which you havesufficient information.

Table5.19

5. Graph m(x) = eX,n(x) = e2x, andp(x) = 2ex on thesame axes and describe how the graphs of n(x) and p(x)

compare with that of m( x).

6. Graph y = h(3x) if h(x) = 2x.

In Exercises '7-9, graph and label f(x), f( ~x), and f( -3x)on the same axes between x = -2 and x = 2.

7. f(x) = eX + X3 - 4X2

8. f(x) = eX+?+ (x - 4)3 - (x + 2)2

9. f(x) = In(x4 + 3X2+ 4)

Problems

10. Using Figure 5.47, match each function to a graph (ifany) that represents it:

(i) Y = f(2x) (ii) y=2f(2x) (iii) y=f(~x)

y y= f(x)

¥"Figure5.47

(a) y (b) y

OL" ~"(c) y (d) y

~" ¥"(e) y (I) y

J\V '"(g)

~"Y (h) Y

/\ /1 +-~-~u I x

,~~ X 'V I-'~(i) Y

, 1L"

11. For the function f (p) an input of 2 yields an output valueof 4. What value of p would you use to have f (3p) = 4?

12. The domain of l(x) is -12 S; x S; 12 and its range is0 S; l(x) S; 3. What are the domain and range of

(a) 1(2x)? (b) l(~x)?

13. The point (a, b) lies on the graph of y = f (x). If thegraph is stretched away from the y-axis by a factor of d(where d > 1), and then translated upward by c units,what are the new coordinates for the point (a, b)?

x -3 -2 -1 0 1 2 3

f(x) -4 -1 2 3 0 -3 -6

fax)

f(2x)

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224 Chapter Five TRANSFORMATIONSOF FUNCTIONSAND THEIRGRAPHS

IIIIn Problems 14-15, graph the transformation of f, the func-tion in Figure 5.48.

I(x)x

2-2 -1-1

-2

1

Figure5.48

14. y = -2f(x - 1) 15. Y = f(x/2) - 1

16. Every day I take the same taxi over the same route fromhome to the train station. The trip is x miles, so the costfor the trip is f(x). Match each story in (a)-(d) to a func-tion in (i)-(iv) representing the amount paid to the driver.

(a) I received a raise yesterday, so today I gave mydriver a fivedollar tip.

(b) I had a new driver today and he got lost. He drovefive extra miles and charged me for it.

(c) I haven't paid my driver all week. Today is Fridayand I'll pay what I owe for the week.

(d) The meter in the taxi went crazy and showed fivetimes the number of miles I actually traveled.

(i) 5f(x)

(iii) f(5x)

(ii) f(x) + 5

(iv) f(x + 5)

17. A companyprojectsa totalprofit,P(t) dollars,in yeart. Explain the economic meaning of r( t) = 0.5P( t) ands(t) = P(0.5t).

18. Let A = f (r) be the area of a circle of radius r.

(a) Write a formula for f(r).(b) Which expression represents the area of a circle

whose radius is increased by 1O%?Explain.

(i) 0.10f(r)(iv) f(l.1r)

(ii) f(r+O.lO) (iii) f(0.10r)(v) f(r)+0.10

(c) By what percent does the area increase if the radiusis increased by 1O%?

In Problems 19-20, state which graph represents

(a) f(x) (b) f( -2x) (c) f( -~x) (d) f(2x)

19. 20.

IVx

x

21. Find a formula for the function in Figure 5.50 as a trans-formation of the function f in Figure 5.49.

y3 B

~ x

Figure5.49

y

~" -3

Figure5.50

22. This problem investigates the effect of a horizontalstretch on the zeros of a function.

(a) Graphf(x) = 4 - X2.Markthe zerosof f on thegraph.

(b) Graphandfinda formulafor g(x) = f(0.5x). Whatare the zeros of g(x)?

(c) Graphand finda formulafor h(x) = f(2x). Whatare the zeros of h(x)?

(d) Without graphing, what are the zeros of f(10x)?

23. In Figure 5.51, the point c is labeled on the x-axis. Onthe y-axis, locate and label output values:

(a) g(c) (b) 2g(c) (c) g(2c)

y

I/y~g(": xc

Figure 5.51