Name:
Group Members:
Exploration 1-1a: Paper Cup Analysis
Date:
Objective: Find an equation for calculating the height of a stack of paper cups.
1. Obtain several paper cups of the same kind. Measure
the height of stacks containing 5, 4, 3, 2, and just
1 cup. Record the heights to the nearest 0.1 cm.
State what kind of cup you used.
5. Let x be the number of cups in a stack, and let y be
the height of the stack, measured in centimeters.
Write an equation for y as a function of x.
Kind:
Number
cm
1
2
6. What is the name of the kind of function whose
equation you wrote in Problem 5?
3
4
7. Show that your equation in Problem 5 gives a height
close to the measured height for a stack of 3 cups.
5
Height (cm)
2. Plot the points in the table on this graph paper.
Show the scale you are using on the vertical axis.
8. Use your equation to predict the height of a stack of
35 cups. Round the answer to 1 decimal place.
1
2
3
4
5
6
7
Number of cups
8
9
10
3. On average, by how much did the stack height
increase for each cup you added? Show how you got
your answer.
9. What are the names of the processes of calculating a
value within the range of the data, as in Problem 7,
and outside the range of data, as in Problem 8?
Within:
Outside:
10. A cup manufacturer wants to package this kind of
cup in boxes that are 45 cm long. What is the
maximum number of cups the box could hold? Show
how you get your answer.
4. How tall would you expect a 10-cup stack to be?
Show how you get your answer. Would this be twice
as tall as a 5-cup stack?
11. What did you learn as a result of doing this
Exploration that you did not know before?
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Name:
Group Members:
Exploration 1-2a: Names of Functions
Date:
Objective: Recall the names of certain kinds of functions.
1. f (x) = 2x + 3 is the equation for a linear function.
Plot the graph and sketch the result here. Give a
reason for the name linear.
2. f (x) = x 2 D 6x + 10 is the equation for a quadratic
function. Plot the graph and sketch the result.
Explain how the word quadratic is related to the
word quadrangle.
5. f (x) = 24
x is the equation for an inverse variation
power function. Plot the graph for x > 0 and sketch
the result. Why do the words “y varies inversely with
x” make sense for this function? Why can the
function be called a power function?
6. f (x) = x 4 D 4x 3 D 43x 2 + 130x + 168 is the equation of
this quartic function. Why do you think the name
quartic is used for this function? Use your grapher to
find the largest value of x at which the graph crosses
the x-axis.
y = f (x)
x
3. f (x) = 3x 0.7 is the equation for a power function. Plot
the graph and sketch the result. Why do you think it
is called a power function?
7. f (x) = xx DD 43 is the equation of a rational function. Plot
the graph and sketch the result. Why do you think it
is called a rational function? What happens to the
graph at x = 3?
4. f (x) = 3 R 0.7x is the equation for an exponential
function. Plot the graph and sketch the result. How
does an exponential function differ from a power
function algebraically? graphically?
8. What did you learn as a result of doing this
Exploration that you did not know before?
32 / Exploration Masters
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Exploration 1-2b: Restricted Domains and
Boolean Variables
Date:
Objective: Use Boolean variables to plot graphs of functions in a restricted domain.
Weight Above and Below Earth’s Surface: If there were a hole all the way through Earth and it
were possible for you to go through it, you would be “weightless” at the center of Earth. This is
because gravity would pull you with the same force in every direction. Between the center and
the surface, your weight would vary directly with the distance from the center.
The linear graph should stop at x = 4000 miles, and the
inverse square function graph should start at x = 4000
miles, the distance from the center to the surface.
Weight (lb)
1. Kevin Vader (Darth’s son) weighs 200 pounds on the
surface of Earth. Earth’s radius is about 4000 miles.
Write the particular equation for Kevin’s weight as a
function of distance from the center when he is
below the surface.
250
200
150
100
50
4000 8000 12,000
Distance (mi)
2. Above the surface, your weight varies inversely with
the square of your distance from the center because
the pull of gravity decreases as you recede from
Earth. Write the particular equation for Kevin’s
weight as a function of distance from the center
when he is above the surface.
Weight (lb)
3. Plot the graphs from Problems 1 and 2 as y1 and y2.
Does your graph agree with this one? Where do the
two graphs cross each other?
4. The Boolean variable (x ≥ 4000) equals 1 if x is
greater than or equal to 4000 and 0 otherwise.
Change your equation for y2 by dividing it by this
Boolean variable, then replot the graph. Explain why
the grapher plots the same graph for y2 when
x ≥ 4000 but plots nothing when x < 4000.
5. What Boolean variable could you divide y1 by so that
the grapher plots it only between 0 and 4000?
Change the equation for y1. Does the complete graph
now match the one above Problem 4?
250
200
150
100
50
4000 8000 12,000
Distance (mi)
6. What word describes the set of x-values for which a
particular function is defined? What word describes
the corresponding set of y-values?
x-values:
y-values:
7. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration Masters / 33
Name:
Group Members:
Exploration 1-3a: Translations and
Dilations, Numerically
Date:
Objective: By calculating values and plotting points, discover the effect on a function
graph of adding and multiplying by constants.
1. The table shows values of a pre-image function
y = f (x). The graph of f is a set of line segments
connecting the points, shown dashed in the figure.
Find values of the image function g(x) = f (x) + 3. For
instance, g(D2) = 2 + 3 = 5. Plot the graph of this
transformed function.
x
f(x)
D2
2
D1
3
0
1
1
D2
2
0
3
1
g(x)
y
f
5. Use the values of f (x) in Problem 1 to make a table of
values of a new image function, g(x) = 2f (x). For
instance, g(D1) = 2f (D1) = 2 • 3 = 6. Plot the image of
this transformed function.
x
g(x) H 2f(x)
f
5
D1
5
x
0
x
5
y
D2
5
5
5
1
2
5
3
5
2. The transformation in Problem 1 is a vertical
translation by 3 units. Give the meaning of a vertical
translation.
6. The transformation in Problem 5 is a vertical
dilation by a factor of 2. Give the meaning of a
vertical dilation, and explain how it differs from a
vertical translation.
7. Use the values of f (x) in Problem 1 to make a table of
values of a new image function, g(x) = f 12x . For
instance,
1
g(D2) = f • (D2) = f (D1) = 3
2
( )
Plot the image of this transformed function.
3. Use the values of f (x) in Problem 1 to make a table of
values of a new image function, g(x) = f (x D 3). For
instance, g(1) = f (1 D 3) = f (D2) = 2. Plot the image of
this transformed function.
x
g(x) H f(x D 3)
y
1
f
5
4
5
5
y
D4
f
5
D2
0
x
5
4
x
5
( )
g(x) H f 12x
5
2
5
2
3
x
5
6
8. The transformation in Problem 7 is a horizontal
dilation. By what factor is the graph dilated? How
is that factor related to the 12 in f 12x ?
( )
6
4. Describe the transformation in Problem 3.
9. What did you learn as a result of doing this
Exploration that you did not know before?
34 / Exploration Masters
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Name:
Group Members:
Exploration 1-3b: Translations and
Dilations, Algebraically
Date:
Objective: Find the effect on a function graph of adding and multiplying by constants.
1. The graph below shows the pre-image function
f (x) = 1 +1 x 2. Plot this graph as y1 on your grapher.
Use the window shown, using GRID ON format.
6. Deactivate y3 from Problem 4. Then plot y4 = 3f (x).
Sketch the result here.
y
4
y
4
x
x
5
5
5
5
2. Plot the graph of y2 = f (x) + 3. Sketch the result on
the graph in Problem 1.
3. The transformation in Problem 2 is a vertical
translation of 3. Give the meaning of a vertical
translation.
7. The transformation in Problem 6 is a vertical
dilation by a factor of 3. Give the meaning of a
vertical dilation, and explain how it differs from a
vertical translation.
8. Deactivate y4 from Problem 7. Then plot y5 = f (3x).
Sketch the result here.
y
4
x
5
4. Deactivate y2 from Problem 2. Then plot y3 = f (x D 3).
Sketch the result here.
5
9. The transformation in Problem 8 is a horizontal
dilation. By what factor is the graph dilated?
y
4
x
5
5
5. What words describe the transformation in
Problem 4?
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1
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10. What did you learn as a result of doing this
Exploration that you did not know before?
Exploration Masters / 35
Name:
Group Members:
Exploration 1-3c: Transformations from Graphs
Date:
Objective: Given the parent and transformed graphs, identify the transformation.
Identify the transformation of f (dotted) to get g (solid).
4. Verbally:
Equation: g(x) =
1. Verbally:
y
Equation: g(x) =
y
10
10
5
5
g
f
x
x
10
f
5
5
g
10
5
5
10
10
5
5
5. Verbally:
Equation: g(x) =
2. Verbally:
y
Equation: g(x) =
y
10
10
Graphs coincide.
Graphs coincide.
5
f
5
f
g
g
x
x
10
5
5
10
5
5
10
10
5
5
6. Verbally:
Equation: g(x) =
3. Verbally:
y
Equation: g(x) =
y
10
g
10
5
5
f
g
f
x
x
10
5
5
10
5
5
10
10
5
5
7. What did you learn as a result of doing this
Exploration that you did not know before?
36 / Exploration Masters
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Name:
Group Members:
Exploration 1-3d: Transformation Review
Date:
Objective: Given the parent and transformed graphs, identify the transformation
and confirm by grapher.
1. The figure shows the graph of f (x) = D0.5x 2 + x + 3.5
in the domain D1 ≤ x ≤ 5. Duplicate this graph on
your grapher. Restrict the domain by dividing by the
Boolean variable (x ≥ D1 and x ≤ 5). Use GRID ON
format to get the dots.
4. Verbally:
Equation: g(x) =
Check:
y
y
5
5
x
x
10
5
5
10
5
10
5
10
5
10
g
f
5
f
5
5. Verbally:
For Problems 2–6, identify the transformation of f (dotted)
to get g (solid), and confirm by grapher.
Equation: g(x) =
Check:
2. Verbally:
y
Equation: g(x) =
5
Check:
x
10
y
5
g
f
5
5
x
10
5
5
g
10
6. Verbally:
f
5
and
Equation: g(x) =
3. Verbally:
Check:
Equation: g(x) =
y
Check:
5
g
y
x
10
g
5
x
10
5
5
5
5
5
5
10
f
10
f
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7. What did you learn as a result of doing this
Exploration that you did not know before?
Exploration Masters / 37
Name:
Group Members:
Exploration 1-4a: Composition of Functions
Date:
Objective: Find the composition of one function with another.
1. The figure shows two linear functions, f and g. Write
the domain and range of each function.
f: Domain:
Range:
g: Domain:
Range:
4. Show in the table an instance where g(x) is defined
but f (g(x)) is not defined.
y
7
5. Plot the values of f (g(x)) on the figure in Problem 1.
If the points do not lie in a straight line, go back and
check your work.
6
5
6. The function in Problem 5 is called the composition
of f with g, which can be written f g . What are the
°
domain and range of f g?
4
y2 = f(x)
3
°
Domain:
2
y1 = g (x)
Range:
7. Find equations for functions f and g.
1
x
1
1
2
3
4
5
6
7
8
9
10
1
2. Read values of g(x) from the graph and write them in
this table. If the value of x is out of the domain,
write “none.”
x
g(x)
0
8. Enter in your grapher the f and g equations as y1 and
y2, respectively. Use Boolean variables to make the
functions have the proper domains. Then plot the
graphs. Does the result agree with the given figure?
1
2
3
4
5
6
9. Enter f g in y3 by entering y1(y2(x)). Plot this graph.
°
Does it agree with the graph you drew in Problem 5?
7
8
9
3. The symbol f (g(x)) is read “f of g of x.” It means find
the value of g(x) first, and then find f of the answer.
For instance, g(5) = 1.5. So f (g(5)) = f (1.5) = 5.5. Put
another column into the table for values of f (g( x)).
Write “none” where appropriate.
10. By suitable algebraic operations on the equations in
Problem 7, find an equation for f (g(x)). Simplify the
equation as much as possible.
11. What did you learn as a result of doing this
Exploration that you did not know before?
38 / Exploration Masters
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Name:
Group Members:
Exploration 1-5a: Inverses of Functions
Date:
Objective: Find the inverse of a function graphically, numerically, or algebraically, and
state whether or not the inverse is a function.
Problems 1–6 refer to the linear function y = 2x D 5.
1. Write the equation for the inverse relation by
interchanging the variables. Then solve the resulting
equation for y in terms of x.
Problems 7 and 8 refer to the quadratic function
y = x 2 D 4x + 7, graphed here.
y
15
10
2. The graph shows y = 2x D 5. Plot the graph of the
inverse relation here.
5
y
x
10
5
10
15
5
x
10
5
5
10
5
7. Plot the line y = x . Then plot the inverse of the
function by reflecting the graph across this line.
8. Explain why the inverse of this function is not a
function.
Problems 9 and 10 refer to the exponential function
f (x) = 2x , graphed here.
10
3. The inverse relation in Problems 1 and 2 is a
function. How can you tell?
y
15
10
5
4. If the equation for the function is written as
f (x) = 2x D 5, how could you write the equation for
the inverse function using the f (x) terminology?
x
5
10
15
9. Find f(0), f(1), f(2), and f(3).
5. Show that f (3) = 1 and f D1(1) = 3. Explain why this is
true, based on the definition of the inverse of a
function.
10. Find f D1(1), f D1(2), f D1(4), and f D1(8). Use these
points to plot the graph of f D1.
6. Plot the line y = x . How are the graphs of f and f D1
related to this line?
11. What did you learn as a result of doing this
Exploration that you did not know before?
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Exploration Masters / 39
Name:
Group Members:
Exploration 1-6a: Translation, Dilation, and Reflection
Date:
Objective: Show that you know the effects of various constants on the graph of a
function.
3. y = f (x) D 4
Name the transformation and sketch the graph.
1. y =
Transformation:
1
f (x)
2
Transformation:
y
10
y
10
x
10
x
10
4. y = f (x D 7)
Transformation:
1
2. y = f x
2
y
10
Transformation:
y
10
x
10
x
10
40 / Exploration Masters
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Name:
Group Members:
Exploration 1-6a: Translation, Dilation, and
Reflection continued
Date:
9. y = |f (x)| is an absolute value transformation. Is it a
transformation in the x-direction or a transformation
in the y-direction?
5. y = f (Dx) = f (D1x)
Transformation (as a dilation):
y
10. Sketch the graph of y = |f (x)|.
10
y
10
x
10
x
10
6. From the results of Problem 5, give another name for
the transformation y = f (Dx).
11. Sketch the graph of y = f (|x |).
y
7. y = Df (x) = D1f (x)
10
Transformation (as a dilation):
y
10
x
10
x
10
12. What did you learn as a result of doing this
Exploration that you did not know before?
8. From the results of Problem 7, give another name for
the transformation y = Df (x).
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1
©2003 Key Curriculum Press
Exploration Masters / 41
Solutions to the Explorations
Chapter 1 • Functions and
Mathematical Models
3.
y
20
x
Exploration 1-1a
40
1. Answers will vary.
2. Answers will vary.
The variable (x in this case) is raised to a power.
3. Answers will vary. The stack height should increase by the
same amount for each additional cup.
4.
y
4. Answers will vary; no, the 10-cup stack would not be twice
as tall as a 5-cup stack.
2
x
5. Answers will vary.
4
6. Linear function
7. Answers will vary. Plug in 3 for x in your equation.
8. Answers will vary.
Algebraically, a power function has the variable as the base
and a constant as the exponent, whereas an exponential
function has a constant as the base and the variable (x in
this case) as an exponent. Graphically, the parent power
function passes through the point (0, 0) and increases to
the right (and increases to the left if the power is an
integer), whereas the parent exponential function passes
through the point (0, 1), increases without bound to the
left, and approaches 0 to the right.
9. Within: Interpolation
Outside: Extrapolation
10. Answers will vary.
11. Answers will vary.
Exploration 1-2a
5.
1.
y
y
8
4
x
x
8
4
The graph is a straight line.
2.
y gets smaller as x gets larger, and vice versa. It can be
written as y = 24xD1, involving the D1 power of x.
6. It is a fourth-degree polynomial. The largest value of x at
which the graph crosses the x-axis is 7.
y
y
4
x
400
4
4
(7, 0)
4
x
Both come from the Latin word for “square,” which in turn
comes from the Latin word for “four,” referring to the four
sides of a square.
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1
©2003 Key Curriculum Press
Solutions to the Explorations / 231
5. The values for g(x) are 4, 6, 2, D4, 0, 2.
7.
y
y
g
4
f
x
4
2
It has a discontinuity and an asymptote at x = 3. It is
p (x)
expressed as q (x)
, where p(x) and q(x) are both functions.
x
2
6. The graph is stretched or squashed vertically but not shifted
vertically as a whole.
7. The values for g(x) are 2, 3, 1, D2, 0, 1.
8. Answers will vary.
y
Exploration 1-2b
f
1. Let y be the weight in pounds. Let x be the distance, in miles,
from center.
x
y=
for 0 ≤ x ≤ 4000
20
2. y =
32 R 108
for x ≥ 4000
x2
3. The graphs look the same. They cross at the point
(4000, 200).
4. y =
32 R 108
x2(x ≥ 4000)
4
x
g
2
8. By a factor of 2. 2 =
1
1
2
9. Answers will vary.
Exploration 1-3b
For x G 4000, the calculator is dividing by 0, which gives no
answer.
1.
y
4
5. y1 = 0.05x/(0 ≤ x and x ≤ 4000)
The complete graph now matches the graph in Problem 4.
y1
x
6. x-values: Domain
y-values: Range
4
2.
Exploration 1-3a
y
y2
1. The values for g(x) are 5, 6, 4, 1, 3, 4.
2
y
g
y1
x
4
f
4
x
3. The graph is slid vertically without changing its shape or
proportions.
2
4.
y
2. The graph is slid vertically without changing its shape or
proportions.
4
3. The values for g(x) are 2, 3, 1, D2, 0, 1.
y3
y1
y
x
2
f
5. Horizontal translation by 3 units (right)
x
4
2
g
6.
y
4
y4
4. Horizontal translation by 3 units (right)
x
y1
232 / Solutions to the Explorations
4
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1
©2003 Key Curriculum Press
7. The graph is stretched or squashed vertically but not shifted
vertically as a whole.
2., 3.
8.
y
2
y1
x
4
y5
9. By a factor of 13
10. Answers will vary.
x
g(x)
f(g(x))
0
None
None
1
None
0
None
3
0.5
None
4
1
6
5
1.5
5.5
6
2
5
7
2.5
4.5
8
Exploration 1-3c
None
2
3
9
4
None
None
1. Vertical translation by D6; g(x) H f(x) D 6
4. The lines in which x H 2 and x H 3
2. Horizontal translation by C10; g(x) = f (x D 10)
5.
y
3. Vertical dilation by 3; g(x) = 3f (x)
fg
f
( )
4. Horizontal dilation by 2; g(x) = f 12 x
4
5. Reflection across the x-axis of that part of the graph that is
below the x-axis; g(x) = |f (x)|
g
x
4
6. Reflection across the y-axis; g(x) = f (Dx)
8
6. Domain: 4 ≤ x ≤ 8
Range: 4 ≤ y ≤ 6
7. Answers will vary.
7. f (x) = Dx + 7
1
g(x) = x D 1
2
Exploration 1-3d
1.
y
8.
5
y
x
5
y1
5
4
5
y2
2. Horizontal translation by D6
g(x) = f (x + 6)
4
Yes
3. Vertical translation by 3
g(x) = 3 + f (x)
9.
4. Vertical dilation by factor of 12
1
g(x) = f (x)
2
y
6. Horizontal dilation by factor of 2 and vertical translation
by 3
1
g(x) = 3 + f x
2
7. Answers will vary.
Exploration 1-4a
Range: 2 ≤ y ≤ 6
Range: 0 ≤ y ≤ 3
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1
©2003 Key Curriculum Press
y3
y1
4
5. Horizontal dilation by factor of 2
1
g(x) = f x
2
1. f : Domain: 1 ≤ x ≤ 5
g: Domain: 2 ≤ x ≤ 8
x
8
y2
4
x
8
Yes
10. f (g(x)) = Dg (x) + 7
1
=D xD1 +7
2
1
=D x+8
2
1
f (g(x)) = D x + 8
2
11. Answers will vary.
Solutions to the Explorations / 233
Exploration 1-5a
Exploration 1-6a
1. x H 2y D 5
x C 5 H 2y
1
y = x + 2.5
2
1. Vertical dilation by 12
10
y
2.
10
x
10
y
5
5
x
10
5
2. Horizontal dilation by 2
5
10
y
3. There is no place where there are two different y’s for the
same x.
x
10
1
4. f D1(x) = x + 2.5
2
5. f(3) H 2(3) D 5 H 1
1
f D1(1) = • 1 + 2.5 = 3
2
f (x) = y ⇒ f D1(y) = x
3. Vertical translation by D4
10
y
6.
10
y
x
10
5
5
5
x
10
5
4. Horizontal translation by C7
10
y
They are reflections in y H x.
7.
x
10
y
15
10
5
x
5
10
15
5. Horizontal dilation by D1 (because
10
1
= D1)
D1
y
8. There are two values of y for the same value of x.
9. f(0) H 1; f(1) H 2; f(2) H 4; f(3) H 8
10. f D1(1) = 0; f D1(2) = 1, f D1(4) = 2; f D1(8) = 3
x
10
y
15
10
6. Reflection across the y-axis
5
x
5
10
15
11. Answers will vary.
234 / Solutions to the Explorations
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1
©2003 Key Curriculum Press
7. Vertical dilation by D1
10
Exploration 2-2a
y
1. θref = 180− D 152− = 28−
v
x
10
152°
28°
u
8. Reflection across the x-axis
9. y-direction
2. θref = 250− D 180− = 70−
10.
v
10
y
Graphs
coincide.
250°
u
x
10
70°
3. Because the angle must be counterclockwise so that its
measure will be positive
11.
10
y
4. Because it must go to the nearest side of the horizontal axis
Graphs
coincide.
5. θref = 360− D 310− = 50−
v
x
10
u
50°
310°
12. Answers will vary.
Chapter 2 • Periodic Functions and
Right Triangle Problems
6. θref = θ
v
Exploration 2-1a
θref = θ
u
1. Horizontal translation by 2
y = g(x) = f (x D 2)
2. Vertical dilation by factor of 3
y = g(x) = 3f (x)
3. Horizontal dilation by factor of 12
y = g(x) = f (2x)
4. Vertical translation by D5
y = g(x) = f (x) D 5
7. θref = 180− + (D150−) = 30−
v
5. Vertical translation by D5; horizontal translation by 2
y = g(x) = f (x D 2)
6. Vertical dilation by factor of 3; horizontal translation by 2
y = g(x) = 3f (x D 2)
u
30°
150°
7. Answers will vary.
Precalculus with Trigonometry: Instructor’s Resource Book, Volume 1
©2003 Key Curriculum Press
Solutions to the Explorations / 235