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 D = {–2, 1, 4}, R = {–1, 2, 3, 5};
2-1 Relations and Functions
CCSS STRUCTURE State the domain and range of each relation. Then determine whether
each relation is a function . If it is a function,
determine if it is one-to-one, onto, both, or
neither.
The relation is not a function because 1 is mapped to
both 2 and 5.
3. SOLUTION: The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
1. D = {–2, 1, 4, 8}, R = {–4, –2, 6};
SOLUTION: The left side of the mapping is the domain and the right side is the range.The members of the domain
are the x-values of the relation while the members of
the range are the y-values.
D = {–2, 5, 6}, R = { –8, 1, 3};
Each element of the domain is paired with exactly
one element of the range. So, the relation is a
function.
Each element of the domain is paired with exactly
one element of the range. So, the relation is a
function.
The function is onto because each element of the
range corresponds to an element of the domain.
4. BASKETBALL The table shows the average
points per game for Dwayne Wade of the Miami
Heat for four years.
The function is both one-to-one and onto because
each element of the domain is paired with a unique
element of the range and each element of the range
corresponds to an element of the domain.
2. SOLUTION: The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
a. Assume that the ages are the domain. Identify the
domain and range.
b. Write a relation of ordered pairs for the data.
c. State whether the relation is discrete or
continuous.
d. Graph the relation. Is this relation a function?
D = {–2, 1, 4}, R = {–1, 2, 3, 5};
SOLUTION: a. Since the ages are the domain, the average points
per game are the range.
D = {24, 25, 26, 27}, R = {24.6, 27.2, 27.4, 30.2}
The relation is not a function because 1 is mapped to
both 2 and 5.
b. In writing ordered pairs for the relation, the
members of the domain are the x-values and the
members of the range are the y -values. {(24, 27.2),
(25, 27.4), (26, 24.6), (27, 30.2)}
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3. SOLUTION: c. The domain is a set of individual points. So the
relation is discrete.
Page 1
d. The relation is a function as each element of the
domain is paired with exactly one element of the
members of the domain are the x-values and the
members of the range are the y -values. {(24, 27.2),
(25, 27.4), and
(26, 24.6),
(27, 30.2)}
2-1 Relations
Functions
c. The domain is a set of individual points. So the
relation is discrete.
d. The relation is a function as each element of the
domain is paired with exactly one element of the
range.
Graph each equation, and determine the domain
and range. Determine whether the equation is a
function, is one-to-one, onto, both, or neither.
Then state whether it is discrete or continuous.
5. SOLUTION: To graph the equation, substitute different values of x
in the equation and solve for y. Then connect the
points.
x
0
1
2
3
-1
-2
-3
y = 5x + 4
4
5
14
19
-1
-6
-11
Graph each equation, and determine the domain
and range. Determine whether the equation is a
function, is one-to-one, onto, both, or neither.
Then state whether it is discrete or continuous.
5. SOLUTION: To graph the equation, substitute different values of x
in the equation and solve for y. Then connect the
points.
The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
D = {all real numbers};
R = {all real numbers};
x
0
1
2
3
-1
-2
-3
y = 5x + 4
4
5
14
19
-1
-6
-11
No vertical line intersects the graph in more than one
point. So the graph is a function.
The function is both one-to-one and onto because
each element of the domain is paired with a unique
element of the range and each element of the range
corresponds to an element of the domain.
The graph of the function is a line. So the function is
continuous.
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eSolutions
The members of the domain are the x-values of the
relation while the members of the range are the y-
SOLUTION: To graph the equation, substitute different values of x
in the equation and solve for y. Then connect thePage 2
points.
correspond to an element of the domain.
corresponds to an element of the domain.
The domain has an infinite number of elements and
the relation can be graphed using a straight line. So
the relation is continuous.
The graph of the function is a line. So the function is
2-1 Relations
continuous.and Functions
6. 7. SOLUTION: To graph the equation, substitute different values of x
in the equation and solve for y. Then connect the
points.
SOLUTION: To graph the equation, substitute different values of x
in the equation and solve for y. Then connect the
points.
x
y = -4x - 2
0
1
2
3
-1
-2
-3
x
-2
-6
-10
-14
2
6
10
y = 3x
0
1
2
3
-1
-2
-3
2
0
3
12
27
3
12
27
The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
D = {all real numbers};
R = {all real numbers};
The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
D = {all real numbers};
R = {all real numbers};
No vertical line intersects the graph in more than one
point. So the graph is a function.
No vertical line intersects the graph in more than one
point. So the graph is a function.
The function is both one-to-one and onto because
each element of the domain is paired with a unique
element of the range and each element of the range
correspond to an element of the domain.
The function is neither one-to-one nor onto because
the elements in the domain do not have unique
images and the negative numbers are left unmapped.
The domain has an infinite number of elements and
the relation can be graphed using a smooth curve. So
the relation is continuous.
The domain has an infinite number of elements and
the relation can be graphed using a straight line. So
the relation is continuous.
8. 7. SOLUTION: To graph the equation, substitute different values of x
in the equation and solve for y. Then connect the
points.
SOLUTION: The graph of the equation is a vertical line through (7,
0). x
y = 3x
2
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0
1
2
0
3
12
Page 3
images and the negative numbers are left unmapped.
The domain has an infinite number of elements and
2-1 Relations
the relationand
canFunctions
be graphed using a smooth curve. So
the relation is continuous.
8. SOLUTION: The graph of the equation is a vertical line through (7,
0). State the domain and range of each relation.
Then determine whether each relation is a
function . If it is a function, determine if it is oneto-one, onto, both, or neither.
11. SOLUTION: The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
In this equation x is always 7 for any value of y.
D = {7}; R = {all real numbers};
The only element in the domain is mapped to all the
elements in the range. So it is not a function.
D = {–0.3, 0.4, 1.2}, R = {–6, –3, –1, 4}
1.2 is mapped to both –1 and 4. So the relation is not
a function.
The domain has a finite number (1) of elements, so
the relation is not continuous.
Evaluate each function.
9. SOLUTION: Replace x by –3.
12. SOLUTION: The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
D = {–8, 2, 4}; R = {–6, –4, 14};
–8 is mapped to both –4 and 14. So the relation is not
a function.
10. 13. {(–3, –4), (–1, 0), (3, 0), (5, 3)}
SOLUTION: Replace x with 5.
SOLUTION: The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
D = {–3, –1, 3, 5}; R = {–4, 0, 3}
Each element of the domain is paired with exactly
one element in the range. So, the relation is a
function.
State the domain and range of each relation.
Then determine whether each relation is a
eSolutions
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by Cognero determine if it is onefunction
. If- Powered
it is a function,
to-one, onto, both, or neither.
The function is onto because each element of the
range corresponds to an element of the domain. Page 4
14. POLITICS The table below shows the population of
a. Scale each axis of the graph by 5. Since population
is on the horizontal axis, these are the x-values of the
relation or the domain. The number of
representatives is the range. Plot the data from the
table as ordered pairs on the graph.
D = {–8, 2, 4}; R = {–6, –4, 14};
2-1 Relations
andtoFunctions
–8 is mapped
both –4 and 14. So the relation is not
a function.
13. {(–3, –4), (–1, 0), (3, 0), (5, 3)}
SOLUTION: The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
D = {–3, –1, 3, 5}; R = {–4, 0, 3}
Each element of the domain is paired with exactly
one element in the range. So, the relation is a
function.
The function is onto because each element of the
range corresponds to an element of the domain.
14. POLITICS The table below shows the population of
several states and the number of U.S.
representatives from those states.
a. Make a graph of the data with population on the
horizontal axis and representatives on the vertical
axis.
b. Identify the domain and range.
c. Is the relation discrete or continuous?
d. Does the graph represent a function? Explain your
reasoning.
b. The members of the domain are the x-values of
the relation while the members of the range are the
y-values. D = {8.07, 12.44, 16.03, 19.00, 20.90,
33.93}; R={13, 19, 25, 29, 32, 53}
c. The domain is a set of individual points. So the
relation is discrete.
d. The relation is a function because each domain
value is paired with only one range value.
CCSS STRUCTURE Graph each equation, and
determine the domain and range. Determine
whether the equation is a function, is one-to-one,
onto, both, or neither. Then state whether it is
discrete or continuous.
15. SOLUTION: a. Scale each axis of the graph by 5. Since population
is on the horizontal axis, these are the x-values of the
relation or the domain. The number of
representatives is the range. Plot the data from the
table as ordered pairs on the graph.
SOLUTION: To graph, substitute values for x into the equation and
solve for y. Draw a smooth curve through these
points.
x
y = -3x + 2
0
1
2
3
-1
-2
-3
2
-1
-4
-7
5
8
11
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Page 5
c. The domain is a set of individual points. So the
relation is discrete.
The domain has an infinite number of elements and
the relation can be graphed with a solid straight line.
So the relation is continuous.
d. The relation
2-1 Relations
and Functions
is a function because each domain
value is paired with only one range value.
CCSS STRUCTURE Graph each equation, and
determine the domain and range. Determine
whether the equation is a function, is one-to-one,
onto, both, or neither. Then state whether it is
discrete or continuous.
16. SOLUTION: To graph, substitute values for x into the equation and
solve for y. Draw a smooth curve through these
points.
15. x
SOLUTION: To graph, substitute values for x into the equation and
solve for y. Draw a smooth curve through these
points.
x
y = -3x + 2
0
1
2
3
-1
-2
-3
y = 0.5x - 3
0
1
2
3
-1
-2
-3
2
-1
-4
-7
5
8
11
-3
-2.5
-2
-1.5
-3.5
-5
-4.5
The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
D = {all real numbers};
R = {all real numbers};
The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
D = {all real numbers};
R = {all real numbers};
No vertical line intersects the graph in more than one
point. So the equation is a function.
The function is both one-to-one and onto because
each element of the domain is paired with a unique
element of the range and each element of the range
corresponds to an element of the domain.
No vertical line intersects the graph in more than one
point. So the equation is a function.
The function is both one-to-one and onto because
each element of the domain is paired with a unique
element of the range and each element of the range
corresponds to an element of the domain.
The domain has an infinite number of elements and
the relation can be graphed with a solid straight line.
So the relation is continuous.
The domain has an infinite number of elements and
the relation can be graphed with a solid straight line.
So the relation is continuous.
17. SOLUTION: To graph, substitute values for x into the equation and
solve for y. Draw a smooth curve through these
points.
16. y = 2x 2
x
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SOLUTION: To graph, substitute values for x into the equation and
solve for y. Draw a smooth curve through these
0
1
2
0
2
8
Page 6
corresponds to an element of the domain.
numbers are left unmapped.
The domain has an infinite number of elements and
the relation can be graphed with a smooth curve. So
the relation is continuous.
The domain has an infinite number of elements and
2-1 Relations
the relationand
canFunctions
be graphed with a solid straight line.
So the relation is continuous.
18. 17. SOLUTION: To graph, substitute values for x into the equation and
solve for y. Draw a smooth curve through these
points.
SOLUTION: To graph, substitute values for x into the equation and
solve for y. Draw a smooth curve through these
points.
x
y = 2x
2
0
1
2
3
-1
-2
-3
x
0
2
8
18
2
8
18
y = -5x
2
0
1
2
3
-1
-2
-3
0
-5
-20
-45
-5
-20
-45
The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
D = {all real numbers};
The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
D = {all real numbers};
No vertical line intersects the graph in more than one
point. So the equation is a function.
No vertical line intersects the graph in more than one
point. So the equation is a function.
The function is not one-to-one because each
element of the domain is not paired with a unique
element of the range.
The function is not one-to-one because each
element of the domain is not paired with a unique
element of the range.
The function is not onto because the negative
numbers are left unmapped.
The function is not onto because the positive
numbers are left unmapped.
The domain has an infinite number of elements and
the relation can be graphed with a smooth curve. So
the relation is continuous.
The domain has an infinite number of elements and
the relation can be graphed with a smooth curve. So
the relation is continuous.
19. 18. SOLUTION: To graph, substitute values for x into the equation and
solve for y. Draw a smooth curve through these
points.
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SOLUTION: To graph, substitute values for x into the equation and
solve for y. Draw a smooth curve through these
points.
Page 7
x
y = -5x
2
x
y = 4x 2 - 8
numbers are left unmapped.
than –8 are left unmapped.
The domain has an infinite number of elements and
the function can be graphed with a smooth curve. So
the function is continuous.
The domain has an infinite number of elements and
2-1 Relations
the relationand
canFunctions
be graphed with a smooth curve. So
the relation is continuous.
20. 19. SOLUTION: To graph, substitute values for x into the equation and
solve for y. Draw a smooth curve through these
points.
SOLUTION: To graph, substitute values for x into the equation and
solve for y. A few of the points on the graph are (0, -
1), (1, -4), (-1, 2),
, (2, -25),
and (-2, 23). Draw a smooth curve through these
points.
2
x
y = 4x - 8
0
1
2
3
-1
-2
-3
-8
-4
8
28
-4
8
28
The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
D = {all real numbers};
R = {all real numbers};
No vertical line intersects the graph in more than one
point. So the equation is a function.
The members of the domain are the x-values of the
relation while the members of the range are the yvalues.
D = {all real numbers};
No vertical line intersects the graph in more than one
point. So the equation is a function.
The domain has an infinite number of elements and
the relation can be graphed with a smooth curve. So
the relation is continuous.
The function is both one-to-one and onto because
each element of the domain is paired with a unique
element of the range and each element of the range
corresponds to an element of the domain.
The function is not one-to-one because each
element of the domain is not paired with a unique
element of the range.
Evaluate each function.
21. The function is not onto because the numbers less
than –8 are left unmapped.
SOLUTION: Replace x with –8.
The domain has an infinite number of elements and
the function can be graphed with a smooth curve. So
the function is continuous.
20. SOLUTION: To graph, substitute values for x into the equation and
solveManual
for y.- Powered
A few of
points on the graph are (0, eSolutions
bythe
Cognero
1), (1, -4), (-1, 2),
, (2, -25),
and (-2, 23). Draw a smooth curve through these
22. SOLUTION: Replace x with 2.5.
Page 8
Replace x with –8.
Replace x with 2.5.
2-1 Relations and Functions
23. DIVING The table below shows the pressure on a
diver at various depths.
22. SOLUTION: a. Write a relation to represent the data.
b. Graph the relation.
c. Identify the domain and range. Is the relation
discrete or continuous?
d. Is the relation a function? Explain your reasoning.
Replace x with 2.5.
SOLUTION: a. Let the depth measurements be the domain and
the pressure be the range.{(0, 1), (20, 1.6), (40, 2.2),
(60, 2.8), (80, 3.4), (100, 4)}
23. DIVING The table below shows the pressure on a
diver at various depths.
a. Write a relation to represent the data.
b. Graph the relation.
c. Identify the domain and range. Is the relation
discrete or continuous?
d. Is the relation a function? Explain your reasoning.
b. Plot each ordered pair from part a on the graph.
Draw a straight line through the points. SOLUTION: a. Let the depth measurements be the domain and
the pressure be the range.{(0, 1), (20, 1.6), (40, 2.2),
(60, 2.8), (80, 3.4), (100, 4)}
b. Plot each ordered pair from part a on the graph.
Draw a straight line through the points. c. The depth begins at 0 and can increase
indefinitely. The pressure begins at 1 and can
increase indefinitely so the domain and range are: D
=
; R=
. The relation is
continuous because the graph represents the
pressure at depths other than the given measures,
such as 10 feet.
d. Each domain value is paired with only one range
value. So the relation is a function.
c. The depth begins at 0 and can increase
indefinitely. The pressure begins at 1 and can
increase indefinitely so the domain and range are: D
=
; R=
. The relation is
continuous because the graph represents the
pressure at depths other than the given measures,
such as 10 feet.
d. Each domain value is paired with only one range
value. So the relation is a function.
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Find each value if
and 24. SOLUTION: f (x) = 3x + 2
Page 9
such as 10 feet.
.
d. Each domain value is paired with only one range
value. So the
is a function.
2-1 Relations
andrelation
Functions
Find each value if
27. and SOLUTION: 24. Replace x with –6.
SOLUTION: f (x) = 3x + 2
Replace x with –5.
28. SOLUTION: 25. Replace x with 3.
SOLUTION: f (x) = 3x + 2
Replace x with 9.
29. SOLUTION: 26. Replace x with 8
.
SOLUTION: Replace x with –3.
.
30. SOLUTION: 27. SOLUTION: Replace x with
.
Replace x with –6.
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28. Page 10
31. .
2-1 Relations and Functions
30. 32. SOLUTION: SOLUTION: Replace x with
.
Replace x with
.
31. SOLUTION: Replace x with
.
33. PODCASTS Chaz has a collection of 15 podcasts
downloaded on his digital audio player. He decides to
download 3 more podcasts each month. The function
P(t) = 15 + 3t counts the number of podcasts P(t) he
has after t months. How many podcasts will he have
after 8 months?
SOLUTION: Replace t with 8.
32. After 8 months Chaz will have 39 podcasts.
SOLUTION: 34. MULTIPLE REPRESENTATIONS In this
problem you will investigate one-to-one and onto
functions.
Replace x with
.
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a. GRAPHICAL Graph each function on a separate
graphing calculator screen.
b. TABULAR Use the graphs to create a tablePage 11
showing the number of times a horizontal line could
intersect the graph of each function. List all
2-1 Relations and Functions
b. TABULAR Use the graphs to create a table
showing the number of times a horizontal line could
intersect the graph of each function. List all
possibilities.
j(x) = x 3
c. ANALYTICAL For a function to be one-to-one,
a horizontal line on the graph of the function can
intersect the function at most once. Which functions
meet this condition? Which do not? Explain your
reasoning.
b.
d. ANALYTICAL For a function to be onto, every
possible horizontal line on the graph of the function
must intersect the function at least once. Which
functions meet this condition? Which do not? Explain
your reasoning.
e . GRAPHICAL Create a table showing whether
each function is one-to-one and/or onto.
SOLUTION: a. f (x) = x
2
c. Placing a pencil on each graph so it’s parallel to
the x-axis and then moving it straight up and down,
only g(x) and j (x) intersect the pencil line once at a
time so they are one-to-one, and f (x) and h(x) are
not
.
d. A horizontal line on the graphs of h(x) and j (x)
will intersect the graph more than once so they are
onto, and f (x) and g(x) are not.
e. g(x) = 2x
35. CCSS CRITIQUE Omar and Madison are finding f
(3d) for the function
Is either of them correct? Explain your reasoning.
h(x) = x 3 - 3x 2 - 5x + 6
j(x) = x 3
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SOLUTION: Sample answer: Both set up the equation correctly,
substituting 3d for x in f (x). However, Madison did
not square the 3 before multiplying by –4 so Omar is
correct.
36. CHALLENGE Consider the functions f (x) and
and g(a) = 33, while f (b) = 31 and g
(b) = 51. If a = 5 and b = 8, find two possible
functions to represent f (x) and g(x).
Page 12
SOLUTION: Sample answer: Organize the given information into a
table. SOLUTION: Sample answer: Both set up the equation correctly,
substituting 3d for x in f (x). However, Madison did
not square and
the 3Functions
before multiplying by –4 so Omar is
2-1 Relations
correct.
36. CHALLENGE Consider the functions f (x) and
and g(a) = 33, while f (b) = 31 and g
(b) = 51. If a = 5 and b = 8, find two possible
functions to represent f (x) and g(x).
SOLUTION: Sample answer: Organize the given information into a
table. f (x)
g(x)
f (a) = 19
g(a) = 33
f (b) = 31
g(b) = 51
a = 5, b = 8
Analyze the information given about f (x).
f (x)
f (a) = f (5) =
4(5) = f (x) = f (a) = 4(5)
19
19
20
4x – 1 – 1 = 20
f (b) = f (8) =
4(8) = f (x) = f (b) = 4(8)
31
31
32
4x – 1 – 1 = 31
If the values of a and b are multiplied by 4, the
product is one more than the value of f (a) and f (b). Next, analyze the information given about g(x).
g(x)
g(a) = 6
g(a) = g(5) = 6(5) = g(x) =
(5) + 3 =
33
33
30
6x + 3
33
g(b) = 6
g(b) = g(8) = 6(8) = g(x) =
(8) + 3 =
51
51
48
4x + 3
51
If the values of a and b are multiplied by 6, the
product is three less than the value of g(a) and g(b).
So the functions are:
.
Never; if the graph crosses the y-axis twice, then
there will be two separate y-values that correspond
to x = 0, which violates the vertical line test.
38. OPEN ENDED Graph a relation that can be used to
represent each of the following.
a. the height of a baseball that is hit into the outfield
b. the speed of a car that travels to the store,
stopping at two lights along the way
c. the height of a person from age 5 to age 80
d. the temperature on a typical day from 6 A.M. to
11 P.M.
SOLUTION: a. Sample answer: let the x-axis be the time the ball
is in the air and the y-axis be the height of the ball.
The height of the ball is zero when the time is zero.
Once the ball is thrown, the height will reach a
maximum point and then decrease eventually landing
on the ground.
b. Sample answer: let the x-axis be the time the car
is being driven and the y-axis be the distance the car
has traveled. At each stop light, time increases but
distance is constant. c. Sample answer: let the x-axis be the age of the
person and the y-axis be the height. Height increases
as a child, then more steeply, finally leveling off and
remaining constant.
37. REASONING If the graph of a relation crosses the
y-axis at more than one point, is the relation
sometimes, always, or never a function? Explain
your reasoning.
SOLUTION: Never; if the graph crosses the y-axis twice, then
there will be two separate y-values that correspond
to x = 0, which violates the vertical line test.
38. OPEN ENDED Graph a relation that can be used to
represent each of the following.
a. the height of a baseball that is hit into the outfield
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b. the speed of a car that travels to the store,
stopping at two lights along the way
c. the height of a person from age 5 to age 80
d. Sample answer: let the x-axis be the hours from 6
to 11 and the y-axis be the temperature. On a typical
day, the morning is cool while the temperature
gradually warms up to reach a maximum. The
temperature gradually decreases as the sun sets.
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d. Sample answer: let the x-axis be the hours from 6
to 11 and the y-axis be the temperature. On a typical
day, the morning
is cool while the temperature
2-1 Relations
and Functions
gradually warms up to reach a maximum. The
temperature gradually decreases as the sun sets.
SOLUTION: Sample answer: A relation is a function if each xvalue only pairs with one y-value. If the vertical line
test fails then there is an x-value that pairs with more
than one y-value, so the relation is not a function.
41. Patricia’s swimming pool contains 19,500 gallons of
water. She drains the pool at a rate of 6 gallons per
minute. Which of these equations represents the
number of gallons of water g, remaining in the pool
after m minutes?
A g = 19,500 – 6m
39. REASONING Determine whether the following
statement is true or false . Explain your reasoning.
If a function is onto, then it must be one-to-one as
well.
SOLUTION: Sample answer: if a function is onto then each
element of the range corresponds to an element of
the domain. A function that is one-to-one has each
element of the domain paired to exactly one unique
element of the range. The statement is false; a
function is onto and not one-to-one if all of the
elements of the domain correspond to an element of
the range, but more than one element of the domain
corresponds to the same element of the range.
40. WRITING IN MATH Explain why the vertical line
test can determine if a relation is a function.
SOLUTION: Sample answer: A relation is a function if each xvalue only pairs with one y-value. If the vertical line
test fails then there is an x-value that pairs with more
than one y-value, so the relation is not a function.
B g = 19,500 + 6m
C D SOLUTION: Number of gallons of water in the pool = 19,500.
Patricia drains the water at a rate of 6 gallons per
minute.
In m minutes, she can drain 6m gallons of water.
So the number of gallons of water remaining in the
pool after m minutes is given by:
g = 19,500 – 6m
The correct choice is A.
42. SHORT RESPONSE Look at the pattern below.
41. Patricia’s swimming pool contains 19,500 gallons of
water. She drains the pool at a rate of 6 gallons per
minute. Which of these equations represents the
number of gallons of water g, remaining in the pool
after m minutes?
If the pattern continues, what will the next term be?
SOLUTION: A g = 19,500 – 6m
Each term of the pattern is obtained by adding
the previous term.
to
B g = 19,500 + 6m
Next term =
C 43. GEOMETRY Which set of dimensions represents a
triangle similar to the triangle shown below?
D SOLUTION: Number of gallons of water in the pool = 19,500.
minute.
F 1 unit, 2 units, 3 units
G 7 units, 11 units, 12 units
H 10 units, 23 units, 24 units
J 20 units, 48 units, 52 units
SOLUTION: Patricia
drains
the water
at a rate
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of 6 gallons per
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Each term of the pattern is obtained by adding
to
the previous term.
2-1 Relations
Next term and
= Functions
The correct choice is C.
Solve each inequality.
43. GEOMETRY Which set of dimensions represents a
triangle similar to the triangle shown below?
45. SOLUTION: F 1 unit, 2 units, 3 units
G 7 units, 11 units, 12 units
H 10 units, 23 units, 24 units
J 20 units, 48 units, 52 units
SOLUTION: No common ratio can be found between the
dimensions of the given triangle and the sets of
dimensions given in the choices F, G, H.
46. SOLUTION: The dimensions of the triangle and the dimensions
given in the choice J are in the ration 1:4.
So the correct choice is J.
44. ACT/SAT If
to g(x + 1)?
A. 1
which expression is equal
2
B. x + 1
2
C. x + 2x + 1
47. SOLUTION: 2
D. x – x
2
E. x + x + 1
SOLUTION: Replace x by x + 1.
This implies:
The correct choice is C.
Solve each inequality.
45. 48. CLUBS Mr. Willis is starting a chess club at his high
school. He sent the advertisement at the right to all
of the homerooms. Write an absolute value inequality
representing the situation.
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SOLUTION: Let x represent the number of members.
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Ling can buy a maximum of 8 shirts.
2-1 Relations and Functions
48. CLUBS Mr. Willis is starting a chess club at his high
school. He sent the advertisement at the right to all
of the homerooms. Write an absolute value inequality
representing the situation.
Solve each equation. Check your solutions.
50. SOLUTION: This implies:
SOLUTION: Let x represent the number of members.
49. SALES Ling can spend no more than $120 at the
summer sale of a department store. She wants to buy
shirts on sale for $15 each. Write and solve an
inequality to determine the number of shirts she can
buy.
51. SOLUTION: SOLUTION: Let x be the number of shirts Ling can buy.
Each shirt costs $15.
So:
Ling can buy a maximum of 8 shirts.
This implies:
52. SOLUTION: Solve each equation. Check your solutions.
50. SOLUTION: This implies:
This implies:
Simplify each expression.
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51. Page 16
53. SOLUTION: 2-1 Relations and Functions
Simplify each expression.
So the solution is x = 6.
57. SOLUTION: 53. SOLUTION: Substitute a = 4 in the original equation.
54. SOLUTION: So the solution is a = 4.
55. 58. SOLUTION: SOLUTION: Solve each equation. Check your solutions.
56. Substitute x = –2 in the original equation.
SOLUTION: So the solution is x = –2.
Substitute x = 6 in the original equation.
59. SOLUTION: So the solution is x = 6.
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57. SOLUTION: Substitute b = –4 in the original equation.
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2-1 Relations
and Functions
So the solution is x = –2.
59. So the solution is x = 3.
61. SOLUTION: SOLUTION: Substitute b = –4 in the original equation.
Substitute y = –4 in the original equation.
So the solution is b = –4.
60. So the solution is y = –4.
SOLUTION: 62. SOLUTION: Substitute x = 3 in the original equation.
Substitute c = 6 in the original equation.
So the solution is x = 3.
So the solution is c = 6.
63. 61. SOLUTION: SOLUTION: eSolutions
Manualy- Powered
Substitute
= –4 inby
theCognero
original equation.
Substitute d = –6 in the original equation.
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2-1 Relations
and Functions
So the solution is c = 6.
63. SOLUTION: Substitute d = –6 in the original equation.
So the solution is d = –6.
64. SOLUTION: Substitute y = 3 in the equation.
So the solution is y = 3.
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