Download State whether each sentence is true or false . If false , replace the

Study Guide and Review - Chapter 9
State whether each sentence is true or false . If false , replace the underlined term to make a true
sentence.
1. The axis of symmetry of a quadratic function can be found by using the equation x =
.
SOLUTION: The shape of the graph of a quadratic function is called a parabola. Parabolas are symmetric about a central line
called the axis of symmetry. The axis of symmetry of a quadratic function can be found by using the equation
. The statement is true.
2. The vertex is the maximum or minimum point of a parabola.
SOLUTION: The axis of symmetry intersects a parabola at only one point, called the vertex. The lowest point on the graph is the
minimum, and the highest point on the graph is the maximum. The vertex is the maximum or minimum point of a
parabola. The statement is true.
3. The graph of a quadratic function is a straight line.
SOLUTION: 2
Quadratic functions are nonlinear and can be written in the form f (x) = ax + bx + c, where a ≠ 0. So, the statement is false. The graph of a quadratic function is a parabola .
2
4. The graph of a quadratic function has a maximum if the coefficient of the x -term is positive.
SOLUTION: 2
When a > 0, the graph of y = ax + bx + c opens upward. The lowest point on the graph is the minimum. So, the
2
statement is false. The graph of a quadratic function has a minimum if the coefficient of the x -term is positive.
5. A quadratic equation with a graph that has two x-intercepts has one real root.
SOLUTION: The solutions or roots of a quadratic equation can be identified by finding the x-intercepts of the related graph. So,
the statement is false. A quadratic equation with a graph that has two x-intercepts has two real roots.
2
6. The expression b − 4ac is called the discriminant.
SOLUTION: 2
In the Quadratic Formula, the expression under the radical sign, b − 4ac, is called the discriminant. The statement is
true.
7. The solutions of a quadratic equation are called roots.
SOLUTION: The solutions of a quadratic equation are called roots. The statement is true.
2
8. The graph of the parent function is translated down to form the graph of f (x) = x + 5.
SOLUTION: 2
If c > 0 in f (x) = x + c, the graph of the parent function is translated
units up. So, the statement is false. The 2
graph of the parent function is translated up 5 units to form the graph of f (x) = x + 5.
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State whether each sentence is true or false . If false , replace the underlined term to make a true
sentence.
Page 1
true.
7. The solutions of a quadratic equation are called roots.
SOLUTION: Study
Guide and Review - Chapter 9
The solutions of a quadratic equation are called roots. The statement is true.
2
8. The graph of the parent function is translated down to form the graph of f (x) = x + 5.
SOLUTION: 2
If c > 0 in f (x) = x + c, the graph of the parent function is translated
units up. So, the statement is false. The 2
graph of the parent function is translated up 5 units to form the graph of f (x) = x + 5.
State whether each sentence is true or false . If false , replace the underlined term to make a true
sentence.
9. The range of the greatest integer function is the set of all real numbers.
SOLUTION: The statement is false. The domain of the greatest integer function is the set of all real numbers. The range is all
integers. State whether each sentence is true or false . If false , replace the underlined term to make a true
sentence.
10. A function that is defined differently for different parts of its domain is called a piecewise-defined function.
SOLUTION: true
Consider each equation.
a. Determine whether the function has a maximum or minimum value.
b. State the maximum or minimum value.
c. What are the domain and range of the function?
2
11. y = x − 4x + 4
SOLUTION: 2
a. For y = x – 4x + 4, a = 1, b = –4, and c = 4. Because a is positive, the graph opens upward, so the function has a
minimum value.
b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is
.
The x-coordinate of the vertex is x = 2. Substitute this value into the function to find the y -coordinate. The minimum value is 0.
c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y|y
≥ 0}.
2
12. y = −x + 3x
SOLUTION: eSolutions
Manual - Powered by Cognero
2
Page 2
a. For y = −x + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has
a maximum value.
The minimum value is 0.
c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y|y
Study
Guide and Review - Chapter 9
≥ 0}.
2
12. y = −x + 3x
SOLUTION: 2
a. For y = −x + 3x, a = –1, b = 3, and c = 0. Because a is negative, the graph opens downward, so the function has
a maximum value.
b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is
.
The x-coordinate of the vertex is x = 1.5. Substitute this value into the function to find the y -coordinate. The maximum value is 2.25.
c. The domain is all real numbers. The range is all real numbers less than or equal to the minimum value, or {y|y ≤ 2.25}.
2
13. y = x − 2x − 3
SOLUTION: 2
a. For y = x − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has
a minimum value.
b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is
.
The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate. eSolutions Manual - Powered by Cognero
The minimum value is –4.
Page 3
The maximum value is 2.25.
c. Guide
Study
and Review
The domain
is all real- Chapter
numbers. 9The range is all real numbers less than or equal to the minimum value, or {y|y ≤ 2.25}.
2
13. y = x − 2x − 3
SOLUTION: 2
a. For y = x − 2x − 3, a = 1, b = –2, and c = –3. Because a is positive, the graph opens upward, so the function has
a minimum value.
b. The minimum value is the y-coordinate of the vertex. The x-coordinate of the vertex is
.
The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate. The minimum value is –4.
c. The domain is all real numbers. The range is all real numbers greater than or equal to the minimum value, or {y|y
≥ –4}.
2
14. y = −x + 2
SOLUTION: 2
a. For y = −x + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a
maximum value.
b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is
.
The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate. eSolutions Manual - Powered by Cognero
The maximum value is 2.
Page 4
The minimum value is –4.
c. Guide
Study
and Review
The domain
is all real- Chapter
numbers. 9The range is all real numbers greater than or equal to the minimum value, or {y|y
≥ –4}.
2
14. y = −x + 2
SOLUTION: 2
a. For y = −x + 2, a = –1, b = 0, and c = 2. Because a is negative, the graph opens downward, so the function has a
maximum value.
b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is
.
The x-coordinate of the vertex is x = 0. Substitute this value into the function to find the y -coordinate. The maximum value is 2.
c. The domain is all real numbers. The range is all real numbers less than or equal to the maximum value, or {y|y ≤ 2}.
2
15. BASEBALL A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t + 32t
gives the height of the rocket t seconds after it is launched.
a. Determine whether the function has a maximum or minimum value.
b. State the maximum or minimum value.
c. State a reasonable domain and range of this situation.
SOLUTION: 2
a. For h = −16t + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function
has a maximum value.
b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is
.
The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate. eSolutions Manual - Powered by Cognero
Page 5
The maximum value is 2.
c. Guide
Study
and Review
The domain
is all real- Chapter
numbers. 9The range is all real numbers less than or equal to the maximum value, or {y|y ≤ 2}.
2
15. BASEBALL A toy rocket is launched with an upward velocity of 32 feet per second. The equation h = −16t + 32t
gives the height of the rocket t seconds after it is launched.
a. Determine whether the function has a maximum or minimum value.
b. State the maximum or minimum value.
c. State a reasonable domain and range of this situation.
SOLUTION: 2
a. For h = −16t + 32t, a = –16, b = 32, and c = 0. Because a is negative, the graph opens downward, so the function
has a maximum value.
b. The maximum value is the y-coordinate of the vertex. The x-coordinate of the vertex is
.
The x-coordinate of the vertex is x = 1. Substitute this value into the function to find the y -coordinate. The maximum value is 16.
c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet,
and then it will return to the ground. Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}. Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest
tenth.
2
16. x − 3x − 4 = 0
SOLUTION: 2
Graph the related function f (x) = x − 3x − 4.
The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.
CHECK: Check each solution in the original equation. eSolutions Manual - Powered by Cognero
Page 6
The maximum value is 16.
c. The rocket will be in the air for a total of 2 seconds. It will go from the ground to a maximum height of 16 feet,
and
then itand
willReview
return to
the ground.
Study
Guide
- Chapter
9 Therefore, a reasonable domain for this situation is D = {t | 0 ≤ t ≤ 2} and a reasonable range is R = {h | 0 ≤ h ≤ 16}. Solve each equation by graphing. If integral roots cannot be found, estimate the roots to the nearest
tenth.
2
16. x − 3x − 4 = 0
SOLUTION: 2
Graph the related function f (x) = x − 3x − 4.
The x-intercepts of the graph appear to be at –1 and 4, so the solutions are –1 and 4.
CHECK: Check each solution in the original equation. Therefore, the solutions are – 1 and 4.
2
17. −x + 6x − 9 = 0
SOLUTION: 2
Graph the related function f (x) = −x + 6x − 9.
The x-intercept of the graph appears to be only at 3, so the only solution is 3.
CHECK: Check the solution in the original equation.
eSolutions Manual - Powered by Cognero
Page 7
Study
Guide and
9 4.
Therefore,
theReview
solutions- Chapter
are – 1 and
2
17. −x + 6x − 9 = 0
SOLUTION: 2
Graph the related function f (x) = −x + 6x − 9.
The x-intercept of the graph appears to be only at 3, so the only solution is 3.
CHECK: Check the solution in the original equation.
Therefore, the solution is 3.
2
18. x − x − 12 = 0
SOLUTION: 2
Graph the related function f (x) = x − x − 12.
The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.
CHECK: Check each solution in the original equation.
eSolutions Manual - Powered by Cognero
Therefore, the solutions are – 3 and 4.
Page 8
Study Guide and Review - Chapter 9
Therefore, the solution is 3.
2
18. x − x − 12 = 0
SOLUTION: 2
Graph the related function f (x) = x − x − 12.
The x-intercepts of the graph appear to be at –3 and 4, so the solutions are –3 and 4.
CHECK: Check each solution in the original equation.
Therefore, the solutions are – 3 and 4.
2
19. x + 4x − 3 = 0
SOLUTION: 2
Graph the related function f (x) = x + 4x − 3.
The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for
the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values.
The function value that is closest to zero is the best approximation for a zero of the function.
–4.9
–4.8
–4.7
1.41
0.84
0.29
–4.4
–4.3
–4.2
x
eSolutions Manual - Powered by Cognero
–1.24
–1.71
–2.16
y
0.1
0.2
0.3
x
x
y
–4.6
–0.24
–4.1
–2.59
0.4
Page 9
Guide and Review - Chapter 9
Study
Therefore, the solutions are – 3 and 4.
2
19. x + 4x − 3 = 0
SOLUTION: 2
Graph the related function f (x) = x + 4x − 3.
The x-intercepts are located between –5 and –4 and between 0 and 1. Make a table using an increment of 0.1 for
the x-values located between –5 and –4 and between 0 and 1. Look for a change in the signs of the function values.
The function value that is closest to zero is the best approximation for a zero of the function.
x
y
x
y
x
y
x
y
–4.9
1.41
–4.4
–1.24
0.1
–2.59
0.6
–0.24
–4.8
0.84
–4.3
–1.71
0.2
–2.16
0.7
0.29
–4.7
0.29
–4.2
–2.16
0.3
–1.71
0.8
0.84
–4.6
–0.24
–4.1
–2.59
0.4
–1.24
0.9
1.41
For each table, the function value that is closest to zero when the sign changes is –0.24. Thus, the solutions are
approximately –4.6 and 0.6.
2
20. x − 10x = −21
SOLUTION: Rewrite the equation in standard form.
2
Graph the related function f (x) = x − 10x + 21.
eSolutions Manual - Powered by Cognero
The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.
CHECK: Check each solution in the original equation.
Page 10
x
y
0.6
–0.24
0.7
0.29
0.8
0.84
0.9
1.41
Study
and Review
- Chapter
9 is closest to zero when the sign changes is –0.24. Thus, the solutions are
ForGuide
each table,
the function
value that
approximately –4.6 and 0.6.
2
20. x − 10x = −21
SOLUTION: Rewrite the equation in standard form.
2
Graph the related function f (x) = x − 10x + 21.
The x-intercepts of the graph appear to be at 3 and 7, so the solutions are 3 and 7.
CHECK: Check each solution in the original equation.
Therefore, the solutions are 3 and 7.
2
21. 6x − 13x = 15
SOLUTION: Rewrite the equation in standard form.
2
Graph the related function f (x) = 6x − 13x – 15.
eSolutions
- Powered
The Manual
x-intercepts
of by
theCognero
graph
appear to be at 3 and between –1 and 0. So, one solution is 3.
CHECK: Check the solution in the original equation.
Page 11
Guide and Review - Chapter 9
Study
Therefore, the solutions are 3 and 7.
2
21. 6x − 13x = 15
SOLUTION: Rewrite the equation in standard form.
2
Graph the related function f (x) = 6x − 13x – 15.
The x-intercepts of the graph appear to be at 3 and between –1 and 0. So, one solution is 3.
CHECK: Check the solution in the original equation.
To find the second solution, make a table using an increment of 0.1 for the x-values located between –1 and 0. Look
for a change in the signs of the function values. The function value that is closest to zero is the best approximation
for a zero of the function.
x
y
x
y
x
x
–0.9
1.56
–0.6
–5.04
–0.3
–10.56
–0.8
–0.76
–0.5
–7
–0.2
–12.16
–0.7
–2.96
–0.4
–8.84
–0.1
–13.64
The function value that is closest to zero when the sign changes is –0.76. Thus, the second root is approximately –
0.8. So, the solutions are about –0.8 and 3.
22. NUMBER THEORY Find two numbers that have a sum of 2 and a product of −15.
SOLUTION: From the given information, write two equations. Let x and y represent the numbers.
x +y = 2
x • y = –15
eSolutions
Manual - Powered by Cognero
Solve the first equation for y.
Page 12
x
x
–0.3
–10.56
–0.2
–12.16
–0.1
–13.64
Study
Guide
andvalue
Review
The
function
that-isChapter
closest to9 zero when the sign changes is –0.76. Thus, the second root is approximately –
0.8. So, the solutions are about –0.8 and 3.
22. NUMBER THEORY Find two numbers that have a sum of 2 and a product of −15.
SOLUTION: From the given information, write two equations. Let x and y represent the numbers.
x +y = 2
x • y = –15
Solve the first equation for y.
Substitute this value for y into the second equation.
2
Graph the related function f (x) = x − 2x – 15.
The x-intercepts of the graph appear to be at –3 and 5, so the solutions for x are –3 and 5.
When x = 3, y = 2 – (–3) or 5. When x = 5, y = 2 – 5 or –3. The sum of –3 and 5 is 2 and the product of –3 and 5 is
–15.
Therefore the two numbers are –3 and 5. 2
Describe how the graph of each function is related to the graph of f (x) = x .
2
23. f (x) = x + 8
SOLUTION: 2
The graph of f (x) = x + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.
So, the graph is shifted up 8 units from the parent function.
2
24. f (x) = x − 3
SOLUTION: eSolutions
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2
Page 13
The graph of f (x) = x + c represents a translation up or down of the parent graph. Since c = –3, the translation is
down. So, the graph is shifted down 3 units from the parent function.
SOLUTION: 2
The graph of f (x) = x + c represents a translation up or down of the parent graph. Since c = 8, the translation is up.
Study
andisReview
Chapter
9 the parent function.
So,Guide
the graph
shifted -up
8 units from
2
24. f (x) = x − 3
SOLUTION: 2
The graph of f (x) = x + c represents a translation up or down of the parent graph. Since c = –3, the translation is
down. So, the graph is shifted down 3 units from the parent function.
25. f (x) = 2x
2
SOLUTION: 2
2
The graph of f (x) = ax stretches or compresses the parent graph vertically. Since a = 2, the graph of y = 2x is the
2
graph of y = x vertically stretched.
2
26. f (x) = 4x − 18
SOLUTION: 2
The function can be written f (x) = ax + c, where a = 4 and c = –18. Since –18 < 0 and
> 1, the graph of y = 4x
2
2
− 18 is the graph of y = x vertically stretched and shifted down 18 units.
27. f (x) =
x
2
SOLUTION: 2
The graph of f (x) = ax stretches or compresses the parent graph vertically. Since a =
, the graph of y =
2
x is
2
the graph of y = x vertically compressed.
28. f (x) =
x
2
SOLUTION: 2
The graph of f (x) = ax stretches or compresses the parent graph vertically. ince a =
, the graph of y =
2
x is 2
the graph of y = x vertically compressed.
29. Write an equation for the function shown in the graph.
SOLUTION: Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have
2
the form y = ax + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this
point to find the value of a.
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Page 14
SOLUTION: 2
The graph of f (x) = ax stretches or compresses the parent graph vertically. ince a =
Study Guide and Review - Chapter 9
2
the graph of y = x vertically compressed.
2
x is , the graph of y =
29. Write an equation for the function shown in the graph.
SOLUTION: Since the graph is a parabola that has only been translated vertically and stretched vertically, the equation must have
2
the form y = ax + c. The y-intercept of the graph is –3, so c = –3. The graph contains the point (1, –1). Use this
point to find the value of a.
2
Write a quadratic equation using a = 2 and c = –3. So, an equation for the function shown in the graph is y = 2x – 3.
2
30. PHYSICS A ball is dropped off a cliff that is 100 feet high. The function h = −16t + 100 models the height h of the
2
ball after t seconds. Compare the graph of this function to the graph of h = t .
SOLUTION: 2
Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at + c,
2
where a = –16 and c = 100. Since 100 > 0 and
> 1, the graph of h = −16t + 100 is the graph of y = x
vertically stretched and shifted up 100 units from the parent function.
2
Solve each equation by completing the square. Round to the nearest tenth if necessary.
2
31. x + 6x + 9 = 16
SOLUTION: The solutions are 1 and –7.
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2
32. −a − 10a + 25 = 25
Page 15
SOLUTION: 2
Because the leading coefficient is negative, the graph opens downward. The function can be written f (t) = at + c,
2
where
a =and
–16Review
and c =-100.
Since 9100 > 0 and
> 1, the graph of h = −16t + 100 is the graph of y = x
Study
Guide
Chapter
vertically stretched and shifted up 100 units from the parent function.
2
Solve each equation by completing the square. Round to the nearest tenth if necessary.
2
31. x + 6x + 9 = 16
SOLUTION: The solutions are 1 and –7.
2
32. −a − 10a + 25 = 25
SOLUTION: The solutions are 0 and –10.
2
33. y − 8y + 16 = 36
SOLUTION: eSolutions Manual - Powered by Cognero
Page 16
Study
Guide and Review - Chapter 9
The solutions are 0 and –10.
2
33. y − 8y + 16 = 36
SOLUTION: The solutions are 10 and –2.
2
34. y − 6y + 2 = 0
SOLUTION: Use a calculator to approximate each value of y.
The solutions are approximately 5.6 and 0.4.
2
35. n − 7n = 5
SOLUTION: eSolutions Manual - Powered by Cognero
Page 17
Guide and Review - Chapter 9
Study
The solutions are approximately 5.6 and 0.4.
2
35. n − 7n = 5
SOLUTION: Use a calculator to approximate each value of n.
The solutions are approximately 7.7 and –0.7.
2
36. −3x + 4 = 0
SOLUTION: Use a calculator to approximate each value of x.
The solutions are approximately 1.2 and –1.2.
37. NUMBER THEORY Find two numbers that have a sum of −2 and a product of −48.
SOLUTION: From the given information, write two equations. Let x and y represent the numbers.
x + yManual
= –2 - Powered by Cognero
eSolutions
x • y = –48
Page 18
Guide and Review - Chapter 9
Study
The solutions are approximately 1.2 and –1.2.
37. NUMBER THEORY Find two numbers that have a sum of −2 and a product of −48.
SOLUTION: From the given information, write two equations. Let x and y represent the numbers.
x + y = –2
x • y = –48
Solve the first equation for y.
Substitute this value for y into the second equation.
Now solve by completing the square.
The solutions are 6 and –8.
Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.
2
38. x − 8x = 20
SOLUTION: Rewrite the equation in standard form.
For this equation, a = 1, b = –8, and c = –20.
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Page 19
Guide and Review - Chapter 9
Study
The solutions are 6 and –8.
Solve each equation by using the Quadratic Formula. Round to the nearest tenth if necessary.
2
38. x − 8x = 20
SOLUTION: Rewrite the equation in standard form.
For this equation, a = 1, b = –8, and c = –20.
The solutions are 10 and –2.
2
39. 21x + 5x − 7 = 0
SOLUTION: For this equation, a = 21, b = 5, and c = –7.
The solutions are approximately 0.5 and –0.7.
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2
40. d − 5d + 6 = 0
SOLUTION: Page 20
Study
Guide
and are
Review
- Chapter
9
The
solutions
10 and
–2.
2
39. 21x + 5x − 7 = 0
SOLUTION: For this equation, a = 21, b = 5, and c = –7.
The solutions are approximately 0.5 and –0.7.
2
40. d − 5d + 6 = 0
SOLUTION: For this equation, a = 1, b = –5, and c = 6.
The solutions are 3 and 2.
2
41. 2f + 7f − 15 = 0
SOLUTION: For this equation, a = 2, b = 7, and c = –15.
eSolutions Manual - Powered by Cognero
Page 21
Study Guide and Review - Chapter 9
The solutions are 3 and 2.
2
41. 2f + 7f − 15 = 0
SOLUTION: For this equation, a = 2, b = 7, and c = –15.
The solutions are 1.5 and –5.
2
42. 2h + 8h + 3 = 3
SOLUTION: Rewrite the equation in standard form.
For this equation, a = 2, b = 8, and c = 0.
The solutions are 0 and –4.
2
43. 4x + 4x = 15
SOLUTION: eSolutions
Manual
byin
Cognero
Rewrite
the- Powered
equation
standard
form.
Page 22
Study Guide and Review - Chapter 9
The solutions are 0 and –4.
2
43. 4x + 4x = 15
SOLUTION: Rewrite the equation in standard form.
For this equation, a = 4, b = 4, and c = –15.
The solutions are 1.5 and –2.5.
44. GEOMETRY The area of a square can be quadrupled by increasing the side length and width by 4 inches. What is
the side length?
SOLUTION: 2
If x = the original side length, then the area of the square = x . If x is increased by 4, the area of the square is
quadrupled. Write an equation to represent the new area of the square.
For this equation, a = –3, b = 8, and c = 16.
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Page 23
Study Guide and Review - Chapter 9
The solutions are 1.5 and –2.5.
44. GEOMETRY The area of a square can be quadrupled by increasing the side length and width by 4 inches. What is
the side length?
SOLUTION: 2
If x = the original side length, then the area of the square = x . If x is increased by 4, the area of the square is
quadrupled. Write an equation to represent the new area of the square.
For this equation, a = –3, b = 8, and c = 16.
Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.
Look for a pattern in each table of values to determine which kind of model best describes the data.
Then write an equation for the function that models the data.
45. SOLUTION: First differences:
Second differences:
Since the second differences are equal, a quadratic function models the data.
2
Write an equation for the function that models the data. The equation has the form y = ax . Use the ordered pairPage
(1, 24
3) to find the value of a.
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Study Guide and Review - Chapter 9
Because the side length cannot be negative, the solution is 4. So, the side length of the square is 4 inches.
Look for a pattern in each table of values to determine which kind of model best describes the data.
Then write an equation for the function that models the data.
45. SOLUTION: First differences:
Second differences:
Since the second differences are equal, a quadratic function models the data.
2
Write an equation for the function that models the data. The equation has the form y = ax . Use the ordered pair (1,
3) to find the value of a.
2
An equation that models the data is y = 3x .
46. SOLUTION: Calculate the first differences.
Calculate the second differences.
Because neither the first difference nor the second differences are equal, the table does not represent a linear or
quadratic function. Compare the ratios of the y-values.
Calculate the ratios.
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The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential
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2
Study
andthat
Review
- Chapter
AnGuide
equation
models
the data is9y = 3x .
46. SOLUTION: Calculate the first differences.
Calculate the second differences.
Because neither the first difference nor the second differences are equal, the table does not represent a linear or
quadratic function. Compare the ratios of the y-values.
Calculate the ratios.
The ratios of successive y-values are equal. Therefore, the table of values can be modeled by an exponential
x
function. The equation has the form y = ab . The constant ratio, or base, is 2. Use the ordered pair (1, 2) to find the
value of a.
x
x
An equation that models the data is y = 1 • 2 or y = 2 .
47. SOLUTION: Calculate the first differences.
Calculate the second differences.
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Since the second differences are equal, a quadratic function models the data.
2
Write an equation for the function that models the data. The equation has the form y = ax . Use the ordered pair (1,
Study Guide and Review - Chapter 9
x
x
An equation that models the data is y = 1 • 2 or y = 2 .
47. SOLUTION: Calculate the first differences.
Calculate the second differences.
Since the second differences are equal, a quadratic function models the data.
2
Write an equation for the function that models the data. The equation has the form y = ax . Use the ordered pair (1,
–1) to find the value of a.
2
An equation that models the data is y = –x .
Graph each function. State the domain and range.
48. SOLUTION: Make a table of values.
x
0
0.5
1
1.5
2
2.5
3
f (x)
0
0
1
1
2
2
3
Because
the
dots and
eSolutions
Manual
- Powered
by circles
Cognero overlap,
The range is all integers.
the graph will cover all possible values of x, so the domain is all real numbers.
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Study Guide and Review - Chapter 9
2
An equation that models the data is y = –x .
Graph each function. State the domain and range.
48. SOLUTION: Make a table of values.
x
0
0.5
1
1.5
2
2.5
3
f (x)
0
0
1
1
2
2
3
Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers.
The range is all integers.
49. f (x) =
SOLUTION: Make a table of values.
x
0
0.25
0.5
1
1.25
1.5
2
2x
0
0.5
1
2
2.5
3
4
f (x)
0
0
1
2
2
3
4
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Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers.
The range is all integers.
Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers.
Study
Guide
- Chapter 9
The
rangeand
is allReview
integers.
49. f (x) =
SOLUTION: Make a table of values.
x
0
0.25
0.5
1
1.25
1.5
2
2x
0
0.5
1
2
2.5
3
4
f (x)
0
0
1
2
2
3
4
Because the dots and circles overlap, the graph will cover all possible values of x, so the domain is all real numbers.
The range is all integers.
50. f (x) = |x|
SOLUTION: Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.
Make a table of values. Be sure to include the domain values for which the function changes.
x
f (x)
−2
2
−1
1
0
0
1
1
2
2
The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 8,
so the range is {y | y 0}.
51. f (x) = |2x − 2|
SOLUTION: eSolutions
Manual - Powered by Cognero
Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.
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The
graphand
willReview
cover all- possible
of x, so the domain is all real numbers. The graph will go no lower than y = 8,
Study
Guide
Chaptervalues
9
so the range is {y | y 0}.
51. f (x) = |2x − 2|
SOLUTION: Since f (x) cannot be negative, the minimum point of the graph is where f (x) = 0.
Make a table of values. Be sure to include the domain values for which the function changes.
x
f (x)
−1
4
0
2
1
0
2
2
3
4
The graph will cover all possible values of x, so the domain is all real numbers. The graph will go no lower than y = 0,
so the range is {y | y 0}.
52. SOLUTION: This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the
function changes.
x
0
1
2
−2
−1
f (x)
3
6
−4
−3
−2
Notice that both functions are linear.
The graph will cover all possible values of x, so the domain is all real numbers. The graph excluded y-values between
−1 and 3. Thus, the range is {y | y < −1 or y 3}.
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SOLUTION: This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the
The
graphand
willReview
cover all- possible
of x, so the domain is all real numbers. The graph excluded y-values between
Study
Guide
Chaptervalues
9
−1 and 3. Thus, the range is {y | y < −1 or y 3}.
53. SOLUTION: This is a piecewise-defined function. Make a table of values. Be sure to include the domain values for which the
function changes.
x
0
1
2
3
−1
f (x)
1
4
−5
−3
−1
Notice that both functions are linear.
The graph will cover all possible values of x, so the domain is all real numbers. The graph does not include y-values
between 1 and 3. Thus, the range is {y | y 1 or y > 3}.
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