Algebra II - Study Guide 1st Semester
Matching
Match each vocabulary term with its definition.
a. linear system
b. system of equations
c. consistent system
d. dependent system
e. inconsistent system
f. independent system
g. system of linear inequalities
h. complex system
i. standard system
____
____
____
____
____
____
____
1.
2.
3.
4.
5.
6.
7.
a system of equations containing only linear equations
a system of equations or inequalities that has no solution
a system of equations that has infinitely many solutions
a set of two or more equations containing two or more variables
a system of inequalities in two or more variables in which all of the inequalities are linear
a system of equations that has exactly one solution
a system of equations or inequalities that has at least one solution
Match each vocabulary term with its definition.
a. parabola
b. quadratic function
c. axis of symmetry
d. vertex of a parabola
e. maximum value
f. standard form
g. vertex form
____
8. a quadratic function written in the form
vertex
, where a, h, and k are constants and (h, k) is the
____ 9.
, where a, b, and c are real numbers and
____ 10. the shape of the graph of a quadratic function
____ 11. the lowest or highest point in a parabola
____ 12. a function that can be written in the form
in the form
, where a, b, and c are real numbers and
, where a, h, and k are real numbers and
Match each vocabulary term with its definition.
a. maximum value
b. minimum value
c. monomial
d. vertex
e. binomial
f. root of an equation
g. trinomial
1
, or
____
____
____
____
____
13.
14.
15.
16.
17.
a polynomial with three terms
any value of the variable that makes the equation true
the y-value of the vertex when a parabola opens downward
the y-value of the vertex when a parabola opens upward
a polynomial with two terms
Short Answer
5
18. Order the numbers 2,  7 ,
, –0.6448,
from least to greatest.
2
19. Find the additive and multiplicative inverse of  3 .
20. Estimate
to the nearest tenth.
21. Simplify the expression
22. Simplify
.
by rationalizing the denominator.
23. Add.
24. Evaluate the expression g + s for g = 9 and s = 3.
25. Simplify the expression
26. For
.
, evaluate
27. Graph the function
.
.
28. Dan paid a total of $25.80 last month for his international calls. He makes international calls only to England.
Dan pays $0.06 per minute in addition to $10.98 fixed monthly payment. How many minutes of international
calls did Dan make last month?
29. Solve
.
30. Solve 3n – 24 = 14 – 30n.
1
31. Graph the line with slope  3 that passes through (–6, –4).
32. Find the intercepts of
, and graph the line.
33. Write the function
34. Graph the inequality
in slope-intercept form. Then graph the function.
.
35. Estimate the value of r for the scatter plot.
2
y
12
10
8
6
4
2
–10 –8
–6
–4
–2
–2
2
4
6
8
10
x
–4
–6
–8
–10
–12
36. Use substitution to determine if (0, 2) is an element of the solution set for the system of equations.
37. Use a graph to solve the system
38. Classify the system
. Check your answer.
, and determine the number of solutions.
39. Jake fills a tank that can hold 200 gallons of water. The tank already has 50 gallons of water in it when Jake
starts filling it at the rate of 10 gallons per minute. Karla fills a tank that can hold 300 gallons of water. That
tank already has 100 gallons of water in it when Karla starts filling it at the rate of 5 gallons per minute. Jake
and Karla start filling the tanks at the same time. How long after they start filling the tanks do the tanks have
the same volume of water? What is that volume of water?
40. Use substitution to solve the system
.
41. Use elimination to solve the system
.
42. Graph the system of inequalities
.
43. Graph the system of inequalities, and classify the figure created by the solution region.
3
44. Graph (–2, 3, 1) in three-dimensional space.
45. Graph the linear equation
in three-dimensional space.
46. A basketball player gets 1 point for each free throw, 2 points for each shot inside the 3-point line, and 3 points
for each shot from outside the 3-point line. A player scores 30 points in a game. Write a linear equation in
three variables to represent this situation. If the player made 3 free throws and 6 shots inside the 3-point line,
how many shots did she make from outside the 3-point line?
47. Use elimination to solve the system of equations
.
48. A teacher prepares 3 different tests. The teacher uses 3 types of questions which are each worth a certain
number of points. The table shows the number of questions of each type on each of the three tests. Find the
number of points each type of question is worth.
Test 1
Test 2
Test 3
Question
Type A
36
3
0
Question
Type B
3
45
1
Question
Type C
0
1
39
Total
Points
150
103
41
49. Three venture capitalists each invested 7 million dollars into three companies: Darnell’s Services, Stochy’s,
and Kammy’s Clothing. Each venture capitalist divided the money differently, as shown in the table. The
table also shows the gain for each venture capitalist for the year. Find the yield per year of each company.
Darnell’s
Services
2 million
4 million
1 million
Investor
B. O’Brian
L. Pham
R. Jackson
50. Graph
Stochy’s
3 million
1 million
2 million
Kammy’s
Clothing
2 million
2 million
4 million
Gain
35 million
31 million
41 million
by using a table.
51. Using the graph of
as a guide, describe the transformations, and then graph the function
.
52. Using the graph of
.
as a guide, describe the transformations, and then graph the function
53. The parent function
is reflected across the x-axis, vertically stretched by a factor of 10, and
translated right 10 units to create g. Use the description to write the quadratic function in vertex form.
4
54. Consider the function
. Determine whether the graph opens up or down. Find the axis of
symmetry, the vertex and the y-intercept. Graph the function.
5
Algebra II - Study Guide 1st Semester
Answer Section
MATCHING
1. ANS:
TOP:
2. ANS:
TOP:
3. ANS:
TOP:
4. ANS:
TOP:
5. ANS:
TOP:
6. ANS:
TOP:
7. ANS:
TOP:
A
PTS: 1
DIF: Basic
3-1 Using Graphs and Tables to Solve Linear Systems
E
PTS: 1
DIF: Basic
3-1 Using Graphs and Tables to Solve Linear Systems
D
PTS: 1
DIF: Basic
3-1 Using Graphs and Tables to Solve Linear Systems
B
PTS: 1
DIF: Basic
3-1 Using Graphs and Tables to Solve Linear Systems
G
PTS: 1
DIF: Basic
3-3 Solving Systems of Linear Inequalities
F
PTS: 1
DIF: Basic
3-1 Using Graphs and Tables to Solve Linear Systems
C
PTS: 1
DIF: Basic
3-1 Using Graphs and Tables to Solve Linear Systems
REF: Page 182
8. ANS:
TOP:
9. ANS:
TOP:
10. ANS:
TOP:
11. ANS:
TOP:
12. ANS:
TOP:
G
PTS: 1
DIF: Basic
5-1 Using Transformations to Graph Quadratic Functions
F
PTS: 1
DIF: Basic
5-2 Properties of Quadratic Functions in Standard Form
A
PTS: 1
DIF: Basic
5-1 Using Transformations to Graph Quadratic Functions
D
PTS: 1
DIF: Basic
5-1 Using Transformations to Graph Quadratic Functions
B
PTS: 1
DIF: Basic
5-1 Using Transformations to Graph Quadratic Functions
REF: Page318
13. ANS:
TOP:
14. ANS:
TOP:
15. ANS:
TOP:
16. ANS:
TOP:
17. ANS:
TOP:
G
PTS: 1
DIF: Basic
REF:
5-3 Solving Quadratic Equations by Graphing and Factoring
F
PTS: 1
DIF: Basic
REF:
5-3 Solving Quadratic Equations by Graphing and Factoring
A
PTS: 1
DIF: Basic
REF:
5-2 Properties of Quadratic Functions in Standard Form
B
PTS: 1
DIF: Basic
REF:
5-2 Properties of Quadratic Functions in Standard Form
E
PTS: 1
DIF: Basic
REF:
5-3 Solving Quadratic Equations by Graphing and Factoring
REF: Page 18
REF: Page 184
REF: Page 182
REF: Page 199
REF: Page 184
REF: Page 183
REF: Page 324
REF: Page 315
REF: Page 318
REF: Page 315
Page 336
Page 334
Page 326
Page 326
Page 336
SHORT ANSWER
18. ANS:
 57 , –0.6448,
, 2,
Convert each number to a decimal to make it easier to compare the numbers:
6
 57 = 0.714285
= 1.4142
= 3.1416
Then use < to compare the numbers.
0.714285 < –0.6448 < 1.4142 < 2 < 3.1416
PTS: 1
DIF: Basic
REF: Page 6
OBJ: 1-1.1 Ordering and Classifying Real Numbers
TOP: 1-1 Sets of Numbers
19. ANS:
additive inverse:
2
3
NAT: 12.1.1.d
;
3
multiplicative inverse:  2
2
2
Since  3 + ( 3 ) 0,
2
3
2
2
3
3
is the additive inverse, or opposite, of  3 . Since  3 (  2 ) 1,  2 is the
2
multiplicative inverse, or reciprocal, of  3 .
PTS: 1
DIF: Basic
REF: Page 14
OBJ: 1-2.1 Finding Inverses
NAT: 12.1.5.e
TOP: 1-2 Properties of Real Numbers
20. ANS:
6.6
36 < 43 < 49
43 lies between the perfect squares 36 and 49.
6<
<7
So
lies between the square roots of 36 and 49.
Square
tenths
between these two integers to find the value closest to 43.
 6.6
will be closer to the square root of the perfect square closest to 43.
PTS: 1
NAT: 12.1.2.d
21. ANS:
5
7
DIF: Average
REF: Page 21
TOP: 1-3 Square Roots
PTS: 1
DIF: Basic
REF: Page 22
OBJ: 1-3.2 Simplifying Square-Root Expressions
TOP: 1-3 Square Roots
22. ANS:
OBJ: 1-3.1 Estimating Square Roots
NAT: 12.5.3.c
Multiply by a form of 1 to get a perfect-square radicand in
the denominator.
Simplify the denominator.
7
PTS: 1
DIF: Average
REF: Page 23
OBJ: 1-3.3 Rationalizing the Denominator
TOP: 1-3 Square Roots
23. ANS:
=
NAT: 12.5.3.c
Combine the like radicals.
PTS: 1
DIF: Basic
REF: Page 23
OBJ: 1-3.4 Adding and Subtracting Square Roots
TOP: 1-3 Square Roots
24. ANS:
12
g+s
=9+3
Substitute 9 for g and 3 for s.
= 12
Simplify.
PTS: 1
DIF: Basic
REF: Page 28
OBJ: 1-4.2 Evaluating Algebraic Expressions
TOP: 1-4 Simplifying Algebraic Expressions
25. ANS:
NAT: 12.5.3.c
NAT: 12.5.2.b
KEY: expression | evaluate
Identify like terms.
Combine like terms.
PTS: 1
NAT: 12.5.3.c
26. ANS:
–27
DIF: Basic
REF: Page 28
OBJ: 1-4.3 Simplifying Expressions
TOP: 1-4 Simplifying Algebraic Expressions
Substitute 5 for x.
Simplify.
PTS: 1
NAT: 12.5.2.b
27. ANS:
DIF: Basic
REF: Page 51
TOP: 1-7 Function Notation
8
OBJ: 1-7.1 Evaluating Functions
KEY: function | input | output | evaluate
y
6
5
4
3
2
1
–6 –5 –4 –3 –2 –1–1
1
2
3
4
5
6
x
–2
–3
–4
–5
–6
Step 1: Make a table.
Step 2: Graph the points.
Step 3: Connect the points with a line.
x
0
1
2
y
6
1
2
3
4
5
4
3
2
1
–6 –5 –4 –3 –2 –1
–1
1
2
3
4
5
6
x
–2
–3
–4
–5
–6
PTS: 1
DIF: Average
REF: Page 52
OBJ: 1-7.2 Graphing Functions
NAT: 12.5.2.b
TOP: 1-7 Function Notation
KEY: graph | function
28. ANS:
247 minutes
Let x represent the number of international call minutes Dan made last month.
number of
fixed monthly
cost per
total monthly
plus
times
international
=
payment
minute
payment
call minutes
10.98
+
0.06
•
x
=
25.80
Solve
PTS: 1
NAT: 12.5.4.a
.
Subtract 10.98 from both sides of the equation.
Divide both sides by 0.06 to find x.
DIF: Average
REF: Page 91
OBJ: 2-1.1 Application
TOP: 2-1 Solving Linear Equations and Inequalities
9
29. ANS:
= –1
Distribute –4.
Add
to both sides.
Divide by
.
PTS: 1
DIF: Basic
REF: Page 91
OBJ: 2-1.2 Solving Equations with the Distributive Property
TOP: 2-1 Solving Linear Equations and Inequalities
30. ANS:
5
n = 1 33
NAT: 12.5.4.a
First, collect all variable terms on one side and all constant terms on the other side. Then, isolate the variable.
PTS:
OBJ:
TOP:
KEY:
31. ANS:
1
DIF: Average
REF: Page 92
2-1.3 Solving Equations with Variables on Both Sides
NAT: 12.5.4.a
2-1 Solving Linear Equations and Inequalities
addition | division | multiplication | multi-step equations | solving | subtraction
y
15
12
9
6
3
–15 –12 –9 –6 –3
–3
3
6
9
12
15 x
–6
–9
–12
–15
1
First, plot the point (–6, –4). The slope  3 indicates a rise of –1 and a run of 3. Move 1 unit down and 3 unit
to the right. Repeat. Then, draw a line through the points.
PTS: 1
DIF: Basic
REF: Page 106
OBJ: 2-3.2 Graphing Lines Using Slope and a Point
NAT: 12.5.4.c
TOP: 2-3 Graphing Linear Functions
KEY: coordinate plane | graph | point | slope
32. ANS:
x-intercept: –2, y-intercept: –2
10
y
10
8
6
4
2
–10 –8 –6 –4 –2
–2
2
4
6
8
10 x
–4
–6
–8
–10
Find the x-intercept:
Substitute 0 for y.
The x-intercept is –2.
Find the y-intercept:
Substitute 0 for x.
The y-intercept is –2.
PTS: 1
DIF: Average
REF: Page 106
OBJ: 2-3.3 Graphing Lines Using the Intercepts
NAT: 12.5.4.c
TOP: 2-3 Graphing Linear Functions
KEY: linear equation | solving | x-intercept | y-intercept
33. ANS:
y
10
8
6
4
2
–10 –8 –6 –4 –2
–2
2
4
6
8
10 x
–4
–6
–8
–10
Solve for y first in
slope:
.
, y-intercept:
11
Plot the point
points.
. Then move down 1 and right 2 to find another point. Draw a line connecting the two
y
10
8
6
4
2
–10 –8
–6
–4
–2
–2
2 4 6
(0, –2)
(2, –3)
–4
8
10
x
–6
–8
–10
PTS: 1
DIF: Average
REF: Page 107
OBJ: 2-3.4 Graphing Functions in Slope-Intercept Form
TOP: 2-3 Graphing Linear Functions
34. ANS:
NAT: 12.5.4.c
y
5
4
3
2
1
–5 –4 –3 –2 –1
–1
1
2
3
4
5
x
–2
–3
–4
–5
The boundary line is
, which has a y-intercept of –2 and a slope of
. Draw the boundary line
dashed because it is not part of the solution. Then shade the region above the boundary line to show
.
PTS:
NAT:
35. ANS:
1.00
If
If
If
1
12.5.4.d
DIF: Basic
REF: Page 124
OBJ: 2-5.1 Graphing Linear Inequalities
TOP: 2-5 Linear Inequalities in Two Variables
, the data correlates well with a line of best fit with positive slope.
, the data set shows little or no correlation with a line of best fit.
, the data correlates well with a line of best fit with negative slope.
12
In this case, the slope of the line of best fit is positive. Moreover, the data set correlates well with this line.
Thus, of the choices available the best estimate for r is 1.00.
PTS: 1
DIF: Advanced
NAT: 12.4.1.c
TOP: 2-7 Curve Fitting by Using Linear Models
36. ANS:
(0, 2) is not a solution of the system.
The point (0, 2) is not a solution of both equations.
PTS: 1
DIF: Basic
REF: Page 182
OBJ: 3-1.1 Verifying Solutions of Linear Systems
TOP: 3-1 Using Graphs and Tables to Solve Linear Systems
37. ANS:
NAT: 12.5.4.g
y
5
4
3
2
1
–5 –4 –3 –2 –1
–1
1
2
3
4
5
x
–2
–3
–4
–5
The solution to the system is (2, 4).
Solve each equation for y.
Then graph each equation.The lines appear to intersect at the point (2, 4). Check by substituting the x- and yvalues into each equation.
PTS: 1
DIF: Average
REF: Page 183
OBJ: 3-1.2 Solving Linear Systems by Using Graphs and Tables
NAT: 12.5.4.g
TOP: 3-1 Using Graphs and Tables to Solve Linear Systems
38. ANS:
This system is consistent. It has infinitely many solutions.
Write both equations in slope-intercept form.
13
7
y= 6x–
2
3
7
y= 6x–
2
3
These are the same line because they have the same slope and the same y-intercept.
PTS: 1
DIF: Average
REF: Page 184
OBJ: 3-1.3 Classifying Linear Systems
NAT: 12.5.4.g
TOP: 3-1 Using Graphs and Tables to Solve Linear Systems
KEY: classifying | systems | consistent | inconsistent
39. ANS:
10 minutes; 150 gallons
Step 1: Write two equations.
Let V1 represent the volume of water in the tank Jake fills, and let V2 represent the volume of water in the tank
Karla fills. Let t represent the time after they start filling the tanks.
Step2: Solve for the value of t when
.
Step3: Find the volume of water in both tanks when t = 10.
PTS: 1
DIF: Advanced
NAT: 12.5.4.g
TOP: 3-1 Using Graphs and Tables to Solve Linear Systems
KEY: multi-step
40. ANS:
( 2 , 3 )
The second equation is solved for y.
Step 1
Step 2
Substitute
for y in the first equation.
Simplify and solve for x.
Step 3
Divide both sides by 4.
x = 2
y = 2
3
Write one of the original equations.
Substitute 2 for x.
Find the value of y.
( 2 , 3 )
Write the solution as an ordered pair.
Step 4
PTS: 1
DIF: Basic
REF: Page 190
OBJ: 3-2.1 Solving Linear Systems by Substitution
TOP: 3-2 Using Algebraic Methods to Solve Linear Systems
14
NAT: 12.5.4.g
41. ANS:
(4, 1)
Step 1
Step 2
PTS:
OBJ:
TOP:
KEY:
42. ANS:
3x
x
4x
x
– 3y = 9
+ 3y = 7
= 16
= 4
The y-terms have opposite coefficients.
Add the equations to eliminate the y terms.
3(4) – 3y = 9
12 – 3y = 9
– 3y = –3
y =1
Substitute for x in one of the original equations.
Simplify and solve for y.
(4, 1)
Write the solution as an ordered pair.
1
DIF: Basic
REF: Page 191
3-2.2 Solving Linear Systems by Elimination
NAT: 12.5.4.g
3-2 Using Algebraic Methods to Solve Linear Systems
linear equations | system of equations | solving | elimination
y
5
4
3
2
1
–5 –4 –3 –2 –1
–1
1
2
3
4
5
x
–2
–3
–4
–5
Graph
and
on the same coordinate plane. The solutions of the system are the
overlapping shaded regions, including the solid boundary line.
PTS: 1
DIF: Basic
REF: Page 199
OBJ: 3-3.1 Graphing Systems of Inequalities
TOP: 3-3 Solving Systems of Linear Inequalities
43. ANS:
15
NAT: 12.5.4.g
y
4
3
2
1
–4
–3
–2
–1
1
2
3
4
x
–1
–2
–3
–4
The shaded region is a rectangle.
Graph the solid boundary lines
and
Graph the solid boundary lines
and
The solution region is a four-sided figure, or quadrilateral.
, and shade below them.
, and shade above them.
The boundary lines
and
have the same slope and are parallel.
The boundary lines
and
have the same slope and are parallel.
The slope of
and
and the slope of
and
are opposite
reciprocals. Thus the two sets of boundary lines are perpendicular.
The solution region is a rectangle.
PTS: 1
NAT: 12.5.4.g
44. ANS:
DIF: Average
REF: Page 201
OBJ: 3-3.3 Application
TOP: 3-3 Solving Systems of Linear Inequalities
Start from the origin. Move 2 units back along the x-axis, 3 units right, and 1 unit up.
PTS: 1
DIF: Basic
REF: Page 214
OBJ: 3-5.1 Graphing Points in Three Dimensions
TOP: 3-5 Linear Equations in Three Dimensions
45. ANS:
16
>
z
(0, 0, 3)
(0, 6, 0)
y
<
(5, 0, 0)
x
Step 1 Find the intercepts.
x-intercept:
y-intercept:
z-intercept:
Step 2 Plot the points (5, 0, 0), (0,
, 0), and (0, 0, 3). Sketch a plane through the three points.
PTS: 1
DIF: Average
REF: Page 215
OBJ: 3-5.2 Graphing Linear Equations in Three Dimensions
TOP: 3-5 Linear Equations in Three Dimensions
46. ANS:
;
5 shots from outside the 3-point line
Step 1 Define the variables.
Let x = the number of free throws, y = the number of shots inside the 3-point line, and z = the number of shots
from outside the 3-point line.
Step 2 Write an equation.
points from free throws + points inside the 3-point line + points outside the 3-point line = 30.
Step 3 Substitute the values for x and y to find the value for z.
She made 3 free throws and 6 shots inside the 3-point line, so x = 3 and y = 6.
PTS: 1
DIF: Average
REF: Page 215
TOP: 3-5 Linear Equations in Three Dimensions
17
OBJ: 3-5.3 Application
47. ANS:
(10, 12, 3)
Eliminate z by adding equations 1 and 3 to create equation 4.
Eliminate z by multiplying equation 3 by 4 and adding it to equation 2 to create equation 5.
Eliminating z creates a 2-by-2 system.
Eliminate y by multiplying the fourth equation by –11 and adding it to the fifth equation..
Solve for x.
x = 10
Use this value to find y in equation 4.
Substitute 10 for x.
Subtract 50 from both sides.
Find z by substituting x and y in one of the original equations.
Substitute 10 for x and 12 for y.
Combine like terms.
Divide both sides by 4.
PTS: 1
DIF: Average
REF: Page 221
OBJ: 3-6.1 Solving a Linear System in Three Variables
TOP: 3-6 Solving Linear Systems in Three Variables
48. ANS:
Question type A is worth 4 points, type B is worth 2 points, and type C is worth 1 point.
Step 1 Let x represent the number of points for a question type A, y for a question type B, and z for a question
type C.
(1)
Test 1 points
(2)
Test 2 points
(3)
Test 3 points
18
Step 2 Use substitution. Solve for y in equation (1).
(1)
Solve for y.
y
Step 3 Substitute for y in equation (2) and (3).
(2)
Substitute
(3)
(4)
(5)
Step 4 Solve equation (4) for z.
(4)
z
Step 5 Substitute for z in equation (5).
(5)
.
Simplify to find a 2-by-2 system.
Solve for z.
Substitute
Solve for x.
for z.
Step 6 Substitute for x to solve for z and then for y.
(4)
(3)
The solution to the system is (4, 2, 1). So, type A question is worth 4 points, type B is worth 2 points, and
type C is worth 1 point.
PTS: 1
DIF: Average
REF: Page 222
OBJ: 3-6.2 Application
TOP: 3-6 Solving Linear Systems in Three Variables
49. ANS:
Darnell’s Services yielded 3 million dollars.
Stochy’s yielded 5 million dollars.
Kammy’s Clothing yielded 7 million dollars.
Step 1 Create a system of equations.
Let x represent the yield from Darnell’s Service, y represent the yield from Stochy’s, and z represent the yield
from Kammy’s Clothing.
Step 2 Solve the system of equations.
Solve for y.
Multiply the first equation by
.
Add the first and second equations together.
19
Multiply the third equation by -4.
Add the second and third equations together.
Create a system of two equations in terms of y and z.
Multiply the first equation by
.
Add the two equations together.
Divide both sides by .
Solve for z.
Substitute y = 5 into either of the two-variable
equations and solve for z.
Solve for x.
Substitute
and
into any of the threevariable equations.
Subtract 29 from both sides.
Divide both sides by 2.
(3, 5, 7) is the solution and Darnell’s Services yielded 3 million dollars, Stochy’s yielded 5 million dollars and
Kammy’s Clothing yielded 7 million dollars.
PTS: 1
DIF: Advanced
NAT: 12.5.4.g
TOP: 3-6 Solving Linear Systems in Three Variables
50. ANS:
y
10
8
6
4
2
–10 –8 –6 –4 –2
–2
2
4
6
8
10 x
–4
–6
–8
–10
20
Make a table.
Plot the ordered pairs and connect with a
smooth curve.
x
y
(x, f(x))
10
–2
(–2, 24)
–1
(–1, 16)
6
0
(0, 10)
4
1
(1, 6)
2
2
(2, 4)
8
–10 –8
–6
–4
–2
–2
2
4
6
8
10
x
–4
–6
–8
–10
PTS: 1
DIF: Average
REF: Page 315
OBJ: 5-1.1 Graphing Quadratic Functions Using a Table
NAT: 12.5.4.c
TOP: 5-1 Using Transformations to Graph Quadratic Functions
51. ANS:
is
translated 6 units left and 2 units down.
y
10
8
6
4
2
–10 –8
–6
–4
g(x)
–2
–2
f(x)
2
4
6
8
10
x
–4
–6
–8
–10
Because
Therefore,
, the graph is translated 6 units left. Because
is
translated 6 units left and 2 units down.
21
, the graph is translated 2 units down.
y
10
8
6
4
f(x)
2
–10 –8
–6
–4
g(x)
–2
–2
2
4
6
8
10
x
–4
–6
–8
–10
PTS: 1
DIF: Average
REF: Page 316
OBJ: 5-1.2 Translating Quadratic Functions
NAT: 12.5.2.d
TOP: 5-1 Using Transformations to Graph Quadratic Functions
52. ANS:
A reflection across the x-axis and a vertical stretch by a factor of 8.
y
4
f(x)
–4
4
x
–4
Because –8 is negative, g is a reflection of f across the x-axis.
Because
, g is a vertical stretch of f by a factor of 8.
PTS: 1
DIF: Basic
REF: Page 317
OBJ: 5-1.3 Reflecting , Stretching , and Compressing Quadratic Functions
NAT: 12.5.2.d
TOP: 5-1 Using Transformations to Graph Quadratic Functions
53. ANS:
Identify how each transformation affects the coefficients in vertex form.
a is negative.
Write the transformed function, using the vertex form g(x) = a(x – h)2 + k.
22
Substitute –10 for a, 10 for h, and 0 for k.
Simplify.
PTS: 1
DIF: Average
REF: Page 318
OBJ: 5-1.4 Writing Transformed Quadratic Functions
NAT: 12.5.2.d
TOP: 5-1 Using Transformations to Graph Quadratic Functions
54. ANS:
The parabola opens downward.
The axis of symmetry is the line
.
The vertex is the point
.
The y-intercept is 10.
y
18
12
6
–12
–6
6
12
x
–6
Because a is
, the graph opens downward.
The axis of symmetry is given by
.
is the axis of symmetry.
The vertex lies on the axis of symmetry, so
.
The y-value is the value of the function at this x-value.
The vertex is
.
Because the last term is 10, the y-intercept is 10.
PTS: 1
DIF: Average
REF: Page 324
OBJ: 5-2.2 Graphing Quadratic Functions in Standard Form
TOP: 5-2 Properties of Quadratic Functions in Standard Form
23
NAT: 12.5.4.c