2-8 Graphing Linear and Absolute Value Inequalities
Graph each inequality.
2. SOLUTION: The boundary of the graph is the graph of x = –6. Since the inequality symbol is ≥, the boundary line is solid.
x is greater than or equal to –6, so the region greater than –6 is shaded.
4. SOLUTION: The boundary of the graph is the graph of 3x + y = –8. Since the inequality symbol is >, the boundary line is dashed.
Test the point (0, 0).
The region that contains (0, 0) is shaded.
Graph each inequality.
6. SOLUTION: The boundary of the graph is the graph of y = |x + 3|.
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This is the absolute value graph
translated 3 units to the left. Since the inequality symbol is ≥, the boundary line is solid.
2-8 Graphing Linear and Absolute Value Inequalities
Graph each inequality.
6. SOLUTION: The boundary of the graph is the graph of y = |x + 3|.
This is the absolute value graph
translated 3 units to the left. Since the inequality symbol is ≥, the boundary line is solid.
Test the point (0, 0).
The region that does not contain (0, 0) is shaded.
Graph each inequality.
8. SOLUTION: The boundary of the graph is the graph of x + 2y = 6. Since the inequality symbol is >, the boundary line is dashed.
Test the point (0, 0).
The region that does not contain (0, 0) is shaded.
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10. Page 2
2-8 Graphing
Linear and Absolute Value Inequalities
Graph each inequality.
8. SOLUTION: The boundary of the graph is the graph of x + 2y = 6. Since the inequality symbol is >, the boundary line is dashed.
Test the point (0, 0).
The region that does not contain (0, 0) is shaded.
10. SOLUTION: The boundary of the graph is the graph of y = 4. Since the inequality symbol is ≤, the boundary line is solid.
Test the point (0, 0).
The region that contains (0, 0) is shaded.
12. SOLUTION: The boundary of the graph is the graph of
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Test the point (0, 0).
. Since the inequality symbol is ≤, the boundary line is solid.
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2-8 Graphing Linear and Absolute Value Inequalities
12. SOLUTION: The boundary of the graph is the graph of
Test the point (0, 0).
. Since the inequality symbol is ≤, the boundary line is solid.
The region that does not contain (0, 0) is shaded.
14. COLLEGE April’s guidance counselor says that she needs a combined score of at least 1700 on her college
entrance exams to be eligible for the college of her choice. The highest possible score is 2400—1200 on the math
portion and 1200 on the verbal portion.
a. The inequality
represents this situation, where x is the verbal score and y is the math score. Graph
this inequality.
b. Refer to your graph. If she scores a 680 on the math portion of the test and 910 on the verbal portion of the test,
will April be eligible for the college of her choice?
SOLUTION: a. The boundary of the graph is the graph of
. Since the inequality symbol is ≥, the boundary line is solid.
Test the point (0, 0).
The region that does not contain (0, 0) is shaded.
Since the highest possible score in math portion is 1200, draw the inequality y ≤ 1200. Similarly the highest possible score in verbal portion is 1200, draw the inequality x ≤ 1200 in the same coordinate plane.
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2-8 Graphing Linear and Absolute Value Inequalities
14. COLLEGE April’s guidance counselor says that she needs a combined score of at least 1700 on her college
entrance exams to be eligible for the college of her choice. The highest possible score is 2400—1200 on the math
portion and 1200 on the verbal portion.
a. The inequality
represents this situation, where x is the verbal score and y is the math score. Graph
this inequality.
b. Refer to your graph. If she scores a 680 on the math portion of the test and 910 on the verbal portion of the test,
will April be eligible for the college of her choice?
SOLUTION: a. The boundary of the graph is the graph of
. Since the inequality symbol is ≥, the boundary line is solid.
Test the point (0, 0).
The region that does not contain (0, 0) is shaded.
Since the highest possible score in math portion is 1200, draw the inequality y ≤ 1200. Similarly the highest possible score in verbal portion is 1200, draw the inequality x ≤ 1200 in the same coordinate plane.
b. The ordered pair (680, 910) is not in the shaded region.
So, April is not eligible for the college of her choice.
Graph each inequality.
16. SOLUTION: The boundary of the graph is the graph of
. The equation can be written
.
This is a translation of the absolute value graph
right 2 units and down 4 units.
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Since the inequality symbol is ≤, the boundary line is solid.
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b. The ordered pair (680, 910) is not in the shaded region.
So, April isLinear
not eligible
the college
of her
choice.
2-8 Graphing
and for
Absolute
Value
Inequalities
Graph each inequality.
16. SOLUTION: The boundary of the graph is the graph of
. The equation can be written
.
This is a translation of the absolute value graph
right 2 units and down 4 units.
Since the inequality symbol is ≤, the boundary line is solid.
Test the point (0, 0).
The region that does not contain (0, 0) is shaded.
18. SOLUTION: The related equation can be written
.
The
stretches the absolute value graph horizontally. The 6 translates the graph 6 units to the left, and the –8 translates the graph 8 units down. The 2 stretches the graph vertically.
The negative sign reflects the absolute value graph
across the y-axis (which has no effect, since it is
symmetric). The 2 compresses the graph horizontally. The 6 translates the graph up 6 units.
The boundary of the graph is the graph of
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dashed.
. Since the inequality symbol is <, the boundary line is
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2-8 Graphing
Linear and Absolute Value Inequalities
18. SOLUTION: The related equation can be written
.
The
stretches the absolute value graph horizontally. The 6 translates the graph 6 units to the left, and the –8 translates the graph 8 units down. The 2 stretches the graph vertically.
The negative sign reflects the absolute value graph
across the y-axis (which has no effect, since it is
symmetric). The 2 compresses the graph horizontally. The 6 translates the graph up 6 units.
The boundary of the graph is the graph of
. Since the inequality symbol is <, the boundary line is
dashed.
Test the point (0, 0).
The region that contains (0, 0) is shaded.
20. SOLUTION: The related equation can be written
.
The –4 translates the graph of
4 units to the right. The 3 compresses the graph horizontally. The negative sign in front reflects the graph across the x-axis.
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The boundary of the graph is the graph of
. Since the inequality symbol is ≤, the boundary line is solid.
Test the point (0, 0).
2-8 Graphing Linear and Absolute Value Inequalities
20. SOLUTION: The related equation can be written
.
The –4 translates the graph of
4 units to the right. The 3 compresses the graph horizontally. The negative sign in front reflects the graph across the x-axis.
The boundary of the graph is the graph of
. Since the inequality symbol is ≤, the boundary line is solid.
Test the point (0, 0).
The region that contains (0, 0) is shaded.
Graph each inequality.
22. SOLUTION: The boundary of the graph is the graph of
.
The –6 translates the absolute value graph
units to the right. The –2 reflects the graph across the y-axis and
compresses it horizontally.
Since the inequality symbol is ≥, the boundary is solid.
Test the point (0, 0).
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The region that does not contain (0, 0) is shaded.
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2-8 Graphing Linear and Absolute Value Inequalities
Graph each inequality.
22. SOLUTION: The boundary of the graph is the graph of
.
The –6 translates the absolute value graph
units to the right. The –2 reflects the graph across the y-axis and
compresses it horizontally.
Since the inequality symbol is ≥, the boundary is solid.
Test the point (0, 0).
The region that does not contain (0, 0) is shaded.
s
24. SOLUTION: The boundary of the graph is the graph of
.
This is the graph of the absolute value function
translated left 4 units, up 3 units, vertically stretched, and reflected across the x-axis.
Since the inequality symbol is >, the boundary is dashed.
Test the point (0, 0).
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The region that contains (0, 0) is shaded.
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s
2-8 Graphing
Linear and Absolute Value Inequalities
24. SOLUTION: The boundary of the graph is the graph of
.
This is the graph of the absolute value function
translated left 4 units, up 3 units, vertically stretched, and reflected across the x-axis.
Since the inequality symbol is >, the boundary is dashed.
Test the point (0, 0).
The region that contains (0, 0) is shaded.
26. SOLUTION: Graph the inequality |x − y| > 5. Here, we want all ordered pairs for which the difference between x and y is greater
than 5.
28. CCSS MODELING Mei is making necklaces and bracelets to sell at a craft show. She has enough beads to make
50 pieces. Let x represent the number of bracelets and y represent the number of necklaces.
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a. Write an inequality that shows the possible number of necklaces and bracelets Mei can make.
b. Graph the inequality.
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2-8 Graphing Linear and Absolute Value Inequalities
28. CCSS MODELING Mei is making necklaces and bracelets to sell at a craft show. She has enough beads to make
50 pieces. Let x represent the number of bracelets and y represent the number of necklaces.
a. Write an inequality that shows the possible number of necklaces and bracelets Mei can make.
b. Graph the inequality.
c. Give three possible solutions for the number of necklaces and bracelets that can be made.
SOLUTION: a. The inequality that shows the possible number of necklaces and bracelets Mei can make is
.
b. The boundary of the graph is the graph of
. Since the inequality symbol is ≤, the boundary is solid.
Test the point (0, 0).
The region that contains (0, 0) is shaded.
c. Sample answers: 0 bracelets and 50 necklaces, 25 necklaces and 25 bracelets, or 30 bracelets and 20 necklaces.
Graph each inequality.
30. SOLUTION: Graph the inequality
.
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c. Sample Linear
answers:and
0 bracelets
and
50 necklaces,
25 necklaces and 25 bracelets, or 30 bracelets and 20 necklaces.
2-8 Graphing
Absolute
Value
Inequalities
Graph each inequality.
30. SOLUTION: Graph the inequality
.
32. SOLUTION: Graph the inequality
.
34. CHALLENGE Graph the following inequality.
SOLUTION: Draw each piece of the graph as a dashed line (since the main inequality is >). Shade the region above the piecewise-defined boundary.
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2-8 Graphing Linear and Absolute Value Inequalities
34. CHALLENGE Graph the following inequality.
SOLUTION: Draw each piece of the graph as a dashed line (since the main inequality is >). Shade the region above the piecewise-defined boundary.
36. REASONING When will it be possible to shade two different areas when graphing a linear absolute value
inequality? Explain your reasoning.
SOLUTION: Sample answer: It will be possible when we have a situation where y and x are both inside an absolute value. An
example is
When this happens, positive and negative values of y will need to be considered, as well as the positive and negative
values of x.
38. EXTENDED RESPONSE Craig scored 85%, 96%, 79%, and 81% on his first four math tests. He hopes to score
high enough on the final test to earn a 90% average. If the final test is worth twice as much as one of the other tests, determine if Craig can earn a 90% average. If so,
what score does Craig need to get on the final test to accomplish this? Explain how you found your answer.
SOLUTION: Sample answer: Craig can earn a 90% average by scoring a 99.5% or higher on his last test. This was determined
by finding the sum of the first 4 tests. By making each original test worth 100 points, then so far he has 341 out of
400 points. If the last test is worth twice as much, then it is worth 200, so overall he has 600 possible points. In order
to earn a 90%, he needs
or 540 points. So, Craig needs 540 – 341 or 199 out of 200 on the last test:
99.5%.
40. SHORT RESPONSE Which theorem of congruence should be used to prove
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by finding the sum of the first 4 tests. By making each original test worth 100 points, then so far he has 341 out of
400 points. If the last test is worth twice as much, then it is worth 200, so overall he has 600 possible points. In order
to earn a 90%, he needs
or 540 points. So, Craig needs 540 – 341 or 199 out of 200 on the last test:
99.5%. Linear and Absolute Value Inequalities
2-8 Graphing
40. SHORT RESPONSE Which theorem of congruence should be used to prove
SOLUTION: By SAS theorem of congruence we can prove
Write an equation for each graph.
42. SOLUTION: The given graph is a combination of transformations of the parent graph
.
When a constant k is added to or subtracted from a parent function, the result
is a translation of the graph up or down.
When a constant h is added to or subtracted from x before evaluating a parent function, the result,
translation left or right.
, is a
So, the equation of the graph is
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SOLUTION: Page 14
translation left or right.
So, the equation of the graph is
2-8 Graphing
Linear and Absolute Value Inequalities
44. SOLUTION: The given graph is a combination of transformations and reflection of the parent graph
When a parent function is multiplied by –1, the result
graph in the x-axis.
The graph was moved 4 units up and 2 units right.
So, the equation of the graph is
.
.
is a reflection of the
Graph each function.
46. SOLUTION: Graph each line on its domain. Use solid dots for endpoints where there is a ≤ or ≥ symbol, and an open circle where
there is a < or > symbol.
Write each equation in standard form. Identify A , B, and C.
48. –6y = 8x – 3
SOLUTION: Rewrite the equation in the standard form
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2-8 Graphing
Linear and Absolute Value Inequalities
Write each equation in standard form. Identify A , B, and C.
48. –6y = 8x – 3
SOLUTION: Rewrite the equation in the standard form
So, A = 8, B = 6 and C = 3.
50. SOLUTION: Rewrite the given equation in the standard form. So,
A=1
B=2
C = 11.
Multiply.
52. (3x – 4)(2x + 1)
SOLUTION: Use the FOIL method to multiply two binomials.
54. (5x + 2)(–2x + 3)
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Use the FOIL method to multiply two binomials.
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2-8 Graphing
Linear and Absolute Value Inequalities
54. (5x + 2)(–2x + 3)
SOLUTION: Use the FOIL method to multiply two binomials.
Graph each linear equation.
56. SOLUTION: Graph a line with slope
and y-intercept 2.
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