pen tool mcgraw-hill ryerson pre-calculus 11 chapter 9 linear and quadratic inequalities click here...
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McGraw-Hill Ryerson
Pre-Calculus 11
Chapter 9
Linear and Quadratic Inequalities
Click here to begin the lesson
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McGraw-Hill Ryerson Pre-Calculus 11
Teacher Notes1. This lesson is designed to help students conceptualize the main ideas of Chapter 9.
2. To view the lesson, go to Slide Show > View Show (PowerPoint 2003).
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Linear InequalitiesThe graph of the linear equation x – y = –2 is referred to as a boundary line. This line divides the Cartesian plane into two regions:For one region, the condition x – y < –2 is true. For the other region, the condition x – y > –2 is true.
Chapter
9
Use the pen to label the conditions below to the corresponding parts of the graph on the Cartesian plane.
x – y < –2
x – y > –2
x – y = –2
x – y
= –2
x – y < –2
x – y > –2
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Linear Inequalities
Click here for the solution.
The ordered pair (x, y) is a solution to a linear inequality if its coordinates satisfy the condition expressed by the inequality.
Chapter
9Which of the following ordered pairs (x, y) are solutions of the linear inequality x – 4y < 4?Click on the ordered pairs to check your answer.
Use the pen tool to graph the boundary line and plot the points on the graph. Then, shade the region that represents the inequality.
32,
2
32,
2
3,2
2
0,0
0,4 0, 4
4,0 4,04 4x y
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Graphing Linear InequalitiesChapter
9 Match the inequality to the appropriate graph of a boundary line below.
Complete the graph of each inequality by shading the correct solution region.
Match Shade
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Click here for the solution.
Chapter
9a)
Graphing a Linear InequalityUse the pen tool to graph the following inequalities. Describe the steps required to graph the inequality.
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Chapter
9 Match each inequality to its graph. Then, click on the graph to check the answer.
Graphing a Linear Inequality
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Linear InequalitiesWrite an inequality that represents each graph.
Chapter
91. 2.
(2, 4)
(0, -2)0
(0, 3)
(2, -1)
0
3 2x y x y 2 3
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Solve an Inequality
Click here for the solution.
Paul is hosting a barbecue and has decided to budget $48 to purchase meat. Hamburger costs $5 per kilogram and chicken costs $6.50 per kilogram.
Chapter
9Let h = kg of hamburger c = kg of chicken
Write an inequality to represent the number of kilograms of each that Paul may purchase.
Write the equation of the boundary line below and draw its graph.
Shade the solution region for the inequality.
Ch
icken
Hamburger
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Click here for the solution.
Chapter
9
1. Can Paul buy 6 kg of hamburger and 4 kg chicken if he wants to stay within his set budget?
2. How many kilograms of chicken can Paul buy if he decides not to buy any hamburger?
3. If Paul buys 3 kg of hamburger, what is the greatest number of kilograms of chicken he can buy?
Solve an Inequality
Hamburger
Ch
icken
h
c
5 6.5 48h c
No
7.38 kg
5.08 kg
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Quadratic InequalitiesChapter
9 Solve x2 – x – 12 > 0. Use the pen tool.
Solve the related equation to determine intervals of numbers that may be solutions of the inequality.
Plot the solutions on a number line creating the intervals for investigation.Pick a number from each interval to test in the original inequality. If the number tested satisfies the inequality, then all of the numbers in that interval are solutions.
State the solution set.
Click here for the solution.
-1 0-2-3-4-5 1 2 3 4 5
-1 0-2-3-4-5 1 2 3 4 5
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Quadratic InequalitiesChapter
9 Solve x2 – x – 12 > 0. Use the pen tool.
Graph the corresponding quadratic function y = x2 – x – 12 to verify your solution from the previous page.
Click here for the solution.
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Sign AnalysisSolve x
2 – 3x – 4 > 0.
Chapter
9Use the pen tool to solve the related quadratic equation to obtain the boundary points for the intervals.
1.
Use the boundary points to mark off test intervals on the number line.
Determine the solution using the number line.
Determine the intervals when each of the factors is positive or negative.
x - 4
(x - 4)(x + 1)
x + 1
Click here for the solution.
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Sign AnalysisSolve x
2 – 3x – 4 > 0.
Chapter
91.Use the pen tool to create a
graph of the related function to confirm your solutions.
Click here for the solution.
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Graphing a Quadratic InequalityChoose the correct shaded region to complete the graph of the
inequality.Circle your choice using the pen tool.
Chapter
9
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Quadratic Inequality in Two Variables
Chapter
9
22y x
2 4 6y x x
2 6 5y x x
22y x 2y x
2 6 10y x x
Match each inequality to its graph using the pen tool. y x y x y x
y x x y x x y x x
2 2 2
2 2 2
2 , 2 ,
4 6, 6 10, 6 5
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The following pages contain solutions for the previous questions.
Click here to return to the start
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(0, 0)(-4, 0)
(0, 4)
(4, 0)
(0, -4)
Solutions
Go back to the question.
0
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Solutions
• Slope of the line is . and the y-intercept is the point (0, 1). • The inequality is less than.
Therefore, the boundary line is a broken line.
• Use a test point (0, 0). The point makes the inequality true.
• Therefore, shade below the line.
• The x-intercept is the point (–2, 0), the y-intercept is the point (0, –4).
• The inequality is greater than and equal to. Therefore, the boundary line is a solid line.
• Use a test point (0, 0). The point makes the inequality true.
• Therefore, shade above the line.
An example method for graphing an inequality would be:
Go back to the question.
1
3
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Let h = kg of hamburger c = kg of chicken
Write an inequality to represent the number of kilograms of each that Paul may purchase.
Graph the boundary line for the inequality.
Hamburger
Solutions
Go back to the question.
Ch
icke
n
c
h
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1. Can Paul buy 6 kg of hamburger and 4 kg chicken if he wants to stay within his set budget?
2. How many kilograms of chicken can Paul buy if he decides not to buy any hamburger?
3. If Paul buys 3 kg of hamburger, what is the greatest whole number of kilograms of chicken he can buy?
Hamburger
Ch
icken
(3, 5)(6, 4)
(0, 7.38)
The point (6, 4) is not within the shaded region. Paul could not purchase 6 kg of hamburger and 4 kg of chicken.
This is the point (0, 7.38). Buying no hamburger would be the y-intercept of the graph.
This would be the point (3, 5). Paul could buy 5 kg of chicken.
Solutions
Go back to the question.
h
c
5 6.5 48h c
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Solve the related equation to determine intervals of numbers that may be solutions of the inequality.
Plot the solutions on a number line creating the intervals for investigation.
Pick a number from each interval to test in the original inequality. If the number tested satisfies the inequality, then all of the numbers in that interval are solutions.
State the solution set.
Solutions
-1 0-2-3-4-5 1 2 3 4 5
Test point -5:
(-5)2 - (-5) - 12 > 0
True
Test point 0:
(0)2 - (0) - 12 > 0
False
Test point 5:
(5)2 - (5) - 12 > 0
True
The solution set is {x | x < –3 or x > 4, x R}.
Go back to the question.
-1 0-2-3-4-5 1 2 3 4 5
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x < –3 or x > 4
The inequality may have been solved by examining the graph of the corresponding function, y = x2 – x – 12. The quadratic inequality is greater than zero where the graph is above the x-axis.
Solutions
Go back to the question.
Solve x2 – x – 12 > 0
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Solve the related quadratic equation to obtain the boundary points for the intervals.
1.
Use the boundary points to mark off test intervals on the number line.
Determine the solution using the number line.
Determine the intervals when each of the factors is positive or negative.
x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0x – 4 = 0 or x + 1 = 0 x = 4 x = – 1
x – 4
(x – 4)(x + 1)
x + 1
––
– +
+ –
+
+
+
x < – 1 or x > 4
Solutions
Go back to the question.
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1.
A graph of the related function may be used to confirm your solutions.
x < – 1 or x > 4
Solutions
Go back to the question.