Lesson 5 Menu : Lesson 5 Menu Five-Minute Check (over Lesson 6-4)
Main Ideas and Vocabulary
California Standards
Key Concept: Solving Absolute Value Equations
Example 1: Solve an Absolute Value Equation
Example 2: Write an Absolute Value Equation
Example 3: Graphing Absolute Value Functions
Lesson 5 MI/Vocab : Lesson 5 MI/Vocab absolute value
absolute value function
piece wise function Solve absolute value equations. Graph absolute value functions.
Lesson 5 CA : Lesson 5 CA Standard 3.0 Students solve equations and inequalities involving absolute values. (CAHSEE)
Lesson 5 KC1 : Lesson 5 KC1
Lesson 5 Ex1 : Lesson 5 Ex1 A. WEATHER The average January temperature in a northern Canadian city is 1 degree Fahrenheit. The actual January temperature for that city may be about 5 degrees Fahrenheit warmer or colder. Solve |t – 1| = 5 to find the range of temperatures. Method 1 Graphing |t – 1| = 5 means that the distance between t and 1 is 5 units. To find t on the number line, start at 1 and move 5 units in either direction.
The distance from 1 to 6 is 5 units.
The distance from 1 to –4 is 5 units.
The solution set is {–4, 6}. Solve an Absolute Value Equation
Lesson 5 Ex1 : Lesson 5 Ex1 Method 2 Compound Sentence Write |t –1| = 5 as t – 1 = 5 or t – 1 = –5. Answer: The solution set is {–4, 6}. The range of temperatures is –4°F to 6°F. Solve an Absolute Value Equation Case 1 Case 2 t – 1 = 5 t – 1 = –5 t – 1 + 1 = 5 + 1 Add 1 to each side. t – 1 + 1 = –5 + 1 Add 1 to each side. t = 6 Simplify. t = –4 Simplify.
Lesson 5 Ex1 : Lesson 5 Ex1 B. Solve |x – 1| = –7. Answer: |x – 1| = –7 means that the distance between x and 1 is –7. Since distance cannot be negative, the solution set is the empty set Ø. Solve an Absolute Value Equation
Slide 8 : A
B
C
D A. {–60, 60}
B. {0, 60}
C. {–45, 45}
D. {30, 60} A. WEATHER The average temperature for Columbus on Tuesday was 45ºF. The actual temperature for anytime during the day may have actually varied from the average temperature by 15ºF. Solve to find the range of temperatures.
Slide 9 : A
B
C
D B. Solve |x – 3| = –5.
Lesson 5 Ex2 : Lesson 5 Ex2 Write an Absolute Value Equation Write an open sentence involving absolute value for the graph. Find the point that is the same distance from –4 as the distance from 6. The midpoint between –4 and 6 is 1.
Lesson 5 Ex2 : Lesson 5 Ex2 Write an Absolute Value Equation The distance from 1 to –4 is 5 units. The distance from 1 to 6 is 5 units. Answer: |y – 1| = 5 So, an equation is |y – 1| = 5.
Slide 12 : A
B
C
D A. |x – 2| = 4
B. |x + 2| = 4
C. |x – 4| = 2
D. |x + 4| = 2 Write an equation involving the absolute value for the graph.
Lesson 5 Ex3 : Lesson 5 Ex3 Graphing Absolute Value Functions A. Graph f(x) = |x + 3|. First, find the minimum point on the graph. Since f(x) cannot be negative, the minimum point of the graph is where f(x) = 0. f(x) = |x + 3| Original function
0 = x + 3 Set f(x) = 0.
–3 = x Subtract 3 from each side.
The minimum point of the graph is at (–3, 0).
Lesson 5 Ex3 : Lesson 5 Ex3 Graphing Absolute Value Functions Next fill out a table of values. Include values for x > –3 and x < –3. Answer:
Lesson 5 Ex3 : Lesson 5 Ex3 Graphing Absolute Value Functions B. Graph f(x) = |x – 5|. f(x) = |x – 5| Original function
0 = x – 5 Set f(x) = 0.
5 = x Add 5 to each side.
The minimum point of the graph is at (5, 0). Answer:
Slide 16 : A
B
C
D A. Graph f(x) = |x + 2|.
Slide 17 : A
B
C
D B. Graph f(x) = |x – 3|.
End of Lesson 4 : End of Lesson 4