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Chapter 2 Linear Functions and Equations. Absolute Value Equations and Inequalities. 2.5. Evaluate and graph the absolute value function Solve absolute value equations Solve absolute value inequalities. Absolute Value Function. The graph of y = | x |. V-shaped - PowerPoint PPT Presentation
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Copyright © 2014, 2010, 2006 Pearson Education, Inc. 1
Chapter 2
Linear Functions and
Equations
2Copyright © 2014, 2010, 2006 Pearson Education, Inc.
Absolute Value Equations and
Inequalities
♦ Evaluate and graph the absolute value function♦ Solve absolute value equations♦ Solve absolute value inequalities
2.5
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 3
Absolute Value Function
The graph of y = |x|.
V-shapedCannot be represented by single linear function
x
x if x 0
x if x 0
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 4
Absolute Value FunctionAlternate Formula
That is, regardless of whether a real number x is positive or negative, the expression equals the absolute value of x.
Examples:
x2 x for all real numbers x
x2
y 2 y x 1 2 x 1
2x 2 2x
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 5
For the linear function f, graph y = f (x) and y = |f (x)| separately. Discuss how the absolute value affects the graph of f.
f(x) = –2x + 4
(For continuity of the solution, it appears completely on the next slide.)
Example: Analyzing the graph of y = |ax + b|
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 6
The graph of y = |–2x + 4| is a reflection of f across the x-axis when y = –2x + 4 is below the x-axis.
Example: Analyzing the graph of y = |ax + b|
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 7
Absolute Value Equations
Solutions to |x| = k with k > 0 are given byx = ±k.
Solutions to |ax + b| = k are given byax + b = ±k.
These concepts can be illustrated visually.
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 8
Absolute Value Equations
Two solutions |ax + b| = k, for k > 0
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Absolute Value Equations
One solution |ax + b| = k, for k = 0
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Absolute Value Equations
No solution |ax + b| = k, for k < 0
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 11
Absolute Value Equations
Let k be a positive number. Then
|ax + b| = k is equivalent to ax + b = ±k.
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 12
Solve the equation |2x + 5| = 2 graphically, numerically, and symbolically.
Graph Y1 = abs(2X + 5) and Y2 = 2Solution
Solutions: –3.5, –1.5
Example: Solving an equation with technology
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 13
Solutions to y1 = y2 are –3.5 and –1.5.
Table Y1 = abs(2x + 5) and Y2 = 2
Example: Solving an equation with technology
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Symbolic: 2x 5 2
2x 5 2
2x 5 2
2x 3
x
3
2
2x 5 2
2x 7
x
7
2
Example: Solving an equation with technology
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 15
Absolute Value Inequalities
Solutions |ax + b| = k labeled s1 and s2 and the
graph of y = |ax + b| is below y = k between s1 and
s2 or when s1 < x < s2. Solution to |ax + b| < k is in
green.
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 16
Absolute Value Inequalities
Solutions |ax + b| = k labeled s1 and s2 and the
graph of y = |ax + b| is above y = k to left of s1 and
right of s2 or x < s1 or x >s2. Solution to |ax + b| > k
is in green.
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 17
Absolute Value Inequalities
Let solutions to |ax + b| = k be s1 and s2, where s1
< s2 and k > 0.
1. |ax + b| < k is equivalent to s1 < x < s2.
2. |ax + b| > k is equivalent to x < s1 or
x > s2.
Similar statements can be made for inequalities involving ≤ or ≥.
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 18
Solve the inequality |2x – 5| ≤ 6. Write the solution set in interval notation.
Solve |2x – 5| = 6 or 2x – 5 = ±6 Solution
2x 5 6 or 2x 5 6
2x 11
x
11
2
x 1
x
1
2
Solution set:
1
2x
11
2, or
1
2,11
2
Example: Solving inequalities involving absolute values symbolically
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Absolute Value InequalitiesAlternative Method
Let k be a positive number.
1. |ax + b| < k is equivalent to –k < ax + b < k.
2. |ax + b| > k is equivalent to ax + b < –k or ax + b > –k.
Similar statements can be made for inequalities involving ≤ or ≥.
Copyright © 2014, 2010, 2006 Pearson Education, Inc. 20
Solve the inequality |4 – 5x | ≤ 3. Write your answer in interval notation.
|4 – 5x| ≤ 3 is equivalent to the three-part inequality
Solution
3 4 5x 3
7 5x 1
7
5x
1
5
In interval notation, solution is .
1
5,7
5
Example: Using an alternative method