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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
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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
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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
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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|
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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|
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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.
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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
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Absolute Value Equations
Let k be a positive number. Then
|ax + b| = k is equivalent to ax + b = ±k.
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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
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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
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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.
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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.
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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 ≥.
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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 ≥.
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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