Download - Chapter 2 Linear Functions and Equations

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


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


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


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|


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|


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.



Two solutions |ax + b| = k, for k > 0



One solution |ax + b| = k, for k = 0



No solution |ax + b| = k, for k < 0



Let k be a positive number. Then

|ax + b| = k is equivalent to ax + b = ±k.


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


Solutions to y1 = y2 are –3.5 and –1.5.

Table Y1 = abs(2x + 5) and Y2 = 2



Symbolic: 2x 5 2

2x 5 2

2x 5 2

2x 3

x

3

2

2x 5 2

2x 7

x

7

2



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.



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.



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 ≥.


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


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 ≥.


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