Chapter 2 Linear Functions and Equations

1. Chapter 2 Linear Functions and Equations

2. Absolute Value Equations and Inequalities 2.5 Evaluate and graph the absolute value function Solve absolute value equations Solve absolute value inequalities

3. Absolute Value Function The graph of y = |x|. V-shaped Cannot be represented by single linear function

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:

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 Example: Analyzing the graph of y = |ax + b| (For continuity of the solution, it appears completely on the next slide.)

6. 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.

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.

8. Absolute Value Equations Two solutions |ax + b| = k, for k > 0

9. Absolute Value Equations One solution |ax + b| = k, for k = 0

10. Absolute Value Equations No solution |ax + b| = k, for k < 0

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

12. Solve the equation |2x + 5| = 2 graphically, numerically, and symbolically. Example: Solving an equation with technology Solution Graph Y1 = abs(2X + 5) and Y2 = 2 Solutions: –3.5, –1.5

13. Example: Solving an equation with technology Table Y1 = abs(2x + 5) and Y2 = 2 Solutions to y1 = y2 are –3.5 and –1.5.

14. Example: Solving an equation with technology Symbolic:

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.

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.

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 orx > s2. Similar statements can be made for inequalities involving ≤ or ≥.

18. Solve the inequality |2x – 5| ≤ 6. Write the solution set in interval notation. Solution set: Example: Solving inequalities involving absolute values symbolically Solution Solve |2x – 5| = 6 or 2x – 5 = ±6

19. 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 orax + b > –k. Similar statements can be made for inequalities involving ≤ or ≥.

20. Solve the inequality |4 – 5x | ≤ 3. Write your answer in interval notation. Example: Using analternative method Solution |4 – 5x| ≤ 3 is equivalent to the three-part inequality In interval notation, solution is .