2. Warm-up 1. Given: y = x 2 – 6x + 3 Find: Vertex, AOS, y-intercept, and graph it 2. Given: y = -2(x – 3) 2 + 4 Find: Vertex, AOS, y-intercept, and graph it 3. Given: y = 2(x – 4)(x + 2) Find: Vertex, AOS, y-intercept, and graph it

3. Chapter 4 Section 4-9 Solving Quadratic Inequalities

4. Objectives I can graph quadratic inequalities with the assistance of a calculator I can solve quadratic inequalities with a calculator or with 3 test method

5. Review The solutions to any quadratic equation are the x-intercepts (where the graph crosses the x-axis) Now, the solutions to any quadratic inequality is where the shaded region crosses the x-axis. (It will usually be a range of values)

6. Review from Linear Inequalities Remember when graphing linear inequalities we used two types of boundaries: Solid Line: Used to include equality. Used when equations contained (  and  ) Dashed Line: Used when equations contained ( ), no equality We will be using same ideas with quadratic graphs.

7. Graphing Quadratic Inequalities To graph any quadratic inequality, you can use the following technique: 1. Graph the boundary (solid or dashed). Use the graphing calculator and data table to find key points. 2. Test a point. Point (0, 0) works best if that point is not on the boudary. 3. Shade the correct region based on the test results.

8. Example 1: y  x 2 – 6x + 2 1 st : Draw the boundary. Shown here in blue. It is solid because of the . Next test a point. I chose (0, 0). 0  (0)2 – 6(0) + 2 0  0 – 0 + 2 0  2 This is true! Since (0,0) tested good, then shade the area outside the boundary which includes (0,0).

9. Solving by Graphing y > x 2 – 9 Graph boundary. Dashed because > Test point. I chose (0,0). 0 > -9 (yes) So shade inside boundary. So solutions are all x-values inside parabola. Solution: (-3, 3) Notice that –3 & 3 are not in the solution because the are on the dashed line.

10. 2 nd Method: Three Test Points A 2 nd method to solve the inequalities is by finding the roots and testing 3 regions. Consider the inequality below: x 2 – x – 12 > 0 (Find solutions using 2 nd Trace in calculator) So x = 4 or x = -3 are boundary points

11. Solve: x 2 - x – 12 > 0 x = 4 or x = -3 -3 4 Test (-5) (-5) 2 - (-5) – 12 > 0 25 + 5 – 12 > 0 30 – 12 > 0 18 > 0 YES Test (0) 0 2 - (0) – 12 > 0 -12 > 0 NO Test (5) 5 2 –(5) – 12 > 0 25 – 5 – 12 > 0 20 – 12 > 0 8 > 0 YES Solutions: (-∞,-3) U (4, ∞)

12. Solve: x 2 - 8x – 33 > 0 x 2 - 8x – 33 > 0 Use calculator to find the solutions! x = 11 or x = -3 Test 3 areas x 11 Solve on next slide.

13. Solve: x 2 - 8x – 33 > 0 x = 11 or x = -3 -3 11 Test (-5) (-5) 2 - 8(-5) – 33 > 0 25 + 40 – 33 > 0 65 – 33 > 0 32 > 0 YES Test (0) 0 2 - 8(0) – 33 > 0 -33 > 0 NO Test (15) 15 2 –8(15) – 33 > 0 225 – 120 – 33 > 0 105 – 33 > 0 72 > 0 YES Solutions: (-∞, -3) U (11, ∞)

14. Solve: 2x 2 - 3x - 4  0 Find solutions w/Calc. -.85, 2.35 -.85 2.35 Test (-1) 2(-1) 2 - 3(-1) – 4  0 2 + 3 – 4  0 5 – 4  0 1  0 NO Test (0) 2(0) 2 - 3(0) – 4  0 -4  0 YES Test (3) 2(3) 2 –3(3) – 4  0 18 – 9 – 4  0 9 – 4  0 5  0 NO Solutions: [-.85, 2.35]

15. Homework WS 4-5 Quiz Wednesday