1. EXAMPLE 1 Graph a quadratic inequality Graph y > x 2 + 3x – 4. SOLUTION STEP 1 Graph y = x 2 + 3x – 4. Because the inequality symbol is >, make the parabola dashed. Test a point inside the parabola, such as (0, 0). STEP 2 y > x 2 + 3x – 4 0 > 0 2 + 3(0) – 4 ? 0 > – 4

2. EXAMPLE 1 Graph a quadratic inequality So, (0, 0) is a solution of the inequality. STEP 3 Shade the region inside the parabola.

3. EXAMPLE 2 Use a quadratic inequality in real life A manila rope used for rappelling down a cliff can safely support a weight W (in pounds) provided Rappelling W ≤ 1480d 2 where d is the rope’s diameter (in inches). Graph the inequality. SOLUTION Graph W = 1480d 2 for nonnegative values of d. Because the inequality symbol is ≤, make the parabola solid. Test a point inside the parabola, such as (1, 2000).

4. EXAMPLE 2 Use a quadratic inequality in real life W ≤ 1480d 2 2000 ≤ 1480(1) 2 2000 ≤ 1480 Because (1, 2000) is not a solution, shade the region below the parabola.

5. EXAMPLE 3 Graph a system of quadratic inequalities Graph the system of quadratic inequalities. y < – x 2 + 4 Inequality 1 y > x 2 – 2x – 3 Inequality 2 SOLUTION STEP 1 Graph y ≤ – x 2 + 4. The graph is the red region inside and including the parabola y = – x 2 + 4.

6. EXAMPLE 3 Graph a system of quadratic inequalities STEP 2 Graph y > x 2 – 2x – 3. The graph is the blue region inside (but not including) the parabola y = x 2 –2x – 3. Identify the purple region where the two graphs overlap. This region is the graph of the system. STEP 3

7. GUIDED PRACTICE for Examples 1, 2, and 3 Graph the inequality. 1. y > x 2 + 2x – 8 STEP 1 Graph y = x 2 + 2x – 8. Because the inequality symbol is >, make the parabola dashed. Test a point inside the parabola, such as (0, 0). STEP 2 y > x 2 + 2x – 8 0 > 0 2 + 2(0) – 8 ? 0 > – 4

8. GUIDED PRACTICE for Examples 1, 2, and 3 So, (0, 0) is a solution of the inequality. STEP 3 Shade the region inside the parabola.

9. GUIDED PRACTICE for Examples 1, 2, and 3 Graph the inequality. y < 2x 2 – 3x + 12. SOLUTION STEP 1 Graph y = 2x 2 – 3x + 1. Because the inequality symbol is <, make the parabola dashed. Test a point inside the parabola, such as (0, 0). STEP 2 y < 2x 2 – 3x + 1 0 < 0 2 – 3(0) + 1 ? 0 < 1

10. GUIDED PRACTICE for Examples 1, 2, and 3 So, (0, 0) is a solution of the inequality. STEP 3 Shade the region inside the parabola.

11. GUIDED PRACTICE for Examples 1, 2, and 3 Graph the inequality. y < – x 2 + 4x + 23. SOLUTION STEP 1 Graph y = – x 2 + 4x + 2. Because the inequality symbol is <, make the parabola dashed. Test a point inside the parabola, such as (0, 0). STEP 2 y < – x 2 + 4x + 2 0 < 0 2 + 4(0) + 2 ? 0 < 2

12. GUIDED PRACTICE for Examples 1, 2, and 3 So, (0, 0) is a solution of the inequality. STEP 3 Shade the region inside the parabola.

13. GUIDED PRACTICE for Examples 1, 2, and 3 4. Graph the system of inequalities consisting of y ≥ x 2 and y < 2x 2 + 5. SOLUTION STEP 1 STEP 2 Graph y < 2x 2 + 5. Graph y > x 2. – Identify the shaded region where the two graphs overlap. This region is the graph of the system. STEP 3

14. EXAMPLE 4 Solve a quadratic inequality using a table Solve x 2 + x ≤ 6 using a table. SOLUTION Rewrite the inequality as x 2 + x – 6 ≤ 0. Then make a table of values. Notice that x 2 + x –6 ≤ 0 when the values of x are between –3 and 2, inclusive. The solution of the inequality is –3 ≤ x ≤ 2. ANSWER

15. EXAMPLE 5 Solve a quadratic inequality by graphing Solve 2x 2 + x – 4 ≥ 0 by graphing. SOLUTION The solution consists of the x -values for which the graph of y = 2x 2 + x – 4 lies on or above the x -axis. Find the graph’s x -intercepts by letting y = 0 and using the quadratic formula to solve for x. 0 = 2x 2 + x – 4 x = – 1+ 1 2 – 4(2)(– 4) 2(2) x = – 1+ 33 4 x 1.19 or x –1.69

16. EXAMPLE 5 Solve a quadratic inequality by graphing Sketch a parabola that opens up and has 1.19 and –1.69 as x -intercepts. The graph lies on or above the x -axis to the left of (and including) x = – 1.69 and to the right of (and including) x = 1.19. The solution of the inequality is approximately x ≤ – 1.69 or x ≥ 1.19. ANSWER

17. GUIDED PRACTICE for Examples 4 and 5 Solve the inequality 2x 2 + 2x ≤ 3 using a table and using a graph. 5. SOLUTION Rewrite the inequality as 2x 2 + 2x – 3 ≤ 0. Then make a table of values. The solution of the inequality is –1.8 ≤ x ≤ 0.82. ANSWER x-3-2-1.8-1.500.50.80.9 2 2 + 2x – 391-0.1-1.5-3 -1.5-0.10.42

18. EXAMPLE 6 Use a quadratic inequality as a model The number T of teams that have participated in a robot- building competition for high school students can be modeled by Robotics T(x) = 7.51x 2 –16.4x + 35.0, 0 ≤ x ≤ 9 Where x is the number of years since 1992. For what years was the number of teams greater than 100?

19. EXAMPLE 6 Use a quadratic inequality as a model T(x) > 100 7.51x 2 – 16.4x + 35.0 > 100 7.51x 2 – 16.4x – 65 > 0 Graph y = 7.51x 2 – 16.4x – 65 on the domain 0 ≤ x ≤ 9. The graph’s x -intercept is about 4.2. The graph lies above the x -axis when 4.2 < x ≤ 9. There were more than 100 teams participating in the years 1997–2001. ANSWER SOLUTION You want to find the values of x for which:

20. EXAMPLE 7 Solve a quadratic inequality algebraically Solve x 2 – 2x > 15 algebraically. SOLUTION First, write and solve the equation obtained by replacing > with =. x 2 – 2x = 15 x 2 – 2x – 15 = 0 (x + 3)(x – 5) = 0 x = – 3 or x = 5 Write equation that corresponds to original inequality. Write in standard form. Factor. Zero product property

21. EXAMPLE 7 Solve a quadratic inequality algebraically The numbers – 3 and 5 are the critical x -values of the inequality x 2 – 2x > 15. Plot – 3 and 5 on a number line, using open dots because the values do not satisfy the inequality. The critical x -values partition the number line into three intervals. Test an x -value in each interval to see if it satisfies the inequality. Test x = – 4: Test x = 1: 1 2 –2(1) 5 –1 >15 Test x = 6: The solution is x 5. ANSWER (– 4) 2 –2(– 4) = 24 >15  6 2 –2(6) = 24 >15 

22. GUIDED PRACTICE for Examples 6 and 7 6. Robotics Use the information in Example 6 to determine in what years at least 200 teams participated in the robot-building competition. SOLUTION You want to find the values of x for which: T(x) > 200 7.51x 2 – 16.4x + 35.0 > 200 7.51x 2 – 16.4x – 165 > 0

23. GUIDED PRACTICE for Examples 6 and 7 Graph y = 7.51x 2 – 16.4x – 165 on the domain 0 ≤ x ≤ 9. There were more than 200 teams participating in the years 1998 – 2001. ANSWER

24. GUIDED PRACTICE for Examples 6 and 7 7. Solve the inequality 2x 2 – 7x = 4 algebraically. SOLUTION First, write and solve the equation obtained by replacing > with 5. Write equation that corresponds to original inequality. Write in standard form. Factor. Zero product property 2x 2 – 7x = 4 2x 2 – 7x – 4 = 0 (2x + 1)(x – 4) = 0 x = – 0.5 or x = 4

25. GUIDED PRACTICE for Examples 6 and 7 The numbers 4 and – 0.5 are the critical x -values of the inequality 2x 2 – 7x > 4. Plot 4 and – 0.5 on a number line, using open dots because the values do not satisfy the inequality. The critical x -values partition the number line into three intervals. Test an x -value in each interval to see if it satisfies the inequality. Test x = – 3 : Test x = 2 : Test x = 5 : – 3 – 4 – 2 – 1 0 1 234 5 6 7– 5 – 6 – 7 2 (– 3) 2 – 7 (– 3) > 4  2 (5) 2 – 7 (3) > 4  2 (2) 2 – 7 (2) > 4 The solution is x 4. ANSWER