Graphing Quadratic Inequalities

1. Graphing Quadratic Inequalities 4.9

2. ? 0> 02 + 3(0)– 4 Graph a quadratic inequality EXAMPLE 1 Graph y > x2 + 3x – 4. SOLUTION STEP 1 Graph y = x2 + 3x – 4. Because the inequality symbol is >, make the parabola dashed. STEP 2 Test a point inside the parabola, such as (0, 0). y> x2 + 3x– 4 0 > – 4

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

4. Use a quadratic inequality in real life EXAMPLE 2 Rappelling A manila rope used for rappelling down a cliff can safely support a weight W(in pounds) provided W ≤ 1480d2 where dis the rope’s diameter (in inches). Graph the inequality. SOLUTION Graph W = 1480d2 for nonnegative values of d. Because the inequality symbol is ≤, make the parabola solid. Test a point inside the parabola, such as (1, 2000).

5. 2000 ≤ 1480 Use a quadratic inequality in real life EXAMPLE 2 W≤ 1480d2 2000≤ 1480(1)2 Because (1, 2000) is not a solution, shade the region below the parabola.

6. Graph a system of quadratic inequalities EXAMPLE 3 Graph the system of quadratic inequalities. y < – x2 + 4 Inequality 1 y > x2 – 2x – 3 Inequality 2 SOLUTION STEP 1 Graph y ≤ –x2 + 4. The graph is the red region inside and including the parabola y = –x2 + 4.

7. Graph a system of quadratic inequalities EXAMPLE 3 STEP 2 Graph y > x2– 2x – 3. The graph is the blue region inside (but not including) the parabola y = x2 –2x – 3. STEP 3 Identify the purple regionwhere the two graphs overlap. This region is the graph of the system.

8. ? 0> 02 + 2(0)– 8 for Examples 1, 2, and 3 GUIDED PRACTICE Graph the inequality. y > x2 + 2x – 8 STEP 1 Graph y = x2 + 2x – 8. Because the inequality symbol is >, make the parabola dashed. STEP 2 Test a point inside the parabola, such as (0, 0). y> x2 + 2x– 8 0 > – 4

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

10. ? 0< 02 – 3(0)+ 1 for Examples 1, 2, and 3 GUIDED PRACTICE Graph the inequality. y < 2x2 – 3x + 1 SOLUTION STEP 1 Graph y = 2x2 – 3x + 1. Because the inequality symbol is <, make the parabola dashed. STEP 2 Test a point inside the parabola, such as (0, 0). y< 2x2 – 3x+ 1 0 < 1

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

12. ? 0< 02 + 4(0)+ 2 for Examples 1, 2, and 3 GUIDED PRACTICE Graph the inequality. y < – x2 + 4x + 2 SOLUTION STEP 1 Graph y = – x2 + 4x + 2. Because the inequality symbol is <, make the parabola dashed. STEP 2 Test a point inside the parabola, such as (0, 0). y< – x2 + 4x+ 2 0 < 2

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

14. for Examples 1, 2, and 3 GUIDED PRACTICE Graph the system of inequalities consisting of y ≥ x2andy < 2x2 + 5. SOLUTION STEP 1 Graph y > x2. – STEP 2 Graph y < 2x2 + 5. STEP 3 Identify the shaded regionwhere the two graphs overlap. This region is the graph of the system.

15. ANSWER The solution of the inequality is –3 ≤ x ≤ 2. Solve a quadratic inequality using a table EXAMPLE 4 Solve x2 + x ≤ 6 using a table. SOLUTION Rewrite the inequality as x2 + x – 6 ≤ 0. Then make a table of values. Notice that x2 + x –6 ≤ 0 when the values of xare between –3 and 2, inclusive.

16. – 1+ 12– 4(2)(– 4) x = – 1+33 2(2) x = 4 x 1.19 orx –1.69 Solve a quadratic inequality by graphing EXAMPLE 5 Solve 2x2 + x – 4 ≥ 0 by graphing. SOLUTION The solution consists of the x-values for which the graph of y = 2x2 + 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 = 2x2 + x – 4

17. Solve a quadratic inequality by graphing EXAMPLE 5 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. ANSWER The solution of the inequality is approximately x ≤ – 1.69 or x ≥ 1.19.

18. ANSWER The solution of the inequality is –1.8 ≤ x ≤ 0.82. for Examples 4 and 5 GUIDED PRACTICE Solve the inequality 2x2 + 2x ≤ 3using a table and using a graph. SOLUTION Rewrite the inequality as 2x2 + 2x –3≤ 0. Then make a table of values.

19. Robotics EXAMPLE 6 Use a quadratic inequality as a model The number Tof teams that have participated in a robot-building competition for high school students can be modeled by T(x) = 7.51x2 –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?

20. ANSWER There were more than 100 teams participating in the years 1997–2001. EXAMPLE 6 Use a quadratic inequality as a model SOLUTION You want to find the values of xfor which: T(x) > 100 7.51x2 – 16.4x + 35.0 > 100 7.51x2 – 16.4x – 65 > 0 Graph y = 7.51x2 – 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.

21. EXAMPLE 7 Solve a quadratic inequality algebraically Solve x2 – 2x > 15 algebraically. SOLUTION First, write and solve the equation obtained by replacing > with = . x2 – 2x = 15 Write equation that corresponds to original inequality. x2 – 2x – 15 = 0 Write in standard form. (x + 3)(x – 5) = 0 Factor. x = – 3 orx = 5 Zero product property

22. (– 4)2–2(– 4) = 24 >15 62 –2(6) = 24 >15 12 –2(1) 5 –1 >15   ANSWER The solution is x < – 3 orx > 5. EXAMPLE 7 Solve a quadratic inequality algebraically The numbers – 3 and 5 are the critical x-values of the inequality x2 – 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: Test x = 6:

23. for Examples 6 and 7 GUIDED PRACTICE 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 xfor which: T(x) > 200 7.51x2 – 16.4x + 35.0 > 200 7.51x2 – 16.4x – 165 > 0

24. ANSWER There were more than 200 teams participating in the years 1998 – 2001. for Examples 6 and 7 GUIDED PRACTICE Graph y = 7.51x2 – 16.4x – 165 on the domain 0 ≤ x ≤ 9.

25. for Examples 6 and 7 GUIDED PRACTICE Solve the inequality 2x2 – 7x = 4 algebraically. SOLUTION First, write and solve the equation obtained by replacing > with 5. 2x2 – 7x = 4 Write equation that corresponds to original inequality. 2x2 – 7x –4= 0 Write in standard form. (2x + 1)(x – 4) = 0 Factor. x = – 0.5 orx = 4 Zero product property

26. 5 – 2 0 1 2 3 4 6 – 6 – 7 – 3 – 1 – 5 7 – 4 2 (– 3)2 – 7 (– 3)> 4 2 (2)2 – 7 (2)> 4 2 (5)2 – 7 (3)> 4   ANSWER The solution is x < – 0.5 orx > 4. for Examples 6 and 7 GUIDED PRACTICE The numbers 4 and –0.5 are the critical x-values of the inequality 2x2 – 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: