1. Solve x2 – 2x – 24 = 0. ANSWER –4 2. Solve –8 < 3x –5 < 7. ANSWER –1 < x < 4

3. A point on a graph moves so that the function y = 2x2 – 3x – 20 describes how the vertical position y is related to the horizontal position x. What is the vertical position when the horizontal position is –3? ANSWER 7

Graph a quadratic inequality EXAMPLE 1 Graph a quadratic inequality 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 > 02 + 3(0) – 4 ? 0 > – 4

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

EXAMPLE 2 Use a quadratic inequality in real life Rappelling A manila rope used for rappelling down a cliff can safely support a weight W (in pounds) provided W ≤ 1480d2 where d is 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).

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

Graph a system of quadratic inequalities EXAMPLE 3 Graph a system of quadratic inequalities 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.

EXAMPLE 3 Graph a system of quadratic inequalities 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 region where the two graphs overlap. This region is the graph of the system.

Test a point inside the parabola, such as (0, 0). YOU TRY. . . for Examples 1, 2, and 3 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 > 02 + 2(0) – 8 ? 0 > – 4

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

Test a point inside the parabola, such as (0, 0). YOU TRY. . . for Examples 1, 2, and 3 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 < 02 – 3(0) + 1 ? 0 < 1

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

Test a point inside the parabola, such as (0, 0). YOU TRY. . . for Examples 1, 2, and 3 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 < 02 + 4(0) + 2 ? 0 < 2

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

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

EXAMPLE 4 Solve a quadratic inequality using a table 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 x are between –3 and 2, inclusive. The solution of the inequality is –3 ≤ x ≤ 2. ANSWER

EXAMPLE 5 Solve a quadratic inequality by graphing 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 x = – 1+ 12– 4(2)(– 4) 2(2) x = – 1+ 33 4 x 1.19 or x –1.69

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. ANSWER The solution of the inequality is approximately x ≤ – 1.69 or x ≥ 1.19.

Solve the inequality 2x2 + 2x ≤ 3 using a table and using a graph. YOU TRY. . . for Examples 4 and 5 Solve the inequality 2x2 + 2x ≤ 3 using a table and using a graph. SOLUTION Rewrite the inequality as 2x2 + 2x – 3 ≤ 0. Then make a table of values. x -3 -2 -1.8 -1.5 -1 0.5 0.8 0.9 22 + 2x – 3 9 1 -0.1 0.42 The solution of the inequality is –1.8 ≤ x ≤ 0.82. ANSWER

EXAMPLE 6 Use a quadratic inequality as a model Robotics The number T of 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?

EXAMPLE 6 Use a quadratic inequality as a model SOLUTION You want to find the values of x for 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. There were more than 100 teams participating in the years 1997–2001. ANSWER

Solve a quadratic inequality algebraically 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 or x = 5 Zero product property

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: (– 4)2 –2(– 4) = 24 >15  12 –2(1) 5 –1 >15 62 –2(6) = 24 >15  The solution is x < – 3 or x > 5. ANSWER

YOU TRY. . . for Examples 6 and 7 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 7.51x2 – 16.4x + 35.0 > 200 7.51x2 – 16.4x – 165 > 0

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

Solve the inequality 2x2 – 7x = 4 algebraically. YOU TRY. . . for Examples 6 and 7 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 or x = 4 Zero product property

The solution is x < – 0.5 or x > 4. ANSWER YOU TRY. . . for Examples 6 and 7 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. – 3 – 4 – 2 – 1 1 2 3 4 5 6 7 – 5 – 6 – 7 Test x = – 3: Test x = 2: Test x = 5: 2 (– 3)2 – 7 (– 3) > 4  2 (2)2 – 7 (2) > 4 2 (5)2 – 7 (3) > 4  The solution is x < – 0.5 or x > 4. ANSWER

KEEP GOING 1. Solve 3x2 + 2x – 1 < 0 by graphing. ANSWER – 1 < x < 1 3 2. Graph y 2x2 – 4 and y < – x2 + 3 as a system of inequalities.  ANSWER

KEEP GOING Solve the inequalities. 3. x2 + 4x 5  ANSWER – 5  x  1 4. 2x2 + x – 15 > 0 ANSWER x < – 3 or x > 5 2