5 7 Solving Quadratic Inequalities Objectives Solve quadratic

5 -7 Solving Quadratic Inequalities Objectives Solve quadratic inequalities by using tables and graphs. Solve quadratic inequalities by using algebra. Holt Algebra 2

5 -7 Solving Quadratic Inequalities Many business profits can be modeled by quadratic functions. To ensure that the profit is above a certain level, financial planners may need to graph and solve quadratic inequalities. Holt Algebra 2

5 -7 Solving Quadratic Inequalities To graph Quadratic Inequalities y < ax 2 + bx + c y > ax 2 + bx + c use a dashed line y ≤ ax 2 + bx + c y ≥ ax 2 + bx + c use a solid line If the parabola opens up: > or < or shade inside shade outside If the parabola opens down: Holt Algebra 2 > or shade outside < or shade inside

5 -7 Solving Quadratic Inequalities Example 1: Graphing Quadratic Inequalities in Two Variables Graph y ≥ x 2 – 7 x + 10. Step 1 Graph the boundary of the related parabola vertex (3. 5, -2. 25) roots (2, 0) and (5, 0) y-intercept (0, 10) reflection of y-intercept Holt Algebra 2 (7, 10)

5 -7 Solving Quadratic Inequalities Example 1 Continued Step 2 Holt Algebra 2 Shade above the parabola because the solution consists of y-values greater than those on the parabola for corresponding x-values.

5 -7 Solving Quadratic Inequalities Example 2 Graph each inequality. y < – 3 x 2 – 6 x – 7 Step 1 Graph the boundary vertex (-1, -4) no roots y-intercept (0, -7) reflect of y-int (-2, -7) with a dashed curve. Holt Algebra 2

5 -7 Solving Quadratic Inequalities Example 2 Step 2 Holt Algebra 2 Shade below the parabola because the solution consists of y-values less than those on the parabola for corresponding x-values.

5 -7 Solving Quadratic Inequalities Quadratic inequalities in one variable, such as ax 2 + bx + c > 0 (a ≠ 0), have solutions in one variable that are graphed on a number line. Reading Math For and statements, both of the conditions must be true. For or statements, at least one of the conditions must be true. Holt Algebra 2

5 -7 Solving Quadratic Inequalities By finding the critical values, you can solve quadratic inequalities algebraically. Holt Algebra 2

5 -7 Solving Quadratic Inequalities Example 3: Solving Quadratic Equations by Using Algebra Solve x 2 – 10 x + 18 ≤ – 3 by using algebra. Step 1 Write the related equation. x 2 – 10 x + 18 = – 3 Step 2 Solve for x to find the critical values. x 2 – 10 x + 21 = 0 (x – 3)(x – 7) = 0 x = 3 or x = 7 The critical values are 3 and 7. The critical values divide the number line into three intervals: x ≤ 3, 3 ≤ x ≤ 7, x ≥ 7. Holt Algebra 2

5 -7 Solving Quadratic Inequalities Example 3 Continued Step 3 Test an x-value in each interval. Critical values x 2 – 10 x + 18 ≤ – 3 – 2 – 1 0 1 2 3 4 5 6 7 Test points (2)2 – 10(2) + 18 ≤ – 3 x Try x = 2. (4)2 – 10(4) + 18 ≤ – 3 Try x = 4. (8)2 – 10(8) + 18 ≤ – 3 x Try x = 8. Holt Algebra 2 8 9

5 -7 Solving Quadratic Inequalities Example 3 Continued Shade the solution regions on the number line. Use solid circles for the critical values because the inequality contains them. The solution is 3 ≤ x ≤ 7 or [3, 7]. – 3 – 2 – 1 Holt Algebra 2 0 1 2 3 4 5 6 7 8 9

5 -7 Solving Quadratic Inequalities Example 4 Solve the inequality by using algebra. x 2 – 6 x + 10 ≥ 2 x 2 – 6 x + 10 = 2 x 2 – 6 x + 8 = 0 (x – 2)(x – 4) = 0 x = 2 or x = 4 The critical values are 2 and 4. The 3 intervals are: x ≤ 2, Holt Algebra 2 2 ≤ x ≤ 4, x ≥ 4.

5 -7 Solving Quadratic Inequalities Example 4 Test an x-value in each interval. Critical values x 2 – 6 x + 10 ≥ 2 – 3 – 2 – 1 0 1 2 3 4 5 Test points (1)2 – 6(1) + 10 ≥ 2 Try x = 1. (3)2 – 6(3) + 10 ≥ 2 x Try x = 3. (5)2 – 6(5) + 10 ≥ 2 Try x = 5. Holt Algebra 2 6 7 8 9

5 -7 Solving Quadratic Inequalities Example 4 Use solid circles for the critical values because the inequality contains them. Shade the solution regions on the number line. The solution is x ≤ 2 or x ≥ 4. – 3 – 2 – 1 Holt Algebra 2 0 1 2 3 4 5 6 7 8 9

5 -7 Solving Quadratic Inequalities Example 5: Problem-Solving Application The monthly profit P of a small business that sells bicycle helmets can be modeled by the function P(x) = – 8 x 2 + 600 x – 4200, where x is the average selling price of a helmet. What range of selling prices will generate a monthly profit of at least $6000? Holt Algebra 2

5 -7 Solving Quadratic Inequalities Example 5 Write the inequality. – 8 x 2 + 600 x – 4200 ≥ 6000 Find the critical values by solving the related equation. – 8 x 2 + 600 x – 4200 = 6000 Write as an equation. – 8 x 2 + 600 x – 10, 200 = 0 Write in standard form. – 8(x 2 – 75 x + 1275) = 0 Holt Algebra 2 Factor out – 8 to simplify.

5 -7 Solving Quadratic Inequalities Example 5 or x ≈ 26. 04 or x ≈ 48. 96 Holt Algebra 2

5 -7 Solving Quadratic Inequalities Example 5 Graph the critical points and test points. Critical values 10 20 30 40 Test points Holt Algebra 2 50 60 70

5 -7 Solving Quadratic Inequalities Example 5 – 8(25)2 + 600(25) – 4200 ≥ 6000 Try x = 25. 5800 ≥ 6000 x – 8(45)2 + 600(45) – 4200 ≥ 6000 Try x = 45. 6600 ≥ 6000 Try x = 50. – 8(50)2 + 600(50) – 4200 ≥ 6000 5800 ≥ 6000 x The solution is approximately Holt Algebra 2 26. 04 ≤ x ≤ 48. 96.

5 -7 Solving Quadratic Inequalities Example 5 For a profit of $6000, the average price of a helmet needs to be between $26. 04 and $48. 96, inclusive. Holt Algebra 2

5 -7 Solving Quadratic Inequalities Lesson Quiz: Part I 1. Graph y ≤ x 2 + 9 x + 14. Solve each inequality. 2. x 2 + 12 x + 39 ≥ 12 x ≤ – 9 or x ≥ – 3 3. x 2 – 24 ≤ 5 x – 3 ≤ x ≤ 8 Holt Algebra 2

5 -7 Solving Quadratic Inequalities Lesson Quiz: Part II 4. A boat operator wants to offer tours of San Francisco Bay. His profit P for a trip can be modeled by P(x) = – 2 x 2 + 120 x – 788, where x is the cost per ticket. What range of ticket prices will generate a profit of at least $500? between $14 and $46, inclusive Holt Algebra 2