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**Solving Quadratic Inequalities**

5-7 Solving Quadratic Inequalities Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2

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**1. Graph the inequality y < 2x + 1.**

Warm Up 1. Graph the inequality y < 2x + 1. Solve using any method. 2. x2 – 16x + 63 = 0 7, 9 3. 3x2 + 8x = 3

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**Objectives Solve quadratic inequalities by using tables and graphs.**

Solve quadratic inequalities by using algebra.

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Vocabulary quadratic inequality in two variables

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**Many business profits can be modeled by quadratic functions**

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. A quadratic inequality in two variables can be written in one of the following forms, where a, b, and c are real numbers and a ≠ 0. Its solution set is a set of ordered pairs (x, y).

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**y < ax2 + bx + c y > ax2 + bx + c**

In Lesson 2-5, you solved linear inequalities in two variables by graphing. You can use a similar procedure to graph quadratic inequalities.

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**Example 1: Graphing Quadratic Inequalities in Two Variables**

Graph y ≥ x2 – 7x + 10. Step 1 Graph the boundary of the related parabola y = x2 – 7x + 10 with a solid curve. Its y-intercept is 10, its vertex is (3.5, –2.25), and its x-intercepts are 2 and 5.

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Example 1 Continued Step 2 Shade above the parabola because the solution consists of y-values greater than those on the parabola for corresponding x-values.

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Example 1 Continued Check Use a test point to verify the solution region. y ≥ x2 – 7x + 10 0 ≥ (4)2 –7(4) + 10 Try (4, 0). 0 ≥ 16 – 0 ≥ –2

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Check It Out! Example 1a Graph the inequality. y ≥ 2x2 – 5x – 2 Step 1 Graph the boundary of the related parabola y = 2x2 – 5x – 2 with a solid curve. Its y-intercept is –2, its vertex is (1.3, –5.1), and its x-intercepts are –0.4 and 2.9.

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**Check It Out! Example 1a Continued**

Step 2 Shade above the parabola because the solution consists of y-values greater than those on the parabola for corresponding x-values.

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**Check It Out! Example 1a Continued**

Check Use a test point to verify the solution region. y < 2x2 – 5x – 2 0 ≥ 2(2)2 – 5(2) – 2 Try (2, 0). 0 ≥ 8 – 10 – 2 0 ≥ –4

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Check It Out! Example 1b Graph each inequality. y < –3x2 – 6x – 7 Step 1 Graph the boundary of the related parabola y = –3x2 – 6x – 7 with a dashed curve. Its y-intercept is –7.

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**Check It Out! Example 1b Continued**

Step 2 Shade below the parabola because the solution consists of y-values less than those on the parabola for corresponding x-values.

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**Check It Out! Example 1b Continued**

Check Use a test point to verify the solution region. y < –3x2 – 6x –7 –10 < –3(–2)2 – 6(–2) – 7 Try (–2, –10). –10 < – – 7 –10 < –7

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Quadratic inequalities in one variable, such as ax2 + bx + c > 0 (a ≠ 0), have solutions in one variable that are graphed on a number line. For and statements, both of the conditions must be true. For or statements, at least one of the conditions must be true. Reading Math

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**Example 2A: Solving Quadratic Inequalities by Using Tables and Graphs**

Solve the inequality by using tables or graphs. x2 + 8x + 20 ≥ 5 Use a graphing calculator to graph each side of the inequality. Set Y1 equal to x2 + 8x + 20 and Y2 equal to 5. Identify the values of x for which Y1 ≥ Y2.

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**The number line shows the solution set.**

Example 2A Continued The parabola is at or above the line when x is less than or equal to –5 or greater than or equal to –3. So, the solution set is x ≤ –5 or x ≥ –3 or (–∞, –5] U [–3, ∞). The table supports your answer. The number line shows the solution set. –6 –4 –

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**Example 2B: Solving Quadratics Inequalities by Using Tables and Graphs**

Solve the inequality by using tables and graph. x2 + 8x + 20 < 5 Use a graphing calculator to graph each side of the inequality. Set Y1 equal to x2 + 8x + 20 and Y2 equal to 5. Identify the values of which Y1 < Y2.

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**The number line shows the solution set.**

Example 2B Continued The parabola is below the line when x is greater than –5 and less than –3. So, the solution set is –5 < x < –3 or (–5, –3). The table supports your answer. The number line shows the solution set. –6 –4 –

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Check It Out! Example 2a Solve the inequality by using tables and graph. x2 – x + 5 < 7 Use a graphing calculator to graph each side of the inequality. Set Y1 equal to x2 – x + 5 and Y2 equal to 7. Identify the values of which Y1 < Y2.

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**Check It Out! Example 2a Continued**

The parabola is below the line when x is greater than –1 and less than 2. So, the solution set is –1 < x < 2 or (–1, 2). The table supports your answer. The number line shows the solution set. –6 –4 –

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Check It Out! Example 2b Solve the inequality by using tables and graph. 2x2 – 5x + 1 ≥ 1 Use a graphing calculator to graph each side of the inequality. Set Y1 equal to 2x2 – 5x + 1 and Y2 equal to 1. Identify the values of which Y1 ≥ Y2.

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**Check It Out! Example 2b Continued**

The parabola is at or above the line when x is less than or equal to 0 or greater than or greater than or equal to 2.5. So, the solution set is (–∞, 0] U [2.5, ∞) The number line shows the solution set. –6 –4 –

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The number lines showing the solution sets in Example 2 are divided into three distinct regions by the points –5 and –3. These points are called critical values. By finding the critical values, you can solve quadratic inequalities algebraically.

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**Example 3: Solving Quadratic Equations by Using Algebra**

Solve the inequality x2 – 10x + 18 ≤ –3 by using algebra. Step 1 Write the related equation. x2 – 10x + 18 = –3

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Example 3 Continued Step 2 Solve the equation for x to find the critical values. x2 –10x + 21 = 0 Write in standard form. (x – 3)(x – 7) = 0 Factor. x – 3 = 0 or x – 7 = 0 Zero Product Property. x = 3 or x = 7 Solve for x. The critical values are 3 and 7. The critical values divide the number line into three intervals: x ≤ 3, 3 ≤ x ≤ 7, x ≥ 7.

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**Step 3 Test an x-value in each interval.**

Example 3 Continued Step 3 Test an x-value in each interval. –3 –2 – Critical values Test points x2 – 10x + 18 ≤ –3 (2)2 – 10(2) + 18 ≤ –3 Try x = 2. x (4)2 – 10(4) + 18 ≤ –3 Try x = 4. (8)2 – 10(8) + 18 ≤ –3 x Try x = 8.

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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 –

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Check It Out! Example 3a Solve the inequality by using algebra. x2 – 6x + 10 ≥ 2 Step 1 Write the related equation. x2 – 6x + 10 = 2

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**Check It Out! Example 3a Continued**

Step 2 Solve the equation for x to find the critical values. x2 – 6x + 8 = 0 Write in standard form. (x – 2)(x – 4) = 0 Factor. x – 2 = 0 or x – 4 = 0 Zero Product Property. x = 2 or x = 4 Solve for x. The critical values are 2 and 4. The critical values divide the number line into three intervals: x ≤ 2, 2 ≤ x ≤ 4, x ≥ 4.

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**Check It Out! Example 3a Continued**

Step 3 Test an x-value in each interval. –3 –2 – Critical values Test points x2 – 6x + 10 ≥ 2 (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.

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**Check It Out! Example 3a Continued**

Shade the solution regions on the number line. Use solid circles for the critical values because the inequality contains them. The solution is x ≤ 2 or x ≥ 4. –3 –2 –

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Check It Out! Example 3b Solve the inequality by using algebra. –2x2 + 3x + 7 < 2 Step 1 Write the related equation. –2x2 + 3x + 7 = 2

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**Check It Out! Example 3b Continued**

Step 2 Solve the equation for x to find the critical values. –2x2 + 3x + 5 = 0 Write in standard form. (–2x + 5)(x + 1) = 0 Factor. –2x + 5 = 0 or x + 1 = 0 Zero Product Property. x = 2.5 or x = –1 Solve for x. The critical values are 2.5 and –1. The critical values divide the number line into three intervals: x < –1, –1 < x < 2.5, x > 2.5.

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**Check It Out! Example 3b Continued**

Step 3 Test an x-value in each interval. –3 –2 – Critical values Test points –2x2 + 3x + 7 < 2 –2(–2)2 + 3(–2) + 7 < 2 Try x = –2. –2(1)2 + 3(1) + 7 < 2 x Try x = 1. –2(3)2 + 3(3) + 7 < 2 Try x = 3.

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Check It Out! Example 3 Shade the solution regions on the number line. Use open circles for the critical values because the inequality does not contain or equal to. The solution is x < –1 or x > 2.5. –3 –2 –

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A compound inequality such as 12 ≤ x ≤ 28 can be written as {x|x ≥12 U x ≤ 28}, or x ≥ 12 and x ≤ 28. (see Lesson 2-8). Remember!

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**Example 4: Problem-Solving Application**

The monthly profit P of a small business that sells bicycle helmets can be modeled by the function P(x) = –8x 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?

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**Understand the Problem**

Example 4 Continued 1 Understand the Problem The answer will be the average price of a helmet required for a profit that is greater than or equal to $6000. List the important information: The profit must be at least $6000. The function for the business’s profit is P(x) = –8x x – 4200.

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Example 4 Continued 2 Make a Plan Write an inequality showing profit greater than or equal to $6000. Then solve the inequality by using algebra.

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Example 4 Continued Solve 3 Write the inequality. –8x x – 4200 ≥ 6000 Find the critical values by solving the related equation. –8x x – 4200 = 6000 Write as an equation. –8x x – 10,200 = 0 Write in standard form. –8(x2 – 75x ) = 0 Factor out –8 to simplify.

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Example 4 Continued Solve 3 Use the Quadratic Formula. Simplify. x ≈ or x ≈ 48.96

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**Example 4 Continued 3 Solve**

Test an x-value in each of the three regions formed by the critical x-values. Critical values Test points

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Example 4 Continued Solve 3 –8(25) (25) – 4200 ≥ 6000 Try x = 25. 5800 ≥ 6000 x –8(45) (45) – 4200 ≥ 6000 Try x = 45. 6600 ≥ 6000 –8(50) (50) – 4200 ≥ 6000 Try x = 50. 5800 ≥ 6000 x Write the solution as an inequality. The solution is approximately ≤ x ≤

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Example 4 Continued Solve 3 For a profit of $6000, the average price of a helmet needs to be between $26.04 and $48.96, inclusive.

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Example 4 Continued Look Back 4 Enter y = –8x x – 4200 into a graphing calculator, and create a table of values. The table shows that integer values of x between and inclusive result in y-values greater than or equal to 6000.

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Check It Out! Example 4 A business offers educational tours to Patagonia, a region of South America that includes parts of Chile and Argentina . The profit P for x number of persons is P(x) = –25x x – The trip will be rescheduled if the profit is less $7500. How many people must have signed up if the trip is rescheduled?

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**Understand the Problem**

Check It Out! Example 4 Continued 1 Understand the Problem The answer will be the number of people signed up for the trip if the profit is less than $7500. List the important information: The profit will be less than $7500. The function for the profit is P(x) = –25x x – 5000.

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**Check It Out! Example 4 Continued**

2 Make a Plan Write an inequality showing profit less than $7500. Then solve the inequality by using algebra.

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**Check It Out! Example 4 Continued**

Solve 3 Write the inequality. –25x x – 5000 < 7500 Find the critical values by solving the related equation. –25x x – 5000 = 7500 Write as an equation. –25x x – 12,500 = 0 Write in standard form. –25(x2 – 50x + 500) = 0 Factor out –25 to simplify.

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**Check It Out! Example 4 Continued**

Solve 3 Use the Quadratic Formula. Simplify. x ≈ or x ≈ 36.18

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**Check It Out! Example 4 Continued**

Solve 3 Test an x-value in each of the three regions formed by the critical x-values. Critical values Test points

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**Check It Out! Example 4 Continued**

Solve 3 –25(13) (13) – 5000 < 7500 Try x = 13. 7025 < 7500 –25(30) (30) – 5000 < 7500 Try x = 30. 10,000 < 7500 x –25(37) (37) – 5000 < 7500 Try x = 37. 7025 < 7500 Write the solution as an inequality. The solution is approximately x > or x < Because you cannot have a fraction of a person, round each critical value to the appropriate whole number.

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**Check It Out! Example 4 Continued**

Solve 3 The trip will be rescheduled if the number of people signed up is fewer than 14 people or more than 36 people.

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**Check It Out! Example 4 Continued**

Look Back 4 Enter y = –25x x – 5000 into a graphing calculator, and create a table of values. The table shows that integer values of x less than and greater than result in y-values less than 7500.

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Lesson Quiz: Part I 1. Graph y ≤ x2 + 9x + 14. Solve each inequality. 2. x2 + 12x + 39 ≥ 12 x ≤ –9 or x ≥ –3 3. x2 – 24 ≤ 5x –3 ≤ x ≤ 8

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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) = –2x 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

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