warm up lesson presentation lesson...

Holt McDougal Algebra 2

2-7 Solving Quadratic Inequalities 2-7 Solving Quadratic Inequalities

Holt Algebra 2

Warm Up

Lesson Presentation

Lesson Quiz



2-7 Solving Quadratic Inequalities

Warm Up 1. Graph the inequality y < 2x + 1.

Solve using any method.

2. x2 – 16x + 63 = 0

3. 3x2 + 8x = 3

7, 9



Solve quadratic inequalities by using tables and graphs.

Solve quadratic inequalities by using algebra.

Objectives



quadratic inequality in two variables Vocabulary



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).



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

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

y ≤ ax2 + bx + c y ≥ ax2 + bx + c



Graph y ≥ x2 – 7x + 10.

Example 1: Graphing Quadratic Inequalities in Two Variables

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.



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.



Example 1 Continued

Check Use a test point to verify the solution region.

y ≥ x2 – 7x + 10

0 ≥ (4)2 –7(4) + 10

0 ≥ 16 – 28 + 10

0 ≥ –2 ü

Try (4, 0).



Graph the inequality.

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.

Check It Out! Example 1a

y ≥ 2x2 – 5x – 2



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

Check It Out! Example 1a Continued




y < 2x2 – 5x – 2

0 ≥ 2(2)2 – 5(2) – 2

0 ≥ 8 – 10 – 2

0 ≥ –4 ü

Try (2, 0).




Graph each inequality.

Step 1 Graph the boundary of the related parabola y = –3x2 – 6x – 7 with a dashed curve. Its y-intercept is –7.

Check It Out! Example 1b

y < –3x2 – 6x – 7



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

Check It Out! Example 1b Continued




y < –3x2 – 6x –7

–10 < –3(–2)2 – 6(–2) – 7

–10 < –12 + 12 – 7

–10 < –7 ü

Try (–2, –10).




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



Solve the inequality by using tables or graphs.

Example 2A: Solving Quadratic Inequalities by Using Tables and 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.



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.

–6 –4 –2 0 2 4 6

The number line shows the solution set.



Solve the inequality by using tables and graph.

Example 2B: Solving Quadratics Inequalities by Using Tables and 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 which Y1 < Y2.



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.

–6 –4 –2 0 2 4 6





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.




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.

–6 –4 –2 0 2 4 6






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.




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, ∞)

–6 –4 –2 0 2 4 6





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.



Solve the inequality x2 – 10x + 18 ≤ –3 by using algebra.

Example 3: Solving Quadratic Equations by Using Algebra

Step 1 Write the related equation.

x2 – 10x + 18 = –3



Example 3 Continued

Write in standard form.

Step 2 Solve the equation for x to find the critical values.

x2 –10x + 21 = 0

x – 3 = 0 or x – 7 = 0 (x – 3)(x – 7) = 0 Factor.

Zero Product Property. Solve for x. 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.



Example 3 Continued

Step 3 Test an x-value in each interval.

(2)2 – 10(2) + 18 ≤ –3

x2 – 10x + 18 ≤ –3

(4)2 – 10(4) + 18 ≤ –3

(8)2 – 10(8) + 18 ≤ –3

Try x = 2.

Try x = 4.

Try x = 8.

–3 –2 –1 0 1 2 3 4 5 6 7 8 9

Critical values

Test points

ü

x

x



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 0 1 2 3 4 5 6 7 8 9

Example 3 Continued



Solve the inequality by using algebra.



x2 – 6x + 10 ≥ 2

x2 – 6x + 10 = 2





x2 – 6x + 8 = 0

x – 2 = 0 or x – 4 = 0 (x – 2)(x – 4) = 0 Factor.

Zero Product Property. Solve for x. x = 2 or x = 4

The critical values are 2 and 4. The critical values divide the number line into three intervals: x ≤ 2, 2 ≤ x ≤ 4, x ≥ 4.





(1)2 – 6(1) + 10 ≥ 2

x2 – 6x + 10 ≥ 2

(3)2 – 6(3) + 10 ≥ 2

(5)2 – 6(5) + 10 ≥ 2

Try x = 1.

Try x = 3.

Try x = 5.


ü

x

ü

–3 –2 –1 0 1 2 3 4 5 6 7 8 9

Critical values

Test points



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 –1 0 1 2 3 4 5 6 7 8 9




Solve the inequality by using algebra.



–2x2 + 3x + 7 < 2

–2x2 + 3x + 7 = 2





–2x2 + 3x + 5 = 0

–2x + 5 = 0 or x + 1 = 0 (–2x + 5)(x + 1) = 0 Factor.

Zero Product Property. Solve for x. x = 2.5 or x = –1

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.





–2(–2)2 + 3(–2) + 7 < 2

–2(1)2 + 3(1) + 7 < 2

–2(3)2 + 3(3) + 7 < 2

Try x = –2.

Try x = 1.

Try x = 3.

–3 –2 –1 0 1 2 3 4 5 6 7 8 9

Critical values

Test points


ü

x

–2x2 + 3x + 7 < 2

ü



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 –1 0 1 2 3 4 5 6 7 8 9

Check It Out! Example 3



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!



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) = –8x2 + 600x – 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?



1 Understand the Problem

Example 4 Continued

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) = –8x2 + 600x – 4200.



2 Make a Plan

Write an inequality showing profit greater than or equal to $6000. Then solve the inequality by using algebra.

Example 4 Continued



Solve 3

Write the inequality.

–8x2 + 600x – 4200 ≥ 6000

–8x2 + 600x – 4200 = 6000

Find the critical values by solving the related equation.

Write as an equation.


Factor out –8 to simplify.

–8x2 + 600x – 10,200 = 0

–8(x2 – 75x + 1275) = 0

Example 4 Continued



Solve 3

Use the Quadratic Formula.

Simplify.

x ≈ 26.04 or x ≈ 48.96

Example 4 Continued



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

10 20 30 40 50 60 70

Critical values

Test points

Example 4 Continued



Solve 3

–8(25)2 + 600(25) – 4200 ≥ 6000

–8(45)2 + 600(45) – 4200 ≥ 6000

–8(50)2 + 600(50) – 4200 ≥ 6000

5800 ≥ 6000 Try x = 25.

Try x = 45.

Try x = 50.

6600 ≥ 6000

5800 ≥ 6000

Write the solution as an inequality. The solution is approximately 26.04 ≤ x ≤ 48.96.

ü

x

x

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.

Example 4 Continued



Look Back 4

Enter y = –8x2 + 600x – 4200 into a graphing calculator, and create a table of values. The table shows that integer values of x between 26.04 and 48.96 inclusive result in y-values greater than or equal to 6000.

Example 4 Continued



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) = –25x2 + 1250x – 5000. The trip will be rescheduled if the profit is less $7500. How many people must have signed up if the trip is rescheduled?

Check It Out! Example 4



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) = –25x2 + 1250x – 5000.

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.




Solve 3

Write the inequality.

–25x2 + 1250x – 5000 < 7500

–25x2 + 1250x – 5000 = 7500

Find the critical values by solving the related equation.

Write as an equation.


Factor out –25 to simplify.

–25x2 + 1250x – 12,500 = 0

–25(x2 – 50x + 500) = 0




Simplify.

x ≈ 13.82 or x ≈ 36.18

Use the Quadratic Formula.

Solve 3




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

5 10 15 20 25 30 35

Critical values

Test points

Solve 3




–25(13)2 + 1250(13) – 5000 < 7500 7025 < 7500

Try x = 13.

Try x = 30.

Try x = 37.

10,000 < 7500

7025 < 7500 Write the solution as an inequality. The solution is approximately x > 36.18 or x < 13.82. Because you cannot have a fraction of a person, round each critical value to the appropriate whole number.

x

ü

ü

–25(30)2 + 1250(30) – 5000 < 7500

–25(37)2 + 1250(37) – 5000 < 7500

Solve 3




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

Solve 3




Look Back 4

Enter y = –25x2 + 1250x – 5000 into a graphing calculator, and create a table of values. The table shows that integer values of x less than 13.81 and greater than 36.18 result in y-values less than 7500.




Lesson Quiz: Part I

1. Graph y ≤ x2 + 9x + 14.

Solve each inequality.

2. x2 + 12x + 39 ≥ 12

3. x2 – 24 ≤ 5x

x ≤ –9 or x ≥ –3

–3 ≤ x ≤ 8



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) = –2x2 + 120x – 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

warm up lesson presentation lesson...

Documents