holt mcdougal algebra 2 2-7 solving quadratic inequalities 2-7 solving quadratic inequalities holt...

Holt McDougal Algebra 2

2-7 Solving Quadratic Inequalities2-7 Solving Quadratic Inequalities

Holt Algebra 2

Warm Up

Lesson Presentation

Lesson Quiz



2-7 Solving Quadratic Inequalities

Warm UpGraph the solution set of the inequality

Graph the solution set of the inequality y < 2x + 1.

3x



Solve quadratic inequalities by using tables and graphs.

Solve quadratic inequalities by using algebra.

Objectives



quadratic inequality in two variables

Vocabulary



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).

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

y ≤ ax2 + bx + c y ≥ ax2 + bx + c



In Algebra I, you graphed the solution set of linear inequalities in two variables. You can use a similar procedure to graph the solution set of quadratic inequalities.



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

Try (4, 0).



Graph the inequality.

Check It Out! Example 1a

y ≥ 2x2 – 5x – 2



Graph the inequality.

Check It Out! Example 1b

y < –3x2 – 6x – 7



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

The number line shows the solution set.




x2 – x + 5 < 7


–6 –4 –2 0 2 4 6




2x2 – 5x + 1 ≥ 1


–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.

–6 –4 –2 0 2 4 6



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

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

–3 –2 –1 0 1 2 3 4 5 6 7 8 9



Solve the inequality by using algebra.


–2x2 + 3x + 7 < 2

–3 –2 –1 0 1 2 3 4 5 6 7 8 9



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



Solve3

Write the inequality.

–8x2 + 600x – 4200 ≥ 6000

–8x2 + 600x – 4200 = 6000

Find the critical values by solving the related equation.

Write as an equation.

Write in standard form.

Factor out –8 to simplify.

–8x2 + 600x – 10,200 = 0

–8(x2 – 75x + 1275) = 0

Example 4 Continued



Solve3

Use the Quadratic Formula.

Simplify.

x ≈ 26.04 or x ≈ 48.96

Example 4 Continued



Solve3

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



Solve3

–8(25)2 + 600(25) – 4200 ≥ 6000

–8(45)2 + 600(45) – 4200 ≥ 6000

–8(50)2 + 600(50) – 4200 ≥ 6000

5800 ≥ 6000Try 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.

x

x

Example 4 Continued



Solve3

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 Back4

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

Homework

pp. 114-117

12-14, 18-20, 24-26, 42-50, 54-58

holt mcdougal algebra 2 2-7 solving quadratic inequalities 2-7 solving quadratic inequalities holt...

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