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Published byMagdalen Martin Modified over 9 years ago
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Objectives Solve quadratic inequalities by using tables and graphs.
Solve quadratic inequalities by using algebra.
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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.
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To graph Quadratic Inequalities
y < ax2 + bx + c y > ax2 + bx + c use a dashed line y ≤ ax2 + bx + c y ≥ ax2 + bx + c use a solid line If the parabola opens up: > or shade inside < or shade outside If the parabola opens down: > or shade outside < or shade inside
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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 vertex (3.5, -2.25) roots (2, 0) and (5, 0) y-intercept (0, 10) reflection of y-intercept (7, 10)
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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 2 Graph each inequality. y < –3x2 – 6x – 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.
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Example 2 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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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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you can solve quadratic inequalities algebraically.
By finding the critical values, you can solve quadratic inequalities algebraically.
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Example 3: Solving Quadratic Equations by Using Algebra
Solve x2 – 10x + 18 ≤ –3 by using algebra. Step 1 Write the related equation. x2 – 10x + 18 = –3 Step 2 Solve for x to find the critical values. x2 –10x + 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.
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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 x (2)2 – 10(2) + 18 ≤ –3 Try x = 2. (4)2 – 10(4) + 18 ≤ –3 Try x = 4. (8)2 – 10(8) + 18 ≤ –3 Try x = 8. x
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Shade the solution regions on the number line.
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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Example 4 Solve the inequality by using algebra. x2 – 6x + 10 ≥ 2 x2 – 6x + 10 = 2 x2 – 6x + 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, 2 ≤ x ≤ 4, x ≥ 4.
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Test an x-value in each interval.
Example 4 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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Shade the solution regions on the number line.
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 –
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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) = –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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Example 5 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. Factor out –8 to simplify. –8(x2 – 75x ) = 0
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Example 5 or x ≈ or x ≈ 48.96
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Graph the critical points and test points.
Example 5 Graph the critical points and test points. Critical values Test points
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Example 5 –8(25) (25) – 4200 ≥ 6000 Try x = 25. 5800 ≥ 6000 x Try x = 45. –8(45) (45) – 4200 ≥ 6000 6600 ≥ 6000 –8(50) (50) – 4200 ≥ 6000 Try x = 50. 5800 ≥ 6000 x The solution is approximately ≤ x ≤
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the average price of a helmet needs to be
Example 5 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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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 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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