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Splash Screen. Five-Minute Check (over Lesson 5–2) CCSS Then/Now New Vocabulary Example 1: Degrees and Leading Coefficients Example 2: Real-World Example: Evaluate a Polynomial Function Example 3: Function Values of Variables Key Concept: End Behavior of a Polynomial Function

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  1. Splash Screen

  2. Five-Minute Check (over Lesson 5–2) CCSS Then/Now New Vocabulary Example 1: Degrees and Leading Coefficients Example 2: Real-World Example: Evaluate a Polynomial Function Example 3: Function Values of Variables Key Concept: End Behavior of a Polynomial Function Key Concept: Zeros of Even- and Odd-Degree Functions Example 4: Graphs of Polynomial Functions Lesson Menu

  3. A. 6m2y3 – 3my B. 6my – 3y C. 3m2y3 – 3my D. 2m2y3 – my 5-Minute Check 1

  4. Simplify (m3 – 3m2 – 18m + 40) ÷ (m + 4). A.m3 + 10 B.m2 + m + 6 C.m2 – 9m + 6 D.m2 – 7m + 10 5-Minute Check 2

  5. Simplify (p3 – 8) ÷ (p – 2). A.p3 + 4 B.p2 + 2p + 4 C.p2 + p + 4 D.p2 + 4 5-Minute Check 3

  6. A.4x4 – x3 – 5x2 + x – 1 B. C. D.x3 – 4x2 – 5x + 1 Simplify (4x4 – x3 – 19x2 + 11x – 3) ÷ (x – 2). 5-Minute Check 4

  7. If the area of a parallelogram is given by x2 – 5x + 4 and the base is x – 1, what is the height of the figure? A.x + 4 B.x – 4 C.x – 2 D.x + 2 5-Minute Check 5

  8. A.x2 + 4x + 3 B.x2 + 2x – 3 C.x2 + 2x – D.x + 3 The volume of a box is given by the expression x3 + 3x2 – x – 3. The height of the box is given by the expression x – 1. Find an expression for the area of the base of the box. 5-Minute Check 6

  9. Content Standards F.IF.4 For a function that models a relationship between two quantities, interpret key features of graphs and tables in terms of the quantities, and sketch graphs showing key features given a verbal description of the relationship. F.IF.7.c Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior. Mathematical Practices 1 Make sense of problems and persevere in solving them. CCSS

  10. You analyzed graphs of quadratic functions. • Evaluate polynomial functions. • Identify general shapes of graphs of polynomial functions. Then/Now

  11. polynomial in one variable • leading coefficient • polynomial function • power function • quartic function • quintic function • end behavior Vocabulary

  12. Degrees and Leading Coefficients A.State the degree and leading coefficient of 7z3 – 4z2 + z. If it is not a polynomial in one variable, explain why. Answer: This is a polynomial in one variable. The degree is 3 and the leading coefficient is 7. Example 1

  13. Degrees and Leading Coefficients B. State the degree and leading coefficient of 6a3 – 4a2 + ab2. If it is not a polynomial in one variable, explain why. Answer: This is not a polynomial in one variable. It contains two variables, a and b. Example 1

  14. Degrees and Leading Coefficients C. State the degree and leading coefficient of 3x5 + 2x2 – 4 – 8x6. If it is not a polynomial in one variable, explain why. Answer: This is a polynomial in one variable. The greatest exponent is 6, so the degree is 6 and the leading coefficient is –8. Example 1

  15. A. Determine whether 3x3 + 2x2 – 3 is a polynomial in one variable. If so, state the degree and leading coefficient. A. degree: 3leading coefficient: 2 B. degree: 3leading coefficient: 3 C. degree: 2leading coefficient: –3 D. not a polynomial in onevariable Example 1

  16. B.Determine whether 3x2 + 2xy – 5 is a polynomial in one variable. If so, state the degree and leading coefficient. A. degree: 2leading coefficient: 3 B. degree: 2leading coefficient: 2 C. degree: 1leading coefficient: –5 D. not a polynomial in one variable Example 1

  17. C.Determine whether 9y3 + 4y6 – 45 – 8y2 – 5y7 is a polynomial in one variable. If so, state the degree and leading coefficient. A. degree: 6leading coefficient: 4 B. degree: 7leading coefficient: –5 C. degree: 7leading coefficient: 5 D. not a polynomial in one variable Example 1

  18. Evaluate a Polynomial Function RESPIRATION The volume of air in the lungs during a 5-second respiratory cycle can be modeled by v(t) = –0.037t3 + 0.152t2 + 0.173t, where v is the volume in liters and t is the time in seconds. This model is an example of a polynomial function. Find the volume of air in the lungs 1.5 seconds into the respiratory cycle. By substituting 1.5 into the function we can find v(1.5), the volume of air in the lungs 1.5 seconds into the respiration cycle. Example 2

  19. Evaluate a Polynomial Function v(t) = –0.037t3 + 0.152t2 + 0.173t Original function v(1.5) = –0.037(1.5)3 + 0.152(1.5)2 + 0.173(1.5)Replace t with 1.5. ≈ –0.1249 + 0.342 + 0.2595Simplify. ≈ 0.4766 Add. Answer: 0.4766 L Example 2

  20. The height of a toy rocket during a 2.35 second flight is predicted by the function h(t) = –4t3 + 6t2 + 8t, where h is the height in meters and t is the time in seconds. This model is an example of a polynomial function. Find the height of the toy rocket 1.25 seconds into the flight. A. 11.6 meters B. 12.1 meters C. 13.5 meters D. 14.2 meters Example 2

  21. Original function Replace m with 2x – 1. Evaluate 2(2x – 1)2. Simplify. Function Values of Variables Find b(2x – 1) – 3b(x) if b(m) = 2m2 + m – 1. To evaluate b(2x – 1), replace the m in b(m) with 2x – 1. Example 3

  22. Original function Replace m with x. Distributive Property Function Values of Variables To evaluate 3b(x), replace m with x in b(m), then multiply the expression by 3. Example 3

  23. Replace b(2x – 1) and 3b(x) with evaluated expressions. Function Values of Variables Now evaluate b(2x – 1) – 3b(x). b(2x – 1) – 3b(x) Distribute. = 2x2 – 9x + 3 Simplify. Answer: 2x2 – 9x + 3 Example 3

  24. Find g(2x + 1) – 2g(x) if g(b) = b2 + 3. A. 1 B. 2x2 + 4x – 2 C. 2x2 + 4x + 10 D. 2x2 – 2 Example 3

  25. Concept

  26. Concept

  27. Graphs of Polynomial Functions • A. For the graph, • describe the end behavior, • determine whether it represents an odd-degree or an even-degree function, and • state the number of real zeros. • Answer: • f(x) → –∞ as x → +∞ • f(x) → –∞ as x → –∞ • It is an even-degree polynomial function. • The graph does not intersect the x-axis, so the function has no real zeros. Example 4

  28. Graphs of Polynomial Functions • B.For the graph, • describe the end behavior, • determine whether it represents an odd-degree or an even-degree function, and • state the number of real zeros. • Answer: • f(x) → +∞ as x → +∞ • f(x) → –∞ as x → –∞ • It is an odd-degree polynomial function. • The graph intersects the x-axis at one point, so the function has one real zero. Example 4

  29. A. For the graph, determine whether it represents an odd-degree or an even-degree function, and state the number of real zeros. A.It is an even-degree polynomial function and has no real zeros. B.It is an even-degree polynomial function and has two real zeros. C.It is an odd-degree polynomial function and has two real zeros. D.It is an odd-degree polynomial function and has no real zeros. Example 4

  30. B. For the graph, determine whether it represents an odd-degree or an even-degree function, and state the number of real zeros. A.It is an even-degree polynomial function and has three real zeros. B.It is an odd-degree polynomial function and has no real zeros. C.It is an odd-degree polynomial function and has three real zeros. D.It is an even-degree polynomial function and has no real zeros. Example 4

  31. End of the Lesson

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