Chapter 2 notes from powerpoints

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1 Chapter 2 notes from powerpoints

2 Synthetic division and basic definitions Sections 1 and 2

3 Definition of a Polynomial Function: Let n be a nonnegative integer and let a n, a n-1,, a 2, a 1, a 0 be real numbers. The following function is called a polynomial function of x with degree n. n f ( x) a x a x... a x a x a n n 1 2 n No! No! In polynomial functions, THE EXPONENTS ON THE VARIABLE CANNOT BE FRACTIONS AND CANNOT BE NEGATIVE.

4 2.1 Polynomials n n 1 2 anx an 1x a2 x a1 x a0 Terms: Coefficients: Leading Term: n a x, a x,..., a x, a x, a Leading Coefficient: Degree: n n n 1 2 n a, a,..., a, a, a n ax n n n a The term containing the highest power of x n The power of x contained ** Note : The polynomial 0 has no degree. in the leading term Constant Term n-1 Term Quadratic Term Linear Term

5 Classifying Polynomials Polynomials are often classified by their degree. The degree of a polynomial is the highest degree of its terms. Degree Name of Polynomial Function Example Zero f(x) = -3 First f(x) = 2x + 5 Second f(x) = 3x 2 5x + 2 Third f(x) = x 3 2x 1 Fourth Constant Linear Quadratic Cubic Quartic f(x) = x 4 3x 3 +7x-6 Fifth f(x) = 2x 5 + 3x 4 x 3 +x 2 Quintic 5

6 Any value of x for which P x = 0 is a root of the equation and a zero of the function. State whether the function is a polynomial function. Give the zeros if they exist. P x = x 2 4 f x 2x 3 32x g x 0 = x 2 4 = (x + 2)(x 2) ±2 are the zeros of the function f is a polynomial function..., f x 2x x 16 2x x 4 x 4 2 the zeros are 0, 4 x 1 g is not a polynomial function x 1 the only zero is 1

7 Main Ideas: Synthetic Division 4 2 Divide 2x 15x 10x 5 by x x 15x 10x x 6x 3x 1 x 3 x 3

8 Main Ideas Synthetic Division Is x 2 a factor? If P x 3x 7x 5x 9x 10, find P The factor If theorem S x states 3x that 5xif a is 9a xconstant, 10 find, then Sx a2 is a factor of polynomial P x if and only if P a = P Therefore, x 2 is a factor of P(x)

9 Main Ideas Synthetic Division 4 2 If S x 3x 5x 9x 10 find, S 2 Is x + 2 a factor? S Therefore, x + 2 is not a factor of S(x) The factor theorem states that if a is a constant, then x a is a factor of polynomial P x if and only if P a = 0

10 Section 2.2 Synthetic Division; The Remainder and Factor Theorems Objective: To use synthetic division and to apply the remainder and factor theorems.

11 Vocabulary Dividend Divisor Quotient Remainder When 23 is divided by 4, the quotient is 5 and the remainder is 3. 4 is not a factor of 23, because if when 23 is divided by 4 the remainder is not zero.

12 The Remainder Theorem When a polynomial P(x) is divided by x a, the remainder is P(a) P( x) ( x a) Remainder =P(a)

13 The Factor Theorem if and only if

14 SECTion 3- graphing polynomials

15 Starts Up, Ends Up Starts Down, Ends Down

16 Starts Down, Ends Up Starts Up, Ends Down

17 2.3 Graphing Polynomial Functions Describe the general shape of a cubic function. How does the sign of the leading coefficient affect the graph? Graph of f x ax 3 bx 2 cx d

18 2.3 Graphing Polynomial Functions You can quickly sketch the graph of a cubic function by hand if the function is in factored form. Sketch f x x 1 x 1 x 2. Zeros: -1, 1, 2

19 2.3 Graphing Polynomial Functions Graph of f x ax 4 bx 3 cx 2 dx e

20 2.3 Graphing Polynomial Functions Sketch the following graphs. Use a calculator to check your graphs. a) y x x 2 x 1 b) y x 3 x 1 x 1 x 2 c) y x x 2 x 1 d) y x x 3 x 1 x 1 x 2

21 2.3 Graphing Polynomial Functions

22 2.3 Graphing Polynomial Functions

23 2.3 Graphing Polynomial Functions

24 2.3 Graphing Polynomial Functions

25 Effect of A Squared Factor 2 P x x c c is a double root of P x 0 The graph of P is tangent to the x-axis at x c. 2 Graph y x 1 x 3 Graph y x 1 x 3 x Graph y x x 1 x 3 2

26 Effect of A Squared Factor 2 Graph y x 1 x 3

27 Homework Section 2.2 Page 61 #1-25 odds

28 Effect of A Squared Factor Graph y x 1 x 3 x 4 2

29 Effect of A Squared Factor 2 2 Graph y x x 1 x 3

30 Effect of A Cubed Factor 3 P x x c c is a triple root of P x 0 P flattens out around x-axis at this point. c,0 and crosses the Graph y x Graph y x 2 3 Graph y x 1 x 3

31 Effect of A Cubed Factor 3 P x x c c is a triple root of P x 0 P flattens out around x-axis at this point. Graph y x 2 3 c,0 and crosses the

32 Effect of A Cubed Factor 3 P x x c c is a triple root of P x 0 P flattens out around x-axis at this point. 3 Graph y x 2 c,0 and crosses the

33 Effect of A Cubed Factor 3 P x x c c is a triple root of P x 0 P flattens out around x-axis at this point. 3 Graph y x 1 x 3 c,0 and crosses the

34 Graphing review For test review: 9/26

35 I can graph polynomials examples Determine end behavior and orientation 1) f x = 6x 7 3x 3 + x ) f x = 3x 6 x + 1 Find the x- and y-intercepts 3) y = 4 x 5 2 (x + 1)(2x 5)

36 Warm Up Determine end behavior and orientation 1) f x = 6x 7 3x 3 + x 2 12 x, f x x, f x 2) f x = 3x 6 x + 1 x, f x x, f x

37 Warm Up Find the x- and y-intercepts 3) y = 4 x 5 2 (x + 1)(2x 5) X-intercepts y-intercept Use Zero Product Property! x 5 = 0 x + 1 = 0 2x 5 = 0 x = 5 x = 1 x = 5/2 Plug in 0 for x y = y = (1)( 5) y = 500

38 y = x 1 2 (x + 1) x y

39 y = (x 1) 2 (x + 1) 2 x y

40 Steps Example: y = 2 x + 1 x 2 (x + 8) 1) Find the roots (x-intercepts) 2) Find the y-intercept x + 1 = 0 x 2 = 0 x + 8 = 0 x = 1 x = 2 x = 8 ( 1,0) (2,0) ( 8,0) y = 2(0 + 1)(0 2)(0 + 8) y = y = 32 (0, 32)

41 Steps Example: y = 2 x + 1 x 2 (x + 8) 3) Determine orientation and end behavior Odd degree, positive leading term As x, f x As x, f x

42 Steps 4) Graph 1,0 (2,0) ( 8,0) (0, 32) As x, f x As x, f x Example: y = 2 x + 1 x 2 (x + 8)

43 Steps 1) Find the roots (x-intercepts) 2) Find the y-intercept Example 2: y = x + 5 (x 1) 2 x + 2 x + 5 = 0 x 1 = 0 x + 2 = 0 x = 5 x = 1 x = 2 (5,0) (1,0) ( 2,0) y = (0 1) y = 5 ( 1) 2 2 y = 10 (0,10)

44 Steps 3) Determine orientation and end behavior Example 2: y = x + 5 (x 1) 2 x + 2 Even degree, negative leading term As x, f x As x, f x

45 Steps 4) Graph 5,0 (1,0) ( 2,0) (0,10) Example: y = x + 5 (x 1) 2 x + 2 As x, f x As x, f x

46 Steps 1) Find the roots (x-intercepts) 2) Find the y-intercept Example 3: y = x x + 1 (x 3) x = 0 x + 1 = 0 x 3 = 0 x = 0 x = 1 x = 3 (0,0) ( 1,0) (3,0) y = y = 0 (1) 3 y = 0 (0,0)

47 Steps Example: y = x x + 1 (x 3) 3) Determine orientation and end behavior Odd degree, negative leading term As x, f x As x, f x

48 I can graph a polynomial Warm Up Graph: y = x x + 1 (x 3) x intercepts: y intercept: End Behavior:

49 Steps 1) Find the roots (x-intercepts) 2) Find the y-intercept Example 3: y = x x + 1 (x 3) x = 0 x + 1 = 0 x 3 = 0 x = 0 x = 1 x = 3 (0,0) ( 1,0) (3,0) y = y = 0 (1) 3 y = 0 (0,0)

50 Steps Example: y = x x + 1 (x 3) 3) Determine orientation and end behavior Odd degree, negative leading term As x, f x As x, f x

51 X Intercepts: (0,0) 3,0 ( 1,0) Y Intercept:(0,0) End Behavior: Odd degree, negative leading term As x, f x As x, f x

52 Objective 2.6 Solving Polynomial Functions by Factoring To solve polynomial equations by various methods of factoring, including the use of the rational root theorem. Grouping Terms 3 2 x x x x x 4x x x x x x x 2 x 2 x 5 0 x 2, 5

53 Grouping best to use if: P(x) is a cubic. P(x) has 4 terms. Pair up terms. Factor within the terms Factor again.

54 Try it: x 3 + 6x 2 4x 24 = 0 x = 6, ±2

55 x 3 + 6x 2 4x 24 = 0 x 3 + 6x 2 + 4x 24 = 0 x 2 x x + 6 = 0 x + 6 x 2 4 = 0 x + 6 x + 2 x 2 = 0 x + 6 = 0, x 2 = 0, x + 2 = 0 x = 6, x = 2, x = 2

56 4 2 Solve: 2x x 3 0 by rewriting the quartic in quadratic form. 6, x i 2

57 2x 4 x 2 3 = 0 Let u = x 2 Then: 2x 4 x 2 3 = 0, becomes: 2u 2 u 3 = 0 2u 3 u + 1 = 0 u = 3, u = 1 2 Remember u = x 2, so solving for x: 3 2 = x2, x = ± 3 2 = 3 2 = = x 2 so x = ±i 2 = ± 6, and 2 2

58 When to use quadratic form The polynomial has 3 terms The degree is an even number (though this is not required) The exponent of the middle term is half the degree (necessary) The 3 rd term is constant Examples: x 8 + x , x 22 + x , x 13 + x

59 Try it: 2x 4 = 7x x = ± 6 2, ±i 5

60 Classwork Class Exercises Page 60 #1-5

61 Homework Page 61 #5,7,13,19,23 Page 66 #3,9,11,13,15,21,23,27,29

62 Homework Page 83 #1, 3, 9, 10, and every 3rd

63 Classwork Page 83 #1-6

Algebra III Chapter 2 Note Packet. Section 2.1: Polynomial Functions

Algebra III Chapter 2 Note Packet Name Essential Question: Section 2.1: Polynomial Functions Polynomials -Have nonnegative exponents -Variables ONLY in -General Form n ax + a x +... + ax + ax+ a n n 1