Math 4 part 1

Answer KeyHow much do you know 1. 2 2. False 3. -3 4. b 5. -4 6. (-1, -5), (0, -2), (1, 1) 7. y = 5x – 2 8. y = -x + 3 9. y = -2x – 8 10. y = 3 x + 14 55Try this outLesson 1 Set A 1. The slope is 4 and the y-intercept is 5. 2. The slope is -7 and the y-intercept is 1. 3. The slope is 4 and the y-intercept is 0. 4. The slope is -4 and the y-intercept is 3. 5. The slope is − 2 and the y-intercept is 2. 7 Set B 1. The slope is 8 and the y-intercept is -9. 2. The slope is -4 and the y-intercept is -6. 3. The slope is 1 and the y-intercept is 0. 4. The slope is − 3 and the y-intercept is 3. 8 5. The slope is 3 and the y-intercept is − 5 . 17 Set C 1. The slope is 1 and the y-intercept is 1 . 2 2. The slope is 9 and the y-intercept is − 3 . 42 3. The slope is -0.3 and the y-intercept is 0.1. 4. The slope is -5.2 and the y-intercept is -4.4. 5. The slope is 9 and the y-intercept is 3 . 82 20

Lesson 2 Set A 1. increasing 2. decreasing 3. decreasing 4. decreasing 5. increasing Set B 1. increasing 2. decreasing 3. decreasing 4. decreasing 5. increasing Set C 1. decreasing 2. increasing 3. decreasing 4. increasing 5. increasingLesson 3 Set A 1. The x-intercept is -2 and the y-intercept is 16. 2. The x-intercept is 1 and the y-intercept is -2. 2 3. The x-intercept is 3 and the y-intercept is 3. 4 4. The x-intercept is 5 and the y-intercept is -5. 2 5. The x-intercept is -2 and the y-intercept is 2. Set B 1. The x-intercept is 1 and the y-intercept is 7. 2 2. The x-intercept is − 3 and the y-intercept is -3. 2 3. The x-intercept is − 3 and the y-intercept is 3. 2 4. The x-intercept is 3 and the y-intercept is -3. 4 21

5. The x-intercept is 7/6 and the y-intercept is - 7 . 11Set C1. The x-intercept is 5 and the y-intercept is − 5 . 32. The x-intercept is − 3 and the y-intercept is 3 . 823. The x-intercept is 1 and the y-intercept is − 1 . 734. The x-intercept is 1 and the y-intercept is 0.1. 35. The x-intercept is − 11 and the y-intercept is -4.4. 13Lesson 4 Set A (Answers may vary) 1. (-1, 5), (0, 6), (1, 7) 2. (-1, 9), (0, 7), (1, 5) 3. (-1, -2), (0, 0), (1, 2) 4. (-1, 7), (0, 4), (1, 1) 5. (-1, 2), (0, 4), (1, 6)Set B (Answers may vary)1. (-1, -9), (0, -4), (1, 2)2. (-1, -2), (0, -6), (1, -8)3. (-1, 1), (0, 3), (1, 1)4.  −1, 1  ,  0, 7  , 1,− 13   11  11  11 5. (−1, −1) ,  0, − 1  , 1, 1  3 3Set C (Answers may vary)1. (-1, 7), (0, 0), (1, -7)2. (-4, 3), (0, 2), (4, 1)3. (−1, −3) ,  0, − 2  , 1, 5  3 34.  −1,− 15  ,  0,− 3  , 1, 3   4   2  45. (-1, 0.4), (0, 0.1), (1, -0.2) 22

Lesson 5 4. f (x) = − 6 x −18 Set A 1. f(x) = 3x + 2 5 2. f(x) = -2x + 4 3. f(x) = 5x – 7 5. f (x) = 3 x + 1 4. f(x) = -8x – 1 5. f(x) = 6x 77Set B Set C1. f(x) = x – 1 1. f (x) = − 1 x + 12. f(x) = -9x 65 53. f (x) = 4 x − 2 2. f (x) = 27x + 3 3. f (x) = − 9 x + 3 3 84. f ( x ) =− 1 x − 1 3 3 4. f (x) = 7 x + 15. f (x) = 7x − 4 82 3 5. f (x) = 39 x + 13Set C 8 121. f (x) = 2 x − 1 Lesson 7 Set A 34 1. f (x) = 3x −12. f (x) =− 2 x − 1 2. f (x) = 4x −16 3. f (x) = 2x +11 53 4. f (x) = −2x −10 5. f (x) = 5x + 353. f(x) = 0.2x + 1 Set B4. f(x) = x – 0.1 1. f (x) = 5x +125. f (x=) 2x + 2 2. f (x) = −3x +10 3. f (x) = x − 7Lesson 6 4. f (x) = −9x − 45 5. f (x) = 1 x − 11 Set A 33 1. f (x) = − 2 x + 2 Set C 3 1. f (x) = 2 x − 11 2. f (x) = 2x + 4 3. f (x) = 3x − 3 33 4. f (x) = − 1 x −1 2. f (x) = − 1 x + 5 8 22 5. f (x) = −2x − 8 3. f (x) = − 2 x + 17Set B 3 181. f (x) = − 6 x + 6 4. f (x) = 0.2x + 0.1 5. f (x) = −0.3x + 0.23 52. f (x) = 3x +123. f (x) = − 4 x −16 3

Lesson 8 2. y = - 3 x + 3 1. y = -x + 4 22 2. y = 2 x - 2 3 3. y = 2x – 11 3. y = 7x + 35 55 4. y = -x + 4 5. y = -3x + 1 4. y = - 3 x + 3 22 Set B 1. y = x + 7 5. y = x 22 What have you learned 2. y = x - 10 1. -5 3. y = x – 9 2. true 3. − 4 22 3 4. y = x - 11 4. a 5. y = 9x - 69 5. 2 3 44 6. (0, -4), (1, -2), (2, 0) 7. y = -3x - 4 Set C 8. y = -x - 2 1. y = 3x – 1 9. y = -4x - 5 10. y = -6x - 17 55 24

Module 2 Polynomial Functions What this module is about This module is about finding the zeros of polynomial functions of degreegreater than 2. In module 1, the factor theorem was introduced to you by simplystating, if zero is obtained as a remainder when c is substituted to the polynomialP(x), then the polynomial x – c is factor of P(x). This time, you will learn differentmethods of finding the zeros of polynomial functions. What you are expected to learn This module is designed for you to find the zeros of polynomial functionsof degree greater than 2 by: a. Factor Theorem b. factoring c. synthetic division d. depressed equations How much do you know 1. How many zeros do the polynomial function f(x) = 2x5 – 3x4 – x3 + 2x2 + x – 3 have? 2. How many roots do the polynomial equation 6x4 + 11x3 + 8x2 – 6x – 4 = 0 have? 3. Determine the zeros of the polynomial function F(x) = x(x – 3)2(x + 1)(2x – 3). 4. What are the possible rational zeros of p(x) = x4 – 9x3 + 23x2 – 15? 5. What are the possible rational roots 3x5 – x4 + 6x3 – 2x2 + 8x – 5 = 0? 6. Find all the zeros of h(x) = x3 – 10x2 + 32x – 32. 7. Solve the polynomial equation x4 – 6x3 – 9x2 + 14x = 0 using synthetic division. 8. Find all zeros of g(x) = x3 – 2x2 – x + 2 using depressed equations.

9. One of the roots of x3 – 12x2 – 8x + 96 = 0 is 2 2 . What are the other roots? 10. One of the zeros of p(x) = 2x4 – x3 + 25x2 – 13x – 13 is −i 13 . Find the other zeros. What you will do Lesson 1 Number of Roots Theorem The Fundamental Theorem of Algebra which is attributed to Karl FreidrichGauss of Germany states that “Every polynomial equation in one variable has atleast one root, real or imaginary”. The next theorem tells us of the exact numberof roots of polynomial equation of degree n: “Every polynomial equation of a degree n ≥ 1 has exactly n roots.”Examples: Determine the number of roots of each polynomial equation. 1. 3x7 + 8x5 – 4x – 1 = 0 3x7 + 8x5 – 4x – 1 = 0 is of the seventh degree. Hence it has 7 roots. 2. (x – 1)(2x + 1)3(2x – 5)2 = 0 (x – 1)(2x + 1)3(2x – 5)2 = 0 is of the sixth degree. Hence it has 6 roots. 3. x(x – a)m(x + b)n = 0 x(x – a)m(x + b)n = 0 is of the (1 + m + n)th degree. Hence it has 1 + m + n roots.Try this out Determine the number of roots of each polynomial equation.Set A 1. x5 + 2x3 – x – 3 = 0 2. -x7 + 2x6 – 4x5 – x2 + 2x – 1 = 0 2

3. 2 + x2 – 3x4 – x6 – x8 – 2x10 = 04. (x – 5)(x + 2)3(2x -1)2 = 05. x2(x + 1)(x – 3)4 = 0Set B 1. 8x3 – 9x + 1 = 0 2. -4x7 – 6x6 + x2 – 2x + 5 = 0 3. (x – 2)(x + 9)3x4 = 0 4. 3x3(x + 8)2(x2 – 4) = 0 5. (x2 – 1)(x3 + 1) = 0Set C1. x  x + 1 3 (x2 + 2)(x + 1) =0 22. x4  9 x − 3  (2x − 3) =0 4 23. (x2 – 2x + 1)(x3 – 1)(x2 + x – 6) = 04. x(x2 + 2)2(x2 + 2x – 1)2 = 05. -9x4(8 – x3)2 = 0 Lesson 2Determining the Zeros of Polynomial Functions in Factored Form Recall that a zero of p(x) is the value of x that will make the function 0.The zeros of a polynomial function in factored form are determined by equatingeach factor to 0 and solving for x.Examples: Determine the zeros of each polynomial function. 1. f(x) = x(x + 3)(x – 2) Equate each factor to 0 and solve for x. x=0 x + 3 = 0, x = -3 x – 2 = 0, x = 2 Therefore, the zeros of f(x) are 0, -3, and 2. 3

2. y = (5x – 2)(2x + 1)(-3x – 4)Equate each factor to 0 and solve for x.5x – 2 = 0, x = 2 52x + 1 = 0, x = − 1 2-3x – 4 = 0, x = − 4 3Therefore, the zeros of f(x) are 2 , − 1 , and − 4 . 52 33. g(x) = (x + 4)3(x – 3)(2x – 1)2 g(x) has 3 factors of (x + 4), 1 factor of (x – 3) and 2 factors of (2x – 1). Thus,the zeros of g(x) are: -4 of multiplicity 3; 3 of multiplicity 1; and 1 of multiplicity 2. 24. h(x) = (x2 – 4)(x2 – 3x – 28)Equate each factor to 0 and solve for x.x2 – 4 = (x + 2)(x – 2) = 0, x = -2 and x = 2x2 – 3x – 28 = (x – 7)(x + 4) = 0, x = 7 and x = -4Thus, the zeros of h(x) are -2, 2, 7, and -4.Try this out Determine the zeros of each polynomial function.Set A 1. f(x) = x(x + 4)(x – 2) 2. g(x) = -x(x + 7)(x – 1) 3. h(x) = (4x – 5)(2x + 3)(x – 3) 4. F(x) = x(4 – 3x)(1 – x) 5. G(x) = x(x + 3)(3x + 1)Set B 1. F(x) = (x – 8)5(x + 2)3 2. G(x) = (5x + 1)6(2x – 7)4 3. H(x) = x(5 – x)(2 – 3x)2 4. f(x) = x2(2x – 3)(x + 4)3(3x – 7) 5. g(x) = -7x3(x – 4)(5x + 2)4(x -1)2 4

Set C 1. h(x) = x(x + 3)(2x – 9)(3x + 1) 2. k(x) = (x2 – 1)(4x2 – 4x + 1) 3. p(x) = (x2 + 7x + 10)(9x2 – 12x) 4. y = (2x3 + 3x2 – 5x)(12x2 + 34x + 14) 5. y = (-2x2 – x + 3)(12x2 + 23x + 5) Lesson 3 The Rational Roots Theorem The next theorem specifies a finite set of rational numbers where the rootsof a polynomial equation can be chosen. “If a rational number L in lowest terms is a root of the polynomial Fequation anxn + an - 1xn – 1 + an - 2xn – 2 + … + a2x2 + a1x + a0 = 0, where an, an – 1,an – 2, …, a2, a1, a0 are integers, then L is a factor of a0 and F is a factor of an.” The Rational Roots Theorem states that “Any rational root of thepolynomial equation xn + an - 1xn – 1 + an - 2xn – 2 + … + a2x2 + a1x + a0 = 0, wherean – 1, an – 2, … , a2, a1, a0 are integers, is an integer and is a factor of a0.”Examples: List all possible zeros of the given polynomial function. 1. f(x) = x3 – 6x2 + 11x – 6 Since the coefficient of the highest degree term is 1, the possible rational zeros of f(x) are the factors of the constant term -6. That is, the possible rational zeros are ±1, ±2, ±3, and ±6. 2. g(x) = x4 – x3 –11x2 + 9x + 18 Since the coefficient of the highest degree term is 1, the possible rational zeros of g(x) are the factors of the constant term 18. That is, the possible rational zeros are ±1, ±2, ±3, ±6, ±9 and ±18. 3. h(x) = 2x4 + 9x3 + 11x2 – 4 If we let L = the factors of -4: ±1, ±2, ±4, and F = the factors of 2: ±1, ±2. 5

Then L are ± 1 =±1, ± 1 , ± 2 =±2 , ± 4 =±4 , and ± 4 =±2 F 1 21 1 2 or L = ±1, ±2, ±4, ± 1 . F24. p(x) = 8x4 + 32x3 + x + 4L = the factors of 4: ±1, ±2, ±4F = the factors of 8: ±1, ±2, ±1, ±4, ±8The possible rational zeros L are ± 1 =±1, ± 1 , ± 1 , ± 1 , ± 2 =±2 , F 1 2481± 2 =±1, ± 2 =± 1 , ± 2 =± 1 , ± 4 =±4 , ± 4 =±2 , ± 4 =±1 and ± 4 =± 12 4 28 41 2 4 82or L = ±1, ± 1 , ± 1 , ± 1 , ±2, and ±4. F 248Try this out List all possible zeros of the given polynomial function.Set A 1. f(x) = x3 – 4x2 – 2x + 5 2. g(x) = x3 – 6x2 + 2x – 6 3. h(x) = x3 – x2 – 5x – 3 4. p(x) = x4 + 2x3 – 8x – 16 5. y = 2x3 + 17x2 + 23x – 42Set B 1. f(x) = x5 + x4 – x – 1 2. g(x) = x4 + 32 3. h(x) = 2x3 + 3x2 – 8x + 3 4. p(x) = 3x3 + 13x2 + 9x + 20 5. y = 4x4 + 16x3 + 9x2 – 32Set C 1. f(x) = x3 – 7x2 – 15 2. g(x) = 2x3 – x2 – 4x + 2 3. h(x) = 3x3 – 2x2 + 3x – 2 4. p(x) = 3x3 + 4x2 + 12x + 16 5. y = 6x4 + x3 – 13x2 – 2x + 2 6

Lesson 4 Determining the Zeros of Polynomial Functions Using the Factor Theorem The Factor Theorem states that “If p(c) = 0, then x – c is a factor of p(x).”This implies that c is a zero of p(x). To determine the rational zeros of a polynomial function from the list of allpossible rational zeros using the Factor Theorem, evaluate the polynomialfunction using these possible zeros one at a time. If a zero was obtained afterevaluating a particular rational zero, then you can say that that number is a zeroof the polynomialExamples: Determine the rational zeros of the given polynomial function using the FactorTheorem. 1. f(x) = x3 + 6x2 + 11x + 6 There are 3 zeros, real or imaginary. According to the Rational Roots Theorem, the possible rational zeros are ±1, ±2, ±3, and ±6. If x = -1, then f(x) = x3 + 6x2 + 11x + 6 becomes f(-1) = (-1)3 + 6(-1)2 + 11(-1) + 6 = -1 + 6 – 11 + 6 =0 -1 is a zero of f(x). If x = 1, then f(x) = x3 + 6x2 + 11x + 6 becomes f(1) = (1)3 + 6(1)2 + 11(1) + 6 = 1 + 6 + 11 + 6 = 24 1 is not a zero of f(x). If x = -2, then f(x) = x3 + 6x2 + 11x + 6 becomes f(-2) = (-2)3 + 6(-2)2 + 11(-2) + 6 = -8 + 24 – 22 + 6 =0 -2 is a zero of f(x). 7

If x = 2, then f(x) = x3 + 6x2 + 11x + 6 becomes f(2) = (2)3 + 6(2)2 + 11(2) + 6 = 8 + 24 + 22 + 6 = 60 -1 is not a zero of f(x). If x = -3, then f(x) = x3 + 6x2 + 11x + 6 becomes f(-3) = (-3)3 + 6(-3)2 + 11(-3) + 6 = -27 + 54 – 33 + 6 =0 -3 is a zero of f(x). f(x) has only 3 zeros and we have already found 3. Thus the zeros are -1, -2 and -3.2. g(x) = x4 – x3 – 11x2 + 9x + 18 There are 4 zeros, real or imaginary. According to the Rational RootsTheorem, the possible rational zeros are ±1, ±2, ±3, ±6, ±9 and ±18. If x = -1, then g(x) = x4 – x3 – 11x2 + 9x + 18 becomes g(-1) = (-1)4 – (-1)3 – 11(-1)2 + 9(-1) + 18 = 1 + 1 – 11 – 9 + 18 =0 -1 is a zero of g(x). If x = 1, then g(x) = x4 – x3 – 11x2 + 9x + 18 becomes g(1) = (1)4 – (1)3 – 11(1)2 + 9(1) + 18 = 1 – 1 – 11 + 9 + 18 = 16 1 is not a zero of g(x). If x = -2, then g(x) = x4 – x3 – 11x2 + 9x + 18 becomes g(-2) = (-2)4 – (-2)3 – 11(-2)2 + 9(-2) + 18 = 16 + 8 – 22 – 18 + 18 =2 -1 is not a zero of g(x). If x = 2, then g(x) = x4 – x3 – 11x2 + 9x + 18 becomes g(2) = (2)4 – (2)3 – 11(2)2 + 9(2) + 18 = 16 – 8 – 44 + 18 + 18 =0 2 is a zero of g(x). 8

If x = -3, then g(x) = x4 – x3 – 11x2 + 9x + 18 becomes g(-3) = (-3)4 – (-3)3 – 11(-3)2 + 9(-3) + 18 = 81 + 27 – 99 – 27 + 18 =0-3 is a zero of g(x).If x = 3, then g(x) = x4 – x3 – 11x2 + 9x + 18 becomes g(3) = (3)4 – (3)3 – 11(3)2 + 9(3) + 18 = 81 – 27 – 99 + 27 + 18 =03 is a zero of g(x). g(x) has only 4 zeros and we have already found 4. Thus the zeros are-1, 2 -3 and 3.3. h(x) = 2x4 + 3x3 + 3x – 2 There are 4 zeros, real or imaginary. The possible rational zeros are ± 1 , 2±1 and ±2.If x = − 1 , then h(x) = 2x4 + 3x3 + 3x – 2 becomes 2h  − 1  = 2  − 1 4 + 3  − 1 3 + 3  − 1  –2 2 2 2 2 = 1 – 3 – 3 –2 88 2 = − 15 4− 1 is not a zero of h(x). 2If x = 1 , then h(x) = 2x4 + 3x3 + 3x – 2 becomes 2h  1  = 2  1 4 + 3  1 3 + 3  1  – 2 2 2 2 2 = 1 + 3 + 3 –2 88 2 =01 is a zero of h(x).2 9

If x = -1, then h(x) = 2x4 + 3x3 + 3x – 2 becomes h(-1) = 2(-1)4 + 3(-1)3 + 3(-1) – 2 =2–3–3–2 = -6 -1 is not a zero of h(x). If x = 1, then h(x) = 2x4 + 3x3 + 3x – 2 becomes h(1) = 2(1)4 + 3(1)3 + 3(1) – 2 =2+3+3–2 =6 1 is not a zero of h(x). If x = -2, then h(x) = 2x4 + 3x3 + 3x – 2 becomes h(-2) = 2(-2)4 + 3(-2)3 + 3(-2) – 2 = 32 – 24 – 6 – 2 =0 -1 is a zero of h(x). If x = 2, then h(x) = 2x4 + 3x3 + 3x – 2 becomes h(2) = 2(2)4 + 3(2)3 + 3(2) – 2 = 32 + 24 + 6 – 2 = 60 2 is not a zero of h(x). g(x) has 4 zeros and we have already used all possible rational zeros. We found only 2 rational zeros. This indicates that there are only 2 rational zeros. The other 2 zeros are not rational; they may be irrational or imaginary. (Irrational and imaginary zeros will be discussed in the succeeding lessons.) Thus the only rational zeros are 1 and -2. 2Try this out Determine the rational zeros of the given polynomial function using the FactorTheorem.Set A 1. f(x) = x3 + 2x2 – 5x – 6 2. g(x) = x3 + 4x2 + x – 6 3. h(x) = x3 + 3x2 – 4x – 12 4. p(x) = x3 – x2 – 10x – 8 5. y = x3 + x2 – x – 1 10

Set B 1. f(x) = x3 – 4x2 + x + 6 2. g(x) = x3 – 5x2 – 2x + 24 3. h(x) = x3 – 3x2 – 4x + 12 4. p(x) = x3 – 6x2 + 5x + 12 5. y = x3 – 5x2 – x + 5Set C 1. f(x) = x3 – x2 – x + 1 2. g(x) = 2x3 – 11x2 – 8x + 12 3. h(x) = 3x3 – 2x2 - 27x + 18 4. p(x) = 4x4 – 5x2 + 1 5. y = 2x4 + 9x3 + 6x2 – 5x – 6 Lesson 5Determining the Zeros of Polynomial Functions by Factoring The zeros of a polynomial function can be determined easily if thepolynomial is in factored form. But the problem arises when the polynomial isexpressed otherwise. The polynomial must be factored (if it is factorable) usingtechniques learned in elementary algebra.Examples:Determine the rational zeros of the given polynomial function by factoring.1. f(x) = x3 – 3x2 – 6x + 8f(x) = x3 – x2 – 2x2 – 6x + 8 Using factoring by grouping. = (x3 – x2) – (2x2 + 6x – 8) = x2(x – 1) – 2(x2 + 3x – 4) = x2(x – 1) – 2(x + 4)(x – 1) = (x – 1)[x2 – 2(x + 4)] = (x – 1)(x2 – 2x – 8) = (x – 1)(x + 2)(x – 4)x-1=0 x+2=0 x–4=0 Equate factors to zero x=1 x = -2 x=4Hence, the zeros of f(x) are 1, -2, and 4 11

2. g(x) = x3 + 2x2 – 5x – 6g(x) = x3 + (x2 + x2) – 5x – 6 Using factoring by grouping. = (x3 + x2) + (x2 – 5x – 6) = x2(x + 1) + (x + 1)(x – 6) = (x + 1)[x2 + (x – 6)] = (x + 1)(x2 + x – 6) = (x + 1)(x + 3)(x – 2)x+1=0 x+3=0 x–2=0 Equate factors to zero x = -1 x = -3 x=2Hence, the zeros of g(x) are -1, -3, and 2.3. h(x) = x4 + 4x3 + x2 – 6xh(x) = x(x3 + 4x2 + x – 6) Using common monomial factoring= x(x3 + 2x2 + 2x2 + x – 6) Factoring by grouping= x[(x3 + 2x2) + (2x2 + x – 6)]= x[x2(x + 2) + (x + 2)(2x – 3)]= x{(x + 2)[x2 + (2x – 3)]}= x[(x + 2)(x2 + 2x – 3)]= x(x + 2)(x + 3)(x – 1)x=0 x+2=0 x+3=0 x–1=0 Equate factors to 0. x=1 x = -2 x=1Hence, the zeros of h(x) are 0, -2, -3, and 1.Try this outDetermine the rational zeros of the given polynomial function by factoring.Set A 1. f(x) = x3 + 3x2 – 4x – 12 2. g(x) = x3 + 2x2 – 5x – 6 3. h(x) = x3 – x2 – 10x – 8 4. p(x) = x3 + 4x2 + x – 6 5. y = x3 + x2 – x – 1Set B 1. f(x) = x3 – 6x2 + 5x + 12 2. g(x)= x3 – 3x2 – 4x + 12 3. h(x) = x3 – 5x2 – 2x + 24 4. p(x) = x3 – 3x2 – 4x + 12 5. y = x3 – x2 – x + 1 12

Set C 1. f(x) = x3 – 4x2 + x + 6 2. g(x) = 3x3 – 2x2 + 3x – 2 3. h(x) = 2x3 – x2 – 4x + 2 4. p(x) = 2x4 + 9x3 + 6x2 – 11x – 6 5. y = 4x4 – 5x2 + 1 Lesson 6 Determining the Zeros of Polynomial Functions by Synthetic Division Synthetic division can also be used in determining the zeros of apolynomial function. Recall the when the remainder of a polynomial function f(x)when divided by x – c is 0, then c is a zero of f(x). Examples: Determine the rational zeros of the given polynomial function using syntheticdivision. 1. f(x) = x3 + 6x2 + 11x + 6 There are 3 zeros, real or imaginary. According to the Rational Roots Theorem, the possible rational zeros are ±1, ±2, ±3, and ±6. If f(x) is divided by x + 1, 1 6 11 6 -1 -1 -5 -6 1560 Since the remainder is 0, -1 is a zero of f(x). If f(x) is divided by x – 1, 1 6 11 6 1 1 7 18 1 7 18 24 Since the remainder is not 0, -1 is not a zero of f(x). If f(x) is divided by x + 2, 13

1 6 11 6 -2 -2 -8 -61430Since the remainder is 0, -2 is a zero of f(x).If f(x) is divided by x – 2,1 6 11 6 2 2 16 541 8 27 60Since the remainder is not 0, 2 is not a zero of f(x).If f(x) is divided by x + 3,1 6 11 6 -3 -3 -9 -61320Since the remainder is 0, -3 is a zero of f(x). f(x) has only 3 zeros and we have already found 3. Thus the zeros are -1,-2 and -3. 2. h(x) = 2x4 + 3x3 + 3x – 2 There are 4 zeros, real or imaginary. The possible rational zeros are ± 1 , 2 ±1 and ±2.If h(x) is divided by x + 1 , 22 3 0 3 -2  − 1 2 -1 -1 1/2 -7/42 2 -1 7/2 − 15 4Since the remainder is not 0, − 1 is not a zero of f(x). 2If h(x) is divided by x – 1 , 2 14

1 2 3 0 3 -2  2 1212 24240 Since the remainder is 0, 1 is a zero of f(x). 2 If h(x) is divided by x + 1, 2 3 0 3 -2 -1 -2 -1 1 -4 2 1 -1 4 -6 Since the remainder is not 0, -1 is not a zero of f(x). If h(x) is divided by x – 1 , 2 3 0 3 -2 1 2558 25586 Since the remainder is not 0, 1 is not a zero of f(x). If h(x) is divided by x + 2 , 2 3 0 3 -2 -2 -4 2 -4 2 2 -1 2 -1 0 Since the remainder is 0, -2 is a zero of f(x). g(x) has 4 zeros and we have already used all possible rational zeros. We found only 2 rational zeros. This indicates that there are only 2 real rational zeros. The other 2 zeros are not real rational; they may be real irrational or imaginary. Thus the real only rational zeros are 1 and -2. 2Try this out Determine the rational zeros of the given polynomial function using syntheticdivision. Leave irrational or imaginary zeros.Set A 1. f(x) = x3 – 4x2 – 2x + 5 2. g(x) = x3 – 6x2 + 11x – 6 15

3. h(x) = 2x3 + 17x2 + 23x – 42 4. p(x) = 8x4 + 32x3 + x + 4 5. y = x4 + 2x3 – 8x – 16Set B 1. f(x) = x3 + 6x2 + 11x + 6 2. g(x) = x3 – 7x + 6 3. h(x) = x3 + x2 – 12x 4. p(x) = 9x3 – 7x + 2 5. y = 5x3 + 4x2 – 31x + 6Set C 1. f(x) = x3 – 7x2 + 17x – 15 2. g(x) = 2x3 + 3x2 – 8x + 3 3. h(x) = 3x3 + 13x2 + 9x + 20 4. p(x) = x4 + x3 – 13x2 – 25x – 12 5. y = 4x5 + 16x4 + 9x3 – 9x2 Lesson 7 Determining the Zeros of Polynomial Functions Using Depressed Equations Consider this division problem x3 + 6x2 + 11x + 6 x +1 Using synthetic division, 1 6 11 6 -1 -1 -5 -6 1 5 6 0 ⇐ 3rd line The 3rd line indicates that x + 1 is a factor of x3 + 6x2 + 11x + 6 since theremainder is 0. Also, the 3rd line gives the quotient to the division problem whichis indicated by the other entries 1, 5, and 6. These are the numerical coefficientsof the quotient. That is, x3 + 6x2 + 11x + 6 = x2 + 5x + 6 x +1 The quotient x2 + 5x + 6 = 0, when equated to 0 is called a depressedequation of x3 + 6x2 + 11x + 6. 16

Depressed equations are factors of a given polynomial. And can be usedto find the roots of polynomial equation or zeros of polynomial function.Examples: Determine the zeros of the given polynomial function using depressedequations.1. f(x) = x3 + 6x2 + 11x + 6The possible rational zeros are ±1, ±2, ±3, and ±6.If f(x) is divided by x + 1, 1 6 11 6 -1 -1 -5 -6 1560Since the remainder is 0, -1 is a zero of f(x).The depressed equation is x2 + 5x + 6 = 0To find the other zeros of f(x), solve the depressed equation. x2 + 5x + 6 = 0 By factoring (x + 3)(x + 2) = 0 x + 3 = 0 or x + 2 = 0 x = -3 and x = -2 Thus, the zeros of f(x) are -1, -3 and -2.2. g(x) = 2x4 + 3x3 + 3x – 2The possible rational zeros are ± 1 , ±1 and ±2. 2If h(x) is divided by x + 2, 2 3 0 3 -2 -2 -4 2 -4 2 2 -1 2 -1 0Since the remainder is 0, -2 is a zero of g(x). 17

The first depressed equation is 2x3 – x2 + 2x – 1 = 0. This depressedequation can be used to find a second depressed equation without affecting theresults.If the depressed equation is divided by x – 1 , 2 2 -1 2 -1  1 2 101 2020Since the remainder is 0, 1 is another zero of g(x). 2The second depressed equation is 2x2 + 2 = 0To find the other zeros of g(x), solve the second depressed equation. 2x2 + 2 = 0 Recall from your lesson in x2 + 1 = 0 quadratic equation, −1 is an x2 = -1 imaginary number = i. x = ± −1 x = i or –iThus the zeros of g(x) are 1 , -2, -i and i. 2Try this out Determine the zeros of the given polynomial function using depressedequations.Set A 1. f(x) = x3 – 4x2 – 2x + 5 2. g(x) = x3 – 6x2 + 11x – 6 3. h(x) = 2x3 + 17x2 + 23x – 42 4. p(x) = 8x4 + 32x3 + x + 4 5. y = x4 + 2x3 – 8x – 16Set B 1. f(x) = x3 + 6x2 + 11x + 6 2. g(x) = x3 – 7x + 6 3. h(x) = x3 + x2 – 12x 4. p(x) = 9x3 – 7x + 2 5. y = 5x3 + 4x2 – 31x + 6Set C 1. f(x) = x3 – 7x2 + 17x – 15 18

2. g(x) = 2x3 + 3x2 – 8x + 3 3. h(x) = 3x3 + 13x2 + 9x + 20 4. p(x) = x4 + x3 – 13x2 – 25x – 12 5. y = 4x5 + 16x4 + 9x3 – 9x2 Lesson 8 Quadratic Surd Roots Theorem One interesting fact about the zeros of polynomial functions or roots ofpolynomial equations of degree n ≥ 2 is that there are some zeros or roots thatoccur in pairs. For instance, x2 – 3 = 0 has roots 3 and - 3 , f(x) = x2 – 6x + 2has zeros 3 + 7 and 3 – 7 . The Quadratic Surd Roots Theorem generalizesthis fact. “If the quadratic surd a + b is a root of a polynomial equation, where aand b are rational numbers, and b is an irrational number, then a – b is also aroot of the polynomial equation.”Examples: If the given quadratic surd is a zero of a polynomial function, give the otherquadratic surd which is also a zero of the polynomial function. 1. 2 + 2 Since 2 + 2 is a zero a polynomial function, 2 – 2 is also a zero of the polynomial function. 2. 4 – 3 5 Since 4 – 3 5 is a zero a polynomial function, 4 + 3 5 is also a zero of the polynomial function. 3. 3 – 7 Since 3 – 7 is a zero a polynomial function, − 3 – 7 is also a zero of the polynomial function. 4. −9 11 + 1 19

Since −9 11 + 1 is a zero a polynomial function, 9 11 + 1 is also a zeroof the polynomial function.Try this out If the given quadratic surd is a zero of a polynomial function, give the otherquadratic surd which is also a zero of the polynomial function.Set A 1. 1 + 9 5 2. -4 – 2 2 3. 3 – 7 4. 3 11 + 1 5. −2 7 – 6Set B 1. -5 + 7 5 2. 14 – 9 2 3. 2 3 – 9 4. -2 13 + 8 5. 7 5 – 6Set C 1. 8 – 2 5 2. -10 + 3 2 3. -3 3 – 3 4. 23 13 – 28 5. -17 5 + 16 Lesson 9Complex Conjugate Roots TheoremComplex conjugate roots behave in the same manner as quadratic surdroots. That is, they also come in pairs. For instance, x2 + 3 = 0 has roots i 3and -i 3 , f(x) = 3x2 – 4x + 5 has zeros 2 + 1 i 11 and 2 – 1 i 11 . The pairs 33 3320

i 3 and -i 3 and 2 + 1 i 11 and 2 – 1 i 11 are examples of complex 33 33conjugates. The Complex Conjugate Roots Theorem generalizes the fact: “If the complex number a + bi is a root of a polynomial equation with realcoefficients, then the complex conjugate a – bi is also a root of the polynomialequation.”Examples: If the given complex conjugate is a zero of a polynomial function, give theother complex conjugate which is also a zero of the polynomial function.1. 2 + 3i Since 2 + 3i is a zero a polynomial function, 2 –3i is also a zero of thepolynomial function.2. 4 – 3i Since 4 – 3i is a zero a polynomial function, 4 + 3i is also a zero of thepolynomial function.3. -i – 7 Since -i – 7 is a zero a polynomial function, -i – 7 is also a zero of thepolynomial function.4. −9i 11 + 1 Since −9i 11 + 1 is a zero a polynomial function, 9i 11 + 1 is also azero of the polynomial function. If the given quadratic surd is a zero of a polynomial function, give the otherquadratic surd which is also a zero of the polynomial function.Try this outSet A 1. 1 + 9i 2. -4 – 2i 3. i 3 – 7 4. 3i 11 + 1 5. −2i 7 – 6 21

Set B 1. -5 + 7i 2. 14 – 9i 3. 2i 3 – 2 4. -2i 13 + 5 5. -7i 5 – 9Set C 1. 9 – 2i 2. -12 + 3i 3. -4i 3 – 8 4. 2i 13 – 2 5. -i 5 + 6 Lesson 10 Shortcuts in Determining the Zeros of Polynomial Functions The different ways of determining the zeros of polynomial functions andthe different theorems concerning roots of polynomial equations can becombined. This is illustrated in the next examples.Examples: Find all the zeros of each polynomial function. 1. p(x) = x3 – 2x2 – x + 2 Possible zeros: ±1 and ±2 1 -2 -1 2 -1 -1 3 -2 1 -3 2 0 Since the remainder is 0, -1 is a zero of p(x) and x2 – 3x + 2 = 0 is a depressed equation. Solving the depressed equation by factoring, x2 – 3x + 2 = 0 (x – 2)(x – 1) = 0 x = 2 or x = 1 Hence, the zeros of p(x) are -1, 2 and 1. 22

2. f(x) = x4 – 3x3 – 4x2 + 12x f(x) = x4 – 3x3 – 4x2 + 12x =x(x3 – 3x2 – 4x + 12) The factored form suggests that one of the zeros is 0. The other zeroscan be found from x3 – 3x2 – 4x + 12 = 12. The possible roots of thisequation are ±1, ±2, ±3, ±4, ±6 and ±12. 1 -3 -4 12 2 2 -2 -12 1 -1 -6 0 Since the remainder is 0, 2 is a zero of f(x) and x2 – x – 6 = 0 is adepressed equation. Solving the depressed equation by factoring, x2 – x – 6 = 0 (x – 3)(x + 2) = 0 x = 3 or x = -2Hence, the zeros of f(x) are 0, 2, 3 and -2.3. g(x) = 6x4 + x3 – 13x2 – 2x + 2Possible zeros: ±1, ±2, ± 1 , ± 1 , ± 1 and ± 2 236 36 1 -13 -2 2  − 1 2 -3 1 6 -26 -2 -12 4 0Since the remainder is 0, − 1 is a zero of g(x) and 6x3 – 2x2 –12x + 4 = 20 is the first depressed equation. Using this depressed equation to findanother zero,6 -2 -12 4  1 3 2 0 -46 0 -12 0Since the remainder is 0, 1 is a zero of g(x) and 6x2 –12 = 0 is the 3second depressed equation. Solving 6x2 –12 = 0, 6x2 –12 = 0 6x2 =12 23

x2 = 2 x = 2 or x = - 2Hence, the zeros of g(x) are − 1 , 1 , 2 and - 2 . 2 3Try this outFind all the zeros of each polynomial function.Set A 1. p(x) = x3 – 13x + 12 2. f(x) = x3 + 9x2 + 23x + 15 3. g(x) = 3x3 + 9x2 – 30x 4. h(x) = x3 – 8 5. y = 2x3 – 13x2 – 26x + 16Set B 1. p(x) = x5 + x4 – 3x3 – x2 + 2x 2. f(x) = 3x3 – 2x2 – 3x + 2 3. g(x) = 4x3 – 13x2 + 11x – 2 4. h(x) = 6x3 + 4x2 – 14x + 4 5. y = 2x3 + 3x2 – 8x + 3Set C 1. p(x) = 6x4 + x3 – 13x2 – 2x + 2 2. f(x) = x5 + 3x4 – 4x3 – 12x2 3. g(x) = 6x4 – 19x3 – 22x2 + 7x + 4 4. h(x) = 2x4 + 22x3 + 46x2 5. y = x4 – 2x3 – 15x2 – 4x + 20 Let’s summarize1. Every polynomial equation of a degree n ≥ 1 has exactly n roots.2. Zeros of polynomial functions in x are determined by equating each factor of the polynomial to 0 and then solving for x.3. If a rational number L in lowest terms is a root of the polynomial equation F anxn + an - 1xn – 1 + an - 2xn – 2 + … + a2x2 + a1x + a0 = 0, where an, an – 1, an – 2, …, a2, a1, a0 are integers, then L is a factor of a0 and F is a factor of an. 24

4. Any rational root of the polynomial equation xn + an - 1xn – 1 + an - 2xn – 2 + … + a2x2 + a1x + a0 = 0, where an – 1, an – 2, … , a2, a1, a0 are integers, is an integer and is a factor of a0.5. The Factor Theorem states that “If p(c) = 0, then x – c is a factor of p(x).” This implies that c is a zero of p(x).6. The zeros of a polynomial function can be determined easily if the polynomial is in factored form.7. Depressed equations are factors of a given polynomial, and can be used to find the roots of polynomial equation or zeros of polynomial function.8. If the quadratic surd a + b is a root of a polynomial equation, where a and b are rational numbers, and b is an irrational number, then a – b is also a root of the polynomial equation9. If the complex number a + bi is a root of a polynomial equation with real coefficients, then the complex conjugate a – bi is also a root of the polynomial equation. What have you learned1. How many zeros do the polynomial function f(x) = x6 – 3x5 – x4 + 2x2 + x – 3 have?2. How many roots do the polynomial equation 2x5 + x4 + 8x2 – 2x – 1 = 0 have?3. Determine the zeros of the polynomial function F(x) = x(x – 2)2(x + 3)(3x – 2).4. What are the possible rational zeros of p(x) = x4 – 4x3 + 2x2 – 9?5. What are the possible rational roots 5x5 – 2x4 + x3 – x2 + 8x – 3 = 0?6. Find all the zeros of h(x) = x3 – 4x2 – 7x + 10.7. Solve the polynomial equation x4 – 2x3 – 15x2 – 4x + 20 = 0 using synthetic division.8. Find all zeros of g(x) = x3 – 4x2 + 5x – 2 using depressed equations.9. One of the roots of x3 – 4x2 + 6x – 4 = 0 is 1 + i. What are the other roots?10. One of the zeros of p(x) = 4x4 + 8x3 – 8x2 – 4x is −3 + 5 . Find the other 2 zeros. 25

Answer KeyHow much do you know 1. 5 2. 4 3. 0, 3 multiplicity 2, -1 and 3 2 4. ±1, ±3, ±5, ±15 5. ±1, ± 1 , ±5, ± 5 33 6. 2 and 4 (multiplicity 2) 7. -2, 0, 1, and 7 8. -1, 1, and 2 9. −2 2 and 12 10. i 13 , − 1 and 1 2Try this outLesson 1Set A 1. 5 2. 7 3. 10 4. 6 5. 7Set B 1. 3 2. 7 3. 8 4. 7 5. 5Set C 1. 7 2. 6 3. 7 4. 9 5. 10Lesson 2Set A 1. -4, 0, 2 2. -7, 0, 1 3. − 3 , 5 , 3 24 26

4. 0, 1, 4 35. -3, − 1 , 0 3Set B 1. -2 multiplicity 3, 8 multiplicity 5 2. − 1 multiplicity 6, 7 multiplicity 4 52 3. 0, 2 multiplicity 2, 5 3 4. -4 multiplicity 3, 0 multiplicity 2, 3 , 7 23 5. − 2 multiplicity 4, 0 multiplicity 3, 4, 1 multiplicity 2 5Set C 1. -3, − 1 , 0, 9 32 2. -1, 1 multiplicity 2, 1 2 3. -5, -2, 0, 4 3 4. − 5 , 0, 1, − 7 , − 1 2 32 5. − 5 , − 3 , − 1 , 1 324Lesson 3Set A1. ±1, ±52. ±1, ±2, ±3, ±63. ±1, ±34. ±1, ±2, ±4, ±8, ±165. ±1, ± 1 , ±2, ±3, , ± 3 , ±6, ±7, ± 7 , ±14, ±21, ± 21 , ±42 2 22 2Set B 1. ±1 2. ±1, ±2, ±4, ±8, ±16, ±32 3. ± 1 , ±1, ±3, ± 3 22 4. ±1, ± 1 , ±2, ± 2 , ±4, ± 4 , ±5, ± 5 , ±10, ± 10 , ±20, ± 20 3333 3 3 27

5. ±1, ± 1 , ± 1 , ±2, ±4, ±8 24Set C1. ±1, ±3, ±5, ±152. ±1, ± 1 , ±2 23. ±1, ± 1 , ±2, ± 2 334. ±1, ± 1 , ± 1 , ±2, ± 2 , ±4, ± 4 , ±8, ± 8 23 3 3 35. ±1, ± 1 , ± 1 , ± 1 , ±2, ± 2 , ± 1 236 33Lesson 4Set A 1. -3, -1, 2 2. -3, -2, 1 3. -3, -2, 2 4. -2, -1, 4 5. -1, -1, 1Set B 1. -1, 2, 3 2. -2, 3, 4 3. -2, 2, 3 4. -1, 3, 4 5. -1, 1, 5Set C 1. -1, 1, 1 2. -2, 3 2, 2 3. –3, 2 , 3 3 4. -1, - 1 , 1 , 1 22 5. -3, -2, - 1 ,1 2Lesson 5Set A 1. -2, 2, -3 2. -2, -1, -3 3. -2, -1, 4 28

4. -3, -2, 1 5. -1, -1, 1Set B 1. -1, 3, 4 2. -2, 2, 3 3. -2, 3, 4 4. -2, 2, 3 5. -1, 1, 1Set C 1. -1, 2, 3 2. 2 , i, -i 3 3. 1 , 2 , − 2 2 4. -3, -2, - 1 , 1 2 5. -1, - 1 , 1 , 1 22Lesson 6Set A 1. 1(the only rational zero), 2 are not rational zeros 2. 1, 2, 3 3. -6, − 14 , 1 4 4. -4, 1 (the 2 rational zeros), 2 are not rational zeros 2 5. -2, 2 (the 2 rational zeros), 2 are not rational zerosSet B 1. -3, -2, -1 2. -3, 1, 2 3. -4, 0, 3 4. -1, 1 , 2 33 5. -3, 1 , 2 5Set C 1. 3 (the only rational zero), 2 are not rational zeros 2. -3, 1 , 1 2 3. -4 (the only rational zero), 2 are not rational zeros 29

4. -3, -1, -1, 45. -3, − 3 , 0, 0, 1 22Lesson 7Set A1. 1, 3 + 29 , 3 − 29 222. 1, 2, 33. -6, − 14 , 1 44. -4, − 1 , 1 + 5 , 1 − 5 24 45. -2, 2, −1 + i 7 , −1 − i 7 22Set B 1. -3, -2, -1 2. -3, 1, 2 3. -4, 0, 3 4. -1, 1 , 2 33 5. -3, 1 , 2 5Set C 1. 3, 2 + i , 2 − i 2. -3, 1 , 1 2 3. -4, −1 + i 59 , −1 − i 59 66 4. -3, -1, -1, 4 5. -3, − 3 , 0, 0, 1 22Lesson 8Set A 1. 1 − 9 5 2. −4 + 2 2 3. − 3 − 7 4. −3 11 + 1 5. 2 7 − 6 30

Set B 1. −5 − 7 5 2. 14 + 9 2 3. −2 3 − 9 4. 2 13 + 8 5. −7 5 − 6Set C 1. 8 + 2 5 2. −10 − 3 2 3. 3 3 − 3 4. −23 13 − 28 5. 17 5 + 16Lesson 9Set A 1. 1 − 9i 2. −4 + 2i 3. −i 3 − 7 4. −3i 11 + 1 5. 2i 7 − 6Set B 1. −5 − 7i 2. 14 + 9i 3. −2i 3 − 2 4. 2i 13 + 8 5. 7i 5 − 9Set C 1. 9 + 2i 2. −12 − 3i 3. 4i 3 − 8 4. −2i 13 − 2 5. i 5 + 6Lesson 10Set A 1. -4, 1, 3 2. -5, -3, -1 31

3. -5, 0, 2 4. 2, −1 + i 3 , −1 − i 3 5. -2, 1 , 8 2Set B 1. -2, -1, 0, 1, 1 2. -1, 2 , 1 3 3. 1 , 1, 2 4 4. -2, 1 , 1 3 5. -3, 1 , 1 2Set C 1. − 1 , 1 , 2 − 2 23 2. -3, -2, 0, 0, 2 3. -1, − 1 , 1 , 4 32 4. 0, 0, −11 + 29 , −11 − 29 22 5. -2, -2, 1, 5What have you learned 1. 6 2. 5 3. -3, 0, 2 , 2, 2 3 4. ±1, ±3, ±9 5. ±1, ± 1 ,± 3 55 6. -2, 1, 5 7. -2, -2. 1, 5 8. 1, 1, 2 9. 1 – i, 2 10. 0, 1, −3 − 5 2 32

Module 2 Quadratic Functions What this module is about This module is about Quadratic Functions. As you go over the exercises,you will develop skills in solving quadratic equations and ability to apply this insolving problems. Treat the lesson with fun and take time to go back if you thinkyou at a loss. What you are expected to learn This module is designed for you to: 1. Draw the graph of a quadratic functions using the • vertex • axis of symmetry • direction of the opening of the graph • given points 2. Analyze the effects on the graph of the changes in a, h, and k in f(x) = a(x-h) 2 + k How much do you know A. Tell the direction or the opening of the graph of the following functions. 1. y = -2x2 + 5 2. y = x2 – 3 3. y = 3x2 – 9x + 2 B. Sketch the graph of the following functions. 4. y = -2x2 + 4x – 3 5. y = x2 – 4x + 2

C. Using f(x) = x2 as the reference graph, which graph opens wider/narrower. 6. y = 3x2 + 2 y = 1 x2 – 5 3 7. y = -4x2 –4 y = -1 x2 + 2 4 D. Which graphs translate to the right or to the left of the origin? 8. f(x) = (x + 3)2 9. f(x) = 3(x –4)2 10. f(x) = -(x –2)2 11. y = 2(x + 2)2 E. Which graphs translate upward or downward considering f(x) = x2 as the reference graph. 12. f(x) = x2 – 4 13. f(x) = 3x2 + 1 14. f(x) = -2(x –3)2 + 3 15. y = (x + 2)2 - 2 What you will do Lesson 1 Graph of Quadratic Function The graph of a quadratic function is a parabola. You can graph using yourprevious knowledge about the characteristics of the graph of a quadratic functionsuch as vertex, axis of symmetry and the direction of the opening.Steps in graphing quadratic functions: 1. find the coordinates of the vertex 2. determine the axis of symmetry 2

3. determine the direction of the opening of the graph4. make the table of values (choose symmetric values with respect to the value of h)Examples: Construct a table of values and graph the following functions:1. f(x) = (x + 1)2 – 2Vertex = (-1, -2)Axis of symmetry: x = -1Direction of opening: upward Table of valuesx (x + 1)2 - 2 f(x)1 (1 + 1)2 - 2 20 (0 + 1)2 - 2 -1-1 (-1 + 1)2 - 2 -2-2 (-2+ 1)2 - 2 -1-3 (-3 + 1)2 - 2 22 f(x) = -(x – 2) 2 + 3 vertex = (2, 3) axis of symmetry: x = 2 direction of opening: Downward Table of valuesx -(x – 2)2 + 3 f(x)4 -(4 – 2)2 +3 -13 -(3 – 2)2 +3 -22 -(2 – 2)2 +3 31 -(1 – 2)2 +3 20 -(0 - 2)2 +3 -1Try this out Draw the graph of each of the following functions by following the stepsmentioned in the given examples.1. f(x) = (x + 2)2 – 3 3

Vertex: ________Axis of symmetry: _________Direction of opening: _________ Table of valuesx (x + 2)2 - 3 f(x)0-1-2-3-42. f(x) = -(x –2)2 + 4 vertex: ________ Axis of symmetry: ________ Direction of opening: ________ Table of valuesx -(x - 2)2 + 4 f(x)3210-13. f(x) = 3(x + 1)2 + 3 vertex:: ________ Axis of symmetry: _________ Direction of the opening: ______ Table of valuesx 3(x + 1)2 + 3 f(x)3210-1 4

4. f(x) = (x - 1)2 + 3 vertex: _________ Axis of symmetry: _________ Direction of opening: __________ Table of valuesx (x - 1)2 + 3 f(x)3210-15. f(x) = 1/3(x – 1)2 + 2 vertex : ________ Axis of symmetry: _________ Direction of opening: _________ Table of valuesx 1/3(x - 1)2 + 2 f(x)3210-1 Lesson 2 Analyze the Effect of the Changes in a in the Graph of the Function f(x) = ax2 The value of a in f(x) = ax2 has an effects on the graph of a quadraticfunction. It determines the width of the parabola with respect to the axis ofsymmetry. If 0 < a < 1, the graph is wider and tends to flatten out. If > 1, thegraph is narrower and steeper. Consider the following functions and f(x) = x2 as reference function. 5

1. y = 2x2 Vertex: (0,0) Axis of symmetry: x = 0 Opening of graph: Upward Table of valuesx 2x2 f(x) y = 2x22 2(2)2 8 y = x2 y = ½ x21 2(1)2 20 2(0)2 0-1 2(-1)2 2-2 2(-2)2 82. f(x) = 1/2x2 vertex: (0,0) Axis of symmetry: x = 0 Opening of the graph: Upward Table of valuesx ½(x) 2 f(x)2 ½(2) 2 21 ½(1) 2 ½0 ½(0) 2 0-1 ½(-1) 2 ½-2 ½(-2) 2 2 You will notice that the graph of f(x) = 2x2 is narrower compared to thegraph of f(x) = x2. While, the graph of f(x) = ½x2 is wider compared to the graphof f(x) = x2.Try this outFor each set of functions, tell which graph is narrower or wider. 1. f(x) = 2x2 f(x) = 3x22. f(x) = -1 x2 2 6

f(x) = -2x23. f(x) = 4x2 f(x) = 1 x2 44. f(x) = 5x2 f(x) = 4x25. f(x) = -3x2 f(x) = -1 x2 36. f(x) = -x2 f(x) = -3x27. f(x) = 2x2 f(x) = 4x28. f(x) = -5x2 f(x = -2x29. f(x) = 2 x2 3 f(x) = 1 x2 210. f(x) = - 1 x2 3 f(x) = - 1 x2 2 Lesson 3 Analyze the Effect on the Changes in h in the Graph of the Function f(x) = a(x-h) 2The graph of the function f(x) = a(x-h) 2 is the same as the graph of 7

f(x) = ax2, except that its vertex is translated horizontally to the right of the originwhen h > 0. The graph of the function f(x) = a(x - h) 2 is the same as the graph of f(x)= ax2 except the vertex is translated to the left of the origin when h < 0.Example:1. y = (x + 1)2 y = (x+1)2vertex : (-1, 0) y = x2Axis of symmetry: x = -1 y = (x – 1)2Opening: Upward Table of valuesx (x + 1)2 f(x)1 (1+ 1)2 40 (0 + 1)2 1-1 (-1 + 1)2 0-2 (-2+ 1)2 1-3 (-3+ 1)2 42. y = (x - 1)2 Table of valuesvertex : (1, 0) x (x - 1)2 f(x)Axis of symmetry: x = 1Opening of the graph: Upward -1 (-1- 1)2 4 0 (0 - 1)2 1 1 (1 - 1)2 0 2 (2 - 1)2 1 3 (3 - 1)2 4 The graph of y = (x + 1) 2 moves to the left of the origin while the graph ofy = (x – 1)2 moves to the right of the origin.Try this out Given are the following functions. Tell the directions of each graph, if ittranslates to the right or to the left of the reference graph f(x) = x2. 1. f(x) = (x + 4)2 2. f(x) = (x – 4)2 3. f(x) = ( x – 2)2 8

4. f(x) = (x + 2)25. f(x) = (x + 5)26. f(x) = 2(x – 5)27. f(x) = 3(x – 4)28. y = (x - 6)29. y = (x + 1)210. y = (x - ½)2 Lesson 4 Analyze the Effect on the Changes in k in the Graph of the Function f(x) = ax 2 + k Using the graph of f(x) = x2 as the reference, the value of k translates thegraph vertically, upward if k > 0 or downward if k < 0.Examples: y = x2 + 31. f(x) = x2 + 2 vertex : ( 0, 2 ) y = x2 + 2 axis of symmetry: x= 0 Opening of the graph: Upwards y = x2 Table of valuesx x2 + 2 f(x)2 (2)2 + 2 61 (1)2 + 2 30 (0)2 + 2 2-1 (-1) 2 + 2 3-2 (-2) 2 + 2 62. f(x) = x2 + 3 Table of values vertex : ( 0, 3) x x2 + 3 f(x) axis of symmetry: x = 0 opening of the graph: upwards 2 (2) 2 + 3 7 1 (1) 2 + 3 4 0 (0) 2 + 3 3 -1 (-1) 2 + 3 4 -2 (-2) 2 + 3 7 9

Try this out Tell the direction of each graph, if it translate upwards or downwards .Consider the graph of f(x) = x2 as your reference graph. 1. y = x2 + 6 2. y = x2 – 4 3. y = (x – 3)2 4. y = (x + 2)2 5. y = (x – 2)2 6. y = x2 - 1 7. y = x2 + 5 8. y = x2 + 8 9. y = (x +3)2 10. y = (x – 1)2 Lesson 5 Analyze the Effects of the Changes in a in the Graph of the Function f(x) = a(x-h) 2 + k. Using your previous knowledge regarding the characteristics and differentforms of quadratic functions lets you analyze how the graph of f(x) = ax2 isaffected by both h and k. This would require translation of the graph from boththe x and y axis.Example 1If a = -2, h = -1 and k = -2 the quadratic function becomesy = -2(x + 1) 2 – 2.Vertex: (-1, -2)Axis of symmetry: x = -1Opening of the graph: Downward Table of values f(x)X -2(x + 1)2 -2 -101 -2(1 + 1)2 -2 -40 -2(0 + 1)2 -2 -2-1 -2(-1 + 1)2 -2 -4-2 -2(-2 + 1)2 -2 -10-3 -2(-3 + 1)2-2 10

Example 2: Write the equation of parabola if the graph of y = 2x2 is shifted 4 units tothe left and 1 unit downward. Answer The equation is y = 2(x + 4)2 –1. Try this out A. Sketch the graph of the following quadratic functions. Complete the table of values (choose points symmetric to the value of h) and other characteristics such as vertex, axis of symmetry and direction of the graph. 1. y = -(x –2)2 + 1 2. y = 2(x + 2)2 –3 3. y = (x – 1)2 + 2 4. y = (x + 1)2 –2 5. y = ( x + 1)2 – 2 B. Write the equation for each parabola described. 1. The graph of y = x2 shifted 5 units upward. 2. The graph of y = x2 shifted 3 units downward 3. The graph of y = 2x2 shifted 2 units above the origin 4. The graph of y = x2 shifted 4 units to the right of the origin 5. The graph of y = 3x2 shifted 2 units to the left of the origin 6. The graph of y = 2x2 shifted 3 units to the left and 5 units upward 7. The graph of y = 3x2 shifted 3 units to the right and 2 units downward 8. the graph of y = x2 shifted 3 units to the left and 4 units downward 9. the graph of y = -2x2 shifted 5 units to the right and 3 units upward 10. the graph of y = -3x2 shifted 2 units to the left and 4 units upward Let’s summarize 1. The graph of a quadratic function is called parabola. 2. Graphing quadratic functions there are steps to be followed a. Find the coordinates of the vertex b. Determine the axis of symmetry c. Determine the direction of the opening of the graph d. Prepare the table of values (choose the values of x symmetric to the value of h) 11

3. The graph of the function of the form f(x) = ax2 , as a increases the graph narrows. 4. The graph of f(x) = a(x –h) 2 has the same shape and direction of opening as the graph of f(x) = ax2. But its position is translated h units to the right or left. 5. As the value of k changes, the graph of f(x) = a(x –h) 2 + k is translated k units up or down. What have you learned A. Tell the direction or the opening of the graph of the following functions. 1. y = 2x2 – 4 2. y = x2 + 3 3. y = -x2 + 2x – 5 B. Sketch the graph of the following functions. 4. y = 3x2 – 9x + 2 5. y = -x2 + 2x – 1C. Using f(x) = x2 as the reference graph , which graph is wider or narrower 6. y = 2x2 – 2 y = x2 + 4 7. y = -3x2 + 3 y = -1 x2 - 2 2D. Which graphs translates to right or to the left of the origin. 8. f(x) = (x – 3)2 9. f(x) = 4(x + 2)2 10. f(x) = - 2(x – 4)2 11. f(x) = 3(x + 3)2E. Which graph translates upwards or downwards from f(x) = x2 . 12. f(x) = 3x2 -4 13. f(x) = -2x2 + 3 14. fX) = -3(x – 3)2 + 5 15. f(x) = (x + 4)2 - 3 12

Answer keyHow much do you know A. 1. Downward 2. Upward 3. Upward B. 4. y = -2x2 + 4x - 35. y = x2 – 4x + 2C. 6. narrower wider 7. narrower wider D. 8. to the left 13