Math 4 part 1

Module 1 Polynomial Functions What this module is about This module is about polynomial functions. In the previous lessons youhave learned about linear and quadratic functions. These two belongs to thefamily of polynomials but whose degrees are 1 and 2. In this module, you willlearn about functions of degree greater than 2. What you are expected to learn This module is designed for you to:1. identify a polynomial function from a given set of relations,2. determine the degree of a given polynomial function,3. find the quotient of polynomials by, • algorithm • synthetic division4. State and illustrate the Remainder Theorem5. Find the value of p(x) for x = k by: • synthetic division • Remainder Theorem6. Illustrate the Factor TheoremHow much do you knowAnswer the following:1. One of the following is not a polynomial function. Which is it?a. f(x) = 4x3 + 3x2 + 4x – 12 c. p(x) = x-4 + 8x3 – x2 + 2x + 8b. f(c) = x3 - 6x2 + 12x + 4 d. f(x) = 7x5 – 9x3 + 5x –2

2. What is the degree of the polynomial function f(x) = 2x4 + 3x3 –x2 + 5x – 4?3. Find the quotient and the remainder if y = 3x4 –x3 + 6x2- 11x + 6 divided by 3x-1.4. If f(x) = x3 + 3x2 + 10x + 5,what will be the value of f(x) at x = 3?5. What will be the value of k such that x –1 is a factor of x3 – 3x + k?6. What must be the value of k such that 3 is the remainder when f(x) = x3 + 4x2 – kx ÷ (x -1)?7. What is the remainder when f(x)= x5 – 2x4 + 3x3 – 2x2 – x + 2 is divided by x+ 1?8. Which of the following binomial is a factor of 2x3 + 5x2 –10x –16?a. x – 2 c. x - 1b. x + 2 d. x + 19. If f(x) = 2x4 – x3 –3x2 + x – 5 , what will be the value of f(x) at x = -3?10. What must be the value of k so that x+ 1 is a factor of f(x) = 3x3 + kx2 - x – 2?What you will do Lesson 1Identify and Determine the Degree of the Polynomial Function from a Given Set of Relations A function defined by p(x) = anxn + an-1xn-1 + an-2xn-2 + … + a1x + a0 where nis a positive integer an, an-1, an-2 are called polynomial functions. The exponent ndenotes the degree of the polynomial function.The functions, p(x) = 3x + 4 is of degree 1. p(x) = 4x2 + 15x + 10 is of degree 2. 2

In this lesson, you will study about polynomial functions of degree greaterthan 2. Remember that there are restrictions to be considered to determine if it arelation is a polynomial function. Looking back at the definition, the exponentshould be positive or the value of n > 0.Examples:1. f(x) = x3 – 3x2 + 4x – 12 Polynomial of degree 32. p(x) = x4 – 4x3 – 13x2 + 3x + 18 Polynomial of degree 4 3. f(x) = x-3 + 4x2 + 2x + 1 Not a polynomial. There is a 4. f(x) = 4x5 – 2x3 + 5x – 1 negative exponent or n < 0. x Not a polynomial. There is aTry this out variable x in the denominator.A. Tell whether the following is a polynomial function or not.1. f(x) = 4x3 + 3x2 + 4x – 122. p(x) = x-4 + 8x3 – x2 + 2x + 83. f(x) = x3 - 6x2 + 12x + 44. f(x) = 7x5 – 9x3 + 5x –25. p(x) = 2x-3 + 3x2 + 5x – 36. p(x) = x2 + 3x + 1+ 5 x7. f(x) = x3 + 138. p(x) = 5x - 6 + 2 x + 7 x29. f(x) = 2 x4 + x10. f(x) = 2x4 + 3x3 + 2x + 1B. Determine the degree of the polynomial function.1. p(x) = x4 + 2x3 + 2x + 12. f(x) = x5 – x4 + 2x3 - 3x2 + 4x – 123. p(x) = x6 + 5x5 – 6x4 + 8x3 + 4x2 - 3x + 14. f(x) = x3 + 6x2 + 3x + 95. p(x) = x8 + 4x4 + 2x2 + 16. p(x) = 16x5 - 67. f(x) = -2x + x2 - 5 - 2x38. p(x) = .10x2 + 5x3 – 29. f(x) = 7x - 2x4 + 110. p(x) = x6 + 5x3 - 6 3

Lesson 2Find the Quotient of Polynomials by Division Algorithm Division algorithm is the division process that you are familiar with.Dividing polynomials are the same as dividing numbers. All you have to do is to follow the steps in dividing a polynomial by anotherpolynomial as illustrated in the example below.Example: Divide: Steps1. ( – x2 + 3x3 - 8x + 5) by (x + 2) 1. Arrange the terms of the dividend and divisor according to degree. (3x3 - x2 - 8x + 5) by (x + 2) 2. Divide the first term of the dividend by 3x2 the first term of the divisor to get the x + 2 3x3 - x2 - 8x + 5 first term of the quotient. 3x2 3. Multiply the result in step 2 by the x + 2 3x3 - x2 - 8x + 5 divisor. 3x3 + 6x2 4. Subtract the result from step 3. Bring down the next term of the dividend. 3x2 -7x x + 2 3x3 - x2 - 8x + 5 5. Repeat the entire process using the result in step 4 as the new dividend. 3x3 + 6x2 - 7x2 - 8x 6. Express the result as: 3x2 - 7x + 6 dividend = quotient + remainder x + 2 3x3 – x2 - 8x + 5 divisor divisor 3x3 + 6x2 - 7x2 - 8x - 7x2 -14x 6x + 5 6x +12 -73x3 – x2 - 8x + 5 = 3x2 – 7x + 6 + -7x+2 x+2 The quotient of – x2 + 3x3 - 8x + 5 by x + 2 is 3x2 – 7x + 6 and theremainder is -7. 4

Check by multiplying the quotient to the divisor. Do not forget to add theremainder.2. (x3 – 13x + 12) by (x + 4) Notice the absence of an x2 term in the dividend. x2 x + 4 x3 + 0x2 - 13x + 12 Add a 0x2 term to the dividend. x2 Follow the steps in the first example.x + 4 x3 + 0x2 - 13x + 12 Do not forget to change the sign of the x3 + 4x2 subtrahend when subtracting. x2 - 4xx + 4 x3 + 0x2 - 13x + 12 x3 + 4x2 - 4x2 - 13x x2 - 4x + 3x + 4 x3 + 0x2 - 13x + 12 x3 + 4x2 - 4x2 -13x - 4x2 -16x 3x +12 3x +12 0x3 – 13x + 12 = x2 - 4x + 3x+4The quotient is x2 - 4x + 3.Try this outFind the quotient by dividing the polynomials using division algorithm:1. (3x3 – x2 - 8x + 5) ÷ (x + 2)2. (4x2 +15x + 10) ÷ (x – 2)3. (x3 - 2x2 + 6x + 3) ÷ (x – 3)4. (x3 - 5x2 - 9x + 3) ÷ (x – 4)5. (2x5 + 4x4 + 8x - 1÷ (x + 2)6. x4 - 3x + 5) ÷ (x + 3)7. (x3 - 2x2 + 4) ÷ (x – 3)8. (-10x + 2x4 - 5x3 + 8) ÷ (x – 3)9. (x3 + 3x - 4x2 - 12) ÷ (x – 4)10. (x5 + 32)÷ (x + 2) 5

Lesson 3Find by Synthetic Division the Quotient and the Remainder When P(x) is Divided by (x - c) Another method of dividing polynomials which has a very short and simpleprocedure is called synthetic division. Unlike the usual division which involvesthe four fundamental operations, this method requires only addition andmultiplication applied to the coefficients. This method is applied when the divisoris of the form x - c.Steps to follow in dividing by synthetic division:1. Arrange the terms of the dividend in descending order of exponent.2. Write the numerical coefficient in a row, with 0 representing any missing term.3. Write the constant term c of the divisor x - c at the left hand side of the of the coefficient.4. Bring down the leading coefficient of the dividend. Multiply it by c and add to the second column.5. Multiply the sum obtained in step 4 by c and add to the 3rd column. Repeat this process until you reach the last column.6. The 3rd rows of numbers are numerical coefficient of the quotient. The degree is one less than that of the dividend. The right member is the remainder.Examples: Find the quotients and the remainder using the steps in synthetic division.Write you answer in the form P(x) = Q(x)D(x) + R where, P(x) is the dividend,Q(x) is the quotient, Q(x) is the divisor, and R is the remainder.1. P(x) = x3 + 4x2 + 3x – 2 by x – 3 1 4 3 -2 x=33 3 21 72 7 24 70 1Since Q(x) = x2 + 7x + 24 and R = 70, then 6

P(x) = (x2 + 7x + 24) (x - 3) + 70 in the form P(x) = Q(x)D(x) + R2. P(x) = 3x 4 – 2x3 + 5x 2 – 4x – 2 by 3x + 13x 4 – 2x3 + 5x 2 – 4x – 2 ; x + 1 Divide both divisor and 33 dividend by 3 then follow steps in synthetic division. 1 -2 5 -4 -2 x=- 1 3 33 3 3 1 -2 2- 1 -1 33 33 2 -2 3 1 -1 0 Q(x) = (x3 - x2 + 2x – 2 ) and R = 0Therefore: P(x) = (x3 - x2 + 2x – 2 ) ( x + 1/3) + 03. P(x) = 2x4 – 18x2 – 7 – x3 by x – 3 Arrange exponent in P(x) = 2x4 – x3 – 18x2 + 0x – 7 descending order and represent the missing 2 - 1 - 18 0 -7 term by 03 6 15 -9 -27 2 5 -3 -9 -34Q(x) = 2x3 + 5x2 –3x – 9 and R = –34Therefore: P(x) = (2x3 + 5x2 – 3x –9) ( x –3) -34Try this outA. Use synthetic division to divide the given polynomial P(x) by the givenpolynomial x – c. Write your answer in the form P(x) = Q(x) (x-c) + R1. P(x) = 4x6 + 21x5 – 26x3 + 27x x+52. P(x) = x5 – 3x4 + 4x + 5 x–23. P(x) = 2x3 – 4x2 – 5x +3 x+34. P(x) = x5 + 5x3 – 3x + 7 x–25. P(x) = x4 – 8 x–26. P(x) = 2x3 + 11x + 12 x+47. P(x) = 2x3 – 3x2 + 3x – 4 x–28. P(x) = x5 + 32 x+29. P(x) = 2x4 – 5x3 – 10x + 8 x–310. P(x) = 6x3 – 19x2 + x + 6 x–3 7

B. Find the quotient and the remainder by synthetic division of the polynomialP(x) for the given polynomial x - c. Write your answer in the form P(x) = Q(x)D(x) + R1. P(x) = 4x4 + 12x3 + 9x2 – 8x – 5 2x + 12. P(x) = 15x3 – 19x2 + 24x – 12 3x – 23. P(x) = -9x4 + 9x3 – 26x2 + 26x – 8 3x - 14. P(x) = 3x4 – x3 + 6x2 – 11x + 6 3x – 15. P(x) = 2x3 – 5x2 + 6x + 1 2x –16. P(x) = 2x4 – x3 + 4x2 –12x + 3 2x –17. P(x) = 2x3 – 9x2 + 10x – 3 2x – 18. P(x) = 6x3 – 2x2 – x – 1 x+29. P(x) = 4x4 – 5x2 + 1 310. P(x) = 2x3 + x2 + 12 x– 1 2 x+2 Lesson 4 State and illustrate the Remainder Theorem In the two previous division processes illustrated, a remainder was notedwhen the polynomial is not exactly divisible by another polynomial. You’ll get azero remainder when a polynomial is exactly divisible by another. By substituting the value of (c) of the divisor x – c in the polynomial P(x),you can also test whether a certain polynomial is exactly divisible by another or isa factor by the Remainder Theorem. The Remainder Theorem states that P(c) is the remainder when thepolynomial p(x) is divided by (x – c). The divisor x – c is then restated as x = c.Examples:1. Find the remainder using the remainder theorem if P(x) = x3 + 4x2 + 3x – 2is divided by x – 3.Solution: Instead of using synthetic division, it is easier to solve by substitution.P(x) = x3 + 4x2 + 3x – 2; x = 3 Substitute 3 for x.P(3) = (3) + 4(3) 2 + 3(3) – 2 = 27 + 36 + 9 – 2P(3) = 70 the remainder 8

Hence, the polynomial P(x) = x3 + 4x2 + 3x – 2 is not exactly divisible by x– 3.2. Find the value of P(x) using the remainder theorem if P(x) = x4 + 3x3 - 5x2 –5x - 2 is divided by x + 2.Solution:P(x) = x4 + 3x3 - 5x2 – 5x - 2 ; x = -2 Substitute -2 for x.P(-2) = (-2) 4 + 3(-2) 3 - 5(-2) 2 – 5(-2) -2 = 16 - 24 + 20 - 10 - 2P(-2) = 0 the remainderHence P(x) = x4 + 3x3 - 5x2 – 5x - 2 is exactly divisible by x + 2.You can also solve an equation using the Remainder Theorem. In the nextexample, the polynomial P(x) is equated to the remainder to solve for the value ofk, the numerical coefficient of the x term.Example: Find the value of k when polynomial 3x2 + kx + 4 is divided by x – 1 and theremainder is 2.Solution: 3x2 + kx + 4 = 2 The polynomial is equal to the remainder 2.3(1) 2 + k(1) + 4 = 2 Substitute 1 for x, then solve for k. 3+ k + 4 = 2 k=2–7 k = -5Check by synthetic division 3x2 - 5x + 4 Substitute k by – 5 in the original expression, 3x2 - kx + 4. 3 -5 41 3 -2 remainder 3 -2 2Try this outA. Find the remainder when P(x) is divided by x – c using the remainder theorem. 9

1. P(x) = (x3 – 7x2 + x + 10) x–22. P(x) = (x4 + 10x3 - 8x –80) x + 103. P(x) = (x5 + 2x4 – 3x3 + 4x2 - 5x + 2) x–14. P(x) = (x3 + 3x2 + 10x + 5) x–35. P(x) = (x3 + 125) x+56. P(x) = (x3 - 4x2 – 3x + 18) x+27. P(x) = (x5 + 5x3 - 3x + 7) x–28. P(x) = (x5 + 5x3 - 3x + 7) x+29. P(x) = (x59 + 3x35 – 5x7 + 9x + 8) x–110. P(x) = (x99 - 2x81 + 3x5 – 5) x+1B. Find the remainder when a polynomial is divided by x - c using the remainder theorem.1. (-x3 + 5x2 – 10x + 3) ÷ (x – 4)2. (-x + 2x3 – 3x + 3) ÷ (x + 2)3. (-2x3 + 3x2 – 3x + 5) ÷ ( x + 1)4. (-9x + 2x3 – 20) ÷ (x + 2)5. (-3x – 15x3 + 4x4 + 20) ÷ (x –3)6. (-5x3 – 12x2 + 10x – 6) ÷ (x + 3)7. (-x3 + 6x2 – 10x + 8) ÷ ( x – 4)8. (-x4 – 3x3 –2x2 + 12x + 72) ÷ ( x + 6)9. (-2x4 – 9x3 + 14x2 + 68) ÷ ( x – 2)10. (-5x5 – 3x4 + 4x + 5) ÷ (x – 1)C. Find the value of the following function using the remainder theorem.1. p(x) = 2x3 - 5x2 + 3x -7 x = -32. p(x) = 5x3 + 7x2 + 8 x = -23. p(x) = 4x4 + 5x3 + 8x2 x=44. p(x) = 3x3 - 7x2 + 5x –2 x = -25. p(x) = 4x3 + 2x + 10 x = -36. p(x) = 5x4 + 6x3 + 10x2 x=57. p(X) = 6x2 + 3x – 9 10

x=1 8. p(x) = 2x3 + 4x2 – 5x + 9 x = -3 9. p(x) = 2x4 - 9x3 + 14x2 – 8 x=2 10. p(x) = 2x4 - 9x3 + 14x2 – 8 x = -2D. Given a condition, determine the value of k. 1. When kx3 – x2 + 2x – 30 is divided by (x – 2), the remainder is 2. 2. When 8x3 – 4x2 – 7x + k is divided by (x – 1), the remainder is 5. 3. When x5 + x4 – 4x3 – 4x2 – 8x + k is divided by x – 2, the remainder is 0. 4. When kx2 – x + 3 divided by x + 1, the remainder is 5. 5. When 6x2 = 4x + k divided by x + 3, the remainder is 2. Lesson 5Find the Value of P(x) for x = c by Synthetic Division and the Remainder Theorem The synthetic division and remainder theorem are two ways used to findthe value of P(x). You have seen in the previous lessons that the last valueobtained in synthetic division is equal to the value of the remainder. Now, how isthis related to the remainder theorem.Let’s find out by comparing the two processes.Examples:1. Use synthetic division and remainder theorem to find the value of P(x) = x4 – 2x3 – x2 – 15x + 2 at x =12Solution:a. By synthetic division: 1 -2 -1 -15 2 the remainder12 12 120 1428 16956 1 10 119 1413 16958 11

b. by remainder theorem: P(12) = x4 – 2x3 – x2 – 15x + 2 = (12) 4 – 2(12) 3 – (12) 2 – 15(12) + 2 = 20,736 – 3456 – 144 – 180 + 2 P(12) = 16,958 Notice that the same value was obtained for the two processes. We cannow say that P(x) = R, and P(x) = 16,958.2. Use synthetic division and remainder theorem to find the value of P(x) = 2x3 + 8x2 + 13x –10 if x = -3Solution:a. by synthetic division: 2 8 13 -10 the remainder-3 -6 -6 -21 2 2 7 -31P(x) = -31b. by remainder theorem:P(-3) = 2 x3 + 8x2 + 13x – 10 = 2 (-3) 3 + 8(-3) 2 + 13(-3) – 10 = -54 + 72 - 39 -10P(-3) = -31 the remainder Again, notice that the value obtained using synthetic division andremainder theorem yield the same value for P(-3).Try this outA. Find the value of the P(x) for the given x using synthetic division andremainder theorem.1. P(x) = x3 – 4x2 + 2x – 6 x=42. P(x) = x5 – 3x2 – 20 x=23. P(x) = 2x3 + 3x2 – x – 79 x=94. P(x) = x3 – 8x2 + 2x + 5 x=35. F(x) = x4 + x3 + x2 + x + 1 x=4 12

6. P(x) = 3x4 + 8x2 – 1 x = -4 7. P(x) = 6x3 + 9x2 – 6x + 2 x=2 8. P(x) = x4 – 2x3 + 4x2 + 6x – 8 x=3 9. P(x) = 4x4 + 3x3 – 2x2 + x + 1 x = -1 10. P(x) = 2x3 + 8x2 – 3x – 1 x = -2B. Using synthetic division or remainder theorem, find the value of the polynomialfor the given value of x.1. x4 - 2x2 - x – 6 x=22. x4 - 4x3 + 3x2 + 12 x = -33. -x4 - x3 + x – 5 x=14. x3 - x2 - x - 5 x= 15. x5 - 6x3 - x – 7 x = -26. x6 - x5 - x4 - x – 3 x=27. 4x5 - 3x + 122 x = -28. x5 - 4x3 - 3x – 2 x=39. x3 - 2x2 - 5x – 6 x = -110. 2x2 - 19x + 35 x=7C. Using synthetic division or remainder theorem, find the value of y for thegiven x.1. y = 6x3 - 17x2 + 14x + 8 , x= 1 32. y = 8x3 - 14x2 - 5x – 1 x=1 23. y = 64x3 + 1 x=-1 44. y = 6x4 - 3x2 + 1 x=-1 25. y = 4x4 + 2x2 + 1 x=1 4 Lesson 6 Illustrate the Factor Theorem In your experience with numbers, you obtain a remainder of zero when anumber is exactly divisible by another number. We can say that the divisor is afactor of the dividend in that case. Same is true with polynomials. A zero remainder obtained when applied using the Remainder Theoremwill give rise to another theorem called the factor theorem. This is a test to find ifa polynomial is a factor of another polynomial. 13

The Factor Theorem states: Let P(x) be a polynomial. If c is a zero of P that is P(c) = 0, then (x - c) isa factor of P(x). Conversely, if (x - c) is a factor of P(x) then, c is a zero of P. Simply, if zero is obtained as a remainder when c is substituted to thepolynomial P(x), then the polynomial x – c is factor of P(x).Examples:1. Show that x - 2 is a factor of x3 + 7x2 + 2x – 40Solution:a. Using the remainder theorem P(x) = x3 + 7x2 + 2x –40 if x = 2 = (2) 3 + 7(2) 2 + 2(2) – 40 = 6 + 28 + 4 - 40 P(x) = 0Since P(x) = 0, then x - 2 is a factor of x3 + 7x2 + 2x – 40.b. Using another method, by synthetic division 1 7 2 -40 the remainder 2 2 18 40 1 9 20 0Since the remainder is 0, then x - 2 is a factor of x3 + 7x2 + 2x –40.2. Determine if ( x – 3) is a factor of (2x4 - x3 – 18x2 – 7)Solution:a. by remainder theorem P(3) = 2x4 - x3 – 18x2 - 7 if x = 3 = 2(3)4 - (3)3 - 18(3)2 - 7 = 2(81) - 27 - 18(9) - 7 = 162 –27 –162 – 7 P(3) = -34 14

Since the P(3) = -34, which is not 0 then, (x – 3) is not a factor of (2x4- x3 – 18x2 – 7).b. Using synthetic division 2 -1 -18 0 -7 remainder 36 15 -9 -27 25 -3 -9 -34Since r = -34, then x – 3 is not a factor of the second polynomial. Again, we can use this knowledge to solve equations. If the polynomial x– c is a factor of P(x), then you can equate P(x) to zero. An example is given toyou below.Example:Find the value of k so that polynomial x – 2 is the factor of 2x3 – kx - 3.Solution:By remainder theorem: 2x3 – kx – 3 = 0 Since x-2 is a factor of the polynomial then equate to 0.2(2) 3 – k(2) – 3 = 0 Substitute x by 2 and perform operations. 2(8) – 2k – 3 = 0 Solve for k. 16 – 2k – 3 = 0 -2k = - 13 k = 13 2Let us check using synthetic division;2x3 – 0x2 – kx – 3 Supply the missing term by 02x3 – 0x2 – 13 x – 3 Substitute k by 13 2 2 2 0 - 13 -3 22 48 324 3 0 2 15

Try this outA. Tell whether the second polynomial is a factor of the first . 1. P(x) = 3x3 – 8x2 + 3x + 2; (x – 2) 2. P(x) = 2x4 + x3 + 2x + 1; (x + 1) 3. P(x) = x3 + 4x2 + x – 6; (x + 3) 4. G(x) = 4x3 – 6x2 + 2x + 1; (2x – 1) 5. H(x) = x3 – 6x2 + 3x + 10; (x – 1)B. Answer the following: 1. Which of the following is a factor of f(x) = x3 – 7x + 6 a. x + 2 b. x – 3 c. x – 1 d. x + 1 2. Which of the following is the factor of f(x) = 2x3 + 3x2 – 3x – 2 a. x + 2 b. x – 3 c. x + 1 d. x – 2 3. Which is a factor of p(x) = 2x4 + 3x3 –4x2 –3x +2 a. x – 2 b. x – 1 c. 2x + 1 d. x - 3 4. Which is a factor of g(x) = x3 – 2x2 – 5x + 6 a. x – 2 b. x + 1 c. x + 3 d. x – 3 5. Which is a factor of p(x) = x3 + 3x2 – 9x – 27 a. x + 3 b. x + 2 c. x – 2 d. x – 3 6. Which is a factor of p(x) = 3x3 + 2x2 – 7x + 2 a. x + 1 b. x – 2 16

c. x – 1 3 d. x + 1 3 7. Which is a factor of p(x) = x4 – 8x3 + 2x2 + 5 a. x + 1 b. x – 5 c. x + 5 d. x – 1 8. Which is a factor of f(x) = x4 – 2x3 – 3x2 + 8x – 4 a. x – 1 b. x + 1 c. x + 3 d. x – 3 9. Which is a factor of f(x) = x4 + 6x3 + 9x2 – 4x – 12 a. x + 3 b. x – 3 c. x + 1 d. x – 2 10. Which is a factor of f(x) = 2x3 + 5x2 + x – 2 a. x + 2 b. x –1 c. x – 2 d. x + 3C. Determine the value of k which is necessary to meet the given condition. 1. (x – 2) is a factor of 3x3 – x2 – 11x + k 2. (x + 3) is a factor of 2x5 + 5x4 + 3x3 + kx2 – 14x + 3 3. (x + 1) is a factor of –x4 + kx3 – x2 + kx + 10 4. (x + 2) is a factor of x3 + x2 + 5x + k 5. (x – 1) is a factor of x3 – x2 – 4x + k 6. (x – 5) is a factor of x3 – 3x2 – kx - 5 7. (x + 1) is a factor of 3x3 + kx2 – x – 2 8. (x + 4) is a factor of kx3 + 4x2 – x - 4 9. ( x + 5) is a factor of kx2 + 4x - 5 10. (x – 2) is factor of x3 + 3x2 – kx + 2 17

Let’s summarize1. Synthetic division is another method in finding the quotient and the remainder.2. Remainder theorem can be used to find the value of a function, that is P( c) is the remainder when a polynomial p(x) is divided by ( x- c).3. Factor theorem: The binomial (x - a) is a factor of the polynomial P( x) if and only if P(x) = 0.What have you learned1. Which of the following is a polynomial function?a. P(x) = 3x-3 – 8x2 + 3x + 2 c. P(x) = 2x4 + x3 + 2x + 1b. P(x) = x3 + 4x2 + 1 – 6 d. G(x) = 4x3 – 6 + 2x + 1 x x22. What is the degree of the polynomial function f(x) = 5x – 3x4 + 1?3. What will be the quotient and the remainder when y = 2x3 – 3x2 – 8x + 4 is divided by (x +2)?a. q(x) = 2x2 – 7x + 6 , R = -8b. q(x) = 2x2 – 7x + 6, R = 8c. q(x) = 2x2 – 7x – 6, R = -8d. q(x) = 2x2 – 7x - 6 , R = 82. If f(a) = 2a3 + a2 + 12, what will be the value of f(a) at a = -2? a. 1 b. -1 c. 0 d. 23. What must be the value of k so that when f(x) = kx2 - x + 3 divide by (x + 1)and the remainder is 5? a. 2 b. -2 c. 0 d. 1 18

4. What must be the value of k in the function f(x) = x4 + x3 – kx2 – 25x – 12 so that (x – 4) is a factor. a. -12 b. -13 c. 13 d. 125. What is the remainder when f(x) = x4 + 3x2 + 4x – 1 divided by (x – 1)? a. 7 b. -7 c. 6 d. 56. Which of the following binomial is a factor of f(x) = x3 – x2 – 5x – 3? a. x + 1 b. x + 2 c. x -3 d. x –27. If f(x) = x3 + 4x2 + 3x – 2, what will be the value f(x) at x = 3? a. -70 b. 70 c. 50 d. –508. For what value of k , when x3 + 4x2 – kx + 1 ÷x + 1 the remainder is 3. a. -1 b. 1 c. 2 d. -2 19

Answer keyHow much do you know 1. c 2. 4 3. Q(x) =(x3 + 2x – 3), R = 3 4. 89 5. 2 6. 2 7. -5 8. a 9. 154 10. 4Try this outLesson 1A. 1. function 2. not function 3. function 4. function 5. not function 6. not function 7. function 8. not function 9. function 10. functionB. 1. fourth 2. fifth 3. sixth 4. third 5. eighth 6. fifth 7. third 8. third 9. fourth 10. sixthLesson 2 1. 3x2 – 7x + 6 + - 7 x+2 20

2. 4x + 23 + 56 x−2 3. x2 + x + 9 + 30 x−3 4. x2 - x -136 + - 49 x−4 5. 2x4 + 8 + - 17 x+2 6. x3 - 3x2 + 9x - 30 + 95 x+3 7. x2 + x + 3 + 13 x−3 8. 2x3 + x2 + 3x - 1 + 5 x−3 9. x2 + 3 10. x4 – 2x3 + 4x2 - 8x + 16Lesson 3A. 1. 4x6 + 21x5 – 26x3 + 27x = (4x5 + x4 –5x3 – x2 + 5x + 2)(x + 5) –10 2. x5 – 3x4 + 4x + 5 = (x4 – x3 – 2x2 –4x –4)(x - 2) – 3 3. 2x3 – 4x2 – 5x +3 = (2x2 – 10x + 25)(x + 3) –72 4. x5 + 5x3 – 3x + 7 = (x4 + 2x3 + 9x + 18x + 33)(x – 2) + 73 5. x4 – 8 = (x3 + 2x2 + 4x + 8)(x – 2) + 8 6. 2x3 + 11x + 12 = (2x2 – 8x + 43)(x + 4) –160 7. 2x3 – 3x2 + 3x –4 = ( 2x2 + x + 5)(x – 2) + 6 8. x5 + 32 = (x4 – 2x3 + 4x2 –8x + 16)(x + 2) + 0 9. 2x4 – 5x3 – 10x + 8 = (2x3 + x2 + 3x – 1)(x –3) + 5 10. 6x3 –19x2 + x + 6 = (6x2 – x –2)(x – 3) + 0B. 1. P(x) = (2x3 + 5x2 + 2x – 5) ( 2x + 1) + 0 2. P(x) = (5x2 – 3x + 6) (3x –2) +0 3. P(x) = (-3x3 + 2x2 –8x + 6) (3x – 1) - 2 4. P(x) = (x3 + 2x –3) (3x –1) +3 5. P(x) = (x2 –2x + 2) (2x –1) + 3 6. P(x) = (x3 + 2x –5) (2x –1) – 2 7. P(x) = (x2 –4x + 3) (2x –1) 8. P(x) = (6x2 – 6x + 3) (x + 2 ) - 3 3 9. P(x) = (4x3 + 2x2 –4x –2) (x – 1 ) 2 10. P(x) = (2x2 – 3x + 6) (x + 2) + 0 21

Lesson 4 A. 1. P(2) = -8 2. P(-10) = 0 3. P(1) = 1 4. P(3) = 89 5. P(-5) 0 6. P(-2) = 0 7. P(2) = 73 8. P(-2) = -59 9. P(1) = 16 10. P(-1) = -7B. 1. - 21 2. -5 3. 13 4. -18 5. -70 6. -9 7. 80 8. -720 9. 20 10. 1C. 1. p( -3) = -115 2. p(-2) = -4 3. p(4) = 1,472 4. p(-2) = -64 5. p(-3) = -104 6. p(5) = 4,125 7. P(1) = 0 8. P(-3) = 6 9. P(2) = 8 10. P(-2) = 152D. 1. k = 4 2. k = 8 3. k = 16 4. k = 1 5. k = -40Lesson 5A. 1. P(x) = x3 – 4x2 + 2x – 6 if x = 4 22

a. by synthetic division 1 -4 2 -6 4 40 8 102 2 b. by remainder theorem x3 – 4x2 + 2x – 6 (4)3 – 4(4) 2 + 2(4) – 6 64 –64 + 8 – 6 =2 2. P(x) = x5 – 3x2 – 20 if x= 2 x5 + 0x4 + 0x3 – 3x2 +0x – 20 a. by synthetic division 1 0 0 -3 0 -20 2 2 4 8 10 20 1 2 4 5 10 0 a. by remainder theorem (2)5 – 3(2)2 – 20 32 -12 - 20 =03. P(x) = 2x3 + 3x2 – x – 79 if x = 9 a. by synthetic division 2 3 -1 - 79 9 18 189 1692 2 21 188 1613 b. by remainder theorem 2(9)3 + 3(9)2 – 9 – 79 2(729) + 3(81) –9 –79 1458 + 243 –88 = 1613 23

4. P(x) = x3 – 8x2 + 2x + 5 if x = 3 a. by synthetic division 1 -8 2 5 3 3 -15 -39 1 -5 -13 -34 b. by remainder theorem (3)3 – 8(3)2 + 2(3) + 5 27 - 72 + 6 + 5 = -345. F(x) = x4 + x3 + x2 + x + 1, if x = 4 a. by synthetic division 111 11 4 4 20 84 340 1 5 21 85 341 b. by remainder theorem (4)4 + (4)3 + (4)2 + 4 + 1 256 + 64 + 16 + 5 = 3416. P(x) = 3x4 + 8x2 – 1, if x = -4 a. by synthetic division 3 0 8 0 -1 -4 -12 48 -224 896 3 -12 56 -224 895 b. by remainder 3(-4)4 + 8(-4)2 – 1 3(256) + 8(16) - 1 768 + 128 - 1 = 895 24

7. P(x) = 6x3 + 9x2 – 6x + 2, if x = 2 a. by synthetic division 6 9 -6 2 2 12 42 72 6 21 36 74 b. by remainder theorem 6(2)3 + 9(2)2 – 6(2) + 2 6(8) + 9(4) - 12 + 2 48 + 36 -10 = 74 8. P(x) = x4 – 2x3 + 4x2 + 6x – 8, if x = 3 a. synthetic division 1 - 2 4 6 -8 3 3 3 21 81 1 1 7 27 73 b. by remainder theorem (3)4 – 2(3)3 + 4(3)2 + 6(3) – 8 81- 54 + 36 + 18 – 8 27 + 54 – 8 = 739. P(x) = 4x4 + 3x3 – 2x2 + x + 1, if x = - 1 a. by synthetic division 4 3 -2 1 1 -1 -4 1 1 -2 4 -1 -1 2 -1 b. by remainder theorem 4(-1)4 + 3(-1) 3– 2(-1)2 + (-1) + 1 4 -3 -2 -1 + 1 = -1 25

10. P(x) = 2x3 + 8x2 – 3x – 1, if x = -2 a. by synthetic division 2 8 -3 -1 -2 -4 -8 22 2 4 -11 21 b. check by remainder theorem 2(-2)3 + 8(-2)2 – 3(-2) – 1 2(-8) + 8(4) + 6 – 1 -16 + 32 + 5 = 21B. 1. 0 2. 228 3. -6 4. -6 5. 11 6. 11 7. 0 8124 9. -4 10. 0C. 1. 11 2. -6 3. 0 4. 5 8 5. 69 64Lesson 6A. 1. is a factor 2. is a factor 3. is a factor 4. is not a factor 5. is not a factorB. 1. c 2. a 26

3. b 4. d 5. a 6. c 7. d 8. a 9. a 10. aC. 1. k = 2 2. k = 13 3. k = 4 4. k = 14 5. k = 4 6. k = 9 7. k = 4 8. k = 1 9. k = 1 10. k = 11What have you learned1. c2. 43. a4. c5. d6. c7. a8. c9. b10. a 27

Module 1 Quadratic Functions What this module is about This module is about identifying quadratic functions, rewriting quadratic functions ingeneral form and standard form, and the properties of its graph. As you go over thediscussion and exercises, you will understand more about this function, and how todifferentiate it from a linear function which you have learned in the previous modules. Enjoylearning and do not hesitate to go back if you think you are at a loss. What you are expected to learn This module is designed for you to: 1. identify quadratic functions f(x) = ax2 + bx + c 2. rewrite a quadratic function f(x) = ax2 + bx + c in the form f(x) = a(x – h)2 + k and vice versa 3. given a quadratic function, determine • highest or lowest point (vertex) • axis of symmetry • direction or opening of the graph How much do you know 1. Tell whether the function is quadratic or linear. a. y = 2x + 3 b. y = 3x2 + 5x – 6 c. f(x) = -2(x + 1)2 - 5 d. f(x) = 7(4x + 5) e. f(x) = 9 – 2x 2. Which of the following table of ordered pairs represents a quadratic function? a. x -2 -1 0 1 2 y 1 -2 -3 -2 1 b. x 0 1 2 3 4 y 2 5 8 11 14 1

c. x -2 -1 0 1 2 y 8 1 0 -1 -8d. x -1 0 1 2 3 y 10 7 4 1 -23. What is f(x) = (x + 1)2 – 3 in general form? a. f(x) = x2 + 2x – 4 b. f(x) = x2 + 2x – 2 c. f(x) = x2 + 2x + 2 d. f(x) = x2 + 2x + 44. How is f(x) = x2 – 6x + 14 written in standard form? a. f(x) = (x + 3)2 + 2 b. f(x) = (x – 3)2 – 5 c. f(x) = (x – 6)2 + 5 d. f(x) = (x – 3)2 + 55. Which of the following is the graph of a quadratic function?a. y c. y x xb. y d. y x x6. What do you call the graph of a quadratic function? a. Parabola b. Line c. Circle d. Curve 2

(x + 2) cm7. Which of the following quadratic functions will open upward? a. y = 2 – 3x – 5x2 b. y = -(x + 4)2 c. y = -(3x2 + 5x – 1) d. y = 4x2 – 12x + 9 8. Determine the vertex of the quadratic function f(x) = 3x2 – 6x + 5. a. (1, 2) b. (-1, 2) c. (1, -2) d. (-1, -2) 9. Which of the following is the axis of symmetry of y = -2x2 + 12x – 23? a. x = -3 b. x = -5 c. x = 3 d. x = 5 10. What is the maximum value of f(x) = -5(x + 1)2 + 4? a. y = -5 b. y = 4 c. y = -1 d. y = -4 What you will do Lesson 1 Identifying Quadratic Functions Consider a rectangle whose width is (x + 2) cm and whose length is (x +5) cm. How do you find the area of this rectangle? The situation above is illustrated in the figure that follows. (x + 5) cm 3

You recall that the formula for the area of a rectangle is A = lw where l isthe length and w is the width. Thus, the area of the given rectangle is – A = [(x + 5) cm] [(x + 2) cm] A = (x + 5)(x + 2) cm2Multiplying the binomial then simplifying, the area of the rectangle is - A = (x2 + 5x + 2x + 10) cm2 A = (x2 + 7x + 10) cm2 If x = 3, the area of the rectangle can be obtained by substituting in theequation above. A = 32 + 7(3) + 10 cm2 A = 9 + 21 + 10 cm2 A = 40 cm2 Notice that in the given example, the area of the rectangle is a function ofits dimension. Thus, the area of the rectangle can also be written in functionalnotation as Area = f(x) = (x2 + 7x + 10) cm2. Observe that the highest exponent is 2. Hence, the degree of f(x) = x2 +7x + 10 is 2 which is called a quadratic function. The following are examples of quadratic functions. 1. f(x) = x2 + 7 2. f(x) = 6x2 – 4x + 3 3. y = 9 + 2x – x2 4. y = x2 – 6x - 16 5. f(t) = t t −1 2 6. y = (x + 7)2 - 9 Why do you think the examples above are called quadratic functions? The following are not quadratic functions. 1. y = 5x + 8 2. f(x) = 5x + 2 3. f(x) = x3 – 27 4. y = x − 2x x + 3 5. y = x(x2 + 7x - 1) Why do you think the examples above are not quadratic functions? 4

Now that you know how to identify a quadratic function given an equation,how will you identify a quadratic function from a given set of ordered pairs or atable of values?Example 1: Consider the ordered pairs of values for the quadratic function f(x) = x2 forthe integers -3 ≤ x ≤ 3. {(-3, 9), (-2, 4), (-1, 1), (0, 0), (1, 1), (2, 4), (3, 9)} The ordered pairs of values above can also be presented using a table ofordered pairs as shown below.. x -3 -2 -1 0 1 2 3 y9 4 1 0149 Observe the characteristics of a quadratic function. when ordered pairsare given. 1 1 1 1 1 1 Differences in x x -3 -2 -1 0 1 2 3 Y = f(x) 9 4 1 0 1 4 9 -5 -3 -1 1 3 5 Differences in y Notice that the differences in x are equal while the differences in y are not.Let us call the differences in y obtained above as first differences in y.Look what happens when second differences in y is obtained. 1 111 1 1 Differences in x x -3 -2 -1 0 1 2 3y = f(x) 9 4 1 0 1 4 9 -5 -3 -1 1 3 5 First differences in y 2 2 22 2 Second differences in y Observe that the second differences in y are equal. Hence, for thequadratic function, f(x) = x2, equal differences in x produce equal seconddifferences in y. The method presented above is called the equal differencesmethod. 5

Example 2:Consider the table of values for the quadratic function f(x) = 5 – 2x2. x -3 -1 13579y = f(x) -13 3 3 -13 -45 -93 -157Solution:Verify if equal differences in x will produce equal second differences in y. 2 2 2 2 2 2 Differences in x x -3 -1 13579y = f(x) -13 3 3 -13 -45 -93 -157 16 0 -16 -32 -48 -64 First differences in y -16 -16 -16 -16 -16 Second differences in y Observe that like in f(x) = x2, the table of values for f(x) = 5 – 2x2 showedthat equal differences in x produced equal second differences in y = f(x). Thisis true for all quadratic functions.Example 3: Determine if the ordered pairs of numbers given in the table belowrepresents a quadratic function of not. x -6 -4 -2 0 2 4 6 8 y -7 -3 1 5 9 13 17 21Solution: Apply the same method as seen in examples 1 and 2 and see if equaldifferences in x will also produce equal second differences in y. 2 2 22222 x -6 -4 -2 0 2 4 6 8 y -7 -3 1 5 9 13 17 21 4444444 Notice that the differences in x produced equal first differences in y.Thus, the ordered pairs of numbers in the given table does not represent aquadratic function. 6

Try this outA. Tell whether the following functions are quadratic functions or not. Explain.1. f(x) = x2 - 92. f(x) = 3x + 153. f(x) = 24 + 5x - x24. f(x) = 27 – 4x25. f(x) = 2(x-6)2 + 16. f(x) = (3x + 2)(x - 5)27. f(x) = 5x2 + x – 28. f(x) = 4-x 59. f(x) = − 5 x2 610. f(x)= 7 x + 3 24B. Using the equal differences method, determine which of the following ordered pairs represent a quadratic function. Justify your answer. 1. {(-1, 11), (0, 6), (1, 3), (2, 2), (3, 3), (4, 6), (5, 11)} 2. {(-3, -35), (-2, -16), (-1, -9), (0, 0), (1, -7), (2, 0), (3, 19) 3. {(1, 5), (3, 13), (5, 29), (7, 53), (9, 85)} 4. {(-2, -13), (-1, -6), (0, -5), (1, -4), (2, 3)} 5. {(-5, 40), (-4, 28), (-3, 18), (-2, 10), (-1, 4), (0, 0)} 6. x -5 -2 1 4 7 y 64 91 100 91 64 7. x -3 -2 -1 0 1 y 56 37 30 29 28 8. x -4 -2 0 2 4 y 39 24 9 -6 -21 9. x -3 -1 1 3 5 y 25 10 -5 -20 -35 10. x -10 -5 0 5 10 y -20 -85 -100 -85 -20 7

C. The sum of two numbers is 12. 1. If one number is represented by x, what is the other number? 2. If f(x) represents their product, express f(x) in terms of x. 3. Give a table of values for this relation for the integers 0 ≤ x ≤ 12. 4. Determine whether the table of values represents a quadratic function or not. 5. Determine the product of each pair of numbers. 6. Which pair of numbers in (a) gives the greatest product? 7. Which pair of numbers in (a) gives the least product? Lesson 2 Rewriting Quadratic Functions from f(x) = ax2 + bx + c to f(x) = a(x-h)2 + k and vice versa In lesson 1, you learned that f(x) = ax2 + bx + c is the standard form of aquadratic function. This function can be written in an equivalent form using the processof completing the square. Study the steps as shown below.f(x) = ax2 + bx + c Standard form of a quadratic function Factor out a from x2 and x termsf(x) = a(x2 + b x) + c a Complete the square by adding and subtracting a b 2  b  b  2  a b  2 x   2a f(x) = a 2 + x + + c −  a  2a    2a f(x) =  x 2 + b x + b2  +c − a b2  Expand the terms added and a a 4a 2  4a 2 subtracted in the previous step  f(x) = a x 2 + b x + b2  2 + c − b2 Simplify a b2  to b2 a 4a 2 4a 4a 2 4af(x) = a x + b 2 + 4ac −b2 Factor the trinomial inside the bracket  2a  4a and simplify the last two terms From the result let b = − h and 4ac −b2 = k . Substituting this to the equation 2a 4aabove will result to f(x) = a(x – h)2 + k. Hence, f(x) = ax2 + bx + c is equivalent to f(x) =a(x – h)2 + k. 8

Examples:Rewrite the following quadratic functions in the form f(x) = a(x- h)2 + k.1. f(x) = x2 - 2x - 152. f(x) = 2x2 - 73. y = 4 + x – 3x24. y = 4x2 + 5xSolution 1: Using completing the square.1. f(x) = x2 - 2x – 15 Factor out 15 in the x terms. f(x) = (x2 - 2x) – 15 Complete the square inside the f(x) = (x2 - 2x + 1) - 15 – 1 parenthesis by adding andf(x) = (x - 1)2 – 16 subtracting 1. Factor the trinomial inside the parenthesis and simplify the last 22. f(x) = 2x2 – 7 terms f(x) = 2x2 - 0x – 7 Write function in the form f(x) = ax2 + bx + c.f(x) = 2 x2 − 0 x  − 7 Factor out 2 in the x terms.  2 Complete the square by addingf(x) = 2  x2 − 0 x + 0 − 7 − 0 and subtracting 0. 2  Factor the trinomial inside thef(x) = 2(x - 0)2 – 7 parenthesis and combine the last two terms.3. y = 4 + x - 3x2 Rewrite the equation in the y = -3x2 + x + 4 form y = ax2 + bx + c.y = -3  x2 − 1 x  + 4 Factor out -3 in the x terms  3  1  1  2  (−3) 1  2 Complete the square by adding 3  and subtracting − 3 − 1 2y = − x 2 − x + − + 4 − −  3  6    6y = − 3 x2 − 1 x + 1  + 4 − (−3) 1   6  3 36   36  Square the added number.y = − 3 x− 1 2 + 4 −  − 1  Factor the trinomial inside the  6  12  parenthesis and reduce the fraction to the lowest term.f(x) = − 3 x− 1 2 + 49  6  12 Add the last two terms 9

4. y = 4x2 + 5x Giveny = 4  x2 + 5 x  Factor out 4 in the x terms  4 Complete the square by addingy = 4  + 5 x +  5  2  − 4 5  2 and subtracting 4 − 5 2 x2   4  8    8   8 Square the added numbery = 4  x2 + 5 x + 25  − 4 25   4 64   64  Factor the trinomial inside the parenthesis and reduce they = 4  x + 5 2 − 25 fraction to the lowest term  8  16Solution 2:Using the formula in solving the values of h and k: In the relation –h = b , the value of h can be obtained using the multiplication 2aproperty of equality so that h = − b . 2a1. f(x) = x2 - 2x - 15. Substitute the values a = 1, b = -2, and c = -15 in the formula. h = −b = −(−2) = 2 = 1 2a 2(1) 2 k = 4ac −b2 = 4(1)(−15)−(−2)2 = −60−4 = −64 = −16 4a 4(1) 4 4 Substituting the values of h and k to f(x) = a(x – h)2 + k. Thus, f(x) = x2 – 2x – 15 is equivalent to f(x) = (x – 1)2 - 16.2. f(x) = 2x2 – 7. Substitute the values a = 2, b = 0 and c = -7 in the formula. h = −b = −0 = 0 =0 2a 2(2) 4 10

k = 4ac − b2 = 4(2)(−7) − 02 = − 56 − 0 = − 56 = −7 4a 4(2) 8 8 Substitute the values of h and k to f(x) = a(x – h)2 + k. Therefore, f(x) = 2x2 – 7 is equivalent to f(x) = 2(x – 0)2 – 7.3. y = 4 + x – 3x2. Here, a = -3, b= 1, c = 4 h = −b = −(1) = −1 = 1 2a 2(−3) -6 6k = 4(−3)(4)−(1)2 = −48−1 = −49 = 49 4(−3) −12 −12 12 Thus, y = − 3 x- 1 2 + 49 .  6  124. y = 4x2 + 5x. Here a = 4, b = 5 and c = 0.h= -b = -5 = -5 2a 2(4) 8k = 4(4)(0) - (5)2 = 0 - 25 = - 25 4(4) 16 16 4x  5  2 + - 25  4 x + 5 2 − 25  8  16   8 16Hence, y = - - or y = Observe that the two solutions resulted to the same answer. Thus, a quadraticfunction in the form f(x) = ax2 + bx + c can be transformed in the form f(x) = a(x – h)2 + kby completing the square or the relation h = − b and k = 4ac − b2 . 2a 4a Now, how will you transform a quadratic function in the form f(x) = a(x – h)2 + k tothe standard form f(x) = ax2 + bx + c?To do this, simply follow the given steps. 1. Expand the square of the binomial indicated in the function. 2. Multiply the result by the value of a. 3. Combine the similar terms.Now, study the examples below. 11

Examples:Transform the following equation to standard form.1. f(x) = (x – 3)2 – 72. f(x) = -2[x – (-5)]2 + 503. y = 5(x + 4)2 – 34. y = − 3 (x − 2)2 + 9 4Solutions:1. f(x) = (x – 3)2 + 7 Square the binomial f(x) = x2 – 6x + 9 + 7 Combine the similar terms f(x) = x2 – 6x + 162. f(x) = -2[x – (-5)]2 + 50 Simplify the term inside the parenthesis f(x) = -2(x + 5)2 + 50 Square the binomial f(x) = -2(x2 + 10x + 25) + 50 Multiply the result by -2 f(x) = -2x2 - 20x -50 + 50 Combine the similar terms f(x) = -x2 - 20x3. y = 5(x + 4)2 – 3 Square the binomial y = 5(x2 + 8x + 16) – 3 Multiply the result by 5 y = 5x2 + 40x + 80 – 3 Combine the similar terms y = 5x2 + 40x + 774. y = − 3 (x − 2)2 + 9 Square the binomial Multiply the result by − 3 4 4 ( )y = − 3 x2 − 4x + 4 + 9 Combine the similar terms 4 y = − 3 x2 + 3x − 3 + 9 4 y = − 3 x2 + 3x + 6 4Try this outA. Rewrite the following quadratic functions to f(x) = a(x – h)2 +k.1. f(x) = 2x2 - 12x + 33 6. f(x) = 1 + 16x - 8x22. f(x) = x2 + 8x3. f(x) = 5x2 - 6 7. f(x) = 12 – x24. f(x) = 3x2 + 24x + 435. f(x) = 0.5x2 + 11 8. f(x) = 3 x2 + 15 2 12

9. f(x) = 5 − 7 x2 310. f(x) = 4x2−5x +7 2B. Transform the following quadratic function to f(x) = ax2 + bx + c.1. (x) = 3(x – 2)2 + 5 8. f(x) = 2 (x − 5)22. f(x) = -7(x + 1)2 – 3 33. f(x) = 5(x - 9)2 + 14. f(x) = -5(x - 3)2 + 2 9. f(x) = − 4 (x – 1)2 + 35. f(x) = (x + 5)2 – 12 76. f(x) = -4(x -1)2 + 14 10. f(x) = 6 (x - 7) 2 - 17. f(x) = 4(x - 1 )2 - 2 7 2 Lesson 3 Properties of the Graph of a Quadratic Function The graph of a quadratic function is called a parabola. It is the set of all pointson the Cartesian Coordinate Plane that satisfies the function defined by f(x) = ax2 + bx +c or the vertex form f(x) = a(x – h)2 + k where (h, k) is the vertex.. Look at the two graphs. What do you notice about the parabolas? In what wayare they similar? In what way are they different? Y Y 10 3 9 f(x) = x2 y = -x2 83 2 2 1 0 1 -4 -3 3 12 34 2 7 -4 -3 -2 -1 -1 0 1 2 3 4 X0 1 1 3 4 -2 0 -3 -4 -3 -2 -1-1 0 -2 -1 -1 0 -4 -2 -2 -3 6 -5 -4 -6 -7 2 -5 -8 -6 -9 -7 -8 5 -10 -9 -10 4 -3 3 -4 2 -5 1 -6 -7 X0 -8 -4 -3 -2 -1-1 0 1 2 3 4 -2 -9 -3 -10 13

The following properties of the parabolas should be observed. 1. The graph of y = x2 opens upward while the graph of y = -x2 opens downward. The direction of opening is indicated by the sign of a in the equation. Note that in y = x2, a is positive or a > 0 while in y = -x2, a is negative or a < 0. 2. The two graphs have turning points. The turning point is called a vertex. The vertex maybe the minimum point or the maximum point of the parabola depending on the direction of opening of the graph. The vertex of y = x2 is at (0, 0) also denoted by V(0, 0). It is the lowest or minimum point on the graph. It is the minimum point if the parabola opens upward The vertex of y = -x2 is also at V(0,0) but it is the highest or maximum point on the graph. It is the maximum point if the parabola opens downward. 3. Drawing a vertical line through the vertices of each graph divide both graphs into two congruent or symmetrical parts such that one part is a mirror image of the other. We call this line axis of symmetry. Thus, the axis of symmetry of both graphs is the y-axis or the line x = 0. Y Y 10 3 9 y = x2 2 y = -x2 83 Vertex 1 X3 2 1 -4 -3 -2 0 2 0 -1 0 1 0 7 -4 -3 -2 -1 -1 0 1 2 3 4 -1 -2 1 2 3 4-4 -3 -2 -1 -1 0 1 2 3 4 -3 -2 -2 -4 -3 -4 Axis of6 -5 -5 -6 -6 -7 -7 -8 -8 -9 -9 -10 5 -10 Symmetry -3 4 -4 Axis of 3 Symmetry -5 2 -6 1 VertexX0 -7 -4 -3 -2 -1-1 0 1 2 3 4 -8 -2 -9 -3 -10Example 2:Draw the graph of f(x) = 2(x - 1)2 – 3 and give the properties the function.Solution: 14

Step 1: Construct a table of values for x and f(x). For this particular example, let us usefor x the values { -1, 0, 1, 2, 3}.Substitute these values in f(x) = 2(x - 1)2 – 3.f(-1) = 2(-1 - 1)2 - 3 = 2(-2)2 - 3 = 2(4) - 7 = 8 - 3 = 5f(0) = 2(0 - 1)2 - 3 = 2(-1)2 - 3 = 2(1) - 3 = 2 - 3 = -1f(1) = 2(1 - 1)2 - 3 = 2(0)2 - 3 = 2(0) - 3 = 0 - 3 = -3f(2) = 2(2 - 1)2 - 3 = 2(1)2 - 3 = 2(1) - 3 = 2 - 3 = -1f(3) = 2(3 - 1)2 - 3 = 2(2)2 - 3 = 2(4) - 3 = 8 - 3 = 5The results in a table. x -1 0 1 2 3 f(x) = 2(x – 1)2 - 3 5 -1 -3 -1 5Step 2: Plot and connect the points on the Cartesian Plane. Y The following properties of the quadratic function can be6 observed from the parabola.5 The functions is f(x) = 2(x – 1)2 – 3 in the form4 f(x) = a(x – h)2 + k3 Here, a = 2, h = 1 and k = -3.2 1. The parabola opens upward because a > 0.1X0 2. The vertex (h, k) is (1, -3), the lowest point of the graph. -2 -1 0 1 2 3 4 3. The axis of symmetry is x = 1, the value of h in the -1 vertex. -2 (1, - 3)-3 4. The minimum value is y = -3 which is the value of k in the vertex.-4Hence, for any quadratic function, its vertex is at the point V(h, k) where h =−b 4ac−b2and k = . You have learned this in the previous lesson.2a 4aExample 3: Determine the direction of opening, vertex, axis of symmetry, and minimum ormaximum point of the quadratic function defined by y = -2x2 +12x – 5. 15

Solution: In the given equation, a = -2, b = 12, and c = -5. a. Opening: downward since a is negative or a < 0. b. Vertex: Since the function is not in the vertex form f(x) = a(x – h)2 + k, we cannot easily determine the vertex (h, k). Use the formula for h and k: h = −b = -12 = −12 = 3 2a 2(−2) −4 k = 4ac-b2 = 4(−2)(−5)-122 = 40-144 = -104 =13 4a 4(-2) -8 −8Hence, the vertex is (3, 13);c. The axis of symmetry is x = 2d. The maximum value is y = 13 since the parabola opens downward.Try this outA. Give the sign of the leading coefficient, the coordinates of the vertex, the axis of symmetry, and the highest/lowest value of the quadratic function represented by the given parabolas. Y Y1. 12 2. 7 11 6 10 5 4 9 3 8 2 7 6 5 1 4 X0 3 -4 -3 -2 -1-1 0 1 2 3 4 2 -2 1 -3X0 -4 -4 -3 -2 -1-1 0 1 2 3 4 -2 Y Y 63. 2 4. 5 1 4X0 3 2 -4 -3 -2 -1 0 1 2 3 4 1 -1 0 -5 -4 -3 -2 -1-1 0 1 2 3 4 5 -2 16 -3 -4 -2

X Y Y5. 11 6. 4 10 3 9 8 2 7 1 6 5 X0 4 3 -2 -1-1 0 1 2 3 4 5 6 2 -2 1 -3 X0 -4 -2 -1-1 0 1 2 3 4 5 6 -2 -5 7. Y -6 3 -7 2 1 Y X0 13 -9 -8 -7 -6 -5 -4 -3 -2 -1-1 0 1 2 3 8. 12 -2 -3 11 -4 10 -5 -6 9 8 7 6 5 4 3 2 1 X0 -4 -3 -2 -1-1 0 1 2 3 4 -2 -39. Y 110. Y 4 8 3 7 6 2 5 4 X 3 1 2 0 -7 -6 -5 -4 -3 -2 -1 0 1 2 3 -1 -2 17 1 -3 0 -6 -4 -2 -1 0 2 4 -4

XB. Determine the direction of opening of the parabola, the vertex, the axis of symmetry and minimum or maximum value of the following quadratic functions.1. f(x) = (x - 3)2 6. f(x) = x2 + 3x – 42. f(x) = -(x +2)2 - 1 7. f(x) = 2x2 + 4x - 73. f(x) = - x2 + 9 8. f(x) = x2 - x - 20 9. f(x) = 5 + 3x – x24. f(x) = 3 (x− 4)2 + 7 10. f(x) = 2 - 5x - 3x2 25. f(x) = -4(x - 6)2 + 5 Let’s Summarize1. A quadratic function is a second degree function in the form f(x) = ax2 + bx + c where a ≠ 0.2. Quadratic functions can be written in two forms- the standard form f(x) = ax2 + bx + c or its equivalent form f(x) = a(x – h)2 + k.3. To rewrite a quadratic function from the form f(x) = ax2 + bx + c to the form f(x) = a(x – h)2 + k, use completing the square; or determine the values of a, b, and c then solve for h and k. Substitute the obtained values in f(x) = a(x – h)2 + k. To find the values of h and k, use the relationships, h = − b and k = 4ac - b2 2a 4a4. To rewrite a quadratic function from the form f(x) = a(x-h)2 + k to the form f(x) = ax2 + bx + c, expand the square of the binomial, multiply by a and add k, then simplify by combining similar terms.5. The graph of a quadratic function is called a parabola.6. A parabola may open upward or downward depending upon the sign of a. If a>0, the parabola opens upward while if a<0, the parabola opens downward. 18

7. The highest or lowest point of a parabola is called the turning point or vertex. It is denoted by the ordered pair V(h, k) where h = − b and k = 4ac - b2 2a 4a x = h is called the axis of symmetry while y = k is the highest or lowest value of f(x).8. The axis of symmetry is the line that passes through the vertex and divides the parabola into two equal parts such that one part is the mirror image of the other.What have you learned1. Which of the following functions is quadratic?a. y = x + 3 c. f(x) = 2x + 1 - 5b. y = 5 – 6x2 d. f(x) = 7(4x + 5)2. Which of the following table of ordered pairs represents a quadratic function? a. x -2 -1 0 1 2 y -8 -4 0 4 8 b. x 0 1 2 3 4 y -2 1 4 7 10 c. x -2 -1 0 1 2 y 9 2 1 0 -7 d. x -1 0 1 2 3 y 13 7 5 7 133. What is f(x) = 2(x - 1)2 + 8 in general form?a. f(x) = 2x2 - 2x + 9 c. f(x) = 2x2 + 4x + 8b. f(x) = 2x2 - x + 10 d. f(x) = 2x2 - 4x + 104. How is f(x) = x2 - 10x + 29 written in standard form?a. f(x) = (x - 5)2 + 4 c. f(x) = (x – 5)2 - 4b. f(x) = (x + 5)2 + 4 d. f(x) = (x + 5)2 - 45. Which of the following is the graph of a quadratic function?y 19

a. c. y x xb. y d. y x x6. What do you call the highest or lowest point in the graph of a quadratic function? a. Axis of symmetry b. Slope c. Vertex d. Major axis7. Which of the following quadratic functions will open downward? a. f(x) = 2x2 – 3x – 5 b. f(x) = (x - 4)2 c. f(x) = 3 + 5x + 2x2 d. f(x) = 4 – 2x - 9x28. Determine the vertex of the quadratic function f(x) = 2x2 – 12x + 11.a. (3, 7) c. (3, -7)b. (-3, -7) d. (-3, -7)9. Which of the following is the axis of symmetry of y = -3x2 - 30x – 75?a. x = 5b. x = -5c. x = 3 20

d. x = -310. What is the minimum value of f(x) = 2(x + 1)2 - 7? a. y = -7 b. y = 7 c. y = 1 d. y = 2 21

Answer KeyHow much do you know d. Linear Function1. a. Linear Function e. Linear Function b. Quadratic Function c. Quadratic Function2. a3. b4. d5. c6. a7. d8. a9. c10. bLesson 1A. 1. Quadratic function since the degree of the function is 2. 2. Not a quadratic function since the degree of the function is 1, 3. Quadratic function since the degree of the function is 2. 4. Quadratic function since the degree of the function is 2. 5. Quadratic function since the degree of the function is 2. 6. Not a quadratic function since the degree of the function is 3. 7. Quadratic function since the degree of the function is 2. 8. Not a quadratic function since the degree of the function is 1, 9. Quadratic function since the degree of the function is 2. 10. Not a quadratic function since the degree of the function is 1,B. 1. Quadratic function; equal differences in x produced equal second differences in y. 2. Not a quadratic function; equal differences in x did not produce equal second differences in y. 3. Quadratic function; equal differences in x produced equal second differences in y. 4. Not a quadratic function; equal differences in x did not produce equal second differences in y. 5. Quadratic function; equal differences in x produced equal second differences in y. 6. Quadratic function; equal differences in x produced equal second differences in y. 7. Not a quadratic function; equal differences in x did not produce equal second differences in y. 8. Not a quadratic function; equal differences in x did not produce equal second differences in y. 9. Not a quadratic function; equal differences in x did not produce equal second differences in y. 22

10.Quadratic function; equal differences in x produced equal second differences in y.C. 1. 12 – x 2. f(x) = x(12 – x) or f(x) = 12x – x2 3. x 0 1 2 3 4 5 6 7 8 9 10 11 12 y 12 11 10 9 8 7 6 5 4 3 2 1 04. No, since equal differences in x did not produce equal second differencesin y.5. x 0 1 2 3 4 5 6 7 8 9 10 11 12 y 12 11 10 9 8 7 6 5 4 3 2 1 0 product 0 11 20 27 32 35 36 35 32 27 20 11 06. (6, 6)7. (0, 12) and (12, 0)Lesson 2A. 1. f(x) = 2(x – 3)2 + 52. f(x) = (x + 4)2 – 163. f(x) = 5(x – 0)2 – 6 or f(x) = 5(x + 0)2 – 64. f(x) = 3(x + 4)2 – 55. f(x) = 0.5(x – 0)2 + 11 of f(x) = 0.5(x + 0)2 + 116. f(x) = -8(x – 1)2 + 97. f(x) = -(x – 0)2 + 12 or f(x) = -(x + 0)2 + 128. f(x) = 3 (x − 0)2 + 15 or f(x) = 3 (x + 0)2 + 15 229. f(x) = - 7(x – 0)2 + 5 or f(x) = -7(x + 0)2 + 5 33( )10. f(x) = 2x−5 2 − 9 8 16B. 1. f(x) = 3x2 – 12x + 17 2. f(x) = -7x2 – 14x – 10 3. f(x) = 5x2 – 90x + 406 4. f(x) = -5x2 + 30x – 43 5. f(x) = x2 + 10x + 13 6. f(x) = 1 + 16x – 8x2 or f(x) = -8x2 + 16x + 1 7. f(x) = 4x2 – 4x – 1 8. f(x) = 2 (x2 − 10x + 25) or f(x) = 2 x2 − 20 x + 50 3 3 33 9. f(x) = − 4 x2 + 8 x − 11 7 77 10. f(x) = 6 x2 − 12x + 41 7 23