Cubic and quartic functions

3 Cubic and quartic functions 3A Epanding 3B Long division of polnomials 3C Polnomial values 3D The remainder and factor theorems 3E Factorising polnomials 3F Sum and difference of two cubes 3G Solving polnomial equations 3H Cubic graphs intercepts method 3I Quartic graphs intercepts method 3J Graphs of cubic functions in power function form 3K Domain, range, maimums and minimums 3L Modelling using technolog 3M Finite differences areas of stud Use of the notation = ƒ() for describing the rule of a function and evaluation of ƒ(a), where a is a real number or a smbolic epression Graphs of polnomial functions to degree 4 Qualitative interpretation of features of graphs, and families of graphs Use of smbolic notation to develop algebraic epressions and represent functions, relations and equations Substitution into and manipulation, epansion and factorisation of algebraic epressions, including the remainder and factor theorems Recognition of equivalent epressions and simplification of algebraic epressions involving functions and relations, including use of eponent laws and logarithm laws Determination of rules of simple functions and relations from given information, including polnomial functions to degree 4 Solution of polnomial equations to degree 4, analticall, numericall and graphicall The connection between factors of ƒ( ), solutions of the equation ƒ() = 0 and the horizontal ais intercepts of the graph of the function ƒ Use of parameters to represent a famil of functions and general solutions of equations involving these functions Development of polnomial models, for eample b the use of finite difference tables or solution of a sstem of simultaneous linear equations obtained from values of a function, or a simple combination of values of a function ebookplus 3a Digital doc 10 Quick Questions This chapter will deal mainl with polnomials of degree 3 (cubics). The general equation of a cubic polnomial is P() = a 3 3 + a 2 2 + a 1 + a 0 or more commonl written as = a 3 + b 2 + c + d. Degree 4 polnomials (quartics) will also be considered. The general equation of a quartic polnomial is P() = a 4 4 + a 3 3 + a 2 2 + a 1 + a 0 or more commonl written as = a 4 + b 3 + c 2 + d + e. epanding If we epand three linear factors, for eample, ( + 1)( + 2)( - 7), we get a cubic polnomial (a polnomial of degree 3) as the following worked eample shows. Cubic and quartic functions 109

Worked Eample 1 Epand: a ( + 2)( - 3) b ( - 1)( + 5)( + 2). Think Write a 1 Write the epression. a ( + 2)( - 3) 2 Epand two linear factors and simplif. = ( 2-3 + 2-6) 3 Multipl b the remaining factor. = ( 2 - - 6) = 3-2 - 6 b 1 Write the epression. b ( - 1)( + 5)( + 2) 2 Epand two linear factors and simplif. = ( - 1)( 2 + 2 + 5 + 10) 3 Multipl b the remaining factor and simplif. = ( - 1)( 2 + 7 + 10) = 3 + 7 2 + 10-2 - 7-10 = 3 + 6 2 + 3-10 Worked Eample 2 Note: Just as there is a shortcut for epanding perfect squares, there is also a shortcut for epanding cubes. We can find the shortcut b epanding (a + b) 3 as usual. (a + b) 3 = (a + b)(a + b)(a + b) = (a + b)(a 2 + 2ab + b 2 ) = a 3 + 2a 2 b + ab 2 + a 2 b + 2ab 2 + b 3 = a 3 + 3a 2 b + 3ab 2 + b 3 Similarl, (a - b) 3 = a 3-3a 2 b + 3ab 2 - b 3. Epand the perfect cube ( - 4) 3 using the appropriate rule. Think 1 Use the rule (a - b) 3 = a 3-3a 2 b + 3ab 2 + b 3. In this case a is and b is 4. Write 2 Simplif. ( - 4) 3 ( - 4) 3 = 3-3 2 4 + 3 4 2-4 3 = 3-12 2 + 48-64 (a + b) 3 = a 3 + 3a 2 b + 3ab 2 + b 3 (a - b) 3 = a 3-3a 2 b + 3ab 2 - b 3 Worked Eample 3 Epand - 2( + 5)( - 12). Think Write/Displa 1 On the Main screen, tap: Action Transformation epand Complete the entr line as: epand( - 2( + 5)( - 12)) Then press E. 110 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

2 Write the answer. Epanding - 2( + 5)( - 12) = - 2 3 + 14 2 + 120 REMEMBER When epanding three linear factors: 1. epand two factors first, then multipl b the remaining linear factor 2. collect like terms at each stage 3. ( + 2) 3 ma be written as ( + 2)( + 2)( + 2). 4. ( a ± b) 3 = a 3 ± 3a 2 b + 3ab 2 ± b 3. Eercise 3A 3B Epanding 1 WE 1a Epand each of the following. a ( + 6)( + 1) b ( - 9)( + 2) c ( - 3)( + 11) d 2( + 2)( + 3) e - 3( - 4)( + 4) f 5( + 8)( + 2) g 2 ( + 4) h - 2 2 (7 - ) i (5)( - 6)( + 9) j - 7( + 4) 2 2 WE 1b Epand each of the following. a ( + 7)( + 2)( + 3) b ( - 2)( + 4)( - 5) c ( - 1)( - 4)( + 8) d ( - 1)( - 2)( - 3) e ( + 6)( - 1)( + 1) f ( - 7)( + 7)( + 5) g ( + 11)( + 5)( - 12) h ( + 5)( - 1) 2 i ( + 2)( - 7) 2 j ( + 1)( - 1)( + 1) 3 Epand each of the following. a ( - 2)( + 7)( + 8) b ( + 5)(3-1)( + 4) c (4-1)( + 3)( - 3) d (5 + 3)(2-3)( - 4) e (1-6)( + 7)( + 5) f 3(7-4)( - 4) g - 9(1-2)(3 + 8) h (6 + 5)(2-7) 2 i (3-4)(2 - )(5 + 9) j 2(7 + 2)( + 3)( + 4) 4 WE2 Epand using the appropriate rule for epanding cubes. a ( + 2) 3 b ( + 5) 3 c ( - 1) 3 d ( - 3) 3 e (2-6) 3 f (3 + 4) 3 5 WE3 Epand using a CAS calculator. a ( + 5)( - 11)( + 2) b 3( + 6)( - 1) c 6( - 5)( - + 15)( + 8) d ( - + 5)( - 12) 2 e - ( + 10) 2 f - ( + 13) 3 Long division of polnomials The reverse of epanding is factorising (epressing a polnomial as a product of its linear factors). Before learning how to factorise cubics, ou must be familiar with long division of polnomials. You ma remember in earlier levels doing long division questions. Cubic and quartic functions 111

Consider 745 3, or 3) 745 The process used is as follows. 3 into 7 goes 2 times. Write 2 at the top. 2 3 = 6. Write down the 6. Subtract to get 1. Bring down the 4 to form 14. 2 3) 745-6 14 3 into 14 goes 4 times. Write 4 at the top. 4 3 = 12. Write down the 12. Subtract to get 2. Bring down the 5 to form 25. 3 into 25 goes 8 times. Write 8 at the top. 8 3 = 24. Write down the 24. Subtract to get 1. Answer: 745 3 = 248 remainder 1 Divisor 248 3 745 ) - 6 14-12 25-24 1 Quotient Dividend Remainder The same process can be used to divide polnomials b polnomial factors. Consider ( 3 + 2 2 3 2-13 + 10) ( - 3) or 3) + 2 13+ 10 into 3 goes 2 times 2 (consider onl the leading terms). 3) 3 2 +2 2 13 + 10 Write 2 at the top. - ( 3-3 2 ) 2 ( - 3) = 3-3 2 5 2-13 Write down the 3-3 2. Subtract. ( 3-3 = 0, 2 2 - - 3 2 = 5 2 ) Bring down the - 13. into 5 2 goes 5 times. Write + 5 at the top. 2 + 5 + 2 3 2 5 ( - 3) = 5 3) 2-15 + 2 13+ 10 Write down the 5 2-15. - ( 3-3 2 ) 5 2-13 Subtract. - (5 2-15) Note: 5 2-5 2 = 0, - 13 - - 15 = +2 2 + 10 Bring down the 10. - (2-6) 16 into 2 goes 2 times. Write + 2 at the top. 2 ( - 3) = 2-6 Write down the 2-6. Subtract to get 16. Answer: ( 3 + 2 2-13 + 10) ( - 3) = 2 + 5 + 2 remainder 16 Quotient Remainder 112 maths Quest 11 mathematical methods Cas for the Casio Classpad

Worked eample 4 Perform the following long divisions and state the quotient and remainder. a (2 3 + 6 2-3 + 2) ( - 6) b ( 3-7 + 1) ( + 5) ebookplus Tutorial int-0281 Worked eample 4 Think WriTe a 1 Write the question in long division format. a 2 + 18 + 105-6)2 Q 2 Perform the long division process. ) 3 + 6 2-3 + 2 - (2-12 ) 18 2-3 -(18 2-108) 105 + 2 - (105-630) 632 R 3 Write down the quotient and remainder. Quotient is 2 2 + 18 + 105, remainder is 632. b 1 Write the question in long division format. b 2-5 + 18 Q Note that there is no 2 term in this equation. + 5) 3 + 0 2-7 + 1 Include 0 2 as a place holder. - ( 3 + 5 2 ) 2 Perform the long division process. - 5 2-7 - ( - 5 2-25) 18 + 1 - (18 + 90) - 89 R 3 Write down the quotient and remainder. Quotient is 2-5 + 18, remainder is - 89. Worked eample 5 Find the quotient and remainder when 4-3 3 + 2 2-8 is divided b + 2. Think 1 Write the question in long division format. Include 0 as a place holder. 2 Divide into 4 and write the result above. 3 Multipl 3 b + 2 and write the result underneath. 4 Subtract and then bring down the net term. 5 Continue to perform the long division process (as ou did for cubic polnomials). 6 Write down the quotient and remainder. WriTe 3-5 2 + 12-24 + 2) 4-3 3 + 2 2 + 0-8 - ( 4 + 2 3 ) - 5 3 + 2 2 - ( - 5 3-10 2 ) 12 2 + 0 - (12 2 + 24) - 24-8 - ( - 24-48) 40 The quotient is 3-5 2 + 12-24. The remainder is 40. Cubic and quartic functions 113

Worked Eample 6 Long division of polnomials can be performed on a CAS calculator. Calculate the quotient and remainder when 3 4 2 7 5 is divided b 1. Think Write/Displa 1 On the Main screen, tap: Action Transformation propfrac Complete the entr line as: propfrac(( 3-4 2-7 - 5)/( - 1)) Then press E. 2 Write the answer given b the calculator. ( 3 4 2 15 7 5) ( 1) = + 2 3 10 1 3 Write the answer required. The quotient is 2 3 10; the remainder is 15. REMEMBER Long division of polnomials is similar to long division with numbers. The highest power term is the main one considered at each stage. Ke steps are: 1. How man? 2. Multipl and write the result underneath. 3. Subtract. 4. Bring down the net term. 5. Repeat the process until no pronumerals remain to be divided. 6. State the quotient and remainder. Eercise 3B Long division of polnomials 1 WE4a Perform the following long divisions, and state the quotient and remainder. a ( 3 + 6 2 + 3 + 1) ( + 3) b ( 3 + 4 2 + 3 + 4) ( + 2) c ( 3 + 2 + + 3) ( + 1) d ( 3 + 2 + 4 + 1) ( + 2) e ( 3 + 2 2-5 - 9) ( - 2) f ( 3 + 2-9 - 5) ( - 2) g ( 3-5 2 + 3-8) ( - 3) h ( 3-9 2 + 2-1) ( - 5) i 3 3-2 + 6 + 5, + 2 j 4 3-4 2 + 10-4, + 1 k 2 3-7 2 + 9 + 1, - 2 l 2 3 + 8 2-9 - 1, + 4 114 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

2 Divide the first polnomial b the second, and state the quotient and remainder. a 6 3-7 2 + 4 + 4, 2-1 b 6 3 + 23 2 + 2-31, 3 + 4 c 8 3 + 6 2-39 - 13, 2 + 5 d 2 3-15 2 + 34-13, 2-7 e 3 3 + 5 2-16 - 23, 3 + 2 f 9 3-6 2-5 + 9, 3-4 3 State the quotient and remainder for each of the following. 3C a c 3 2 6 7 16 + 1 3 2 2 + 9 + 17 + 15 2 + 1 b 3 2 3 + 7 + 10 15 3 3 2 d 4 20 + 23 2 2 + 3 4 WE4b State the quotient and remainder for each of the following. a ( 3-3 + 1) ( + 1) b ( 3 + 2 2-7) ( + 2) c ( 3-5 2 + 2) ( - 4) d ( - 3-7 + 8) ( - 1) e (5 2 + 13 + 1) ( + 3) f (2 3 + 8 2-4) ( + 5) g ( - 2 3 - + 2) ( - 2) h ( - 4 3 + 6 2 + 2) (2 + 1) 5 WE5 Find the quotient and remainder for each of the following. a ( 4 + 3 + 3 2 7) ( 1) b ( 4 13 2 + 36) ( 2) c (6 4 3 + 2 2 4) ( 3) 6 WE6 Calculate the quotient and remainder for each of the following. a ( 3 + 9 2 + 11 + 25) ( + 15) b (2 3 18 2 + 5 9) ( 31) c (12 3 + 32 9) (3 + 4) d (18 4 + 3 3 + 45) (2 + 7) Polnomial values Consider the polnomial P() = 3-5 2 + + 1. The value of the polnomial when = 3 is denoted b P(3) and is found b substi tuting = 3 into the equation in place of. That is, P(3) = (3) 3-5(3) 2 + (3) + 1 = 27-5(9) + 3 + 1 = 27-45 + 4 = - 14. Worked Eample 7 If P() = 2 3 + 2-3 - 4, determine: a P(1) b P( - 2) c P(a) d P( + 1). Think Write a 1 Write the epression. a P() = 2 3 + 2-3 - 4 2 Replace with 1. P(1) = 2(1) 3 + (1) 2-3(1) - 4 = 2 + 1-3 - 4 3 Simplif. = - 4 b 1 Write the epression. b P() = 2 3 + 2-3 - 4 2 Replace with - 2. P( - 2) = 2( - 2) 3 + ( - 2) 2-3( - 2) - 4 = 2( - 8) + (4) + 6-4 3 Simplif. = - 16 + 4 + 6-4 = - 10 Cubic and quartic functions 115

c 1 Write the epression. c P() = 2 3 + 2-3 - 4 2 Replace with a. P(a) = 2a 3 + a 2-3a - 4 No further simplification is possible. d 1 2 3 Write the epression. Replace with ( + 1). Epand the right-hand side and collect like terms. Use the rules for epanding cubics and quadratics. d P() = 2 3 + 2-3 - 4 P( + 1) = 2( + 1) 3 + ( + 1) 2-3( + 1) - 4 = 2( 3 + 3 2 + 3 + 1) + 2 + 2 + 1-3 - 3-4 = 2 3 + 6 2 + 6 + 2 + 2 - - 6 = 2 3 + 7 2 + 5-4 Using a CAS calculator, it is best to define the polnomial first. Then finding the value of the polnomial for different values of can be done ver efficientl. Worked Eample 8 Use a CAS calculator to determine the following, considering P() = 16 4 + 3 3 22 + 17. a P( 14) b P( + 7) Think Write/displa a 1 On the Main screen, tap: a Action Command Define Complete the entr line as: Define p() = 16 4 + 3 3 22 + 17 P( - 14) Then press E after each entr. 2 Write the answer. P( 14) = 606 749 b 1 Complete the entr line as: b p( + 7) Then press E. 2 Write the answer. P( + 7) = 16 4 + 451 3 + 4767 2 + 22 371 + 39 308 116 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

remember eercise 3C ebookplus Digital doc Spreadsheet 016 Cubic valuer ebookplus Digital doc WorkSHEET 3.1 P(a) means the value of P() when is replaced b a and the polnomial is evaluated. polnomial values 1 We7 If P() = 2 3-3 2 + 2 + 10, determine the following. a P(0) b P(1) c P(2) d P(3) e P( - 1) f P( - 2) g P( - 3) h P(a) i P(2b) j P( + 2) k P( - 3) l P( - 4) 2 We8 Use a CAS calculator to determine the following considering P() = 16 4 + 3 3-22 + 17. a P( - 11) b P(102) c P(2 + 9) d P( 3 + 2) 3 Cop the following table. Column 1 Column 2 Column 3 Column 4 Column 5 P() P(1) P(2) P( - 1) P( - 2) a b c Column 6 Rem. when divided b ( - 1) Column 7 Rem. when divided b ( - 2) Column 8 Rem. when divided b ( + 1) Column 9 Rem. when divided b ( + 2) d Complete columns 2 to 5 of the table for each of the following polnomials. a P() = 3 + 2 + + 1 b P() = 3 + 2 2 + 5 + 2 c P() = 3-2 + 4-1 d P() = 3-4 2-7 + 3 4 Find the remainder when each polnomial in question 2 is divided b ( - 1) and complete column 6 of the table. 5 Find the remainder when each polnomial in question 2 is divided b ( - 2) and complete column 7 of the table. 6 Find the remainder when each polnomial in question 2 is divided b ( + 1) and complete column 8 of the table. 7 Find the remainder when each polnomial in question 2 is divided b ( + 2) and complete column 9 of the table. 8 Cop and complete using our answers to questions 3 to 7 to find the pattern. a A quick wa of finding the remainder when P() is divided b ( + 8) is to calculate. b A quick wa of finding the remainder when P() is divided b ( - 7) is to calculate. c A quick wa of finding the remainder when P() is divided b ( - a) is to calculate. Cubic and quartic functions 117

3d The remainder and factor theorems The remainder theorem In the previous eercise, ou ma have noticed that: The remainder when P() is divided b ( - a) is equal to P(a). That is, R = P(a). This is called the remainder theorem. We could have derived this result as follows. If 13 is divided b 4, the quotient is 3, and the remainder is 1. That is, 13 4 = 3 + 1 4 and 13 = 4 3 + 1. Similarl, if P() = 3 + 2 + + 1 is divided b ( - 2), the quotient is 2 + 3 + 7 and the remainder is 15. That is, ( 3 + 2 + + 1) ( - 2) = 2 15 + 3 + 7 + 2 ( 3 + 2 + + 1) = ( 2 + 3 + 7)( - 2) + 15. In general, if P() is divided b ( - a), and the quotient is Q(), and the remainder is R, we can write R P() ( - a) = Q() + and ( a) P() = ( - a)q() + R. Substituting = a into this last epression ields P(a) = (a - a)q() + R = 0 Q() + R = R as before. and Worked Eample 9 Without actuall dividing, find the remainder when 3-7 2-2 + 4 is divided b: a - 3 b + 6. Think Write a 1 Name the polnomial. a Let P() = 3-7 2-2 + 4 b 2 The remainder when P() is divided b ( - 3) is equal to P(3). The remainder when P() is divided b ( + 6) is equal to P( - 6). R = P(3) = 3 3-7(3) 2-2(3) + 4 = 27-7(9) - 6 + 4 = 27-63 - 6 + 4 = - 38 b R = P( - 6) = ( - 6) 3-7( - 6) 2-2( - 6) + 4 = - 216-7(36) + 12 + 4 = - 216-252 + 12 + 4 = - 452 118 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

Worked Eample 10 The remainder when 3 + k 2 + - 2 is divided b ( - 2) is equal to 20. Find the value of k. Think Write 1 Name the polnomial. Let P() = 3 + k 2 + - 2. 2 The remainder when P() is divided b R = P(2) ( - 2) is equal to P(2). = 2 3 + k(2) 2 + 2-2 = 8 + 4k 3 We are given R = 20. Put 8 + 4k = 20. 4 Solve for k. Since R = 20. 8 + 4k = 20 4k = 12 k = 3 The factor theorem The remainder when 12 is divided b 4 is zero, since 4 is a factor of 12. Similarl, if the remainder (R) when P() is divided b ( - a) is zero, then ( - a) must be a factor of P(). Since R = P(a), all we need to do is to find a value of a that makes P(a) = 0, and we can sa that ( - a) is a factor. If P(a) = 0, then ( - a) is a factor of P(). This is called the factor theorem. Imagine P() could be factorised as follows: P() = ( - a)q(), where Q() is the other factor of P(). Then we have P(a) = (a - a)q(a) = 0 Q(a) = 0. So if P(a) = 0, ( - a) is a factor. Worked Eample 11 Appl the factor theorem to determine which of the following is a factor of 4 4 3 43 2 + 58 + 240. a ( + 2) b ( 1) Think Write a 1 Name the polnomial. a Let P() = 4 4 3 43 2 + 58 + 240 2 To find the remainder when P() is divided b ( a), find P(a). P( 2) = ( 2) 4 4( 2) 3 43( 2) 2 + 58( 2) + 240 = 16 4( 8) 43(4) 116 + 240 = 16 + 32 172 116 + 240 = 0 3 State the answer. As P( 2) = 0, the remainder when P() is divided b ( + 2) is zero so ( + 2) is a factor. b 1 To find the remainder when P() is divided b ( a), find P(a). b P(1) = (1) 4 4(1) 3 43(1) 2 + 58(1) + 240 = 1 4 43 + 58 + 240 = 252 2 State the answer. As P(1) = 252, the remainder when P() is divided b ( 1) is 252; therefore, ( 1) is not a factor. Cubic and quartic functions 119

remember 1. Remainder R = P(a) when P() is divided b a. 2. If P(a) = 0, then ( - a) is a factor of P(). eercise 3d ebookplus Digital doc Spreadsheet 016 Cubic valuer The remainder and factor theorems 1 We9 Without actuall dividing, find the remainder when 3 + 3 2-10 - 24 is divided b: a - 1 b + 2 c - 3 d + 5 e - 0 f - k g + n h + 3c. 2 Find the remainder when the first polnomial is divided b the second without performing long division. a 3 + 2 2 + 3 + 4, - 3 b 3-4 2 + 2-1, + 1 c 3 + 3 2-3 + 1, + 2 d 3-2 - 4-5, - 1 e 2 3 + 3 2 + 6 + 3, + 5 f - 3 3-2 2 + + 6, + 1 g 3 + 2 + 8, - 5 h 3-3 2-2, - 2 i - 3 + 8, + 3 j 3 + 2 2, - 7 3 a We 10 The remainder when 3 + k + 1 is divided b ( + 2) is - 19. Find the value of k. b The remainder when 3 + 2 2 + m + 5 is divided b ( - 2) is 27. Find the value of m. c The remainder when 3-3 2 + 2 + n is divided b ( - 1) is 1. Find the value of n. d The remainder when a 3 + 4 2-2 + 1 is divided b ( - 3) is - 23. Find the value of a. e The remainder when 3 - b 2-2 + 1 is divided b ( + 1) is 0. Find the value of b. f The remainder when - 4 2 + 2 + 7 is divided b ( - c) is - 5. Find a possible whole number value of c. g The remainder when 2-3 + 1 is divided b ( + d) is 11. Find the possible values of d. h The remainder when 3 + a 2 + b + 1 is divided b ( - 5) is - 14. When the cubic polnomial is divided b ( + 1), the remainder is - 2. Find a and b. 4 We 11 Appl the factor theorem to determine which of the following are factors of 3 + 2 2-11 - 12. a ( - 1) b ( - 3) c ( + 1) d ( + 2) 5 Prove that each of the following are linear factors of 3 + 4 2-11 - 30 b substi tuting values into the cubic function: ( + 2), ( - 3), ( + 5). 6 Use the factor theorem to show that the first polnomial is eactl divisible b the second (that is, the second polnomial is a factor of the first). a 3 + 5 2 + 2-8, - 1 b 3-7 2 - + 7, - 7 c 3-7 2 + 4 + 12, - 2 d 3 + 2 2-9 - 18, + 2 e 3 + 3 2-9 - 27, + 3 f - 3 + 2 + 9-9, - 1 g - 2 3 + 9 2 - - 12, - 4 h 3 3 + 22 2 + 37 + 10, + 5 7 mc a When 3 + 2 2-5 - 5 is divided b ( + 2), the remainder is: A - 5 B - 2 C 0 D 2 E 5 b Which of the following is a factor of 2 3 + 15 2 + 22-15? A ( - 1) B ( - 2) C ( + 3) D ( - 5) E ( + 4) c When 3-13 2 + 48-36 is divided b ( - 1), the remainder is: A - 3 B - 2 C - 1 D 0 E 1 d Which of the following is a factor of 3-5 2-22 + 56? A ( - 2) B ( + 2) C ( - 5) D ( + 5) E ( + 7) 120 maths Quest 11 mathematical methods Cas for the Casio Classpad

3e 8 Find one factor of each of the following cubic polnomials. a 3-3 2 + 3-1 b 3-7 2 + 16-12 c 3 + 2-8 - 12 d 3 + 3 2-34 - 120 9 For the polnomial P() = 6 3 + 7 2 - - 2, find: ( ) iii P ( ) 2 3 a i P( - 1) ii P 1 2 b i Factorise P() as the product of ( + 1) and a quadratic factor. ii Further factorise so P() is written as the product of three linear factors. c Eplain how the other two linear factors relate to what ou found in parts b and c. d Cop and complete the following: In general if (a + b) is a factor, then P( ) = 0. Factorising polnomials Using long division ebookplus Digital doc SkillSHEET 3.1 Reviewing the discriminant Once one factor of a polnomial has been found (using the factor theorem as in the previous section), long division ma be used to find other factors. Worked eample 12 Use long division to factorise 3-19 + 30. Think 1 Name the polnomial. Note: There is no 2 term, so include 0 2. 2 Look at the last term in P(), which is 30. This suggests it is worth tring P(5) or P( - 5). Tr P( - 5). P( - 5) = 0 so ( + 5) is a factor. 3 Divide ( + 5) into P() using long division to find a quadratic factor. 4 Write P() as a product of the two factors found so far. WriTe P() = 3-19 + 30 P() = 3 + 0 2-19 + 30 P( - 5) = ( - 5) 3-19 ( - 5) + 30 = - 125 + 95 + 30 = 0 So ( + 5) is a factor. 2-5 + 6 + 5) 3 + 0 2-19 + 30 -( 3 + 5 2 ) - 5 2-19 -( - 5 2-25) 6 + 30 -(6 + 30) 0 P() = ( + 5)( 2-5 + 6) 5 Factorise the second bracket if possible. P() = ( + 5)( - 2)( - 3) ebookplus Tutorial int-0282 Worked eample 12 Note: In this eample, P() ma have been factorised without long division b finding all three values of which make P() = 0, and hence three factors, then checking that the three factors multipl to give P(). Using short division The process of long division can take a lot of time (and space). One short division method is shown here; it ma take a little longer to understand, but it is quicker than long division once mastered. Consider P() = 3 + 2 2-13 + 10. Using the factor theorem, we can find that ( - 1) is a factor of P(). So, P() = ( - 1)(a 2 + b + c). Cubic and quartic functions 121

Actuall, we know more than this: as P() begins with 3 and ends with +10, we could write: P() = ( - 1)( 2 + b - 10) Imagine epanding this version of P(). Our 2 terms give - 1 2 + b 2. Since P() = 3 + 2 2-13 + 10, we need +2 2. That is, we need - 1 2 + 3 2. To get this, the b must be 3, as when in the first bracket is multiplied b 3 in the second bracket, +3 2 results. That is, we have deduced P() = ( - 1) ( 2 + 3-10). Factorising the second bracket gives P() = ( - 1)( + 5)( - 2) Worked Eample 13 Factorise the following using short division where possible. a 3-5 2-2 + 24 b 4 + 3-13 2-25 - 12 Think Write a 1 Name the polnomial. a Let P() = 3-5 2-2 + 24 2 Look for a value of such that P() = 0. Tr P( - 2). P( - 2) does equal 0, so ( + 2) is a factor. 3 Write the original polnomial as the found factor multiplied b a 2 + b + c. The first term in the brackets must be 2, and the last term must be 12. 4 Imagine the epansion of the epression in step 3. We have 2 2, and require - 5 2. We need an etra - 7 2. So b = - 7. P( - 2) = ( - 2) 3-5 ( - 2) 2-2 ( - 2) + 24 = - 8-20 + 4 + 24 = - 28 + 28 = 0 So ( + 2) is a factor. P() = 3-5 2-2 + 24 P() = ( + 2) (a 2 + b + c) = ( + 2) ( 2 + b + 12) 2 2 + b 2 = - 5 2 b = - 7 P() = ( + 2)( 2-7 + 12) 5 Factorise the second bracket if possible. P() = ( + 2)( - 3)( - 4) b 1 Name the polnomial. b Let P() = 4 + 3 13 2 25 12 2 Look for a value of such that P() = 0. P( 1) = 0 Tr P( 1). So ( + 1) is a factor. 3 It is difficult to factorise a quartic using short division so we will use long division here. 4 Name the cubic factor, and tr to find another factor using the factor theorem. 3 + 0 2-13 - 12 + 1) 4 + 3-13 2-25 - 12 - ( 4 + 3 ) 0-13 2-25 - ( - 13 2 + 13) - 12-12 - ( - 12-12) 0 Let Q() = 3 13 12. Q( 3) = 0 So ( + 3) is a factor. 5 Factorise the cubic using short division. Q() = 3 13 12 = ( + 3)( 2 3 4) 122 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

6 Factorise the quadratic if possible. Q() = ( + 3)( 4)( + 1) 7 Write the original polnomial in factorised form. Epressions can also be factorised using a CAS calculator. P() = 4 + 3 13 2 25 12 = ( + 1)( + 3)( 4)( + 1) = ( + 1) 2 ( + 3)( 4) Worked Eample 14 Use a CAS calculator to factorise the epression 20 3 + 4 2 199 210. Think Write/displa 1 On the Main screen, tap: Action Transformation factor Complete the entr line as: factor(20 3 + 4 2-199 - 210) Then press E. 2 Write the answer. Factorising 20 3 + 4 2 199 210 = (2 7)(2 + 5)(5 + 6) REMEMBER To factorise a polnomial: 1. Let P() = the given polnomial. 2. Use the factor theorem to find a linear factor (tr factors of the constant term). 3. Use long or short division to find another factor. 4. Repeat steps 2 and 3, or factorise b inspection if possible. Alternativel, use the factor function on our CAS calculator. Eercise 3E Factorising polnomials 1 WE 12 Use long division to factorise each dividend. 3 2 a + 1) + 10 + 27+ 18 b 3 2 + 2) + 8 + 17+ 10 3 2 3 2 c + 9) + 12 + 29+ 18 d + 1) + 8 + 19+ 12 e 3 2 3 2 + 3) + 14 + 61+ 84 f + 7) + 12 + 41+ 42 Cubic and quartic functions 123

g 3 2 3 2 + 2) + 4 + 5+ 2 h + 3) + 7 + 16+ 12 3 2 i + 5) + 14 + 65+ 100 j ) 3 + 13 2 + 40 3 2 3 2 k ) + 7 + 12 l + 5) + 10 + 25 3 2 3 2 m + 1) + 6 + 5 n + 6) + 6 ebookplus Digital doc Spreadsheet 096 Polnomials zero search 3F 2 We 12, 13 Factorise the following as full as possible. a 3 + 2 - - 1 b 3-2 2 - + 2 c 3 + 7 2 + 11 + 5 d 3 + 2-8 - 12 e 3 + 9 2 + 24 + 16 f 3-5 2-4 + 20 g 3 + 2 2 - - 2 h 3-7 - 6 i 3 + 3 2-4 j 3 + 2 + + 6 k 3 + 8 2 + 17 + 10 l 3 + 2-9 - 9 m 3-2 - 8 + 12 n 3 + 9 2-12 - 160 o 4 + 4 3 + 3 2-4 - 4 p 4 + 3 3-6 2-28 - 24 q 4 + 6 3 + 8 2-6 - 9 r 4-5 3-17 2 + 21 3 We 14 Use a CAS calculator to factorise the following as full as possible. a 3 3-2 - 10 b 4 3 + 2 2-2 c 3 3-6 2-24 d - 2 3-12 2-18 e 6 3-6 2 f - 3-7 2-12 g - 3-3 2 + + 3 h - 2 3 + 10 2-12 i - 6 3-5 2 + 12-4 j - 5 3 + 24 2-36 + 16 k - 5-4 + 21 3 + 49 2-8 - 60 l 24 4-53 3-71 2 + 152 + 20 4 Factorise the following as full as possible. a 2 3 + 5 2 - - 6 b 3 3 + 14 2 + 7-4 c 3 3 + 2 2-12 - 8 d 4 3 + 35 2 + 84 + 45 e 5 3 + 9 2 + 3-1 f 3 + 2 + 4 g 4 3 + 16 2 + 21 + 9 h 6 3-23 2 + 26-8 i 10 3 + 19 2-94 - 40 j 7 3 + 12 2-60 + 16 k 2 4-3 - 11 2-11 - 3 l 6 4 + 11 3-22 2 - + 6 sum and difference of two cubes Two special cases of cubic polnomials, called sum and difference of cubes are discussed in this section. There are shortcuts for factorising such cubic epressions. Eamples of each are shown in the table below. Sum of cubes Difference of cubes 3 + 2 3 3-27 125 + 64b 3 3 1000-813 3 3 + 1 w 6-1 (2 + 1) 3 + 8 216 - (uv) 3 124 maths Quest 11 mathematical methods Cas for the Casio Classpad

Consider the following epansions. (a + b)(a 2 - ab + b 2 ) and (a - b)(a 2 + ab + b 2 ) = a 3 - a 2 b + ab 2 + ba 2 - ab 2 + b 3 = a 3 + a 2 b + ab 2 - ba 2 - ab 2 - b 3 = a 3 - a 2 b + ab 2 + a 2 b - ab 2 + b 3 = a 3 + a 2 b + ab 2 - a 2 b - ab 2 - b 3 = a 3 + b 3 = a 3 - b 3 These epansions show that: a 3 + b 3 = (a + b)(a 2 - ab + b 2 ) and a 3 - b 3 = (a - b)(a 2 + ab + b 2 ). That is, we have two formulas which ma be used to factorise sums and differences of cubes. Worked eample 15 Factorise the following using the sum or difference of cubes formula. a 3-1000 3 b 2( + 6) 3 + 16 Think WriTe ebookplus Tutorial int-0283 Worked eample 15 a 1 Write the epression. a 3-1000 3 2 Recognise a difference of cubes. 3 Identif a and b for use with the formula a 3 - b 3 = (a - b)(a 2 + ab + b 2 ). 4 Use the formula to factorise. 5 Simplif. = 3 - (10) 3 a =, b = 10 b 1 Write the epression. b 2( + 6) 3 + 16 2 Take out a common factor of 2 to produce a sum of cubes. 3 Identif a and b for use with the formula a 3 + b 3 = (a + b)(a 2 - ab + b 2 ). 4 Appl the sum of cubes formula. 5 Simplif. = ( - 10)[ 2 + (10) + (10) 2 ] = ( - 10)( 2 + 10 + 100 2 ) = 2[( + 6) 3 + 8] = 2[( + 6) 3 + 2 3 ] a = ( + 6), b = 2 = 2[( + 6) + 2][( + 6) 2 - ( + 6)(2) + 2 2 ] = 2( + 8)( 2 + 12 + 36-2 - 12 + 4) = 2( + 8)( 2 + 10 + 28) remember Sum of two cubes: a 3 + b 3 = (a + b)(a 2 - ab + b 2 ) Difference of two cubes: a 3 - b 3 = (a - b)(a 2 + ab + b 2 ) eercise 3F sum and difference of two cubes 1 Identif a and b (as used in the above sum and difference of cubes epressions) in each of the following (do not factorise). a 3 + 6 3 b 8 3 + z 3 c ( + 5) 3 + 27 d 1-64h 3 e 1 8 c3 - e 3 g 3 f t 3 - u3 216 2 We 15a Factorise the following using the sum or difference of cubes formula. a 3-125 b j 3 + k 3 c 3-8 d 27 3 + 3 e 64t 3-216u 3 f 3-1 Cubic and quartic functions 125

g 3 729 + 8p3 h 27r 3-1 i (3k) 3-1 8 j s 3 t 3 + g 6 3 WE15b Factorise: a (a - 1) 3 + a 3 b ( + 2) 3-8 c (2 + 3) 3 + 1 d (w - 5) 3 - w 3 e (2m + p) 3 + (3m - p) 3 f 27 3 - ( + 3) 3 g (2 + 7) 3 + ( - 2) 3 h (3 + ) 3 + ( - 4) 3 i (2-4p) 3 - (p + 1) 3 j (5-9) 3 - (7 - ) 3 k 6 + 9 l 2 3-54 m 3a 3 + 3 n 6( 2 + 1) 3 + 162 4 When m 3 n 3 is full factorised it gives (3 )(9 2 + 3 + 2 ). What are the values of m and n? 5 a Write 3 3 + m 3 in the form a 3 + b 3. b Identif the values of a and b. c Factorise using the rule for sum of cubes. 3G Solving polnomial equations Isolating Cubic equations of the form a( - b) 3 + c = 0 ma be solved b isolating as follows. a( - b) 3 = - c ( - b) 3 = c 3 8 - b = 3 a = b + Unlike a square root, a cube root can be onl positive or negative, but not both, for eample, = - 3 2, 8 = 2. c a 3 c a Worked Eample 16 Solve 3( + 2) 3 + 192 = 0 b isolating. Think Write 1 Write the equation. 3( + 2) 3 + 192 = 0 2 Subtract 192 from both sides. 3 Divide both sides b 3. 4 Take the cube root of both sides. 5 Subtract 2 from both sides and simplif. 3( + 2) 3 = - 192 ( + 2) 3 = - 64 + 2 = - 4 = - 4-2 = - 6 126 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

Factorising to solve polnomial equations The Null Factor Law applies to cubic and quartic equations just as it does for quadratics. If P() = ( - a)( - b)( - c) = 0, then solutions are: = a, = b and = c. If P() = k(l - a)(m - b)(n - c) = 0, solutions are found b solving the following equations: l - a = 0, m - b = 0 and n - c = 0. Worked eample 17 a 4 = 16 2 b 2 3-11 2 + 18-9 = 0. Think WriTe a 1 Write the equation. a 4 = 16 2 2 Rearrange so that all terms are on the left. 4-16 2 = 0 3 Take out a common factor of. 2 ( 2-16) = 0 4 Factorise the brackets using a difference of squares. 2 ( + 4)( - 4) = 0 5 Use the Null Factor Law to solve. = 0, + 4 = 0 or - 4 = 0 so = 0, = - 4 or = 4 b 1 Name the polnomial. b Let P() = 2 3-11 2 + 18-9. 2 Use the factor theorem to find a factor (search for a value a such that P(a) = 0). Consider factors of the constant term (that is, factors of 9 such as 1, 3). The simplest value to tr is 1. 3 Use long or short division to find the other factors of P(). 4 Factorise the brackets. 5 Consider the factorised equation to solve. 6 Use the Null Factor Law to solve. P(1) = 2-11 + 18-9 = 0 So ( - 1) is a factor. 2 2-9 + 9-1)2 3-11 2-18 - 9 2 3-2 2-9 2 + 18-9 2 + 9 9-9 9-9 0 P() = ( - 1)(2 2-9 + 9) P() = ( - 1)(2-3)( - 3) For ( - 1)(2-3)( - 3) = 0-1 = 0, 2-3 = 0 or - 3 = 0 so = 1, = 3 or = 3 2 Polnomial equations can also be solved using a CAS calculator, which is useful when the solutions are not rational. The solutions can be found b using the solve function on a calculator page, or b constructing a graph. Cubic and quartic functions 127

Worked Eample 18 Use a CAS calculator to solve 3 + 3 2 8 5 = 0. Think Write/Displa Method 1: Algebraicall a 1 On the Main screen, tap: Action Advanced solve Complete the entr line as: solve( 3 + 3 2-8 - 5 = 0, ) Then press E. 2 Write the answer. Solving 3 + 3 2 8 5 = 0 for gives = - 4.524, - 0.536, 2.060. Method 2: Graphicall b 1 Construct the graph of = 3 + 3 2-8 - 5 on the Graph & Tab screen. Adjust the window appropriatel. We want to find where the curve intersects with = 0, which is the -ais. Tap: Analsis G-Solve Root To find the other roots use the left and right arrow kes. 2 Write the answer. Solving 3 + 3 2 8 5 = 0 for gives = - 4.524, - 0.536, 2.060. REMEMBER To solve a polnomial equation: 1. Rewrite the equation so it equals zero. 2. Factorise the polnomial as much as possible. 3. Let each linear factor equal zero and solve for in each case. 128 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

eercise 3G ebookplus Digital doc WorkSHEET 3.2 3h solving polnomial equations 1 We 16 Solve the following b isolating. a 2( - 1) 3-250 = 0 b 3( + 2) 3 + 81 = 0 c ( - 4) 3-1000 = 0 d ( + 7) 3-8 = 0 e - 2( - 5) 3-2 = 0 f - ( + 3) 3 + 1 = 0 g (2 + 3) 3-27 = 0 h 4(3-1) 3 + 500 = 0 2 i 3 ( + 1)3 1 + 18 = 0 j 2 (5 - )3-32 = 0 k ( - 5) 3 = 343 l 4-4 5 ( + 8)3 = 104 2 Find all solutions of the following equations. a ( - 1)( - 2)( - 5) = 0 b ( + 3)( + 4)( + 7) = 0 c ( - 5)( + 2)( - 9) ( - 1) = 0 d (2-4)( + 1)( - 3) ( + 2) = 0 e (3 + 12)( - 4)( + 4) = 0 f (2 - )( + 2)(1 - )(1 + ) = 0 g ( + 5)( - 8) 2 = 0 h ( - 1) 3 = 0 i 2 ( + 1)( - 1) = 0 j - 3( - 9) 3 = 0 k (6 - ) 2 (2 + ) = 0 l 2 (2 + 7) = 0 m (5-6)(2 + 3) (6-7) = 0 n (3-4) 2 (5-1) = 0 3 We 17a Solve. a 3-4 = 0 b 3-16 = 0 c 2 4-50 2 = 0 d - 3 4 + 81 = 0 e 3 + 5 2 = 0 f 3-2 2 = 0 g - 4 3 + 8 = 0 h 12 3 + 3 2 = 0 i 4 3-20 4 = 0 j 4-5 3 + 6 2 = 0 k 3-8 2 + 16 = 0 l 3 + 6 2 = 7 m 9 2 = 20 + 3 n 3 + 6 = 4 2 4 We 17b Use the Null Factor Law to solve the following. a 3-2 - 16 + 16 = 0 b 3-6 2 - + 30 = 0 c 3-2 - 25 + 25 = 0 d 3 + 4 2-4 - 16 = 0 e 3-4 2 + + 6 = 0 f 3-4 2-7 + 10 = 0 g 4-3 3-7 2 + 15 = - 18 h 4 + 2 3-13 2 + 10 i 2 3 + 15 2 + 19 + 6 = 0 j - 4 3 + 16 2-9 - 9 = 0 k - 2 3-9 2-7 + 6 = 0 l 2 3 + 4 2-2 - 4 = 0 5 We 18 Use a CAS calculator to find all solutions to the following equations. a 3-17 2-56 + 1153 = 0 b 3 + 12 2-49 - 588 = 0 c - 3 + 17 2 + 65-1521 = 0 d 3 48 2 + 768-4096 = 0 e 3 + 6 2-6 + 2 = 0 f 3 14 2 4 + 13 = 0 g 3 2 + 2 + 1 = 3 h - 2 3 3 2 + 2 + 0.5 = 0 i 4 + 4 3-5 2 + 15 = 0 j 2 4 + 25 3 + 17 = 9 6 mc Which of the following is a solution to 3-7 2 + 2 + 40? A - 5 B - 4 C - 2 D 1 E 2 7 mc A solution of 3-9 2 + 15 + 25 = 0 is = 5. How man other (distinct) solutions are there? A 0 B 1 C 2 D 3 E 4 Cubic graphs intercepts method A good sketch graph of a function shows: 1. - and -intercepts, 2. the behaviour of the function at etreme values of, that is, as approaches infinit ( + ) and as approaches negative infinit ( - ), and 3. the general location of turning points. Cubic and quartic functions 129

Note that for cubic functions, humps are not smmetrical as the are for parabolas, but are skewed to one side. The graphs below show the two main tpes of cubic graph. Turning points A positive cubic A negative cubic For positive cubic graphs, as positive values of become larger and larger, -values also become larger. For negative cubic graphs, as positive values of become larger and larger, -values become smaller. Sometimes instead of two turning points there is a point of inflection, where the graph changes from a decreasing gradient to zero to an increasing gradient (or vice versa). When this occurs, there is onl one -intercept. Consider the general factorised cubic f () = ( - a)( - b)( - c). The Null Factor Law tells us that f () = 0 when = a or Point of inflection = b or = c. Cubic with a point of inflection The -intercept occurs when = 0, that is, the -intercept is f (0) = (0 - a)(0 - b)(0 - c) = - abc c b a abc Worked Eample 19 Sketch the following, showing all intercepts: a = ( - 2)( - 3)( + 5) b = ( - 6) 2 (4 - ) c = ( - 2) 3. Think a 1 Note that the function is alread factorised and that the graph is a positive cubic. Write/draw a = ( - 2)( - 3)( + 5) 130 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

2 The -intercept occurs where = 0. Substitute = 0 into the equation. 3 Use the Null Factor Law to find the -intercepts. (Put each bracket = 0 and solve a mini-equation.) 4 Combine information from the above steps to sketch the graph. -intercept: if = 0, = ( - 2)( - 3)(5) = 30 Point: (0, 30) -intercepts: if = 0, - 2 = 0, - 3 = 0 or + 5 = 0 = 2, = 3 or = - 5 Points: (2, 0), (3, 0), ( - 5, 0) 30 5 2 3 b 1 The graph is a negative cubic (the - in the last factor produces a negative 3 coefficient if the RHS is epanded). b = ( - 6) 2 (4 - ) 2 Substitute = 0 to find the -intercept. -intercept: if = 0, = ( - 6) 2 (4) = 144 Point: (0, 144) 3 Use the Null Factor Law to find the -intercepts. (Put each bracket = 0 and solve a mini-equation.) 4 Combine all information and sketch the graph. Note the skimming of the -ais indicative of a repeated factor, in this case the ( - 6) 2 part of the epression. -intercepts: if = 0, - 6 = 0 or 4 - = 0 = 6 or = 4 Points: (6, 0), (4, 0) 144 4 6 c 1 Positive cubic. c = ( - 2) 3 2 Substitute = 0 to find the -intercept. -intercept: if = 0, = ( - 2) 3 = - 8 3 Use the Null Factor Law to find the -intercept. (Put each bracket = 0 and solve a mini-equation.) -intercept: if = 0, - 2 = 0 = 2 Cubic and quartic functions 131

4 Combine all information and sketch the graph. The cubed factor, ( - 2), indicates a point of inflection and onl one -intercept. 8 2 If a cubic function is not in the form f () = ( - a)( - b)( - c), we ma tr to factorise to find the -intercepts. We can use the factor theorem and division of pol nomials to achieve this. Worked eample 20 Sketch the graph of = 2 3-11 2 + 7 + 20 showing all intercepts. Think 1 Write the equation, and name the polnomial, P(). WriTe/draW = P() = 2 3-11 2 + 7 + 20 ebookplus Tutorial int-0284 Worked eample 20 2 Note the graph is a positive cubic. 3 Let = 0 to find the -intercept. Note: All terms involving are equal to zero. 4 Factorise P() to find -intercepts. ( - 1) is not a factor. 5 Use long or short division to factorise P(). Here, short division has been used. 6 Write down the -intercepts (determined b putting each bracket = 0 and solving for ). 7 Use all available information to sketch the graph. -intercept: if = 0, = 20 Point: (0, 20) P(1) = 2 1 3-11 1 2 + 7 1 + 20 = 2-11 + 7 + 20 = 18 0 P( - 1) = 2 ( - 1) 3-11 ( - 1) 2 + 7 ( - 1) + 20 = - 2-11 - 7 + 20 = 0 So ( + 1) is a factor. P() = ( + 1)(2 2-13 + 20) = ( + 1)(2-5)( - 4) -intercepts: if = 0, = - 1, 5, 4 2 Points: ( - 1, 0), ( 5, 0), (4, 0) 2 20 1 5 2 4 132 maths Quest 11 mathematical methods Cas for the Casio Classpad

Worked Eample 21 Sketch the graph of = 3 3 + 2 2 + 5 using a CAS calculator. Find all intercepts and stationar points. Think Write/Displa 1 On the Graph & Tab screen, complete the entr line as: 1 = 3 3 + 2-2 + 5 Tick the 1 bo and tap!. Create an appropriate viewing window; then tap: Analsis G-Solve Root 2 To find the maimum turning point, tap: Analsis G-Solve Ma 3 To find the minimum turning point, tap: Analsis G-Solve Min Cubic and quartic functions 133

4 Sketch the graph showing all relevant points. ( 0.595, 5.912) = 3 3 + 2 2 + 5 (0, 5) (0.37, 4.55) ( 1.51, 0) 0 REMEMBER To sketch a cubic function of the form = ( + 1)(2-5)( - 4) f () = A 3 + B 2 + C + D: 1. determine if the epression is a positive or negative 20 cubic (that is, if A is positive or negative) 2. find the -intercept (let = 0) 3. factorise if necessar and/or possible, for eample, obtain an epression in the form f () = ( - a)( - b)( - c) 1 5 2 4 4. find the -intercepts (let factors of f () equal 0) 5. use all available information to sketch the graph. Eercise 3h Cubic graphs intercepts method 1 WE19 Sketch the following, showing all intercepts. a = ( - 1)( - 2)( - 3) b = ( + 6)( + 1)( - 7) c = ( + 8)( - 11)( + 1) d = (2-5)( + 4)( - 3) e = (4-3)(2 + 1)( - 4) f = ( - 3) 2 ( - 6) 2 Sketch the following. a = (2 - )( + 5)( + 3) b = ( + 8)( - 8)(2 + 3) c = ( + 1)( - 2) d = 3( + 1)( + 10)( + 5) e = 4 2 ( + 8) f = (6-1) 2 ( + 7) 3 WE20 Sketch each of the following as full as possible. a = 3 + 2 2 - - 2 b = 3 + 6 2 + 11 + 6 c = 3 + 7 2 + 14 + 8 d = 3-2 - 14 + 24 e = 3-3 2-25 - 21 f = 3 3 + 17 2 + 28 + 12 g = 6 3-17 2 + 6 + 8 h = - 2 3-18 2 4 WE21 Sketch the following using a CAS calculator. Find the intercepts and the coordinates of all turning points, correct to 2 decimal places. a = - 3-8 2-5 + 14 b = - 3 + 8 2 + 13-140 c = 3 3-3 2-15 - 9 d = 7 3 + 29 2 + 32 + 4 134 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

ebookplus Digital doc Spreadsheet 014 Cubic graphs general form 5 mc Which of the following is a reasonable sketch of = ( + 2)( - 3)(2 + 1)? A D 3 1 2 2 B E 2 1 2 3 C 3 1 2 2 2 1 3 2 1 2 2 3 6 mc The graph shown on the right could be that of: A = 2 ( + 2) B = ( + 2) 3 C = ( - 2)( + 2) 2 D = ( - 2) 2 ( + 2) E = ( - 2) 3 2 2 7 mc The graph below has the equation: 8 3 1 2 6 A = ( + 1)( + 2)( + 3) B = ( + 1)( - 2)( + 3) C = ( - 1)( + 2)( + 3) D = ( + 1)( + 2)( - 3) E = ( - 1)( - 2)( - 3) 8 mc If a, b and c are positive numbers, the equation of the graph shown below is: A = ( - a)( - b)( - c) B = ( + a)( - b)( + c) C = ( + a)( + b)( - c) D = ( + a)( + b)( + c) E = ( - a)( + b)( - c) b c a 9 mc Which of the following has onl two distinct -intercepts when graphed? A = ( + 1)( + 2) B = ( + 1)( + 2)( + 3) C = 3 D = ( + 1)( + 2) 2 E = ( + 1)( - 1) Cubic and quartic functions 135

3I 10 a Sketch the graph of = 3-2 + 3 + 5 showing all intercepts. b Factorise = 3-2 + 3 + 5 epressing our answer in the form of = ( + a)( 2 + b + c). c Hence, show that = 3-2 + 3 + 5 has onl one real solution. (Hint: Consider the discriminant.) d If = ( + a)( 2 + kb + c) where k is a constant, find the values of k such that the cubic has: i two real solutions ii three real solutions. Note: a, b and c are the same values from part b. Quartic graphs intercepts method Quartic functions are polnomial functions of degree 4. The graphs shown below are the main tpes of quartic graphs. = 4 = ( a) 3 ( b) = ( a)( b)( c)( d) 0 2 (2, 16) a repeated factor ( a) 3 b a b c d Negative quartics are reflected across the -ais. Consider the general factorised quartic, f () = ( a)( b)( c) ( d). As for the cubic functions, the Null Factor Law tells us that f () = 0 (that is, an -intercept occurs) when = a, = b, = c or = d. The -intercept occurs when = 0, therefore the -intercept is f (0) = (0 a)(0 b)(0 c)(0 d) = abcd abcd a b c d Worked Eample 22 Sketch the following graphs, showing all intercepts. a = ( - 2) ( - 1) ( + 1) ( + 3) b = ( - 3) 2 ( + 1) ( + 5) c = (2-1)( + 1) 3 Think Write/Draw a 1 The -intercept occurs when = 0. a = ( - 2)( - 1)( + 1)( + 3) = ( - 2)( - 1)(1)(3) = 6 Point: (0, 6) 2 Use the Null Factor Law to find the -intercepts. (Make each bracket equal to 0 and solve a mini-equation.) 3 The graph has a positive 4 coefficient, so large positive values for result in large positive values for. -intercepts: if = 0, - 2 = 0, - 1 = 0, + 1 = 0 or + 3 = 0 = 2, = 1, = - 1 or = - 3 Points: (2, 0), (1, 0), ( - 1, 0), ( - 3, 0) Shape: 136 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

4 Combine information from steps 1 to 3 to sketch the graph. 6 3 1 0 1 2 b 1 The -intercept occurs when = 0. b = ( - 3) 2 ( + 1)( + 5) = ( - 3) 2 (1)(5) = 45 Point: (0, 45) 2 Use the Null Factor Law to find the -intercepts. (Make each bracket equal to 0 and solve a mini-equation.) 3 The graph has a positive 4 coefficient. (Large positive values for result in large positive values for.) 4 Combine all the information above to sketch the graph. Note that it touches the -ais where there are repeated squared factors ( - 3) 2. -intercepts: if = 0, - 3 = 0, + 1 = 0 or + 5 = 0 = 3, = - 1 or = - 5 Points: (3, 0), ( - 1, 0), ( - 5, 0) Shape: 5 1 0 3 45 c 1 The -intercept occurs when = 0. c = (2-1)( + 1) 3 2 Use the Null Factor Law to find the -intercepts. (Make each bracket equal to 0 and solve a mini-equation.) 3 The graph has a positive 4 coefficient. (Large positive values for result in large positive values for.) 4 Combine all the information from above to sketch the graph. Note that the graph has a point of inflection where it crosses the -ais with repeated cubic factors ( + 1) 3. = ( - 1)(1) 3 = - 1 Point: (0, - 1) -intercepts: if = 0, 2-1 = 0 or + 1 = 0 = 1 2 or = - 1 Points: ( 1 2, 0) and (- 1, 0) Shape: 1 0 1 2 1 Follow the instructions in Worked eample 21 to draw cubic graphs using CAS. Cubic and quartic functions 137

REMEMBER To sketch a quartic function in the form f () = ( a) ( b) ( c) ( d): 1. Find the -intercept (let = 0). 2. Find the -intercepts (let factors of f () = 0). 3. Use all available information to sketch the graph. The graph shown below is = ( 7)( 3)( + 5)( + 8). 840 8 5 0 3 7 Eercise 3I Quartic graphs intercepts method 1 WE 22 Sketch the following graphs, showing all intercepts. a = ( 3)( 2)( + 1)( + 2) b = ( 2) 2 ( + 1)( + 2) c = ( + 5)( 1) 3 d = ( 1) 4 e = (2 1)( 3)( + 3) f = ( 2) 2 ( + 1) 2 g = (1 3) 4 h = ( + 5) 3 (1 ) 2 Compare our answers to question 1 to those found using a CAS calculator. 3 If the graph of = ( + a)( + 3)( + 1)( 3) has four distinct -intercepts, and has a -intercept at (0, 45), find the value of a. 4 A quartic graph has onl two -intercepts, at = a and = b, and a -intercept at the point (0, 81). If a = b: a find the equation of the quartic graph b sketch the graph, labelling all intercepts. 5 MC Which of the following has two distinct -intercepts when graphed? A = ( + 7)( 7)( + 2) B = 2 ( + 3)( 3) C = 3 ( + 27) D = ( + 7) 2 ( 7) E = 4 6 MC If a, b and c are positive numbers, the equation of the graph shown is: b abc a c A = ( + a) 2 ( + b)( + c) B = ( a) 2 ( b)( c) C = ( + a) 2 ( b)( + c) D = ( a)( + b)( c) E = ( a) 2 ( + b)( c) 7 MC For the graph of the quadratic equation = ( 2)( + 1)( + 3) 2, the -intercept occurs at: A 6 b 6 C 12 D 18 E 18 138 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

8 Use a CAS calculator to help ou sketch the following quartics showing all intercepts and turning points correct to 2 decimal places. a = 2 4 + 3-5 2 + 7 + 2 b = 3 4-9 3-8 2 + 12 + 9 c = - 4-2 3 + 5 2 + 4 d = 4-5 3-45 2 + 8 + 120 e = 3 4-10 2-3 f = - 8 4-10 3 + 120 2 + 15 + 35 3J Graphs of cubic functions in power function form Remember the power form or turning point form for quadratic graphs = a( - b) 2 + c which was related to transformations of the basic parabola? The same understanding of transformations can be used to sketch cubic functions. Cubic functions are also power functions. Power functions are functions of the form f () = n, n R. The value of the power, n, determines the tpe of function. When n = 1, f () =, and the function is linear. When n = 2, f () = 2, and the function is quadratic. When n = 3, f () = 3, and the function is cubic. When n = 4, f () = 4, and the function is quartic. Other power functions will be discussed later. Under a sequence of transformations of f () = n, n R, the general form of a power function is f () = a( - b) n + c, (where a, b, c and n R). While all linear and quadratic polnomials are also linear and quadratic power functions, this is not the case for cubic functions (or quartic functions). For eample, a cubic power function in the form of f () = a( - b) n + c has eactl one -intercept and one stationar point of inflection. A cubic polnomial in the form f () = a 3 - b 2 + c + d can have one, two or three -intercepts and is therefore not a power function. All cubic power functions are also cubic polnomials, but not all cubic polnomials are cubic power functions. For eample, the cubic function = 2( - 3) 3 + 1 is a polnomial and a power function. It is the graph of = 3 under a sequence of transformations. Such graphs have a stationar point of inflection at (b, c). A stationar point of inflection is where a graph levels off to have a zero gradient at one point with the same sign gradient either side. = 3 = 3 Stationar point of inflection Summar of transformations = a( - b) 3 + c Dilation factor from the -ais (-stretch) -translation -translation (b, c) b c Stationar point of inflection Cubic and quartic functions 139

The effect of a is illustrated below. = 2 3 = 3 = 1 3 2 = 2 3 = 3 = 1 2 3 Positive a Negative a Intercepts Intercepts ma be found b substituting = 0 (to find the -intercepts) and = 0 (to find the -intercepts) into the equation. Worked Eample 23 Sketch the graph of each of the following, showing the stationar point of inflection and intercepts. a = 3( - 2) 3 + 3 b = - 2 3 + 54 c = - 2(1-2) 3-16 Think Write/draw a 1 Compare the equation with = a( - b) 3 + c, having stationar point of inflection (b, c). 2 Note the values that match, namel a = 3, b = 2 and c = 3. State the stationar point of inflection (b, c). a = 3( - 2) 3 + 3 Stationar point of inflection (2, 3) 3 Find the -intercept. If = 0, = 3(0-2) 3 + 3 = 3( - 8) + 3 = - 21 4 Find the -intercept. If = 0, 0 = 3( - 2) 3 + 3 3( - 2) 3 = 3 ( - 2) 3 = - 1-2 = - 1 = 1 5 Note that the equation is for a positive cubic. Shape: 6 Sketch, showing the stationar point of inflection and intercepts. (2, 3) 1 2 21 140 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

b 1 Manipulate into = a( - b) 3 + c form. b = - 2 3 + 54 = - 2( - 0) 3 + 54 2 Note the graph is a negative cubic with stationar point of inflection (0, 54). Stationar point of inflection (0, 54) 3 Find the -intercept. If = 0, = - 2(0) + 54 = 54 4 Find the -intercept. If = 0, 0 = - 2 3 + 54 2 3 = 54 3 = 27 = 3 5 Sketch, showing the stationar point of inflection and intercepts. (0, 54) 3 c 1 Manipulate into = a( - b) 3 + c form. c = - 2(1-2) 3-16 = - 2[ - 2( - 1 2 )]3-16 = - 2[ - 8( - 1 2 )3 ] - 16 = 16( - 1 2 )3-16 2 Note the graph is a positive cubic with stationar point of inflection ( 1 2, - 16). Stationar point of inflection ( 1 2, - 16) 3 Find the -intercept. If = 0, = - 2(1-0) 3-16 = - 2(1) - 16 = - 18 4 Find the -intercept. If = 0, 0 = - 2(1-2) 3-16 2(1-2) 3 = - 16 (1-2) 3 = - 8 1-2 = - 2 2 = - 3 = 3 2 5 Sketch, showing stationar point of inflection and intercepts. 3 2 1 18 ( 2, 16) Cubic and quartic functions 141

remember Cubic power functions with -dilation of a and stationar point of inflection at (b, c) = a( - b) 3 + c a > 0 a < 0 (b, c) (b, c) Positive a Negative a eercise 3J ebookplus Digital doc Spreadsheet 015 Cubic graphs basic form Graphs of cubic functions in power function form 1 Without sketching graphs for each of the following, state: i the dilation factor ii the coordinates of the stationar point of inflection. a = 2( - 1) 3 + 3 b = 3( + 5) 3-2 c = - 2( - 6) 3-8 d = - 7( + 4) 3 + 1 e = ( - 9) 3 + 4 f = 3-7 g = - ( + 1) 3-1 i = 1 4 ( - 3)3 + 2 h = 1 ( + 2)3 2 j = 4 3 k = 1 3 3 l = - 2 3-2 2 We23 Sketch the graph of each of the following, showing the stationar point of inflection and intercepts. a = 2( - 2) 3 + 2 b = - 3( + 3) 3 + 81 c = 4( - 4) 3-32 d = - 5( - 1) 3 + 5 e = - 3-8 f = 3-1 g = ( + 2) 3 + 27 h = 1 2 ( + 5)3-32 i = 1 3 3)3-9 j = 1 4 ( + 1)3 + 2 k = 1( + 5 2)3 + 25 l = - 2 3 m = 5 3 n = 3 3-3 3 Sketch the following, showing the stationar point of inflection. Intercepts are not required. Use a CAS calcu lator to verif answers. a = (4 - ) 3 + 1 b = 3(5 - ) 3-3 c = 2(4-1) 3 d = 5(3-2) 3 + 1 e = 2(1-5 )3 f = 1 7 (3-4)3-2 g = - (4 - ) 3 + 3 h = (9-5) 3-7 i = 8 3 (6 - )3 + 4 j = - 2(5-2) 3-1 4 mc The basic cubic graph = 3 undergoes a dilation factor of 6 from the -ais and is translated right 4 units and down 3 units. The equation for this graph is: A = 6( - 4) 3-3 B = 3( - 4) 3-6 C = - 6( - 3) 3-4 D = 4( + 6) 3 + 3 E = - 4( + 3) 3 + 6 5 mc The graph of = 5(2 - ) 3 + 9 has a stationar point of inflection at: A (5, 2) B (5, 9) C ( - 2, 9) D (2, - 9) E (2, 9) 142 maths Quest 11 mathematical methods Cas for the Casio Classpad

6 Suggest a possible equation for each of the following, given that each is a cubic with a dilation factor of 1 or - 1 from the -ais. a b (1, 5) ( 2, 2) c d ( 3, 4) ( 3, 0) 7 Write an equation for a cubic with: a a dilation factor of 4 from the -ais, stationar point of inflection (2, 3) b a dilation factor of 2 from the -ais, stationar point of inflection ( - 5, 1) c a dilation factor of 1 from the -ais, stationar point of 4 inflection (1, - 2) d a dilation factor of 1 from the -ais, stationar point of 2 inflection (0, 4). ebookplus Digital doc Investigation Graphs of the form = a ( - b ) n + c 3k domain, range, maimums and minimums The domain of a function is the set of -coordinates of points on its graph. The range is the set of -coordinates of points on the graph. Normall, the domain and range of a cubic function are the set of all real numbers, or R for short, as such graphs etend indefinitel in both positive and negative ais directions. The domain and range of a restricted cubic function ma be a smaller set of numbers. Actual maimum (within given domain) Local maimum 10 Range = [ 8, 10] 5 3 8 Domain = [ 5, 3] Cubic and quartic functions 143

The restricted graph has a domain of -values between - 5 and 3, denoted [ - 5, 3]. The range is [ - 8, 10]. Square brackets are used to indicate that an end value is included, and correspond to a small coloured-in circle on the graph. If an end value is not included, a curved bracket is used. We show such points on a graph using a hollow circle. Range = [ 8, 10] 10 5 3 8 Function notation When we wish to conve information about the domain of a function, the following notation ma be used: f :[ - 4, 1] R, where f () = ( - 1)( + 2)( + 4) } Domain = [ 5, 3] } The name of the function The domain The co-domain. The range is within this set. The rule for the function Note: The range is not necessaril equal to R; the range is within R. Worked eample 24 For the function f: [ - 4, 1] R where f () = ( - 1)( + 2)( + 4), sketch the graph of f (), showing intercepts and the coordinates of an local maimum or local minimum, and state the range. Think WriTe/displa 1 Write the ke information from the question. f : [ - 4, 1] R where f () = ( - 1)( + 2)( + 4) 2 Find the -intercept. If = 0, f () = ( - 1)(2)(4) = - 8. 3 Use the Null Factor Law (or a CAS If f () = 0, = 1, - 2 or - 4. calculator) to determine -intercepts. 4 Use a CAS calculator to determine the local maimum and local minimum. On the Graph & Tab screen, complete the entr line as: 1 = ( - 1)( + 2)( + 4) - 4 1 Tick the 1 bo and tap!. Tap: Analsis G-Solve Ma / Min 144 maths Quest 11 mathematical methods Cas for the Casio Classpad

5 Show the information in steps 1 to 4 on a sketch. Take care to draw the graph onl within the required domain. The end points should be clearl visible. ( 3.12, 4.06) ( 4, 0) (1, 0) ( 0.21, 8.21) 6 Use the graph to state the range. Range = [ - 8.21, 4.06]. The maimum and minimum of a graph within a certain domain are not necessaril the values of the local ma imum or minimum. Sometimes an etreme value is simpl the -coordinate of an end point of a graph. Local maimum Absolute maimum Local minimum Absolute minimum Worked Eample 25 Sketch f: [0, 7) R where f () = ( - 5) 2, showing intercepts, end points, the local maimum and minimum, and state the range. Think 1 Write the ke information from the question. Write/Displa f: [0, 7) R where f () = ( - 5) 2 2 Find the -intercept. If = 0, f () = (0)(0-5) 2 = 0. 3 Use the Null Factor Law (or a CAS calculator) to determine intercepts. 4 Graph the function in the usual manner adjusting the viewing window settings. To determine the -coordinate for = 7, tap: Analsis G-Solve -Cal Enter the -value of 7 and tap: OK If f () = 0, = 0, or = 5. Cubic and quartic functions 145

5 Using a CAS calculator we can establish that the absolute minimum and local minimum are at = 0. We can also show a local maimum at = 1 2 3. Show the above information on a sketch. (1.67, 18.52) (7, 28) (0, 0) (5, 0) 6 Use the graph to state the range. Range = [0, 28). Domain and range will be discussed in more detail in chapter 4, Relations and functions. A method of finding maimums and minimums without a CAS calculator will be covered in the stud of calculus later in this book. remember The absolute maimum or minimum is either the -value at a local maimum or minimum or the -value at an end of the domain. eercise 3k ebookplus Digital doc SkillSHEET 3.2 Interval notation domain, range, maimums and minimums A CAS calculator is required for this eercise. 1 State the domain and range of the sections of graph shown in each case. a b ( 2, 5) (1, 7) ( 5, 2) (4, 0) ( 2, 3) (2, 2) (4, 2) c d (6, 10) ( 4, 2) (4, 1) ( 3, 5) (2, 3) ( 2, 2) 146 maths Quest 11 mathematical methods Cas for the Casio Classpad

e ( 1, 2) (5, 5) f (2, 0) ( 3, 0) (3, 0) ( 3, 8) (4, 5) ( 1, 9) 2 WE24, 25 For each of the following, sketch the graph (showing local maimums and minimums, and intercepts) and state the range. a f: [ - 1, 4] R where f () = ( - 3)( - 4)( + 1) b f: [ - 5, 1] R where f () = ( + 2)( + 5)( - 1) c f: [1, 3) R where f () = ( - 2) 2 ( - 1) d f: ( - 3, 0] R where f () = ( + 3)( + 1) 2 e f: [ - 8, 2) R where f () = (2-3)( + 1)( + 7) f f: [0, 4] R where f () = 3-6 2 + 9-4 g f: [ - 4, - 1.442] R where f () = 3 + 6 2 + 11 + 6 h f: ( - 2, 2.1) R where f () = 3 + 2 2-5 - 6 i f: [ - 3, 5.1] R where f () = - 3 + 2 + 17 + 15 j f: ( - 3, 2 1 3 ) R where f () = 33 + 5 2-19 - 21 3 MC The range of the function shown below is: ( 0.786, 8.209) 6 (4, 18) 2 1 3 (2.120, 4.061) a [ - 4.061, 18] B [ - 4.061, 8.209] C (0, 18] D ( - 2, 3] E ( - 2, 4] 4 MC Point A on the curve is: a an intercept b a local minimum c an absolute minimum d a local maimum e an absolute maimum A Cubic and quartic functions 147

5 A roller-coaster ride is modelled b the function f () = 0.001( - 10)( + 20)( - 40). a What is the height above ground level of the track at = 50? b How far apart verticall are points A and B? 20 20 A B ground level 40 6 The course of a river as marked on a map follows the curve defined b the function f () = 1.5 3-2.7 2 + - 1. Find the coordinates of the southernmost point on the river between = 0 and = 2. N 5 1.5 1.5 5 ebookplus Digital doc Investigation Modelling the path of a roller-coaster 3l modelling using technolog Scientists, economists, doctors and biologists often wish to find an equation that closel matches, or models a set of data. For eample, the wombat population of a particular island ma var as recorded in the following table. ebookplus Interactivit int-0262 Modelling data using polnomials Year of stud () 0 1 2 3 4 5 6 7 8 9 10 Wombat population (W) 59 62 69 83 81 76 70 66 52 49 41 148 maths Quest 11 mathematical methods Cas for the Casio Classpad

The graph below shows these data, with a possible model for the wombat population super imposed. Wombat population 90 80 70 60 50 40 30 20 10 = 0.0888 3 2.4598 2 + 14.196 + 55.063 We will eamine polnomial models up to degree 3, that is, models of the form: = a 3 3 + a 2 2 + a 1 + a 0 where a 0, a 1, a 2 and a 3 are constants. Several technological options are avail able to assist in obtaining models for data, including a CAS calculator, spread sheets and computer algebra sstems such as Mathcad. Man of these applica tions use a method involving minimising the sum of the squares of the vertical dis tances of the data points from the graph of the function known as the least squares method. modelling using a Cas calculator 0 2 4 6 8 10 Year The CAS calculator can be used to find a model or regression for a set of data. The following eample emplos cubic regression, but the general approach is the same for all tpes of regression. 35 30 25 20 15 10 5 = 2.6636 + 7.3182 0 2 4 6 8 10 12 Sum of squares of these lengths is minimised. Worked eample 26 Fit a cubic model to the following data using a CAS calculator. Write the equation and draw a rough sketch of the graph. 0 1 2 3 4 5 6 7 8 9 10 627 545 580 528 436 318 238 229 134 169 139 ebookplus Tutorial int-0285 Worked eample 26 Think WriTe/displa 1 On the Statistics screen, enter the -values into list1 and -values into list2. Name them cord and cord respectivel. Cubic and quartic functions 149

2 To fit a cubic model, tap: Calc Cubic Reg Set: XList: main\cord YList: main\cord Cop Formula: 1 OK 3 To draw the graph of the cubic regression curve and determine its equation, enter the Graph & Tab screen and the cubic regression equation should appear as 1. To graph the equation, tap:! 4 Write the equation. = 1.37 3 + 19.14 2 + 8.55 + 607.48 5 Draw a rough sketch of the graph. 600 0 10 remember An equation to model data ma be obtained using CAS: 1. Enter data into the spreadsheet. 2. Find the regression equation (linear, quadratic etc.). eercise 3l ebookplus Digital doc Spreadsheet 076 Modelling modelling using technolog Use a CAS calculator or other technolog to answer the questions in this eercise. 1 We26 Find a linear model for each of the following sets of data, and draw a rough sketch of the graph. a 0 1 2 3 4 5 6 7 8 9 10-30 0 5-9 28 29 50 68 73 77 84 150 maths Quest 11 mathematical methods Cas for the Casio Classpad

b 0 1 2 3 4 5 6 7 8 9 10-15 - 12-26 - 27-12 - 20-39 - 46-50 - 40-67 c 0 1 2 3 4 5 6 7 8 9 10 11 8 9 14 19 18 29 29 28 32 39 d 0 1 2 3 4 5 6 7 8 9 10 53 44 39 42 35 32 30 29 23 27 19 2 Find a quadratic model for each of the following sets of data, and draw a rough sketch of the graph. a 0 1 2 3 4 5 6 7 8 9 10 19 4 48 60 36 88 126 116 159 168 122 b 0 1 2 3 4 5 6 7 8 9 10 65 33 80 12 50 248 228 252 496 439 694 c 0 1 2 3 4 5 6 7 8 9 10-14 16 32 36 37 51 57 56 55 54 56 d 0 1 2 3 4 5 6 7 8 9 10 70-27 92 2-148 - 327-447 - 639-733 - 910-1204 3 Find the cubic model for each of the following sets of data, and draw a rough sketch of the graph. a 0 1 2 3 4 5 6 7 8 9 10 627 545 580 528 436 318 238 229 134 169 139 b 0 1 2 3 4 5 6 7 8 9 10 21 28 91 182 81 203 345 397 730 873 1205 c 0 1 2 3 4 5 6 7 8 9 10 55 84 64 29 10 4-17 35 182 400 631 d 0 1 2 3 4 5 6 7 8 9 10 45 26 109 201 399 466 621 755 800 868 854 4 For the following data set, find and sketch: a a linear model b a quadratic model c a cubic model. 0 1 2 3 4 5 6 7 8 9 10 537 681 536 624 632 763 686 885 1090 1230 1451 5 Which of the models in question 4 fits best? 6 Use the model from question 1a to predict the value of when = 20. 7 Use the model from question 2a to predict the value of to the nearest unit when = 5.5. 8 Use the model from question 3a to predict the value of to the nearest unit when = 12. Cubic and quartic functions 151

9 The value of shares in the compan Mathsco is plotted b a sharemarket analst over a 12-month period as shown. Month Share price 0 J 1 J 2 A 3 S 4 O 5 N 0.50 0.58 0.53 0.76 1.00 1.50 1.55 2.20 3.06 3.83 4.79 4.40 6 D 7 J 8 F 9 M 10 A 11 M ebookplus Digital doc Investigation Modelling 3m a Find and sketch a quadratic model for the data. b Use our model to predict the share price 2 months later. c Give reasons wh such a prediction ma not be accurate. 10 The population of a colon of ellow-bellied sap-suckers on an isolated island is studied over a number of ears. The population at the start of each ear is shown in the table below. Year 0 1 2 3 4 5 6 7 8 9 10 Population 250 270 310 375 410 395 335 290 290 320 325 Find and sketch a cubic model for the population, and use it to estimate the population at the start of Year 11. Finite differences If pairs of data values in a set obe a polnomial equation, that equation or model ma be found using the method of finite differences. Consider a difference table for a general polnomial of the form = a 3 3 + a 2 2 + a 1 + a 0. We begin the difference table b evaluating the polnomial for values of 0, 1, 2 etc. The differences between successive -values (see table) are called the fi rst differences. The differences between successive first differences are called second differences. The differences between successive second differences called the third differences. We will call the first shaded cell (nearest the top of the table) stepped cell 1, the second shaded cell stepped cell 2 and so on. (= a 3 3 + a 2 2 + a 1 + a 0 ) First differences 0 a 0 a 3 + a 2 + a 1 Second differences Third differences 1 a 3 + a 2 + a 1 + a 0 6a 3 + 2a 2 7a 3 + 3a 2 + a 1 6a 3 2 8a 3 + 4a 2 + 2a 1 + a 0 12a 3 + 2a 2 19a 3 + 5a 2 + a 1 6a 3 3 27a 3 + 9a 2 + 3a 1 + a 0 18a 3 + 2a 2 37a 3 + 7a 2 + a 1 6a 3 4 64a 3 + 16a 2 + 4a 1 + a 0 24a 3 + 2a 2 61a 3 + 9a 2 + a 1 5 125a 3 + 25a 2 + 5a 1 + a 0 152 maths Quest 11 mathematical methods Cas for the Casio Classpad

If a 3 0, the above polnomial equation represents a cubic model, and the third differences are identical (all equal to 6a 3 ). If a 3 = 0, a 2 0 and the polnomial reduces to = a 2 2 + a 1 + a 0, that is, a quadratic model, and the second differences become identical (all equal to 2a 2 ). If a 3 = 0 and a 2 = 0, the polnomial becomes = a 1 + a 0, that is, a linear model, and the first differences are identical (all equal to a 1 ). 1. Stepped cell 1 = a 0 2. Stepped cell 2 = a 1 + a 2 + a 3 3. Stepped cell 3 = 2a 2 + 6a 3 4. Stepped cell 4 = 6a 3 Worked Eample 27 Complete a finite difference table based on the data below, and use it to determine the equation for in terms of. 0 1 2 3 4 5-1 0 7 20 39 64 Think Write 1 Place the data in columns as shown, allowing space for 3 difference columns. 2 Calculate the first differences and place them in the net column. The first differences are 0-1 not constant, so we need to find the second differences. 3 Calculate these and place them in the net column. The second differences are constant, so our table is complete. Showing the third differences is optional. The curve is a quadratic. Differences 1st 2nd 3rd 1 0 6 1 7 0 2 7 6 13 0 3 20 6 19 0 4 39 6 5 64 25 4 Recall the stepped cell equations, and equate them to the shaded cells as shown: Stepped cell 1 = a 0 Stepped cell 2 = a 1 + a 2 + a 3 Stepped cell 3 = 2a 2 + 6a 3 Stepped cell 4 = 6a 3 Here, * is used to denote solved values. 5 [1] gives a 0 = - 1 and [4] gives a 3 = 0. Substitute this information into [2] and [3]. a 0 = - 1* [1] a 1 + a 2 + a 3 = 1 [2] 2a 2 + 6a 3 = 6 [3] 6a 3 = 0 [4] So a 3 = 0* Sub a 3 = 0 into [2]: Cubic and quartic functions 153

6 Substitute a 2 = 0 into [5] to find a 1. a 1 + 3 = 1 a 1 = - 2* 7 Substitute our values for a 0, a 1, a 2 and a 3 into the equation: = a 3 3 + a 2 2 + a 1 + a 0. a 1 + a 2 + 0 = 1 a 1 + a 2 = 1 [5] Sub a 3 = 0 into [3]: 2a 2 + 6 0 = 6 2a 2 = 6 a 2 = 3* Sub a 2 = 3 into [5]: = a 3 3 + a 2 2 + a 1 + a 0 becomes = (0) 3 + (3) 2 + ( - 2) + ( - 1) = 3 2-2 - 1 The stepped equations work onl if the finite differences table begins with = 0, and increases in steps of 1. It ma be necessar on occasions to adjust the table to achieve this, as the following eample shows. Worked Eample 28 Complete a finite difference table based on these data and use it to determine the equation for in terms of. Think 1 2 3 4 5-3 5 13 21 29 Write 1 Construct a difference table, leaving room for the = 0 row. 2 Calculate and fill in the first differences where possible. 0-11 3 Note the first differences are constant, so last two columns are optional. The relationship is linear. 4 Working backwards, the first stepped cell must be - 11 in order for the difference between it and the net cell to be 8. Differences 1st 2nd 3rd 1-3 0 8 8 0 2 5 0 8 0 3 13 0 8 0 4 21 0 5 29 8 5 Recall the stepped cell equations, and equate them to the shaded cells as shown: Stepped cell 1 = a 0 Stepped cell 2 = a 1 + a 2 + a 3 Stepped cell 3 = 2a 2 + 6a 3 Stepped cell 4 = 6a 3 Here, an asterisk (*) is used to denote solved values. a 0 = - 11* [1] a 1 + a 2 + a 3 = 8 [2] 2a 2 + 6a 3 = 0 [3] 6a 3 = 0 [4] 154 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

6 [1] gives a 0 = - 11, [4] ields a 3 = 0, and hence [3] ields a 2 = 0. Substitute this information into [2]. 7 Use the asterisked values to build the equation: = a 3 3 + a 2 2 + a 1 + a 0. So a 3 = 0* and a 2 = 0* Sub a 2 = 0 and a 3 = 0 into [2]: a 1 + 0 + 0 = 8 a 1 = 8* = a 3 3 + a 2 2 + a 1 + a 0 becomes = (0) 3 + (0) 2 + (8) + ( - 11) = 8-11 Using simultaneous equations to find a polnomial model The method of fitting a polnomial to a set of data using finite differences requires the data to be sequential. Often this is not the case. Simultaneous equations can be used to find a polnomial model when the data are not sequential. The number of simultaneous equations required to find the rule of a degree n polno mial is n + 1. For eample, to find a quadratic model, 2 + 1 = 3 points are required as a quadratic is a degree 2 polnomial. Each of the points are substituted into the general equation of the quadratic polnomial, = a 2 + b + c to generate 3 simultaneous equations. These can be solved using elimination or b using a CAS calculator. Worked Eample 29 Using simultaneous equations, find a quadratic model for the points ( - 2, - 9), (3, 1) and (1, 9). Think Write 1 Write down the general rule of a quadratic. = a 2 + b + c 2 Substitute each point into the general equation to get three simultaneous equations. 3 Solve equations [1], [2] and [3] using elimination. Equation [1] - [2] to eliminate c. Equation [2] - [3] to eliminate c. Equation [4] 8 Equation [5] 5 Add equations [6] and [7] to eliminate a and solve for b. Substitute b = 4 into equation [4] to find a. Substitute a = - 2 and b = 4 into equation [3] to find c. 4 Write the rule. = - 2 2 + 4 + 7 a( - 2) 2 + b( - 2) + c = - 9 4a - 2b + c = - 9 [1] a(3) 2 + b(3) + c = 1 9a + 3b + c = 1 [2] a(1) 2 + b(1) + c = 9 a + b + c = 9 [3] - 5a - 5b = - 10 [4] 8a + 2b = - 8 [5] - 40a - 40b = - 80 [6] 40a + 10b = - 40 [7] - 30b = - 120 b = 4-5a - 5(4) = - 10-5a = 10 a = - 2-2 + 4 + c = 9 c = 7 Cubic and quartic functions 155

Worked Eample 30 Using simultaneous equations, find a cubic model for the points ( - 2, - 10), (1, 2), (3, - 20) and (6, 22). Use a CAS calculator to solve the simultaneous equations. Think Write/Displa 1 Write down the general rule of a cubic. = a 3 + b 2 + c + d 2 Substitute each point into the general equation to get four simultaneous equations. 3 To solve the simultaneous equations using a CAS calculator, on the Main screen and using the soft keboard, tap: ) {N Complete the entr lines as: - 8a + 4b - 2c + d = - 10 27a + 9b + 3c + d = 2 27a + 9b + 3c + d = - 20 216a + 36b + 6c + d = 22 Tpe in the variables to be solved in the bottom right corner, a, b, c, d, and then press E. 4 Write the rule. = 3-5 2-4 + 10 5 An alternative wa to solve simultaneous equations with a CAS calculator is the use the Define command. Tap: Action Command Define Complete the entr line as: Define f () = a 3 + b 2 + c + d. Press E, and then tap: ) {N Complete the entr lines as shown on the right. Tpe in the variables to be solved in the bottom right corner, a, b, c, d, and then press E. a( - 2) 3 + b( - 2) 2 + c( - 2) + d = - 10-8a + 4b - 2c + d = - 10 [1] a(1) 3 + b(1) 2 + c(1) + d = 2 a + b + c + d = 2 [2] a(3) 3 + b(3) 2 + c(3) + d = - 20 27a + 9b + 3c + d = - 20 [3] a(6) 3 + b(6) 2 + c(6) + d = 22 216a + 36b + 6c + d = 22 [4] The values are a = 1, b = - 5, c = - 4 and d = 10. 156 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

REMEMBER 1. To determine the polnomial equation using the method of finite differences: (a) set up a table and find the first, second and third differences (b) equate the stepped cell equations to the appropriate cells in the table (c) find the coefficients a 0, a 1, a 2, a 3 for the equation = a 3 3 + a 2 2 + a 1 + a 0. Stepped cell 1: a 0 Stepped cell 2: a 1 + a 2 + a 3 Stepped cell 3: 2a 2 + 6a 3 Stepped cell 4: 6a 3 2. Alternativel, simultaneous equations can be used to find a polnomial model. Simultaneous equations can be solved with or without a CAS calculator. Eercise 3M Finite differences 1 WE27, 28 For each of the following, complete a finite difference table based on the data below, and use it to determine the equation of in terms of. a 0 1 2 3 4 5 6 17 28 39 50 61 b 0 1 2 3 4 5 100 74 48 22-4 - 30 c 0 1 2 3 4 5-4 7 32 71 124 191 d 1 2 3 4 5 1-9 - 13-11 - 3 e 0 1 2 3 4 5-7 - 10-1 26 77 158 f 0 1 2 3 4 5 16 17 20 31 56 101 g 0 1 2 3 4 5-23 - 11 5 25 49 77 h 1 2 3 4 5 28 27 32 31 12 i 0 1 2 3 4 5-27 - 18-9 0 9 18 j 1 2 3 4 5-7 - 3-3 - 7-15 Cubic and quartic functions 157

k 0 1 2 3 4 5-66 - 5 58 99 94 19 l 0 1 2 3 4 5 43 35 27 19 11 3 2 Triangular numbers ma be illustrated as shown at right. If is the number of dots on the base of each diagram, and is the total number of dots: a complete the table below b find an equation linking and (base dots) 0 1 2 3 4 5 (total dots) c find the total number of cans in the supermarket displa shown at right using the equation found in b, and check our answer b counting the cans. = 1 = 3 = 2 = 4 3 The diagonals in polgons of various tpes are shown below in red. Find the relationship between the number of dots () and the number of diagonals (n). (Hint: Continue patterns in a difference table so that it is completed back to = 0.) 4 If n is the number of different squares that can be found within a square grid of edge length, find an equation for n in terms of, and use this equation to find the number of different squares on a chessboard. = 1 n = 1 n = = 2 5 = 3 n =? 5 Find a linear model for the following sets of points. a (2, 1) ( - 1, - 23) b ( - 4, 6) (8, - 3) 6 WE29 Using simultaneous equations, find a quadratic model for the following sets of points. a ( - 2, - 13) (6, - 37) ( - 4, - 57) b ( - 1, 4) (1, - 2) (4, 19) c (4, 8) (0, 8) ( - 4, 24) d ( - 5, - 360) ( - 2, - 96) (6, - 272) 7 WE30 Using simultaneous equations, find a cubic model for the following sets of points. Use a CAS calculator to solve the simultaneous equations. a ( - 6, 3) ( - 3, - 27) (3, - 33) (2, 3) b ( - 2, - 39) (1, 6) (4, 141) ( - 3, - 118) c (4, - 10) (6, 90) (8, 302) ( - 2, 2) d ( - 1, - 4) (1, - 8) (4, - 314) (0, - 6) 8 Using simultaneous equations, find a quartic model for the following sets of points. Use a CAS calculator to solve the simultaneous equations. a (1, 2) ( - 3, 354) (4, 1313) (2, 79) ( - 1, - 2) b ( - 4, 73) (0, 1) (2, - 11) ( - 2, 13) (6, - 707) 158 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

Summar Epandi ng When epanding three linear factors: 1. epand two factors first, then multipl the result b the remaining linear factor 2. collect like terms at each stage 3. ( + 2) 3 ma be written as ( + 2)( + 2)( + 2). Long division of polnomials Long division of polnomials is similar to long division with numbers. The highest power term is the main one considered at each stage. Ke steps are: 1. How man? 2. Multipl and write the result underneath. 3. Subtract. 4. Bring down the net term. 5. Repeat until no pronumerals remain to be divided. 6. State the quotient and the remainder. Polnomial values P(a) means the value of P() when is replaced b a and the polnomial is evaluated. The remainder and factor theorems Remainder R = P(a), when P() is divided b - a. If P(a) = 0, then ( - a) is a factor of P(). Factorising polnomials To factorise a polnomial: 1. let P() = the given polnomial 2. use the factor theorem to find a linear factor 3. use long or short division to find another factor 4. repeat steps 2 and 3, or factorise b inspection if possible. Alternativel, use the factor function on our CAS calculator. Sum and difference of two cubes a3 + b 3 = (a + b)(a 2 - ab + b 2 ) a3 - b 3 = (a - b)(a 2 + ab + b 2 ) Solving polnomial equations To solve a polnomial equation: 1. rewrite the equation so it equals zero 2. factorise the polnomial as much as possible 3. let each linear factor equal zero and solve for in each case. Cubic graphs intercepts method To sketch a cubic function of the form f () = A 3 + B 2 + C + D: 1. determine if the epression is a positive or negative cubic (that is, if A is positive or negative) 2. find the -intercept (let = 0) 3. factorise if necessar or possible, for eample, obtain an epression in the form f () = ( - a)( - b)( - c) Cubic and quartic functions 159

4. find the -intercepts (let factors of f () equal 0) 5. use all available information to sketch the graph. 20 1 5 2 4 Quartic graphs intercepts method To sketch a quartic function in the form f () = ( a) ( b)( c) ( d): 1. find the -intercept (f(0) = abcd) 2. find the -intercepts (let factors of f () = 0) 3. use all available information to sketch the graph. 840 8 5 0 3 7 Graphs of cubic functions in power function form Cubic with -dilation of a and stationar point of inflection at (b, c) = a( - b) 3 + c a > 0 a < 0 (b, c) (b, c) Positive a Negative a Domain, range, maimums and minimums The absolute maimum or minimum is either the -value at a local maimum or minimum, or the -value at an end of the domain. Modelling using technolog An equation to model data ma be obtained using a CAS calculator. 1. Enter data as lists. 2. Find the regression equation (linear, quadratic etc.). 160 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

Finite differences To use the method of finite differences. 1. Set up a table as shown, and find differences b subtracting successive values (value previous value). Circle or shade the stepped cells. 0 Stepped cell 1 First differences Stepped cell 2 Second differences 1 Stepped cell 3 2 Third differences Stepped cell 4 3 Etc. 4 Previous value 5 Value Etc. Value previous value 2. Use the following equations to determine the polnomial model s coefficients: 1. Stepped cell 1 = a 0 2. Stepped cell 2 = a 1 + a 2 + a 3 } Equation of the polnomial model is 3. Stepped cell 3 = 2a 2 + 6a 3 = a 3 3 + a 2 2 + a 1 + a 0 4. Stepped cell 4 = 6a 3 Simultaneous equations Simultaneous equations can be used to find a polnomial model, given a number of points. The number of simultaneous equations required to find the rule of a degree n polnomial is n + 1. Simultaneous equations can be solved with or without a CAS calculator. Cubic and quartic functions 161

ChapTer review short answer 1 Epand: a ( - 2) 2 ( + 10) b ( + 6)( - 1)( + 5) c ( - 7) 3 d (5-2)(1 + )( + 2). 2 Find the quotient and remainder when the first polnomial is divided b the second in each case. a 3 + 2 2-16 - 3, + 2 b 3 + 3 2-13 - 7, - 3 c - 3 + 2 + 4-7, + 1 3 If P() = - 3 3 + 2 2 + - 4, find: a P(1) b P( - 4) c P(2a). 4 Without dividing, find the remainder when 3 + 3 2-16 + 5 is divided b - 1. 5 Show that + 3 is a factor of 3-2 2-29 - 42. 6 Factorise 3 + 4 2-100 - 400. 7 Factorise: a 1-125 3 b ( - 2) 3 + ( + 3) 3. 8 Solve: a 5( + 5) 3 + 5 = 0 b (2 + 1) 2 ( - 3) 2 = 0 c 3-9 2 + 26-24 = 0. 9 Sketch: a = ( - 2)( + 11) b = 3 + 6 2-15 + 8 c = - 2 3 + 2. 10 Sketch: a = ( 7) ( 2) ( + 4) b = (2 1) ( + 1) ( + 4) 2 c = - ( + 5) 3. 11 Sketch = 1 8 ( + 1)3 + 8. 12 Find the range of f :[ - 6, 3] R, where f () = ( + 1)(2 - )( + 5). 13 Complete a finite difference table, and use it to determine the equation for in terms of for the following data set. 0 1 2 3 4 5 8 7 8 17 40 83 14 The following series of diagrams show the maimum number of regions produced b drawing chords in a circle. = 0 r = 1 = 1 r = 2 = 2 r = 4 = 3 r = 7 Find a relationship between the number of chords () and the maimum number of regions (r). 15 Using simultaneous equations, find a cubic model for the points ( - 1, - 10), (2, - 4), ( - 3, - 104) and (0, - 2). 16 Use the remainder theorem to determine if 2 2-3 3 + 7 + 11 is eactl divisible b ( + 1). eam Tip A surprisingl large number of students did not know the remainder theorem. Man used long division and received no marks as the had not responded to the eplicit instruction to use the remainder theorem. Others used both without indicating which was the solution to the question asked. P(1) appeared on occasions. Some students found P( - 1) and then made no statement as to what this meant. [VCAA Assessment report 1 2003] multiple ChoiCe [VCAA 2003] 1 The epansion of ( + 5)( + 1)( - 6) is: A 3-30 B 3-6 2 + 5-6 C 3 + 12 2-31 + 30 D 3-31 - 30 E 3 + 5 2-36 - 30 2 3 + 5 2 + 3-9 is the epansion of: A ( + 3) 3 B ( + 3)( - 3) C ( - 1)( + 3) 2 D ( - 1)( + 1)( + 3) E ( + 1)( + 2)( - 3) Questions 4 and 5 refer to the following long division. 2 + + 2 + 4) 3 + 5 2 + 6-1 3 + 4 2 2 + 6 2 + 4 2-1 2 + 8-9 162 maths Quest 11 mathematical methods Cas for the Casio Classpad

3 The quotient is: a - 9 b 4 c + 4 d 2 + + 2 E 3 + 5 2 + 6-1 4 The remainder is: a - 9 b 2 c 4 d 2-1 E 2 + 8 5 If P() = 3-3 2 + 7 + 1, then P( - 2) equals: a - 34 b - 33 c - 9 d 7 e 35 6 The remainder when 3-7 is divided b - 1 is: a - 6 b 1 c 6 D 7 e 8 7 Which of the following is a factor of 3-3 2-18 + 40? a ( - 4) b ( - 2) c ( + 1) d ( + 3) e (2-1) 8 3 + 6 2-15 + 8 factorises to: a ( - 1) 2 ( + 8) B ( + 1) 2 ( + 8) c ( + 2) 3 d ( + 1)( + 2)( + 4) e ( - 1)( + 2)( + 4) 9 64 3-3 factorises to: a (4 - )(16 2 + 4 + 2 ) b (4 - )(16 2-4 + 2 ) c (4 - )(16 2 + 8 + 2 ) d (4 + )(16 2-8 + 2 ) E (4 + )(16 2-4 + 2 ) 10 Which of the following is the solution to - ( - 4) 3-2 = 6? a - 6 B - 2 c 2 D 4 e 6 11 Which of the following is a solution to ( - 11)(3 + 5)(7-3)(2 + 5) = 0? a - 11 B 3 5 c 3 7 13 The equation for the graph shown below could be: a = ( 3) 2 ( + 3) 2 b = ( 3) ( + 3) 3 c = ( 3) 3 ( + 3) d = ( 3) 2 ( + 3) 2 e = ( + 3) 4 3 81 0 3 14 Which of the following shows the graph of = - 2( + 5) 3-12? a b c ( 5, 12) ( 5, 12) (5, 12) d 5 3 E 7 3 12 The equation for this graph could be: a = ( - 5)( + 1)( + 3) b = ( - 3)( - 1)( + 5) c = ( - 3)( + 1)( + 5) d = (3-1)( + 1)( - 5) e = (5 - )(1 + )(3 + ) d (5, 12) e 3 1 5 ( 5, 12) Cubic and quartic functions 163

Questions 15 and 16 refer to the following graph (below). ( 7, 252) (4.813, 60.370) 5 2 7 ( 2.147, 108.222) 15 The domain of the graph is: a [ - 108.222, 252) b ( - 2.147, 4.183] C ( - 108.222, 60.370] D ( - 7, 7] E ( - 6, 7] 16 The range of the graph is: a [ - 108.222, 252) b [ - 2.147, 4.183] c [ - 108.222, 60.370] d [7, 252) e [0, 252) 17 The data below obe which tpe of relationship? 0 1 2 3 4 5 0 4 16 66 208 520 A Linear B Quadratic C Cubic D Quartic E None of the above 18 Which of the following points lies on the curve of the quadratic model that fits the points (1, 0), (0, - 7) and (2, 11)? A (3, - 4) B (0, 7) C ( - 2, - 7) D ( - 1, - 10) E ( - 1, - 14) 19 A polnomial function p has degree three. A part of its graph, near the point on the graph with coordinates (2, 0), is shown below. 1.8 1.9 2 2.1 2.2 Which one of the following could be the rule for the third degree polnomial p? A p() = ( + 2) 2 B p() = ( - 2) 3 C p() = 2 ( - 2) D p() = ( - 1)( - 2) 2 E p() = - ( - 2) 2 [VCAA 2003] 20 Let f be a polnomial function of degree 3. The graph of the curve with rule = f () either intersects or touches the -ais at eactl two points (a, 0) and (b, 0). A possible rule for f could be: A f () = ( - a)( - b) B f () = ( - a)( + b) 2 C f () = ( - a)( - b) 2 D f () = ( + a) 2 ( - b) E f () = ( + a) 2 ( + b) [VCAA 2004] 21 Which one of the following is a complete set of linear factors of the third-degree polnomial a 3 3 ba b, where a and b are positive real numbers? A, a 2 b B, a b, a + b C, a b D, a b, a + b E, a b, a + b [VCAA 2004] 22 Part of the graph = a 3 + b 2 + c + d is shown below. 50 40 30 20 10 4 3 2 1 0 10 20 30 40 1 2 3 4 5 The values of a, b and c respectivel could be: a b c d A 1-2 - 11 12 B 2-4 - 22 24 C 1 2-22 12 D - 2-6 - 2 24 E 3-1 1 24 [VCAA 2006] 164 Maths Quest 11 Mathematical Methods CAS for the Casio ClassPad

Etended response 1 For P() = 5 3-3 2 6 22, find P(3) and P( ). 2 Find the value of m if + 3 is a factor of 2 3 15 2 + m 21. 3 Factorise 3 2 2 9 + 18. Sketch the graph of f () = 3-2 2-9 +18. 4 Factorise (3 2) 3 + ( + 5) 3. 5 Determine the - and -intercepts of the cubic graph = (2 3)(4 + 1)(2 7). Hence, sketch the graph. 6 The graph = 3 has been moved parallel to the -ais 5 units to the left and moved upwards 2 units from the -ais. What is the equation of the translated graph and what are the coordinates of the point of inflection? Sketch the translated graph. 7 Sketch the graph of = ( + 2) 2 ( 3)( 4), showing all intercepts. 8 The polnomial P() = 3 + a 2 + b + 54 is eactl divisible b 9 and also eactl divisible b 6. a Find the values of a and b. b Find the third factor. c Hence, sketch the graph of the polnomial = 3 + a 2 + b + 54. 9 Factorise 3-2 2-3 + 6 over the real number field. Sketch the graph of = 3-2 2-3 + 6. 10 Find the points of intersection between = 3 2 19 13 and 3 + 7 = 0. 11 Use the method of finite differences to fit a polnomial model to the following data. 0 1 2 3 4 4 16 25 30 30 12 A diagram of a proposed waterslide based on a cubic function appears below. Find: a the height, h 1, of the top of the slide b the coordinates of point A (where the slide enters the water) c the length, L, of the ladder d the height, h 2, of the mini-hump to the nearest centimetre. = 0.008( 3 30 2 + 285 900) L h 1 ( 5, 0) (8, 0.22) 13 An innovative local council decides to put a map of the district on a web site. Part of the map involves two ke features the Cubic River and the Linear Highwa. A mathematicall able web site designer has found the following equations for these features: Cubic River: = 3 + 2-4 - 4 Linear Highwa: = 5 + 5. a Sketch the river and highwa, showing - and -ais intercepts. b Find the coordinates of the points of intersection of the highwa and the river. c A fun-run organiser wishes to arrange checkpoints at the closest points of intersection. Find the distance between the proposed checkpoints. A h 2 Cubic and quartic functions 165

14 A cubic function in the form f () = a 3 + b 2 + c + d has the following values. 0 1 2 3 4 5 42 36 20 0-18 - 28 a Use finite differences to find the values of a, b, c and d. b State one factor of f (), giving our reasoning. c Using long or short division, factorise f (). d Sketch the graph of f (), labelling all intercepts. 15 The height (in centimetres) of a wave above a 1-metre pole is measured over an interval of 8 seconds. The wave s height has been found to approimate the function H = t 3 13t 2 + 48t. a Find the initial height. b Using a CAS calculator, sketch the function and find the local maimum and minimum height of the wave. c The height of a later wave is found to approimate the function K = t 3 14t 2 + 53t 40. Show at what times the height of this wave is eactl the same height as the pole. ebookplus Digital doc Test Yourself 166 maths Quest 11 mathematical methods Cas for the Casio Classpad

ebookplus activities Chapter opener Digital doc 10 Quick Questions: Warm up with ten quick questions on cubic and quartic functions (page 109) 3B Long division of polnomials Tutorial We 4 int-0281: Watch how to perform long division of polnomials (page 113) 3C Polnomial values Digital docs Spreadsheet 016: Investigate solutions to cubic equations (page 117) WorkSHEET 3.1: Review the discriminant (page 117) 3D The remainder and factor theorems Digital doc Spreadsheet 016: Investigate solutions to cubic equations (page 120) 3E Factorising polnomials Digital docs SkillSHEET 3.1: Practise using calculating and using the discriminant (page 121) Spreadsheet 096: Investigate zeros of cubics (page 124) Tutorial We 12 int-0282: Use long division to factorise a cubic (page 121) 3F Sum and difference of two cubes Tutorial We 15 int-0283: Watch how to factorise epressions using the sum or difference of two cubes formulae (page 125) 3G Solving polnomial equations Digital doc WorkSHEET 3.2: Factorising cubics and quartics using long division, appling the Null Factor Law to determine -intercepts and sketching cubics and quartics (page 129) 3H Cubic graphs intercepts method Tutorial We 20 int-0284: Sketch the graph of a cubic showing aial intercepts (page 132) Digital doc Spreadsheet 014: Investigate the effect of changing coefficients of cubics in general form on its graph (page 135) 3J Graphs of cubic functions in power function form Digital docs Spreadsheet 015: Investigate the graphs of cubic functions in power form (page 142) Investigation: Graphs of the form = a( - b) n + c (page 143) 3K Domain, range, maimums and minimums Digital docs SkillSHEET 3.2: Practice epressing intervals using varing notation (page 146) Investigation: Modelling the path of a roller-coaster (page 148) 3L Modelling using technolog Interactivit int-0262 Modelling data using polnomials: Use the interactivit to consolidate our understanding of how to fit a polnomial model to data (page 148) Tutorial We 26 int-0285 int-0286: Watch how to fit a cubic model to a set of data using a CAS calculator (page 149) Digital docs Spreadsheet 076: Investigate the best model for a set of data (page 150) Investigation: Fitting a model eactl (page 152) Chapter review Digital doc Test Yourself: Take the end-of-chapter test to test our progress (page 166) To access ebookplus activities, log on to www.jacplus.com.au Cubic and quartic functions 167