SAMPLE. Polynomial functions

Objectives C H A P T E R 4 Polnomial functions To be able to use the technique of equating coefficients. To introduce the functions of the form f () = a( + h) n + k and to sketch graphs of this form through the use of transformations. To divide polnomials. To use the factor theorem to solve cubic equations and quartic equations. To use the remainder theorem. To draw and use sign diagrams. To find equations for given graphs of polnomials. To appl polnomial functions to problem solving. 4.1 Polnomials In an earlier chapter, linear functions were discussed. This famil of functions is a member of a larger famil of polnomial functions. A function with rule P() = a + a 1 + a + +a n n,where a, a 1, a, a 3,...,a n are real constants and n is a positive integer, is called a polnomial in over the real numbers. The numbers a, a 1, a, a 3,...,a n are called the coefficients of the polnomial. Assuming a n =, the term a n n is called the leading term. The integer n of the leading term is the degree of the polnomial. Foreample: f () = a + a 1 is a degree one polnomial (a linear function). f () = a + a 1 + a is a degree two polnomial (a quadratic function). f () = a + a 1 + a + a 3 3 is a degree three polnomial (a cubic function). f () = a + a 1 + a + a 3 3 + a 4 4 is a degree four polnomial (a quartic function). The polnomials above are written with ascending powers of. The can also be written with descending powers of, for eample: f () = 3 + 1, f () = + + 3, f () = 3 + 4 + 3 + 1 Polnomials are often written in factorised form, e.g. (3 + ),4( 1) 3 +. Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard 18

Chapter 4 Polnomial functions 19 Although it is fairl simple to epand such polnomials when the degree is small, sa two or three, the binomial theorem discussed in Appendi A facilitates the epansion of polnomials of larger degree. Foreample: ( + 3) 4 = ( 4 ) () 4 (3) + ( 4 1 ) () 3 (3) 1 + ( 4 ) () (3) + ( 4 3 ) () 1 (3) 3 + ( 4 4 ) () (3) 4 Equating coefficients = 16 4 + 96 3 + 16 + 16 + 81 If P() = a + a 1 + a + +a n n and Q() = b + b 1 + b + +b m m are equal, i.e. if P() = Q() for all, then the degree of P() = degree of Q() and a = b, a 1 = b 1, a = b,...,etc. Eample 1 If + 6 + 4 = a( + 3) + b for all R, find the values of a and b. Epanding the right-hand side of the equation gives: a( + 3) + b = a( + 6 + 9) + b = a + 6a + 9a + b If + 6 + 4 = a + 6a + 9a + b for all R, then b equating coefficients: (coefficient of ) (coefficient of ) (coefficient of, the constant term of the polnomial) Hence a = 1 and b = 5. Eample 1 = a 6 = 6a 4 = 9a + b a If 3 + 3 + 3 + 8 = a( + 1) 3 + b for all R, find the values of a and b. b Show that 3 + 6 + 6 + 8 cannot be written in the form a( + c) 3 + b for real numbers a, b and c. a The epansion of the right-hand side of the equation gives: a( + 1) 3 + b = a( 3 + 3 + 3 + 1) + b = a 3 + 3a + 3a + a + b Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

13 Essential Mathematical Methods 3&4CAS b If 3 + 3 + 3 + 8 = a 3 + 3a + 3a + a + b for all R, equating coefficients gives: (coefficient of 3 ) (coefficient of ) (coefficient of ) (coefficient of, the constant term of the polnomial) Hence a = 1 and b = 7. The epansion of the right-hand side of the equation gives: a( + c) 3 + b = a( 3 + 3c + 3c + c 3 ) + b = a 3 + 3ca + 3c a + c 3 a + b 1 = a 3 = 3a 3 = 3a 8 = a + b If 3 + 6 + 6 + 8 = a 3 + 3ca + 3c a + c 3 a + b for all R: (coefficient of 3 ) 1 = a (1) (coefficient of ) 6 = 3ca () (coefficient of ) 6 = 3c a (3) (coefficient of, the constant term of the polnomial) 8 = c 3 a + b (4) From (1), a = 1 From (), c = For (3), the right-hand side 3c a equals 1, which is a contradiction. Division of polnomials The division of polnomials was introduced in Essential Mathematical Methods 1&CAS. The general result for polnomial division is: For non-zero polnomials, P() and D(), if P() (the dividend) is divided b D() (the divisor), then there are unique polnomials, Q() (the quotient) and R() (the remainder), such that P() = D()Q() + R() Either the degree of R() < D(), or R() = When R() =, then D() iscalled a divisor of P() and P() = D()Q() The following eample illustrates the process of dividing. Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 131 Eample 3 Divide 3 4 9 + 7 8b. ) 3 3 + 6 + 3 + 33 3 4 + 3 9 + 7 8 3 4 6 3 6 3 9 6 3 1 Thus 3 + 7 3 6 33 8 33 66 58 3 4 9 + 7 8 = ( )(3 3 + 6 + 3 + 33) + 58 or, equivalentl, 3 4 9 + 7 8 = 3 3 + 6 + 3 + 33 + 58 In this eample 3 4 9 + 7 8isthe dividend, isthe divisor and the remainder is 58. Using the TI-Nspire Use propfrac (b 371)asshown. Using the Casio ClassPad Enter and highlight (3 4 9 + 7 8)/( ) and tap Interactive > Transformation > propfrac. Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

13 Essential Mathematical Methods 3&4CAS The remainder theorem and factor theorem The following two results are recalled from Essential Mathematical Methods 1&CAS. The remainder theorem ( ) b If a polnomial P()isdivided b a + b the remainder is P. a This is easil proved b observing if P() = D()(a + b) + R() and = b ( ) ( )( a b b P = D a b ) ( ) b a a a + b + R a and thus ( ) ( ) b b P = R a a The factor theorem An immediate consequence of the remainder theorem is the factor theorem. ( ) b a + b is a factor of the polnomial P() ifandonl if P =. a Eample 4 Find the remainder when P() = 3 3 + + + 1isdivided b + 1. B the remainder theorem the remainder is: ( P 1 ) ( = 3 1 3 ( + ) 1 ( + ) 1 ) + 1 Eample 5 = 3 8 + 4 1 + 1 = 3 8 + 1 = 5 8 Given that + 1 and are factors of 6 4 3 + a 6 + b, find the values of a and b. Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Let P() = 6 4 3 + a 6 + b. B the factor theorem, P( 1) = and P() =. Hence, 6 + 1 + a + 6 + b = (1) and 96 8 + 4a 1 + b = () Rearranging gives: a + b = 13 (1 ) 4a + b = 76 ( ) Subtract (1 ) from ( ): 3a = 63 Therefore a = 1 and, from (1 ), b = 8. Solving polnomial equations The factor theorem ma be used in the solution of equations. Eample 6 Factorise P() = 3 4 11 + 3 and hence solve the equation 3 4 11 + 3 = P(1) = 1 4 11 + 3 = P( 1) = 1 4 + 11 + 3 = P() = 8 16 + 3 = isafactor. Dividing 3 4 11 + 3 b reveals that: P() = ( )( 15) = ( )( 5)( + 3) Therefore = or 5 = or + 3 = = or = 5 or or = 3 Using the TI-Nspire Use factor( ) (b 3) and solve( ) (b 31)asshown. Chapter 4 Polnomial functions 133 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

134 Essential Mathematical Methods 3&4CAS Using the Casio ClassPad Enter and highlight 3 4 11 + 3 then tap Interactive > Transformation > factor. Cop and paste the answer to the net entr line and tap Interactive > Equation/inequalit > solve and ensure the variable set is. Eercise 4A 1 Find the values of A and B such that A( + 3) + B( + ) = 4 + 9 for all real numbers. Find the values of A, B and C in each of the following if: a 4 + 1 = A( + B) + C for all R b 4 1 + 14 = A( + B) + C for all R c 3 9 + 7 = A( + B) 3 + C for all R 3 For each of the following, divide the first term b the second: a 3 7 + 15 3, 3 b 5 5 + 13 4 6, + 1 c 4 9 3 + 5 8, 4 a Find the remainder when 3 + 3 isdivided b +. b Find the value of a for which (1 a) + 5a + (a 1)(a 8) is divisible b ( ) but not b ( 1). 5 Given that f () = 6 3 + 5 17 6: a Find the remainder when f () isdivided b. b Find the remainder when f () isdivided b +. c Factorise f () completel. 6 a Prove that the epression 3 + (k 1) + (k 9) 7isdivisible b + 1 for all values of k. b Find the value of k for which the epression has a remainder of 1 when divided b. 7 The function f () = 3 + a b + 3 has a factor ( + 3). When f () isdivided b ( ), the remainder is 15. a Calculate the values of a and b. b Find the other two factors of f (). Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 135 8 The epression 4 3 + a 5 + b leaves remainders of 8 and 1 when divided b ( 3) and ( 3) respectivel. Calculate the values of a and b. 9 Find the remainder when ( + 1) 4 is divided b 1 Let P() = 5 3 4 + 3 + 3 + 1 a Show that neither ( 1) nor ( + 1) is a factor of P() b Given that P() can be written in the form ( 1)Q() + a + b where Q() isa polnomial and a and b are constants, hence or otherwise, find the remainder when P()isdivided b 1 11 a Show that both ( 3) and ( + 3) are factors of 4 + 3 3 6 b Hence write down one quadratic factor of 4 + 3 3 6, and find a second quadratic factor. 1 Solve each of the following equations for : a ( )( + 4)( )( 3) = b 3 ( ) = c ( 1) 3 ( ) = d ( + ) 3 ( ) = e 4 4 = f 4 9 = g 1 4 + 11 3 6 + + = h 4 + 3 3 4 + 4 = i 6 4 5 3 + 5 6 = 13 Find the -ais intercepts and the -ais intercepts of the graphs of each of the following: a = 3 b = 3 5 + 6 c = 3 4 + + 6 d = 3 5 + + e = 3 + f = 3 3 4 13 6 g = 5 3 + 1 36 16 h = 6 3 5 + 1 i = 3 3 9 3 14 The epressions p 4 5 + q and 4 3 p q 8have acommon factor. Find the values of p and q. 15 Find the remainder when f () = 4 3 + 5 + 4 36 is divided b + 1. 16 Factorise each of the following polnomials using a calculator to help find at least one linear factor: a 3 11 15 + 187 b 3 9 11 + 189 c 3 9 4 + 189 d 4 3 367 + 187 17 Factorise each of the following: a 4 3 43 + + 4 b 4 + 4 3 7 18 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

136 Essential Mathematical Methods 3&4CAS 18 Factorise each of the following polnomials, using a calculator to help find at least one linear factor: a 4 5 3 + 57 + 9 + 45 b 4 + 13 3 + 4 + 81 + 45 c 4 + 3 3 4 + 3 135 d 4 + 4 3 35 78 + 36 4. Quadratic functions The following is a summar of material assumed to have been covered in Essential Mathematical Methods1&CAS. The general epression of a quadratic function is = a + b + c, R. To sketch the graph of a quadratic function (called a parabola) use the following: If a >, the function has a minimum value. If a <, the function has a maimum value. The value of c gives the -ais intercept. b The equation of the ais of smmetr is = a The -ais intercepts are determined b solving the equation a + b + c = A quadratic equation ma be solved b: factorising e.g., + 5 1 = ( 3)( + 4) = = 3 or 4 completing the square e.g., + 4 = ( ) b Add and subtract to complete the square. + + 1 1 4 = ( + 1) 5 = ( + 1) = 5 + 1 =± 5 = 1 ± 5 b ± b 4ac using the general quadratic formula = a e.g., 3 1 7 = = ( 1) ± ( 1) 4( 3)( 7) ( 3) = 6 ± 15 3 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 137 Using the discriminant of the quadratic function f () = a + b + c If b 4ac >, the graph of the function has two -ais intercepts. If b 4ac =, the graph of the function touches the -ais. If b 4ac <, the graph of the function does not intersect the -ais. Eample 7 Sketch the graph of f () = 3 1 7busing the quadratic formula to calculate the -ais intercepts. Since c = 7 the -ais intercept is (, 7). Find the turning point coordinates. Ais of smmetr, = b a = ( 1) 3 = and f ( ) = 3( ) 1( ) 7 = 5 turning point coordinates are (, 5). Calculate the -ais intercepts. 3 1 7 = 6 = b ± b 4ac a 4 = ( 1) ± ( 1) 4( 3)( 7) ( 3) 15 15 + 3 3 = 1 ± 6 4 3 1 6 = 1 ± 15 6 4 = 3 1 7 = 6 ± 15 3 6 7 6 ± 3.87 (to nd decimal place) 8 3 = 3.9 or.71 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

138 Essential Mathematical Methods 3&4CAS Use of completing the square to sketch quadratics For a >, the graph of the function f () = a is obtained from the graph of f () = badilation of factor a from the -ais. The graphs on the right are those of =, = and = 1, i.e. a = 1, and 1. For h >, the graph of f () = ( + h) is obtained from the graph of f () = batranslation of h units in the negative direction of the -ais. For h <, the graph of f () = ( + h) is obtained from the graph of f () = batranslation of h units in the positive direction of the -ais. The graphs of = ( + ) and = ( ) are shown: For k >, the graph of f () = + k is obtained from the graph of f () = b atranslation of k units in the positive direction of the -ais. For k <, the translation is in the negative direction of the -ais. Foreample, the graph of the function f () = ( 3) + is obtained b translating the graph of the function f () = b 3units in the positive direction of the -ais and units in the positive direction of the -ais. Note that the verte is at (3, ). 1 = = 1 = (1, ) 1 (, 4) (1, 1) (1,.5) = ( + ) = ( ) = + = (, 11) = ( 3) + The graph of the function f () = ( ) + 3 is obtained from the graph of f () = b the following transformations: dilation of factor from the -ais translation of units in the positive direction of the -ais translation of 3 units in the positive direction of the -ais. Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard (, 11) (, 3) (3, )

Chapter 4 Polnomial functions 139 B completing the square, an quadratic function can be written in the form: = a( + h) + k Eample 8 Solve 4 5 = bepressing 4 5inthe form = a( + h) + k Use this to help ou sketch the graph of f () = 4 5 f () = ( 4 5 = 5 ) ( = + 1 1 5 ) ( ) (adding b and subtracting to complete [ = ( + 1) 7 ] the square ) [ = ( 1) 7 ] = ( 1) 7 Therefore, the graph of f () = 4 5ma be obtained from the graph of = b the following transformations: dilation of factor from the -ais translation of 1 unit in the positive direction of the -ais translation of 7 units in the negative direction of the -ais. The -ais intercepts can also be determined b this process. To solve = 4 5 = ( 1) 7 ( 1) = 7 7 1 =± 7 = 1 ± 7 7 = 1 + or = 1.87 or.87 This information can now be used to sketch = 4 5 the graph. Since c = 5 the -ais 1 1 3 intercept is (, 5). 1 + 7 1 7 Turning point coordinates are (1, 7). 4 -ais intercepts are (.87, ) and 5 (.87, ) to two decimal places. 6 7 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

14 Essential Mathematical Methods 3&4CAS Eercise 4B 1 Without sketching the graphs of the following functions, determine whether the cross, touch or do not intersect the -ais: a f () = 5 + b f () = 4 + 1 c f () = 6 + 9 d f () = 8 3 e f () = 3 + + 5 f f () = 1 Sketch the graphs of the following functions: a f () = ( 1) b f () = ( 1) c f () = ( 1) ( d f () = 4 ( + 1) e f () = 4 + + 1 f f () = ( + 1) ) 1 g f () = 3( ) 4 h f () = ( + 1) 1 i f () = 5 1 j f () = ( + 1) 4 3 Epress each of the following functions in the form = a( + h) + k and hence find the maimum or minimum value and the range in each case: a f () = + 3 b f () = 6 + 8 c f () = + 8 6 d f () = 4 + 8 7 e f () = 5 f f () = 7 3 g f () = + 9 + 11 4 Sketch the graphs of the following functions, clearl labelling the intercepts and turning points: a = + b = 6 + 8 c = + 8 6 d = 5 6 e f () = + 3 f f () = + 4 7 g f () = 5 1 1 h f () = + 4 1 i =.5 + 3 +.3 j =.6 1.3.1 5 a Which of the graphs shown could represent the graph of the equation = ( 4) 3? b Which graph could represent = 3 ( 4)? A B 13 C 4 6 13 4 6 D 13 4 6 13 4 6 6 Which of the curves shown could be defined b each of the following? a = 1 3 ( + 4)(8 ) b = + 1 c = 1 + ( 1) d = 1 (9 ) Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 141 A B 1 4 5 C 1 5 5 5 51 D 1 5 4 5 1 5 5 = 1 4 4 7 Forwhich values of m does the equation m m + 3 = have: a two solutions for? b one solution for? 8 Show that the equation (k + 1) k = has a solution for all values of k. 9 Forwhich values of k does the equation k k = 5have: a two solutions for? b one solution for? 1 Show that the equation a (a + b) + b = has a solution for all values of a and b. 4.3 Determining the rule for a parabola Given sufficient information about a curve, a rule for the function of the graph ma be determined. Foreample, if the coordinates of three points on a parabola of the form = a + b + c are known, the rule for the parabola ma be found, i.e. the values of a, b and c ma be found. Sometimes a more specific rule is known. For eample, the curve ma be a dilation of =. It is then known to be of the = a famil, and the coordinates of one point (with the eception of the origin) will be enough to determine the value for a. In the following it is assumed that each of the graphs is that of a parabola, and each rule is that of a quadratic function in. 1 This is of the form = a (since the graph has its verte at the origin). When =, = 5 5 = a() a = 5 4 The rule is = 5 4 (, 5) Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

14 Essential Mathematical Methods 3&4CAS This is of the form = a + c (since the graph is smmetric about the -ais). For (, 3) 3 = a() + c implies c = 3 3 4 For ( 3, 1) 1 = a( 3) + 3 1 = 9a + 3 a = 9 the rule is = 9 + 3 This is of the form = a( 3) As the point ( 1, 8) is on the parabola: 8 = a( 1 3) And hence 4a = 8 Therefore a = The rule is = ( 3) This is of the form = a + b + c As the point with coordinates ( 1, ) is on the parabola: = a b + c (1) The -ais intercept is and therefore c = () As the point with coordinates (1, ) is on the parabola: = a + b + c (3) Substitute c = in(1) and (3) = a b + From (1) = a b (1a) From (3) = a + b (3a) Subtract (3a) from (1a) = b b = 1 Substitute b = 1 and c = in(1). ( 1, 8) 1 ( 3, 1) (, 3) 3 1 = a 1 + = a + 1 a = 1 the quadratic rule is = + + Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 143 Eercise 4C 1 Determine the equation of each of the following parabolas: a c e g 4 5 4 1 4 ( 1, ) (3, ) (1, ) (1, 3) b d f h ( 1, 3) 6 4 1 5 (, ) 4 Find quadratic epressions representing the two curves shown in the diagram, given that the coefficient of in each case is 1. A is (4, 3), B is (4, 1), C is (, 5) and D is (, 1). 4 A 3 1 D B 4 5 C 5 3 The graph of the quadratic function f () = A( + b) + B has a verte at (, 4) and passes through the point (, 8). Find the values of A, b and B. Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

144 Essential Mathematical Methods 3&4CAS 4.4 Functions of the form f: R R, f() = a( + h) n + k, where n is a natural number In the previous section it was shown that ever quadratic polnomial can be written in the form a( + h) + k. This is not true for polnomials of higher degree. This was shown in Eample of this chapter. However, there are man polnomials that can be written in this form. In Chapter 3 the famil of power functions was introduced. In this section the sub-famil of power functions with rules of the form f () = n where n is a natural number are considered. f() = n where n is an odd positive integer The diagrams below shows the graphs of = 3 and = 5 ( 3, 7) = 3 (3, 7) ( 3, 43) = 5 (3, 43) The diagram on the right shows both functions graphed on the one set of aes for a smaller domain. The following properties can be observed for a = 3 (1, 1) function f () = n where n is an odd integer: f () = = 5 f (1) = 1 and f ( 1) = 1 ( 1, 1) f () = f ( ), i.e. f is an odd function. f ( ) = ( ) n = ( 1) n () n = f () asn is odd. As, f () As, f () From Essential Mathematical Methods 1&CASou will recall that the gradient function of f () = n has rule n n 1. Hence the gradient is when =. As n 1isevenforn odd, n n 1 > for all non-zero. That is, the gradient of the graph of = f () ispositive for all non-zero and zero when =. Recall that the stationar point at =, for functions of this form, is called a stationar point of infleion. Comparing the graphs of = n and = m where n and m are odd and n > m n < m when < < 1 ( n m = m ( n m 1) < ) n = m when = n > m when 1 < < ( n m = m ( n m 1) > asn m is even, m is odd and is negative and greater than 1) Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

n = m for = 1 and n = m for = 1 n > m for > 1 n < m for < 1 It should be noted that the appearance of graphs is dependent on the scales on the - and -aes. Odd power functions are often depicted as shown. Chapter 4 Polnomial functions 145 Transformations of graphs of functions with rule f() = n where n is an odd positive integer Transformations of these graphs result in graphs with rules of the form = a( + h) n + k where a, h and k are real constants. Eample 9 Sketch the graph of f () = ( ) 3 + 1 This can be done b noting that a translation (, ) ( +, + 1) maps the graph of = 3 to = ( ) 3 + 1 Note: The point with coordinates (, 1) is a point of zero gradient. For the aes intercepts, consider this: When =, = ( ) 3 + 1 = 7 When =, = ( ) 3 + 1 1 = ( ) 3 1 = and hence = 1 The reflection in the -ais described b the transformation rule (, ) (, ) and applied to = 3 results in the graph of = 3 (, 1) (1, ) (, 7) = f() Eample 1 = 3 Sketch the graph of = ( 1) 3 + Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

146 Essential Mathematical Methods 3&4CAS A reflection in the -ais followed b a translation with rule (, ) ( + 1, + ) applied to the graph of = 3 results in the graph = ( 1) 3 + Note: (1, ) is a point of zero gradient. For the aes intercepts consider this: When =, = ( 1) 3 + = 3 When =, = ( 1) 3 (, 3) + (1, ) ( 1) 3 = 1 = 1 (.6, ) 3 and hence = 1 + 1 3.6 Eample 11 Sketch the graph of = ( + 1) 3 + A dilation of factor from the -ais followed b the translation with rule (, ) ( 1, + ) maps the graph of = 3 to the graph of = ( + 1) 3 + Note: ( 1, ) is a point of zero gradient. For the aes intercepts consider this: When =, = + = 4 When =, = ( + 1) 3 + 1 = ( + 1) 3 1 = + 1 and hence = ( 1, ) (, ) Eample 1 The graph of = a( + h) 3 + k has a point of zero gradient at (1, 1) and passes through the point (, 4). Find the values of a, h and k. h = 1 and k = 1 Therefore as the graph passes through (, 4): 4 = a + 1 a = 3 (, 4) Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 147 a b Eample 13 Find the rule for the image of the graph of = 5 under the following sequence of transformations: reflection in the -ais dilation of factor from the -ais translation of units in the positive direction of the -ais and 3 units in the positive direction of the -ais Find a sequence of transformations which takes the graph of = 5 to the graph of = 6 ( + 5) 5 a (, ) (, ) (, ) ( +, + 3) Let (, )bethe image of (, ) under this transformation. Hence = + and = + 3 Thus =, which implies = and = 3 ( ) Therefore, the graph of = 5 maps to the graph of 5 3 =, i.e., to the graph of = 1 3 ( )5 + 3 b Rearrange = 6 ( + 5) 5 to 6 = ( + 5) 5 Therefore, = 6 and = + 5 and = + 6 and = 5 The sequence of transformations is: reflection in the -ais dilation of factor from the -ais translation of 5 units in the negative direction of the -ais and 6 units in the positive direction of the -ais. f() = n where n is an even positive integer The graphs of = and = 4 are shown on the one set of aes. ( 1, 1) = 4 (1, 1) = The following properties can be observed for a function f () = n where n is an even integer: Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

148 Essential Mathematical Methods 3&4CAS f () = f (1) = 1 and f ( 1) = 1 f () = f ( ), i.e. f is an even function, f ( ) = ( ) n = ( 1) n () n = f () asn is even As, f () As, f () From Essential Mathematical Methods 1&CASou will recall that the gradient function of f () = n has rule n n 1. Hence the gradient is when =. As n 1isodd for n even, n n 1 > for > and n n 1 < for <. That is, the gradient of the graph of = f () is positive for positive, negative for negative and zero when =. Comparing the graphs of = n and = m where n and m are even and n > m n < m when < < 1 ( n m = m ( n m 1) < ) n = m when = n < m when 1 < < ( n m = m ( n m 1) < asn m is even, m is even and is negative and greater than 1) n = m for = 1 and n = m for = 1 n > m for > 1 n > m for < 1 It should be noted that the appearance of graphs is dependent on the scales on the - and -aes. Power functions of even degree are often depicted as shown. Eample 14 The graph of = a( + h) 4 + k has a turning point at (, ) and passes through the point (, 4). Find the values of a and h and k. h = and k = Therefore as the graph passes through (, 4): 4 = 16a + a = 1 8 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 149 Eercise 4D 1 Sketch the graphs of each of the following. State the coordinates of the point of zero gradient and the aes intercepts. a f () = 3 b g() = 3 c h() = 5 + 1 d f () = 3 4 e f () = ( + 1) 3 8 f f () = ( 1) 3 g g() = ( 1) 3 + h h() = 3( ) 3 4 i f () = ( 1) 3 + j h() = ( 1) 3 4 k f () = ( + 1) 5 3 l f () = ( 1) 5 The graph of = a( + h) 3 + k has a point of zero gradient at (, 4) and passes through the point (1, 1). Find the values of a, h and k. 3 The graph of = a( + h) 4 + k has a turning point at (1, 7) and passes through the point (, 3). Find the values of a, h and k. 4 Find the equation of the image of = 3 under each of the following transformations: a a dilation of factor 3 from the -ais b a translation with rule (, ) ( 1, + 1) c a reflection in the -ais followed b a translation with rule d e (, ) ( +, 3) a dilation of factor from the -ais followed b a translation with rule (, ) ( 1, ) a dilation of factor 3 from the -ais. 5 B appling suitable transformations to = 4,sketch the graph of each of the following: a = 3( 1) 4 b = ( + ) 4 c = ( ) 4 6 d = ( 3) 4 1 e = 1 ( + 4) 4 f = 3( ) 4 3 4.5 The general cubic function Not all cubic functions are of the form f () = a( + h) 3 + k. In this section the general cubic function is considered. The form of a general cubic function is: f () = a 3 + b + c + d It is impossible to full investigate cubic functions without the use of calculus. Cubic functions will be revisited in Chapter 9. The shapes of cubic graphs var. Below is a galler of cubic graphs demonstrating the variet of shapes that are possible. Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

15 Essential Mathematical Methods 3&4CAS 1 f () = 3 + Note: (, ) is not a point of zero gradient. There is one root,. f () = 3 Note: The turning points do not occur smmetricall between consecutive -ais intercepts as the do for quadratics. Differential calculus must be used to determine them. There are three roots: 1, and 1. 3 f () = 3 3 Note: There are two roots: 1 and. 1 4 4 1 1 1 1 4 f () = 3 3 + 5 The graphs of f () = 3 and f () = 3 + 3 + are shown. The are the reflection in the -ais of the graphs of 1 and 4 respectivel. = 3 1 1 4 4 1 1 1 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 151 Sign diagrams A sign diagram is a number line diagram that shows when an epression is positive or negative. For a cubic function with rule f () = ( )( )( ) when > >, the sign diagram is as shown: Eample 15 + α β γ Draw a sign diagram for the epression 3 4 11 + 3 From Eample 6, f () = ( )( 5)( + 3) We note: f () > for > 5 f () < for < < 5 f () > for 3 < < f () < for < 3 Also, f () = f (5) = f ( 3) = Hence the sign diagram ma be drawn. Eample 16 + 3 5 For the cubic function with rule f () = 3 + 19 3: a Sketch the graph of = f () using a calculator to find the values of the coordinates of the turning points, correct to two decimal places. b Sketch the graph of = 1 f ( 1) a f () = 3 + 19 3 = (3 )( )( + 5) = ( + 5)( )( 3) We note: f () < for > 3 f () > for < < 3 f () < for 5 < < f () > for < 5 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

15 Essential Mathematical Methods 3&4CAS b Also, f ( 5) = f () = f (3) = Hence the sign diagram ma be drawn. + 5 3 The -ais intercepts are at = 5, = and = 3 and the -ais intercept is at = 3. The window has to be adjusted carefull to see the -ais intercepts and the turning points. The local minimum turning point ma be found b selecting 3:Minimum from the F5 menu and then entering the lower bound 5 and the upper bound and pressing ENTER. The local minimum is found to occur at the point with coordinates (.5, 61.88) with values given correct to two decimal places. The local maimum turning point ma be found b selecting 4:Maimum from the F5 menu and then entering the lower bound and the upper bound 3 and pressing ENTER. The local maimum is found to occur at the point with coordinates (.5, 1.88) with values given correct to two decimal places. The rule for the transformation is (, ) ( + 1, 1 ) B the transformation ( 5, ) ( 4, ), (, ) (3, ), (3, ) (4, ), (, 3) (1, 15), (.5, 1.88) (3.5,.94) and (.5, 61.88) ( 1.5, 3.94) 5 3 (.5, 61.88) (.5, 1.88) (3.51,.94) 4 3 4 4 ( 1.5, 3.94) 3 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 153 Eercise 4E 1 Draw sign diagrams for each of the following epressions: a (3 )( 1)( 6) b (3 + )( 1)( + 6) c ( 5)( + 1)( 6) d (4 )(5 )(1 ) a Use a calculator to plot the graph of = f () where f () = 3 + 1 b On the same screen plot the graphs of: i = f ( ) ii = f ( + ) iii = 3 f () 4.6 Polnomials of higher degree The techniques that have been developed for cubic functions ma now be applied to quartic functions and in general to functions of higher degree. It is clear that a polnomial P() of degree n has at most n solutions for the equation P() =. It is possible for the graphs of polnomials of even degree to have no -ais intercepts, for eample P() = + 1, but graphs of polnomials of odd degree have at least one -ais intercept. The general form for a quartic function is: f () = a 4 + b 3 + c + d + e Agaller of quartic functions produced with a graphing package is shown below. 1 f () = 4 f () = 4 +. 1.. 1. 1.. 3 f () = 4 4... 1. 1... 1.. 1. 1.. 4 f () = ( 1) ( + ) 4.. 3.. 1. 1.. Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

154 Essential Mathematical Methods 3&4CAS 5 f () = ( 1) 3 ( + ) 15. 1. 5.. 1. 1.. 3. 5. 1. Eample 17 Draw a sign diagram for the quartic epressions: a ( )( + )( 3)( 5) b 4 + a + 3 5 b Let P() = 4 + P(1) = 1 + 1 = 1isafactor. ) 3 + + + 1 4 + 3 + + 4 3 3 + 3 + P() = ( 1)( 3 + + + ) = ( 1)[ ( + 1) + ( + 1)] = ( 1)( + 1)( + ) + 1 1 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 155 Eample 18 Find the coordinates of the points where the graph of = p(), p() = 4 + 1, crosses the - and -aes, and hence sketch the graph. 1isafactor. p(1) = 1 + 1 = p() = ( 1)( 3 + 1) = ( 1)[ ( + 1) ( + 1)] = ( 1)( + 1)( 1) = ( 1) ( + 1) Alternativel, note that: p() = ( 1) = [( 1)( + 1)] = ( 1) ( + 1) Therefore, the -ais intercepts are (1, ) and ( 1, ). When =, = 1. Eercise 4F (, 1) ( 1, ) (1, ) 1 a Use a calculator to plot the graph of = f () where f () = 4 3 + + 1 b On the same screen plot the graphs of: ( ) i = f ( ) ii = f () iii = f The graph of = 9 4 is as shown. Sketch the graph of each of the following b appling suitable transformations. a = 9( 1) ( 1) 4 b = 18 4 c = 18( + 1) ( + 1) 4 d = 9 4 81 4 e = 9 4 + 1 (Do not find the -ais intercepts for e.) 9 81, 4 3 9 81, 4 3 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

156 Essential Mathematical Methods 3&4CAS 4.7 Determining rules for the graphs of polnomials It is first worth noting that the graph of a polnomial function of degree n is completel determined b an n + 1 points on the curve. Foreample, for a cubic function with rule = f (), if it is known that f (a 1 ) = b 1, f (a ) = b, f (a 3 ) = b 3, f (a 4 ) = b 4, then the rule can be determined. Finding the rule for a parabola has been discussed in Section 4.3 of this chapter. Eample 19 For the cubic function with rule f () = a 3 + b + c + d, itisknown that the points with coordinates ( 1, 18), (, 5), (1, 4) and (, 9) lie on the graph of the cubic. Find the values of a, b, c and d. The following equations can be formed. a + b c + d = 18 (1) d = 5 () a + b + c + d = 4 (3) 8a + 4b + c + d = 9 (4) Adding (1) and (3) gives b + d = and as d = 5, b = 6. There are now onl two unknowns. (3) becomes a + c = 7 (3 ) and (4) becomes 8a + c = (4 ) Multipl (3 )band subtract from (4 ). 6a = 6which gives a = 1 and c = 6 Using the TI-Nspire Enter solve( a + b c + d = 18 and d = 5 and a + b + c + d = 4 and 8a + 4b + c + d = 9, {a, b, c, d}). Alternativel, Define f () = a 3 + b + c + d and then solve( f ( 1) = 18 and f () = 5 and f (1) = 4 and f () = 9, {a, b, c, d}). Both methods are shown on the screen to the right. Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 157 Using the Casio ClassPad Press and select menu (if necessar tap ). Tap three times to produce the template to enter four simultaneous equations. Enter the equations and set the variables in the variable bo to a,b,c,d then tap. An alternative method is to define f () = a 3 + b + c + d then enter f ( 1) = 18, f () = 5, f (1) = 4 and f () = 9asthe simultaneous equations to be solved with variables a,b,c,d as above. The following eamples provide further procedures for finding the rules of cubic functions. It should be noted that a similar facilit is available for quartics. Eample The graph shown is that of a cubic function. Find the rule for this cubic function. From the graph the function can be seen to be of the form = a( 4) ( 1) ( + 3). 3 (, 4) It remains to find the value of a. The point with coordinates (, 4) is on the graph. Hence: 4 = a( 4)( 1)3 a = 1 3 and = 1 ( 4)( 1)( + 3) 3 1 4 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

158 Essential Mathematical Methods 3&4CAS Eample 1 The graph shown is that of a cubic function. Find the rule for this cubic function. From the graph it can be seen to be of the form = k( 1)( + 3).Itremains to find the value of k.asthe value (, 9) is on the graph: 9 = k( 1)(9) k = 1 and = ( 1)( + 3) (, 9) 3 1 Note: Because the graph reveals that there are two factors, ( + 3) and ( 1), there must be a third linear factor (a + b), but this implies that there is an intercept b a. Thus a + b = k 1 ( + 3) or k ( 1). A consideration of the signs reveals that ( + 3) is a repeated factor. Eample The graph of a cubic function passes through the point with coordinates (, 1), (1, 4), (, 17) and ( 1, ). Find the rule for this cubic function. The cubic function will have a rule of the form: = a 3 + b + c + d The values of a, b, c and d have to be determined. As the point (, 1) is on the graph, d = 1. B using the points (1, 4), (, 17) and ( 1, ), three simultaneous equations are produced: 4 = a + b + c + 1 17 = 8a + 4b + c + 1 = a + b c + 1 These become: 3 = a + b + c (1) 16 = 8a + 4b + c () 1 = a + b c (3) Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 159 Add (1) and (3) to find: b = 4 i.e., b = Substitute in (1) and (): 1 = a + c (4) 8 = 8a + c (5) Multipl (4) b and subtract from (5). Hence 6 = 6a and a = 1 From (4) c = Thus = 3 + + 1 Eercise 4G 1 Determine the rule for the cubic function with graph as shown: Determine the rule for the cubic function with graph as shown: 5 (, 11) (, 5) ( 1, ) 3 Find the rule for the cubic function that passes through the following points: a (, 1), (1, 3), ( 1, 1) and (, 11) b (, 1), (1, 1), ( 1, 1) and (, 7) c (, ), (1, ), ( 1, 6) and (, 1) 6 3 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

16 Essential Mathematical Methods 3&4CAS 4 Find epressions which define the following cubic curves: a b d.5 1 1 1 1 e 1 (, 3) 18 4 4 1 (, 3) (1,.75) 5 Find the equation of the cubic function for which the graph passes through the points with coordinates: a (, 135), (1, 156), (, 115), (3, ) b (, 3), (, 13), (1, 5), (, 11) 6 Find the equation of the quartic function for which the graph passes through the points with coordinates: a ( 1, 43), (, 4), (, 7), (6, 1618), (1, 67) b ( 3, 119), (, 3), ( 1, 9), (, 8), (1, 11) c ( 3, 6), ( 1, ), (1, ), (3, 66), (6, 17) 4.8 of literal equations and sstems of equations Literal equations Solving literal linear equations and simultaneous equations was undertaken in Section.. In this section other non-linear epressions are considered. The certainl can be solved with a CAS calculator but full setting out is shown here. Eample 3 Solve each of the following literal equations for : a + k + k = b 3 3a + a = c ( a) = 3 c 1 1 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 161 a Completing the square gives + k + k 4 + k k ( 4 = + k ) = k 4 k + k =± k 4k 4 = k ± k 4k A real solution eists onl for k 4k ; that is for k 4ork b 3 3a + a = ( 3a + a ) = ( a)( a) = = or = a or = a c ( a) = implies = or = a or = a In the following, the propert that for suitable values of a, b and an odd natural number n, b n = a is equivalent to b = a 1 n and also b = n a. Also b p q q = a is equivalent to b = a p suitable values of a and b and integers q and p. Care must be taken with the application of these. For eample = isequivalent to =± and 3 = 4isequivalent to = 8or = 8. Eample 4 Solve each of the following equations for : a a 3 b = c b 3 5 = a c 1 n = a where n is a natural number and is a positive real number d a ( + b) 3 = c e a 1 5 = b f 5 c = d a a 3 b = c a 3 = b + c 3 = b + c a b + c = 3 a ( ) 1 b + c 3 or = a b 3 5 = a is equivalent to = a 5 3 c 1 n = a is equivalent to = a n d a ( + b) 3 = c for ( + b) 3 = c ( a c ) 1 3 + b = a ( c ) 1 3 = b a Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

16 Essential Mathematical Methods 3&4CAS e a 1 5 = b f 1 5 = b ( a b = a ) 5 5 c = d 5 = c + d = (c + d) 1 5 Simultaneous equations In this section, methods for finding the coordinates of the points of intersection of different graphs are discussed. Eample 5 Find the coordinates of the points of intersection of the parabola with equation = with the straight line with equation = + 4 Consider + 4 = Then = 3 6 = 3 ± 9 4 6 1 = 3 ± 33 The points of intersection have coordinates ( 3 33 A, 11 ) ( 33 3 + 33 and B, 11 + ) 33 Using the TI-Nspire Use solve( ) from the Algebra menu (b 31)asshown. The word and can be tped directl or found in the catalog (b 31). Use the up arrow ( )tomoveuptothe answer and use the right arrow ( )to displa the remaining part of the answer. 4 A 4 = + 4 B = Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 163 Using the Casio ClassPad In the simultaneous equation entr screen, enter the equations = and = + 4 and set the variables as,. Note the at the end of the answer line indicates that ou must scroll to the right to see all the solutions. Eample 6 Find the coordinates of the points of intersection of the circle with equation ( 4) + = 16 and the line with equation = Rearrange = tomake the subject. Substitute = into the equation of the circle. i.e., ( 4) + = 16 8 + 16 + = 16 i.e., 8 = ( 4) = = or = 4 The points of intersection are (, ) and (4, 4). Eample 7 Find the point of contact of the line with equation 1 9 + = and the curve with equation 3 = 1 Rewrite the equations as = 1 9 + 3 and = 1 Consider 1 9 + 3 = 1 + 6 = 9 and + 6 = 9 Therefore 6 + 9 = and ( 3) = i.e. = 3 The point of intersection is ( 3, 1 3 ). = 1 + 9 3 (4, ) = (4, 4) 1 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

164 Essential Mathematical Methods 3&4CAS Using the TI-Nspire Use solve( ) from the Algebra menu (b 3 1)asshown. Note that the multiplication sign is required between and. Using the Casio ClassPad In the simultaneous equation entr screen, enter the equations 9 + = 3 and = 1 and set the variables as,. Note that for this eample, since the fraction entr is on the same screen as the simultaneous equation in the keboard menu, the fractions can be entered using the fraction entr ke. Eercise 4H 1 Solve each of the following literal equations for : a k + + k = b 3 7a + 1a = c ( 3 a) = d k + k = e 3 a = f 4 a 4 = g ( a) 5 ( b) = h (a ) 4 (a 3 )( a) = Solve each of the following equations for : a a 3 b = c b 3 7 = a c 1 n + c = a where n is a natural number and is a positive real number d a ( + b) 3 = c e a 1 3 = b f 3 c = d 3 Find the coordinates of the points of intersection for each of the following: a = = b = = c = = + 1 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 165 4 Find the coordinates of the points of intersection for each of the following: a + = 178 + = 16 d + = 97 + = 13 b + = 15 + = 15 e + = 16 = 4 c + = 185 = 3 5 Find the coordinates of the points of intersection for each of the following: a + = 8 = 187 b + = 51 = 518 c = 5 = 16 6 Find the coordinates of the points of intersection of the straight line with equation = and the circle with equation ( 5) + = 5 7 Find the coordinates of the points of intersection of the curves with equation = 1 + 3 and = 8 Find the coordinates of the points of intersection of the line with equation 4 5 = 1 and the circle with equation + 4 + = 1 9 Find the coordinates of the points of intersection of the curve with equation = 1 3 and the line with equation = + 1 Find the coordinates of the point where the line 4 = 9 + 4 touches the parabola with equation = 9 11 Find the coordinates of the point where the line with equation = + 3 5 touches the circle + = 9 1 Find the coordinates of the point where the straight line with equation = 1 4 + 1 touches the curve with equation = 1 13 Find the coordinates of the points of intersection of the curve with equation = and the line = 1 14 Solve the simultaneous equations: a 5 4 = 7 and = 6 b + 3 = 37 and = 45 c 5 3 = 18 and = 4 Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

166 Essential Mathematical Methods 3&4CAS 15 What is the condition for + a + b to be divisible b + c? 16 Solve the simultaneous equations = + and = 16 17 a Solve the simultaneous equations = m and = 1 + 5 for in terms of m. b Find the value of m for which the graphs of = m and = 1 + 5 touch and give the coordinates of this point. c Forwhich values of m do the graphs not meet? Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 167 Chapter summar A function P() = a + a 1 + +a n n,where a, a 1,...,a n are constants, is called a polnomial in. The numbers a, a 1,...a n are the coefficients. Assuming a n =,the term a n n is the leading term. The integer n is the degree of the polnomial. Degree one polnomials are called linear functions. Degree two polnomials are called quadratic functions. Degree three polnomials are called cubic functions. Degree four polnomials are called quartic functions. The ( remainder theorem:ifthe polnomial P() isdivided b a + b, the remainder is P b ) a ( The factor theorem:(a + b)isafactor of P() ifandonl if P b ) = a To sketch the graph of a quadratic function f () = a + b + c (called a parabola), use the following: If a >, the function has a minimum value. If a <, the function has a maimum value. The value of c gives the -ais intercept. b The equation of the ais of smmetr is = a The -ais intercepts are determined b solving the equation a + b + c = A quadratic equation ma be solved b: factorising completing the square b ± b 4ac using the general quadratic formula = a Using the discriminant of the quadratic function f () = a + b + c: If b 4ac >, the graph of the function has two -ais intercepts. If b 4ac =, the graph of the function touches the -ais. If b 4ac <, the graph of the function does not intercept the -ais. B completing the square, an quadratic function can be written in the form = A( + b) + B From this it can be seen that the graph of an quadratic function ma be obtained b a composition of transformations applied to the graph of =. Multiple-choice questions 1 The equation 5 1 inturning point form a( + h) + k b completing the square is: A (5 + 1) + 5 B (5 1) 5 C 5( 1) 5 D 5( + 1) E 5( 1) 7 Review Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

168 Essential Mathematical Methods 3&4CAS Review The value(s) of m that will give the equation m + 6 3 = two real roots is/are: A m = 3 B m = 3 C m = D m > 3 E m < 3 3 3 + 7 factorised over R is equal to: A ( + 3) 3 B ( 3) 3 C ( + 3)( 6 + 9) D ( 3)( + 3 + 9) E ( + 3)( 3 + 9) 4 The equation of the graph shown is: 6 4 4 6 A = ( )( + 4) B = ( + )( 4) C = ( + ) ( 4) D = ( + )( 4) E = ( + ) ( 4) 5 1isafactor of 3 + 3 a + 1. The value of a is: A B 5 C 5 D 5 E 5 6 6 8 8 is equal to: A (3 + )( 4) B (3 )(6 + 4) C (6 4)( + ) D (3 )( + 4) E (6 + )( 8) 7 A part of the graph of the third-degree polnomial function f (), near the point (1, ), is shown below. 1 Which of the following could be the rule for f ()? A f () = ( 1) B f () = ( 1) 3 C f () = ( 1) D f () = ( 1) E f () = ( + 1) 8 The coordinates of the turning point of the graph of the function p() = 3(( ) + 4) are: A (, 1) B (, 4) C (, 1) D (, 4) E (, 1) 9 The diagram below shows part of the graph of a polnomial function. c b Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 169 A possible equation for the rule of the function is: A = ( + c)( b) B = ( b)( c) C = ( c)(b ) D = ( c)(b ) E = ( + b) ( c) 1 The number of roots to the equation ( + a)( b)( + c) =, where a, b and c R +, is: A B 1 C D 3 E 4 Short-answer questions (technolog-free) 1 Sketch the graphs of each of the following quadratic functions. Clearl indicate coordinates of the verte and aes intercepts. a h() = 3( 1) + b h() = ( 1) 9 c f () = + 6 d f () = 6 e f () = + 5 f h() = 1 The points with coordinates (1, 1) and (, 5) lie on a parabola with equation of the form = a + b.find the values of a and b. 3 Solve the equation 3 1 = busing the quadratic formula. 4 Sketch the graphs of each of the following. State the coordinates of the point of zero gradient and the aes intercepts: a f () = ( 1) 3 16 b g() = ( + 1) 3 + 8 c h() = ( + ) 3 1 d f () = ( + 3) 3 1 e f () = 1 ( 1) 3 5 Draw a sign diagram for each of the following: a = ( + )( )( + 1) b = ( 3)( + 1)( 1) c = 3 + 7 + 14 + 8 d = 3 3 + 1 + 6 6 Without actuall dividing, find the remainder when the first polnomial is divided b the second: a 3 + 3 4 +, + 1 b 3 3 + 6, c 3 + 3 3, + 7 Determine the rule for the cubic function shown in the graphs: 3 (, 4) 7 Review Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

17 Essential Mathematical Methods 3&4CAS Review 8 The graph of f () = ( + 1) 3 ( ) is shown. Sketch the graph of: a = f ( 1) b = f ( + 1) c = f () d = f () + 1 (, ) 5, 187 4 56 9 Find the rule for the cubic function, the graph of which passes through the points (1, 1), (, 4), (3, 9) and (, 6). Etended-response questions 1 The rate of flow of water, R ml/min, into a vessel is described b the quartic epression R = kt 3 ( t), t, where t minutes is the time elapsed from the beginning of the flow. The graph is shown. a Find the value of k. b Find the rate of flow when t = 1. c The flow is adjusted so that a new epression for flow is: R new = kt 3 ( t), t d R (15, ) i Sketch the graph of R new against t for t. ii Find the rate of flow when t = 1. Water is allowed to run from the vessel and it is found that the rate of flow from the vessel is given b R out = k(t ) 3 (4 t) for t 4 i Sketch the graph of R out against t for t 4. ii Find the rate of flow when t = 3. Hint: Note that the graph of R new against t is given b a dilation of factor from the -ais. The graph of R out against t is given b a translation with rule (t, R) (t +, R), followed b a reflection in the t-ais. A large gas container is being deflated. The volume V (in m 3 )attime t hours is given b: V = 4(9 t) 3 where t 9 a Find the volume when: i t = ii t = 9 b Sketch the graph of V against t for t 9. c At what time is the volume 51 m 3? t Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 171 3 A reinforced bo is made b cutting congruent squares of side length cm from the four corners of a rectangular piece of cardboard that measures 48 cm b 96 cm. The flaps are folded up. 48 cm 96 cm a Find an epression for V, the volume of the bo formed. b Agraph of V against is as shown. i What is the domain of the V (cm 3 ) function V? ii From the graph, find the maimum volume of the bo and the value 15 of for which this occurs (approimate values are required). 1 c Find the volume of the bo when = 1. d It is decided that 5. 5 Find the maimum volume possible. e If 5 15, what is the minimum volume of the bo? 4 8 1 16 4 4 A hemispherical bowl of radius 6 cm contains water. The volume of water in the hemispherical bowl where the depth of the water is cm and is given b: V = 1 3 (18 )cm 3 a Find the volume of water when: i = ii = 3 iii = 4 b Find the volume when the hemispherical bowl is full. c Sketch the graph of V against. d Find the depth of water when the volume is equal to 35 cm 3. 3 5 A metal worker is required to cut a circular clinder from a solid sphere of radius 5 cm. A cross-section of the sphere and the clinder is shown in the diagram. a Epress r in terms of h,where r cm is the radius of the clinder and h cm is the height of the clinder. Hence show that the volume, V cm 3,ofthe clinder is given b V = 1 4 h(1 h ). 5 r h Review Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

17 Essential Mathematical Methods 3&4CAS Review b Sketch the graph of V against h for < h < 1. The coordinates of the maimum point are (5.77, 3.3). c Find the volume of the clinder if h = 6. d Find the height and radius of the clinder if the volume of the clinder is 48 cm 3. 6 An open tank is to be made from a sheet of cm cm metal 84 cm b 4 cm b cutting congruent squares of side length cm from each of the corners. cm cm a Find the volume V cm 3 of the bo in cm cm terms of. cm cm b State the maimal domain for V when it is considered as a function of. c Plot the graph of V against using a calculator. d Find the volume of the tank when: i = ii = 6 iii = 8 iv = 1 e Find the value(s) of, correct to two decimal places, for which the capacit of the tank is 1 litres. f Find, correct to two decimal places, the maimum capacit of the tank in cubic centimetres. 7 The rectangle is defined b vertices B and C on the curve with equation = 16 and vertices A and D on the -ais. B C(, ) a i Find the area, A,ofthe rectangle in = 16 terms of. ii State the implied domain for the A D function defined b the rule given in i. b Find: i the value of A when = 3 ii the value, correct to two decimal places, of when A = 5 c A cuboid has volume V given b the rule V = A i Find V in terms of. ii Find the value, correct to two decimal places, of such that V = 1. 8 The plan of a garden adjoining a wall is shown. The rectangle BCEF is of length m and width m. The borders of the A m F m two end sections are quarter circles of radius m and centres at E and F.Afence B is erected along the curves AB and CD and the straight line CB. m E m m C D Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard

Chapter 4 Polnomial functions 173 9 a Find the area A of the garden in terms of and. b If the length of the fence is 1 m, find: i in terms of ii A in terms of iii the maimal domain of the function for which the rule has been obtained in ii c Find, correct to two decimal places, the value(s) of if the area of the garden is to be 1 m. d It is decided to build the garden up to a height of metres. If the length of the fence is 5 1 m, find correct to two decimal places: i the volume V m 3 of soil needed in terms of ii the volume V m 3 of soil needed for a garden of area 1 m iii the value(s) of for which 5 m 3 of soil is required A mound of earth is piled up against a wall. h (m) The cross-section is as shown. The coordinates of several points on the surface are given. C(4, 3) a Find the equation for the cubic function for which the graph passes through the B(3, ) A(1, 1) points O, A, B and C. (m) b Forwhat value of is the height of the 1 3 4 mound 1.5 metres? c The coefficient of 3 for the function is small. Consider the quadratic formed when the 3 term is deleted. Compare the graph of the resulting quadratic function with the graph of the cubic function. d The mound moves and the curve describing the cross-section now passes through the points O(, ), A(1,.3), B(3,.7) and D(4,.8). Find the equation of the cubic function for which the graph passes through these points. e Let = f ()bethe function obtained in a. i Sketch the graph of the hbrid function: { f () 4 g() = f (8 ) 4 < 8 ii Comment on the appearance of the graph of = g() Review Cambridge Universit Press Uncorrected Sample Pages 8 Evans, Lipson, Jones, Aver, TI-Nspire & Casio ClassPad material prepared in collaboration with Jan Honnens & David Hibbard