SAMPLE. Quadratics. Objectives

Objectives C H A P T E R 4 Quadratics To recognise and sketch the graphs of quadratic relations. To determine the maimum or minimum values of a quadratic relation. To solve quadratic equations b factorising, completing the square and using the general formula. To appl the discriminant to determine the nature and number of roots of quadratic relations. To appl quadratic relations to solving problems. A polnomial function has a rule of the tpe = a n n + a n 1 n 1 +...+ a 1 + a 0 (n N) where a 0, a 1...a n are numbers called coefficients. The degree of a polnomial is given b the value of n, the highest power of with a non-zero coefficient. Eamples i = + 3isapolnomial of degree 1. ii = + 3 isapolnomial of degree. iii = 3 + 3 + 9 7isapolnomial of degree 3. First degree polnomials, otherwise called linear relations, have been discussed in Chapter. In this chapter second degree polnomials will be investigated. These are called quadratics. 4.1 Epanding and collecting like terms Finding the -ais intercepts, if the eist, is a requirement for sketching graphs of quadratics. To do this requires the solution of quadratic equations, and as an introduction to the methods of solving quadratic equations the basic algebraic processes of epansion and factorisation will be reviewed. 93

94 Essential Mathematical Methods 1&CAS Eample 1 Simplif ( 5) 3( + 5), b first epanding. Eample Epand (3 ) + 3( ). ( 5) 3( + 5) = 10 3 15 Epand each bracket. = 3 10 15 Collect like terms. = 5 (3 ) + 3( ) = 6 4 + 3 6 = 9 10 Forepansions of the tpe (a + b)(c + d) proceed as follows: Eample 3 (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd Epand the following: a ( + 3)( 3) b ( 3)( ) a ( + 3)( 3) = ( 3) + 3( 3) = 3 + 6 9 = + 3 9 Eample 4 Epand ( 1)(3 + + 4). b ( 3)( ) = ( ) 3( ) = 6 + 6 = ( + 6) + 6 ( 1)(3 + + 4) = (3 + + 4) 1(3 + + 4) = 6 3 + 4 + 8 3 4 = 6 3 + + 6 4

Chapter 4 Quadratics 95 Using the TI-Nspire Use Epand( ) from the Algebra menu (b 33)toepand the epression ( 1)(3 + + 4). Using the Casio ClassPad Enter the epression ( 1)(3 + + 4) into, then highlight and select Interactive Transformation epand. Consider the epansion of a perfect square, ( + a). ( + a) = ( + a)( + a) = ( + a) + a( + a) = + a + a + a = + a + a Thus the general result can be stated as: ( + a) = + a + a

96 Essential Mathematical Methods 1&CAS Eample 5 Epand (3 ). (3 ) = (3) + (3)( ) + ( ) = 9 1 + 4 Consider the epansion of ( + a)( a). ( + a)( a) = ( a) + a( a) = a + a a = a Thus the epansion of the difference of two squares has been obtained: Eample 6 ( + a)( a) = a Epand: a ( 4)( + 4) b ( 7)( + 7) a ( 4)( + 4) = () (4) = 4 16 Eample 7 Epand (a b + c)(a b c). (a b + c)(a b c) = ((a b) + c)((a b) c) = (a b) c = 4a 4ab + b c Eercise 4A 1 Epand each of the following: a ( 4) b ( 4) c 3( 4) d 3(4 ) e ( 1) f ( 5) b ( 7)( + 7) = ( 7) = 8

Chapter 4 Quadratics 97 Eample 1 Eample Eample 3 Eample 5 Collect like terms in each of the following: a + 4 + 1 b 6 + c 3 + 1 d + 3 + 4 3 Simplif each of the following b epanding and collecting like terms: a 8( 3) ( + 4) b ( 4) 3 c 4( 3) + 4(6 ) d 4 3(5 ) 4 Simplif each of the following b epanding and collecting like terms: a ( 4) 3 b ( 5) + ( 5) c ( 10 3) d 3( 3 + ) e 3 ( ) f 3(4 ) 6 5 Simplif each of the following b epanding and collecting like terms: a (3 7)( + 4) b ( 10)( 1) c (3 1)(1 + 4) d (4 5)( 3) e ( 3)( ) f ( 5)( + 5) 6 Simplif each of the following b epanding and collecting like terms: a ( 4) b ( 3) c (6 ) d ( ) 1 e ( 5) f ( 3) 7 Simplif each of the following b epanding and collecting like terms: a ( 3)(3 + 4) b ( 1)( + + 1) c (6 3 )(4 ) d ( 3)( + 3) e ( 4)( + 4) f (9 11)(9 + 11) g (5 3)( + ) ( 3)( + 3) h ( + 3)(3 ) (4 + )(4 ) i ( + z)( z) j ( )(a b) 8 Find the area of each of the following b: i ii a finding the area of the four non overlapping rectangles, two of which are squares, and adding multipling length b width of the undivided square (boundar in blue) b 1 cm 1 cm 3 cm 4 1 cm 1 cm 1 cm 1 3 4 cm cm

98 Essential Mathematical Methods 1&CAS 4. Factorising Four different tpes of factorisation will be considered. 1 Removing the highest common factor (HCF) Eample 8 Factorise 9 + 81. Eample 9 Factorise a 8a. Eample 10 Factorise 7 35. Grouping of terms 9 + 81 = 9 + 9 9 = 9( + 9) removing the HCF a 8a = a a a 4 = a(a 4 ) 7 35 = 7( 5) This can be used for epressions containing four terms. Eample 11 Factorise 3 + 4 3 1. where 7 is the HCF The terms in this epression can be grouped as follows: 3 + 4 3 1 = ( 3 + 4 ) (3 + 1) = ( + 4) 3( + 4) removing the HCF = ( 3)( + 4) 3 Difference of two squares (DOTS) a = ( + a)( a)

Chapter 4 Quadratics 99 Eample 1 Factorise 3 75. 3 75 = 3( 5) = 3( + 5)( 5) Eample 13 Factorise 9 36. This is a difference of two squares: Eample 14 Factorise ( ) 16. 9 36 = 9( 4) = 9( )( + ) ( ) 16 = ( ) (4) 3isthe HCF = ( + 4)( 4) = ( + 3)( 5) 4 Factorising quadratic epressions Eample 15 Factorise 8. 8 = ( + a)( + b) = + (a + b) + ab The values of a and b are such that ab = 8 and a + b = Values of a and b which satisf these two conditions are a = 4 and b = 8 = ( 4)( + )

100 Essential Mathematical Methods 1&CAS Eample 16 Factorise 6 13 15. There are several combinations of factors of 6 and 15 to Factors of 6 Factors of 15 Cross-products add to give 13 consider. Onl one combination is correct. 6 +5 +5 Factors of 6 13 15 = (6 + 5)( 3) Using the TI-Nspire Use Factor( ) from the Algebra menu (b 3)tofactorise the epression 6 13 15. Using the Casio ClassPad Enter the epression 6 13 15 into, then highlight and select Interactive Transformation factor. 3 18 13

Chapter 4 Quadratics 101 Eamples 8, 9 Eample 10 Eample 11 Eercise 4B 1 Factorise each of the following: a + 4 b 4a 8 c 6 3 d 10 e 18 + 1 f 4 16 Factorise: a 4 b 8a + 3 c 6ab 1b d 6 + 14 e + f 5 15 g 4 16 h 7 + 49 i j 6 9 k 7 6 l 8 + 6 3 Factorise: a 3 + 5 + + 5 b + 1 c a + a + b + b d a 3 3a + a 3 e 3 b a + a b Eample 14 g 4 Factorise: a 36 b 4 81 c 98 d 3a 7a e ( ) 16 f 5 ( + ) 3( + 1) 1 h ( ) ( + 3) Eamples 1, 13 Eample 15 Eample 16 5 Factorise: a 7 18 b 19 + 48 c 3 7 + d 6 + 7 + e a 14a + 4 f a + 18a + 81 g 5 + 3 + 1 h 3 1 36 i 18 + 8 j 4 36 + 7 k 3 + 15 + 18 l a + 7a + 1a m 5 3 16 + 1 n 48 4 + 3 3 o ( 1) + 4( 1) + 3 4.3 Quadratic equations In this section the solution of quadratic equations b simple factorisation is considered. There are three steps to solving a quadratic equation b factorisation. Step 1 Write the equation in the form a + b + c = 0. Step Factorise the quadratic epression. Step 3 Use the result that ab = 0 implies a = 0orb = 0 (or both) (the null factor theorem). Foreample = 1 1 = 0 Step 1 ( 4)( + 3) = 0 Step 4 = 0 or + 3 = 0 Step 3 = 4 or = 3

10 Essential Mathematical Methods 1&CAS Eample 17 Solve + 11 + 4 = 0. +4 +11 +3 +3 +8 +8 +11 Factorising we obtain ( + 3)( + 8) = 0 + 3 = 0 or + 8 = 0 = 3 or = 8 i.e. both = 8 and = 3 are solutions of + 11 + 4 = 0 To verif, substitute in the equation. When = 8 ( 8) + 11( 8) + 4 = 0 = 3 ( 3) + 11( 3) + 4 = 0 Eample 18 Solve + 5 1 = 0. Eample 19 1 +5 3 3 +4 +8 +5 Factorising gives ( 3)( + 4) = 0 3 = 0 or + 4 = 0 = 3 or = 4 The perimeter of a rectangle is 0 cm and its area is 4 cm. Calculate the length and width of the rectangle. Let cm be the length of the rectangle and cm the width. Then ( + ) = 0 and thus = 10. The area is 4 cm and therefore (10 ) = 4. i.e. 10 = 4 This implies 10 4 = 0 ( 6)( 4) = 0 Thus the length is 6 cm or 4 cm. The width is 4 cm or 6 cm.

Chapter 4 Quadratics 103 Eample 17 Eample 18 Eample 19 Eample 19 Eercise 4C 1 Solve each of the following for : a ( )( 3) = 0 b ( 4) = 0 c ( 4)( 6) = 0 d (3 )( 4) = 0 e ( 6)( + 4) = 0 f ( 1) = 0 g (5 )(6 ) = 0 h = 16 Use a CAS calculator to solve each of the following equations. Give our answer correct to decimal places. a 4 3 = 0 b 4 3 = 0 c 4 + 3 = 0 3 Solve for in each of the following: a 6 + 8 = 0 b 8 33 = 0 c ( + 1) = 64 d + 5 14 = 0 e + 5 + 3 = 0 f 4 8 + 3 = 0 g = 5 + 4 h 6 + 13 + 6 = 0 i = 6 j 6 + 15 = 3 k 3 9 = 0 l 10 11 + 3 = 0 m 1 + = 6 n 4 + 1 = 4 o ( + 4) = 5 p 1 7 = 3 7 q + 8 = 15 r 5 = 11 4 The bending moment, M,ofasimple beam used in bridge construction is given b the formula M = wl w. If l = 13 m, w = 16 kg/m and M = 88 kg m, calculate the value of. 5 Calculate the value of. cm 6 cm Area = 30 cm cm 7 cm 6 The height, h, reached b a projectile after t seconds travelling verticall upwards is given b the formula h = 70t 16t. Calculate t if h is 76 metres. n(n 3) 7 A polgon with n sides has diagonals. How man sides has a polgon with 65 diagonals? 8 Foraparticular electric train the tractive resistance, R, atspeed, v km/h is given b R = 1.6 + 0.03v + 0.003v.Find v when the tractive resistance is 10.6. 9 The perimeter of a rectangle is 16 cm and its area is 1 cm. Calculate the length and width of the rectangle. 10 The altitude of a triangle is 1 cm shorter than the base. If the area of the triangle is 15 cm, calculate the altitude.

104 Essential Mathematical Methods 1&CAS 11 Tickets for a concert are available at two prices. The more epensive ticket is $30 more than the cheaper one. Find the cost of each tpe of ticket if a group can bu 10 more of the cheaper tickets than the epensive ones for $1800. 1 The members of a club hire a bus for $100. Seven members withdraw from the club and the remaining members have to pa $10 more each to cover the cost. How man members originall agreed to go on the bus? 4.4 Graphing quadratics A quadratic relation is defined b the general rule = a + b + c where a, b and c are constants and a 0. This is called polnomial form. The simplest quadratic relation is =. If a table of values is constructed for = for 3 3, 3 1 0 1 3 9 4 1 0 1 4 9 these points can be plotted and then connected to produce a continuous curve. Features of the graph of = : The graph is called a parabola. The possible -values are all positive real numbers and 0. (This is called the range of 10 Ais of smmetr the quadratic and is discussed in a more 8 = general contet in Chapter 6.) 6 It is smmetrical about the -ais. The line 4 about which the graph is smmetrical is called the ais of smmetr. The graph has a verte or turning point at the origin (0, 0). 3 1 0 1 3 Verte or turning point The minimum value of is 0 and it occurs at the turning point. B a process called completing the square (to be discussed later in this chapter) all quadratics in polnomial form = a + b + c ma be transposed into what will be called the turning point form: = a( h) + k The effect of changing the values of a, h and k on our basic graph of = will now be investigated. Graphs of the form = a( h) + k are formed b transforming the graph of =.Amore formal approach to transformations is undertaken in Chapter 6.

i Changing the value of a First consider graphs of the form = a.inthis case both h = 0 and k = 0. In the basic graph of =, a is equal to 1. The following graphs are shown on the same set of aes. = = (a = ) = 1 ( a = 1 ) = (a = ) Chapter 4 Quadratics 105 8 6 4 1 0 1 4 6 8 = = = = If a > 1 the graph is narrower, if a < 1 the graph is broader. The transformation which produces the graph of = from the graph of = is called a dilation of factor from the -ais. When a is negative the graph is reflected in the -ais. The transformation which produces the graph of = from the graph of = is called a reflection in the -ais. ii Changing the value of k (a = 1 and h = 0) On this set of aes are the graphs of = + 1 = 4 = (k = ) = = + 1 (k = 1) 3 As can be seen, changing k moves the basic graph of = in a vertical direction. When k = the graph is 1 = translated units in the negative 1 1 0 1 direction of the -ais. The verte is now (0, ) and the range is now all real numbers greater than or equal to. When k = 1 the graph is translated 1 unit in the positive direction of the -ais. The verte is now (0, 1) and the range is now all real numbers greater than or equal to 1. All other features of the graph are unchanged. The ais of smmetr is still the -ais. 1

106 Essential Mathematical Methods 1&CAS iii Changing the value of h (a = 1 and k = 0) On this set of aes are the graphs of = ( + 3) = = ( ) = = ( ) (h = ) 8 = ( + 3) (h = 3) As can be seen, changing h moves 4 the graph in a horizontal direction. When h = the graph is translated units in the positive direction of the -ais. 3 1 0 1 3 The verte is now (, 0) and the ais of smmetr is now the line = ; however, the range is unchanged and is still all non-negative real numbers. When h = 3the graph is translated 3 units in the negative direction of the -ais. The verte is now ( 3, 0) and the ais of smmetr is now the line = 3; however, again the range is unchanged and is still all non-negative real numbers. All these effects can be combined and the graph of an quadratic, epressed in the form = a( h) + k, can be sketched. Eample 0 Sketch the graph of = 3. The graph of = 3isobtained from the graph of = batranslation of 3 units in the negative direction of the -ais The verte is now at (0, 3). The ais of smmetr is the line with equation = 0. The -ais intercepts are determined b solving the equation 3 = 0 = 3 =± 3 -ais intercepts are ± 3. 4 0 3 = = 3

Chapter 4 Quadratics 107 Eample 1 Sketch the graph of = ( + 1). The graph of = ( + 1) is obtained from the graph of = bareflection in the -ais followed b a translation of 1 unit in the negative direction of the -ais. The verte is now at ( 1, 0). The ais of smmetr is the line with equation = 1. The -ais intercept is ( 1, 0). = ( + 1) Eample Sketch the graph of = ( 1) + 3. 1 5 0 = = The graph of = is translated 1 unit in the positive direction of the -ais and 3 units in the positive direction of the -ais. 3 (1, 3) The verte has coordinates (1, 3). The ais of smmetr is the line = 1. The graph will be narrower than =. 0 1 The range will be 3. To add further detail to our graph, the -ais intercept and the -ais intercepts (if an) can be found. -ais intercept: Let = 0 = (0 1) + 3 = 5 -ais intercept(s): In this eample the minimum value of is 3, so cannot be 0. this graph has no -ais intercepts. Note: Let = 0 and tr to solve for. 0 = ( 1) + 3 3 = ( 1) 3 = ( 1) As the square root of a negative number is not a real number, this equation has no real solutions.

108 Essential Mathematical Methods 1&CAS Eample 3 Sketch the graph of = ( + 1) + 4. The verte has coordinates ( 1, 4). The ais of smmetr is the line = 1. -ais intercept: Let = 0 = (0 + 1) + 4 = 3 -ais intercepts: Let = 0 0 = ( + 1) + 4 ( + 1) = 4 + 1 =± =± 1 -ais intercepts are (1, 0) and ( 3, 0) Eercise 4D ( 3, 0) ( 1, 4) 0 (0, 3) Find: i the coordinates of the turning point ii the ais of smmetr iii the -ais intercepts (if an) (1, 0) of each case and use this information to help sketch the following graphs. Eample 0 a = 4 b = + c = + 3 Eample 1 d = + 5 e = ( ) f = ( + 3) Eample g = ( + 1) h = 1 ( 4) i = ( ) 1 Eample 3 j = ( 1) + k = ( + 1) 1 l = ( 3) + 1 m = ( + ) 4 n = ( + ) 18 o = 3( 4) + 3 p = 1 ( + 5) q = 3( + ) 1 r = 4( ) + 8 4.5 Completing the square In order to use the above technique for sketching quadratics, it is necessar for the quadratic to be epressed in turning point form. To transpose a quadratic in polnomial form we must complete the square. Consider the perfect square ( + a) which when epanded becomes + a + a

Chapter 4 Quadratics 109 The last term of the epansion is the square of half the coefficient of the middle term. Now consider the quadratic = + 3 This is not a perfect square. We can however find the correct last term to make this a perfect square. ( ) 1 If the last term is 1 =, then = + + 1 = ( + 1) a perfect square. In order to keep our original quadratic intact, we both add and subtract the correct last term. Foreample: = + 3 becomes = ( + + 1) 1 3 This can now be simplified to = ( + 1) 4 Hence the verte (turning point) can now be seen to be the point with coordinates ( 1, 4). In the above eample the coefficient of was 1.Ifthe coefficient is not 1, this coefficient must first be factored out before proceeding to complete the square. Completing the square for + is represented in the following diagram. The diagram to the left shows + 1 +. The small rectangle to the right is 1 1 1 moved to the base of the b 1 square. The red square of area + 1 1 unit is added. Thus + + 1 = ( + 1). The process of completing the square can also be used for the solution of equations. Eample 4 Solve each of the following equations for b first completing the square: a 3 + 1 = 0 b k + 1 = 0 c 3 1 = 0

110 Essential Mathematical Methods 1&CAS a 3 + 1 = 0 Completing the square: ( ) 3 ( ) 3 3 + + 1 = 0 ( 3 ) 5 4 = 0 ( Therefore 3 ) = 5 4 and 3 5 =± Hence = 3 5 ± = 3 ± 5. c 3 1 = 0 ( 3 1 ) = 0 3 ( ) 3 + 1 ( ) 3 4 = 0 4 ( 3 ) = 17 4 16 Therefore 3 17 4 =± 4 and = 3 17 4 ± = 3 ± 17 4 4 Eample 5 b k + 1 = 0 Completing the square: k + k + 1 k = 0 ( k) = k 1 Therefore k =± k 1 And = k ± k 1 Note: If k =±1 then =±1. If k > 1ork < 1 then there are two solutions. If 1 < k < 1 then there are no solutions. Divide both sides b and then complete the square. Find the coordinates of the verte b completing the square and hence sketch the graph of = + 6 8. = + 6 8 = ( 3 + 4) is factored out ( ( ) 3 ( ) 3 = 3 + + 4) { ( = 3 ) } 9 4 + 4 { ( = 3 ) } + 7 4 ( = 3 ) 7 3 4 8 0 1 3, 7

Eample 4 Eample 5 Chapter 4 Quadratics 111 ( ) 3 The verte is, 7. -ais intercept = 8. Graph has maimum value of 7, there are no -ais intercepts. The ais of smmetr is = 3. Eercise 4E 1 Epand each of the following: a ( 1) b ( + ) c ( 3) d ( + 3) ( e ( ) f ( 5) g 1 ( h ) 3 Factorise each of the following: a 4 + 4 b 1 + 36 c + 4 4 d 8 + 8 e + 1 18 f + 1 4 g 3 + 9 4 h + 5 + 5 4 3 Solve each of the following equations for b first completing the square: a 1 = 0 b 4 = 0 c 6 + = 0 d 5 + = 0 e 4 + 1 = 0 f 3 5 = 0 g + + k = 0 h k + + k = 0 i 3k + 1 = 0 4 Epress each of the following in the form = a( h) + k. Hence state the coordinates of the turning point and sketch the graph in each case. a + 3 b + 4 + 1 c 3 + 1 d 8 + 1 e f + 4 g + 4 + 1 h 1 1 i 3 6 + 1 4.6 Sketching quadratics in polnomial form It is not essential to convert quadratics to turning point form in order to sketch the graph. We can find the - and -ais intercepts and the ais of smmetr from polnomial form b alternative methods and use these details to sketch the graph. -ais intercept: Letting = 0inthe general equation = a + b + c we have = a(0) + b(0) + c = c ) -ais intercepts: i.e. the -ais intercept is alwas equal to c. Letting = 0inthe general equation we have 0 = a + b + c

11 Essential Mathematical Methods 1&CAS In order to solve such an equation it is necessar to factorise the right-hand side and use the null factor theorem. Once the -ais intercepts have been found, the turning point can be found b using the smmetr properties of the parabola. Eample 6 Find the - and -ais intercepts and the turning point, and hence sketch the graph of = 4. Step 1 c = 0, -ais intercept is (0, 0) Step Set = 0 and factorise right-hand side of equation: 0 = 4 0 = ( 4) = 0or = 4 -ais intercepts are (0, 0) and (4, 0) Step 3 Ais of smmetr is the line with equation = 0 + 4 i.e. = When =, = () 4() = 4 turning point has coordinates (, 4). Eample 7 4 0 4 (, 4) Find the - and -ais intercepts and the coordinates of the turning point, and hence sketch the graph of = 9. Step 1 c = 9, -ais intercept is (0, 9). Step Set = 0 and factorise right-hand side: 0 = 9 0 = ( + 3)( 3) = 3or = 3 3 0 9 3 (0, 9) -ais intercepts are ( 3, 0) and (3, 0) Step 3 Ais of smmetr is the line with equation = 3 + 3 i.e. = 0 When = 0, = (0) 9 = 9 turning point has coordinates (0, 9).

Chapter 4 Quadratics 113 Eample 8 Find the - and -ais intercepts and the turning point, and hence sketch the graph of = + 1. Step 1 c = 1, -ais intercept is (0, 1) Step Set = 0 and factorise the right-hand side: Step 3 0 = + 1 0 = ( + 4)( 3) = 4or = 3 -ais intercepts are ( 4, 0) and (3, 0) 4 1 1, 1 1 4 0 1 Due to the smmetr of the parabola, the ais of smmetr will be the line bisecting the two -ais intercepts. the ais of smmetr is the line with equation = 4 + 3 = 1. When = 1 (, = ) 1 ( + 1 ) 1 = 1 1 4 ( the turning point has coordinates 1 ), 11. 4 Using the TI-Nspire To graph the quadratic relation with rule = + 1, enter the rule in the Entr Line of a Graphs & Geometr application as shown and press enter to obtain the graph. 3

114 Essential Mathematical Methods 1&CAS Eamples 6, 7 Using the Casio ClassPad To graph the quadratic relation with rule = + 1, enter the rule in the screen, tick the bo and click.itma be necessar to change the view window using. Eercise 4F 1 a A parabola has -ais intercepts 4 and 10. State the -coordinate of the verte. b A parabola has -ais intercepts 6 and 8. State the -coordinate of the verte. c A parabola has -ais intercepts 6 and 8. State the -coordinate of the verte. a A parabola has verte (, 6) and one of the -ais intercepts is at 6. Find the other -ais intercept. b A parabola has verte (, 6) and one of the -ais intercepts is at 4. Find the other -ais intercept. c A parabola has verte (, 6) and one of the -ais intercepts is at the origin. Find the other -ais intercept. 3 Sketch each of the following parabolas, clearl showing the ais intercepts and the turning point: a = 1 b = + 6 c = 5 d = 4 e = + 3 f = 4 g = 3 h = + 1 4 Sketch each of the following parabolas, clearl showing the ais intercepts and the turning point: Eample 8 a = + 3 10 b = 5 + 4 c = + 3 d = + 4 + 3 e = 1 f = 6 g = 5 6 h = 5 4

Chapter 4 Quadratics 115 4.7 The general quadratic formula Not all quadratics can be factorised b inspection and it is often difficult to find the -ais intercepts this wa. If the general epression for a quadratic in polnomial form is considered, a general formula can be developed b using the completing the square technique. This can be used to solve quadratic equations. To solve a + b + c = 0 + b a + c = 0 Divide all terms b a. a + b ( ) b ( ) b a + + c a a a = 0 Complete the square b adding and + b ( ) b ( ) b a + = c ( ) b subtracting. a a a a ( + b ) = b a 4a c a = b 4ac 4a + b a =± b 4ac 4a = b a ± b 4ac 4a = b ± b 4ac a Note: The use of the quadratic formula is an alternative method to completing the square for solving quadratic equations, but is probabl not as useful as completing the square for curve sketching, which gives the turning point coordinates directl. It should be noted that the equation of the ais of smmetr can be derived from this general formula. The ais of smmetr is the line with equation = b a. Also, from the formula it can be seen that: if b 4ac > 0 there are two solutions if b 4ac = 0 there is one solution if b 4ac < 0 there are no real solutions. This will be further eplored in the net section. A CAS Calculator gives the following result.

116 Essential Mathematical Methods 1&CAS Eample 9 Solve each of the following equations for b using the quadratic formula: a 1 = 0 b k 3 = 0 a 1 = 0 a = 1, b = 1and c = 1 The formula gives = ( 1) ± ( 1) 4 1 ( 1) 1 = 1 ± 5 b k 3 = 0 a = 1, b = k and c = 3 The formula gives = ( k) ± ( k) 4 1 ( 3) = k ± 4k + 1 4 = k ± k + 3 Using the TI-Nspire Use Solve( ) from the Algebra menu (b 31)tosolve the equation k 3 = 0 for.

Chapter 4 Quadratics 117 Using the Casio ClassPad To solve the equation k 3 = 0 for, enter and highlight the equation (use VARonthe keboard to enter the variables). Select Interactive Equation/ Inequalit Solve and set the variable to. Eample 30 Sketch the graph of = 3 1 7bfirst using the quadratic formula to calculate the -ais intercepts. Since c = 7the -ais intercept is (0, 7). Turning point coordinates Ais of smmetr: = b a = ( 1) 3 = When =, = 3( ) 1( ) 7 = 5 turning point coordinates are (, 5).

118 Essential Mathematical Methods 1&CAS -ais intercepts: Eamples 9, 30 3 1 7 = 0 Eercise 4G = b ± b 4ac a = ( 1) ± ( 1) 4( 3)( 7) ( 3) = 1 ± 60 6 = 1 ± 15 6 = 6 ± 15 3 = 6 ± 3.87 (to decimal places) 3 = 3.9 or 0.71 (, 5) 3.9 0.71 5 0 1 For each of the following the coefficients a, b and c of a quadratic = a + b + c are given. Find: i b 4ac ii b 4ac in simplest surd form a a =, b = 4 and c = 3 b a = 1, b = 10 and c =18 c a = 1, b = 10 and c = 18 d a = 1, b = 6 and c = 15 e a = 1, b = 9 and c = 7 Simplif each of the following: a + 5 b 9 3 5 5 + 5 5 c d 6 + 1 6 10 6 3 Solve each of the following for.give eact answers. a + 6 = 4 b 7 3 = 0 c 5 + = 0 d + 4 7 = 0 e + 8 = 1 f 5 10 = 1 g + 4 1 = 0 h + = 3 i.5 + 3 + 0.3 = 0 j 0.6 1.3 = 0.1 k k 4 + k = 0 l (1 k) 4k + k = 0 4 The surface area, S,ofaclindrical tank with a hemispherical top is given b the formula S = ar + brh, where a = 9.4 and b = 6.8. What is the radius of a tank of height 6mwhich has a surface area of 15.6 m? h r 7

Eample 9 Chapter 4 Quadratics 119 5 Sketch the graphs of the following parabolas. Use the quadratic formula to find the -ais intercepts (if the eist) and the ais of smmetr and, hence, the turning point. a = + 5 1 b = 3 1 c = 3 + 1 d + 4 = + e = 4 + 5 + 1 f = 3 + 4 4.8 Iteration Quadratic equations can be solved using a process of simple iteration. Inthis section the process is discussed. Knowledge of sequence notation from General Mathematics is useful but not essential for this topic. Consider the quadratic equation + 3 5 = 0. This can be rearranged to the equivalent equations ( + 3) = 5 and = 5 + 3 Solving the equation + 3 5 = 0isequivalent to solving the simultaneous equations = and = 5 + 3 The equation = 5 can be used to form a difference equation (iterative equation) + 3 n+1 = 5 n + 3 5 or n = n 1 + 3 The elements of the sequence are called iterations. Using the TI-Nspire Method 1 Choose a starting value near the positive solution of + 3 5 = 0. Let 1 =. In a Lists & Spreadsheet application, enter 1 in cell A1 and. in cell B1. Enter = a1 + 1incell A and = 5/(b1 + 3) in cell B. (Entering. rather than in B1 ensures the iterations are displaed as decimal numbers.)

10 Essential Mathematical Methods 1&CAS Highlight the cells A and B using g and the Nav Pad and use Fill Down (b 33)togenerate the sequence of iterations. To better see the values of the iterations use Maimize Column Width (b 1) for column B. Tr other starting values, for eample 1 = 400 or 1 = 00. Convergence is alwas towards the solution = 3 + 9. The convergence can be illustrated with a cobweb diagram. The graphs of = 5 and = are sketched on the one set of aes and the + 3 path of the sequence is illustrated. Method = 5 + 3 1 0 = 1 (1, 1.5) (1, 1) (1.5, 1.1764) (1.5, 1.5) (, 1) 0 1 Another method which can be used to generate the sequence is to use ans (/ v) in 5 a Calculator application. To generate the sequence n = first tpe. And n 1 + 3 5 press enter. Then tpe and press enter repeatedl to generate the sequence of ans + 3 iterations.

Chapter 4 Quadratics 11 Using the Casio ClassPad Method 1 Choose a starting value near the positive solution of + 3 5 = 0. Let 1 =. In enter column headings n, (n), (n + 1). Enter the numbers 1 to 10 in the n column. Enter the value for 1 into cell B. In cell C enter the formula = 5/(B + 3) for n+1. The formula will appear in the formula bar towards the bottom of the screen as ou enter it and the answer will appear in cell C. Tpe the formula =C into cell B3 to set the answer to the first iteration as the n value for the net iteration. Click on cell B3 and drag to select cells B3 to B11. Select Edit Fill Range and OK. The formula is copied and the column fills with zeroes as there are no values in cells C3 downwards. Click on cell C and drag to select cells C to C11. Select Edit Fill Range and OK to cop the formula. You can tr some other starting points b replacing the 1 value in cell B; for eample, set 1 = 400 or 1 = 00. Convergence is alwas towards the solution = 3 + 9. Method Another method which can be used to generate the sequence is to enter the first value as variable.turn on the keboard and remember to use from the VAR menu or from the keboard. Enter the formula for n+1 in the net entr line and send its value to as shown. Press EXE repeatedl, enter the previous value and calculate the net value in the iterative sequence.

1 Essential Mathematical Methods 1&CAS Other arrangements of the equation + 3 5 = 0 can be tried to find the other solution. Foreample: A = + 5 ields the sequence n+1 = ( n) + 5. This converges to 3 + 9 again 3 3 if 1 = isused for the start. 1 = 9ields no convergence. B = 5 3 ields the sequence n+1 = 5 3. This converges to 3 9 for the start n 1 = 9. C = 5 3 ields the sequence n+1 = 5 3 n. This converges to 3 9 for the start 1 = 10. Eercise 4H Solve the following equations b using iteration: a + 5 10 = 0 using n+1 = 10 n + 5 b 3 5 = 0 using n+1 = 5 n 3 c 7 + 7 = 0 using n+1 = n + d 5 + 5 = 0 using n+1 = n + 4.9 The discriminant When graphing quadratics it is apparent that the number of -ais intercepts a parabola ma have is: i zero the graph is either all above or all below the -ais ii one the graph touches the -ais; the turning point is the -ais intercept or iii two the graph crosses the -ais. B considering the formula for the general solution to a quadratic equation, a + b + c = 0, we can establish whether a parabola will have zero, one or two -ais intercepts.

Chapter 4 Quadratics 13 The epression b 4ac,which is part of the quadratic formula, is called the discriminant. i If the discriminant b 4ac < 0, then the equation a + b + c = 0 has zero solutions and the corresponding parabola will have no -ais intercepts. ii If the discriminant b 4ac = 0, then the equation a + b + c = 0 has one solution and the corresponding parabola will have one -ais intercept. (We sometimes sa the equation has two coincident solutions.) iii If the discriminant b 4ac > 0, then the equation a + b + c = 0 has two solutions and the corresponding parabola will have two -ais intercepts. Eample 31 Find the discriminant of each of the following quadratics and state whether the graph of each crosses the -ais, touches the -ais or does not intersect the -ais. a = 6 + 8 b = 8 + 16 c = 3 + 4 Note: Discriminant is denoted b the smbol. a Discriminant =b 4ac = ( 6) (4 1 8) = 4 As > 0 the graph intersects the -ais at two distinct points, i.e. there are two distinct solutions of the equation 6 + 8 = 0. b =b 4ac = ( 8) (4 1 16) = 64 64 = 0 As =0the graph touches the -ais, i.e. there is one solution of the equation 8 + 16 = 0.

14 Essential Mathematical Methods 1&CAS Eample 31 c =b 4ac = ( 3) (4 4) = 3 As < 0 the graph does not intersect the -ais, i.e. there are no real solutions for the equation 3 + 4 = 0. The discriminant can be used to assist in the identification of the particular tpe of solution for a quadratic equation. For a, b and c rational: If =b 4ac is a perfect square, which is not zero, then the quadratic equation has two rational solutions. If =b 4ac = 0 the quadratic equation has one rational solution. If =b 4ac is not a perfect square then the quadratic equation has two irrational solutions. Eercise 4I 1 Determine the discriminant of each of the following quadratics: a + 4 b + + 4 c + 3 4 d + 3 4 e + 3 + 4 Without sketching the graphs of the following quadratics, determine whether the cross or touch the -ais: a = 5 + b = 4 + 1 c = 6 + 9 d = 8 3 e = 3 + + 5 f = 1 3 B eamining the discriminant, find the number of roots of: a + 8 + 7 = 0 b 3 + 8 + 7 = 0 c 10 3 = 0 d + 8 7 = 0 e 3 8 7 = 0 f 10 + 3 = 0 4 B eamining the discriminant, state the nature and number of roots for each of the following: a 9 4 + 16 = 0 b 5 6 = 0 c 4 = 0 d 5 0 + 4 = 0 e 6 3 = 0 f + 3 + = 0 5 Find the discriminant for the equation 4 + (m 4) m = 0, where m is a rational number and hence show that the equation has rational solution(s). Eercise 4J contains more eercises involving the discriminant.

Chapter 4 Quadratics 15 4.10 Solving quadratic inequations Eample 3 Solve + 1 > 0. Step 1 Solve the equation + 1 = 0 ( + 4)( 3) = 0 = 3or = 4 Step Sketch the graph of the quadratic = + 1. Step 3 Use the graph to determine the values of for which + 1 > 0. From the graph it can be seen that + 1 > 0when < 4 or when > 3. { : + 1 > 0} = { : < 4} { : > 3} Eample 33 Find the values of m for which the equation 3 m + 3 = 0 has: a 1 solution b no solution c distinct solutions For the quadratic 3 m + 3, =4m 36. a For1solution b Fornosolution =0 4m 36 = 0 m = 9 m =±3 4 0 4 8 < 0 i.e. 4m 36 < 0 From the graph 3 < m < 3 c Fortwo distinct solutions > 0 i.e. 4m 36 > 0 From the graph it can be seen that m > 3orm < 3 1 3 3 0 3 m 36

16 Essential Mathematical Methods 1&CAS Eample 3 Eample 33 Eample 34 Eample 34 Show that the solutions of the equation 3 + (m 3) m are rational for all rational values of m. =(m 3) 4 3 ( m) = m 6m + 9 + 1m = m + 6m + 9 = (m + 3) 0 for all m Furthermore is a perfect square for all m. Eercise 4J 1 Solve each of the following inequalities: a + 8 0 b 5 4 < 0 c 4 1 0 d 3 9 > 0 e 6 + 13 < 6 f 5 6 0 g 1 + > 6 h 10 11 3 i ( 1) 0 j 4 + 5p p 0 k 3 + < 0 l + 3 Find the values of m for which each of the following equations: i has no solutions ii has one solution iii has two distinct solutions a 4m + 0 = 0 b m 3m + 3 = 0 c 5 5m m = 0 d + 4m 4(m ) = 0 3 Find the values of p for which the equation p + ( p + ) + p + 7 = 0 has no real solution. 4 Find the values of p for which the equation (1 p) + 8p ( + 8p) = 0 has one solution. 5 Find the values of p for which the graph of = p + 8 + p 6 crosses the -ais. 6 Show that the equation ( p + 1) + pq + q = 0 has no real solution for an values of p and q (q 0). 7 For m and n rational numbers show that m + (m + n) + n = 0 has rational solutions.

Chapter 4 Quadratics 17 4.11 Solving simultaneous linear and quadratic equations As discussed in Section 1.3, when solving simultaneous linear equations we are actuall finding the point of intersection of the two linear graphs involved. If we wish to find the point or points of intersection between a straight line and a parabola we can solve the equations simultaneousl. It should be noted that depending on whether the straight line intersects, touches or does not intersect the parabola we ma get two, one or zero points of intersection. 0 0 0 Two points of intersection One point of intersection No point of intersection If there is one point of intersection between the parabola and the straight line then the line is a tangent to the parabola. As we usuall have the quadratic equation written with as the subject, it is necessar to have the linear equation written with as the subject (i.e. in gradient form). Then the linear epression for can be substituted into the quadratic equation. Eample 35 Find the points of intersection of the line with the equation = + 4 and the parabola with the equation = 8 + 1. At the point of intersection 8 + 1 = + 4 6 8 = 0 ( )( 4) = 0 Hence = or = 4 When =, = () + 4 = 0 = 4, = (4) + 4 = 4 Therefore the points of intersection are (, 0) and (4, 4). The result can be shown graphicall. 1 4 4 = 8 + 1 (, 0) 4 (4, 4) = + 4

18 Essential Mathematical Methods 1&CAS Using the TI-Nspire Use Solve( ) from the Algebra menu (b 41)tosolve the simultaneous equations = + 4 and = 8 + 1. Using the Casio ClassPad In turn on the keboard, select D (and at bottom left if necessar), then click the simultaneous equations entr button. Enter the simultaneous equations = + 4 and = 8 + 1 into the two lines and the variables, as the variables to be solved. Eample 36 Prove that the straight line with the equation = 1 meets the parabola with the equation = 3 + once onl.

Chapter 4 Quadratics 19 Eample 35 Eample 36 At the point of intersection: 3 + = 1 + 1 = 0 ( 1) = 0 Therefore = 1 When = 1, = 0 The straight line just touches the parabola. This can be shown graphicall. Eercise 4K 1 Solve each of the following pairs of equations: a = + 8 = d = + 6 + 6 = + 3 b = 3 = 4 7 e = 6 = 1 0 1 c = + 5 = f = + + 6 = 6 + 8 = 3 + = 1 Prove that, for the pairs of equations given, the straight line meets the parabola onl once: a = 6 + 8 = + 4 c = + 11 + 10 = 3 + b = + 6 = 4 3 d = + 7 + 4 = 1 3 Solve each of the following pairs of equations: a d = 6 = 8 + = 3 + 3 = 3 3 b = 3 + 9 = 3 e = 1 = 3( 4) c = 5 + 9 = 1 f = 11 = 1 6 4 a Find the value of c such that = + c is a tangent to the parabola = 1. (Hint: Consider the discriminant of the resulting quadratic.) b i Sketch the parabola with equation = 6 +. ii Find the values of m for which the straight line = m + 6istangent to the parabola. (Hint: Use the discriminant of the resulting quadratic.)

130 Essential Mathematical Methods 1&CAS 5 a Find the value of c such that the line with equation = + c is tangent to the parabola with equation = + 3. b Find the possible values of c such that the line with equation = + c twice intersects the parabola with equation = + 3. 6 Find the value(s) of a such that the line with equation = is tangent to the parabola with equation = + a + 1. 7 Find the value of b such that the line with equation = is tangent to the parabola with equation = + + b. 8 Find the equation of the straight line(s) which pass through the point (1, ) and is (are) tangent to the parabola with equation =. 4.1 Determining quadratic rules It is possible to find the quadratic rule to fit given points, if it is assumed that the points lie on a suitable parabola. Eample 37 Determine the quadratic rule for each of the following graphs, assuming each is a parabola. a b 0 ( 3, 1) 0 (0, 3) (, 5) a This is of the form = a When =, = 5, thus 5 = 4a Therefore a = 5 4 and the rule is = 5 4 b This is of the form = a + c For (0, 3) 3 = a(0) + c Therefore c = 3 For ( 3, 1) 1 = a( 3) + 3 1 = 9a + 3 Therefore a = 9 and the rule is = 9 + 3

Chapter 4 Quadratics 131 d c ( 1, 8) 0 3 e (0, 8) 0 (1, 6) 1 0 1 c This is of the form = a( 3) For ( 1, 8) 8 = a( 1 3) 8 = 4a Therefore a = and the rule is = ( 3) = 6 d This is of the form = k( 1) + 6 When = 0, = 8 8 = k + 6 and k = = ( 1) + 6 and the rule is = ( + 1) + 6 = 4 + 8 e This is of the form = a + b + c For ( 1, 0) 0 = a b + c (1) For (0, ) = c () For (1, ) = a + b + c (3) Substitute c = in(1)and (3) Substract (3a) from (1a) 0 = a b + = a b (1a) 0 = a + b (3a) = b Therefore b = 1 Substitute b = 1 and c = in(1) Therefore 0 = a 1 + 0 = a + 1 and hence a = 1 Thus the quadratic rule is = + +

13 Essential Mathematical Methods 1&CAS Eample Eample 37a 37b Using the TI-Nspire The equation = a + b + and the two points ( 1, 0) and (1, ) are used to generate equations in a and b. These equations are then solved simultaneousl to find a and b. Using the Casio ClassPad The equation = a + b + c is used to generate equations in a and b. These equations are then solved simultaneousl to find a and b. Note: To generate the equation in a and b when = 1, the smbol is found b clicking mth and OPTN on the keboard. Remember to use VAR to enter the variables a and b. Eercise 4L 1 A quadratic rule for a particular parabola is of the form = a. The parabola passes through the point with coordinates (, 8). Find the value of a. A quadratic rule for a particular parabola is of the form = a + c. The parabola passes through the points with coordinates ( 1, 4) and (0, 8). Find the value of a and of c.

Chapter 4 Quadratics 133 Eample Eample 37c 37d 3 A quadratic rule for a particular parabola is of the form = a + b. The parabola passes through the points with coordinates ( 1, 4) and one of its -ais intercepts is 6. Find the value of a and of b. 4 A quadratic rule for a particular parabola is of the form = a( b) + c. The parabola has verte (1, 6) and passes through the point with coordinates (, 4). Find the values of a, b and c. 5 Determine the equation of each of the following parabolas: a b c d 5 0 4 e ( 3, 9) 0 ( 1, 5) f 7 0 6 3 ( 3, 0) (1, 0) (, ) 0 1 3 0 0 4 6 A parabola has verte with coordinates ( 1, 3) and passes through the point with coordinates (3, 8). Find the equation for the parabola. (4, 4) 7 A parabola has -ais intercepts 6 and 3 and passes through the point (1, 10). Find the equation of the parabola. 8 A parabola has verte with coordinates ( 1, 3) and -ais intercept 4. Find the equation for the parabola. 9 Assuming that the suspension cable shown in the diagram forms a parabola, find the rule which describes its shape. The minimum height of the cable above the roadwa is 30 m. 75 m 180 m

134 Essential Mathematical Methods 1&CAS 10 Which of the curves could be most nearl defined b each of the following? a = 1 3 ( + 4) (8 ) b = + c = 10 + ( 1) d = 1 (9 ) A 5 0 4 B 10 5 0 4 C 5 10 0 5 5 10 D 0 11 A parabola has the same shape as = but its turning point is (1, ). Write its equation. 1 A parabola has its verte at (1, ) and passes through the point (3, ). Write its equation. 13 A parabola has its verte at (, ) and passes through (4, 6). Write its equation. 14 Write down four quadratic rules that have graphs similar to those in the diagram. 15 Find quadratic epressions which could represent the two curves in this diagram, given that the coefficient of is 1 in each case. A is (, 3), B is (, 1), C is (0, 5) and D is (0, ) (b) 8 6 4 (a) 8 4 0 4 6 8 10 1 1 4 6 0 3 D 5 C A (6, 6) (b) B (c) (d) (a) 1 3 4 5 16 The rate of rainfall during a storm t hours after it began was 3 mm per hour when t = 5, 6mmper hour when t = 9 and 5 mm per hour when t = 13. Assuming that a quadratic model applies, find an epression for the rate, r mm per hour, in terms of t.

Chapter 4 Quadratics 135 17 a Which of the graphs shown below could represent the graph of the equation = ( 4) 3? b Which graph could represent = 3 ( 4)? A B C D 13 0 4 6 4.13 Quadratic models Eample 38 13 0 4 6 0 13 4 6 0 13 4 6 Jenn wishes to fence off a rectangular vegetable garden in her backard. She has 0 m of fencing wire which she will use to fence three sides of the garden, with the eisting timber fence forming the fourth side. Calculate the maimum area she can enclose. Let A = area of the rectangular garden = length of the garden width = 0 = 10 ( A = 10 ) = 10 = 1 ( 0 + 100 100) (completing the square) A 50 40 30 0 10 timber fence 10 = 1 ( 0 + 100) + 50 = 1 ( 10) + 50 the maimum area is 50 m when = 10. The graph of the relation is shown. 0 5 10 15 0

136 Essential Mathematical Methods 1&CAS Eample 39 A cricket ball is thrown b a fielder. It leaves his hand at a height of metres above the (5, 15) ground and the wicketkeeper takes the ball 60 metres awa again at a height of metres. It is known that after the ball has gone 5 metres it is 15 metres above the ground. 0 60 The path of the cricket ball is a parabola with equation = a + b + c. a Find the values of a, b, and c. b Find the maimum height of the ball above the ground. c Find the height of the ball 5 metres horizontall before it hits the wicketkeeper s gloves. a The data can be used to obtain three equations: = c (1) 15 = (5) a + 5b + c () = (60) a + 60b + c (3) Substitute equation (1) in equations () and (3). 13 = 65a + 5b (1 ) 0 = 3600a + 60b ( ) Simplif ( )bdividing both sides b 60. 0 = 60a + b ( ) Multipl this b 5 and subtract from equation (1 ). 13 = 875a a = 13 and b = 156 875 175 = 13 875 + 156 175 + b The maimum height occurs when = 30 and = 538 35. maimum height is 538 35 m. c When = 55, = 13 35 height of ball is 13 35 m.

Chapter 4 Quadratics 137 Eample 38 Eercise 4M 1 Afarmer has 60 m of fencing with which to construct three sides of a rectangular ard connected to an eisting fence. a If the width of paddock is m and the area inside the ard A m, write down the rule connecting A and. b Sketch the graph of A against. c Determine the maimum area that can be formed for the ard. The efficienc rating, E,ofaparticular spark plug when the gap is set at mm is said to be 400( ). a Sketch the graph of E against for 0 1. b What values of give a zero efficienc rating? c What value of gives the maimum efficienc rating? d Use the graph, or otherwise, to determine the values of between which the efficienc rating is 70 or more. eisting fence 3 A piece of wire 68 cm in length is bent into the shape of a rectangle. a b c Gap mm If cm is the length of the rectangle and A cm is the area enclosed b the rectangular shape, write down a formula which connects A and. Sketch the graph of A against for suitable -values. Use our graph to determine the maimum area formed. 4 A construction firm has won a contract to build cable-car plons at various positions on the side of a mountain. Because of difficulties associated with construction in alpine areas, the construction firm will be paid an etra amount, C($), given b the formula C = 40h + 100h,where h is the height in km above sea level. a Sketch the graph of C as a function of h. Comment on the possible values of h. b Does C have a maimum value? c What is the value of C for a plon built at an altitude of 500 m? 5 A tug-o-war team produces a tension in a rope described b the rule T = 90(8t 0.5t 1.4) units when t is the number of seconds after commencing the pull. a b Sketch a graph of T against t, stating the practical domain. What is the greatest tension produced during a heave?

138 Essential Mathematical Methods 1&CAS Eample 39 6 A cricketer struck a cricket ball such that its height, d metres, after it had travelled metres horizontall was given b the rule d = 1 + 3 5 1 50, 0. a Use a CAS calculator to graph d against for values of ranging from 0 to 30. b i What was the maimum height reached b the ball? ii If a fielder caught the ball when it was mabove the ground, how far was the ball from where it was hit? iii At what height was the ball when it was struck? 7 Find the equation of the quadratic which passes through the points with coordinates: a (, 1), (1, ), (3, 16) b ( 1, ), (1, 4), (3, 10) c ( 3, 5), (3, 0), (5, 57) 8 An arch on the top of a door is parabolic in shape. The point A is 3.1 m above the bottom of the door. The equation = a + b + c can be used to describe the arch. Find the values of a, b, and c. 0.5 m 1.5 m 9 It is known that the dail spending of a government department follows a quadratic model. Let t be the number of das after 1 Januar and s be the spending in hundreds of thousands of dollars on a particular da, where s = at + bt + c. t 30 150 300 s 7. 1.5 6 a Find the values of a, b and c. b Sketch the graph for 0 t 360. (Use a CAS calculator.) c Find an estimate for the spending when: i t = 180 ii t = 350 A

Chapter 4 Quadratics 139 Chapter summar The general epression for a quadratic function is = a + b + c. Epansions e.g. i ( 3) = 6 ii ( 3)(3 + 4) = (3 + 4) 3(3 + 4) = 6 + 8 9 1 = 6 1 iii ( + a) = + a + a iv ( a)( + a) = a Factorising Tpe 1 Removing the highest common factor e.g. 9 3 + 7 = 9 ( + 3) Tpe Difference of two squares: a = ( a)( + a) Tpe 3 Tpe 4 e.g. 16 36 = (4 6)(4 + 6) Grouping of terms e.g. 3 + 4 3 1 = ( 3 + 4 ) (3 + 1) = ( + 4) 3( + 4) = ( 3)( + 4) Factorising quadratic epressions e.g. i + 8 = ( + 4)( ) ii 6 13 15 6 15 13 6 +5 +5 3 18 13 6 13 15 = (6 + 5)( 3) Review

140 Essential Mathematical Methods 1&CAS Review The graphs of a quadratic ma be sketched b first epressing the rule in the form = a( h) + k. The graph of a quadratic of this form is obtained b translating the graph of = a h units in the positive direction of the -ais and k units in the positive direction of the -ais (h, k positive). e.g. for = ( 1) + 3 = (0, 5) (0, 0) 0 A quadratic equation ma be solved b: i Factorising e.g. + 5 1 = 0 ii iii ( 3)( + 4) = 0 = 3 or 4 Completing the square e.g. + 4 = 0 ( ) b Add and subtract to complete the square + + 1 1 4 = 0 ( + 1) 5 = 0 ( + 1) = 5 + 1 =± 5 = 1 ± 5 (1, 3) Using the general quadratic formula, = b ± b 4ac a e.g. 3 1 7 = 0 = ( 1) ± ( 1) 4( 3)( 7) ( 3) = 6 ± 15 3 = ( 1) + 3

Chapter 4 Quadratics 141 The following can be used for sketching the graph of the quadratic relation = a + b + c: i If a > 0, the function has a minimum value. ii If a < 0, the function has a maimum value. iii The value of c gives the -ais intercept. iv The equation of the ais of smmetr is = b a. v The -ais intercepts are determined b solving the equation a + b + c = 0. From the general solution for the quadratic equation a + b + c = 0, the number of -ais intercepts can be determined: i If b 4ac > 0, the equation has two distinct real roots a and b. ii If b 4ac = 0, there is one root, b a. iii If b 4ac < 0, the equation has no real roots. To find a quadratic rule to fit given points: i iii 0 = a ii 0 = a + c a can be calculated if Two points are needed to one point is known. determine a and c. iv 0 = a + b 0 = a + b + c Two points required to Three points required to determine a and b. determine a, b and c. Multiple-choice questions 1 The linear factors of 1 + 7 1 are A 4 3 and 3 + 4 B 3 4 and 4 + 3 C 3 and 4 + 6 D 3 + and 4 6 E 6 + 4 and 3 The solutions of the equation 5 14 = 0 are A = 7onl B = 7, = C =, = 7 D =, = 7 E = onl Review

14 Essential Mathematical Methods 1&CAS Review 3 For = 8 +, the maimum value of is A 3 1 B 5 1 C 9 D 9 1 E 10 4 4 4 If the graph of = k + 3 touches the -ais then the possible values of k are A k = ork = 3 B k = 1 C D k = 1ork = 3 E k = 6ork = 6 k = 3ork = 1 5 The solutions of the equation 56 = are A = 8or7 B = 7or8 C = 7or8 D = 9or6 E = 9or 6 6 The value of the discriminant of + 3 10 is A 5 B 5 C 49 D 7 E 7 The coordinates of the turning point of the graph with equation = 3 + 6 1 are ( ) 1 A 3, B ( 13 ), C ( 13 ), 4 D (1, 4) E ( 1, 4) 8 The equation 5 10 inturning point form a( h) + k, bcompleting the square, is A (5 + 1) + 5 B (5 1) 5 C 5( 1) 5 D 5( + 1) E 5( 1) 7 9 The value(s) of m that will give the equation m + 6 3 = 0two real roots is (are) A m = 3 B m = 3 C m = 0 D m > 3 E m < 3 10 6 8 8 is equal to A (3 + )( 4) B (3 )(6 + 4) C (6 4)( + ) D (3 )( + 4) E (6 + )( 8) Short-answer questions (technolog-free) 1 Epress each of the following in the form (a + b) : a + 9 + 81 b 4 + 18 + 81 c 4 5 + 4 5 d + b + b e 9 6 + 1 f 5 + 0 + 4 Epand each of the following products: a 3( ) b a( a) c (7a b)(7a + b) d ( + 3)( 4) e ( + 3)( 4) f ( + )( ) g (a b)(a + ab + b ) h ( + )(3 + ) i (3a + 1)(a ) j ( + ) ( ) k u(v + ) + v(1 u) l (3 + )( 4) + (4 )(6 1)

Chapter 4 Quadratics 143 3 Epress each of the following as a product of factors: a 4 8 b 3 + 8 c 4a 3 d 4 e au + av + 3aw f 4a b 9a 4 g 1 36 a h + 1 i + j + 3 k 6 + 7 + l 3 8 3 m 3 + n 6a a o 6 7 + 4 Sketch the graphs of each of the following: a = + 3 b = + 3 c = ( ) + 3 d = ( + ) + 3 e = ( 4) 3 f = 9 4 g = 3( ) h = ( ) + 3 5 Epress in the form = a( h) + k and hence sketch the graphs of the following: a = 4 5 b = 6 c = 8 + 4 d = + 8 4 e = 3 1 + 9 f = + 4 + 5 6 Find: i the - and -intercepts ii the ais of the smmetr iii the turning point and hence sketch the graphs of each of the following: a = 7 + 6 b = + 1 c = + 5 + 14 d = 10 + 16 e = + 15 f = 6 13 5 g = 9 16 h = 4 5 7 Use the quadratic formula to solve each of the following: a + 6 + 3 = 0 b + 9 + 1 = 0 c 4 + = 0 d + 7 + = 0 e + 7 + 4 = 0 f 3 + 9 1 = 0 8 Find the equation of the quadratic, the graph of which is shown. 0 5 9 A parabola has the same shape as = 3 but its verte is at (5, ). Find the equation corresponding to this parabola. 10 Find the rule of the quadratic relation which describes the graph. (6, 10) 0 (, 10) 11 Find the coordinates of the point of intersection of the graphs with equations: a = + 3 and = b = 8 + 11 and = c = 3 + 7 and = d = and = 3 (1, 5) Review