Chapter 3 3Quadratics Objectives To recognise and sketch the graphs of quadratic polnomials. To find the ke features of the graph of a quadratic polnomial: ais intercepts, turning point and ais of smmetr. To determine the maimum or minimum value of a quadratic function. 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 solutions of a quadratic equation. To appl quadratic functions to solving problems. A polnomial function has a rule of the tpe = a n n + a n 1 n 1 + + a 1 + a 0 where n is a natural number or zero, and 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. For eample: = 2 + 3 is a polnomial function of degree 1 = 2 2 + 3 2 is a polnomial function of degree 2. This chapter deals with polnomials of degree 2. These are called quadratic polnomials. The graph of a linear polnomial function, = m + c, is a straight line and the graph of a quadratic polnomial function, = a 2 + b + c, a 0, is a parabola. Polnomials of higher degree will be studied in Chapter 8. Knowledge check See the online test of required knowledge, with links to revision lessons.
3A Epanding and collecting like terms 3A Epanding and collecting like terms 81 In order to sketch graphs of quadratics, we need to find the -ais intercepts (if the eist), and to do this we need to solve quadratic equations. As an introduction to the methods of solving quadratic equations, the first two sections of this chapter review the basic algebraic processes of epansion and factorisation. An algebraic epression is the sum of its terms. For eample: The terms of the linear epression 3 1 are 3 and 1. The terms of the quadratic epression 2 2 + 3 4 are 2 2, 3 and 4. Eample 1 Simplif 2( 5) 3( + 5) b first epanding. Eplanation 2( 5) 3( + 5) = 2 10 3 15 Epand each bracket. Eample 2 = 2 3 10 15 Collect like terms. = 25 Epand 2(3 2) + 3( 2). 2(3 2) + 3( 2) = 6 2 4 + 3 2 6 = 9 2 10 For epansions of the tpe (a + b)(c + d), proceed as follows: (a + b)(c + d) = a(c + d) + b(c + d) Eample 3 Epand the following: = ac + ad + bc + bd a ( + 3)(2 3) b ( 3)(2 2 2) a ( + 3)(2 3) = (2 3) + 3(2 3) = 2 2 3 + 6 9 = 2 2 + 3 9 Eplanation Each term in the second pair of brackets is multiplied b each term in the first.
82 Chapter 3: Quadratics b ( 3)(2 2 2) = (2 2 2) 3(2 2 2) = 2 2 2 2 6 + 6 2 = 2 2 (2 2 + 6) + 6 2 Eample 4 Epand (2 1)(3 2 + 2 + 4). (2 1)(3 2 + 2 + 4) = 2(3 2 + 2 + 4) 1(3 2 + 2 + 4) Perfect squares = 6 3 + 4 2 + 8 3 2 2 4 = 6 3 + 2 + 6 4 Consider the epansion of a perfect square, (a + b) 2 : (a + b) 2 = (a + b)(a + b) = a(a + b) + b(a + b) = a 2 + ab + ab + b 2 = a 2 + 2ab + b 2 Thus the general result can be stated as: (a + b) 2 = a 2 + 2ab + b 2 Be careful with negative signs. You can also complete binomial epansions with a table; this emphasises the terms. 3 2 2 2 6 2 2 2 2 6 2 You add the terms to complete the epansion. That is, to epand (a + b) 2 take the sum of the squares of the terms and add twice the product of the terms. Eample 5 Epand (3 2) 2. Eplanation (3 2) 2 = (3) 2 + 2(3)( 2) + ( 2) 2 Use the epansion (a + b) 2 = a 2 + 2ab + b 2. = 9 2 12 + 4 Here a = 3 and b = 2.
Difference of two squares Consider the epansion of (a + b)(a b): (a + b)(a b) = a(a b) + b(a b) = a 2 ab + ab b 2 = a 2 b 2 Thus the epansion of the difference of two squares has been obtained: (a + b)(a b) = a 2 b 2 Eample 6 3A Epanding and collecting like terms 83 a Epand (2 4)(2 + 4). b Epand ( 2 7)( + 2 7). a (2 4)(2 + 4) = (2) 2 (4) 2 b ( 2 7)( + 2 7) = 2 (2 7) 2 Eample 7 = 4 2 16 Epand (2a b + c)(2a b c). (2a b + c)(2a b c) = ( (2a b) + c )( (2a b) c ) Section summar = (2a b) 2 c 2 = 4a 2 4ab + b 2 c 2 A polnomial function has a rule of the tpe = a n n + a n 1 n 1 + + a 1 + a 0 = 2 28 where n is a natural number or zero, and 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. A polnomial function of degree 2 is called a quadratic function. The general rule is of the form = a 2 + b + c, where a 0. General binomial epansion: (a + b)(c + d) = a(c + d) + b(c + d) = ac + ad + bc + bd Perfect square epansion: Difference of two squares epansion: (a + b) 2 = a 2 + 2ab + b 2 (a + b)(a b) = a 2 b 2
84 Chapter 3: Quadratics 3A Eample 1 Eample 2 Eample 3 Eample 4 Eample 5 Eample 6 Eample 7 Eercise 3A 1 Epand each of the following: a 2( 4) b 2( 4) c 3(2 4) d 3(4 2) e ( 1) f 2( 5) 2 Collect like terms in each of the following: 3 4 5 6 7 8 9 a c 2 + 4 + 1 3 + 1 2 b d 2 6 + + 2 3 + 4 Simplif each of the following b epanding and collecting like terms: a 8(2 3) 2( + 4) b 2( 4) 3 c 4(2 3) + 4(6 ) d 4 3(5 2) Simplif each of the following b epanding and collecting like terms: a 2( 4) 3 b 2( 5) + ( 5) c 2( 10 3) d 3(2 3 + 2 2 ) e 3 2(2 ) f 3(4 2) 6 Simplif each of the following b epanding and collecting like terms: a (3 7)(2 + 4) b ( 10)( 12) c (3 1)(12 + 4) d (4 5)(2 3) e ( 3)( 2) f (2 5)( + 5) g (3 2 7)( + 7) h (5 3)( + 2 2) i ( 5 3)( 5 32 2) Simplif each of the following b epanding and collecting like terms: a (2 3)(3 2 + 2 4) b ( 1)( 2 + + 1) c (6 2 3 2 )(4 2) d (5 3)( + 2) (2 3)( + 3) e (2 + 3)(3 2) (4 + 2)(4 2) Simplif each of the following b epanding and collecting like terms: a ( 4) 2 b (2 3) 2 c (6 2) 2 ( 1 ) 2 d e ( 5) 2 f ( 2 3) 2 2 Simplif each of the following b epanding and collecting like terms: a ( 3)( + 3) b (2 4)(2 + 4) c (9 11)(9 + 11) d (2 3)(2 + 3) e (2 + 5)(2 5) f ( 5)( + 5) g (2 + 3 3)(2 3 3) h ( 3 7)( 3 + 7) Simplif each of the following b epanding and collecting like terms: a ( + z)( z) b (2a b + c)(2a b c) c (3w 4z + u)(3w + 4z u) d (2a 5b + c)(2a + 5b + c) SF
3A 3B Factorising 85 10 Find the area of each of the following b: i adding the areas of the four non-overlapping rectangles (two of which are squares) ii multipling length b width of the undivided square (boundar in blue). a 1 3 cm 3B Factorising 2 cm 4 1 cm 1 cm Four different tpes of factorisation will be considered. Factorisation using common factors b 1 cm 1 cm 1 2 2 3 4 cm cm If each term in an algebraic epression to be factorised contains a common factor, then this common factor is a factor of the entire epression. To find the other factor, divide each term b the common factor. The common factor is placed outside the brackets. This process is known as taking the common factor outside the brackets. Eample 8 a Factorise 9 2 + 81. b Factorise 2a 2 8a 2. a 9 2 + 81 = 9 + 9 9 = 9( + 9) b 2a 2 8a 2 = 2a a 2a 4 2 = 2a(a 4 2 ) Eplanation The common factor 9 is taken out of the brackets. The common factor 2a is taken out of the brackets. Note: The answers can be checked b epanding. In general, take out as man common factors as possible. Eample 9 Factorise 7 2 35 2. 7 2 35 2 = 7( 5) Eplanation The common factor 7 is taken out of the brackets. CF
86 Chapter 3: Quadratics Grouping of terms This method can be used for epressions containing four terms. Eample 10 Factorise 3 + 4 2 3 12. 3 + 4 2 3 12 = ( 3 + 4 2 ) (3 + 12) = 2 ( + 4) 3( + 4) = ( 2 3)( + 4) Difference of two squares Eplanation You will recall the following identit from the previous section: (a + b)(a b) = a 2 b 2 We can now use the result the other wa in order to factorise: a 2 b 2 = (a + b)(a b) Eample 11 a Factorise 3 2 75. b Factorise 9 2 36. The terms in this epression can be grouped as shown. The common factor ( + 4) is taken out of the brackets. Eplanation a 3 2 75 = 3( 2 25) First take out the common factor 3. = 3( + 5)( 5) Use the difference of squares identit. b 9 2 36 = 9( 2 4) First take out the common factor 9. Eample 12 Factorise ( ) 2 16 2. = 9( 2)( + 2) Use the difference of squares identit. ( ) 2 16 2 = ( ) 2 (4) 2 = ( + 4)( 4) = ( + 3)( 5) Eplanation Use the difference of squares identit a 2 b 2 = (a + b)(a b) with a = ( ) and b = 4.
Factorising quadratic polnomials 3B Factorising 87 A quadratic polnomial is an epression of the form a 2 + b + c with a 0. We have seen in the previous section that we can epand a product of two binomial factors to obtain a quadratic epression. For eample: ( + 2)( 4) = ( 4) + 2( 4) = 2 4 + 2 8 = 2 2 8 We want to be able to reverse this process. That is, we want to start from the epanded epression and obtain the factorised form. We have alread done this for epressions that are differences of squares. We now turn our attention to the general case. Eample 13 Factorise 2 2 8. Using the method described in the eplanation opposite, we can factorise without an further setting out: 2 2 8 = ( 4)( + 2) Alternativel, we can reverse the process we used for epanding: 2 2 8 = 2 4 + 2 8 = ( 4) + 2( 4) = ( 4)( + 2) Eplanation We want 2 2 8 = ( + a)( + b) = 2 + (a + b) + ab The values of a and b are such that ab = 8 and a + b = 2. Values of a and b which satisf these two conditions are a = 4 and b = 2. A quadratic polnomial is called a monic quadratic polnomial if the coefficient of 2 is 1. The quadratic polnomial 2 2 8 factorised in the previous eample is monic. Factorising non-monic quadratic polnomials involves a slightl different approach. We need to consider all possible combinations of factors of the 2 term and the constant term. The net eample and the following discussion give two methods. Eample 14 Factorise 6 2 13 15. There are several combinations of factors of 6 2 and 15 to consider. Onl one combination is correct. Factors of 62 Factors of 15 Cross-products add to give 13 6 +5 +5 6 2 13 15 = (6 + 5)( 3) 3 18 13
88 Chapter 3: Quadratics Here is a second method for factorising 6 2 13 15 which still requires some trial and error but is more sstematic. It is the reverse process of epanding ( 3)(6 + 5). We let a 2 + b + c = (α + γ)(β + δ) Epanding the right-hand side gives a 2 + b + c = αβ 2 + (γβ + αδ) + γδ Note that ac = αβγδ and b = γβ + αδ. We now appl this to factorising 6 2 13 15. First we look for two numbers that multipl together to give ac and add to give b. That is, we look for two numbers whose product is 6 ( 15) = 90 and whose sum is 13. The two numbers are 18 and 5. We write: 6 2 13 15 = 6 2 18 + 5 15 Eample 15 Factorise 8 2 + 2 15. = 6( 3) + 5( 3) = ( 3)(6 + 5) 8 2 + 2 15 = 8 2 + 12 10 15 = 4(2 + 3) 5(2 + 3) = (4 5)(2 + 3) Eplanation Here ac = 8 ( 15) = 120 and b = 2. We look for two numbers whose product is 120 and whose sum is 2. The two numbers are 12 and 10. So we write 2 = 12 10. It is sometimes possible to take out a common factor first to simplif the factorisation. Eample 16 Factorise 2 2 + 6 20. 2 2 + 6 20 = 2( 2 + 3 10) = 2( + 5)( 2) Eplanation The common factor 2 is taken out first.
3B 3B Factorising 89 Eample 17 Eample 8 Eample 9 Eample 10 Factorise ( + 1) 2 2( + 1) 3. ( + 1) 2 2( + 1) 3 = a 2 2a 3 Section summar = (a 3)(a + 1) = ( + 1 3)( + 1 + 1) = ( 2)( + 2) Eplanation Difference of two squares identit: a 2 b 2 = (a + b)(a b). The substitution a = + 1 makes it easier to recognise the required factorisation. Factorisation of monic quadratics: To factorise a quadratic of the form 2 + b + c, find two numbers whose sum is the coefficient of and whose product is the constant term. Factorisation of general quadratics: To factorise a quadratic of the form a 2 + b + c, find two numbers e and f whose product is ac and whose sum is b. Split the middle term b as e + f and then factorise b grouping. Eercise 3B 1 Factorise each of the following: 2 3 4 a 2 + 4 b 4a 8 d 2 10 e 18 + 12 Factorise: a d 4 2 2 6 + 14 2 b e 8a + 32 2 + 2 c f c f 6 3 24 16 6ab 12b 5 2 15 g 4 2 16 h 7 + 49 2 i 2 2 Factorise: Eample 11 5 a 6 3 2 + 12 2 2 Factorise: b 7 2 6 2 c 8 2 2 + 6 2 a 3 + 5 2 + + 5 b + 2 + 3 + 6 c 2 2 2 2 + 1 d a + a + b + b e a 3 3a 2 + a 3 f 2ab 12a 5b + 30 g 2 2 2 + 5 5 h 3 4 + 2 2 8 Factorise: i 3 b 2 a 2 + a 2 b a 2 36 b 2 81 c 2 a 2 d 4 2 81 e 9 2 16 f 25 2 2 g 3 2 48 h 2 2 98 i 3a 2 27a j a 2 7 k 2a 2 5 l 2 12 SF
90 Chapter 3: Quadratics 3B Eample 12 Eample 13 Eample 14, 15 Eample 16 Eample 17 6 7 8 9 10 Factorise: a ( 2) 2 16 b 25 (2 + ) 2 c 3( + 1) 2 12 d ( 2) 2 ( + 3) 2 e (2 3) 2 (2 + 3) 2 f (2 1) 2 (3 + 6) 2 Factorise: a 2 7 18 b 2 19 + 48 c a 2 14a + 24 d a 2 + 18a + 81 e 2 5 24 f 2 2 120 Factorise: a 3 2 7 + 2 b 6 2 + 7 + 2 c 5 2 + 23 + 12 d 2 2 + 9 + 4 e 6 2 19 + 10 f 6 2 7 3 g 12 2 17 + 6 h 5 2 4 12 Factorise: i 5 3 16 2 + 12 a 3 2 12 36 b 2 2 18 + 28 c 4 2 36 + 72 d 3 2 + 15 + 18 Factorise: a ( 1) 2 + 4( 1) + 3 b 2( 1) 2 + 5( 1) 3 c (2 + 1) 2 + 7(2 + 1) + 12 3C Quadratic equations e a 2 + 7a + 12a f 48 24 2 + 3 3 This section looks at the solution of quadratic equations b simple factorisation. There are three steps to solving a quadratic equation b factorisation: Step 1 Write the equation in the form a 2 + b + c = 0. Step 2 Step 3 Factorise the quadratic epression. Use the result that mn = 0 implies m = 0 or n = 0 (or both); this is known as the null factor theorem. For eample, to solve the equation 2 = 12: 2 = 12 2 12 = 0 (Step 1) ( 4)( + 3) = 0 (Step 2) 4 = 0 or + 3 = 0 (Step 3) = 4 or = 3 In the simplest cases, the first two steps ma have been done alread. SF CF
3C Quadratic equations 91 Eample 18 Solve 2 + 11 + 24 = 0. Eplanation 2 +24 +11 +3 +3 +8 +8 Factorising gives +11 2 + 11 + 24 = 0 ( + 3)( + 8) = 0 + 3 = 0 or + 8 = 0 = 3 or = 8 The quadratic can also be factorised in the following wa: 2 + 11 + 24 = 2 + 8 + 3 + 24 = ( + 8) + 3( + 8) = ( + 8)( + 3) Note: We can check the answer for this eample b substituting into the equation: Eample 19 Solve 2 2 + 5 12 = 0. 2 2 12 +5 2 3 3 +4 +8 Factorising gives ( 3) 2 + 11( 3) + 24 = 0 ( 8) 2 + 11( 8) + 24 = 0 +5 2 2 + 5 12 = 0 (2 3)( + 4) = 0 2 3 = 0 or + 4 = 0 = 3 2 or = 4 Eplanation The quadratic can also be factorised in the following wa: 2 2 + 5 12 = 2 2 + 8 3 12 = 2( + 4) 3( + 4) = (2 3)( + 4)
92 Chapter 3: Quadratics Using the TI-Nspire CX non-cas To solve the quadratic equation 2 2 + 5 12 = 0, use a Calculator application. Select menu > Algebra > Polnomial Tools > Find Roots of Polnomial. Select OK in the initial pop-up screen, and then complete as shown opposite. Hence = 4 or = 3 2. Note: You can also use menu > Algebra > Numerical Solve. But ou need to be careful, as quadratic equations can have two solutions. You ma need to tr several guess values using the snta: nsolve(equation, = guess value). Using the Casio To solve the quadratic equation 2 2 + 5 12 = 0: Press MENU 1 to select Run-Matri mode. Select the numerical solver b going to Calculation OPTN F4, then SolveN F5. Enter the equation: 2 X,θ,T 2 + 5 X,θ,T 1 2 SHIFT. 0 ) EXE Press EXIT. Hence = 4 or = 3 2.
3C 3C Quadratic equations 93 Applications of quadratic equations Problems involving the solution of quadratic equations arise in man situations. We will meet more such problems in Section 3L. Eample 20 The perimeter of a rectangle is 20 cm and its area is 24 cm 2. Calculate the length and width of the rectangle. Let cm be the length of the rectangle and cm the width. Then 2( + ) = 20 and thus = 10. The area is 24 cm 2 and therefore = (10 ) = 24. i.e. 10 2 = 24 2 10 + 24 = 0 ( 6)( 4) = 0 Thus the length is 6 cm or 4 cm. The width is 4 cm or 6 cm. Section summar To solve a quadratic equation b factorisation: Step 1 Write the equation in the form a 2 + b + c = 0. Step 2 Step 3 Eercise 3C Factorise the quadratic polnomial. Use the result that mn = 0 implies m = 0 or n = 0 (or both). Skillsheet 1 Solve each of the following for : Eample 18 a ( 2)( 3) = 0 b (2 4) = 0 c ( 4)(2 6) = 0 d (3 )( 4) = 0 e (2 6)( + 4) = 0 f 2( 1) = 0 g (5 2)(6 ) = 0 h 2 = 16 2 Use a calculator to solve each of the following equations. Give our answer correct to two decimal places. 3 a 2 4 3 = 0 b 2 2 4 3 = 0 c 2 2 4 + 3 = 0 Solve for in each of the following: a 2 72 = 0 b 2 6 + 8 = 0 c 2 8 33 = 0 d ( + 12) = 64 e 2 + 5 14 = 0 f 2 = 5 + 24 SF
94 Chapter 3: Quadratics 3C Eample 19 Eample 20 4 Solve for in each of the following: a 2 2 + 5 + 3 = 0 b 4 2 8 + 3 = 0 c 6 2 + 13 + 6 = 0 d 2 2 = 6 g 10 2 11 + 3 = 0 h 12 2 + = 6 j ( + 4) = 5 k 1 7 2 = 3 7 m 5 2 = 11 2 5 Calculate the value of. cm e 6 2 + 15 = 23 f 2 2 3 9 = 0 6 cm 7 cm Area = 30 cm 2 cm i l 4 2 + 1 = 4 2 + 8 = 15 6 The bending moment, M, of a simple beam used in bridge construction is given b the formula M = wl 2 w 2 2 If l = 13 m, w = 16 kg/m and M = 288 kg m, calculate the value of. 7 The height, h metres, reached b a projectile after t seconds travelling verticall upwards is given b the formula h = 70t 16t 2. Calculate t if h is 76 metres. n(n 3) 8 A polgon with n sides has diagonals. How man sides has a polgon with 2 65 diagonals? 9 For a particular electric train, the tractive resistance R at speed v km/h is given b R = 1.6 + 0.03v + 0.003v 2. Find v when the tractive resistance is 10.6. 10 The perimeter of a rectangle is 16 cm and its area is 12 cm 2. Calculate the length and width of the rectangle. 11 The altitude of a triangle is 1 cm shorter than the base. If the area of the triangle is 15 cm 2, calculate the altitude. 12 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. 13 The members of a club hire a bus for $2100. 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? SF CF
3D Graphing quadratics 95 3D Graphing quadratics A quadratic polnomial function is defined b the general rule = a 2 + b + c where a, b and c are constants and a 0. This is called polnomial form. The parabola = 2 The simplest quadratic function is = 2. If a table of values is constructed for = 2 for 3 3, these points can be plotted and then connected to produce a continuous curve. 3 2 1 0 1 2 3 9 4 1 0 1 4 9 Features of the graph of = 2 : The graph is called a parabola. The possible -values are all positive real numbers and 0. (This is called the range of the quadratic and is discussed in a more general contet in Chapter 7.) The graph is smmetrical about the -ais. The line 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). The minimum value of is 0 and it occurs at the turning point. Transformations of = 2 3 2 1 10 Ais of smmetr 8 6 4 2 = 2 0 1 2 3 Verte or turning point B a process called completing the square (to be discussed in Section 3E), all quadratics in polnomial form = a 2 + b + c ma be transposed into what will be called the turning point form: = a( h) 2 + k We first consider the effect of changing the value of a for our basic graph of = 2. We then consider the effect of changing h and k for graphs of the form = a 2. Graphs of the form = a( h) 2 + k are formed b translating the graph of = a 2. The graph of = a( h) 2 + k is eactl the same shape as = a 2. All of these graphs are indeed congruent to = a 2 and each other.
96 Chapter 3: Quadratics Graphs of = a 2 We first consider graphs of the form = a 2. In this case both h = 0 and k = 0. In the basic graph of = 2, the value of a is 1. The following graphs are shown on the same set of aes: = 2 = 2 2 (a = 2) = 1 2 2 (a = 1 2 ) = 2 2 (a = 2) 8 6 4 2 2 1 0 1 2 2 4 6 8 = 22 = 2 = = 22 If a > 1, the graph is narrower. If 0 < a < 1, the graph is broader. The transformation which produces the graph of = 2 2 from the graph of = 2 is called a dilation of factor 2 from the -ais. When a is negative, the graph is reflected in the -ais. The transformation which produces the graph of = 2 from the graph of = 2 is called a reflection in the -ais. Graphs of = 2 + k On this set of aes are the graphs of = 2 = 2 2 (k = 2) = 2 + 1 (k = 1) As can be seen, changing k moves the basic graph of = 2 in a vertical direction. 2 4 3 2 1 1 0 1 2 1 2 1 2 2 = 2 + 1 = 2 = 2 2 When k = 2 the graph is translated 2 units in the negative direction of the -ais. The verte is now (0, 2) and the range is now all real numbers greater than or equal to 2. 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.
3D Graphing quadratics 97 Graphs of = ( h) 2 On this set of aes are the graphs of = 2 = ( 2) 2 (h = 2) = ( + 3) 2 (h = 3) As can be seen, changing h moves the graph in a horizontal direction. When h = 2 the graph is translated 2 units in the positive direction of the -ais. The verte is now (2, 0) and the ais of smmetr is now the line = 2. = ( + 3) 2 = 2 = ( 2) 2 9 4 2 3 2 1 0 1 2 3 When h = 3 the 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. In both cases, the range is unchanged and is still all non-negative real numbers. Eamples of transformations B combining dilations, reflections and translations, we can sketch the graph of an quadratic epressed in the form = a( h) 2 + k: The verte is the point (h, k). The ais of smmetr is = h. If h and k are positive, then the graph of = a( h) 2 + k is obtained from the graph of = a 2 b translating h units in the positive direction of the -ais and k units in the positive direction of the -ais. Similar results hold for different combinations of h and k positive and negative. Eample 21 Sketch the graph of = 2 3. The graph of = 2 3 is obtained from the graph of = 2 b translating 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. To find the -ais intercepts, let = 0: 0 = 2 3 2 = 3 = ± 3 Hence the -ais intercepts are ± 3. 4 2 0 2 3 = 2 = 2 3
98 Chapter 3: Quadratics Eample 22 Sketch the graph of = ( + 1) 2. The graph of = ( + 1) 2 is obtained from the graph of = 2 b a reflection 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. Eample 23 Sketch the graph of = 2( 1) 2 + 3. The graph of = 2 2 is translated 1 unit in the positive direction of the -ais and 3 units in the positive direction of the -ais. The verte has coordinates (1, 3). The ais of smmetr is the line = 1. The graph will be narrower than = 2. The range will be 3. To add further detail to our graph, we can find the ais intercepts: -ais intercept When = 0, = 2(0 1) 2 + 3 = 5. 1 5 3 0 (1, 3) 0 1 2 = 2 = 2 -ais intercepts In this eample, the minimum value of is 3, and so cannot be 0. Therefore this graph has no -ais intercepts. Note: Another wa to see this is to let = 0 and tr to solve for : 0 = 2( 1) 2 + 3 3 = 2( 1) 2 3 2 = ( 1)2 As the square root of a negative number is not a real number, this equation has no real solutions.
3D Graphing quadratics 99 Eample 24 Sketch the graph of = ( + 1) 2 + 4. The verte has coordinates ( 1, 4) and so the ais of smmetr is the line = 1. -ais intercept When = 0, = (0 + 1) 2 + 4 = 3 the -ais intercept is 3. -ais intercepts When = 0, ( + 1) 2 + 4 = 0 ( + 1) 2 = 4 + 1 = ±2 = ±2 1 the -ais intercepts are 1 and 3. Section summar ( 3, 0) ( 1, 4) 0 (0, 3) (1, 0) The graph of = 2 is called a parabola. The verte (or turning point) is the point (0, 0) and the ais of smmetr is the -ais. The graph of = 2 is the reflection of the graph of = 2 in the -ais. For = a 2 and a > 1, the graph is narrower than the graph of = 2. For = a 2 and 0 < a < 1, the graph is broader than the graph of = 2. All quadratic functions in polnomial form = a 2 + b + c ma be transposed into the turning point form = a( h) 2 + k. The graph of = a( h) 2 + k is a parabola congruent to the graph of = a 2. The verte (or turning point) is the point (h, k). The ais of smmetr is = h. If h and k are positive numbers, then the graph of = a( h) 2 + k is obtained from the graph of = a 2 b translating h units in the positive direction of the -ais and k units in the positive direction of the -ais. Similar results hold for different combinations of h and k positive and negative.
100 Chapter 3: Quadratics 3D Eercise 3D Sketch the graph of each of the following quadratics b first finding: i ii iii the coordinates of the turning point the ais of smmetr the -ais intercepts (if an). Eample 21 1 a = 2 4 b = 2 + 2 c = 2 + 3 d = 2 2 + 5 e = 2 + 4 f = 3 2 9 Eample 22 2 a = ( 2) 2 b = ( + 3) 2 c = ( + 1) 2 d = 1 2 ( 4)2 Eample 23, 24 3 a = ( 2) 2 + 1 b = ( 2) 2 1 c = ( 1) 2 + 2 d = ( + 1) 2 1 e = ( 3) 2 + 1 f = ( + 2) 2 4 g = 2( + 2) 2 18 h = 3( 4) 2 + 3 i = 1 2 ( + 5)2 2 j = 3( + 2) 2 12 k = 4( 2) 2 + 8 l = 1 3 ( 1)2 3 3E Completing the square and turning points To sketch the graph of a quadratic using the techniques from the previous section, the quadratic must be epressed in turning point form. This can be done using two different but related methods: b completing the square and b using the equation of the ais of smmetr. Completing the square To transpose a quadratic in polnomial form we can complete the square. Consider the epansion of a perfect square: ( + a) 2 = 2 + 2a + a 2 The last term of the epansion is the square of half the coefficient of the middle term. Now consider the quadratic polnomial 2 + 2 3 This is not a perfect square. However, b adding and subtracting a new term, we can form a perfect square as part of a new epression for the same polnomial. We have that 2 + 2 + 1 = ( + 1) 2 which is a perfect square. In order to keep our original quadratic intact, we both add and subtract the correct new term. For eample: 2 + 2 3 = ( 2 + 2 + 1) 1 3 = ( + 1) 2 4 SF
3E Completing the square and turning points 101 Hence the quadratic = 2 + 2 3 is epressed in turning point form as = ( + 1) 2 4, and so the verte (turning point) of its graph is the point with coordinates ( 1, 4). In the above eample, the coefficient of 2 was 1. If the coefficient is not 1, this coefficient must first be factored out before proceeding to complete the square. A geometric representation of completing the square Completing the square for 2 + 2 is represented in the following diagrams. The diagram on the left shows 2 + 2. The small rectangle to the right is moved to the base of the b square. The red square of area 1 unit is added. Thus 2 + 2 + 1 = ( + 1) 2. + 1 1 1 Solving equations b completing the square + 1 The process of completing the square can also be used for the solution of equations. Eample 25 Solve each of the following equations for b first completing the square: a 2 3 + 1 = 0 b 2 2 3 1 = 0 a Completing the square: 2 3 + 1 = 0 ( 3 ) 2 ( 3 ) 2 2 3 + + 1 = 0 2 2 ( 3 2 2) 5 4 = 0 Therefore and so ( 3 ) 2 = 5 2 4 3 5 2 = ± 2 = 3 5 2 ± 2 = 3 ± 5 2 Eplanation 1 1 2 ( 3) = 3 2 ( We add and subtract 3 2 = 2) 9 on the 4 left-hand side of the equation. This gives an equivalent epression to the epression of the left-hand side. Solve the equation as shown. 1
102 Chapter 3: Quadratics b Completing the square: 2 2 3 1 = 0 ( 2 2 3 2 1 ) = 0 2 2 3 ( 3 2 ( 3 ) 2 2 4) + 1 4 2 = 0 ( 3 ) 2 = 17 4 16 Therefore and so 3 17 4 = ± 4 = 3 17 4 ± = 3 ± 17 4 4 Divide both sides b 2 before completing the square. 1 ( 2 3 ) = 3 2 4 ( We add and subtract 3 ) 2 = 9 4 16 on the left-hand side of the equation. Sketching the graph of a quadratic polnomial after completing the square Completing the square enables the quadratic rule to be written in turning point form. We have seen that this can be used to sketch the graphs of quadratic polnomials. Eample 26 Find the coordinates of the verte b completing the square and hence sketch the graph of = 2 2 + 6 8. Take out 2 as a common factor and then complete the square: = 2 2 + 6 8 = 2( 2 3 + 4) ( ( 3 ) 2 ( 3 ) 2 = 2 2 3 + + 4) 2 2 (( = 2 3 ) 2 + 7 2 4) ( = 2 3 ) 2 7 2 2 ( 3 Therefore the verte is 2 2), 7 and the ais of smmetr is = 3 2. The -ais intercept is 8. The graph has maimum value of 7, and so there are no -ais intercepts. 2 3 4 8 0 1 2 3, 7 2 2
3E Completing the square and turning points 103 The equation for the ais of smmetr of a parabola We first complete the square for = a 2 + b + c: = a 2 + b + c ( = a 2 + b a + c ) a = a ( 2 + b a + b2 4a b2 2 4a + c ) 2 a ( = a + b ) 2 b2 2a 4a + c Ais of smmetr of a parabola completing the square For a quadratic function written in polnomial form = a 2 + b + c, the ais of smmetr of its graph has the equation = b 2a. Therefore the -coordinate of the turning point is b. Substitute this value into the 2a quadratic polnomial to find the -coordinate of the turning point. Eample 27 Use the ais of smmetr to find the turning point of the graph and hence epress in turning point form: a = 2 4 + 3 b = 2 2 + 12 7 a The -coordinate of the turning point is 2. When = 2, = 4 8 + 3 = 1. The coordinates of the turning point are (2, 1). Hence the equation is = ( 2) 2 1. b The -coordinate of the turning point is 3. When = 3, = 2 3 2 + 12 3 7 = 11. The coordinates of the turning point are (3, 11). Hence the equation is = 2( 3) 2 + 11. Eplanation Here a = 1 and b = 4, so the ais of ( 4 ) smmetr is = = 2. 2 For the turning point form = a( h) 2 + k, we have found that a = 1, h = 2 and k = 1. Here a = 2 and b = 12, so the ais of ( 12 ) smmetr is = = 3. 4 For the turning point form = a( h) 2 + k, we have found that a = 2, h = 3 and k = 11.
104 Chapter 3: Quadratics 3E Skillsheet Eample 25 Eample 26 Section summar Quadratic equations can be solved b completing the square. This method allows us to deal with all quadratic equations, even though some have no solutions. To complete the square of 2 + b + c: Take half the coefficient of (that is, b b2 ) and add and subtract its square 2 4. To complete the square of a 2 + b + c: First take out a as a factor and then complete the square inside the bracket. The ais of smmetr of the graph of = a 2 + b + c has equation = b 2a. To convert the quadratic function = a 2 + b + c into turning point form using the ais of smmetr: 1 The -coordinate h of the verte of the parabola is b 2a. 2 Find the -coordinate k of the verte b substituting in = a 2 + b + c. 3 Substitute these values for h and k in = a( h) 2 + k. Eercise 3E 1 Epand each of the following: a ( 1) 2 b ( + 2) 2 c ( 3) 2 d ( + 3) 2 ( e ( 2) 2 f ( 5) 2 1 ) 2 ( g 3 ) 2 h 2 2 2 Factorise each of the following: 3 4 a 2 4 + 4 b 2 12 + 36 c 2 + 4 4 d 2 2 8 + 8 e 2 2 + 12 18 f 2 + 1 g 2 3 + 9 h 2 + 5 + 25 4 4 4 Solve each of the following equations for b first completing the square: a 2 2 1 = 0 b 2 4 2 = 0 c 2 6 + 2 = 0 d 2 5 + 2 = 0 e 2 2 4 + 1 = 0 f 3 2 5 2 = 0 g 2 + 2 + k = 0 h k 2 + 2 + k = 0 i 2 3k + 1 = 0 Eample 27 Epress each of the following in the form = a( h) 2 + k b completing the square. Hence state the coordinates of the turning point and sketch the graph in each case. a = 2 2 + 3 b = 2 + 4 + 1 c = 2 3 + 1 5 Epress each of the following in the form = a( h) 2 + k b completing the square. Hence state the coordinates of the turning point and sketch the graph in each case. 6 a = 2 2 2 5 b = 4 2 + 8 + 8 c = 3 2 6 4 Epress each of the following in the form = a( h) 2 + k using the ais of smmetr. Hence state the coordinates of the turning point and sketch the graph in each case. a = 2 8 + 12 b = 2 2 c = 2 2 + 4 2 d = 2 + 4 + 1 e = 2 2 12 12 f = 3 2 6 + 12 SF
3F Graphing quadratics in polnomial form 3F Graphing quadratics in polnomial form 105 It is not alwas essential to convert a quadratic to turning point form in order to sketch its graph. We can sometimes find the - and -ais intercepts and the ais of smmetr from polnomial form b other methods and use these details to sketch the graph. Step 1 Step 2 Step 3 Step 4 Find the -ais intercept Let = 0. For the general quadratic = a 2 + b + c, this gives = a(0) 2 + b(0) + c = c Hence the -ais intercept is alwas equal to c. Find the -ais intercepts Let = 0. In general, this gives 0 = a 2 + b + c In order to solve such an equation it is necessar to factorise the right-hand side and then use the null factor theorem. Find the equation of the ais of smmetr Once the -ais intercepts have been found, the equation of the ais of smmetr can be found b using the smmetr properties of the parabola. The ais of smmetr is the perpendicular bisector of the line segment joining the -ais intercepts. Find the coordinates of the turning point The ais of smmetr gives the -coordinate of the turning point. Substitute this into the quadratic polnomial to obtain the -coordinate. Eample 28 Find the - and -ais intercepts and the turning point, and hence sketch the graph of = 2 4. Step 1 c = 0. Therefore the -ais intercept is 0. Step 2 Let = 0. Then 0 = 2 4 0 = ( 4) = 0 or = 4 The -ais intercepts are 0 and 4. 4 0 2 4 (2, 4) Step 3 The ais of smmetr is the line with equation = 0 + 4, that is, = 2. 2 Step 4 When = 2, = (2) 2 4(2) = 4. The turning point has coordinates (2, 4).
106 Chapter 3: Quadratics Eample 29 Find the - and -ais intercepts and the turning point, and hence sketch the graph of = 2 9. Step 1 c = 9. Therefore the -ais intercept is 9. Step 2 Step 3 Let = 0. Then 0 = 2 9 0 = ( + 3)( 3) = 3 or = 3 The -ais intercepts are 3 and 3. The ais of smmetr is the line with equation = 3 + 3, that is, = 0. 2 Step 4 When = 0, = (0) 2 9 Eample 30 = 9 The turning point has coordinates (0, 9). 3 0 9 3 (0, 9) Find the - and -ais intercepts and the turning point, and hence sketch the graph of = 2 + 12. Step 1 c = 12. Therefore the -ais intercept is 12. Step 2 Step 3 Let = 0. Then 0 = 2 + 12 0 = ( + 4)( 3) = 4 or = 3 The -ais intercepts are 4 and 3. The ais of smmetr is the line with equation = 4 + 3 = 1 2 2 Step 4 When = 1 2, = ( 1 2 )2 + ( 1 2 ) 12 = 12 1 4 The turning point has coordinates ( 1 2, 12 1 4 ). 4 1 1 1, 12 2 4 0 12 3
3F Graphing quadratics in polnomial form 107 Using the TI-Nspire CX non-cas To graph the quadratic function with rule = 2 + 12: Enter the rule in the entr line of a Graphs application as shown, and press enter. Using menu > Window/Zoom > Window Settings, select the window settings 10 10 and 15 15 to obtain the graph shown. Using the Casio To graph the quadratic function with rule = 2 + 12: Press MENU 5 to select Graph mode. Enter the epression 2 + 12 in Y1: X,θ,T 2 + X,θ,T 1 2 EXE Select Draw F6. Adjust the View Window SHIFT F3 if required. To find the -ais intercepts, go to G-Solve and select Root: SHIFT F5 F1 EXE EXE To find the -ais intercept, go to G-Solve and select -Intercept: SHIFT F5 F4 EXE To find the turning point, go to G-Solve and select Minimum: SHIFT F5 F3 EXE
108 Chapter 3: Quadratics 3F Section summar Eample 28, 29 Eample 30 Steps for sketching the graph of a quadratic function given in polnomial form: Step 1 Step 2 Step 3 Step 4 Eercise 3F Find the -ais intercept. Find the -ais intercepts. Find the equation of the ais of smmetr. Find the coordinates of the turning point. 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. 2 a A parabola has verte (2, 6) and one of the -ais intercepts is at 6. Find the other -ais intercept. 3 4 b A parabola has verte (2, 6) and one of the -ais intercepts is at 4. Find the other -ais intercept. c A parabola has verte (2, 6) and one of the -ais intercepts is at the origin. Find the other -ais intercept. Sketch each of the following parabolas, clearl showing the ais intercepts and the turning point: a = 2 1 b = 2 + 6 c = 25 2 d = 2 4 e = 2 2 + 3 f = 2 2 4 g = 2 2 3 h = 2 + 1 Sketch each of the following parabolas, clearl showing the ais intercepts and the turning point: a = 2 + 3 10 b = 2 5 + 4 c = 2 + 2 3 d = 2 + 4 + 3 e = 2 2 1 f = 6 2 g = 2 5 6 h = 2 5 24 SF
3G Solving quadratic inequalities 109 3G Solving quadratic inequalities In Chapter 1 we looked at solving linear inequalities. The situation is a little more comple for quadratic inequalities. We suggest one possible approach. To solve a quadratic inequalit (for eample, 2 + 12 > 0): Step 1 Solve the corresponding equation (for eample, 2 + 12 = 0). Step 2 Sketch the graph of the quadratic polnomial (for eample, = 2 + 12). Step 3 Eample 31 Use the graph to determine the set of -values which satisf the inequalit. Solve 2 + 12 > 0. Step 1 Step 2 Step 3 Solve the equation 2 + 12 = 0 ( + 4)( 3) = 0 = 4 or = 3 Sketch the graph of the quadratic = 2 + 12. From the graph it can be seen that 2 + 12 > 0 when < 4 or > 3. Using the TI-Nspire CX non-cas To solve the inequalit 2 + 12 > 0: In a Graphs application, plot the graph of f 1() = 2 + 12. Use menu > Analze Graph > Zero to find the intersections with the -ais. From the graph, we see that f 1() > 0 when < 4 or > 3. Using the Casio 4 To solve the inequalit 2 + 12 > 0: Plot the graph of = 2 + 12 and find the -ais intercepts, as shown in Section 3F. From the graph, we see that > 0 when < 4 or > 3. 0 4 8 12 3
110 Chapter 3: Quadratics 3G Section summar Skillsheet When solving quadratic inequalities of the form a 2 + b + c 0 (or with, > or <), it is best to sketch the graph of = a 2 + b + c. Eercise 3G 1 a Solve the equation 2 2 8 = 0. Eample 31 b Sketch the graph of = 2 2 8. c Solve the inequalit 2 2 8 0. d Solve the inequalit 2 2 8 > 0. 2 Solve each of the following inequalities: a ( 3)( + 2) 0 b ( + 4)( + 3) < 0 c (2 1)( + 4) 0 d ( 6)(2 4) > 0 e (2 6)(2 4) < 0 f (7 2)(2 3) 0 g (2 + 7)(2 4) < 0 h (3 + 6)(2 5) 0 i (5 2)(5 + ) < 0 j (7 2)( + 2) 0 k (7 2)(5 2) < 0 l (11 2)(5 2) 0 3 Solve each of the following inequalities: a (5 )(5 + ) < 0 b 4 9 2 0 c 16 2 < 0 d 36 25 2 0 e 1 16 2 0 f 25 36 2 < 0 4 Solve each of the following inequalities: a 2 + 2 8 0 b 2 5 24 < 0 c 2 4 12 0 d g 2 2 3 9 > 0 12 2 + > 6 j 4 + 5p p 2 0 k 3 + 2 2 < 0 5 Solve each of the following inequalities: e 6 2 + 13 < 6 f 2 5 6 0 h 10 2 11 3 i ( 1) 20 l 2 + 3 2 a 2 + 3 5 0 b 2 5 + 2 < 0 c 2 2 3 1 0 d 8 3 2 > 0 e 2 2 + 7 + 1 < 0 f 2 2 8 + 5 0 6 Eplain wh ( 3) 2 0 for all. 7 Eplain wh ( 1) 2 0 for all. 8 Complete the square for 2 + 2 + 7 and hence show that 2 + 2 + 7 6 for all. 9 Complete the square for 2 2 7 and hence show that 2 2 7 6 for all. SF CF
3H The general quadratic formula 111 3H The general quadratic formula Not all quadratics can be factorised b inspection, and it is often difficult to find the -ais intercepts this wa. There is a general formula for finding the solutions of a quadratic equation in polnomial form. This formula comes from completing the square for the general quadratic. In Section 3E we showed that = a 2 + b + c ( = a + b ) 2 b2 2a 4a + c We can use this to solve the general quadratic equation: a 2 + b + c = 0 ( a + b ) 2 b2 2a 4a + c = 0 ( a + b ) 2 = b2 2a 4a c Now divide both sides b a: ( + b ) 2 = b2 2a 4a c 2 a = b2 4ac 4a 2 + b 2a = ± b 2 4ac 4a 2 = b 2a ± b 2 4ac = b ± b2 4ac 4a 2 2a The solutions of the quadratic equation a 2 + b + c = 0, where a 0, are given b the quadratic formula = b ± b 2 4ac 2a Note: The quadratic formula provides an alternative method for solving quadratic equations to completing the square, but it is probabl not as useful for curve sketching as completing the square, 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 2a. Also, from the formula it can be seen that: If b 2 4ac > 0, there are two solutions. If b 2 4ac = 0, there is one solution. If b 2 4ac < 0, there are no real solutions. This will be further eplored in the net section.
112 Chapter 3: Quadratics Eample 32 Solve each of the following equations for b using the quadratic formula: a 2 1 = 0 b 2 2k 3 = 0 a 2 1 = 0 b 2 2k 3 = 0 Here a = 1, b = 1 and c = 1. Here a = 1, b = 2k and c = 3. The formula gives The formula gives = b ± b 2 4ac 2a = ( 1) ± ( 1) 2 4 1 ( 1) 2 1 = 1 ± 5 2 Eample 33 = b ± b 2 4ac 2a = ( 2k) ± ( 2k) 2 4 1 ( 3) 2 1 = 2k ± 4k 2 + 12 2 = k ± k 2 + 3 Note that k 2 + 3 0 for all values of k, since k 2 0. Sketch the graph of = 3 2 12 7. Use the quadratic formula to calculate the -ais intercepts. Since c = 7, the -ais intercept is 7. Ais of smmetr = b ( 2a = 12 ) = 2 2 ( 3) Turning point When = 2, = 3( 2) 2 12( 2) 7 = 5. The turning point coordinates are ( 2, 5). -ais intercepts 3 2 12 7 = 0 = b ± b 2 4ac 2a = ( 12) ± ( 12) 2 4( 3)( 7) 2( 3) = 12 ± 60 6 = 12 ± 2 15 6 The -ais intercepts are 2 1 3 15 and 2 + 1 3 15. 1 2 15 3 ( 2, 5) 2 1 2 + 15 3 5 7 0
3H 3H The general quadratic formula 113 Section summar Eample 32 Eample 33 The solutions of the quadratic equation a 2 + b + c = 0, where a 0, are given b the quadratic formula = b ± b 2 4ac 2a From the formula it can be seen that: If b 2 4ac > 0, there are two solutions. If b 2 4ac = 0, there is one solution. If b 2 4ac < 0, there are no real solutions. Eercise 3H 1 For each of the following, the coefficients a, b and c of a quadratic = a 2 + b + c are given. Find: i b 2 4ac ii b2 4ac in simplest surd form a a = 2, 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 = 27 2 Simplif each of the following: 3 4 a 2 + 2 5 2 b 9 3 5 6 c 5 + 5 5 10 Solve each of the following for. Give eact answers. a 2 + 6 = 4 b 2 7 3 = 0 c 2 2 5 + 2 = 0 d 2 2 + 4 7 = 0 e 2 2 + 8 = 1 f 5 2 10 = 1 g 2 2 + 4 1 = 0 h 2 2 + = 3 i 2.5 2 + 3 + 0.3 = 0 j 0.6 2 1.3 = 0.1 k 2k 2 4 + k = 0 l 2(1 k) 2 4k + k = 0 d 6 + 12 2 6 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 = 2 + 5 1 b = 2 2 3 1 c = 2 3 + 1 d + 4 = 2 + 2 e = 4 2 + 5 + 1 f = 3 2 + 4 2 g = 2 + 5 + 6 h = 4 2 3 + 2 i = 3 2 4 SF
114 Chapter 3: Quadratics 3I The discriminant In the previous section we found that the solutions to the quadratic equation a 2 + b + c = 0 are given b = b ± b 2 4ac 2a The epression under the square root sign is called the discriminant. We write = b 2 4ac The number of -ais intercepts There are three different possibilities for the number of -ais intercepts of a parabola: zero the graph is either all above or all below the -ais one the graph touches the -ais and the turning point is the -ais intercept two the graph crosses the -ais. For a parabola = a 2 + b + c, we can use the discriminant = b 2 4ac to determine when each of these three situations occur. If the discriminant b 2 4ac < 0, then the equation a 2 + b + c = 0 has no solutions and the corresponding parabola will have no -ais intercepts. If the discriminant b 2 4ac = 0, then the equation a 2 + b + c = 0 has one solution and the corresponding parabola will have one -ais intercept. (We sometimes sa the equation has two coincident solutions.) If the discriminant b 2 4ac > 0, then the equation a 2 + b + c = 0 has two solutions and the corresponding parabola will have two -ais intercepts.
3I The discriminant 115 Eample 34 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 = 2 6 + 8 b = 2 8 + 16 c = 2 2 3 + 4 a c Discriminant = b 2 4ac = ( 6) 2 (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 2 6 + 8 = 0. = b 2 4ac = ( 3) 2 (4 2 4) = 23 As < 0, the graph does not intersect the -ais, i.e. there are no real solutions for the equation 2 2 3 + 4 = 0. Eample 35 b = b 2 4ac = ( 8) 2 (4 1 16) = 0 Find the values of m for which the equation 3 2 2m + 3 = 0 has: a one solution b no solution For the quadratic 3 2 2m + 3, the discriminant is = 4m 2 36. a For one solution: i.e. = 0 4m 2 36 = 0 m 2 = 9 m = ±3 As = 0, the graph touches the -ais, i.e. there is one solution of the equation 2 8 + 16 = 0. c For no solution: i.e. two distinct solutions. < 0 4m 2 36 < 0 c For two distinct solutions: i.e. > 0 4m 2 36 > 0 From the graph it can be seen that m > 3 or m < 3 b From the graph, this is equivalent to 3 < m < 3 3 0 3 m 36
116 Chapter 3: Quadratics 3I The nature of the solutions of a quadratic equation The discriminant can be used to assist in the identification of the particular tpe of solution for a quadratic equation a 2 + b + c = 0. For a, b and c rational numbers: If = b 2 4ac is a perfect square and 0, then the quadratic equation has two rational solutions. If = b 2 4ac = 0, then the quadratic equation has one rational solution. If = b 2 4ac is not a perfect square and > 0, then the quadratic equation has two irrational solutions. Eample 36 Show that the solutions of the equation 3 2 + (m 3) m = 0 are rational for all rational values of m. = (m 3) 2 4 3 ( m) = m 2 6m + 9 + 12m = m 2 + 6m + 9 = (m + 3) 2 0 for all m Furthermore, is a perfect square for all m. Section summar The discriminant of a quadratic polnomial a 2 + b + c is = b 2 4ac For the equation a 2 + b + c = 0: If > 0, there are two solutions. If = 0, there is one solution. If < 0, there are no real solutions. For the equation a 2 + b + c = 0 where a, b and c are rational numbers: Skillsheet If is a perfect square and 0, then the equation has two rational solutions. If = 0, then the equation has one rational solution. If is not a perfect square and > 0, then the equation has two irrational solutions. Eercise 3I 1 Determine the discriminant of each of the following quadratics: a 2 + 2 4 b 2 + 2 + 4 c 2 + 3 4 d 2 2 + 3 4 e 2 2 + 3 + 4 SF
3I 3I The discriminant 117 Eample 34 Eample 35 Eample 36 2 Without sketching the graphs of the following quadratics, determine whether the cross or touch the -ais: a = 2 5 + 2 b = 4 2 + 2 1 c = 2 6 + 9 d = 8 3 2 2 e = 3 2 + 2 + 5 f = 2 1 3 B eamining the discriminant, find the number of distinct solutions of: a 2 + 8 + 7 = 0 b 3 2 + 8 + 7 = 0 c 10 2 3 = 0 d 2 2 + 8 7 = 0 e 3 2 8 7 = 0 f 10 2 + 3 = 0 4 B eamining the discriminant, state the nature and number of distinct solutions for each of the following: 5 6 a 9 2 24 + 16 = 0 b 2 5 6 = 0 c 2 4 = 0 d 25 2 20 + 4 = 0 e 6 2 3 2 = 0 f 2 + 3 + 2 = 0 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 2 4m + 20 = 0 b m 2 3m + 3 = 0 c 5 2 5m m = 0 d 2 + 4m 4(m 2) = 0 For m and n rational numbers show that m 2 + (2m + n) + 2n = 0 has rational solutions. 7 Find the values of p for which the equation p 2 + 2(p + 2) + p + 7 = 0 has no solution. 8 Find the values of p for which the equation (1 2p) 2 + 8p (2 + 8p) = 0 has one solution. 9 Find the value(s) of p for which: a c p 2 6 + p = 0 has one solution 3 2 = 2 + p 1 has two solutions b d 2 2 4 + 3 = p has two solutions 2 2 + 2 = p has two solutions. 10 Find the values of p for which the graph of = p 2 + 8 + p 6 crosses the -ais. 11 Show that the equation (p 2 + 1) 2 + 2pq + q 2 = 0 has no real solution for an values of p and q (q 0). 12 a Find the discriminant of 2 + 4m + 24m 44. b Show the equation 2 + 4m + 24m 44 = 0 has a solution for all values of m. 13 a Find the discriminant of 4m 2 + 4(m 1) + m 2. b Show the equation 4m 2 + 4(m 1) + m 2 = 0 has a solution for all values of m. 14 Find the discriminant of the equation 4 2 + (m 4) m = 0, where m is a rational number, and hence show that the equation has rational solution(s). 15 Find the discriminant of the equation 2 (m + 2n) + 2mn = 0, where m and n are rational numbers, and hence show that the equation has rational solution(s). 16 If both a and c are positive, what can be said about the graph of = a 2 + b c? SF CF 17 If a is negative and c is positive, what can be said about the graph of = a 2 + b + c?
118 Chapter 3: Quadratics 3J Solving simultaneous linear and quadratic equations As discussed in Section 2H, when solving simultaneous linear equations we are actuall finding the point of intersection of the two corresponding linear graphs. 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. Then the linear epression for can be substituted into the quadratic equation. Eample 37 Find the points of intersection of the line with equation = 2 + 4 and the parabola with equation = 2 8 + 12. At the point of intersection: 2 8 + 12 = 2 + 4 2 6 + 8 = 0 ( 2)( 4) = 0 Hence = 2 or = 4. When = 2, = 2(2) + 4 = 0. When = 4, = 2(4) + 4 = 4. Therefore the points of intersection are (2, 0) and (4, 4). The result can be shown graphicall. = 2 8 + 12 12 4 4 (2, 0) 2 4 (4, 4) = 2 + 4
3J Solving simultaneous linear and quadratic equations 119 Using the TI-Nspire CX non-cas To find the intersection points of the two graphs: In a Graphs application, plot the graphs of f 1() = 2 + 4 and f 2() = 2 8 + 12. Adjust the Window Settings as required. Use menu > Analze Graph > Intersection to find the intersection points. Note: Alternativel, using menu > Geometr > Points & Lines > Intersection Point(s) will find both intersection points at once. Using the Casio To find the intersection points of the two graphs: Press MENU 5 to select Graph mode. Enter = 2 + 4 in Y1: ( ) 2 X,θ,T + 4 EXE Enter = 2 8 + 12 in Y2: X,θ,T 2 8 X,θ,T + 1 2 EXE Select Draw F6. Adjust the View Window SHIFT F3 if required. Go to the G-Solve menu and select Intersection: SHIFT F5 F5 EXE EXE Eample 38 Prove that the straight line with the equation = 1 meets the parabola with the equation = 2 3 + 2 once onl. At the point of intersection: 2 3 + 2 = 1 2 2 + 1 = 0 ( 1) 2 = 0 Therefore = 1 and = 1 1 = 0. The straight line just touches the parabola at (1, 0). This can be illustrated graphicall. = 2 3 + 2 2 1 0 1 2 = 1
120 Chapter 3: Quadratics 3J Section summar Eample 37 Eample 38 To find the points of intersection of a straight line = m + c 2 and a parabola = a 2 + b + c 1 : Form the quadratic equation a 2 + b + c 1 = m + c 2 Rearrange the equation so that the right-hand side is zero: a 2 + (b m) + (c 1 c 2 ) = 0 Solve the equation for and substitute these -values into the equation of the line to find the corresponding -values. The discriminant applied to the second equation, a 2 + (b m) + (c 1 c 2 ) = 0, can be used to determine the number of intersection points: If > 0, there are two intersection points. If = 0, there is one intersection point. If < 0, there are no intersection points. Eercise 3J 1 a Find the points of intersection of the line with equation = 2 and the parabola with equation = 2 6. b Find the points of intersection of the line with equation + = 6 and the parabola with equation = 2. c Find the points of intersection of the line with equation 5 + 4 = 21 and the parabola with equation = 2. d Find the points of intersection of the line with equation = 2 + 1 and the parabola with equation = 2 + 3. 2 Solve each of the following pairs of equations: 3 a = 2 + 2 8 b = 2 3 c = 2 + 5 = 2 = 4 7 = 2 d = 2 + 6 + 6 e = 6 2 f = 2 + + 6 = 2 + 3 = 2 2 = 6 + 8 Prove that, for each of the following pairs of equations, the straight line meets the parabola onl once: a = 2 6 + 8 b = 2 2 + 6 = 2 + 4 = 4 3 c = 2 2 + 11 + 10 d = 2 + 7 + 4 = 3 + 2 = 12 SF
3J 3K Families of quadratic polnomial functions 121 4 Solve each of the following pairs of equations: a = 2 6 = 8 + b = 3 2 + 9 = 32 c = 5 2 + 9 = 12 2 d = 3 2 + 32 e = 2 2 12 f = 11 2 = 32 3 = 3( 4) = 21 6 5 a Find the value of c such that = + c is a tangent to the parabola = 2 12. Hint: Consider the discriminant of the resulting quadratic. b i Sketch the parabola with equation = 2 2 6 + 2. ii Find the values of m for which the straight line = m + 6 is tangent to the parabola. Hint: Use the discriminant of the resulting quadratic. 6 a Find the value of c such that the line with equation = 2 + c is tangent to the parabola with equation = 2 + 3. b Find the possible values of c such that the line with equation = 2 + c twice intersects the parabola with equation = 2 + 3. 7 Find the value(s) of a such that the line with equation = is tangent to the parabola with equation = 2 + a + 1. 8 Find the value of b such that the line with equation = is tangent to the parabola with equation = 2 + + b. 9 Find the equation of the straight line(s) which pass through the point (1, 2) and is (are) tangent to the parabola with equation = 2. 3K Families of quadratic polnomial functions In Chapter 2 we considered the information that is necessar to determine the equation of a straight line and we also studied families of straight lines. In this section these two ideas are etended for our stud of quadratic polnomials. Families of quadratics Here are some eamples of families of quadratic polnomial functions: = a 2, a 0 The parabolas with their vertices at the origin. = a( 2) 2 + 3, a 0 The parabolas with turning point at (2, 3). = a( 2)( + 5), a 0 The parabolas with -ais intercepts 2 and 5. = a( h)( 2), a 0 The parabolas with -ais intercept 2. = a 2 + b, a 0 and b 0 The parabolas with two -ais intercepts, one of which is the origin. SF CF
122 Chapter 3: Quadratics The letters a, b and h used to define these families are called parameters. Varing the values of the parameters produces different parabolas. For eample, for = a 2 + b some possible curves are shown below. Eample 39 = 2 + 2 = 2 2 = 22 + 4 0 = 22 + 4 = 2 3 A famil of parabolas have rules of the form = a 2 + c. For the parabola in this famil that passes through the points (1, 7) and (2, 10), find the values of a and c. When = 1, = 7 and when = 2, = 10. 7 = a + c (1) 10 = 4a + c (2) Subtract (1) from (2): 3 = 3a and hence a = 1. Substitute in (1): 7 = 1 + c and therefore c = 6. The equation is = 2 + 6. Eplanation Substitute = 1, = 7 in the equation = a 2 + c to obtain (1). Substitute = 2, = 10 in the equation = a 2 + c to obtain (2). Eample 40 A famil of parabolas have rules of the form = a 2 + b + 2, where a 0. a For a parabola in this famil with its turning point on the -ais, find a in terms of b. b If the turning point is at (4, 0), find the values of a and b.
3K Families of quadratic polnomial functions 123 a The discriminant = b 2 8a. We have = 0 and therefore a = b2 8. b We have b = 4, which implies b = 8a. 2a From part a, we have a = b2 8. Hence a = 64a2 = 8a 2. 8 Thus a(1 8a) = 0 and, since a 0, a = 1 8. Substituting for a in b = 8a gives b = 1. Determining quadratic rules Eplanation The discriminant of a 2 + b + c is = b 2 4ac. In this case c = 2. The discriminant = 0 since the parabola touches the -ais at its turning point. The ais of smmetr has equation = b 2a. At the beginning of this section we looked at different families of quadratic polnomial functions. We now consider three important such families which can be used as a basis for finding a quadratic rule from given information. These are certainl not the onl useful forms. You will see others in the worked eamples. 1 = a( e)( f ) This can be used if two -ais intercepts and the coordinates of one other point are known. 2 = a( h) 2 + k This can be used if the coordinates of the turning point and one other point are known. 3 = a 2 + b + c This can be used if the coordinates of three points on the parabola are known. Eample 41 A parabola has -ais intercepts 3 and 4 and it passes through the point (1, 24). Find the rule for this parabola. = a( + 3)( 4) When = 1, = 24. Therefore 24 = a(1 + 3)(1 4) 24 = 12a a = 2 The rule is = 2( + 3)( 4). Eplanation Two -ais intercepts are given. Therefore use the form = a( e)( f ).
124 Chapter 3: Quadratics Eample 42 The coordinates of the turning point of a parabola are (2, 6) and the parabola passes through the point (3, 3). Find the rule for this parabola. = a( 2) 2 + 6 When = 3, = 3. Therefore 3 = a(3 2) 2 + 6 3 = a + 6 a = 3 The rule is = 3( 2) 2 + 6. Eample 43 Eplanation The coordinates of the turning point and one other point on the parabola are given. Therefore use = a( h) 2 + k. A parabola passes through the points (1, 4), (0, 5) and ( 1, 10). Find the rule for this parabola. = a 2 + b + c When = 1, = 4. When = 0, = 5. When = 1, = 10. 4 = a + b + c (1) 5 = c (2) 10 = a b + c (3) Substitute from equation (2) into equations (1) and (3): 1 = a + b (1 ) 5 = a b (3 ) Add (1 ) and (3 ): 4 = 2a a = 2 Using equation (1 ), we obtain b = 3. The rule is = 2 2 3 + 5. Eplanation The coordinates of three points on the parabola are given. Therefore substitute values into the polnomial form = a 2 + b + c to obtain three equations in three unknowns.
3K Families of quadratic polnomial functions 125 Using the TI-Nspire CX non-cas To find the rule for the parabola passing through the points (1, 4), (0, 5) and ( 1, 10): Open a Lists & Spreadsheet application. Enter the coordinates in lists named and. Insert a Calculator page ( ctrl I > Add Calculator). Use menu > Statistics > Stat Calculations > Quadratic Regression. Complete the pop-up screen as shown. Select OK to displa the values of a, b and c. Hence the rule is = 2 2 3 + 5. Using the Casio To find the rule for the parabola passing through the points (1, 4), (0, 5) and ( 1, 10): Press MENU 2 to select Statistics mode. Enter the -coordinates in List1 and the -coordinates in List2: 1 EXE 0 EXE ( ) 1 EXE 4 EXE 5 EXE 1 0 EXE Select Quadratic Regression b going to Calculation F2, Regression F3, then X 2 F3. Hence the rule is = 2 2 3 + 5.
126 Chapter 3: Quadratics Eample 44 Determine the quadratic rule for each of the following parabolas: a b (0, 3) c ( 1, 8) a 0 (2, 5) 0 3 This is of the form = a 2 For (2, 5): Hence the rule is 5 = 4a a = 5 4 = 5 4 2 d b ( 3, 1) (0, 8) 0 0 (1, 6) This is of the form = a 2 + c For (0, 3): 3 = a(0) + c c = 3 For ( 3, 1): 1 = a( 3) 2 + 3 1 = 9a + 3 a = 2 9 Hence the rule is = 2 9 2 + 3 c This is of the form = a( 3) d This is of the form = a( 1) 2 + 6 For ( 1, 8): 8 = a( 1 3) 8 = 4a a = 2 Hence the rule is = 2( 3) = 2 2 6 For (0, 8): 8 = a + 6 a = 2 Hence the rule is = 2( 1) 2 + 6 = 2( 2 2 + 1) + 6 = 2 2 4 + 8
3K 3K Families of quadratic polnomial functions 127 Section summar Skillsheet Eample 39 To find a quadratic rule to fit given points, first choose the best form of quadratic epression to work with. Then substitute in the coordinates of the known points to determine the unknown parameters. Some possible forms are given here: i iii 0 = a2 One point is needed to determine a. 0 = a2 + b Two points are needed to determine a and b. Eercise 3K 1 ii iv 0 = a 2 + c Two points are needed to determine a and c. 0 = a 2 + b + c Three points are needed to determine a, b and c. A famil of parabolas have rules of the form = a 2 + c. For the parabola in this famil that passes through the points ( 1, 2) and (0, 6), find the values of a and c. Eample 40 2 A famil of parabolas have rules of the form = a 2 + b + 4, where a 0. Eample 41 3 a a Find the discriminant of the quadratic polnomial a 2 + b + 4. b For a parabola in this famil with its turning point on the -ais, find a in terms of b. c If the turning point is at ( 4, 0), find the values of a and b. Eample 42 Eample 43 b c A parabola has -ais intercepts 2 and 6 and it passes through the point (1, 30). Find the rule for this parabola. The coordinates of the turning point of a parabola are ( 2, 4) and the parabola passes through the point (3, 46). Find the rule for this parabola. A parabola passes through the points (1, 2), (0, 3) and ( 1, 6). Find the rule for this parabola. 4 A quadratic rule for a particular parabola is of the form = a 2. The parabola passes through the point with coordinates (2, 8). Find the value of a. CF
128 Chapter 3: Quadratics 3K Eample 44 5 A quadratic rule for a particular parabola is of the form = a 2 + b. The parabola passes through the point with coordinates ( 1, 4) and one of its -ais intercepts is 6. Find the values of a and b. 6 A quadratic rule for a particular parabola is of the form = a( b) 2 + c. The parabola has verte (1, 6) and passes through the point with coordinates (2, 4). Find the values of a, b and c. 7 Determine the equation of each of the following parabolas: a b c d 3 5 0 4 0 1 3 e ( 3, 9) ( 3, 0) 0 ( 1, 5) 0 (1, 0) f 6 7 0 (2, 2) 0 4 8 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) 9 A parabola has -ais intercepts 6 and 3 and passes through the point (1, 10). Find the equation of the parabola. 10 A parabola has verte with coordinates ( 1, 3) and -ais intercept 4. Find the equation for the parabola. 11 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 12 A parabola has the same shape as = 2 2, but its turning point is (1, 2). Write its equation. 13 A parabola has its verte at (1, 2) and passes through the point (3, 2). Write its equation. CF
3K 3K Families of quadratic polnomial functions 129 14 Which of the curves could be most nearl defined b each of the following? a = 1 3 ( + 4)(8 ) b = 2 + 2 c = 10 + 2( 1) 2 d = 1 2 (9 2 ) A 5 2 0 2 4 B 10 5 2 0 2 4 C 5 10 0 5 5 10 15 A famil of parabolas satisfies the rule = a 2 + 2 + a. D 2 0 2 a Epress a 2 + 2 + a in the form a( + b) 2 + c for real numbers b and c. b Give the coordinates of the turning point of the graph of = a 2 + 2 + a in terms of a. c For which values of a is a 2 + 2 + a a perfect square? d For which values of a does the graph of = a 2 + 2 + a have two -ais intercepts? 16 A parabola has its verte at (2, 2) and passes through (4, 6). Write its equation. 17 Write down four quadratic rules that have graphs similar to those in the diagram. 18 Find quadratic epressions which could represent the two curves in this diagram, given that the coefficient of is 1 in each case. The labelled points are A(2, 3), B(2, 1), C(0, 5) and D(0, 2). (b) 2 8 6 (a) 4 2 4 2 0 2 4 6 8 (6, 6) (b) (c) 2 4 6 8 10 12 A 3 2 D B 1 0 1 2 3 19 The rate of rainfall during a storm t hours after it began was 3 mm per hour when t = 5, 6 mm per hour when t = 9 and 5 mm per hour when t = 13. Assuming that a quadratic model applies, find an epression for the rate of rainfall, r mm per hour, in terms of t. 5 C 4 (d) (a) 5 CF
130 Chapter 3: Quadratics 3K 20 a Which of the graphs shown below could represent the equation = ( 4) 2 3? b Which graph could represent = 3 ( 4) 2? A 13 0 2 4 6 B 13 0 2 4 6 C 0 13 2 4 6 D 0 13 2 4 6 21 Find the equation of the quadratic which passes through the points with coordinates: a ( 2, 1), (1, 2), (3, 16) b ( 1, 2), (1, 4), (3, 10) c ( 3, 5), (3, 20), (5, 57) 3L Quadratic models In this section it is shown how quadratics can be used to solve worded problems, including problems which involve finding the maimum or minimum value of a quadratic polnomial that has been used to model a practical situation. Eample 45 Jenn wishes to fence off a rectangular vegetable garden in her backard. She has 20 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 Then width = 20 = 10 2 2 Therefore ( A = 10 ) 2 = 10 2 2 = 1 2 (2 20 + 100 100) = 1 2 (2 20 + 100) + 50 (completing the square) timber fence 10 2 = 1 2 ( 10)2 + 50 0 5 10 15 20 Hence the maimum area is 50 m 2 when = 10. A 50 40 30 20 10 CF
3L Quadratic models 131 Eample 46 A cricket ball is thrown b a fielder. It leaves his hand at a height of 2 metres above the ground and the wicketkeeper takes the ball 60 metres awa again at a height of 2 metres. It is known that after the ball has gone 25 metres it is 15 metres above the ground. The path of the cricket ball is a parabola with equation = a 2 + b + c. a Find the values of a, b and c. b Find the maimum height of the ball above the ground. (25, 15) 2 0 60 c Find the height of the ball when it is 5 metres horizontall before it hits the wicketkeeper s gloves. a The data can be used to obtain three equations: 2 = c (1) 15 = (25) 2 a + 25b + c (2) 2 = (60) 2 a + 60b + c (3) Substitute equation (1) in equations (2) and (3): 13 = 625a + 25b (2 ) 0 = 3600a + 60b (3 ) Simplif (3 ) b dividing both sides b 60: 0 = 60a + b (3 ) Multipl this b 25 and subtract from equation (2 ): 13 = 875a a = 13 875 and b = 156 175 The path of the ball has equation = 13 875 2 + 156 175 + 2 b The maimum height occurs when = 30 and = 538 35. maimum height is 538 35 m. c When = 55, = 213 35. height of the ball is 213 35 m.
132 Chapter 3: Quadratics 3L Skillsheet Eample 45 Eercise 3L 1 A farmer 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 the ard is m and the area inside the ard is A m 2, 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. eisting fence 2 A rectangle has a perimeter of 20 m. Let m be the length of one side. Find a formula for the area A of the rectangle in terms of. Hence find the maimum area A. 3 The efficienc rating, E, of a particular spark plug when the gap is set at mm is said to be 400( 2 ). 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. 4 A piece of wire 68 cm in length is bent into the shape of a rectangle. a If cm is the length of the rectangle and A cm 2 is the area enclosed b the rectangular shape, write down a formula which connects A and. b Sketch the graph of A against for suitable -values. c Use our graph to determine the maimum area formed. Gap mm 5 A piece of wire 80 cm long is to be cut into two pieces. One piece is to be bent into a square and the other into a rectangle four times as long as it is wide. a Let cm be the length of a side of the square and cm be the width of the rectangle. Write a formula connecting and. b Let A cm 2 be the sum of the areas of the square and the rectangle. i Find a formula for A in terms of. ii Find the length of both pieces of wire if A is to be a minimum. 6 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 for each plon, given b the formula C = 240h + 100h 2, 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 2500 m? CF
3L 3L Quadratic models 133 Eample 46 7 A tug-o-war team produces a tension in a rope described b the rule 8 T = 290(8t 0.5t 2 1.4) units where t is the number of seconds after commencing the pull. a Sketch a graph of T against t, stating the practical domain. b What is the greatest tension produced during a heave? 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 2, 0. a Use a 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 2 m above the ground, how far was the ball from where it was hit? iii At what height was the ball when it was struck? 9 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 2 + b + c can be used to describe the arch. Find the values of a, b and c. 2.5 m 0 A 1.5 m 10 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 2 + bt + c. t 30 150 300 s 7.2 12.5 6 a Find the values of a, b and c. b Sketch the graph for 0 t 360. (Use a calculator.) c Find an estimate for the spending when: i t = 180 ii t = 350 CF