UNCORRECTED SAMPLE PAGES. Australian curriculum NUMBER AND ALGEBRA

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1 7A 7B 7C 7D 7E 7F 7G 7H 7I 7J 7K Chpter Wht ou will lern 7Prols nd other grphs Eploring prols Skething prols with trnsformtions Skething prols using ftoristion Skething ompleting the squre Skething using the qudrti formul (0A) Applitions involving prols Intersetion of lines nd prols (0A) Grphs of irles Grphs of eponentils Grphs of hperols Further trnsformtions of grphs (0A) Austrlin urriulum NUMBER AND ALGEBRA Ptterns nd lger Ftorise lgeri epressions tking out ommon lgeri ftor. Epnd inomil produts nd ftorise moni qudrti epressions using vriet of strtegies. Sustitute vlues into formuls to determine n unknown. Liner nd non-liner reltionships Eplore the onnetion etween lgeri nd grphil representtions of reltions suh s simple qudrtis, irles nd eponentils, using digitl tehnolog s pproprite. Solve simple qudrti equtions using rnge of strtegies. (0A) Desrie, interpret nd sketh prols, hperols, irles nd eponentil funtions nd their trnsformtions. (0A) Ftorise moni nd non-moni qudrti epressions nd solve wide rnge of qudrti equtions derived from vriet of ontets. UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

2 Solr refletors Solr power is one of the world s developing renewle energ soures. Colleting solr power involves refleting the Sun s rs nd pturing nd trnsferring the het generted. ne of the most effetive tpes of solr power plnt uses proli refletors. The proli shpe of these refletors n e desried qudrti reltion. Consequentl the diret ll the rs refleted the mirror to the sme ple. In proli dis sstem the Sun s rs re onentrted onto glss tue reeiver. This reeiver is positioned t the fol point of the prol. The fluid in the reeiver is heted up nd trnsferred to n engine nd onverted to eletriit. ther tpes of solr power plnts lso use proli trough refletors nd ollet the Sun s rs to generte eletriit. nline resoures Chpter pre-test Videos of ll worked emples Intertive widgets Intertive wlkthroughs Downlodle HTsheets Aess to HTmths Austrlin Curriulum ourses UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

3 6 Chpter 7 Prols nd other grphs 7A Eploring prols ne of the simplest nd most importnt non-liner grphs is the prol. When ll is thrown or wter strems up nd out from grden hose, the pth followed hs proli shpe. The prol is the grph of qudrti reltion with the si rule =. Qudrti rules, suh s = ( ), = nd = ( + ) 7, lso give grphs tht re prols nd re trnsformtions of the grph of =. Let s strt: To wht effet? To see how different qudrti rules ompre to the grph of =, omplete this tle nd plot the grph of eh eqution on the sme set of es. 0 = 9 = 9 = ( ) = For ll the grphs, find suh fetures s the: turning point is of smmetr -interept -interepts. Disuss how eh of the grphs of, nd ompre to the grph of =. Compre the rule with the position of the grph UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

4 Numer nd Alger 65 A prol is the grph of qudrti reltion. The si prol hs the rule =. The verte (or turning point) is (0, 0). It is minimum turning point. Ais of smmetr is = 0. -interept is 0. -interept is 0. Simple trnsformtions of the grph of = inlude: diltion refletion = (, ) = = (, ) (, ) trnsltion = ( + ) = = (, ) (, ) = = = + = = ( ) Emple Identifing ke fetures of prols Determine the following ke fetures of eh of the given grphs. i turning point nd whether it is mimum or minimum ii is of smmetr iii iv -interepts -interept (, ) = UNCRRECTED Ke ides Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

5 66 Chpter 7 Prols nd other grphs SLUTIN EXPLANATIN i Turning point is minimum t (, ). ii Ais of smmetr is =. iii -interepts re nd. iv -interept is. i turning point is mimum t (, 0). ii Ais of smmetr is =. iii -interept is. iv -interept is. Emple Trnsforming prols Cop nd omplete the tle for the following grphs. = = ( + ) Formul = = ( + ) = + SLUTIN Formul = Mimum or minimum Mimum or minimum Refleted in the -is (es/no) Lowest point of grph is t (, ). Line of smmetr is through the -oordinte of the turning point. -interepts lie on the -is ( = 0) nd the -interept on the -is ( = 0). Grph hs highest point t (, 0). Line of smmetr is through the -oordinte of the turning point. Turning point is lso the one -interept. Turning point vlue Refleted in the Turning when -is (es/no) point = = vlue when = Wider or nrrower thn = = minimum no (0, 0) nrrower = ( + ) minimum no (, 0) 9 sme = + mimum es (0, ) sme = = + Wider or nrrower thn = UNCRRECTED EXPLANATIN Red fetures from grphs nd onsider the effet of eh hnge in eqution on the grph. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

6 Numer nd Alger 67 Eerise 7A, (½) (½) Emple Complete this tle nd grid to plot the grph of =. 0 9 Write the missing fetures for this grph. The prol hs (mimum or minimum). The oordintes of the turning point re. The -interept is. d The -interepts re nd. e The is of smmetr is. Determine these ke fetures of the following grphs. i turning point nd whether it is mimum or minimum ii is of smmetr iii -interepts (, 5) iv (, ) -interept UNCRRECTED 6 UNDERSTANDING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

7 68 Chpter 7 Prols nd other grphs 7A Emple d e 6 (, ) Cop nd omplete the tle elow for the following grphs. = d 8 6 Formul = = = d = e = f = = = = = Mimum or minimum e 8 6 = Refleted in the -is (es/no) = = 6 6 Turning point f f vlue when = 8 6 = 6(½) = = = UNCRRECTED Wider or nrrower thn = UNDERSTANDING FLUENCY Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

8 Numer nd Alger 69 UNCRRECTED FLUENCY Emple Emple 5 6 Cop nd omplete the tle elow for the following grphs. = ( + ) = = ( + ) = ( ) = ( ) d = ( + ) = = ( ) d = 8 6 = ( + ) = ( ) = Formul Turning point Ais of smmetr -interept ( = 0) -interept Cop nd omplete the tle for the following grphs. = + = = + d = = = + = = = = + Formul Turning point -interept ( = 0) vlue when = d = = 7A Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

9 70 Chpter 7 Prols nd other grphs 7A 7 9(½) 7 9(½), Write down the eqution of the is of smmetr for the grphs of these rules. = = + 7 = d = e = f = ( ) g = ( + ) h = ( + ) i = j = + k = 6 l = ( + ) 8 Write down the oordintes of the turning point for the grphs of the equtions in Question 7. 9 Find the -interept (i.e. when = 0) for the grphs of the equtions in Question 7. 0 Mth eh of the following equtions to one of the grphs elow. = = 6 = ( + ) d = 5 e i iv vii = ( ) 9 (, 5) f = g = + h = ( ) + ii v (, 5) iii vi 6 (, 5) (, ) (, ) (, ) UNCRRECTED viii 7 PRBLEM-SLVING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

10 Numer nd Alger 7 Using tehnolog, plot the following pirs of grphs on the sme set of es for 5 5 nd ompre their tles of vlues. i iii v = nd = = nd = 6 = nd = 7 ii iv vi = nd = = nd = = nd = 5 Suggest how the onstnt in = trnsforms the grph of =. Using tehnolog, plot the following sets of grphs on the sme set of es for 5 5 nd ompre the turning point of eh. i =, = ( + ), = ( + ), = ( + ) ii =, = ( ), = ( ), = ( ) Eplin how the onstnt h in = ( + h) trnsforms the grph of =. Using tehnolog, plot the following sets of grphs on the sme set of es for 5 5 nd ompre the turning point of eh. i =, = +, = +, = + ii =, =, =, = 5 Eplin how the onstnt k in = + k trnsforms the grph of =. Write down n emple of qudrti eqution whose grph hs: two -interepts one -interept Finding the rule 5 Find qudrti rule tht stisfies the following informtion. turning point (0, ) nd nother point (, ) turning point (0, ) nd nother point (, ) turning point (, 0) nd -interept d turning point (, 0) nd -interept e turning point (0, 0) nd nother point (, 8) f turning point (0, 0) nd nother point (, ) g turning point (, ) nd -interept h turning point (, ) nd -interept 0 no -interepts (), UNCRRECTED 6 Plot grph of the prol = for nd desrie its fetures. 5, 6 REASNING ENRICHMENT 7A Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

11 7 Chpter 7 Prols nd other grphs 7B Ke ides Skething prols with trnsformtions Previousl we hve eplored simple trnsformtions of the grph of = nd plotted these on numer plne. We will now formlise these trnsformtions nd sketh grphs showing ke fetures without the need to plot ever point. Let s strt: So where is the turning point? Consider the qudrti rule = ( ) + 7. Disuss the effet of the negtive sign in = ompred with =. Disuss the effet of in = ( ) ompred with =. Disuss the effet of +7 in = + 7 ompred with =. Now for = ( ) + 7, find: the oordintes of the turning point the is of smmetr the -interept. Wht would e the oordintes of the turning point in these qudrtis? = ( h) + k = ( h) + k Sketh prols drwing proli urve nd lelling ke fetures. turning point is of smmetr -interept (sustitute = 0) For =, diltes the grph of =. Turning point is (0, 0). -interept nd -interept re oth 0. Ais of smmetr is = 0. When > 0, the prol is upright. When < 0, the prol is inverted. For = ( h), h trnsltes the grph of = horizontll. When h > 0, the grph is trnslted h units to the right. When h < 0, the grph is trnslted h units to the left. For = + k, k trnsltes the grph of = vertill. When k > 0, the grph is trnslted k units up. When k < 0, the grph is trnslted k units down. The turning point form of qudrti is = ( h) + k. The turning point is (h, k). The is of smmetr is = h. The gtew rh in St Louis, USA, is prol 9 m high. UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

12 Numer nd Alger 7 Emple Skething with trnsformtions Sketh grphs of the following qudrti reltions, lelling the turning point nd the -interept. = = + = ( ) SLUTIN (, ) (, ) Emple Using turning point form EXPLANATIN = is upright nd nrrower thn =. Turning point nd -interept re t the origin (0, 0). Sustitute = to lel seond point. = + is inverted (i.e. hs mimum) nd is trnslted units up ompred with =. Turning point is t (0, ) nd -interept (i.e. when = 0) is. = ( ) is upright (i.e. hs minimum) nd is trnslted units right ompred with =. Thus, the turning point is t (, 0). Sustitute = 0 for -interept: = (0 ) Sketh the grphs of the following, lelling the turning point nd the -interept. = ( ) = ( + ) + = ( ) UNCRRECTED = Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

13 7 Chpter 7 Prols nd other grphs SLUTIN EXPLANATIN = ( ) 7 (, ) = ( + ) + (, ) Emple 5 Finding rule from simple grph In = ( h) + k, h = nd k =, so the verte is (, ). Sustitute = 0 to find the -interept: = (0 ) = 9 = 7 The grph is inverted sine =. h = nd k =, so the verte is (, ). When = 0: = (0 + ) + = + Find rule for this prol with turning point (0, ) nd nother point (, ). UNCRRECTED SLUTIN (, ) = + When =, = so: = () + = So = +. EXPLANATIN = In = ( h) + k, h = 0 nd k = so the rule is = +. We need = when =, so =. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

14 Numer nd Alger 75 Eerise 7B (½) (½) Emple Emple Give the oordintes of the turning point for the grphs of these rules. = = = + d = e = + 7 f = ( ) g = ( + 5) h = i = Sustitute = 0 to find the -interept of the grphs with these equtions. = + = = d = 6 e = ( ) f = ( + ) g = ( + ) h = ( 5) i = ( + ) j = ( + ) + k = ( ) l = ( + 7) Choose the word: left, right, up or down to suit. Compred with the grph of =, the grph of = + is trnslted. Compred with the grph of =, the grph of = ( ) is trnslted. Compred with the grph of =, the grph of = ( + ) is trnslted. d Compred with the grph of =, the grph of = 6 is trnslted. e Compred with the grph of =, the grph of = is trnslted. f Compred with the grph of =, the grph of = ( + ) is trnslted. g Compred with the grph of =, the grph of = ( ) is trnslted. h Compred with the grph of =, the grph of = + is trnslted. Sketh grphs of the following qudrtis, lelling the turning point nd the -interept. If the turning point is on the -is, lso lel point where = = = = d = e = + f = g = + h = i = ( + ) j = ( ) k = ( + ) l = ( ) 5 Stte the oordintes of the turning point for the grphs of these rules. = ( + ) + = ( + ) = ( ) + d = ( ) e = ( ) 5 f = ( ) + g = ( ) + h = ( ) + 6 i = ( + ) + j = ( ) 5 k = ( + ) l = ( ) 0 6 6(½) 6(½) Sketh grphs of the following qudrtis, lelling the turning point nd the -interept. = ( + ) + = ( + ) = ( + ) + d = ( ) + e = ( ) + f = ( ) g = ( ) + h = ( ) + i = ( + ) j = ( ) + k = ( ) l = ( + ) + 6(½) UNCRRECTED UNDERSTANDING FLUENCY Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

15 76 Chpter 7 Prols nd other grphs 7B Emple 5 7 8(½) 7 8(½), 9 7 Write the rule for eh grph when = is trnsformed the following. 8 d e f g h i refleted in the -is trnslted units to the left trnslted 5 units down trnslted units up trnslted unit to the right refleted in the -is nd trnslted units up refleted in the -is nd trnslted units left trnslted 5 units left nd units down trnslted 6 units right nd unit up Determine the rule for the following prols. d g (, 6) e h (, 5) (, ) f i 9 (, 7) (, ) 8(½), 9 UNCRRECTED PRBLEM-SLVING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

16 Numer nd Alger 77 9 The pth of sketll is given = ( 5) + 5, where metres is the height nd metres is the horizontl distne. d e Is the turning point mimum or minimum? Wht re the oordintes of the turning point? Wht is the -interept? Wht is the mimum height of the ll? Wht is the height of the ll t these horizontl distnes? i = ii = 7 iii = 0 0 Rell tht = ( h) + k nd = ( h) + k oth hve the sme turning point oordintes. Stte the oordintes of the turning point for the grphs of these rules. = ( ) = ( + ) = ( + ) d = e = 5 f = + 5 g = 6( + ) h = ( + ) + i = ( 5) + j = ( + ) + k = ( + ) 5 l = 5( ) Desrie the trnsformtions tht tke = to: = ( ) = ( + ) = d = + 7 e = f = ( + ) g = ( 5) + 8 h = ( + ) i = + 6 For = ( h) + k, write: the oordintes of the turning point Skething with mn trnsformtions 0(½) the -interept. 0 (½) Sketh the grph of the following, showing the turning point nd -interept. = ( ) + = ( + ) + 5 = ( ) + d = ( + ) e = ( ) + f = ( ) + g = h = i = 5 j = + ( ) k = ( + ) l = ( ) (½), UNCRRECTED PRBLEM-SLVING REASNING ENRICHMENT 7B Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

17 78 Chpter 7 Prols nd other grphs Using lultors to sketh prols Sketh the grph of the fmil = + k, using k = {,, 0,, }. Sketh grph of = nd show the -interepts nd the turning point. Using the TI-Nspire: In grphs nd geometr pge, tpe the rule in f () using the given smol (l) found in the smol plette. Enter the rule f () =. Chnge the sle using the window settings. Selet menu, Tre, Grph Tre nd shift left or right to show signifint points. Using the ClssPd: In the Grph&Tle pplition enter the rule = + {,, 0,, } followed EXE. Tp to see the grph. Enter the rule =. Tp 6 nd set n pproprite sle. Tp Anlsis, G-Solve, -Cl to lote the -interepts. Tp Anlsis, G-Solve, Min to lote the turning point. UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

18 Numer nd Alger 79 7C Skething prols using ftoristion A qudrti reltion written in the form = + + differs from tht of turning point form, = ( h) + k, nd so the trnsformtions of the grph of = to give = + + re less ovious. To ddress this, we hve numer of options. We n first tr to ftorise to find the -interepts nd then use smmetr to find the turning point or, lterntivel, we n omplete the squre nd epress the qudrti reltion in turning point form. The seond of these methods will e studied in the net setion. -interept -interepts turning point Let s strt: Wh does the turning point of = hve oordintes (, )? Ftorise =. Hene, find the -interepts. Disuss how smmetr n e used to lote the turning point. Hene, onfirm the oordintes of the turning point. To sketh grph of = + + : Find the -interept sustituting = 0. Find the -interept(s) sustituting = 0. Ftorise where possile nd use the null ftor lw. ne the -interepts re known, the turning point n e found using smmetr. The is of smmetr (lso the -oordinte of the turning point) lies hlfw etween the -interepts: = + 5 =. = UNCRRECTED 5 Sustitute this -oordinte into the rule to find the -oordinte of the turning point. Ke ides Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

19 80 Chpter 7 Prols nd other grphs Emple 6 Using the -interepts to find the turning point Sketh the grph of the qudrti = nd determine the oordintes of the turning point, using smmetr. SLUTIN -interept t = 0: = 5 -interepts t = 0: 0 = interepts re nd 5. 0 = ( 5)( ) 5 = 0 or = 0 = 5 or = Turning point t = + 5 =. = () 6 () + 5 = = Turning point is minimum t (, ). 5 5 (, ) EXPLANATIN Identif ke fetures of the grph: -interept (when = 0), -interepts (when = 0), then ftorise nd solve ppling the null ftor lw. Rell tht if = 0, then = 0 or = 0. Using smmetr the -oordinte of the turning point is hlfw etween the -interepts. Sustitute = into = to find the -oordinte of the turning point. It is minimum turning point sine the oeffiient of is positive. Lel ke fetures on the grph nd join points in the shpe of prol. UNCRRECTED Computer-ontrolled milling mhines like this n e progrmmed with equtions to rete omple shpes. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

20 Numer nd Alger 8 Emple 7 Skething perfet squre Sketh the grph of the qudrti = SLUTIN -interept t = 0: = 9 -interepts t = 0: 0 = = ( + ) + = 0 = -interept is. Turning point is t (, 0). 9 Emple 8 Finding turning point from grph EXPLANATIN For -interept sustitute = 0. For -interepts sustitute = 0 nd ftorise: ( + )( + ) = ( + ). Appl the null ftor lw to solve for. As there is onl one -interept, it is lso the turning point. Lel ke fetures on grph. The eqution of this grph is of the form = ( + )( + ). Use the -interepts to find the vlues of nd, then find the oordintes of the turning point. SLUTIN = nd =. = ( + )( ) -oordinte of the turning point is + UNCRRECTED =. -oordinte is ( + )( ) = ( ) =. Turning point is (, ). EXPLANATIN Using the null ftor lw, ( + )( ) = 0 gives = nd =, so = nd =. Find the verge of the two -interepts to find the -oordinte of the turning point. Sustitute = into the rule to find the vlue of the turning point. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

21 8 Chpter 7 Prols nd other grphs Eerise 7C (½), Emple 6 Emple 7 Use the null ftor lw to find the -interepts ( = 0) for these ftorised qudrtis. = ( + )( ) = ( )( ) = ( + )( + 5) d = ( ) e = ( 5) f = ( + ) g = ( + 5)( 5) h = ( + 7)( 7) i = ( + )( ) Ftorise nd use the null ftor lw to find the -interepts for these qudrtis. = + + = + 8 = d = e = 6 f = + 0 g = 9 h = 5 i = 0 Find the -interept for the qudrtis in Question. A prol hs the rule = 8. 5 Ftorise 8. Find the -interepts of = 8. Hene, stte the eqution of the is of smmetr. d Find the oordintes of the turning point. Sketh the grphs of the following qudrtis. = = 8 + = d = 6 6 e = 8 f = g = h = Sketh grphs of the following qudrtis. = = = + d = e = f = + + g = h = i = 5 j = + k = l = Sketh first finding -interepts. = + = + 6 = d = 5 e = + f = (½) Sketh grphs of the following perfet squres. = + + = = d = (½) 5 9(½) UNCRRECTED UNDERSTANDING FLUENCY 9 Sketh grphs of the following qudrtis tht inlude differene of perfet squres. = 9 = 6 = Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

22 Numer nd Alger 8 Emple 8 0 Determine the turning points of the following qudrtis. = ( 7 + 0) = ( 7 + 0) = d = + + e = ( 9) f = ( 9) g = 6 + h = i = + 0 The equtions of these grphs re of the form = ( + )( + ). Use the -interepts to find the vlues of nd, nd then find the oordintes of the turning point. 0(½), 5 Stte the -interepts nd turning point for these qudrtis. = = = 50 0 (½) 6 Sketh grph of these qudrtis. = 9 = = d = e = f = , (½), If the grph of = ( + )( ) psses through the point (, 6), determine the vlue of nd the oordintes of the turning point for this prol. 5 Eplin wh = ( )( 5) nd = ( )( 5) oth hve the sme -interepts. 6 Eplin wh = + hs onl one -interept. Eplin wh = + hs zero -interepts. 7 Consider the qudrtis = 8 nd = Show tht oth qudrtis hve the sme -interepts. Find the oordintes of the turning points for oth qudrtis. Compre the positions of the turning points. 8 A qudrti hs the rule = +. Give: the -interept the -interepts the oordintes of the turning point. 5 5, 6 UNCRRECTED 7, 8 PRBLEM-SLVING REASNING 7C Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

23 8 Chpter 7 Prols nd other grphs 7C More rules from grphs 9 9 Determine the eqution of eh of these grphs in ftorised form; for emple, = ( )( + ). d g j -9 (, ) (, ) e h k (, ) f i 6 9 (, 9) UNCRRECTED l ENRICHMENT Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

24 Numer nd Alger 85 7D Skething ompleting the squre We hve lernt previousl tht the turning point of prol n e red diretl from rule in the form = ( h) + k. This form of qudrti n e otined ompleting the squre. Let s strt: I forgot how to omplete the squre! ( ) 6 To mke + 6 perfet squre we need to dd 9 (from ) sine = ( + ). So to omplete ( ) 6 ( ) 6 the squre for we write = ( + ) 7. Disuss the rules for ompleting the squre nd eplin how eomes ( + ) 7. Wht does the turning point form of tell us out its grph? How n ou use the turning point form of to help find the -interepts of = + 6 +? B ompleting the squre, ll qudrtis in the form = + + n e epressed in turning point form; i.e. = ( h) + k. To sketh qudrti in the form = ( h) + k, follow these steps. Determine the oordintes of the turning point (h, k). When is positive, the prol hs minimum turning point. When is negtive, the prol hs mimum turning point. Determine the -interept sustituting = 0. Determine the -interepts, if n, sustituting = 0 nd ftorising to solve the eqution. Use the null ftor lw to help solve the eqution. Emple 9 Finding ke fetures of qudrtis in turning point form For = ( ) + 6: Determine the oordintes of its turning point nd stte whether it is mimum or minimum. Determine the -interept. UNCRRECTED Determine the -interepts (if n). SLUTIN EXPLANATIN Turning point is mimum t (, 6). For = ( h) + k the turning point is t (h, k). As = is negtive, the prol hs mimum turning point. -interept t = 0: = (0 ) + 6 = + 6 = -interept is. Sustitute = 0 to find the -interept. Rell tht (0 ) = ( ) =. Ke ides Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

25 86 Chpter 7 Prols nd other grphs -interepts t = 0: 0 = ( ) = ( ) 0 = ( ) () 0 = ( )( + ) 0 = ( )( + ) = 0 or + = 0 = or = -interepts re nd. Emple 0 Skething ompleting the squre Sustitute = 0 for -interepts. Divide oth sides. Use = ( )( + ) with = nd = to write in ftorised form. Simplif nd ppl the null ftor lw to solve for. Note: Chek tht the -interepts re evenl sped either side of the turning point. Alterntivel, solve ( ) = 0 ( ) = Sketh these grphs ompleting the squre, giving the -interepts in et form. = = SLUTIN Turning point form: = ( ) ( ) = = ( + ) + 6 Turning point is minimum t (, 6). -interept t = 0: = (0) + 6(0) + 5 = 5 -interept is 5. -interepts t = 0: 0 = ( + ) + 6 There is no solution nd there re no -interepts. 5 EXPLANATIN = ± =, To hnge the eqution into turning point form, omplete the squre dding nd ( ) 6 sutrting = 9. Red off the turning point, whih is minimum, s = is positive. For the -interept, sustitute = 0 into the originl eqution. For the -interepts, sustitute = 0 into the turning point form. This nnot e epressed s differene of perfet squres nd, hene, there re no ftors nd no -interepts. Note lso tht the turning point ws minimum with lowest -oordinte of 6, telling us there re no -interepts. Sketh the grph, showing the ke points. UNCRRECTED (, 6) Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

26 Numer nd Alger 87 Turning point form: = ( ) = ( ) + ( = ) ( ) Turning point is minimum t,. -interept t = 0: = (0) (0) = -interept is. -interepts t = 0: ( 0 = ) ( 0 = ( ) ) ( 0 = )( ) + = +, = ), ) + Complete the squre to write in turning point ( ) form: = 9 nd 9 = 9 =. Sustitute = 0 to find the -interept. Sustitute = 0 to find the -interepts. Ftorise using differene of perfet squres. = = Solve for using the null ftor lw: = 0 or + = 0 The solutions n e lso epressed s = ±. Lel ke fetures on grph, using et vlues. UNCRRECTED The stud of qudrti equtions, strted nient Greek mthemtiins looking t the properties of irles nd urves, remined prinipll prt of pure mthemtis until round 600, when it ws relised the ould e pplied to the desription of motion, inluding the orits of the plnets. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

27 88 Chpter 7 Prols nd other grphs Eerise 7D (½) (½) Emple 9 Emple 9 Emple 9 Cop nd omplete the squre, then stte the oordintes of the turning point (TP). = + 5 = + = + + = ( ) TP = (, ) = + + = ( ) TP = (, ) = d = 7 = = TP = (, ) = = TP = (, ) Solve these equtions for, using DPS nd giving et nswers. 9 = 0 = 0 ( ) = 0 d ( + ) 6 = 0 e ( + ) = 0 f ( 6) 5 = 0 5 Stte whether the turning points of the following re mimum or minimum nd give the oordintes. = ( ) + 5 = ( ) + = ( + ) d = 6( + ) 5 e = ( + 5) + 0 f = ( 7) + g = 5( ) + 8 h = ( ) 7 7(½) 8(½) Determine the -interept of eh of the following. = ( + ) + 5 = ( + ) 6 = ( ) d = ( ) 7 e = ( + 5) + 9 f = ( 7) 6 g = h = i = j = + 8 k = 5 + l = 5 Determine the -interepts (if n) of the following. = ( ) = ( + ) 9 = ( ) 6 d = ( + ) 0 e = ( ) + 0 f = ( 5) g = ( ) h = ( + 6) i = ( 7) + 8 j = ( ) k = ( ) + 5 l = ( ) + 0 8(½) UNCRRECTED 6 Determine the -interepts (if n) first ompleting the squre nd rewriting the eqution in turning point form. Give et nswers. = = = d = + 6 e = + f = 5 UNDERSTANDING FLUENCY Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

28 Numer nd Alger 89 Emple 0 Emple 0 7 Sketh the grphs of the following. Lel the turning point nd interepts. = ( ) = ( + ) 9 = ( + ) d = ( ) e = ( + 8) + 6 f = ( + 7) + g = ( ) + h = ( ) + 6 i = ( 5) j = ( + ) 9 k = ( + 9) + 5 l = ( ) Sketh these grphs ompleting the squre. Lel the turning point nd interepts. = + + = = d = e = 8 f = 5 g = h = i = + Sketh these grphs ompleting the squre. Lel the turning point nd interepts with et vlues. = + + = = + 6 d = e = + f = + g = h = i = (½) 0 Complete the squre nd deide if the grphs of the following qudrtis will hve zero, one or two -interepts. = + = + = d = e = + f = Tke out ommon ftor nd omplete the squre to find the -interepts for these qudrtis. = + 0 = + 9 = d = + 6 e = f = 6 + 9(½) 9 0(½) To sketh grph of the form = + + we n omplete the squre tking out ftor of. Here is n emple. = + 5 = ( + 5) ( ) ( ) = = (( + ) 6) = ( + ) + 6 So the turning point is mimum t (, 6). (½) (½), UNCRRECTED Sketh the grph of these qudrtis using the tehnique ove. = + = + + = + 6 d = e = 5 f = 5 +, FLUENCY PRBLEM-SLVING REASNING 7D Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

29 90 Chpter 7 Prols nd other grphs 7D For wht vlues of k will the grph of = ( h) + k hve: zero -interepts? one -interept? two -interepts? ( Show tht + + = + ). Completing the squre with non-monis 5 This emple shows how to omplete the squre with non-moni qudrtis of the form = + +. = Ke fetures: ( = + + ) The turning point is (, ). ( ( ) = ( ) ) -interept is interepts: ( = ( + ) ) 0 = ( + ) ( + ) = = ( + ) = ± Use this tehnique to sketh the grphs of these non-moni qudrtis. = = + 0 = + + d = + e = 7 + f = g = h = i = 5 + j = k = 9 + l = + 0 UNCRRECTED ne of the most importnt pplitions of qudrti equtions is in modelling elertion, first formulted Glileo in the erl 600s, nd shown in tion here in the present d. 5 REASNING ENRICHMENT Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

30 Numer nd Alger 9 7E Skething using the qudrti formul? 0A So fr we hve found -interepts for prols ftorising (nd using the null ftor lw) nd ompleting the squre. An lterntive method is to use the qudrti formul, whih sttes tht if + + = 0, then = ±. The disriminnt = determines the numer of solutions to the eqution. If = 0, then = 0. The solution to the eqution eomes =. There is one solution nd one -interept. -interept Let s strt: No working required If > 0, then > 0. The solution to the eqution eomes = ±. There re two solutions nd two -interepts. -interepts Three students set to work to find the -interepts for = + : Student A finds the interepts ftorising. Student B finds the interepts ompleting the squre. Student C uses the disriminnt in the qudrti formul. If < 0, then < 0. Squre roots eist for positive numers onl. There re no solutions or -interepts. Tr the method for student A. Wht do ou notie? Tr the method for student B. Wht do ou notie? Wht is the vlue of the disriminnt for student C? Wht does this tell them out the numer of -interepts for the qudrti? Wht dvie would student C give students A nd B? UNCRRECTED To sketh the grph of = + +, find the following points. -interept t = 0: = (0) + (0) + = -interepts when = 0: For 0 = + +, use the qudrti formul: = or + Alterntivel, = ±. Ke ides Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

31 9 Chpter 7 Prols nd other grphs Ke ides To determine if there re zero, one or two solutions/-interepts, use the disriminnt =. If < 0 no -interepts. If = 0 one -interept. If > 0 two -interepts. Turning point: The -oordinte lies hlfw etween the -interepts, so =. The -oordinte is found sustituting the -oordinte into the originl eqution. = is the is of smmetr. = + Emple Using the disriminnt nd using = to find the turning point Consider the prol given the qudrti eqution = Determine the numer of -interepts. Determine the -interept. Use = to determine the turning point. SLUTIN = = ( 6) ()(5) = < 0, so there re no -interepts. EXPLANATIN UNCRRECTED Use the disriminnt = to find the numer of -interepts. In 6 + 5, =, = 6 nd = 5. Interpret the result. -interept is 5. Sustitute = 0 for the -interept. = = ( 6) () = = () 6() + 5 = Turning point is t (, ). For the -oordinte of the turning point use = with = nd = 6, s ove. Sustitute the -oordinte into = to find the orresponding -oordinte of the turning point. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

32 Numer nd Alger 9 Emple Skething grphs using the qudrti formul Sketh the grph of the qudrti = +, lelling ll signifint points. Round the -interepts to two deiml ples. SLUTIN -interept is. -interepts ( = 0): + = 0 = ± = ± ()( ) () = ± 0 = ± 0 = ± 0 = 0.58,.58 (to d.p.) -interepts re 0.58,.58. Turning point is t = = () () = nd = ( ) + ( ) = 5. Turning point is t (, 5) (, 5) EXPLANATIN Identif ke fetures; i.e. - nd -interepts nd the turning point. Sustitute = 0 for the -interept. Use the qudrti formul to find the -interepts. For = +, =, = nd =. Simplif 0 = 0 = 0 = 0, then nel the ommon ftor of. Use lultor to round to two deiml ples. Sustitute = into = + to find the -oordinte of the turning point. Lel the ke fetures on the grph nd sketh. UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

33 9 Chpter 7 Prols nd other grphs Eerise 7E Emple Emple Emple A grph hs the rule = + +. Determine the numer of -interepts it will hve if: > 0 < 0 = 0 Give the et vlue of ± when: =, =, = =, =, = 5 =, =, = d =, = 6, = For the following grphs, stte whether the disriminnt of these qudrtis would e zero, positive or negtive (½) Use the disriminnt to determine the numer of -interepts for the prols given the following qudrtis. = + + = + 5 = + + d = e = + f = + 8 g = h = + 5 i = j = + 5 k = l = 5 + Determine the -interept for the prols given the following qudrtis. = + + = + 5 = + d = 5 e = f = g = + 8 h = i = 5 7 Use = to determine the oordintes of the turning points for the prols defined the following qudrtis. = + + = + = + d = + 6 e = + f = g = + h = i = 9 j = + k = l = 5 + 7(½) 7(½) UNCRRECTED UNDERSTANDING FLUENCY Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

34 Numer nd Alger 95 Emple 7 Sketh the grph of these qudrtis, lelling ll signifint points. Round the -interepts to two deiml ples. = = + 6 = d = 9 e = 8 f = g = h = + 7 i = j = + k = l = (½) 8 Sketh the grphs of these qudrtis. = = = 0 5 d = e = + 8 f = + 6 g = + + h = Give the et -interepts of the grphs of these prols. Simplif n surds. = 6 = + = d = + 6 e = f = Find rule in the form = + + tht mthes this grph. Write down two rules in the form = + + tht hve: two -interepts one -interept 8 9(½) 6, (, 6) no -interepts 8 9(½), Eplin wh the qudrti formul gives onl one solution when the disriminnt = 0. UNCRRECTED Write down the qudrti formul for moni qudrti equtions (i.e. where = ). Some proof Sustitute = into = + + to find the generl rule for the -oordinte of the turning point in terms of, nd. 5 Prove the formul = ± solving + + = 0. (Hint: Divide oth sides nd omplete the squre.),, 5 FLUENCY PRBLEM-SLVING REASNING ENRICHMENT 7E Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

35 96 Chpter 7 Prols nd other grphs 7F Ke ides Applitions involving prols Qudrti equtions nd their grphs n e used to solve rnge of prtil prolems. These ould involve, for emple, the pth of projetile or the shpe of ridge s rh. We n relte quntities with qudrti rules nd use their grphs to illustrte ke fetures. For emple, -interepts show where one quntit () is equl to zero, nd the turning point is where quntit is mimum or minimum. Let s strt: The ivil engineer Belind, ivil engineer, designs model for the urved le of 6 m suspension ridge using the eqution h = (d ) +, where h metres is the height of the hnging les ove the rod for distne d metres from the left pillr. Wht re the possile vlues for d? Sketh the grph of h = (d ) + for pproprite vlues of d. Wht is the height of the pillrs ove the rod? Wht is the minimum height of the le ove the rod? Disuss how ke fetures of the grph hve helped to nswer the questions ove. UNCRRECTED Appling qudrtis to solve prolems m involve: defining vriles forming equtions solving equtions deiding on suitle rnge of vlues for the vriles skething grphs showing ke fetures finding the mimum or minimum turning point. 6 m rod Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

36 Numer nd Alger 97 Emple Appling qudrtis in prolems A piee of wire mesuring 00 m in length is ent into the shpe of retngle. Let m e the width of the retngle. Use the perimeter to write n epression for the length of the retngle in terms of. Write n eqution for the re of the retngle (A m ) in terms of. Deide on the suitle vlues of. d Sketh the grph of A versus for suitle vlues of. e f Use the grph to determine the mimum re tht n e formed. Wht will e the dimensions of the retngle to hieve its mimum re? SLUTIN length + = 00 length = 00 Length = 50 EXPLANATIN P = 00 m m 00 m of wire will form the perimeter. Length is hlf of (00 width). A = (50 ) Are of retngle = length width. d e f Length nd width must e positive, so we require: > 0 nd 50 > 0 i.e. > 0 nd 50 > i.e. 0 < < 50 A (5, 65) 50 The mimum re tht n e formed is 65 m. Mimum ours when width = 5 m, so Length = 50 5 = 5 m Dimensions tht give mimum re re 5 m 5 m, whih is, in ft, squre. Require eh dimension to e positive, solve for. Sketh the grph, lelling the interepts nd turning point, whih hs -oordinte hlfw etween the -interepts; i.e. = 5. Sustitute = 5 into the re formul to find the mimum re: A = 5(50 5) = 65. UNCRRECTED Red from the grph. The mimum re is the -oordinte of the turning point. From turning point, = 5 gives the mimum re. Sustitute to find the orresponding length. Length = 50. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

37 98 Chpter 7 Prols nd other grphs Eerise 7F A ll is thrown upwrds from ground level nd rehes height of h metres fter t seonds, given the formul h = 0t 5t. Sketh grph of the rule for 0 t finding the t-interepts (-interepts) nd the oordintes of the turning point. Wht mimum height does the ll reh? How long does it tke the ll to return to ground level? The pth of jvelin thrown Jo is given the formul h = 6 (d 0) + 9, where h metres is the height of the jvelin ove the ground nd d metres is the horizontl distne trvelled. Sketh the grph of the rule for 0 d finding the interepts nd the oordintes of the turning point. Wht is the mimum height the jvelin rehes? Wht horizontl distne does the jvelin trvel? A wood turner rves out owl ording to the formul d = 7, where d m is the depth of the owl nd m is the distne from the entre of the owl. Sketh grph for 9 9, showing -interepts nd the turning point. Wht is the width of the owl? Wht is the mimum depth of the owl? The eqution for the rh of prtiulr ridge is given h = 500 ( 00) + 0, where h m is the height ove the se of the ridge nd m is the distne from the left side. d e, 5 6 UNCRRECTED Determine the oordintes of the turning point of the grph. Determine the -interepts of the grph. Rihmond Bridge, Tsmni, is the oldest Sketh the grph of the rh for pproprite vlues ridge in Austrli tht is still in use. of. Wht is the spn of the rh? Wht is the mimum height of the rh? 5 7 UNDERSTANDING FLUENCY Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

38 Numer nd Alger 99 UNCRRECTED FLUENCY Emple 5 A 0 m piee of wire is ent to form retngle. Let m e the width of the retngle. Use the perimeter to write n epression for the length of the retngle in terms of. Write n eqution for the re of the retngle (A m ) in terms of. Deide on suitle vlues of. d Sketh the grph of A versus for suitle vlues of. e Use the grph to determine the mimum re tht n e formed. f Wht will e the dimensions of the retngle to hieve its mimum re? 6 A frmer hs 00 m of fening to form retngulr pddok with river on one side (tht does not require fening), s shown. d e f Use the perimeter to write n epression for the length of the pddok in terms of the width, metres. Write n eqution for the re of the pddok (A m ) in terms of. river Deide on suitle vlues of. Sketh the grph of A versus for suitle vlues of. Use the grph to determine the mimum pddok re tht n e formed. Wht will e the dimensions of the pddok to hieve its mimum re? 7 The sum of two positive numers is 0 nd is the smller numer. Write the seond numer in terms of. Write rule for the produt, P, of the two numers in terms of. Sketh grph of P vs. d Find the vlues of when: i P = 0 ii P is mimum. e Wht is the mimum vlue of P? m 7F Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

39 500 Chpter 7 Prols nd other grphs 7F 8 The eqution for support spn is given h = 0 ( 0), where h m is the distne elow the se of ridge nd m is the distne from the left side. Determine the oordintes of the turning point of the grph. Sketh grph of the eqution using 0 0. d Wht is the width of the support spn? Wht is the mimum height of the support spn? 9 A rok is tossed from the top of 0 metre high liff nd its height (h metres) ove the se is given h = 0 5t, where t is in seonds. Find the et time it tkes for the rok to hit the wter. Sketh grph of h vs t for pproprite vlues of t. Wht is the et time it tkes for the rok to fll to height of 0 metres? 0 A ird dives into the wter to th fish. It follows pth given h = t 8t + 7, where h is the height in metres ove se level nd t is the time in seonds. d Sketh grph of h vs t, showing interepts nd the turning point. Find the time when the ird: i enters the wter ii eits the wter iii Wht is the mimum depth to whih the ird dives? At wht times is the ird t depth of 8 metres? FP 8, 9 8, 9 9, 0 rehes mimum depth. UNCRRECTED PRBLEM-SLVING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

40 Numer nd Alger 50 The height, h metres, of fling kite is given the rule h = t 6t + 0 for t seonds. Find the minimum height of the kite during this time. Does the kite ever hit the ground during this time? Give resons. The sum of two numers is 6. Show tht their produt hs mimum of 0. A retngulr frmed piture hs totl length nd width of 0 m nd 0 m, respetivel. The frme hs width m. Find the rule for the re (A m ) of the piture inside. Wht re the minimum nd mimum vlues of? Sketh grph of A vs using suitle vlues of. d Eplin wh there is no turning point for our grph, using suitle vlues of. e Find the width of the frme if the re of the piture is m. 0 m m 0 m A dolphin jumping out of the wter follows pth desried h = ( 0 + 6), where h is the vertil height, in metres, nd metres is the horizontl distne trvelled. How fr horizontll does the dolphin trvel out of the wter? Does the dolphin ever reh height of 5 metres ove wter level? Give resons. UNCRRECTED, REASNING 7F Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

41 50 Chpter 7 Prols nd other grphs 7F The highw nd the river nd the loed ll 5, 6 5 The pth of river is given the rule = ( 00) nd ll units re given in metres. 0 A highw is to e uilt ner or over the river on the line =. Sketh grph of the pth of the river, showing ke fetures. For the highw with eqution =, deide how mn ridges will need to e uilt if: i = 0 ii = 00 Lote the oordintes of the ridge, orret to one deiml ple, if: i = 00 ii = 0 d Desrie the sitution when = A tennis ll is loed from ground level nd must over horizontl distne of m if it is to lnd just inside the opposite end of the ourt. If the opponent is stnding m from the seline nd he n hit n ll less thn m high, wht is the lowest mimum height the lo must reh to win the point? m m proli pth of ll m tennis net UNCRRECTED ENRICHMENT Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

42 Numer nd Alger 50 8pt 7B Progress quiz Sketh grphs of the following qudrti reltions, lelling the turning point nd the -interept. (The -interepts re not required.) = = + = ( ) d = ( + ) 8pt 7B Find rule for this prol with turning point (0, ) nd nother point (, 5). 8pt 7C 8pt 7C 8pt 7D 8pt 7D 5 (, 5) Sketh grphs of the following qudrtis, lel the - nd -interepts nd determine the oordintes of the turning point, using smmetr. = = + The eqution of this grph is of the form = ( + )( + ). Use the -interepts to find the vlues of nd, then find the oordintes of the turning point. For = ( ) + 8 determine the: oordintes of the turning point nd stte whether it is mimum or minimum UNCRRECTED -interept -interepts (if n). 6 Sketh these grphs ompleting the squre to write the eqution in turning point form. Lel the et - nd -interepts nd turning point on the grph. = + = 6 Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

43 50 Chpter 7 Prols nd other grphs 8pt 7E 0A 8pt 7E 0A 8pt 7F For the prols given the following qudrti equtions: i ii Use the disriminnt to determine the numer of -interepts. Determine the -interept. iii Use = to determine the turning point. = + 5 = = 8 6 Sketh the grph of the qudrti = 8 + 5, lelling ll signifint points. Give the -interepts, rounded to two deiml ples. A frmer hs metres of fening to uild three sides of retngulr niml pen with the wll of the frm shed forming the other side. The width of the pen is metres. Write n eqution for the re of the pen (A m ) in terms of. Sketh the grph of A versus for suitle vlues of. Determine the mimum re tht n e formed nd stte the dimensions. UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

44 Numer nd Alger 505 7G Intersetion of lines nd prols 0A We hve seen previousl when solving pir of liner equtions simultneousl tht there is, t most, one solution. Grphill, this represents the point of intersetion for the two stright lines. When the lines re prllel there will e no solution; tht is, there is no point of intersetion. For the intersetion of prol nd line we n hve either zero, one or two points of intersetion. As we hve done for liner simultneous equtions, we n use the method of sustitution to solve liner eqution nd non-liner eqution simultneousl. Let s strt: How mn times does line ut prol? Use omputer grphing softwre to plot grph of =. B plotting lines of the form = h, determine how mn points of intersetion vertil line will hve with the prol. B plotting lines of the form =, determine how mn points of intersetion horizontl line ould hve with the prol. B plotting stright lines of the form = + k for vrious vlues of k, determine the numer of possile intersetions etween line nd prol. Stte some vlues of k for whih the line ove intersets the prol: twie never Cn ou find the vlue of k for whih the line intersets the prol etl one? When solving pir of simultneous equtions involving prol nd line, we m otin zero, one or two solutions. Grphill, this represents zero, one or two points of intersetion etween the prol nd the line. A line tht intersets urve twie is lled sent. A line tht intersets urve in etl one ple is lled tngent. UNCRRECTED zero points of intersetion one point of intersetion two points of intersetion The method of sustitution is used to solve the equtions simultneousl. Sustitute one eqution into the other. Rerrnge the resulting eqution into the form + + = 0. Solve for ftorising nd ppling the null ftor lw or use the qudrti formul = ±. Sustitute the vlues into one of the originl equtions to find the orresponding vlue. Ke ides Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

45 506 Chpter 7 Prols nd other grphs Ke ides After sustituting the equtions nd rerrnging, we rrive t n eqution of the form + + = 0. Hene, the disriminnt,, n e used to determine the numer of solutions (i.e points of intersetion) of the two equtions. zero solutions < 0 one solution = 0 Emple Finding points of intersetion of prol nd horizontl line Find n points of intersetion of these prols nd lines. = = + + = SLUTIN B sustitution: = = 0 ( )( + ) = 0 = 0 or + = 0 = or = The points of intersetion re t (, ) nd (, ). B sustitution: + + = = 0 = ± = ± () ()(6) () = ± 0 There re no points of intersetion. = EXPLANATIN two solutions > 0 Sustitute = from the seond eqution into the first eqution. Write in the form + + = 0 sutrting from oth sides. Ftorise nd ppl the null ftor lw to solve for. As the points re on the line =, the -oordinte of the points of intersetion is. UNCRRECTED Sustitute = into the first eqution. Appl the qudrti formul to solve = 0, where =, = nd = 6. 0 hs no rel solutions. The prol = + + nd the line = do not interset. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

46 Numer nd Alger 507 Emple 5 Solving simultneous equtions involving line nd prol Solve the following equtions simultneousl. = = + 6 = + = SLUTIN B sustitution: = = 0 ( ) = 0 = 0 or = 0 = 0 or = When = 0, = (0) = 0. When =, = () =. The solutions re = 0, = 0 nd =, =. B sustitution: + 6 = + 7 ( + )( + ) = = = = = 0 = + 7 = When = (, = ) + 7 = or 5 The onl solution is =, =. EXPLANATIN = Sustitute = into =. Rerrnge the eqution so tht it is equl to 0. Ftorise removing the ommon ftor. Appl the null ftor lw to solve for. Sustitute the vlues into = to otin the orresponding vlue. Alterntivel, the eqution = n e used to find the vlues or it n e used to hek the vlues. The points (0, 0) nd (, ) lie on oth the line = nd the prol =. Sustitute = + 7 into the first eqution. When rerrnging the eqution equl to 0, gther the terms on the side tht mkes the oeffiient of positive, s this will mke the ftorising esier. Hene, dd to oth sides, then dd to oth sides nd sutrt 6 from oth sides. Ftorise nd solve for. UNCRRECTED Sustitute the vlue into = + 7 (or = + 6 ut = + 7 is simpler eqution). Finding onl one solution indites tht this line is tngent to the prol. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

47 508 Chpter 7 Prols nd other grphs B sustitution: ( + ) = = = = = 0 ( + )( ) = 0 + = 0 or = 0 = or = ( When =, = = 9 + = 0 9 When =, = () + = ) + The solutions re =, = 0 9 nd =, =. Reple in = with +, mking sure ou inlude rkets. Epnd the rkets nd then rerrnge into the form + + = 0. Ftorise nd solve for. Sustitute the vlues into one of the two originl equtions to solve for. The line nd prol interset in two ples. UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph

48 Numer nd Alger 509 Emple 6 Solving simultneous equtions with the qudrti formul Solve the equtions = nd = simultneousl. Round our vlues to two deiml ples. SLUTIN B sustitution: = + 5 = 0 Using the qudrti formul: = ± () ()( 5) () = ± = ± 9 = or ( ± ) 9 In et form, = = = ± 9 The solutions re =.9, =.9 nd =.9, = 8.9 (to d.p.). EXPLANATIN Rerrnge into stndrd form. + 5 does not ftorise with whole numers. Qudrti formul: If + + = 0, then = ±. Here, =, = nd = 5. Use lultor to evlute 9 nd + 9. Rell tht if the numer under the squre root is negtive, then there will e no rel solutions. Sustitute et vlues into =. Round our vlues to two deiml ples, s required. Emple 7 Determining the numer of solutions of simultneous equtions Determine the numer of solutions (points of intersetion) of the following pirs of equtions. = + = + 8 = SLUTIN B sustitution: + = + + = 0 Using the disriminnt: = () ()() = = 0 There is one solution to the pir of equtions. = 5 EXPLANATIN UNCRRECTED ne the eqution is in the form + + = 0, the disriminnt n e used to determine the numer of solutions. Here, =, = nd =. Rell: > 0 mens two solutions. = 0 mens one solution. < 0 mens no solutions. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 0 ISBN Ph