UNCORRECTED. Australian curriculum NUMBER AND ALGEBRA

Transcription

1 0A 0B 0C 0D 0E 0F 0G 0H Chpter Wht ou will lern Qudrti equtions (Etending) Solving + = 0 nd = d (Etending) Solving + + = 0 (Etending) Applitions of qudrti equtions (Etending) The prol Skething = with diltions nd refletions Trnsltions of = Skething prols using interept form (Etending) to qudrti equtions nd grphs 0Introdution Austrlin urriulum NUMBER AND ALGEBRA Liner nd non-liner reltionships Sketh simple non-liner reltions with nd without the use of digitl tehnologies UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

2 CSIR Prkes servtor The CSIR Prkes servtor in New South Wles is home to the Prkes Rdio Telesope, whih is fmous for providing the fi rst pitures to the world of the Apollo moonwlk in 969. The telesope s gignti dish hs dimeter of 6 metres nd reeives rdio nd mirowve signls from outer spe. The shpe of the dish is proli, mening tht its urvture n e desried qudrti eqution. The proli shpe gurntees tht ll the rdio nd mirowve signls oune off the dish surfe nd re refl eted to the reeiver, whih sits t speil point (lled the fous) ove the dish. The Prkes Rdio Telesope is still in use tod s prt of the Austrli Telesope Ntionl Filit. 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., 05 ISBN Ph

3 66 Chpter 0 Introdution to qudrti equtions nd grphs 0A Qudrti equtions EXTENDING Ke ides Qudrti equtions re ommonple in theoretil nd prtil pplitions of mthemtis. The re used to solve prolems in geometr nd mesurement s well s in numer theor nd phsis. The pth tht projetile tkes while fling through the ir, for emple, n e nlsed using qudrti equtions. A qudrti eqution n e written in the form + + = 0 where, nd re onstnts nd 0. Emples inlude + = 0, 5 The trjetor of this rrow n e modelled using qudrti = 0 nd equtions = 0. Unlike liner eqution whih hs single solution, qudrti equtions n hve zero, one or two solutions. For emple: = nd = re solutions to the qudrti eqution = 0 sine = 0 nd ( ) ( ) = 0. ne method for finding the solutions to qudrti equtions involves the use of the Null Ftor Lw where eh ftor of ftorised qudrti epression is equted to zero. Let s strt: Eploring the Null Ftor Lw = is not solution to the qudrti eqution = 0 sine 0. Use tril nd error to find t lest one of the two numers whih re solutions to = 0. Rewrite the eqution in ftorised form. = 0 eomes ( )( ) = 0 Now repet the first tsk ove to find solutions to the eqution using the ftorised form. Ws the ftorised form esier to work with? Disuss. UNCRRECTED A qudrti eqution n e written in the form + + = 0 This is lled stndrd form., nd re onstnts nd 0. The Null Ftor Lw sttes tht if the produt of two numers is zero then either or oth of the two numers is zero. If = 0 then = 0 or = 0. If ( + )( ) = 0 then + = 0 (so = ) or = 0 (so = ). + nd re the liner ftors of ( + )( ) (whih equls ). Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

4 Numer nd Alger 67 Emple Writing in stndrd form Write these qudrti equtions in stndrd form. = + 7 ( ) = 5 7 = SLUTIN = = 0 ( ) = 5 6 = = 0 7 = + 7 = 0 Emple Testing for solution EXPLANATIN We require the form + + = 0. Sutrt nd 7 from oth sides to move ll terms to the left-hnd side. First epnd rkets then sutrt 5 from oth sides. Add to oth sides. Sustitute the given vlue into the eqution nd s whether or not it is solution. + 6 = 0 ( = ) + 5 = 0 ( = ) SLUTIN + 6 = + 6 = 6 6 = 0 = is solution + 5 = ( ) + 5( ) = 6 + ( 0) = 0 = 9 = is not solution EXPLANATIN Sustitute = into the eqution to see if the left-hnd side equls zero. = stisfies the eqution so = is solution. UNCRRECTED Sustitute =. Rell tht ( ) = 6 nd 5 ( ) = 0. The eqution is not stisfied so = is not solution. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

5 68 Chpter 0 Introdution to qudrti equtions nd grphs Emple Using the Null Ftor Lw Use the Null Ftor Lw to solve these equtions. ( + ) = 0 ( )( + 5) = 0 ( )(5 + ) = 0 SLUTIN ( + ) = 0 = 0 or + = 0 = 0 or = ( )( + 5) = 0 = 0 or + 5 = 0 = or = 5 ( )(5 + ) = 0 = 0 or 5 + = 0 = or 5 = = or = 5 Eerise 0A EXPLANATIN The ftors re nd ( + ). Solve eh liner ftor equl to zero. Chek eh solution sustituting into the orginl eqution. Solve eh ftor ( ) nd ( + 5) equl to zero. Chek our solutions using sustitution. The two ftors re ( ) nd (5 + ). Eh one results in two-step liner eqution. Chek our solutions using sustitution. Evlute these qudrti epressions sustituting the vlue given in the rkets. + ( = ) 5 ( = ) + ( = ) d + ( = ) e ( = ) f + ( = ) g + ( = 5) h 5 + ( = ) i + ( = ) Deide if the following equtions re qudrtis. + = 0 = = 0 d 5 + = 0 e = 0 f + = 0 g 5 + = 0 h = + i + = Solve these liner equtions. = 0 + = 0 + = 0 d + = 0 e 9 = 0 f = 0 Cop nd omplete the following working whih uses the Null Ftor Lw. ( 5) = 0 ( + )( ) = 0 = 0 or = 0 = 0 or = (½) (½), (½) UNCRRECTED = 0 or = 0 = or = = or = UNDERSTANDING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

6 Numer nd Alger 69 Emple 5 Write these qudrti equtions in stndrd form ( + + = 0). + = 5 5 = = d = 7 + e ( + ) + = 0 f ( ) = g = h = i 5 = ( + 5) Emple Emple, Emple 6 Sustitute the given vlue into the qudrti eqution nd s whether or not it is solution. = 0 ( = ) 5 = 0 ( = 5) = 0 ( = ) d + = 0 ( = 0) e 9 = 0 ( = ) f + + = 0 ( = ) g = 0 ( = 5) h + = 0 ( = ) i 5 + = 0 ( = ) 7 Sustitute = nd = 5 into the eqution 0 = 0. Wht do ou notie? 8 Sustitute = nd = into the eqution = 0. Wht do ou notie? (½), 7, 9 0(½) 5 6(½), 8, 9 0(½) Use the Null Ftor Lw to solve these equtions. ( + ) = 0 ( + 5) = 0 ( ) = 0 d ( 7) = 0 e ( + )( ) = 0 f ( )( + ) = 0 ( g ( + 7)( ) = 0 + )( ) h = 0 i ( + 5) = 0 ( 5 ) ( j = 0 + ) k = 0 l ( + ) = 0 5 Use the Null Ftor Lw to solve these equtions. ( )( + ) = 0 ( + )( ) = 0 (5 + )( + ) = 0 d ( )( ) = 0 e ( + 5)(7 + ) = 0 f ( )(5 + ) = 0 g ( 7)( ) = 0 h ( + 9)( 7) = 0 i ( )(7 + ) = 0 Find the numers whih stisf the given ondition. d e, (½) The produt of nd numer more thn is zero. The produt of nd numer 7 less thn is zero. The produt of numer less thn nd numer more thn is zero. The produt of numer less thn twie nd 6 more thn is zero. The produt of numer more thn twie nd less thn twie is zero. 5 6(½), 8, 9 0(½) (½) UNCRRECTED FLUENCY PRBLEM-SLVING 0A Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

7 60 Chpter 0 Introdution to qudrti equtions nd grphs 0A Write these equtions s qudrtis in stndrd form. Remove n rkets nd frtions. 5 + = = ( ) = + d ( ) = ( ) e = f = ( ) g 5 + = h + 5 = These qudrti equtions hve two integer solutions etween 5 nd 5. Use tril nd error to find them. = 0 + = 0 = 0 d + = 0 e + = 0 f 6 = 0 Consider the qudrti eqution ( + ) = 0. Write the eqution in the form ( )( ) = 0. Use the Null Ftor Lw to find the solutions to the eqution. Wht do ou notie? Now solve these qudrti equtions. i ( + ) = 0 ii ( 5) = 0 iii ( ) = 0 iv (5 7) = 0 5 Consider the eqution ( )( + ) = 0. d First divide oth sides of the eqution. Write down the new eqution. Solve the eqution using the Null Ftor Lw. Compre the given originl eqution with the eqution found in prt. Eplin wh the solutions re the sme. Solve these equtions. i 7( + )( ) = 0 ii ( + ) = 0 iii ( + )( ) = 0 6 Consider the eqution ( ) + = 0., 5 UNCRRECTED Sustitute these vlues to deide if the re solutions to the eqution. i = ii = iii = 0 iv = Do ou think the eqution will hve solution? Eplin wh. 5, 6 PRBLEM-SLVING REASNING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

8 Numer nd Alger 6 Polnomils 7 Polnomils re sums of integer powers of. The re given nmes ording to the highest power of in the polnomil epression. Emple Polnomil Nme Constnt Liner Qudrti Cui Qurti Quinti Nme these polnomil equtions. = 0 + = = 0 d 5 + = 0 e + = 0 f 5 5 = + 8 Solve these polnomil equtions using the Null Ftor Lw. ( + )( )( + ) = 0 ( )( 5)( + ) = 0 ( )( + )(5 ) = 0 d ( + )(5 + )(7 + 0)( ) = 0 UNCRRECTED A CNC (omputer numerill-ontrolled) milling mhine uts mehnil omponents out of solid steel. The softwre m hve to solve thousnds of polnomils to ut omple shpes. 7, 8 ENRICHMENT 0A Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

9 6 Chpter 0 Introdution to qudrti equtions nd grphs 0B Solving + = 0 nd = d EXTENDING Ke ides When using the Null Ftor Lw we notie tht equtions must first e epressed s produt of two ftors. Hene, n eqution not in this form must first e ftorised. Two tpes of qudrti equtions re studied here. The first is of the form + = 0 where is ommon ftor nd the seond is of the form = d. Let s strt: Whih ftoristion tehnique? Rememer null mens zero. These two equtions m look similr ut the re not the sme: 9 = 0 nd 9 = 0. Disuss how ou ould ftorise eh epression on the left-hnd side of the equtions. How does the ftorised form help to solve the equtions? Wht re the solutions? Are the solutions the sme for oth equtions? B rerrnging 9 = 0 into = 9, n ou eplin how to get the two solutions from ove? When solving n eqution of the form 8 = 0 + = 0, ftorise tking out ommon ( ) = 0 ftors inluding. = 0 or = 0 = 0 or = When solving n eqution of the form = d, 5 = 0 divide oth sides nd then tke the squre = root of oth sides. = or = = d ould lso e rerrnged to d = 0 nd then ftorised s differene of perfet squres. Emple Solving when is ommon ftor Solve eh of the following equtions. + = 0 SLUTIN + = 0 ( + ) = 0 = 0 or + = 0 = 0 or = = 8 EXPLANATIN or = 0 ( + )( ) = 0 = or = UNCRRECTED Ftorise tking out the ommon ftor. Using the Null Ftor Lw, set eh ftor, nd ( + ), equl to 0. Solve for. Chek our solutions using sustitution. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

10 Numer nd Alger 6 = 8 8 = 0 ( ) = 0 = 0 or = 0 = 0 or = Emple 5 Solving equtions of the form = d Solve eh of these equtions. = 6 = 8 SLUTIN = 6 = or = Altentive solution: = 6 6 = 0 ( + )( ) = 0 + = 0 or = 0 = or = = 8 = 6 = 6 or = 6 Alterntive solution: = 8 8 = 0 ( 6) = 0 ( + 6)( 6) = = 0 or 6 = 0 = 6 or = 6 Mke the right-hnd side equl to zero sutrting 8 from oth sides. Ftorise tking out the ommon ftor of nd ppl the Null Ftor Lw to solve. EXPLANATIN Tke the squre root of oth sides. = or sine = 6 nd ( ) = 6. Rerrnge into stndrd form. Ftorise using = ( + )( ) then use the Null Ftor Lw to find the solutions. Divide oth sides nd then tke the squre root of oth sides. Sine 6 is not squre numer leve nswers in et form s 6 nd 6. UNCRRECTED Alterntivel, epress in stndrd form nd then note the ommon ftor of. Tret 6 = ( 6) s differene of squres. Appl the Null Ftor Lw nd solve for. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

11 6 Chpter 0 Introdution to qudrti equtions nd grphs Eerise 0B (½) (½) Emple Emple Emple 5 Emple 5 Write down the highest ommon ftor of these pirs of terms. nd 5 nd 0 nd 6 d 6 nd e nd f nd 7 g nd 6 h 9 nd 5 Ftorise these epressions full first tking out the highest ommon ftor d e f + 7 g h 5 5 i 6 + j 9 7 k 6 l Use the Null Ftor Lw to write down the solutions to these equtions. ( ) = 0 ( + ) = 0 7( + ) = 0 Solve eh of these equtions. + = = 0 + = 0 d 5 = 0 e 8 = 0 f = 0 g + = 0 h = 0 5 Solve these equtions first tking out the highest ommon ftor. 6 = 0 = = 0 d 6 8 = 0 e = 0 f 8 = (½) 8(½) Solve eh of the following equtions. = 5 = 0 = 6 d = 9 e = 8 f 7 = Solve eh of the following equtions. = 9 = 6 = 5 d = 0 e 8 = 0 f 00 = 0 Solve eh of the following equtions noting the ommon ftor first. 7 = 8 5 = 5 = 50 d 6 = e = f = 5 g 5 5 = 0 h 8 = 0 9 Rerrnge these equtions then solve them. = + 5 = 0 = 00 d = e = 0 f = 0 g = 8 h + = 7 i 9 = 8(½) UNCRRECTED 9 9, 0 0, UNDERSTANDING FLUENCY PRBLEM-SLVING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

12 Numer nd Alger 65 0 Remove rkets or frtions to help solve these equtions. = 0 6 = 0 = d 5( + ) = + 7 Write n eqution nd solve it to find the numer. e ( ) = + f (5 ) = (7 ) The squre of the numer is 7 times the sme numer. The differene etween the squre of numer nd 6 is zero. times the squre of numer is equl to times the numer. Consider the eqution + = 0. Eplin wh it nnot e written in the form ( + )( ) = 0. Are there n solutions to the eqution + = 0? Wh/Wh not? An eqution of the form + = 0 will lws hve two solutions if nd re not zero. Eplin wh one of the solutions will lws e = 0. Write the rule for the seond solution in terms of nd. More qudrti eqution forms Note for emple tht = 9 eomes = 9 nd then = or =, sine 9 =. Now solve these equtions. 9 = 6 5 = 6 = 00 d 8 5 = 0 e 6 = 0 f 9 + = 0 5 Note for emple tht ( ) = 0 eomes ( ) = with = or =. Now solve these equtions. ( ) = 9 ( + 5) = 6 ( + ) = d (5 ) 5 = 0 e ( ) 9 = 0 f ( 7) 00 = 0 UNCRRECTED Using CAS lultor 0B: Solving qudrti equtions This tivit is in the intertive tetook in the form of printle PDF.,, 5 PRBLEM-SLVING REASNING ENRICHMENT 0B Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

13 66 Chpter 0 Introdution to qudrti equtions nd grphs 0C Solving + + = 0 EXTENDING Ke ides Erlier in Chpter 8 we lernt to ftorise qudrti trinomils with three terms. For emple, ftorises to ( + )( + ). This mens tht the Null Ftor Lw n e used to solve equtions of the form + + = 0. Let s strt: Rememering how to ftorise qudrti trinomils First epnd these qudrtis using the distriutive lw: Distriutive lw ( + )( + d) = + d + + d. ( + )( + ) ( )( + ) ( 5)( ) Now ftorise these epressions Disuss our method for finding the ftors of eh qudrti ove. Solve qudrtis of the form + + = 0 8 = 0 ftorising the qudrti trinomil. ( 7)( + ) = 0 Ask Wht ftors of dd to give? 7 = 0 or + = 0 Then use the Null Ftor Lw. + + is lled moni qudrti = 7 or = sine the oeffiient of is. Perfet squres will give onl one solution = 0 Emple 6 Solving equtions with qudrti trinomils ( )( ) = 0 = 0 = Solve these qudrti equtions = 0 8 = = 0 SLUTIN = 0 ( + )( + ) = 0 + = 0 or + = 0 = or = EXPLANATIN nd 7 = 8 Ftors of whih dd to 7 re nd. =, + = = UNCRRECTED Use the Null Ftor Lw to solve the eqution. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

14 Numer nd Alger 67 8 = 0 ( )( + ) = 0 = 0 or + = 0 = or = = 0 ( 5)( ) = 0 5 = 0 or = 0 = 5 or = Ftors of 8 whih dd to re nd. = 8, + =, Emple 7 Solving with perfet squres nd other trinomils Solve these qudrti equtions = 0 = + 6 SLUTIN = 0 ( )( ) = 0 = 0 = = = 0 ( )( + ) = 0 = 0 or + = 0 = or = Eerise 0C Finish using the Null Ftor Lw. The ftors of 5 must dd to give 8. 5 ( ) = 5 nd 5 + ( ) = 8, so 5 nd re the two numers. EXPLANATIN Ftors of 6 whih dd to 8 re nd. ( )( ) = ( ) is perfet squre so there is onl one solution. First mke the right-hnd side equl zero sutrting nd 6 from oth sides. This is stndrd form. Ftors of 6 whih dd to re nd. Deide wht two ftors of the first numer dd to give the seond numer. 6, 5 8, 6, d 0, 7 e, f 5, g, h, Cop nd omplete the working to solve eh eqution = 0 = 0 ( + 5)( ) = = 0 or = 0 = or =, (½) UNCRRECTED ( 6)( ) = 0 6 = 0 or = 0 = or = + 5 = 0 d = 0 ( + 9)( ) = 0 ( 8)( ) = = 0 or = 0 8 = 0 or = 0 = or = = or = Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph UNDERSTANDING

15 68 Chpter 0 Introdution to qudrti equtions nd grphs 0C Emple 6 Emple 7 Emple 7 5 (½), 5(½) Solve these qudrti equtions = = = 0 d + 5 = 0 e + = 0 f = 0 g + = 0 h = 0 i 0 + = 0 j 5 = 0 k 6 6 = 0 l 5 = 0 m 0 + = 0 n = 0 o = 0 p + + = 0 q 6 7 = 0 r + 0 = 0 Solve these qudrti equtions whih inlude perfet squres = = = 0 d + + = 0 e = 0 f = 0 g + 6 = 0 h = 0 i = 0 Solve these qudrti equtions first rerrnging to stndrd form. = + 0 = 7 0 = 6 9 d = e 5 = f + 6 = 8 g = h 6 = 5 i 5 = 8 j 6 6 = k 6 = + 8 l 7 = 8 6 Solve these equtions tking out ommon ftor s the first ftorising step. = = 0 6 = 0 d 0 + = 0 e = 0 f = 0 g = 0 h + = 8 i 5 = 5 0 5(½) 7 Remove rkets, deimls or frtions nd write in stndrd form to help solve these equtions. = 5(.) ( ) = 9 ( + 0) = 5( + ) d = 5 + e = f = 8 Write down qudrti eqution in stndrd form whih hs the following solutions. = nd = = nd = = nd = d = nd = 0 e = 5 onl f = onl 9 The temperture in C inside room t sientifi se in Antrti fter 0:00 m is given the epression t 9t + 8 where t is in hours. Find the times in the d when the temperture is 0 C. 6(½), 9 6(½), (½), 9 UNCRRECTED FLUENCY PRBLEM-SLVING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

16 Numer nd Alger 69 0 Consider this eqution nd solution. Eplin wht is wrong with the solution. Find the orret solutions to the eqution. Eplin wh + = 0 onl hs one solution. 0 0, = 6 Write down qudrti eqution tht hs these solutions. Write in ftorised form. = onl = nd = Coeffiient of other thn, ( )( + ) = 6 = 6 or + = 6 = or = 9 Mn qudrti equtions hve oeffiient of not equl to. These re lled non-moni qudrtis. Ftoristion n e trikier ut is still possile. You m hve overed this in Chpter 8. Here is n emple. + 5 = = 0 ( ) + ( ) = 0 ( )( + ) = 0 = 0 or + = 0 = or = (ftors of ( ) = 6 tht dd to give 5 re nd 6.) (6 = 5) Solve these qudrti equtions = 0 + = = 0 d = 0 e = 5 7 f = 9 g = h 7 = 5 i 5 = 9 0 UNCRRECTED REASNING ENRICHMENT 0C Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

17 60 Chpter 0 Introdution to qudrti equtions nd grphs 0D Applitions of qudrti equtions EXTENDING Ke ides When using mthemtis to solve prolems we often rrive t qudrti eqution. In solving the qudrti eqution we otin the solutions to the prolem. Setting up the originl eqution nd then interpreting the solution re importnt prts of the prolem-solving proess. Surveors work on mn prolems tht involve qudrti equtions. Let s strt: Solving for the unknown numer The produt of positive numer nd 6 more thn the sme numer is 6. Using s the unknown numer, write n eqution desriing the given ondition. Solve our eqution for. Are oth solutions fesile (llowed)? Disuss how this method ompres to the method of tril nd error. UNCRRECTED When using qudrti equtions to solve prolems, follow these steps. Define our vrile. Write Let e Write n eqution desriing the sitution. Solve our eqution using the Null Ftor Lw. Chek tht our solutions re fesile. Some prolems m not llow solutions tht re negtive numers or frtions. Answer the originl question in words nd hek tht our nswer seems resonle. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

18 Numer nd Alger 6 Emple 8 Solving re prolems The length of ook is m more thn its width nd the re of the fe of the ook is 0 m. Find the dimensions of the fe of the ook. SLUTIN Let m e the width of the ook fe. Length = ( + ) m Are: ( + ) = = 0 ( + 0)( 6) = = 0 or 6 = 0 = 0 or = 6 = 6 sine > 0 width = 6 m nd length = 0 m Eerise 0D EXPLANATIN Define vrile for width then write the length in terms of the width. Write n eqution to suit the given sitution. Epnd nd sutrt 0 from oth sides. Solve using the Null Ftor Lw ut note tht width of 0 m is not fesile. Finish writing the dimensions; width nd length s required. Length = + = 6 + = 0 m,, This retngle hs n re of 8 m nd length whih is m more thn its width. Using length width = re, write n eqution. Solve our eqution epnding nd sutrting 8 from oth sides. Then use the Null Ftor Lw. Whih of our two solutions is fesile for the width of the retngle? d Write down the dimensions (width nd length) of the retngle. This retngle hs n re of m nd length whih is 5 m more thn its width. Using length width = re, write n eqution. Solve our eqution epnding nd sutrting from oth sides. Then use the Null Ftor Lw. Whih of our two solutions is fesile for the width of the retngle? d Write down the dimensions (width nd length) of the retngle. ( + ) m 8 m UNCRRECTED ( + 5) m m Solve these equtions for first epnding nd produing zero on the right-hnd side. ( + ) = 8 ( ) = 0 ( )( + ) = 6 m m UNDERSTANDING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

19 6 Chpter 0 Introdution to qudrti equtions nd grphs 0D Emple 8 The produt of numer nd more thn the sme numer is 8. Write n eqution nd solve to find the two possile solutions. 5 The produt of numer nd 7 less thn the sme numer is 60. Write n eqution nd solve to find the two possile solutions. 6 The produt of numer nd less thn the sme numer is 0. Write n eqution nd solve to find the two possile solutions , 7, 8 The length of retngulr rohure is 5 m more thn its width nd the re of the fe of the rohure is 6 m. Find the dimensions of the fe of the rohure. 8 The length of smll kindergrten pl re is 0 metres less thn its width nd the re is 69 m. Find the dimensions of the kindergrten pl re. 9 A squre of side length 0 metres hs squre of side length metres removed from one orner. Write n epression for the re remining fter the squre of side length metres is removed. Hint: use sutrtion. Find the vlue of if the re remining is to e 6 m. 0 An isoseles tringle hs height (h) equl to hlf its se length. Find the vlue of h if the re is 5 squre units. A retngulr frm shed ( m 5 m) is to e etended to form n L shpe s shown. Write n epression for the totl re of the etended frm shed. Find the vlue of if the totl re is to e 99 m. Use Pthgors theorem to find the vlue of in these right-ngled tringles , 0 9 m 0 m UNCRRECTED h m m m 5 m FLUENCY PRBLEM-SLVING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

20 Numer nd Alger 6 The eqution for the re of this retngle is + 8 = 0 whih hs solutions = 8 nd = 6. Whih solution do ou ignore nd wh? ( + ) m 8 m The produt of n integer nd one less thn the sme integer is 6. The eqution for this is 6 = 0. How mn different solutions re possile nd wh? 5 This tle shows the sum of the first n positive integers. If n = then the sum is + + = 6. m n 5 6 sum 6 Write the sum for n =, n = 5 nd n = 6. n(n + ) The epression for the sum is given. Use this epression to find the sum if: i n = 7 ii n = 0 Use the epression to find n for these sums. Write n eqution nd solve it. i sum = 5 ii sum = 0 6 The numer of digonls in polgon with n sides is given n (n ). Shown here re the digonls for qudrilterl nd pentgon., n Digonls 5 Use the given epression to find the two missing numers in the tle. Find the vlue of n if the numer of digonls is: i 0 ii 5 Piture frmes 7 A squre piture is to e edged with order of width m. The inside piture hs side length of 0 m. Write n epression for the totl re. Find the width of the frme if the totl re is to e 600 m. 0 m 8 A squre piture is surrounded retngulr frme s shown. The totl re is to e 0 m. Find the side length of the piture. 6 UNCRRECTED 6 m 8 m m 6 m 7, 8 8 m m REASNING ENRICHMENT 0D Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

21 6 Chpter 0 Introdution to qudrti equtions nd grphs Progress quiz 8pt 0A Et 8pt 0A Et 8pt 0A Et 8pt 0B Et 8pt 0B Et 8pt 0C Et 8pt 0C Et 8pt 0D Et 8pt 0D Et 8pt 0C Et Write these qudrti equtions in stndrd form. = 8 ( ) = ( 5) = 7 d ( ) = 5 Sustitute the given vlue into the eqution nd s whether or not it is solution. 6 = 0 ( = ) 6 = 0 ( = ) 9 = 0 ( = ) d 9 = 0 ( = ) Use the Null Ftor Lw to solve these equtions. ( 7) = 0 ( 5)( + ) = 0 ( + ) = 0 d ( )( ) = 0 5 Solve eh of the following equtions. + = 0 8 = 0 0 = 0 d 5 = 0 Solve eh of these equtions. = 9 = 0 = d 7 6 = 0 Solve these qudrti equtions. + + = = 0 5 = 0 d = 0 Solve these qudrti equtions = = = 0 d = 5 A retngulr lwn is m longer thn it is wide nd hs n re of 5 m. Write n eqution nd solve it to find the dimensions of this lwn. UNCRRECTED The produt of numer nd 5 more thn the sme numer is 8. Write n eqution nd solve to find the two possile vlues of the numer. Write down qudrti eqution in stndrd form whih hs the following solutions. = nd = = 6 nd = = nd = d = onl Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

22 Numer nd Alger 65 0E The prol Reltions tht hve rules in whih the highest power of is, suh s =, = or = +, re lled qudrtis nd their grphs re lled prols. Proli shpes n e seen in mn modern d ojets or situtions suh s the rhes of ridges, the pths of projetiles nd the surfe of refletors. Jets of wter ngled w from the vertil show proli trjetories. Let s strt: Finding fetures A qudrti is given the eqution =. Complete these tsks to disover its grphil fetures. Use the rule to omplete this tle of vlues. 0 Plot our points on op of the es shown t right nd join them to form smooth urve. Desrie these fetures: minimum turning point is of smmetr oordintes of the -interept oordintes of the -interepts 5 UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

23 66 Chpter 0 Introdution to qudrti equtions nd grphs Ke ides The grph of qudrti reltion is lled prol. Its si shpe is shown here. The si qudrti rule is =. The generl eqution of qudrti is = + +. A prol is smmetril out line lled the is of smmetr nd it hs one turning point ( verte), whih m e mimum or minimum. Here is n emple of prol with eqution = +, showing ll the ke fetures. Emple 9 Identifing fetures For this grph stte the: eqution of the is of smmetr d e tpe of turning point oordintes of the turning point -interepts -interept Ais of smmetr Verte (minimum turning point) Verte (mimum turning point) -interept Ais of smmetr -interepts Turning point (, ) Ais of smmetr = UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

24 Numer nd Alger 67 SLUTIN EXPLANATIN = mimum turning point turning point is (, 5) d -interepts: 5 nd e -interept: 5 Grph is smmetril out the vertil line =. Grph hs its highest -oordinte t the turning point, so it is mimum point. Highest point: turning point = (, 5) -interepts Ais of smmetr: = The trils of the projetiles shot out this firework n e modelled qudrti equtions. -interept UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

25 68 Chpter 0 Introdution to qudrti equtions nd grphs Emple 0 Plotting prol Use the qudrti rule = to omplete these tsks. Complete this tle of vlues. 0 Drw set of es using sle tht suits the numers in our tle. Then plot the points to form prol. Stte these fetures. i Tpe of turning point iii v SLUTIN Coordintes of the turning point The -interepts i Minimum turning point ii = 0 is the is of smmetr iii turning point (0, ) iv -interept: v -interepts: nd Eerise 0E Choose word from this list to omplete eh sentene. lowest, prol, verte, highest, interepts, zero ii iv Ais of smmetr The -interept EXPLANATIN Sustitute eh vlue into the rule to find eh vlue, e.g. =, = ( ) = 9 = 5. Plot eh oordinte pir nd join to form smooth urve. The turning point t (0, ) hs the lowest -oordinte for the entire grph. The vertil line = 0 divides the grph like mirror line. The -interept is t = 0. The -interepts re t = 0 on the -is. UNCRRECTED,, (½) (½) A mimum turning point is the point on the grph. The grph of qudrti is lled. The - re the points where the grph uts the -is. d The is of smmetr is vertil line pssing through the. e A minimum turning point is the point on the grph. f The -interept is t equls. Write down the eqution of vertil line (e.g. = ) tht psses through these points. (, 0) (, 5) (, ) d ( 5, 0) Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph (½) UNDERSTANDING

26 Numer nd Alger 69 Emple 9 For eh of the following grphs, stte: i the eqution of the is of smmetr ii the tpe of turning point (mimum or minimum) d g j iii the oordintes of the turning point v the -interept. 5 e h k iv the -interepts f i UNCRRECTED l UNDERSTANDING 0E Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

27 650 Chpter 0 Introdution to qudrti equtions nd grphs 0E Emple 0 Use the qudrti rule = to omplete these tsks. Complete the tle of vlues. 0 Drw set of es using sle tht suits the numers in our tle. Then plot the points to form prol. Stte these fetures. i Tpe of turning point ii Ais of smmetr iii Coordintes of the turning point iv The -interept v The -interepts 5 Use the qudrti rule = 9 to omplete these tsks. Complete the tle of vlues. 0 Drw set of es using sle tht suits the numers in our tle. Then plot the points to form prol. Stte these fetures. i Tpe of turning point ii Ais of smmetr iii Coordintes of the turning point iv The -interept v The -interepts 6 Use the qudrti rule = + to omplete these tsks. Complete the tle of vlues. 0 Drw set of es using sle tht suits the numers in our tle. Then plot the points to form prol. Stte these fetures. i Tpe of turning point ii Ais of smmetr iii Coordintes of the turning point iv The -interept v The -interepts 7 Use the qudrti rule = + + to omplete these tsks. Rell tht =. Complete the tle of vlues UNCRRECTED, 6, 7 Drw set of es using sle tht suits the numers in our tle. Then plot the points to form prol. Stte these fetures. i Tpe of turning point ii Ais of smmetr iii Coordintes of the turning point iv The -interept v The -interepts FLUENCY Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

28 Numer nd Alger 65 8 This grph shows the height of riket ll, metres, s funtion of time t seonds. i At wht times is the ll t height of 9 m? ii Wh re there two different times? i At wht time is the ll t its gretest height? ii iii Wht is the gretest height the ll rehes? After how mn seonds does it hit the ground? 9 The grph gives the height, h m, t time t seonds, of roket whih is fired up in the ir. From wht height is the roket lunhed? Wht is the pproimte mimum height tht the roket rehes? For how long is the roket in the ir? d Wht is the differene in time for when the roket is going up nd when it is going down? h A prol hs two -interepts t nd. The -oordinte of the turning point is. Wht is the eqution of its is of smmetr? Wht re the oordintes of the turning point? A prol hs turning point t (, ) nd one -interept t 0. Wht is the eqution of its is of smmetr? Wht is the other -interept? 8, 9 8, 0, 0 UNCRRECTED t t PRBLEM-SLVING 0E Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

29 65 Chpter 0 Introdution to qudrti equtions nd grphs 0E Write the rule for these prols. Use tril nd error to help nd hek our rule sustituting known point. d e f 5 6 Is it possile to drw prol with the following properties? If es, drw n emple. Two -interepts ne -interept No -interepts d No -interept Ml lultes the vlue for = using = + nd gets = 8. Eplin his error. Mi lultes the vlue for = using = nd gets = 6. Eplin her error. 5 This grph shows the prol for =. For wht vlues of is = 0? For wht vlue of is =? How mn vlues of give vlue whih is: i greter thn? ii equl to? iii less thn?, 5, 5, 6 UNCRRECTED PRBLEM-SLVING REASNING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

30 Numer nd Alger 65 6 This tle orresponds to the rule =. 0 Use this tle to solve these equtions i 0 = ii = d How mn solutions would there e to the eqution 8 =? Wh? How mn solutions would there e to the eqution =? Wh? How mn solutions would there e to the eqution =? Wh? Using softwre to onstrut prol 7 Follow the steps elow to onstrut prol using dnmi geometr pkge. Step. Show the oordinte es sstem seleting Show Aes from the Drw toolo. Step. Construt line whih is prllel to the -is nd psses through point F on the -is ner the point (0, ). Step. Construt line segment AB on this line s shown in the digrm. Step. Hide the line AB nd then onstrut: point C on the line segment AB A F point P on the -is ner the point (0, ). Step 5. Construt line whih psses through the point C nd is perpendiulr to AB. Step 6. Construt the point D whih is P equidistnt from point P nd segment AB. D Hint: use the perpendiulr isetor of PC. Step 7. Selet Tre from the Displ toolo nd lik on the point D. A F C Step 8. Animte point C nd oserve wht hppens. Step 9. Selet Lous from the Construt toolo nd lik t D nd then t C. Step 0. Drg point P nd/or segment AB P ( drgging F). (Cler the tre points D seleting Refresh drwing from the Edit menu.) Wht do ou notie? A F C UNCRRECTED B 7 B B REASNING ENRICHMENT 0E Using CAS lultor 0E: Skething prols This tivit is in the intertive tetook in the form of printle PDF. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

31 65 Chpter 0 Introdution to qudrti equtions nd grphs 0F Skething = with diltions nd refletions In geometr we know tht shpes n e trnsformed ppling refletions, rottions, trnsltions nd diltions (enlrgement). The sme tpes of trnsformtions n lso e pplied to grphs inluding prols. Altering the vlue of in = uses oth diltions nd refletions. Let s strt: Wht is the effet of? This tle nd grph show numer of emples of = with vring vlues of. The ould lso e produed using tehnolog. 0 = 0 = = 0 = 0 = = = = Disuss how the different vlues of ffet the -vlues in the tle. Disuss how the different vlues of ffet the shpe of the grph. How would the grphs of the following rules ompre to the grphs shown ove? = = = UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

32 Numer nd Alger 655 The eqution = desries fmil of prols inluding =, =, =, = et. The ontin the following fetures: the verte (or turning point) is (0, 0) the is of smmetr is = 0 if > 0 the grph is upright (or onve up) nd hs minimum turning point if < 0 the grph is inverted (or onve down) nd hs mimum turning point. If > or < : the grph ppers nrrower thn either = or =. For emple: = or = If < < : = = = = the grph ppers wider thn either = or =. For emple: = or = = = UNCRRECTED = = For = we s tht the grph of = is dilted from the -is ftor of. For = we s tht the grph of = is refleted in the -is. Ke ides Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

33 656 Chpter 0 Introdution to qudrti equtions nd grphs Emple Compring grphs of =, > 0 Complete the following for =, = nd =. Drw up nd omplete tle of vlues for. Plot their grphs on the sme set of es. Write down the eqution of the is of smmetr nd the oordintes of the turning point. d i Does the grph of = pper wider or nrrower thn =? ii Does the grph of = pper wider or nrrower thn =? SLUTIN = 0 0 = = 0 0 = = = is of smmetr: -is ( = 0) turning point: minimum t (0, 0) d i The grph of = ppers nrrower thn the grph of =. ii The grph of = ppers wider thn the grph of =. EXPLANATIN Sustitute eh vlue into =, = nd =. e.g. for =, if =, = () = () = 8 If =, = ( ) = () = Plot the points for eh grph using the oordintes from the tles nd join them with smooth urve. UNCRRECTED Look t grphs to see smmetr out the -is nd minimum turning point t the origin. For eh vlue of, is twie tht of ; hene, the grph (-vlues) of rises more quikl. For eh vlue of, is hlf tht of ; hene, the grph of rises more slowl. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

34 Numer nd Alger 657 Emple Compring grphs of =, < 0 Complete the following for =, = nd =. Drw up nd omplete tle of vlues for. Plot their grphs on the sme set of es. Write down the the eqution of the is of smmetr nd the oordintes of the turning point. d i Does the grph of = pper wider or nrrower thn =? ii Does the grph of = pper wider or nrrower thn =? SLUTIN = 0 0 = 0 0 = = = = is of smmetr: -is ( = 0) turning point: mimum t (0, 0) d i The grph of = ppers nrrower thn the grph of =. EXPLANATIN Sustitute eh -vlue into =, = nd =. e.g. for =, if =, = () = () = If =, = ( ) = () = Plot the oordintes for eh grph from the tles nd join them with smooth urve. UNCRRECTED ii The grph of = ppers wider thn the grph of =. Grphs re smmetril out the -is with mimum turning point t the origin. For eh vlue of, is three times tht of ; hene, the grph of gets lrger in the negtive diretion more quikl. For eh vlue of, is hlf tht of. Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

35 658 Chpter 0 Introdution to qudrti equtions nd grphs Eerise 0F Shown here re the grphs of =, =, =, =, = nd =. d e f Write the rules of the three grphs whih hve minimum turning point. = = = = = = Write the rules of the three grphs whih hve mimum turning point. Wht re the oordintes of the turning point for ll the grphs? Wht is the eqution of the is of smmetr for ll the grphs? Write the rule of the grph whih is: i upright (onve up) nd nrrower thn = ii upright (onve up) nd wider thn = iii inverted (onve down) nd nrrower thn = iv inverted (onve down) nd wider thn =. Write the rule of the grph whih is: i ii iii refletion of = in the -is refletion of = in the -is UNCRRECTED refletion of = in the -is. Selet the word positive or negtive to suit eh sentene. The grph of = will e upright (onve up) with minimum turning point if is. The grph of = will e inverted (onve down) with mimum turning point if is. Write the rule of grph whih is refletion in the -is of the grph of the rule =. Write the rule of grph whih is refletion in the -is of the grph of the rule =. Write the rule of grph whih is refletion in the -is of the grph of the rule = 5. d Write the rule of grph whih is refletion in the -is of the grph of the rule =. UNDERSTANDING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

36 Numer nd Alger 659 Emple Complete the following for =, = nd =. Emple Drw up nd omplete tle of vlues for. Plot their grphs on the sme set of es. Write down the oordintes of the turning point nd the eqution of the is of smmetr. d i Does the grph of = pper wider or nrrower thn =? ii Does the grph of = pper wider or nrrower thn =? 5 For the equtions given elow, omplete these tsks. 6 i Drw up nd omplete tle of vlues for. ii iii iv Plot the grphs of the equtions on the sme set of es. Write down the oordintes of the turning point nd the eqution of the is of smmetr. Determine whether the grphs of the equtions eh pper wider or nrrower thn the grph of =. = = 5 = d = 5 For the equtions given elow, omplete these tsks. i Drw up nd omplete tle of vlues for. ii iii iv Plot the grphs of the equtions on the sme set of es. List the ke fetures for eh grph, suh s the is of smmetr, turning point, -interept nd -interept. Determine whether the grphs of the equtions eh pper wider or nrrower thn the grph of =. = = 7 Here re eight qudrtis of the form =., 5 6(½), 5 6(½) = A = 6 B = 7 C = d 7 7, 8 = UNCRRECTED D = 9 E = F = 0. 7 G =.8 H = 0.5 Whih rule would give grph whih is upright (onve up) nd the nrrowest? Whih rule would give grph whih is inverted (onve down) nd the widest?, 6 7, 8 FLUENCY PRBLEM-SLVING 0F Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

37 660 Chpter 0 Introdution to qudrti equtions nd grphs 0F 8 Mth eh of the following prols with the pproprite eqution from the list elow. Do mentl hek sustituting the oordintes of known point. i = ii = iii = 5 d iv = v = 5 vi = 5 e The grph of = n e otined from = onduting these trnsformtions: refleting in the -is dilting ftor of from the -is. In the sme w s ove, desrie the two trnsformtions whih tke: = to = = to = 6 = to = d = to = e = to = f = to = f 5 9 9, 0 = Refletion Diltion = = 0 UNCRRECTED PRBLEM-SLVING REASNING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

38 Numer nd Alger 66 0 Write the rule for the grph fter eh set of trnsformtions. d The grph of = is refleted in the -is then dilted ftor of from the -is. The grph of = is refleted in the -is then dilted ftor of from the -is. The grph of = is refleted in the -is then dilted ftor of from the -is. The grph of = is refleted in the -is then dilted ftor of from the -is. The grph of = is refleted in the -is nd dilted from the -is given ftor. Does it mtter whih trnsformtion is ompleted first? Eplin. The grph of the rule = is refleted in the -is. Wht is the new rule of the grph? Sustitute to find the rule If rule is of the form = nd it psses through point, s (, ), we n sustitute this point to find the vlue of. So =, = () = nd = Use this method to determine the eqution of qudrti reltion if it hs n eqution of the form = nd psses through: (, 5) (, 7) (, ) d (, 7) e ( 5, ) f (, 6) g (, 80) h (, 5) (, ) This photo shows the proli les of the Golden Gte Bridge. The rule of the form = desries the shpe of the proli les. If the le is entred t (0, 0) nd the top of the right plon hs the oordintes (9, 67), find possile eqution tht desries this shpe. The numers given re in metres. UNCRRECTED, REASNING ENRICHMENT 0F Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

39 66 Chpter 0 Introdution to qudrti equtions nd grphs 0G Trnsltions of = Ke ides Added to refletion nd diltion is third tpe of trnsformtion lled trnsltion. This involves shift of ever point on the grph horizontll nd/or vertill. Unlike refletions nd diltions, trnsltion lters the oordintes of the turning point. The shpe of the urve is unhnged ut horizontl shift hnges the eqution of the is of smmetr. Let s strt: Whih w: left, right, up or down? This tle nd grph shows the qudrtis =, = ( ), = ( + ), = nd = +. The tle ould lso e produed using tehnolog. Disuss wht effet the different numers in the rules hd on the vlues in the tle. Also disuss wht effet the numers in the rules hve on eh grph. How re the oordintes of the turning point hnged? Wht onlusions ould ou drw on the effet of h in the rule = ( h)? Wht onlusions ould ou drw on the effet of k in the rule = + k? Wht if the rule ws = + or = ( + )? Desrie how the grphs would look. 0 = = ( ) = ( + ) = = = ( + ) A trnsltion of grph involves shift of ever point horizontll nd/or vertill. Vertil trnsltions: = + k If k > 0, the grph is trnslted k units up (red urve). If k < 0, the grph is trnslted k units down (green urve). The turning point is (0, k) for ll urves. The is of smmetr is the line = 0 for ll urves. The -interept is k for ll urves. = + = = = ( ) UNCRRECTED = + = = Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

40 Numer nd Alger 66 Horizontl trnsltions: = ( h) If h > 0, the grph is trnslted h units to the right. = 5 If h < 0, the grph is trnslted h units to the left. = = ( ) = ( + ) The turning point is (h, 0) in oth ses. The is of smmetr is the line = h in oth ses. The -interept is h in oth ses. The turning point form of qudrti is given : = ( h) + k > 0 upright (onve up) grph < 0 grph inverted (onve down) Ais of smmetr: = h Turning point = (h, k) k Trnsltes the grph left or right: h > 0 h < 0 Turning point (h, k) h To sketh grph of qudrti eqution in turning point form, follow these steps. Drw nd lel set of es. Identif importnt points inluding the turning point nd -interept. Sketh the urve onneting the ke points nd mking the urve smmetril. Trnsltes the grph up or down k > 0 k < 0 UNCRRECTED = ( ) (, ) Ke ides Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

41 66 Chpter 0 Introdution to qudrti equtions nd grphs Emple Skething with horizontl nd vertil trnsltions Sketh the grphs of these rules showing the -interept nd the oordintes of the turning point. = + = = ( ) d = ( + ) SLUTIN = + Turning point is (0, ) -interept: = (0) + = (0, ) = Turning point is (0, ) -interept: = (0) = = ( ) (0, ) Turning point is (, 0) -interept: = (0 ) 9 (, 0) = ( ) = 9 EXPLANATIN For = + k, k = so the grph of = is trnslted units up. The point (0, 0) shifts to (0, ). The grph of = is refletion of the grph of = in the -is. For = + k, k = so the grph of = is trnslted down unit. For = ( h), h = so the grph of = is trnslted units to the right. The -interept is found sustituting = 0 into the rule. UNCRRECTED Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

42 Numer nd Alger 665 d = ( + ) Turning point is (, 0) -interept: = (0 + ) (, 0) = = Emple Skething with omined trnsltions For = ( h), h = sine ( ( )) = ( + ). So the grph of = is trnslted units to the left. The -interept is found sustituting = 0. The negtive sign in front mens the grph is inverted. Sketh these grphs on the sme set of es showing the -interept nd the oordintes of the turning point. = ( ) = ( + ) + SLUTIN = ( ) Turning point is (, ) -interept: = (0 ) = = = ( + ) + Turning point is (, ) -interept: = (0 + ) + = 9 + = 7 (, ) (, ) EXPLANATIN For = ( h) + k, h = nd k = so the grph of = is shifted to the right nd down. Sustitute = 0 for the -interept. For = ( h) + k, h = nd k = so the grph of = is shifted to the left nd up. UNCRRECTED First, position the oordintes of the turning point nd -interept then join to form eh urve. 7 Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

43 666 Chpter 0 Introdution to qudrti equtions nd grphs Eerise 0G, This digrm shows the grphs of =, =, = ( + ) nd = ( ). Stte the turning point of the grph of: i = ( + ) ii = ( ) Stte the -interept for the grph of: i = ( + ) ii = ( ) Compred to the grph of =, whih w hs the grph of = ( + ) een trnslted (left or right)? d Compred to the grph of = whih w hs the grph of = ( ) een trnslted (left or right)? d Stte the -interept for the grph of: i = ii = + = ( + ) Compred to the grph of =, whih w hs the grph of = een trnslted (up or down)? Compred to the grph of =, whih w hs the grph of = + een trnslted (up or down)? = 9 = = ( ) This digrm shows the grphs of =, =, = nd = +. Stte the turning point of the grph of: = i = ii = + = = = + UNCRRECTED UNDERSTANDING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

44 Numer nd Alger 667 Emple, Emple, d Emple This digrm shows the grphs of =, =, = ( ) + nd = ( + ). Stte the turning point of the grph of: i = ( ) + ii = ( + ) Stte the -interept for the grph of: i = ( ) + ii = ( + ) Stte the missing words nd numers. i Compred to the grph of =, the grph of = ( + ) hs to e trnslted unit to the nd unit. ii = ( + ) (, ) = = (, ) = ( ) + Compred to the grph of =, the grphof = ( ) + hs to e trnslted units to the nd units. Sustitute = 0 to find the -interept for these rules. = + = = ( ) d = ( + 5) (½) 5 7(½) Sketh the grphs of these rules showing the -interept nd the oordintes of the turning point. = + = + = d = + e = + f = 5 Sketh the grphs of these rules showing the -interept nd the oordintes of the turning point. = ( ) = ( ) = ( + ) d = ( ) e = ( + ) f = ( + 6) Sketh eh grph showing the -interept nd the oordintes of the turning point. = ( ) + = ( ) = ( + ) d = ( + ) + 7 e = ( ) + f = ( 5) + g = ( + ) h = ( + ) 5 i = ( ) 6 5 7(½) UNCRRECTED UNDERSTANDING FLUENCY 0G Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

45 668 Chpter 0 Introdution to qudrti equtions nd grphs 0G 8 Mth eh of the following prols with the pproprite eqution from the list elow. i = ii = ( + ) iii = iv = v = ( ) vi = + vii = viii = ( + ) i = ( 5) = + i = ( + ) ii = + d g j 8 6 e h k , f i UNCRRECTED l PRBLEM-SLVING Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

46 Numer nd Alger Write the rule for eh grph in turning point form. 6 d (, ) (, ) e (, ) 6 (, ) f (, ) (, ) 0 A ike trk n e modelled pproimtel omining two different qudrti equtions. The first prt of the ike pth n e modelled the eqution = ( ) + 9 for 5. The seond prt of the ike trk n e modelled the eqution = ( 7) for 5 0. Find the turning point of the grph of eh qudrti eqution. Sketh eh grph on the sme set of es. n our sketh of the ike pth ou need to show the oordintes of the strt nd finish of the trk nd where it rosses the -is. Written in the form = ( h) + k, the rule = ( + ) ould e rerrnged to give = ( + ) +. Rerrnge these rules nd write in the form = ( h) + k. i = ( + ) ii = + ( + ) iii = + ( ) iv = 7 ( 5) v = vi = 6 + Write down the oordintes of the turning point for eh of the ove qudrtis. A qudrti hs the rule = ( h) + k. Wht re the oordintes of the turning point? Write n epression for the -interept., UNCRRECTED, PRBLEM-SLVING REASNING 0G Investigte nd eplin how the grph of: = ( ) ompres to the grph of = ( ) = ( ) ompres to the grph of = ( ). Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph

47 670 Chpter 0 Introdution to qudrti equtions nd grphs 0G Finding rules Find the eqution of the qudrti reltion whih is of the form = + nd psses through: (, ) (, 5) (, ) d (, ) 5 Find the eqution of the qudrti reltion whih is of the form = + nd psses through: (, ) (, ) (, 5) d (, 6) 6 Find the possile equtions of eh of the following qudrtis if their eqution is of the form = ( h) nd their grph psses through the point: (, 6) (, ) (, 9) d (, 9) 7 Find the rule for eh of these grphs with the given turning point (TP) nd -interept. TP = (, ), -int = 0 TP = (, 0), -int = TP = (, 0), -int = 9 d TP = (, ), -int = 7 e TP = (, ), -int = 5 f TP = (, 9), -int = 0 7 UNCRRECTED ENRICHMENT Unorreted rd smple pges Cmridge Universit Press Greenwood et l., 05 ISBN Ph