A Math Mindmaps

TURNINGPoiNT-⑭ OFQUADRA TICFUNCTIONS i -Forgraphofy= ACxIh12IK, i l Form ·whena>o, y= ax2+ bx+c minturningpoint=(Fh,Ik) merisedForm ·whena<O, mapsectaroe maxturningpoint=(Th,Ik) e. g. y= 2(1-312-5 y= a(x= n)2=k solt: minturningpt=(3,-5) HOW?e. g. express2x2-6x+5intheformacx-h)+K. e. g. y=- 7(x+ 5)+ 87 ensurethatcoefficientof12is1. solution 2(x2-3xt) 1-maxturningpt=(- 5,8) 2Tocomplete((1bx1c= (x=2)=c- (2) 2[(x- )+ =- (- 2)] pationstocurveToLie Completely Aree = 2[(x- 2)+ I] ·whena>O= 2(x- 2)+I# · minimumvalueofcurveis ive 2Below-axis- ·whena<O ·maximumvalueofcurveisgive

SIQUADRA TICSEQUATIONS ⑰ - REOFROOTnddistinctroots1 ByFactorisation 2Bycompletingthesquare 3Byquadraticformulai.e. F 49c 7 > 2a b2- 49c(D)>0 CASE22equalandrealroots EQUADRA TICSIWEQUALITIES >x IensurethatonesideoftheequationisequaltoZERO 77 2Factorisequadraticexpression. b2- 4ac(D)= 0 3Sketchadiagram& sderequiredregion. CASE3Norealroots -> 7 ' as2+bxtc<O -> 7 P 9 P·1/11q I "II / 'Ils,, - & ⑧. b2- 4ac(D)<O Ia>O a<O graphisalwayspositiveigraphisalwaysSOl: andliecompletelyabovenegativeandlie- - x-axis. completelybelowP<x<q soltip<a Intersectionbtwline&curve x-axis AC2+bactc>0 ①Linemeetscurveatpoints...b2- 4aC>0 "unI IIIIIIP q Pq 2Linemeetscurveatpointclinetargenttocurves 9>0 a<O ..b2- 49= 0 SOl: SO17: 3Lineduesnotintersectcurve. - Sporx79 - P(x ..b" - 4acO

#OFSURDS 1 xB= 5 2NaxXa= a 3 =4PraI qua: (p19)a SMYINGSURDS -ALIZINGSURDS e. g. 150= 25x2 yourationalizewhenthedenominatorisinsurdform. = -25x52 HOW?multiplywithitsconjugatesurd. = 552# e. g. **=- 2 eg. IEE= as943 =2ax3 5(3)15 ast-e = 653X 2(1--2)-- =a(1)2-(WI)2 4-thatte53+ 21

DEGREEOFAPOL YNOMIAL -IInFRAa properfraction. - inthehighestpowerinthepolynomial - *degreef(ic)<degreeg(x)· An eisouxquotient+remainder, Ifdegreef(x)Idegreeg(),uselongdivisionto makeitproperfirst. REMAINDERTHEOREM ②checkwhetheryoucancompletelyfactoriseg(x). - f(x)dividedbyalineardivisor(c-a),thenf(a)=remainder. ③Expressproperalgebraicfractionsaccording tothecasesbelow. ETHEOREMf(x)dividedbyalinearfactor(c-al,thenf(a)= 0. CASEDenominatorg(x)FractionsPartialFractions adistinctlinearfactorstalandA+Asaremainderof12whenitisdividedby(ct2),findf(x)indescending 2Repeatedlinearfactors - e. G. Acubicequationf(x)=0hasroots- 3,-1&4.. Giventhatfix)leaves & caltb)2acctb(ax+b)2 powersof&theremainderwhenfix)isdividedbysc2-2x+1. uti= - 3,=- 1,x=4arerose 3Linear&quadrat,actiotealcan talac (c2+ d) ...f(x)=k(x+ 3)(x+1)(x- 4) ④solveforunknownconstantsbysubstitutingsuitablevalues Sincef(-2)=12 oforcomparingcoefficients. k(1)(- 1)- 6)=12 6k= 12 k= 2 CUBICEQUA TIONS :.f(x)= 2(1+33(1+1)(x- 4) IFindthefirstlinearfactorbytrialanderror. => 2(x+3)(x2- 3x- 4) = 2(x3-3x2- 4x+ 3x2- 9x-12) 2FindtheremainingquadraticfactorusingDIVISION => 2(x3- 13x- 12)= 2x3- 26x-24 3Cross-factorisethequadraticformula&solve. Bylongdivision: -zict1("sy *ifquadraticfactorcannotbefactorised, UseQUADRATICFORMULAtosolveit. - (2x3-4x2+212 - ae Thus,remainder=-20x-28#

MALTHERE+C(a)-(b)+C. Ca(b)+.... Cn(a)"- Ie- - 1stterm 1astterm RALTERMFormerre FACTORIAL #1andn!= nx(n-1)x(n-2)x... . x3x2 x 1 ...C= (v)= rhrt=itheerependentofIconstantterm => powerofe= 0· findthetermindependentofin(x+ct). solution ... 8 -2r= xo- GeneralFormula: (i)(x)(2)= (8) (x)()"(it)"8-2r= 0 =(2)(x) -() )- r= 4 =(2)()"(x)-(x-Y)subr=4intocoefficient. = )(x8-2) (4)(2)*= #. coefficientsincetermindependentofx=x

F INDICES -OFLOGARITHMS 9Mxa"= am+n 2 -. 109aK+109aY= 109a(y)<PRODUCT(Aw] am:a = am-n · loga-logay= 1)[QuoTIENT(Aw] Cam= amn · logqC= rlOgaK[PowERLAw] (ab)" q"b" riesop109a a= 1 (b)= En · 109a1= 0 90= 1 Ex= 109aY · Igy= 109,Y *a-= 1 Les · Iny=logeyan "Fire-logab=1gcb(changeofBASEm = anexam\** HS 109:9 i am= aA iexponential e. g. Solve~Y =9"where971 (a)109,x= 4log,5+3 (b)1095(5-4x)= log(2-x) e. g. Solve9"+2(3): 3**-12 y=qk solution solution ~ - - solution where02a1 log=(= 4[ ]+3 1095(5-4x)= [I⊥- 1 (34"+ 2(3")= (3(37-12 O >x 1095x= 4[10i]+ 3 1095(5-4x)= 1st(3"+ 2(3)= (3)(37-12logarithmic Letybelog](. 1095(5-4x)= -) Letybe3 14 y=109axwhere...y= 4(5)+3 1095(5-4x)= 2log=(2-) y2+ zy= qy- 12 a)1 y2= 4+ 3y 1095(5-4x)= 1095(2-x)2 y2- 3y- 4= 0 ..(5- 4x)= (2-x) y2- 7y+12= 0 (y-4)(y+1)= 0 5432= 4-4x+x2 >x x2- 1=0(y-3)(y- 4)= 0 0 I ! ..y= 4 ory=-1 (x-1)(x+1)= 0 y= 3ory= 4 I y=logakwhere1095)= 4;10957= 1 ..x= 1or1=- 1#3x= 3 3x= 4 0>AL1. x= 54 x= 5! # can3= 1n4x=1H ic= I= 1.26(357)# x= 625#x= 5 #

GRADIENTy i.YeN- straightline graphs ANDARDFORSee⑧ m= - x x,-72 wherev=radius;centre= (a,b)Ex,Y,) y,- Yz ⑧ yes anNigy+ILENGTH- -((,-x2)+(y,-yz) wherecentre=(-g,-f)⑭MIDPOINT -radius= g2+f2-c , 4) MALELLINES IfI,isparallelto12,thentheirgradientsareAL. IFM2/ MPERPENDICULARLINES - - Ife,isperpendicularto2, thentheproductofgradients--1. -M,xM2=- 1 ⊥RECTILINEARFIGURE anticlockwisedirection- 1,147I mustgoin- (x,Y ~mustendwiththestartingcoordinates. egy..." C2 CX,-In ↑↓Psy14↓I↓↓↓↓ I YzY3Y4Y, cniyal,Ye = ((,y+ x-Y+ x(zy4+xyy,)- (x,yy+)yy,+ )yy+ )y)]

STOTAKENOTE isvariablesXandYcannotcontainunknownconstantssuchas1andb.- ↑constantsmandcatcontainvariablesand1. SINGNowIWEARFUNUT=me -solutions TakeIgonbothsides. Igy= Ig(aby) Igy= Iga+ lgb" 1gy= dga+xlgb 1gy= 1gb(x)+ 19a⊥ ww un y= m XC e. g. Thediagramshowsast. lineobtainedbyplottingIagainst2 EXPress yintermsofC. Nationy> 13,6) ①Findgradient. M= !3)= E (-3,1) ②Findy-intercept. Y= mx+c ⊥x 6= =(3)+ c c= 5:Y= =X+ E ③subY=YandX= x 2, : = x2+ j= 2 :.y= 5+ 2

ITr i g o30°)45")60°special · onsAmplitudeacceofthemaximumandminimumvalueeH I -sin I - 2 Ang definit - COS- I · eg - 2 I W~rivdeasuredforonecompletecya2 tal H 1 ⑮5- - ⊥ERIGOGRAPHS Amplitude= 1 ⊥mentaryanglesA NEGRAS"for0013 Period=3600 sin(90"-E)= cost Max: 1: Min=- 1- COS190")= sint 18r> 1 x Range:-12Sin=7 tan190"-E)=ant amplitudealisoo oo K N -I-Aeveangles Iperiod sinc-O)=-sinb-oh tanc-R)=-tant x period= 360°y en 1- ...............................- - COS(-8)= WS S* b I EGRAY0 x 13 Amplitude= 1 AllpositiveSint=tre- · Fory= asin(bx)+< 7amplitudeI 2 Max= 1: Min= 1 - (2nd)(1st)>x y= a cos(bx)+c - abiszobo >Range:-11=1 (3rd)(4th) Lant=treCOSO= tre ⊥amplitudea E .................... period= 1 period Nant=in ⊥max-atc;min= -atC ENGRAPO0=3e Noamplitude· Fory= atan(bx)+C period= 1800 &Sect= 1 ⊥amplitude=NIL Nomax;No min period: 1800 Tx ③cosect= sino I ICOSO hL. iisn se ⊥: positive;graphshifts ④cott=tan= 8 wards - ⊥c=negative;graphshifts downwards-

SOLVINGTRIGOEQUATIONS 190°sinic1900,where-1[1=1 I ADDITIONFORMULAE- ①Expressequationintheformofsin,cosortan. -(i.e. sin(UE)=a;cos(n8)= a;tan(n=a) 00cos-1<180%where-1<x=1 · sin(A1B)=sinAcusBIcosAsinB ②Calculatethebasicangle,2.*usingpositivevalue · cos(AlB)= cosAcosBIsinAsinB& ofa. - 90°Stan-'<900,whereisanyrealnumber- · +an(AIB)= tanAltanB (i.e. X= sin'(a);x=cos- (a);x=tan-call Ak IFtanAtanB ③Identifywhichquadrantsitlies. e. g.Findtheprincipalvalueofsint'(-). TANCOS - sin30°= I iSINALL ·mo sin(-30%=- I IDOUBLEANGLEFORMULA:principalvalue=-30° · Sin2A= 2sinACOSA TRIGOEQUATIONS · COS2A=POS2A-sinA ④checktherangeofangles.(nE). Articl =1-2 sinA ⑤Findangle(no)andsolvefor8. =ICSA-1 E. g. Solvecosec(x-10% )= 2.5for0x1(270 -· tan2A= 2tanA 1-tan2A #11sinttior= e. g. GiventhatsinA=Iandtanx<o, sin(x-10%= 0.4 findthevalueof #21X= Sin20.4) ANDIDENTITIES (A)COS2A (b)cos#= 23.5780 liesin1st92ndquad. x3(a)COSCA: 1- 2SinA !- 100- 100260 ↓*R- FORMULAE ⊥ =1- 2[]= I&5/x -100= 23.578%180"- 23.578% · asindIbcost=Rsin(t!x) (b)cOS2/E): 2cOS(E)-1x= 33.578%166.4220 ·aCOSOIbsint=RCos(0FX) COSA= 2 cos(E)-1= 33.60,166. 4"(1dp) where R= +b=and<= tan-1( ) *2 cos(E)-1 I= 2c0s(() · Sin2&+CUS28=1 Howtodeterminemaximumandminimumvolume? to= cos(E)B· Ittan?t= sec2* ~ForasintIbcost, cos()=I(rei-Fo-28 =COSCC28 Maxvalue= Ab2,whensin(tIX)= 1. Minvalue=- +2,whensince1x=-1. = · ForacosObsint, Maxvalue= +b,whencosc8=x)= 1 Minvalue=- B,whencos(8IX)=- 1.