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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#

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