chapter 1- polynomial functions - jensenmath.ca 1 unit package teacher.pdf · 1.4 l3 factored form...
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Chapter1-Polynomial
Functions
LessonPackage
MHF4U
Chapter1OutlineUnitGoal:Bytheendofthisunit,youwillbeabletoidentifyanddescribesomekeyfeaturesofpolynomialfunctions,andmakeconnectionsbetweenthenumeric,graphical,andalgebraicrepresentationsofpolynomialfunctions.
Section Subject LearningGoals CurriculumExpectations
L1 PowerFunctions-describekeyfeaturesofgraphsofpowerfunctions-learnintervalnotation-beabletodescribeendbehaviour
C1.1,1.2,1.3
L2 CharacteristicsofPolynomialFunctions
-describecharacteristicsofequationsandgraphsofpolynomialfunctions-learnhowdegreerelatedtoturningpointsand𝑥-intercepts
C1.1,1.2,1.3,1.4
L3 FactoredFormPolynomialFunctions
-connecthowfactoredformequationrelatedto𝑥-interceptsofgraphofpolynomialfunction-givengraph,determineequationinfactoredform
C1.5,1.7,1.8
L4 TransformationsofPolynomialFunctions
-understandhowtheparameters𝑎, 𝑘, 𝑑,and𝑐transformpowerfunctions
C1.6
L5 SymmetryinPolynomialFunctions
-understandthepropertiesofevenandoddpolynomialfunctions C1.9
Assessments F/A/O MinistryCode P/O/C KTACNoteCompletion A P PracticeWorksheetCompletion F/A P
Quiz–PropertiesofPolynomialFunctions F P
PreTestReview F/A P Test-Functions O C1.1,1.2,1.3,1.4,1.5,1.6,1.7,
1.8,1.9 P K(21%),T(34%),A(10%),C(34%)
L1–1.1–PowerFunctionsLessonMHF4UJensenThingstoRememberAboutFunctions
• Arelationisafunctionifforevery𝑥-valuethereisonly1corresponding𝑦-value.Thegraphofarelationrepresentsafunctionifitpassestheverticallinetest,thatis,ifaverticallinedrawnanywherealongthegraphintersectsthatgraphatnomorethanonepoint.
• TheDOMAINofafunctionisthecompletesetofallpossiblevaluesoftheindependentvariable(𝑥)
o Setofallpossible𝑥-valesthatwilloutputreal𝑦-values
• TheRANGEofafunctionisthecompletesetofallpossibleresultingvaluesofthedependentvariable(𝑦)
o Setofallpossible𝑦-valueswegetaftersubstitutingallpossible𝑥-values
• Forthefunction𝑓 𝑥 = (𝑥 − 1)) + 3
o D: 𝑋 ∈ ℝ}
o R:{𝑌 ∈ ℝ|𝑦 ≥ 3}
• Thedegreeofafunctionisthehighestexponentintheexpressiono 𝑓 𝑥 = 6𝑥5 − 3𝑥) + 4𝑥 − 9hasadegreeof3
• AnASYMPTOTEisalinethatacurveapproachesmoreandmorecloselybutnevertouches.
Thefunction𝒚 = 𝟏
𝒙;𝟑hastwoasymptotes:
VerticalAsymptote:Divisionbyzeroisundefined.Thereforetheexpressioninthedenominatorofthefunctioncannotbezero.Thereforex≠-3.Thisiswhytheverticallinex=-3isanasymptoteforthisfunction.HorizontalAsymptote:Fortherange,therecanneverbeasituationwheretheresultofthedivisioniszero.Thereforetheliney=0isahorizontalasymptote.Forallfunctionswherethedenominatorisahigherdegreethanthenumerator,therewillbyahorizontalasymptoteaty=0.
PolynomialFunctionsApolynomialfunctionhastheform
𝑓 𝑥 = 𝑎>𝑥> + 𝑎>?@𝑥>?@ + 𝑎>?)𝑥>?) + ⋯+ 𝑎)𝑥) + 𝑎@𝑥@ + 𝑎B
• 𝑛Isawholenumber• 𝑥Isavariable• thecoefficients𝑎B, 𝑎@, … , 𝑎>arerealnumbers• thedegreeofthefunctionis𝑛,theexponentofthegreatestpowerof𝑥• 𝑎>,thecoefficientofthegreatestpowerof𝑥,istheleadingcoefficient• 𝑎B,thetermwithoutavariable,istheconstantterm• ThedomainofapolynomialfunctionisthesetofrealnumbersD: 𝑋 ∈ ℝ}• Therangeofapolynomialfunctionmaybeallrealnumbers,oritmayhavealowerboundoran
upperbound(butnotboth)• Thegraphofpolynomialfunctionsdonothavehorizontalorverticalasymptotes• Thegraphsofpolynomialfunctionsofdegree0arehorizontallines.Theshapesofothergraphs
dependsonthedegreeofthefunction.Fivetypicalshapesareshownforvariousdegrees:Apowerfunctionisthesimplesttypeofpolynomialfunctionandhastheform:
𝑓 𝑥 = 𝑎𝑥>• 𝑎isarealnumber• 𝑥isavariable• 𝑛isawholenumber
Example1:Determinewhichfunctionsarepolynomials.Statethedegreeandtheleadingcoefficientofeachpolynomialfunction.a)𝑔 𝑥 = sin 𝑥b)𝑓 𝑥 = 2𝑥Kc)𝑦 = 𝑥5 − 5𝑥) + 6𝑥 − 8d)𝑔 𝑥 = 3N
Thisisatrigonometricfunction,notapolynomialfunction.
Thisisapolynomialfunctionofdegree4.Theleadingcoefficientis2
Thisisapolynomialfunctionofdegree3.Theleadingcoefficientis1.
Thisisnotapolynomialfunctionbutanexponentialfunction,sincethebaseisanumberandtheexponentisavariable.
IntervalNotationInthiscourse,youwilloftendescribethefeaturesofthegraphsofavarietyoftypesoffunctionsinrelationtoreal-numbervalues.Setsofrealnumbersmaybedescribedinavarietyofways:1)asaninequality−3 < 𝑥 ≤ 52)interval(orbracket)notation(−3, 5]3)graphicallyonanumberlineNote:
• Intervalsthatareinfiniteareexpressedusing∞(infinity)or−∞(negativeinfinity)• Squarebracketsindicatethattheendvalueisincludedintheinterval• RoundbracketsindicatethattheendvalueisNOTincludedintheinterval• Aroundbracketisalwaysusedatinfinityandnegativeinfinity
Example2:Belowarethegraphsofcommonpowerfunctions.Usethegraphtocompletethetable.
PowerFunction
SpecialName Graph Domain Range
EndBehaviouras𝒙 →−∞
EndBehaviouras
𝒙 → ∞
𝒚 = 𝒙 Linear
(−∞,∞) (−∞,∞)
𝑦 → −∞
Startsinquadrant3
𝑦 → ∞
Endsinquadrant1
𝒚 = 𝒙𝟐 Quadratic
(−∞,∞) [0,∞)
𝑦 → ∞
Startsinquadrant2
𝑦 → ∞
Endsinquadrant1
𝒚 = 𝒙𝟑 Cubic
(−∞,∞) (−∞,∞)
𝑦 → −∞
Startsinquadrant3
𝑦 → ∞
Endsinquadrant1
PowerFunction
SpecialName Graph Domain Range
EndBehaviouras𝒙 →−∞
EndBehaviouras
𝒙 → ∞
𝒚 = 𝒙𝟒 Quartic
(−∞,∞) [0,∞)
𝑦 → ∞
Startsinquadrant2
𝑦 → ∞
Endsinquadrant1
𝒚 = 𝒙𝟓 Quintic
(−∞,∞) [−∞,∞)
𝑦 → −∞
Startsinquadrant3
𝑦 → ∞
Endsinquadrant1
𝒚 = 𝒙𝟔 Sextic
(−∞,∞) [0,∞)
𝑦 → ∞
Startsinquadrant2
𝑦 → ∞
Endsinquadrant1
KeyFeaturesofEVENDegreePowerFunctionsWhentheleadingcoefficient(a)ispositive Whentheleadingcoefficient(a)isnegative
Endbehaviour
as𝑥 → −∞, 𝑦 → ∞andas𝑥 → ∞,𝑦 → ∞
Q2toQ1
Endbehaviour
as𝑥 → −∞, 𝑦 → −∞andas𝑥 → ∞,𝑦 → −∞
Q3toQ4
Domain
(−∞,∞)
Domain (−∞,∞)
Range [0,∞) Range
[0, −∞)
Example:
𝑓 𝑥 = 2𝑥K
Example:𝑓 𝑥 = −3𝑥)
LineSymmetryAgraphhaslinesymmetryifthereisaverticalline𝑥 = 𝑎thatdividesthegraphintotwopartssuchthateachpartisareflectionoftheother.Note:Thegraphsofevendegreepowerfunctionshavelinesymmetryabouttheverticalline𝑥 = 0(they-axis).
KeyFeaturesofODDDegreePowerFunctionsWhentheleadingcoefficient(a)ispositive Whentheleadingcoefficient(a)isnegative
Endbehaviour
as𝑥 → −∞, 𝑦 → −∞and
as𝑥 → ∞,𝑦 → ∞
Q3toQ1
Endbehaviour
as𝑥 → −∞, 𝑦 → ∞andas𝑥 → ∞,𝑦 → −∞
Q2toQ4
Domain
(−∞,∞)
Domain (−∞,∞)
Range (−∞,∞) Range
(−∞,∞)
Example:
𝑓 𝑥 = 3𝑥[
Example:𝑓 𝑥 = −2𝑥5
PointSymmetryAgraphhaspointpointsymmetryaboutapoint(𝑎, 𝑏)ifeachpartofthegraphononesideof(𝑎, 𝑏)canberotated180°tocoincidewithpartofthegraphontheothersideof(𝑎, 𝑏).Note:Thegraphofodddegreepowerfunctionshavepointsymmetryabouttheorigin(0,0).
Example3:Writeeachfunctionintheappropriaterowofthesecondcolumnofthetable.Givereasonsforyourchoices.𝑦 = 2𝑥 𝑦 = 5𝑥^ 𝑦 = −3𝑥) 𝑦 = 𝑥_𝑦 = − )
[𝑥` 𝑦 = −4𝑥[ 𝑦 = 𝑥@B 𝑦 = −0.5𝑥b
EndBehaviour Functions Reasons
Q3toQ1
𝑦 = 2𝑥
𝑦 = 𝑥_
Oddexponent
Positiveleadingcoefficient
Q2toQ4𝑦 = −
25 𝑥
`
𝑦 = −4𝑥[
Oddexponent
Negativeleadingcoefficient
Q2toQ1𝑦 = 5𝑥^
𝑦 = 𝑥@B
Evenexponent
Positiveleadingcoefficient
Q3toQ4𝑦 = −3𝑥)
𝑦 = −0.5𝑥b
Evenexponent
Negativeleadingcoefficient
Example4:Foreachofthefollowingfunctionsi)Statethedomainandrangeii)Describetheendbehavioriii)Identifyanysymmetry
a)b)c)
𝒚 = −𝒙
𝒚 = 𝟎. 𝟓𝒙𝟐
𝒚 = 𝟒𝒙𝟑
i)Domain:(−∞,∞) Range:(−∞,∞) ii)As𝑥 → −∞,𝑦 → ∞andas𝑥 → ∞,𝑦 → −∞Thegraphextendsfromquadrant2to4iii)Pointsymmetryabouttheorigin(0,0)
i)Domain:(−∞,∞) Range:[0,∞) ii)As𝑥 → −∞,𝑦 → ∞andas𝑥 → ∞,𝑦 → ∞Thegraphextendsfromquadrant2to1iii)Linesymmetryabouttheline𝑥 = 0(they-axis)
i)Domain:(−∞,∞) Range:(−∞,∞) ii)As𝑥 → −∞,𝑦 → −∞andas𝑥 → ∞,𝑦 → ∞Thegraphextendsfromquadrant3to1iii)Pointsymmetryabouttheorigin(0,0)
L2–1.2–CharacteristicsofPolynomialFunctionsLessonMHF4UJensenInsection1.1welookedatpowerfunctions,whicharesingle-termpolynomialfunctions.Manypolynomialfunctionsaremadeupoftwoormoreterms.Inthissectionwewilllookatthecharacteristicsofthegraphsandequationsofpolynomialfunctions.NewTerminology–LocalMin/Maxvs.AbsoluteMin/MaxLocalMinorMaxPoint–Pointsthatareminimumormaximumpointsonsomeintervalaroundthatpoint.AbsoluteMaxorMin–Thegreatest/leastvalueattainedbyafunctionforALLvaluesinitsdomain.Investigation:GraphsofPolynomialFunctionsThedegreeandtheleadingcoefficientintheequationofapolynomialfunctionindicatetheendbehavioursofthegraph.Thedegreeofapolynomialfunctionprovidesinformationabouttheshape,turningpoints(localmin/max),andzeros(x-intercepts)ofthegraph.Completethefollowingtableusingtheequationandgraphsgiven:
Inthisgraph,(-1,4)isalocalmaxand(1,-4)isalocalmin.Thesearenotabsoluteminandmaxpointsbecausethereareotherpointsonthegraphofthefunctionthataresmallerandgreater.Sometimeslocalminandmaxpointsarecalledturningpoints.
Onthegraphofthisfunction…Thereare3localmin/maxpoints.2arelocalminand1isalocalmax.Oneofthelocalminpointsisalsoanabsolutemin(itislabeled).
EquationandGraph DegreeEvenorOdd
Degree?
LeadingCoefficient EndBehaviour
Numberofturningpoints
Numberofx-intercepts
𝑓 𝑥 = 𝑥$ + 4𝑥 − 5
𝑓 𝑥 = 3𝑥* − 4𝑥+ − 4𝑥$ + 5𝑥 + 5
𝑓 𝑥 = 𝑥+ − 2𝑥
𝑓 𝑥 = −𝑥* − 2𝑥+ + 𝑥$ + 2𝑥
𝑓 𝑥 = 2𝑥- − 12𝑥* + 18𝑥$ + 𝑥 − 10
𝑓 𝑥 = 2𝑥1 + 7𝑥* − 3𝑥+ − 18𝑥$ + 5
SummaryofFindings:
• Apolynomialfunctionofdegree𝑛hasatmost𝒏 − 𝟏localmax/minpoints(turningpoints)• Apolynomialfunctionofdegree𝑛mayhaveupto𝒏distinctzeros(x-intercepts)• Ifapolynomialfunctionisodddegree,itmusthaveatleastonex-intercept,andanevennumberofturning
points• Ifapolynomialfunctionisevendegree,itmayhavenox-intercepts,andanoddnumberofturningpoints
• Anodddegreepolynomialfunctionextendsfrom…o 3rdquadrantto1stquadrantifithasapositiveleadingcoefficiento 2ndquadrantto4thquadrantifithasanegativeleadingcoefficient
• Anevendegreepolynomialfunctionextendsfrom…o 2ndquadrantto1stquadrantifithasapositiveleadingcoefficiento 3rdquadrantto4thquadrantifishasanegativeleadingcoefficient
EquationandGraph DegreeEvenorOdd
Degree?
LeadingCoefficient EndBehaviour
Numberofturningpoints
Numberofx-intercepts
𝑓 𝑥 = 5𝑥1 + 5𝑥* − 2𝑥+ + 4𝑥$ − 3𝑥
𝑓 𝑥 = −2𝑥+ + 4𝑥$ − 3𝑥 − 1
𝑓 𝑥 = 𝑥* + 2𝑥+ − 3𝑥 − 1
Note:OdddegreepolynomialshaveOPPOSITEendbehaviours
Note:EvendegreepolynomialshaveTHESAMEendbehaviour.
Example1:Describetheendbehavioursofeachfunction,thepossiblenumberofturningpoints,andthepossiblenumberofzeros.Usethesecharacteristicstosketchpossiblegraphsofthefunctiona)𝑓 𝑥 = −3𝑥1 + 4𝑥+ − 8𝑥$ + 7𝑥 − 5ThedegreeisoddandtheleadingcoefficientisnegativesothefunctionmustextendfromQ2toQ4As𝑥 → −∞, 𝑦 → ∞As𝑥 → ∞, 𝑦 → −∞Thefunctioncanhaveatmost5𝑥-intercepts(1,2,3,4,or5)Thefunctioncanhaveatmost4turningpoints(0,2,or4)Possiblegraphsof5thdegreepolynomialfunctionswithanegativeleadingcoefficient:b)𝑔 𝑥 = 2𝑥* + 𝑥$ + 2Thedegreeisevenandtheleadingcoefficientispositivesothefunctionmustextendfromthesecondquadranttothefirstquadrant.As𝑥 → −∞, 𝑦 → ∞As𝑥 → ∞, 𝑦 → ∞Thefunctioncanhaveatmost4𝑥-intercepts(0,1,2,3,or4)Thefunctioncanhaveatmost3turningpoints(1,or3)Possiblegraphsof4thdegreepolynomialfunctionswithapositiveleadingcoefficient:Example2:Filloutthefollowingchart
Degree Possible#of𝒙-intercepts Possible#ofturningpoints1 1 02 0,1,2 13 1,2,3 0,24 0,1,2,3,4 1,35 1,2,3,4,5 0,2,4
Note:Odddegreefunctionsmusthaveanevennumberofturningpoints.
Note:Evendegreefunctionsmusthaveanoddnumberofturningpoints.
Example3:Determinethekeyfeaturesofthegraphofeachpolynomialfunction.Usethesefeaturestomatcheachfunctionwithitsgraph.Statethenumberof𝑥-intercepts,thenumberoflocalmax/minpoints,andthenumberofabsolutemax/minpointsforthegraphofeachfunction.Howarethesefeaturesrelatedtothedegreeofeachfunction?a)𝑓 𝑥 = 2𝑥+ − 4𝑥$ + 𝑥 + 1b)𝑔 𝑥 = −𝑥* + 10𝑥$ + 5𝑥 − 4c)ℎ 𝑥 = −2𝑥1 + 5𝑥+ − 𝑥d)𝑝 𝑥 = 𝑥- − 16𝑥$ + 3a)Thefunctioniscubicwithapositiveleadingcoefficient.ThegraphextendsfromQ3toQ1.They-interceptis1.Graphiv)correspondstothisequation.Thereare3x-interceptsandthedegreeis3.Thefunctionhasonelocalmaxandonelocalmin,whichisatotaloftwoturningpoints,whichisonelessthanthedegree.Thereisnoabsolutemaxorminpoint.b)Thefunctionisquarticwithanegativeleadingcoefficient.Thegraphextendsfromquadrant3to4.They-interceptis-4.Graphi)correspondstothisequation.Thereare4x-interceptsandthedegreeis4.Thefunctionhastwolocalmaxandonelocalmin,whichisatotalof3turningpoints,whichisonelessthanthedegree.Thefunctionhasoneabsolutemaxpoint.c)Thefunctionisquinticwithanegativeleadingcoefficient.Thegraphextendsfromquadrant2to4.They-interceptis0.Graphiii)correspondstothisequation.Thereare5x-interceptsandthedegreeis5.Thefunctionhastwolocalmaxandtwolocalmin,whichisatotalof4,whichisonelessthanthedegree.Thefunctionhasnoabsolutemaxorminpoints.d)Thefunctionisdegree6withapositiveleadingcoefficient.Thegraphextendsfromquadrant2to1.They-interceptis3.Graphii)correspondstothisequation.Thereare4x-interceptsandthedegreeis6.Thefunctionhastwolocalmaxandonelocalmin,whichisatotalof3,whichisthreelessthanthedegree.Thefunctionhastwoabsoluteminpoints.FiniteDifferencesForapolynomialfunctionofdegree𝑛,where𝑛isapositiveinteger,the𝑛?@differences…
• areequal• havethesamesignastheleadingcoefficient• areequalto𝑎 ∙ 𝑛!,where𝑎istheleadingcoefficient
Note:𝑛!isreadas𝑛factorial.𝑛! = 𝑛×(𝑛 − 1)×(𝑛 − 2)×…×2×15! = 5×4×3×2×1 = 120
Example4:Thetableofvaluesrepresentsapolynomialfunction.Usefinitedifferencestodetermine
a) thedegreeofthepolynomialfunctionb) thesignoftheleadingcoefficientc) thevalueoftheleadingcoefficient
a)Thethirddifferencesareconstant.So,thetableofvaluesrepresentsacubicfunction.Thedegreeofthefunctionis3.b)Theleadingcoefficientispositivesince6ispositive.c) 6 = 𝑎 ∙ 𝑛! 6 = 𝑎 ∙ 3! 6 = 𝑎 ∙ 6 -
-= 𝑎
𝑎 = 1Therefore,theleadingcoefficientofthepolynomialfunctionis1.Example5:Forthefunction2𝑥* − 4𝑥$ + 𝑥 + 1whatisthevalueoftheconstantfinitedifferences?𝐹𝑖𝑛𝑖𝑡𝑒𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒𝑠 = 𝑎 ∙ 𝑛! = 2 ∙ 4! = 2 ∙ 24 = 48Therefore,thefourthdifferenceswouldallbe48forthisfunction.
𝒙 𝒚 Firstdifferences Seconddifferences Thirddifferences
-3 -36 -2 -12 −12 − −36 = 24 -1 -2 −2 − −12 = 10 10 − 24 = −14 0 0 0 − −2 = 2 2 − 10 = −8 −8 − −14 = 61 0 0 − 0 = 0 0 − 2 = −2 −2 − −8 = 62 4 4 − 0 = 4 4 − 0 = 4 4 − −2 = 63 18 18 − 4 = 14 14 − 4 = 10 10 − 4 = 64 48 48 − 18 = 30 30 − 14 = 16 16 − 10 = 6
L3–1.3–FactoredFormPolynomialFunctionsLessonMHF4UJensenInthissection,youwillinvestigatetherelationshipbetweenthefactoredformofapolynomialfunctionandthe𝑥-interceptsofthecorrespondinggraph,andyouwillexaminetheeffectofrepeatedfactoronthegraphofapolynomialfunction.FactoredFormInvestigationIfwewanttographthepolynomialfunction𝑓 𝑥 = 𝑥$ + 3𝑥' + 𝑥( − 3𝑥 − 2accurately,itwouldbemostusefultolookatthefactoredformversionofthefunction:𝑓 𝑥 = 𝑥 + 1 ( 𝑥 + 2 𝑥 − 1 Letsstartbylookingatthegraphofthefunctionandmakingconnectionstothefactoredformequation.
Graphof𝑓(𝑥): Fromthegraph,answerthefollowingquestions… a)Whatisthedegreeofthefunction? b)Whatisthesignoftheleadingcoefficient? c)Whatarethe𝑥-intercepts? d)Whatisthe𝑦-intercept?
e)The𝑥-interceptsdividethegraphintointofourintervals.Writetheintervalsinthefirstrowofthetable.Inthesecondrow,chooseatestpointwithintheinterval.Inthethirdrow,indicatewhetherthefunctionispositive(abovethe𝑥-axis)ornegative(belowthe𝑦-axis).Interval (−∞,−2) (−2,−1) (−1, 1) (1,∞)
TestPoint
𝑓 −3 = −3 + 1 ( −3 + 2 −3 − 1 = −2 ((−1)(−4)= 16
𝑓 −1.5 = −1.5 + 1 ( −1.5 + 2 −1.5 − 1 = −0.5 ((0.5)(−2.5)= −0.3125
𝑓 0 = 0 + 1 ( 0 + 2 0 − 1 = 1 ((2)(−1)= −2
𝑓 3 = 3 + 1 ( 3 + 2 3 − 1 = 4 ((5)(2)= 160
Signof𝒇(𝒙) + − − +f)Whathappenstothesignoftheof𝑓(𝑥)neareach𝑥-intercept?
Thehighestdegreetermis𝑥$,thereforethefunctionisdegree4(quartic)
Theleadingcoefficientis1,thereforetheleadingcoefficientisPOSITIVE
The𝑥-interceptsare(−2, 0)oforder1,(−1, 0)oforder2,and(1, 0)oforder1
The𝑦-interceptisthepoint(0,−2)
At(-2,0)whichisorder1,itchangessignsAt(-1,0)whichisorder2,thesigndoesNOTchangeAt(1,0)whichisorder1,itchangessigns
Conclusionsfrominvestigation:The𝑥-interceptsofthegraphofthefunctioncorrespondtotheroots(zeros)ofthecorrespondingequation.Forexample,thefunction𝑓 𝑥 = (𝑥 − 2)(𝑥 + 1)has𝑥-interceptsat2and-1.Thesearetherootsoftheequation 𝑥 − 2 𝑥 + 1 = 0.Ifapolynomialfunctionhasafactor(𝑥 − 𝑎)thatisrepeated𝑛times,then𝑥 = 𝑎isazeroofORDER𝑛.Order–theexponenttowhicheachfactorinanalgebraicexpressionisraised.Forexample,thefunction𝑓 𝑥 = 𝑥 − 3 ((𝑥 − 1)hasazeroofordertwoat𝑥 = 3andazerooforderoneat𝑥 = 1.Thegraphofapolynomialfunctionchangessignatzerosofoddorderbutdoesnotchangesignatzerosofevenorder.Shapesbasedonorderofzero:
𝑓 𝑥 = 0.01(𝑥 − 1) 𝑥 + 2 ((𝑥 − 4)'
ORDER2(-2,0)isan𝑥-interceptoforder2.Therefore,itdoesn’tchangesign.“Bouncesoff”𝑥-axis.Parabolicshape.
ORDER1(1,0)isan𝑥-interceptoforder1.Therefore,itchangessign.“Goesstraightthrough”𝑥-axis.LinearShape
ORDER3(4,0)isan𝑥-interceptoforder3.Therefore,itchangessign.“S-shape”through𝑥-axis.Cubicshape.
Three𝑥-interceptsoforder1,sotheleastpossibledegreeis3.ThegraphgoesfromQ2toQ4sotheleadingcoefficientisnegative.
The𝑥-interceptsare-5,0,and3.Thefactorsare(𝑥 + 5),𝑥,and(𝑥 − 3)
Two𝑥-interceptsoforder1,andone𝑥-interceptoforder2,sotheleastpossibledegreeis4.ThegraphgoesfromQ2toQ1sotheleadingcoefficientispositive.
The𝑥-interceptsare-2,1,and3.Thefactorsare(𝑥 + 2),(𝑥 − 1),and(𝑥 − 3)(
Example1:AnalyzingGraphsofPolynomialFunctionsForeachgraph,
i) theleastpossibledegreeandthesignoftheleadingcoefficientii) the𝑥-interceptsandthefactorsofthefunctioniii) theintervalswherethefunctionispositive/negative
a) i) ii) iii)b) i) ii) iii)
Example2:AnalyzeFactoredFormEquationstoSketchGraphs
Degree LeadingCoefficient EndBehaviour 𝒙-intercepts 𝒚-interceptTheexponenton𝑥whenallfactorsof𝑥aremultipliedtogether
OR
Addtheexponentsonthefactorsthatincludean𝑥.
Theproductofallthe𝑥coefficients
Usedegreeandsignofleadingcoefficienttodeterminethis
Seteachfactorequaltozeroandsolvefor𝑥
Set𝑥 = 0andsolvefor𝑦
Sketchagraphofeachpolynomialfunction:a)𝑓 𝑥 = (𝑥 − 1)(𝑥 + 2)(𝑥 + 3)
Degree LeadingCoefficient EndBehaviour 𝒙-intercepts 𝒚-interceptTheproductofallfactorsof𝑥is:
𝑥 𝑥 𝑥 = 𝑥'Thefunctioniscubic.
DEGREE3
Theproductofallthe𝑥coefficientsis:1 1 1 = 1
LeadingCoefficientis1
Cubicwithapositiveleadingcoefficientextendsfrom:
Q3toQ1
The𝒙-interceptsare1,-2,and-3(1,0)(-2,0)(-3,0)
Set𝑥equalto0andsolve:𝑦 = 0 − 1 0 + 2 0 + 3 𝑦 = (−1)(2)(3)𝑦 = −6The𝒚-interceptis
at(0,-6)
b)𝑔 𝑥 = −2 𝑥 − 1 ((𝑥 + 2)
Degree LeadingCoefficient EndBehaviour 𝒙-intercepts 𝒚-interceptTheproductofallfactorsof𝑥is:
𝑥( 𝑥 = 𝑥'Thefunctioniscubic.
DEGREE3
Theproductofallthe𝑥coefficientsis:−2 1 ( 1 = −2
LeadingCoefficientis−𝟐
Cubicwithanegativeleadingcoefficientextendsfrom:
Q2toQ4
The𝒙-interceptsare1(order2),and-2.(1,0)(-2,0)
Set𝑥equalto0andsolve:𝑦 = −2 0 − 1 ( 0 + 2 𝑦 = (−2)(1)(2)𝑦 = −4The𝒚-interceptis
at(0,-4)
c)ℎ 𝑥 = − 2𝑥 + 1 '(𝑥 − 3)
Degree LeadingCoefficient EndBehaviour 𝒙-intercepts 𝒚-interceptTheproductofallfactorsof𝑥is:
𝑥' 𝑥 = 𝑥$Thefunctionisquartic.
DEGREE4
Theproductofallthe𝑥coefficientsis:−1 2 ' 1 = −8
LeadingCoefficientis−𝟖
Aquarticwithanegativeleadingcoefficientextendsfrom:
Q3toQ4
The𝒙-interceptsare−𝟏
𝟐(order3),
and3.−A(, 0
(3, 0)
Set𝑥equalto0andsolve:𝑦 = − 2 0 + 1 '[0 − 3]𝑦 = (−1)(1)(−3)𝑦 = 3The𝒚-interceptis
at(0,3)
d)𝑗 𝑥 = 𝑥$ − 4𝑥' + 3𝑥(
𝑗 𝑥 = 𝑥((𝑥( − 4𝑥 + 3)𝑗 𝑥 = 𝑥((𝑥 − 3)(𝑥 − 1)
Degree LeadingCoefficient EndBehaviour 𝒙-intercepts 𝒚-interceptTheproductofallfactorsof𝑥is:
𝑥( 𝑥 (𝑥) = 𝑥$Thefunctionisquartic.
DEGREE4
Theproductofallthe𝑥coefficientsis:1 ( 1 (1) = 1
LeadingCoefficientis𝟏
Aquarticwithapositiveleadingcoefficientextendsfrom:
Q2toQ1
The𝒙-interceptsare0(order2),3,and1.(0, 0)(3, 0)(1, 0)
Set𝑥equalto0andsolve:𝑦 = 0 ((0 − 3)(0 − 1)𝑦 = (0)(−3)(−1)𝑦 = 0The𝒚-interceptis
at(0,0)
Example3:RepresentingtheGraphofaPolynomialFunctionwithitsEquationa)Writetheequationofthefunctionshownbelow:
Note:mustputintofactoredformtofind𝑥-intercepts
Steps:1)Writetheequationofthefamilyofpolynomialsusingfactorscreatedfrom𝑥-intercepts2)Substitutethecoordinatesofanotherpoint(𝑥, 𝑦)intotheequation.3)Solvefor𝑎4)Writetheequationinfactoredform
Thefunctionhas𝑥-interceptsat-2and3.Bothareoforder2.𝑓(𝑥) = 𝑘(𝑥 + 2)((𝑥 − 3)(
4 = 𝑘(2 + 2)((2 − 3)(
4 = 𝑘(4)((−1)(
4 = 16𝑘
𝑘 =14
𝒇(𝒙) =𝟏𝟒 (𝒙 + 𝟐)
𝟐(𝒙 − 𝟑)𝟐
b)Findtheequationofapolynomialfunctionthatisdegree4withzeros−1(order3)and1,andwitha𝑦-interceptof−2.
𝑓(𝑥) = 𝑘(𝑥 + 1)'(𝑥 − 1)
−2 = 𝑘(0 + 1)'(0 − 1)−2 = 𝑘(1)'(−1)−2 = −1𝑘𝑘 = 2
𝒇(𝒙) = 𝟐(𝒙 + 𝟏)𝟑(𝒙 − 𝟏)
L4–1.4–TransformationsLessonMHF4UJensenInthissection,youwillinvestigatetherolesoftheparameters𝑎, 𝑘, 𝑑,and𝑐inpolynomialfunctionsoftheform𝒇(𝒙) = 𝒂 𝒌 𝒙 − 𝒅 𝒏 + 𝒄.Youwillapplytransformationstothegraphsofbasicpowerfunctionstosketchthegraphofitstransformedfunction.Part1:TransformationsInvestigationInthisinvestigation,youwillbelookingattransformationsofthepowerfunction𝑦 = 𝑥4.Completethefollowingtableusinggraphingtechnologytohelp.Thegraphof𝑦 = 𝑥4isgivenoneachsetofaxes;sketchthegraphofthetransformedfunctiononthesamesetofaxes.Thencommentonhowthevalueoftheparameter𝑎, 𝑘, 𝑑,or𝑐transformstheparentfunction.Effectsof𝑐on𝑦 = 𝑥4 + 𝑐TransformedFunction Valueof𝒄 Transformationsto𝒚 = 𝒙𝟒 Graphoftransformedfunction
comparedto𝒚 = 𝒙𝟒
𝑦 = 𝑥4 + 1 𝑐 = 1 Shiftup1unit
𝑦 = 𝑥4 − 2 𝑐 = −2 Shiftdown2units
Effectsof𝑑on𝑦 = (𝑥 − 𝑑)4TransformedFunction
Valueof𝒅 Transformationsto𝒚 = 𝒙𝟒 Graphoftransformedfunctioncomparedto𝒚 = 𝒙𝟒
𝑦 = (𝑥 − 2)4 𝑑 = 2 Shiftright2units
𝑦 = (𝑥 + 3)4 𝑑 = −3 Shiftleft3units
Effectsof𝑎on𝑦 = 𝑎𝑥4TransformedFunction
Valueof𝒂 Transformationsto𝒚 = 𝒙𝟒 Graphoftransformedfunctioncomparedto𝒚 = 𝒙𝟒
𝑦 = 2𝑥4 𝑎 = 2 Verticalstretchbyafactorof2
𝑦 =12𝑥
4 𝑎 =12
Verticalcompressionbyafactorof:
;
𝑦 = −2𝑥4 𝑎 = −2 Verticalstretchbyafactorof2andaverticalreflection.
Effectsof𝑘on𝑦 = (𝑘𝑥)4TransformedFunction
Valueof𝒌 Transformationsto𝒚 = 𝒙𝟒 Graphoftransformedfunctioncomparedto𝒚 = 𝒙𝟒
𝑦 = (2𝑥)4 𝑘 = 2Horizontalcompressionbya
factorof:;
𝑦 =13𝑥
4
𝑘 =13
Horizontalstretchbyafactorof3
𝑦 = (−2𝑥)4 𝑘 = −2Horizontalcompressionbyafactorof:
;andahorizontalreflection
Summaryofeffectsof𝑎, 𝑘, 𝑑,and𝑐inpolynomialfunctionsoftheform𝑓(𝑥) = 𝑎 𝑘 𝑥 − 𝑑 = + 𝑐
Valueof𝒄in𝒇(𝒙) = 𝒂 𝒌 𝒙 − 𝒅 𝒏 + 𝒄
𝑐 > 0 Shift𝑐unitsup
𝑐 < 0 Shift𝑐unitsdown
Valueof𝒅in𝒇(𝒙) = 𝒂 𝒌 𝒙 − 𝒅 𝒏 + 𝒄
𝑑 > 0 Shift𝑑unitsright
𝑑 < 0 Shift|𝑑|unitsleft
Valueof𝒂in𝒇(𝒙) = 𝒂 𝒌 𝒙 − 𝒅 𝒏 + 𝒄
𝑎 > 1or𝑎 < −1 Verticalstretchbyafactorof|𝑎|
−1 < 𝑎 < 1 Verticalcompressionbyafactorof|𝑎|
𝑎 < 0 Verticalreflection(reflectioninthe𝑥-axis)
Valueof𝒌in𝒇(𝒙) = 𝒂 𝒌 𝒙 − 𝒅 𝒏 + 𝒄
𝑘 > 1or𝑘 < −1 Horizontalcompressionbyafactorof :|B|
−1 < 𝑘 < 1 Horizontalstretchbyafactorof :|B|
𝑘 < 0 Horizontalreflection(reflectioninthe𝑦-axis)
Note:𝑎and𝑐causeVERTICALtransformationsandthereforeeffectthe𝑦-coordinatesofthefunction.𝑘and𝑑causeHORIZONTALtransformationsandthereforeeffectthe𝑥-coordinatesofthefunction.Whenapplyingtransformationstoaparentfunction,makesuretoapplythetransformationsrepresentedby𝑎and𝑘BEFOREthetransformationsrepresentedby𝑑and𝑐.
Part2:DescribingTransformationsfromanEquationExample1:Describethetransformationsthatmustbeappliedtothegraphofeachpowerfunction,𝑓(𝑥),toobtainthetransformedfunction,𝑔(𝑥).Then,writethecorrespondingequationofthetransformedfunction.Then,statethedomainandrangeofthetransformedfunction.a)𝑓 𝑥 = 𝑥4,𝑔 𝑥 = 2𝑓 :
D𝑥 − 5
𝑎 = 2;verticalstretchbyafactorof2(2𝑦)𝑘 = :
D;horizontalstretchbyafactorof3(3𝑥)
𝑑 = 5;shift5unitsright(𝑥 + 5)
𝑔 𝑥 = 213 𝑥 − 5
4
b)𝑓 𝑥 = 𝑥G,𝑔 𝑥 = :
4𝑓 −2 𝑥 − 3 + 4
𝑎 = :
4;verticalcompressionbyafactorof:
4 I4
𝑘 = −2;horizontalcompressionbyafactorof:
;andahorizontalreflection J
K;
𝑑 = 3;shiftright3units(𝑥 + 3)𝑐 = 4;shift4unitsup(𝑦 + 4)
𝑔 𝑥 =14 −2 𝑥 − 3 G + 4
Part3:ApplyingTransformationstoSketchaGraphExample2:Thegraphof𝑓(𝑥) = 𝑥Distransformedtoobtainthegraphof𝑔(𝑥) = 3 −2 𝑥 + 1 D + 5.a)Statetheparametersanddescribethecorrespondingtransformations𝑎 = 3;verticalstretchbyafactorof3(3𝑦)
𝑘 = −2;horizontalcompressionbyafactorof:;andahorizontalreflection J
K;
𝑑 = −1;shiftleft1unit(𝑥 − 1)𝑐 = 5;shiftup5units(𝑦 + 5)b)Makeatableofvaluesfortheparentfunctionandthenusethetransformationsdescribedinparta)tomakeatableofvaluesforthetransformedfunction.
c)Graphtheparentfunctionandthetransformedfunctiononthesamegrid.
𝑓 𝑥 = 𝑥D𝒙 𝒚-2 -8
-1 -1
0 0
1 1
2 8
𝒈(𝒙) = 𝟑 -𝟐 𝒙 + 𝟏 𝟑 + 𝟓𝒙K𝟐− 𝟏 𝟑𝒚 + 𝟓0 −19
−0.5 2
−1 5
−1.5 8
−2 29
Note:Whenchoosingkeypointsfortheparentfunction,alwayschoose𝑥-valuesbetween-2and2andcalculatethecorrespondingvaluesof𝑦.
Example3:Thegraphof𝑓(𝑥) = 𝑥4istransformedtoobtainthegraphof𝑔(𝑥) = − :D𝑥 + 2
4− 1.
a)Statetheparametersanddescribethecorrespondingtransformations
𝑔 𝑥 = −13 𝑥 + 6
4
− 1𝑎 = −1;verticalreflection(−1𝑦)
𝑘 = :D;horizontalstretchbyafactorof3(3𝑥)
𝑑 = −6;shiftleft6units(𝑥 − 6)𝑐 = −1;shiftdown1unit(𝑦 − 1)b)Makeatableofvaluesfortheparentfunctionandthenusethetransformationsdescribedinparta)tomakeatableofvaluesforthetransformedfunction.
c)Graphtheparentfunctionandthetransformedfunctiononthesamegrid.
𝑓 𝑥 = 𝑥4𝒙 𝒚-2 16
-1 1
0 0
1 1
2 16
𝑔 𝑥 = −13𝑥 + 6
4− 1
3𝑥 − 6 −𝑦 − 1−12 −17
−9 −2
−6 −1
−3 −2
0 −17
Note:𝑘valuemustbefactoredoutintotheform[𝑘(𝑥 + 𝑑)]
Part4:DetermininganEquationGiventheGraphofaTransformedFunctionExample4:Transformationsareappliedtoeachpowerfunctiontoobtaintheresultinggraph.Determineanequationforthetransformedfunction.Thenstatethedomainandrangeofthetransformedfunction.a) b)
Noticethetransformedfunctionisthesameshapeastheparentfunction.Therefore,ithasnotbeenstretchedorcompressed.𝑑 = −3;ithasbeenshiftedleft3units𝑐 = −5;ithasbeenshifteddown5units𝒈(𝒙) = (𝒙 + 𝟑)𝟒 − 𝟓Domain:(−∞,∞) Range:[−𝟓,∞)
Noticethetransformedfunctionisthesameshapeastheparentfunction.Therefore,ithasnotbeenstretchedorcompressed.𝑎 = −1;ithasbeenreflectedvertically𝑑 = 5;ithasbeenshiftedright5units𝒈(𝒙) = −(𝒙 − 𝟓)𝟑Domain:(−∞,∞) Range:(−∞,∞)
f(1)
f(-1)
f(1)f(-1)
𝑓(𝑥) = 2𝑥' + 3𝑥* − 2Notice:𝑓(1) = 3𝑓(−1) = 3∴ 𝑓(1) = 𝑓(−1)
L5–1.3–SymmetryinPolynomialFunctionsMHF4UJensenInthissection,youwilllearnaboutthepropertiesofevenandoddpolynomialfunctions.SymmetryinPolynomialFunctionsLineSymmetry–thereisaverticallineoverwhichthepolynomialremainsunchangedwhenreflected.Pointsymmetry/RotationalSymmetry–thereisapointaboutwhichthepolynomialremainsunchangedwhenrotated180°Section1:PropertiesofEvenandOddFunctionsApolynomialfunctionofevenorodddegreeisNOTnecessarilyandevenoroddfunction.Thefollowingarepropertiesofallevenandoddfunctions:
EvenFunctions OddFunctionsAnevendegreepolynomialfunctionisanEVENFUNCTIONif:
• Linesymmetryoverthe𝑦-axis• Theexponentofeachtermiseven• Mayhaveaconstantterm
AnodddegreepolynomialfunctionisanODDFUNCTIONif:
• Pointsymmetryabouttheorigin(0,0)• Theexponentofeachtermisodd• Noconstantterm
Rule:𝑓 −𝑥 = 𝑓(𝑥)
Rule:−𝑓 𝑥 = 𝑓(−𝑥)
Example:
Example:
𝑓(𝑥) = 2𝑥2 + 3𝑥Notice:𝑓(1) = 5𝑓(−1) = −5∴ −𝑓(1) = 𝑓(−1)
Example1:Identifyeachfunctionasanevenfunction,oddfunction,orneither.Explainhowyoucantell.a)𝑦 = 𝑥2 − 4𝑥b)𝑦 = 𝑥2 − 4𝑥 + 2c)𝑦 = 𝑥' − 4𝑥* + 2
Thisisanoddfunctionbecause:
• Ithaspointsymmetryabouttheorigin• Alltermsintheequationhaveanoddexponentandthereisno
constantterm
NeitherThisfunctionhaspointsymmetry.However,theoriginisnotthepointaboutwhichthefunctionissymmetrical.Therefore,itisnotanoddorevenfunction.FromtheequationwecantellitisNOTanoddfunctionbecausethereisaconstantterm.
Thisisanevenfunctionbecause:
• Ithaslinesymmetryaboutthe𝑦-axis• Alltermsintheequationhaveanevenexponent.Evenfunctionsare
allowedtohaveaconstantterm.
d)𝑦 = 3𝑥' + 𝑥2 − 4𝑥* + 2e)𝑦 = −3𝑥* − 6𝑥Example2:Chooseallthatapplyforeachfunctiona) b)
NeitherThisfunctiondoesnothavelineorpointsymmetry.FromtheequationwecantellitisNOTanevenoroddfunctionbecausethereisamixofevenandoddexponents.
NeitherThisfunctionhaslinesymmetry.However,the𝑦-axisisnotthelineaboutwhichthefunctionissymmetrical.Therefore,itisnotanoddorevenfunction.FromtheequationwecantellitisNOTanevenoroddfunctionbecausethereisamixtureofevenandoddexponents.
i)nosymmetry
ii)pointsymmetry
iii)linesymmetry
iv)oddfunction
v)evenfunction
i)nosymmetry
ii)pointsymmetry
iii)linesymmetry
iv)oddfunction
v)evenfunction
c)𝑃 𝑥 = 5𝑥2 + 3𝑥* + 2 d)𝑃 𝑥 = 𝑥8 + 𝑥* − 11
e) f)g)𝑃 𝑥 = 5𝑥9 − 4𝑥2 + 8𝑥Example3:Withoutgraphing,determineifeachpolynomialfunctionhaslinesymmetryaboutthey-axis,pointsymmetryabouttheorigin,orneither.Verifyyourresponsealgebraically.a)𝑓 𝑥 = 2𝑥' − 5𝑥* + 4Thefunctionisevensincetheexponentofeachtermiseven.Thefunctionhaslinesymmetryaboutthe𝑦-axis.Verify𝑓 𝑥 = 𝑓(−𝑥)𝑓 −𝑥 = 2 −𝑥 ' − 5 −𝑥 * + 4𝑓 −𝑥 = 2𝑥' − 5𝑥* + 4𝑓 −𝑥 = 𝑓(𝑥)
i)nosymmetry
ii)pointsymmetry
iii)linesymmetry
iv)oddfunction
v)evenfunction
i)nosymmetry
ii)pointsymmetry
iii)linesymmetry
iv)oddfunction
v)evenfunction
i)nosymmetry
ii)pointsymmetry
iii)linesymmetry
iv)oddfunction
v)evenfunction
i)nosymmetry
ii)pointsymmetry
iii)linesymmetry
iv)oddfunction
v)evenfunction
i)nosymmetry
ii)pointsymmetry
iii)linesymmetry
iv)oddfunction
v)evenfunction
Note:allcubicfunctionshavepointsymmetry
(-1, 3)
b)𝑓 𝑥 = −3𝑥9 + 9𝑥2 + 2𝑥Thefunctionisoddsincetheexponentofeachtermisodd.Thefunctionhaspointsymmetryabouttheorigin.Verify– 𝑓 𝑥 = 𝑓(−𝑥)−𝑓 𝑥 = − −3𝑥9 + 9𝑥2 + 2𝑥 −𝑓 𝑥 = 3𝑥9 − 9𝑥2 − 2𝑥∴ −𝑓 𝑥 = 𝑓(−𝑥)c)𝑥8 − 4𝑥2 + 6𝑥* − 4Someexponentsareevenandsomeareodd,sothefunctionisneitherevennorodd.Itdoesnothavelinesymmetryaboutthe𝑦-axisorpointsymmetryabouttheorigin.Section2:ConnectingfromthroughouttheunitExample4:Usethegivengraphtostate:a)𝑥-intercepts−2(order2),0(order1),and2(order2)b)numberofturningpoints2localminand2localmax4turningpointsc)leastpossibledegreeLeastpossibledegreeis5b)anysymmetrypresentPointsymmetryabouttheorigin.Therefore,thisisanoddfunction.c)theintervalswhere𝑓 𝑥 < 0(0, 2) ∪ (2,∞)
𝑓(−𝑥) = −3(−𝑥)9 + 9(−𝑥)2 + 2(−𝑥)𝑓(−𝑥) = 3𝑥9 − 9𝑥2 − 2𝑥
d)Findtheequationinfactoredform𝑃 𝑥 = 𝑘(𝑥)(𝑥 + 2)*(𝑥 − 2)*3 = 𝑘(−1)(−1 + 2)*(−1 − 2)*3 = 𝑘 −1 1 * −3 *3 = −9𝑘
𝑘 = −13
𝑃 𝑥 = −13𝑥(𝑥 + 2)
*(𝑥 − 2)*