Chapter 1- Polynomial Functions - jensenmath.ca 1 unit package TEACHER.pdf · 1.4 L3 Factored Form...

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)*