Polynomial Function for Thesis

8/3/2019 Polynomial Function for Thesis

1/27

WELCOME TO THEART OF

POLYNOMIAL

EQUATIONS


2/27

POLYNOMIAL EQUATIONS


3/27

Definitionofthe Polynomial

Letn beanonnegativeintegerandletan,an-1,a2,

a1 anda0 berealnumbers withan 0. Thefunction

defined by

f(x) = anxn +an-1x

n-1 ++a2x2+a1x+a0

iscalleda polynomialfunctionofn. Thenumberan,thecoefficientofthevariabletothehighest power,

iscalledtheleadingcoefficient.


4/27


5/27

Question 3: Is quadraticfunction,f(x) = ax2 +

bx +c, wherea0,a polynomial

function?

Answer: Yes. Itisclassifiedasa polynomial

function withdegree2.


6/27

Characteristicsofthe Graphsof

Polynomial Functions

Smooth Itmeansthatthegraphof polynomial

functionscontainonlyroundedcurves with

nosharp corners.

C

ontinuous

Itmeansthatthegraphsof polynomial

functionhaveno breaksandcan be

drawn withoutliftingthe pencilonthe

rectangularcoordinatesystem. Thus,its

domainisallrealnumbersas wellas

range [Domain:(,), Range:(,)].


7/27

Example

Whichofthefollowinggraphsareclassifiedas

Polynomial Functions?

A CB D


8/27

Answer

The polynomialfunctionsare A andBsincethegraphs

havesmoothandroundedcorners,andcontinuous,

thatfollowsthecharacteristicsofa polynomial

graph.

CandDarenotclassifiedas polynomialfunctions

sincegraphCisdiscontinuousandgraphDhassharp corners.


9/27

The Importanceofthe Leading

Coefficient

As wedefinedthe polynomialfunctionas:

f(x) = anxn

+an-1xn-1

++a2x2

+a1x+a0,(a0n0)

Whenthevalueofx increasesordecreases without

bound,thegraphof polynomialfunctioneventually

risesorfalls.


10/27


Coefficient

Ifthehighestdegree

(n)isoddandtheleadingcoefficientis

positive(+),the

graphof polynomial

willfallstotheleftandrisestotheright.

falls

rises


11/27


Coefficient

Ifthehighestdegree


negative(-),the

graphof polynomial

willfallstotherightandrisestotheleft.

fallsrises


12/27


Coefficient

Ifthehighestdegree

(n)isevenandtheleadingcoefficientis

positive(+),the

graphof polynomial

risestotheleftandtotheright.

risesrises


13/27


Coefficient

Ifthehighestdegree


negative(-),the

graphof polynomial

fallstotheleftandtotheright.

fallsfalls


14/27

Zeros/Rootsof Polynomial Functions

Iff isa polynomialfunction,thenthevaluesofx for

whichf(x) isequalto0areallcalled zerosoff.

Thesevaluesofx arerootsorsolutionsofthe

polynomialequationf(x) = 0. Eachrealrootofpolynomialappearsasanx-interceptofthegraph

ofthe polynomialfunction.

Insimple wayitisthevalueofx thatmakesf(x) = 0.Tofinditletf(x) = 0.


15/27

Example

1. Findtherootsoff(x) = x3 +2x2 x 2.

Solution:

0 = x3 +2x2 x 2

= x2(x+2) (x+2)

= (x2-1)(x+2)

x+2 = 0 andx2 1 = 0

x = -2 x2 = 1

x2=1

The zeros/rootsoffare-2,-1,and1. Thisindicatesthatthegraphpassedthroughthe points(-2,0),(-1,0)and(1,0) whicharethexinterceptsofthe polynomial.

Let f(x) = 0

Factorx2 fromthefirsttwoterms and-1fromthelasttwo

terms

A commonfactorofx+2isfactoredfromtheexpression.

Seteachfactorequalto0.

Solveforx

Rememberthat if x2 = d,thenx =


16/27

Example

2. Findthe zerooff(x) = - x4 +2x3 x2.

Solution:

0 = - x4 +2x3 x2

-1(0 = - x4 +2x3 x2)-1

1. = x4 2x3 +x2

= x2(x2 2x+1)

= x2(x-1)2

x2 = 0 andx-1 = 0

x = 0 x = 1

Therootsoffare0and2. Thisindicatesthatthegraphsx-interceptsare(0,0)and(2,0) whereinthegraph passesthrough.

Letx = 0

Multiply bothsides by 1.

Factorcompletely

Factoroutx2


Solveforx


17/27

Example

3. Findtherootsoff(x) = x3 +4x2 4x 16.

Solution:

0 = x3 +4x2 4x 16

= x2(x+4) 4(x+4)

= (x2 4)(x+4)

x2 4 = 0 andx+4 = 0

x = 2 x = -4

Therootsoffare(-4,0),(-2,0)and(2,0). Thisindicates

thatthegraph passesthroughthethese points.

Letf(x) = 0

Factorx2 fromthefirsttwoterms

and-4fromthelasttwoterms

A commonfactorofx+4isfactored

fromtheexpression.


Solveforx


18/27

Example

4. Findtherootsoff(x) = x4 9x2.

Solution:

0 = x4 9x2

= x2 (x2 9)

x2 = 0andx2 9 = 0

x = 0andx = 3

Therootsoffare(-3,0),(0,0)and(3,0). Thisindicatesthatthegraph passesthroughthese points

Letf(x) = 0

Factoroutx2


Solveforx


19/27

MultiplicityofZerosof Polynomial

Functions

Ifr isa zeroofevenmultiplicity,thenthegraphtouchesx axisandturnsaroundatr. Ifr isa zeroofoddmultiplicity,thenthegraphcrossesthex-axisatr. Regardlessof whetherthemultiplicityofazeroisevenorodd,graphstendtop flattenoutat zeros withmultiplicitygreaterthanone(1).

Say, f(x) = -x4 +4x3 4x2

-1(0= -x4 +4x3 4x2)-1

= x4 - 4x3 +4x2.

= x2(x2 - 4x+4)

=x2(x-2)2

Observethateachfactoroccurstwice. Inthefollowing,theequationforthe polynomialfunctionf,ifthesamefactorx - r occursk times(x-r)k, butnotk + 1 times, wecallthisasa zero/root withmultiplicityk.Thus,forthe polynomialfunctionf(x) = x2(x-2)2,0and2are bothzeros/roots withmultiplicity2. Therefore,thegraphoff touches(0,0)and(2,0) pointsandturnaround.


20/27

Example

Findtherootsoff(x) = (x+2)(x-3)2andgiveitsmultiplicity. Stateifthegraphcrossesthex-axisortouchesthex-axisandturnsaroundateach zero.

Solution:

(x+2)(x-3)2= 0

x+2 = 0andx3 = 0

x = -2and x = 3

The zeros/rootsoffare-2and 3, withmultiplicityof1

and2,respectively.Becausethemultiplicityof-2isodd,thegraphcrossesthex axisatthisroot.Becausethemultiplicityof 3 even,thegraphtouchesthex-axisandturnaroundatthisroot.

Letf(x) = 0


Solveforx


21/27

Whento Usethe Intermediate Value

Theorem?

Definition. Letfbea polynomialfunction withrealcoefficients. If f(a) andf(b) haveoppositesigns,thenthereisatleastonevalueofc betweena andb for whichf(c)= 0. Equivalently,theequationf(x) = 0hasatleastonerealroot betweena andb.

Thistheoremisveyusefulifthe polynomialfunctionhasnofactor butthex-intercept/sexistsonthegraph.

b

a

c

(b,f(b))

f(b)>0

(a,f(a))

f(a)


22/27

Example

Show thatthe polynomialhasareal zero between2and 3.

given:f(x) = x3 2x 5

Solution:

Evaluatefat2andat 3. Iff(2) andf(3) haveoppositesigns,thenthereisatleastonereal zero between2and 3. Usingf(x) = x3 2x 5,theobtainvalue

f(2) = (2)3 2(2) 5 = -1

and

f(3

)=

(3

)

3

2(3

) 5

=16

Becausef(2) = -1andf(3) = 16,thesignchangeshowsthatthe polynomialfunctionhasreal zero between2and 3. Thiszeroisactuallyirrationalandisapproximated.

f(2)isnegative

f(3)is positive


23/27

Turning Pointsof Polynomial Function

Ineachturning pointofthe polynomialgraph,the

changesitsdirectionfromincreasingtodecreasing

orviceversa. Andtodeterminethe possiblenumber

ofturning points withoutgraphingthefunction weneedtofollow thecondition, iff isa polynomial

functionofdegreen,thenthegraphoffhasat

mostn-1 turning points.


24/27

Easy Graphingof Polynomial Function

withoutusingany Graphing Utility

1. Usetheleadingcoefficienttesttodeterminethe

graphsend behavior.

2.Determinethex-intercept bysettingf(x) = 0and

solvingtheresulting polynomialequation. Ifthereis

anx-interceptatr asaresultof(x-r)k inthe

completefactorizationoff(x),thenthefollowing

must beconsidered:

a. Ifk iseven,thegraphtouchesthex-axisatr and

turnsaround.


25/27



b. Ifk isodd,thegraphcrossesthex-axisatr.

c. If k>1,thegraphflattensoutat(r,0).

3.Determinethey-intercept bycomputingf(0) orlet

x = 0.

4. Usesymmetry,ifapplicable,tohelp draw the

graph:

a.y-axissymmetry:f(-x) = f(x)

b.originsymmetry:f(-x) = -f(x)


26/27



5. Usethefactthatthemaximumnumberofturning

pointsofthegraphisn-1 tocheck whetheritis

drawncorrectly.


27/27

Exampleof Graphing Polynomial

Functions

Pleaserefertotheembedded

videosforexamples

Date post:	06-Apr-2018
Category:	Documents
Upload:	lpuelearning
View:	221 times
Download:	0 times

Polynomial Function for Thesis

Documents