3.1 Characteristics of Polynomial Functions R12 (p.106-113)

Polynomial Function = a function of the form where

f (x) = ax n + an_ 1 xn-1 + an_ 2 xn-2 + a2x2 + ai x + a0

where n = a whole number 01 I / 1 3) "‘

X = a variable an and (40 are real numbers

Examples:

f (x) = 3x•-5

g(x) = x2 + 3x —17

h(x) = x3 +2x2 x — 2

y = +7x3 —1

ty 'Id up ilrik) dOtet

behoyt'our ofc y-ValucS

(1 .S IX I 6eCo VY\ ior ere

• degree +Ix 1 Cirj,zsf eX olle-rvf o1 X.

• constant term (a() ) w4-hou x ) 7-1

• leading coefficient =fk Coe-1. 41'c Its) 1- of ±h kr 144:1

cxpowif.

Vocabulary loco

• end behaviour

Exl: Identify the functions that are not polynomials and state why. For each, state the degree, leading coefficient, and the constant term of each polynomial function.

a) g (x) =

b) y = Ix' ob,SoLtAbk ya Lux_ -Pt) h n co 1) hofk rcte(6-Ht n 0,3 x

c) f (x) = 3x 4 Pyr)e01-1)'0 1 -Ptv

(go d (1)3 coc -hti'cr)f

d) y = 2x3 + 3x2 — 4x — 1

DolyhOrr,o'cil for) cftor of dQJ r c 3 / 3r4 rce P°1

(1: "1,, oc fjc,cr-of 1), 4-erm (s —1

Your turn:

a) h(x) = r-a-1-1.ona( -Cu n c47.0 n cat) be

LqAf ho+ whok

+ cr b wrii + kr) ca x wWch

) hos b cxpo htni-, no+ a WAoic #-

'on Of sk rc.c.

is 3 Con f-rt

rot, po iynohfo

krrn o f 0.

ANY)

b) y = 3x2 2x5 + 4 po lyno o

(Qi c: 3 c ccckr f rec. 5' 514 cit

rm 1,2 lit

y = — 4 x3 — 4 x + 3 poljnoho'Pf of de rex; 3

Cocf-ko'cin Cop,c-h7r) (p ArA

r 3 c rcc po ly

( 7

Ndi'cA/ fonc47.0n

f(x) = x2 f(x) = -x2 f(x) = x2 - x 4- 6

f(x) = -x2 - 8x - 7 quadratic quadratic quadratic quadratic

J

f(x) = x 4

f(x) = -x 4

f(x) = x4 - 4x3 + x2 + 7x - 3 f(x) = -x4 + 7x2 - quartic quartic quartic quartic

4

Hi

4

+bl\A

f(x) = x3 2x2 - 2x + 6 f(x) = -x3 + 7x cubic cubic

f(x) =x5 quintic

f(x) -xs - 4x4 + 40x3 + 160x2 - 144x - 576

quintic

f(x) = x linear

111

f(x) = x3 cubic

f(x) = -x + 4 linear

11

f(x) = -x3 cubic

Look at the graphs on page 109 of your text and list any generalizations or patterns you notice.

Characteristics of Polynomial Functions Compare the graphs of even and odd functions. How does the leading term affect the general behaviour of the graph?

a) The equations and graphs of severat ee Elyao—mTaTgare shown below. Study these 4 S.V.4.41iMagi.4

graphs and oeneralize the end behaviour of even-degree polynomials.

c) irrkrcepit ehd b bori‘ot) tioroot t, rah e 47 of -/-tAms

,....,..,.......—...._-_,....-64,..„..,,,„..„...-- b) The equations and graphs of severatiodt-.ckgreepotynomiajts are shown below. Study these ,,,,,„,,,,.. . ,,..._.... mom. mammon AVERMAI.14.1..+AW, . 5

graphs and generalize the end behayiour of odd-degree p9tynomiats. it ,,) '. i rig 41) C„hd tcheW6)14A r 40 fro i A rani,:: C,,,, - Of' ill,

Ex2: Match a Polynomial function with its graph. Look for general patterns.

a 1. y = —x4 + 6x2 + x — 5 chtei: Tifritrce

q 2. y = x3 + 2x2 — x — 2

6 3. y = —x3 + 2x2 + 4x -- 3

C 4. y = x.4 + 5x3 + 5x2 — Sx — 6

046ir-ve,

A.

-4

.

r

f

4

1

, 1

,-,.... 4 ,f

.6.

I

ff.

r

s..<

, 1 ,

A

-e- .&- • •

1.

, i

f.,..3. —

4 x

•

L

Odd —y Vre,

B.

,

i

,

t

i

(

! (.

-,4

,

,

1

C

3 f

i ! ,

4., t ,

1

,

, .

.....- - S.-

■ ■

"

')

, i t ,

? i

,

,. .,

,

1

1

1

tVZ-fl . -1-Ve,

C.

,

-a

, ,

,

„

,

,.,

:4„.. t

g ,

,

/ \

t

t ,

, !A 1----8

x a

.

Ii

CYen —Ved

D.

,....

t

,

8

'

, .4. ,

1 , , i !

.

.

i

i

Example 2

A toaster oven is built in the shape of a rectangular prism. Its volume, V in cubic inches, is related to the height, h, in inches, of the oven door by the function V(h) = h3 +10112 + 31)* + 30.

a) What is the volume, in cubic inches, of the toaster oven if the door height is 8 in?

h V- ( a) = ( 8) 3+(0( ) f 31(4)+30

5/P- -F- 640 21-/-E f 30

/43o /6 3

Homework:

Page 114 #1, 2, 3, 4, 6, C 1 &C2

Applications of a Polynomial Function

Example 1:

A bank vault is built in the shape of a rectangular prism. Its volume, V, is related to the width, w, in metres, of the vault doorway by the function V(w) = w3 +13w2 + 54w + 72.

a) What is the volume, in cubic metres, of the vault if the door is lm wide?

W 7-":" I hi

V(') (1 )34- 13(i)211- 54i (1)

4 13 4, s Li 4- -72.,

Pip

f VOIMAR 01 V ,4 rt 1 4:1 0 rn

b) What is the least volume of the vault? What is the width of the door for this volume? Is this realistic?

(2- P4 VOt1-441-2- 143 LAU) dbot- 14 0 ir

V(0) o) 3 / 3 (6)'1(o j4 7:10./

whf'ci) tiv-1 constaii .frfir)

tea's/ Volurnc, of V1L kiwi-I-- 7;a.rtda'

bf fhcrt woutd b il(y belt-704'1C iltn

+kit V1kL,4 4 would hoi- hew( a door

3.2 The Remainder Theorem

Long Division

Divide the following expression:

R11 (p.118-123)

2 + 8x +15

Dividend = polynomial Divisor = binomial (x — a)

Quotient = answer x+3

We can divide the expression above by using long division:

x + 3 x2 +8x+15 x2- + 3x

5 , tis 5)rt-rs

After you divide, your answer can be written in two forms:

1) Dividend

uotient +remainder

Divisor OR Divisor

2) Dividend = Divisor(Quotient)± remainder

Answer:

Note: Since the remainder is 0, this tells us that (x+3) is a factor of the polynomial .X 2 + 8x +15 .

Note: The restriction on the variable is x # —3 since division by 0 is not defined.

Note: To verify, multiply the divisor by the quotient and add the remainder.

k-/- 5 YA- f FiS v F:01 L

Exl: Divide:

,

Restriction on the variable:

Answer:

I 0

or (x2--Xf5- x--1

Note: This quotient has a remainder which means that (x factor of the polynomial.

) IS NOT a

Verification:

t--

Th 1

it.

—2x 2 6x-12 Divide: x-1

1 V

EM,

Y"- ... •

- et A

. . ' , • ••

0 •

440

Rt +(lc_410 — 3

Note: x3 — 2x2 6x —12 =

Exl: Divide: (2x3 5x2 4,4k.

=

R/‹.

„7) 4

Synthetic division

Synthetic division is an alternate form of long division that we can use to divide polynomials.

•

,

Cr,

Note: There is a remainder of 0 which tells us that (x+3) IS a factor of this polynomial.

3 b) x 3 it•") 44, x + 1

--

W.A. •

Aka

i„,,,,

01,641,17

(3)+ii 1;t

•Nyftte

+ow, Walrar.

Ittnelitt der is 3 3

Remainder Theorem

x — a is a factor of polynomial P (x) if P (a) =

Exl: P (x) = — 5x2 —17x + 21 t

Ex2: Completely factor the polynomial x3 — 7x2 — 6x + 72 if x — 4 is one of the factors.

Y!' rer,

,c7

,

Homework: Page 124 41, 2, 3 (choose 2), 4 (choose 3), 5 (choose 2), 6 (choose 3), 8 (choose 2), 9, 11, C2

,

• x — 3

Lt

Why do you think these 6 binomials were selected? ,

Irr,"

Pc rd

,

3113 The Factor Theorem

Which of the following are factors of P(x)= .X3 7X + 6 ? fr731L htl ifed

R11 (p.126-133)

x -- 1 and so P(1)—(1)-7(1)± = .• •

3 e

q+17, orms1

x

•

TM. •mrrf!'

The factor theorem states that (x - a) is a factor of P (x)

if and only if P (a) = 0 .

Find the factors of f (x) = + 2x 2 — 5x — 6 .

Let's start at 0. -11,1 r )iic y

5

'

P (1) = f

21"

P(-1)= 3 I-- R(-1)

q't Si6c€

-I 5 - 7 -3 - o (x+r)

...and so on. Eventually, P(2)

and (x +3) are factors. X

Thus, x3 + 2x 2 — x —

= 0 and P (-3) = . Therefore (x +0 (x-2) -„ 9;),)

1,31'

#

0

a>( ) AhN

Integral Zero Theorem Expand the following expression:

r

1XX + 2XX 5) = (ej YX.). ) (A/ ,r'' I A :77 '";). )( = 5

y 3*.'' Li ,V '''":7 X + i C.)

Note: The factors of the polynomial are x — 1, x ± 2 and x —5. The zeroes of the polynomial are 1, —2, and 5.

Note: When we multiply all of the factors, the constant is ± 10 which means that only factors of 10 can be factors of the polynomial. This is known as the integral zero theorem. f

Exl: a) Find the possible factors of the following polynomial:

f (4 = 3x2 ?IC n tr,i c:f

b) Completely factor the polynomial above.

- C

k

roof%

b) Determine the )(1.

With a = 2:

t-

)( -3 )( t r9

CZX J

Ex2: a) Expand (2x -1)(2x +3)(2x -

g)(3 4 v—aok

foc-br c4 (Pgri cot

212,rb +11torcm irN rot xivti*ifs)

Note that none of the roots are factors of 15.

To find the possible zeroes of a function when the value of a is not ±1, we must divide each possible zero by the value of a.

er

Ex3: a) Determine all of the possible zeroes of the following polynomial:

f(x)= —2x 3 +3x 2 +8x

Without a = 2: Coh &to = I 1,2) ± t Li) ± ± I

A. II posqb0...) faczr

b) Factor the polynomial.

'P(/) -Q(1) +3(1) z- f- (0-12 3 + 4# 0

EA )71 a,+3 o

-r(D) = —r4g) -----z

---(DL- 3 — I

I —4 1 rd—#

—

24- a 7, 7

:--2(55e° 1.0 „or'

t Arr.

VI

a 20 —a 4 0 „„)..

o

19() zr1 S (z-) + '24 + ao (:),)

- Ds)

ct,X)

it?) Ex4: Factoc,x

7 — 5x3 2x2 20x — 24 .

rai

5o :s

c ,

p 747,7

vos,

,

' , , , . . ..,.„.. ' • . .

e.v rot so ry-34

tt;1')

rtAtkli

) (x— a_d)

p ) 10,0100051Wha

,

3 W Ivor ci

A ia

boo lit ci

Homework: Page 133 #1, 2a, 2f, 3c, 3e, 4-6 (choose 3 each), 7, 11, Cl, C2

Exl : a) Determine the zeroes of the following cubic function:

x — 2)(x 3 )(x — 1)

= x I

) ( o

fLrWo

y ,fve,

CIO

3.4 Equations and Graphs of Polynomial Functions R12 (p 136-147)

b) Determine de y-intercept of the function.

c) Summarize what we know about this function.

degree yd ,..1 u c re c po cv no rYli. a I

leading coefficient i

end behaviour down in 0 EE, up in GT i

zeroes — 0...., 3) 1 )

y-intercept 6

intervals (sign diagram) -y- k - V (0 u X aA)14.

— -t- ---3 i

— 1– 42_,

' vi:10 -3-.. bc 144 tit'

N-ctk )3

d) Sketch the graph.

441

'1' V I, -

Ex2: a) Determine the roots of the following function:

P(x)= —x3 +4x2 +x — 4

Note: The first step is to factor out the -1 from the function.

p (k) (x Ltx x Li) I I I NO (I - L4 —11 0 V I I 3 t.1

b) Determine the y-intercept of the function.

4- 0 () Li-

t+

c) Summarize what we know about this function.

degree 3 d leading coefficient - 1

end behaviour tlpfl L ,

zeroes .

y-intercept mo.c...WC

intervals (sign diagram) -1 _ ................. I 44--

_ ±-+ -+

d) Sketch the graph.

ii(xi( I;ij 4-

)C 2- f-,Vc•2x--2),.- -45( 141 71 41

b) y = ± 2) )

-3 -2

4`

/

-3

Exl: Sketch the following graphs:

(x-1)2 y-()(-1)0(-1)

Multiplicity of a Zero

If P(x) has factor (x — a) n times, we say that x = a is a zero of multiplicity n.

For example,

y = (x — 2)(x —

Multiplicity represents the number of times a factor is repeated.

(Multiplicity 1 Multiplicity 2 Multiplicity 3

Note: The effect of an even multiplicity is bounce on the x-axis. The effect of an odd multiplicity is hard to see without the use of technology.

{

x = —1 is a zpro of multiplicity 2

x = 2 is a zero of multiplicity 1 x = 4 is a zero of multiplicity 3

-1.y

Ex2: Sketch the following graphs:

a) y = (x -1)2 (x + 2)

degree a k

P-, )(

3rd

leading coefficient A

end behaviour idol" Dir wp QI

zeroes (- -- i moitipi

y-intercept (-- OH ( Z) c).,

intervals intervals (sign diagram)

i -I- acl"..e.........r.....4%,,,W, p..........1÷,,m... )1.

a 1 44 +

b) y = -x(x

degree - ti

leading coefficient —

end behaviour down ( 4 „

zeroes - '1;? 0

y-intercept 0

intervals (sign diagram)

-

, 0 i i i

Ex.3 The zeroes of a polynomial function are -2, 3, and 5. Write an equation to represent this function. x x 3 0><9 - S

Homework: Page 147 #1, 3, 4, 7, 8, 10, Cl, C2, C3

Sketching polynomial functions

Sketch the following functions. Be sure to clearly indicate the x and y-intercepts for each.

g (x) = (x — 1)(x + 4)(x + 2)

A 10

: (-0) (.2,) Y3 xi ; '

-10 I I I I I I J>

10 x

2. y = — x (x + 1) (x + 5) )( 3 y14; 0

°

10

3. f (x) = (x — 6)(x — 2)(x + 3)(x + 8) )( 41 X 14 ; —21 )̀ (lb

2

5. f (x) = X 2 (x + X 3 ,X-44;

0

^

6. f 7,)

't)4-

10

v

S

s 09- 1 h

E :-I-— x G

I

11—x) ('--i.- ,x ) (k

1--1- X -b 4%7,0 (I —X) —

0 E 17 1

E 17 1 11 Z- I- Z I

A ( - I- El- 11 ) - qqf (.-- >(-1-c X -+ X)

01,-

Ic

E x+ zxE — s x— = (x)1 .8

x 01- < III

0 4V ,o i)

, V f -.X ,

I- 4 x ) ( ) p---)0

oL

— x+ zxt + E x =

Sketching polynomial functions

Sketch the following functions. Be sure to clearly indicate the x and y-intercepts for each.

— )1:x + .2;)

2. y = —x(ix ÷ (x-1--

Page 21

3. 17) = 8)

4. y =(t— 1)(x 3)2

4,0

Page 22

5. f = 2(z

6. y=

Page 23

7. v x

YA-1-3j 8. f = -x3 - X+ 3 =z

Page 24

3.4 Applications of Polynomial Functions

Exl : The volume of air flowing into the lungs during one breath can be represented by the polynomial function V(t) = -0.041t3 + 0.181t2 + 0.202t

where V is the volume in litres and t is the time in seconds.

This situation can be represented by the graph below.

4 u .,

vir ,

/,/ „ 3 2'i!

/* /

,,,,, ..

.. .. / ',I... .

6 -5/4 -3/2 ,-1 7

2/ Th

3/4/, I(

/ /

/ / /2/ / , i e },

iv 4,, , ,/' I

e ,,..., - ' _3

4 ,

' / '

„„,,,f. ,. :, 4

,. i / 0.

Q , , ,s ,.„

-

5' I / / / , /

, , ( : )4/ , ,, '

What does the x-axis represent?

il;rnt fri

What does the y-axis represent? Vo't

\ecktA o'r(rLkincy

Determine any restrictions on the variables.

catoo4 Ilavc -V C V Ott-4 "VI( ,

C'C hi)01' ha if -VC

114t-c,

Using the graph above, answer the following questions:

a) Determine the maximum volume of air inhaled into the lungs. At what time during the breath does this occur?

ri\coe

O7L q vsre.,t krotA

b) How many seconds does it take for one complete breath?

pprolim-kly L,, 5 ,P(c on&

c) What percentage of the breath is spent inhaling? 3•D'

0:73•1 3 11.5.

(;, -1-17 '3(0

-1 I I 36

- I

1 1 ---3(1

y ai fity -34

-H,V+36 )

x ± (11) - 11( "‘,

= lit - Ithr sokA-1-con

3.4 Applications of Polynomial Functions

Exl: A block of snow measures 3m by 4m by 5m. The block melts in such a way that each dimension decreases in size at the same rate. At the end of a warm, sunny day, the block has a volume of 24m3.

a) Draw the initial block of snow.

b) Write a polynomial function to represent this situation.

V = Wime) In ni3

4 0

c) Determine algebraically the new dimensions of the block.

0----- I x 117 )ci

I '-P" X -(i)3 4- 12(1) 21-1i1 (li 1-36

Thus) •-z-71 oh ly pcD1bk ,s-O+°' •

Cifi‘nch,s\t‘otry block_ Q-1:\ zhol4 arc,

5-1,

wiv z: rn 3 trOef 2iry)

;ac

hot

•

r,

. r ' ,

PF,

'8 (3

,

20

„

+.

(

- 2(4 LL.

d 1

h 6 U.ir-har\

Ex2: The length of a swimming pool is 3m larger than the depth. The width of the pool is 10 in larger than the depth.

The City of Winnipeg charges $2 per in of water used to fill the pool. The bill to fill the above pool is $240.

a) Represent this situation algebraically in terms of the pool's depth, d. (A) 1 ci+h ci V(cI) (c,1--3)(ci + 10)Cc:0 flçTf/d+(O d 13 d

(le p 1-) (itzi 1 -4

b) Determine all possible values for the depth of this pool.

12,0

15 CIO 0

Gt 2 4--Is d 1-00)

c) Determine the real dimensions of the pool.

Hs t (00,o

i en b

v6k) 0 ioo )(1-41,3r 21

*we..

I (Xx 'f 000 )('

dx

2c) crf

11,

„ Homework: Page 150 #12, 15,16, 18

ch'vide Gulf- ai 'Pt/

x I 0 -25 +/SD -a-SO

X -1 0 3-.250) 21-1- /57) -250 J -.2s0

y =5 — 5) 2 c(c- ) 4' 157) (s) ---25D

I 25" 5.0 -230

-ZO 5-0 0

- ) Cx- )

/000

0 — 3-100)(21 71- 00 -1000

-25X'11 iSO X -

4 er:—...--

'-- ' : ;0:10

o'' )( '''. 5- ti> ir Z ci 3 — )1 67 0 A 8

........................._

...„ . 1r, 44,,MC,Cr11.- . •

5c h. ..) Y aochl ) ......_,.........10,1, ) ) „.._,....„

3O- (/ii) 09 3 o ,

Q . c

cm3

v - boo ye. # I I q

(-i000 /2

..,,,' c" 3

X 1 0 4,

400.-- ;;Ic

Ex3: A box is assembled by cutting the corners of a piece of cardboard and then folding up the remaining sides.

A piece of cardboard has a length of 30cm and a width of 20cm.

A square with sides measuring x cm is cut from each of the corners of the cardboard as shown in the diagram below.

'Jo crsr‘ •■••,. +••.

30-,Dx

ax

a) Write an algebraic expression that represents the volume of this box.

(s_h 30- Px

het

b) We would like a box with a volume of 1000cm3. Determine the dimensions of the box that could be created with this piece of cardboard.