Senior 3 Pre-Calculus Mathematics Module 8, Lesson 4 37

Lesson 4Polynomial Functions and Their Graphs

Objectives

1. You will learn to identifr a polynomial function.2. You will learn to relate its factors to its zeros.3. You will learn to graph a polynomial function.A polynomial function is an expression that can be written inthe form ax’ + +.. a2x2 + a1x + a0, where n is a nonnegative integer.

Each part of the expression is called a term and each term hasa coefficient associated with it. The terms are usually writtenin descending powers of x. The term containing the highestpower of x is called the leading term and its coefficient isreferred to as the leading coefficient. The power ofxcontained in the leading term is called the degree of thepolynomial. Polynomials of the first few degrees have specialnames as listed in the table below.

begree Name Example of Function

0 constant fix) = 21 linear fix> = x +22 quadratic fix) = x2 — 5x + 63 cubic fix)=x3—4x2+2x—64 quartic fix) = —3x4 + x5 quintic fix) = 5x5+4x3—6

Polynomial functions with higher degrees are named by theirdegree. For example, fix) = + 6x7 + 4x2 is called an eighthdegree polynomial.

Every polynomial defines a function. Any value for x for whichfix) = 0 is the root of the equation and the zero of a function.In Lesson 2, you learned how to factor polynomials. You cannow extend that to find the zeros of the polynomial.

38 Module B, Lesson 4 Senior 3 Pre-Caiculus Mathematics

Example 1

Given fix) = (x — l)(x + 2)(x — 3).

Solution

To find its zeros, let fix) = 0.

(x—1)(x+2)(x—3)=0

:.x—l=O x+2=0 x—3=0

x=l x=—2 x=3

The polynomial has 3 zeros or 3 x-intercepts at —2, 1, and 3.

This will be useful when you are sketching graphs of polynomial

functions.

Example 22

Givenfix)=(x—2)(x—l).

Solution

In this particular eaniple,

fix)=’Oand (x—.2)2(x--1)=0

(x — 2)(x — 2)(x +1) =0

x=2 x=2 x=—l

You say that 2 is a double root when it repeats twice. Therefore,

the zeros, will be at —land 2.

You have investigated graphs of linear and quadratic functions

(polynomial functions of degrees 1 and 2 respectively). You can

explore the graphs of quadratic functions with higher degrees.

Graph the following third degree polynomial functions using a

graphing calculator. Make a sketch of each graph on graph

paper and label your sketch with the equation.

Set window at: X. 6X6

xsc11Y6

‘max 6

YscI 1

I

_______________

Senior 3 Pre-Calculus Mathematics

3a) y=x

Module 8, Lesson 4 39

,7t

3 2c) y=x —2x —x—2

y

/

)

. •‘—•

3b) y=—xy

x

/x

40 Module 8, Lesson 4 Senior 3 Pre-Calculus Mathematics

d) y=—x3+2x2-i-x-t-2

y

I I I I I I I X

e) y=x3—3x+2

I

I

II I

f) y=-.-x3+3x—2

\Y

Senior 3 Pre-Calculus Mathematics Module 8 Lesson 4 41

How are these graphs alike? How are they different? How arethe graphs with positive leading coefficients different from thegraphs with negative leading coefficients?Graph the following fourth degree polynomial functions. Make asketch of each graph and label it on. Use the same window asbefore.

4a) y=x

x

b) y=—x4

y

x

42 Module 8, Lesson 4 SenIor 3 Pre-Calculus Mathematics

c) y =x4— 5x2 +4 [1

__

__

0I I X LI

0

42 .

d)y=—x+5x—4 .

— [

e) y=x4—3x3—x2+3x-i-3 r

Senior 3 Pre-Caiculus Mathematics Module B, Lesson 4 43

f) y=—x4+3x3+x2—3x—3

y

I- I I-- -I I

.pJDescribe the similarities and differences in these graphs.The following features of polynomial functions, f, of degree, n,should be noted.

1. The graph of a polynomial function is continuous. Thismeans the graph has no breaks — you could sketch thegraph without lifting your pencil from the paper.

2. The graph of a polynomial function has only smooth turns.The graph offhas at most (n —1) turning points. Turningpoints are points at which the graph changes from increasing

-• to decreasing or vice versa.

A function, f, is increasing in the interval if for any x1 and x2in the interval, x1 <x2 implies fix1)<fix2).

•1

44 Module 8, Lesson 4 Senior 3 Pro-Calculus Mathematics

For Points A (2, 1) and B (4, 4), 2 <4 and 1 <4 in theinterval so the function increases between A and B.

A function, f, is decreasing in the interval for any x1 and x2 in

the interval x1 <x2 implies fx1)> fix2). For point D (—2, 4)

and E (0, 1), —2 <0 and 4> 1 in the interval so the function

decreases between D and E.

A function, f, is constant in the interval if for any x1 and x2 in

the interval f(x1) = f(x2). [For points E (0, 1) and A (2, 1), the two y-coordinates are the

same so the function between E and A is constant. [For the graphs that you investigated, the cubic equation will

have at most (3 — 1) turns or two turns. For the graphs that

you investigated, the quartic equations will have at most

(4 — 1) turns or 3 turns.

3. a) When the degree, n, of a polynomial is odd. [If the leading coefficient is positive(>0), then the graph falls tothe left and rises to the right.

______ _________

If the leading coefficient is negative(<0), then the graph rises tothe left and falls to the right.

____ _______

E

b) When the degree, n, of a polynomial is even. [If the leading coefficient is positive(>0), then the graph rises tothe left and right.

_____ ________

E

C

SenIor 3 Pre-Calculus Mathematics Module 8, Lesson 445

If the leading coefficient is negative(<0), then the graph falls to theleft and right.

________

4. The function, f, has at most n.real roots. Ifyou have a cubicfunction you can expect at most three roots.When you have a quartic function, you can expect it to haveat most four roots, and so on.

Y f?x)——x3+4xcubic — at most3 roots

I.

.Ix

4 2

f(x)=x—5x+4

quartic — atmosst 4 roots

. .

.

.

flp

46 Module 8, Lesson 4 Senior 3 Pro-Calculus Mathematics Li

5. Generally speaking, the graph of a cubic function is shaped

like a “sideways S” as shown.3 2

Graphoffix)=ax ÷bx +cx+d

a>O a<O

Generally spepking, the graph of a quartic equation has a “W

shape” or a “M shape.”

Graph offix) = ax4 + bx3 + cx2 + dx + e

I *\a>O a<O

6. If a polynomial fix) has a squared factor such as (x — c)2, then [x = c is a double root offix) =0. In this case, the graph of

y = fix) is tangent to the x-axis at x = c, as shown in

figures (1), (2), and (3) below.

I FigurelI Cubic1 2

y = (x — 1)(x — 3)

Jx

Senior 3 Pre-Caiculus Mathematics Module 8, Lesson 447

•Y Figure2

y = (x — 1)(x — 3)2(x —4)

y Figure3Quintic (5th degree)

I I I

I

y=x2(x—1)(x—3)2

If a polynomial P(x) has a cubed factor such as (x — c)3, thenx = c is a triple root ofP(x) =0. In this case, the graph ofy = P(x) flattens out around Cc, 0) and crosses the x-axis atthis point, as shown in figures (4), (5), and (6) below.Y Figure 4

Cubic)7

U

48 Module 8, Lesson 4 SenIor 3 Pre-Calculus Mathematics

Y Figure5Cubicy=—(x—2)3 0

I 0

Figure 6Quarticy = (x — 1) (x —3)

[

7. When you studied quadratic functions, you were able to

determine exactly one maximum or minimum value by

determining the coordinates of the vertex. If the graph of the

function opened upward, the function had a minimum value.

If the graph of the function opened downward, it had a

maximum value.

minimum value L

/

Senior 3 Pre-Calculus Mathematics ModuleS, Lesson 449

maximum valuey2y = —x +2)

Cubic functions have neither a mimum or a minimum valuebut they may have relative mrximum or relative minimumvalues in certain intervals.

Y y=x3—3x+2

-——.1 B

In this figure, you can see that it does not have an absolutemdmum y-value (largest y-value) or an absolute minimumy-value (smallest y-value).However, it does have a peak at A where a relative maximumvalue fory is found in a specific interval. In the interval between—2 and 1 the relative maximum is when y =4.The same graph has a valley at B. This is representing thesmallest y-value within a specific interval and is called arelative minimum. Its relative minimum is at y = 0 and occurswhen x = 1.

50 Module 8, Lesson 4 Senior 3 Pre-Calculus Mathematics

Quartic functions have either an absolute maximum or an

absolute minimum value. They can also have relative maximum

andlor relative minimum values within intervals.

y=x4—5x2+4

\JIt has an absolute minimum value occurring at points A and B

wherey —2.2.

It has a relative maximum value at point C where y =4.

You can use the trace feature on the graphing calculator for the

approximations.

Another feature that you can use to sketch the graph of a

polynomial function is to use the sign analysis feature by testing

one x-value from each of the intervals determined by the zeros.

ExampleSketch the graph of the factored cubic function [

fix)=(x+ 1)(x—1)(x—2)

Remember to find the zero, let fix) = 0

(x + 1)(x — 1)(x —2) = 0

x=—1,x=1,x=2 [The zeros are at —1, 1, and 2.

CL

.

CC

SenIor 3 Ike-Calculus MathematIcs Module 8 Lesson 451

+ — +

— I 1 4 I . 4 -I—3 —2 —1 0 1 2 3

Interval:x>2,testx=3f3) = (3 + 1)(3 — 1)(3 —2) = + product

+ + +

meaning the y-values are above the lineInterval: 1 <x <2, test x = 1.5flL5) = (1.5 + 1)(1.5 — 1)(1.5 —2) = — product

+ +

meaning the y-values are below the lineInterval: —1 <x <1, testx = 0f0) = (0 + 1)(0 — 1)(0 —2) = + product

÷ —

meaning they-values are above the lineInterval: x <—1, test x = —2f(—2) = (—2 + 1)(—2 — 1)(—2 —2) = — product

meaning the y-values are below the lineTo get the position of the curve you could plot points within theinterval but if it is a sketch and you know it is cubic, rememberit will be a “sideways S.” The graph will be like the one below.

Y y—_(x+1)(x—1)(x—2)

52Module 8, Lesson 4 SenIor 3 Pre-Calculus Mathematics

Technology will save you time if you know the equation of

the function and can use the trace function to find the zeros,

relative meirirnum or minimum, etc., and are familiar with

the features of the polynomial function.

8. You can use transformations to advantage for some of your

sketching of graphs.

The polynomial functions that have the simplest graphs are

the monomial functions fix) = ax’. When n is even the graph

is similar to the graph offix) = x2. When n is odd the graph is

similar to the graph offix) = x3. Moreover, the greater the

value of n, the flatter the graph of a monomial is on the

interval—i x 1.

For n even, the graph of

y = x’ is tangent to thex-axis at the origin.

(—1,1)

5y =x

Uj

y

(1

/ /.*.-yx2

4y =x

(1,1)

x

IUU50JE[

F

c

[

5y=x —

y

(1, 1)

For n odd, the graph of

y = x’ crosses the x-axis at

the origin.

SenIor 3 Pre-Calculüs Mathematics Module 8, Lesson 453

• Reflection: To sketch the graph ofg(x) = —x5 reflect thegraph offix) = x5 in the x-axis.

y

(—1,1)I I I I I I

x•(1,-4)

g(x)=—x5

Vertical shift: To sketch the graph ofg(x) = x4 + 1 shift thegraph offix) = x up one unit.

4 g(x)=—x4+1

I I I

‘(Ui)

x

Horizontal shift: To sketch the graph ofg(x) = (x + 1) shiftthe graph offix) = x3 one unit to the left.

g(x)= (x+ i)

I1 I

They-intercept of the graph of a function occurs when x = 0.The x-intercepts are the zeros of the function. At an—

x-intercept,the value of the function is zero.

54 Module 8, Lesson 4 Senior 3 Pre-Calculus Mathematics

Assignment 4

1. Explain what is meant by a continuous graph?

2. Name a feati.ire of the graph offix) = I x that is not shared

by the graphs of polynomial functions.

3. Does the graph off(x) = 2x4 — 3x rise or fall to the right? How

can you tell? What happens to the left?

4. State the maximum number of turns in the following graphs: r3

La) fix)=x .—4x

6 2b) g(x)=x —4x

c) fix)=—x2--5x-i-6

d) g(x).=x —4x +64

e) fix)=—3x —5x-i-6

5. Determine the right and left behaviour of the following

graphs:3

a) fix)=—x +3x

b) fix)=2x4—5x2-i-4

c) fix)=(x—1)(x+3)(x—1)4 2

d) /tx)=—x +x

e) /(x)=—2x5+x4—2x

e) ftx)= 3x5 -t- x3 —2

6. Match the polynomial function with the correct graph:

a) f(x)=—3x+5

b) f(x)=x2—2x

c) f(x)=—2x2—9x—9

d) f(x)=3x3—9x÷1

13 2e) f(x)=—x ÷x—

f) f(x)=_.x4+2x2

g) f(x) = 3x4 ÷ 4x3

h) f(x) = — 5x3 + 4x

Senior 3 Pre-Calcutus Mathematics ModuleS, Lesson 4 55

i) y

Ii(-1

I-- i-- I IIi

1—i

[ii)

-IlLIl-IlIl-Il IIIlI,IIIIIia. [‘N

•• iii)

y

I I I 1 I. I

UUUU

56 Module 8, Lesson 4 Senior 3 Pre-Calculus Mathematics

LC

C

iv) y

fEE

v)

[C

1:CCI

x

x

vi)

x

Senior 3 Pre-Calculu Mathematics Module 8, Lesson 4 57

7. Find 1) the zeros of the following functions, ii) the number ofterms, iii) left-right behaviour, iv) sketch the graph of thefunction, and v) the maximum and/or minimum values.a) f(x)=x3—x2—6x

b) f(x)=—x3+4x

c) f(x)=x4—2x3—3x2+4x+4

d) f(x)=—x5—2

I

3e) f(x)=(x—4)

0 f(x)=x3+5x2+2x—8

vii)

viii)

7y

x

[1

58 Module 8, Lesson 4 Senior 3 Pie-Calculus Mathematics Li

8. An open box is to be made from a 10 cm by 12 cm piece of

cardboard by cutting x cm squares from each corner and

folding up the sides. Write the function given the volume of

the box in terms of x and use it to approximate the value ofx

that produces the greatest volume.

LFr

C

p

:.

Senior 3 Pre-Calculus Mathematics Module 8, Lesson 4, Answer Key 23

Answer Key

Lesson 41. It is a graph that you can sketch without raising your pencil

from the paper.

2. The graph of f(x) = x) is a V shaped curve so the turn issharp and not smooth.

3. The graph rises to the right when the leading coefficient (2)is positive. If the degree is an even number and the leadingcoefficient is positive it rises to the left.

4. The maximum number of turns is (n —1) where n is thedegree of the polynomial function.a) Since the degree, n, is 3, then there are (3 — 1) or 2 turnsb) Since n = 6, there are 6— 1 or 5 turns possiblec) Since n = 2, there is 2 — 1 or 1 turn possibled) since ii = , there are 5-. 1 or 4 turns possiblee) Since n = 4, there are 4— 1 or 3 turns possible

5. a) Because the degree is odd and the leading coefficient isnegative, the graph rises to the left and falls to the right.

b) Because the degree is even and the leading coefficient ispositive, the graph rises to the left and right.

c) You can determine that the degree is 3 if you multiply thethree factors together. Because the degree is odd but theleading coefficient is positive, the graph rises to the rightbut falls to the left.

d) Because the degree is even and the leading coefficient isnegative, the graph falls to the left and right.

e) Because the degree is odd and the leading coefficient isnegative, the graph falls to the right and rises to the left.

f) Because the degree is odd and the leading coefficient ispositive,the graph rises to the right and falls to the left.

-

fi24 ModuleS, Lesson 4, Answer Key SenIor 3 Pre-Calculus Mathematics

6. a)—e)b)—c)c) — b)d)—f)

e)—a)

f)—g)

g)—d)

[7. a) i) f(x) = x3 — — 6x common factor

f(x)=x(x—3)(x+2)

.Letf(x)=O

.‘. x(x—3)(x+2)=O rx=O,x=3,orx=—2

ii) (n, — 1) turns where n = 3 is 3— 1 or 2 turns at most

iii) Because the degree of the function is odd and the leadingcoefficient is positive, the graph falls to the left and rises L

to the right.

iv) Y

A/)

II I I I I I I

3 2f(x)=x —x —6x

Using sign analysis you can determine that:

• in the interval x < —2, the curve is below the x-axis,

• in the interval —2 <x < 0, above the x-axis• in the interval 0 <x <2, below the x-axis• in the interval x> 2, above the x-axis. [Choose points between —2 and 0 to find the approximate

peak (y = 4) and between 0 and 3 to find the approximate

value(y7.875).... -

Senior 3 Pre-Calculus Mathematics Module B, Lesson 4, Answer Key25

v) No absolute maximum or minimum values. However,there is a local maximum at A where y = 4 and at Bwhere

b) 1) f(x)=—x3+4x

f(x) = —4)

f(x)=—x(x—2)(x+2)

Let f(x) =0 to find the zeros—x(x — 2)(x +2) = 0x =0, x =2, or x = —2

ii) Number of terms (n— 1)(3— 1)2 turns at mostiii) Because the degree of the function is odd and the leadingcoefficient is negative, the graph falls to the right andrises to the left.

iv)

f(x) = _‘: 4x

v) There is no absolute maximum or minimum. There is alocal maximum at A where y = 3 and a local minimum atB where y = —3.

26 Module 8, Lesson 4, Answer Key Senior 3 Pre-Calculus Mathematics

c) i) f(x)x4—2x3—3x2+4x+4

Use the remainder and. factor theorem to find the factor.

possible zeros of 4: ±1, ±2, ±4

If f(—1) = 0, then x + 1 is a factor and —1. is a zero.

f(x) = — 2x3 = 3x2 + 4x +4

= (—i)4 — 2(—1) — 3(_1)2 + 4(—1) ÷4

=1÷2—3—4+4=0

Use synthetic division to get a new quotient to factor:

—1)1 —2 —3 +4 +4 0: :

New quotient x3 — 3x2 + 4

f(x)= x3 —3x2+4

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

4

=—1—3+4

=0

x+ lisafactor andx=—lisalsoaroot.

Use synthetic division to get a new quotient:

—4)1 —3 +0 4

—1 4 —4

1 —4 4 0

New quotient is x2 — 4x + 4 which factors as (x — 2)2.

(x—2)2 (x+1)20

(x — 2)2 = 0 (x + 1)2 = 0 rx=2 x—1

Double zeros of 2 and —1.

ii) Number of turns (n — 1) or (4— 1) or 3 turns at most.

iii) Because the degree of the function is even and the

leading coefficient is positive, the graph rises to the right

and to the left.

Senior 3 Pre-Calcutus Mathematics Module 8, Lesson 4, Answer Key 27

• iv)• 4

I I I I I

• —1 2

v) Absoluteminimumwhenx—1 andy=Oandatx=2when y 0. Relative maximum when x = 0.51 andy= 5.1.

d) i) Zeros

Letf(x)=0

—x5—2=0

x5=—2x—t15

ii) Number of turns (n — 1) or 4 turns.iii) Sine the degree of the function is odd an the leading

coefficient is negative, the graph rises to the left and fallsto the right.

iv) You can use a reflection ofy = over the x-axis and avertical transformation of f(x) = —x5 shifted two unitsdownward.

y

I I II I x

v) There is no maximum of minimum.

U

28 Module 8, Lesson 4, Answer Key Senior 3 Pre-Calcutus Mathematics

e) i) Letf(x)=0

(x—4) =0

x—4=0

x=4

ii) Number of turns is 3— 1 or 2 turns.

iii) Since the degree of the function is odd and the leading

coefficient is positive, the graph of the function rises to

the right and falls to the left.

iv) y

UI I I X j

2y = (x —4)

This is a horizontal transformation of y = x3 shifted

4 units to the right.. ..

v) There is no maximum or minimum value.

f) i) f(x)=x3+5x2+2x—8

Possible zeros are the factors of 8: ±1, ±2, ±4, ±8

If f(r) = 0, then x — r is a factor and r is a zero.

f(1)=1-i-5(1)2-i-2(1)—8

=1+5÷2—8

=0

x — 1 is a factor and x = 1 is a zero

Senior 3 Pre-Calculus Mathematics Module 8, Lesson 4, Answer Key 29

Use synthetic division to get a new quotient:

1)1 5 2 —8

16 8

168 0

.. x2 + 6x + 8 is the new quotient which factors into(x+4)(x+2)0(x — 1)(x + 4)(x +2) = 0zero = — 2, — 4, 1

ii) Number of turns is 3— 1 or 2 turns.iii) Since the degree of the function is odd and the leading

coefficient is positive, the graph of the function rises tothe right and falls to the left.

iv) y

________________

I14 I I I III I x

y=x3+5x2+2x—8

v) There is no absolute maximum or minimum values. A isa local maximum achieved at y — 4.1 where x is —3 and Bis a local minimum achieved when x 0.26 and y = —8.2.

cir

0

30 Module B, Lesson 4, Answer Key Senior 3 Pre-Calculus Mathematics

8. Volume = height x length x width

= x(12 — 2x)(1O — 2x)

x 10 0xJ

12—2x 12

U

Fxx x

Use your graphing caJ.culator and the trace function to find

the maximum volume. The value of x is 1.81 cm and the

maximum vollume is 96.770564 cm3.

i...

4f .

‘ -A . : .

- :-:-_

;-:r