Polynomials-graphs Summary of polynomial graphs and solving them luxvis 11/19/2012

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POLYNOMIAL FUNCTIONS

A polynomial can be represented in the general form by the following expression:

n

n xaxaxaxaxaaxP ...)( 4

4

3

3

2

210

Where 0a , 1a , 2a , are constants

n

n xa Is called the leading term and n is called the degree of the polynomial.

Polynomials of different

degrees

1 1 degree of polynomial a called is This

13)( 1 xxP

linear functions

2 2 degree of polynomial a called is This

132)( 12 xxxP

Quadratic functions

3 3 degree of polynomial a called is This

1354)( 123 xxxxP

Cubic functions

Review of Polynomials of degree 3

Polynomials of degree 3

General form 3axy

Properties Point of inflexion at (0, 0)

if a>0

If a < 0 then the graph

looks like this

Page 2 of 13

khxay 3)(

a h k Point of

inflexion

minus

inverts the

graph

a-dilation, graph

steeper if a > 1

Flatter if 0 < a <1

horizontal

translation of

h to right

vertical

translation of

k

( , )h k

Example 3

Sketch the graph of 2)1(3 3 xy

Also did you notice that compared with the general equation, here a > 0 therefore graph is in quadrant 1 and 3.

Polynomials of the form ))()(( cxbxaxy

X intercepts: (a, 0) (b, 0) and (c, 0)

Y intercepts: (0, -abc)

To find the turning points we will need to differentiate the function and set the derivative to zero and then

solve for x

Polynomials of the form )()( 2 bxaxy

X intercepts: (a, 0) (b, 0)

Y intercepts: (0, -a2b)

One of the turning points will be at (a, 0) because (x-a) is a repeated factor.

Polynomials of the form dcxbxaxy 23

To sketch this type of polynomial we must use the factor theorem and factorize the polynomial to find the x

intercepts.

If the expression does not factorize, the x intercepts might then be found using a graphics calculator.

Y intercept: (0, d)

To find the turning points we need to differentiate the function and equate the derivative to zero and then

solve for x.

x-6 -4 -2 2 4

y

-2

2

4

6

8

3(x – 1)3

+ 2

Point of Horizontal Inflection

( 1 , 2 )

x Intercept

( 0.12642 , 0 )

y Intercept

( 0 , -1 )

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Graphs of the various polynomials

Example 1

3)2( 3 xy

Compare this with khxay 3)(

Notice negative sign (-) meaning graph is inverted

h = 2 and k = -3 therefore (h, k) is (2, -3) is the point of inflection

so now you could draw the graph roughly

Let us work out the X intercept let y=0

5577.0

4422.12

32

3)2(

3)2(

03)2(

3

3

3

3

x

x

x

x

x

x

x-10 -5 5 10

y

-10

-5

5

10

y = – (x – 2)3– 3

x Intercept

( 0.55775 , 0 )

Point of Horizontal Inflection

( 2 , -3 )

y Intercept

( 0 , 5 )

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Example 2

Sketch 652 23 xxxy

Obviously we could just use our graphics calculator but we would like to use algebra to solve and sketch this

graph.

Use the factor theorem to find the X intercepts

factor a is 1)-(x Therefore

0)1(

6521)1(

6)1(5)1(2)1()1(

652)(

23

23

P

P

P

xxxxP

Now we do long division

652 23 xxx )2)(3)(1()6)(1( 2 XXXXXX

Therefore the x intercepts are x= 1, x=3 and x= -2

Y intercepts (0, 5)

To find the turning points we take the derivative

)21.8,79.( and )0606.4,1196.2(

thereforeare points Turning

786.0or 1196.2

formula theusing quadratic theSolve

0543

543

2

2

xx

xx

xxdx

dy

Now sketching the graph

6

652)1(

2

3

xx

xxxx

Page 5 of 13

Obviously it is faster to work it out with a graphics calculator but there will be questions where you will be

required to work things out from first principles

More information on polynomials that you might have missed out in previous years

POLYNOMIALS GRAPHS

We have already seen polynomial graphs- for example straight line graphs and quadratic graphs are a type of

polynomial graphs but now we will examine very briefly some more aspects of polynomial graphs and their

behaviour.

A polynomial can have constants, variables and the exponents such as 0,1,2,3,4,5, etc

Polynomials can be combined using addition , subtraction but not by division!

Examples of polynomials- notice the exponents are

all positive!

33 2 2x x , 3 25 2 4xy x y

These are not polynomials- we cannot have division

or negative exponents!

2

2x

33x

Monomial , Binomial , Trinomial

Monomial- polynomials with 1 term 34x y

Binomial-polynomials with 2 terms 6 2x

Trinomial-polynomials with3 terms 32 2 5y x

x-10 -5 5 10

y

-10

-5

5

10

x3– 2x

2– 5x + 6

x Intercept

( 3 , 0 )

x Intercept

( 1 , 0 )Local Minimum

( 2.1196 , -4.0607 )

Local Maximum

( -0.7863 , 8.2088 )

y Intercept

( 0 , 6 )

x Intercept

( -2 , 0 )

Page 6 of 13

Quick Note about the Fundamental Theorem of algebra

The fundamental theorem of algebra is not the start of algebra , though it might sound like it is , but it does say

something interesting about polynomials and that is the following:

Any polynomial of degree n has up to n roots

Let us explain the idea behind this and you will see the connection with graphs!

Example:3 23 5 2x x x

How many terms does this polynomial have It has4 terms

What is the degree of the polynomial It is the higher power which in this case is 3

So if the degree is 3 then that means that the

polynomial will have 3 roots

A root is the same as where it cuts the x axis.

A root is the where the polynomial is equal to zero

So how does the graph look like? Notice that it crosses the x axis three times.

A polynomial can be rewritten to look like this sometimes - 1 2 3 4Polynomial ( )( )( )( )a x r x r x r x r

where 1 2 3 4( ),( ),( ),( )x r x r x r x r are called factors of the polynomial.

For example 2 9 ( 3)( 3)x x x

Therefore the roots are 3 and -3

x-20 -10 10 20

y

-20

-10

10

20

x Intercept

( 4.100432 , 0 )

x Intercept

( 0.33888 , 0 )

x Intercept

( -1.439312 , 0 )

x3– 3x

2– 5x + 2

Page 7 of 13

Multiplicity

When a factor appears more than once this is called multiplicity. For example2( 3) ( 3)( 3)x x x , so here

( 3)x has a multiplicity of 2 (it appears twice)

We need to factor in multiplicity when we looking at the degree of polynomial and how many roots it has.

For example- 4 3x x

How many roots would we expect? 4, would be the logical answer ( up to 4)

However when you factorize it , it looks like this 3( 1)x x , and you will notice that x has a multiplicity of 3 , as

the root is repeated 3 times.

There should be 4 roots and of course 4 factors but what happens is we get the following3( 1)x x , which leads

to

x = 0 and x = -1 as been the only roots and factors! ( but if you count the multiplicities there are actually 4 after

all)

So the point is that sometimes if we do not account for the multiplicity it might appear that we way short when it

comes to roots.

Multiplicity 3x 3

( 1)x 1

Total 4

Summary of ideas so far

1. A polynomial of degree n has n roots

2. Sometimes a factor appears more than once and that is its multiplicity.

3. Solving a polynomial means finding its roots i.e. that is where the function is equal to zero

4. If you know the roots you can also find the factors of a polynomial and the opposite is true also

5. If you have found the root put that in the place of x and the polynomial should be equal to zero. So if the

root is 2x then when we put it into the equation 3( ) ( 2)f x x x , then we would obtain a zero. So

3(2) 2 (2 2) 0f . Remember that 2x , then ( 2)x is a factor of the original polynomial

equation.

Page 8 of 13

Various names of common polynomials

Degree Name Example What the graph look like

0 Constant 4

1 Linear 2 1x

2 Quadratic 2 4 2x x

3 Cubic 3 2x x

4 Quartic 4 3x x

Interesting graphs and multiplicity of a root and turning points

We can determine the shape of polynomials quickly if we know the multiplicity of the polynomial

x-10 -5 5 10

y

-10

-5

5

10

4

x-10 -5 5 10

y

-10

-5

5

10

2x + 1

x-10 -5 5 10

y

-10

-5

5

10

x2– 4x + 2

x-10 -5 5 10

y

-10

-5

5

10

x3

+ x2

x-10 -5 5 10

y

-10

-5

5

10

x4

+ x + 3

Page 9 of 13

Multiplicity What it does

Even-2-4-6-etc powers Touches the x -axis at that point

Odd-1-3-5- Crosses the x axis at that point

Turning points = Degree - 1 or less-(watch out for inflexion points! )

Let us put that into practice- 2( 1)( 3)y x x

Analysis

Look at the factor 1( 1)x this has a multiplicity of 1 It is an odd power so the graph cross at 1x

Look at the factor 2( 3)x , this has a multiplicity of

2

This is an even power so the graph will touch at

3x

Then put 0x into the equation will give us a

6y

This provides us with the y intercept and now we can

quickly sketch this graph

How many turning points will it have? Find the

degree which is the highest power and subtract 1

from it

Now the degree is 3, so the number of turning points

is 3-1 = 2 at most

Now let us try another example using this method

Let us put that into practice- 3 2( 1) ( 3)y x x

Analysis

Look at the factor 3( 1)x this has a multiplicity of 3 It is an odd power so the graph cross at 1x

Look at the factor 2( 3)x , this has a multiplicity of

2

This is an even power so the graph will touch at

3x

Then put 0x into the equation will give us a

6y

This provides us with the y intercept and now we can

quicly sketch this graph

What is the degree of this polynomial? 5 So 5-1 = 4 means they should be at most 4 turning

points

x-10 -5 5 10

y

-10

-5

5

10

(x – 1)(x – 3)2

Local Minimum

( 3 , 0 )x Intercept

( 1 , 0 )

Page 10 of 13

Factorizing Polynomials-Review

The factor theorem and the remainder theorem applications

The factor theorem

Consider a polynomial xP that has x as a linear factor. Then xQxxP , where xQ is a polynomial

one degree lower than xP , obtained by expanding and comparing coefficients, or dividing xP by x .

Replacing x by in xQxxP ,

QP . Hence 0P .

Conversely, for any polynomial xP , if 0P , then x is a factor of xP . This statement is known as the

factor theorem, and can be used to find the linear factors of a polynomial if other methods failed.

The factor theorem is best used for polynomials with linear factors of rational coefficients. The value(s) of is

found by trial and error. The possible values of for trying depend on the first and last coefficients of P(x)

nn

nn axaxaxaxP

1

1

10 ...... , 0...

...

aoffactora

aoffactora n .

If all these values give 0P , it does not necessarily mean that xP has no linear factors, because the

coefficients of the linear factor(s) may be irrational.

Example 1: Use the factor theorem to find a linear factor of 2723 23 xxx , then find the quadratic factor and

hence all the linear factors.

x-10 -5 5 10

y

-10

-5

5

10

x Intercept

( 1 , 0 )

Point of Horizontal Inflection

( 1 , 0 )

Local Minimum

( 3 , 0 )

Local Maximum

( 2.2 , 1.1059 )

Page 11 of 13

Let 2723 23 xxxxP . The possible values of for testing are 3,1

2,1 , i.e. 2,1,

3

2,

3

1 .

02171213123

P .

02171213123

P , 1x is a factor.

Divide xP by 1x to find the quadratic factor.

253 2 xx

1x 2723 23 xxx

23 33 xx

xx 75 2

xx 55 2

22 x

22 x

0

Hence 21312531 2 xxxxxxxP .

Another possible outcome:

023

17

3

12

3

13

3

123

P ,

3

1x is a factor.

633 2 xx

31x 2723 23 xxx

233 xx

xx 73 2

xx 23

26 x

26 x

0

Hence 23

13633

3

1 22

xxxxxxxP

123

13

xxx . It is equivalent to the previous result.

Example 2: Given 1x is a factor of 2993 234 xxxx , find the cubic factor xQ such that

xQxxxxx 12993 234 .

Let dcxbxaxxQ 23, then

dcxbxaxxxxxx 23234 12993 .

Expand to obtain 2993 234 xxxx

dxdcxcbxbaax 234.

Page 12 of 13

Compare the coefficients on both sides, 3a , 1 ba , 9 cb , 9 dc and 2d . Hence 2b , 7c .

2723 23 xxxxQ .

This example illustrates an alternative method in finding xQ to long division shown in the last example.

Example 3: Use the factor theorem to find the linear factors of 23234 xxxx .

Let 23234 xxxxxP . Test 2,1 .

02131111234

P

02131111234

P , 1x is a factor.

Hence xQxxP 1 .

Use long division or comparing coefficients to find 23 xxxQ .

Hence 21 3 xxxxP . Use the factor theorem on 23 xxxQ to find its linear factor. Test 2,1 .

021113

Q , 1x is a factor of xQ .

Hence xTxxxP 11 . xT is quadratic and found by long division of xP by the expansion of

11 xx , or comparison of coefficients as discussed in example 2, 22 xxxT .

211 2 xxxxxP .

The remainder theorem When a polynomial xP is divided by a linear binomial x , the remainder can be

found quickly without actually carrying out the division.

Since

( )Q x Quotient

ax ( )P x Dividend (polynomial)

Divisor

R Remainder

RxQxxP .

When

x , RRxQP

,

i.e. the remainder

PR when xP is divided by x . This is known as the remainder theorem.

Example 4: Find the remainder when 1152 34 xxxxP is divided by (i) 5x , (ii) 32 x , (iii) ax 2 .

(i) 13391155552534

PR

(ii) 4

1311

2

35

2

3

2

32

2

334

PR

(iii) 111083211252222 3434 aaaaaaaPR

Example 5: Given that 2x is a factor of 23 23 qxpxx and the remainder is 20 when the cubic

polynomial is divided by 2x . Find the values of p and q.

Use the factor theorem and the remainder theorem to set up two simultaneous equations for p and q.

Let 23 23 qxpxxxP .

2x is a factor of xP , 02 P ,

Page 13 of 13

02222323

qp , 112 qp ……(1)

The remainder is 20 when divided by 2x , 202 P ,

202222323

qp , 32 qp ……(2)

Solve eqs (1) and (2) for p and q.

(1) + (2), 84 p , 2 p

(1) (2), 142 q , 7q

Special Notes 2 2 ( )( )x y x y x y

2 22 ( )( )x xy y x y x y

2 22 ( )( )x xy y x y x y

3 3 2 2( )( )x y x y x xy y

3 3 2 2( )( )x y x y x xy y

3 2 2 3 33 3 ( )x x y xy y x y

3 2 2 3 33 3 ( )x x y xy y x y