3.5 – Graphs of

Symmetry of Polynomial FunctionsThis information is a review of symmetry from the unit on graphs of functions. We will be considering two types of symmetry in this lesson; symmetry about the y Even Functions: An even function is a function that is symmetric to the yfunctions with all even exponents are symmetric to the y

33)( 24 +−= xxxf Although it is clear by inspection that the above polynomial functions are even, such is not always the case. To determine if a function is even we may substitute the x

)()( xfxf −= . Example: Show that the function f Solution: Replace all x’s with –x’s.

)()(

33)(

(3)(3)(

33)(

24

4

24

xfxf

xxxf

xxf

xxxf

=−+−=−

−−=−+−=

Since )()( xfxf −= the function is even and is therefore symmetric about the y-axis. Example: Show that the function h Solution: Replace all x’s with –x’s.

)()(

75)(

)(5)(

75)(

26

6

26

xhxh

xxxh

xxh

xxxh

−=−=−

−−=−−=

Since )()( xhxh −= the function is even and is therefore symmetric about the y-axis.

Polynomial Functions

of Polynomial Functions: This information is a review of symmetry from the unit on graphs of functions. We will be considering two types of symmetry in this lesson; symmetry about the y- axis and symmetry about the origin.

An even function is a function that is symmetric to the y-axis. It gets its name from the fact that polynomial functions with all even exponents are symmetric to the y-axis. Some examples of even functions are:

2 1)( 2 −= xxg )(xh

Although it is clear by inspection that the above polynomial functions are even, such is not always the case. To determine if a function is even we may substitute the x-values with (–x)-values and determine if

233)( 24 +−= xxxf is even.

x’s.

2

2)(

22x

++−

the function is even and is therefore

26 75)( xxxh −= is even.

x’s.

)(72

2x−

the function is even and is therefore

Polynomial Functions

This information is a review of symmetry from the unit on graphs of functions. We will be considering two about the origin.

axis. It gets its name from the fact that polynomial axis. Some examples of even functions are:

26 75) xx −=

Although it is clear by inspection that the above polynomial functions are even, such is not always the case. ues and determine if

Odd Functions: An odd function is a function that is symmetric to the origin. It gets its name from the fact that polynomial functions with all odd exponents are symmetric to the origin. This type of symmetry is also called a 180 degree rotation since all points on the graph are rotated 180 degrees about the origin. Some examples of odd functions are:

xxxf 3)( 3 −= Although it is clear by inspection that the above polynomial functions are odd, such To determine if a function is odd we may substitute the x(–x)-values and determine if )( xf − Example: Show that the function h Solution: Replace all x’s with –x’s.

)()(

75()(

75)(

(7)(5)(

275)(

57

57

7

57

xhxh

xxxh

xxxh

xxh

xxxxh

−=−−−=−

−+−=−−−−=−

+−=

Since )()( xhxh −= the function is odd and is therefore symmetric about the origin. Example: Show that the function g Solution: Replace all x’s with –x’s.

)()(

12()(

12)(

12)()(

12)(

5

5

5

35

xgxg

xxg

xxg

xxg

xxxg

−=−−−=−

+−=−−−=−

−=

Since )()( xgxg −= the function is odd and is therefore symmetric about the origin.

An odd function is a function that is symmetric to the origin. It gets its name from the fact that polynomial functions with all odd exponents are symmetric to the origin. This type of symmetry is also called a 180

ts on the graph are rotated 180 degrees about the origin. Some examples of odd

35 12)( xxxg −= xh )( =

Although it is clear by inspection that the above polynomial functions are odd, such To determine if a function is odd we may substitute the x-values with

)(xf−= .

xxxxh 275)( 57 +−= is odd.

x’s.

)2

2

)(2)5

x

x

xx

x

+

−−+−

the function is odd and is therefore

35 12)( xxxg −= is odd.

x’s.

)12

12

)(12

3

3

3

x

x

x−

the function is odd and is therefore

An odd function is a function that is symmetric to the origin. It gets its name from the fact that polynomial functions with all odd exponents are symmetric to the origin. This type of symmetry is also called a 180

ts on the graph are rotated 180 degrees about the origin. Some examples of odd

xxx 275 57 +−

Although it is clear by inspection that the above polynomial functions are odd, such is not always the case.

Behavior at the X-Intercepts:When a polynomial equation touches the xthe x-axis. This behavior is determined from the factors of the polynomial as follows:

1. The graph crosses the x-axis is the factor the root comes from is raised to an odd power (odd multiplicity).

2. The graph does not cross the xeven power (even multiplicity).

Example: Determine the behavior of the polynomial Solution: Since this polynomial is already factored, we simply derive the roots from the factors.

Factor Root K x-2 x=2 1 x+1 x=-1 2

Remember that K represents the multiplicity of the factor or how many times it occurs. Example: Determine the behavior of the polynomial Solution: Since this polynomial is already factored, we simply derive the roots from the factors.

Factor Root K x+2 x =-2 3 x-3 x=3 4

Remember that K represents the multiplicity of the factor or how many times it occurs.

Intercepts: When a polynomial equation touches the x-axis, it will do one of two things; cross the x

is determined from the factors of the polynomial as follows:

axis is the factor the root comes from is raised to an odd power (odd

The graph does not cross the x-axis (turns around) if the factor the root comes froeven power (even multiplicity).

Determine the behavior of the polynomial 2)1)(2()( +−= xxxP at its roots.

Since this polynomial is already factored, we simply derive the roots from the factors.

Behavior Crosses

Not Cross

Remember that K represents the multiplicity of the factor or how many times it occurs.

Determine the behavior of the polynomial 43 )3()2()( −+= xxxP at its roots.

Since this polynomial is already factored, we simply derive the roots from the factors.

Behavior Crosses

Not Cross

Remember that K represents the multiplicity of the factor or how many times it occurs.

axis, it will do one of two things; cross the x-axis or not cross is determined from the factors of the polynomial as follows:

axis is the factor the root comes from is raised to an odd power (odd

if the factor the root comes from is raised to an

at its roots.

at its roots.

Example: Determine the behavior of the polynomial Solution: After using synthetic division to determine the factors, the polynomial may be written as

2)1)(5()( −−= xxxP . Now we simply derive the roots from the facto

Factor Root K x-5 x=5 1 x-1 x=1 2

Remember that K represents the multiplicity of the factor or how many times it occurs.

Turning Points: A polynomial function of degree n will have at most n Example: The function 2)( 3= xxfpoints. Example: The function )( 5 −= xxfpoints.

Determine the behavior of the polynomial 5107)( 23 −+−= xxxxP at its roots.

After using synthetic division to determine the factors, the polynomial may be written as. Now we simply derive the roots from the factors.

Behavior Crosses

Not Cross

Remember that K represents the multiplicity of the factor or how many times it occurs.

A polynomial function of degree n will have at most n-1 turning points.

153 +− x is a 3rd degree polynomial, so it will have at most 2 turning

186 3 ++− xx is a 5th degree polynomial, so it will have at most 4 turning

at its roots.

After using synthetic division to determine the factors, the polynomial may be written as

degree polynomial, so it will have at most 2 turning

degree polynomial, so it will have at most 4 turning

End Behavior of a Polynomial FunctionA polynomial function has very predictable behavior as the xand is refered to as the functions end behaviorincreasing without bound. Likewise, to state that x approaches negative infinity decreasing without bound. As ∞→x , the y-values of the polynomial function will increase or decrea

−∞→x , the y-values of the polynomial function will increase or decrease without bound. This behavior can be predicted by looking at the leading term of the polynomial. Specifically we need to look at the signand the power of the leading term. Example: Determine the end behavior of the polynomial function Solution: We only need to look at the leading termSince the term is positive and the power is odd, we have:

Behavior of a Polynomial Function: A polynomial function has very predictable behavior as the x-values approach positive or negative infinityand is refered to as the functions end behavior. To state that x approaches infinity increasing without bound. Likewise, to state that x approaches negative infinity (

values of the polynomial function will increase or decrease without bound. Likewise, asvalues of the polynomial function will increase or decrease without bound. This behavior

can be predicted by looking at the leading term of the polynomial. Specifically we need to look at the sign This is called the Leading Term Test:

behavior of the polynomial function 2)( 3 −= xxxP

We only need to look at the leading term3x . Since the term is positive and the power is odd, we

values approach positive or negative infinity . To state that x approaches infinity ( ∞→x ) means that x is

( −∞→x ) means that x is

se without bound. Likewise, asvalues of the polynomial function will increase or decrease without bound. This behavior

can be predicted by looking at the leading term of the polynomial. Specifically we need to look at the sign

1+x

Example: Determine the end behavior of the polynomial function Solution: We only need to look at the leading termsince the term is negative and the power is odd, we have: Example: Determine the end behavior of the polynomial function Solution: We only need to look at the leading termSince the term is positive and the power is

Example: Determine the end behavior of Solution: We only need to look at the leading termSince the term is negative and the power is

Sketching the Graph of a Polynomial

behavior of the polynomial function )( 5 −−= xxP

We only need to look at the leading term5x− negative and the power is odd, we have:

behavior of the polynomial function 23)( 4 −= xxP

We only need to look at the leading term4x . tive and the power is even, we have:

end behavior of the polynomial function 4)( xxP −−=

We only need to look at the leading term4x− . Since the term is negative and the power is even, we have:

Sketching the Graph of a Polynomial

2−x

232 3 −+ xx

32x .

Function: To sketch the graph of a polynomial we will do the following:

1. Determine number of roots and turning points from leading term. 2. Find the roots of the equation and their multiplicity 3. Determine the end behavior of the function 4. Determine the symmetry, if any. 5. Graph the equation.

Example: Sketch the graph of the polynomial function 485)( 23 −+−= xxxxP Solution: Step 1: Find the zeros of the polynomial equation. Using synthetic division the equation may be written in terms of its factors as:

2

1

0)2)(1( 2

==

=−−

x

x

xx

Step 2: Determine nature of function at its roots

Factor Root K Behavior x-1 x =1 1 Crosses x-2 x=2 2 Not Cross

Step 3: Determine nature of function as ∞→x and −∞→x . Since the leading term is positive and odd we

have:

The Derivative: A useful mathematical tool borrowed from calculus is the derivative. To find the derivative of a polynomial function, multiply the coefficient of each term by the exponent and reduce the value of the exponent by one. If the original function is P(x) then the new function is P’(x) (pronounced P prime of x)..

Example: Find the derivative of the function 485)( 23 −+−= xxxxP Solution: Multiply the coefficient of each term by the exponent and reduce the value of the exponent by one.

8103)(

485)(2'

23

+−=−+−=

xxxP

xxxxP

Example: Find the derivative of the function 1432)( 235 −+−−= xxxxxP Solution: Multiply the coefficient of each term by the exponent and reduce the value of the exponent by one.

42910)('

1432)(24

235

+−−=−+−−=

xxxxP

xxxxxP

Example: Find the derivative of the function 12847)( 245 −+−+−= xxxxxP Solution: Multiply the coefficient of each term by the exponent and reduce the value of the exponent by one.

2161635)('

12847)(34

245

+−+−=−+−+−=

xxxxP

xxxxxP

Example: Find the derivative of the function 234961312)( 356 ++−−= xxxxxP Solution: Multiply the coefficient of each term by the exponent and reduce the value of the exponent by one.

9186572)('

234961312)(245

356

+−−=++−−=

xxxxP

xxxxxP

Example: Find the derivative of the function 3

78

2

5

3

2)( 23 ++−= xxxxP

Solution: Multiply the coefficient of each term by the exponent and reduce the value of the exponent by one.

852)('

3

78

2

5

3

2)(

2

23

+−=

++−=

xxxP

xxxxP

Local Maximum and Minimum A useful method for finding the local minimum values of a higher degree polynomial function is to use the derivative. This method has steps:

1. Find the derivative.

2. Set the derivative equal to zero equation.

3. Evaluate the x-values in the original function to find the y-values.

Example: Find the local maximum and minimum values of the function Solution: Find the derivative.

Set the derivative equal to zero and solve the quadratic equation

5

2

)2)(5(60

)103(60

601860

60186)('

2

2

2

=−=

+−=−−=−−=

−−=

x

x

xx

xx

xx

xxxP

Evaluate the x-values in the original function to find the y-values.

68)2(

1203616)2(

)2(60)2(9)2(2)2(

6092)(23

23

=−+−−=−

−−−−−=−−−=

P

P

P

xxxxP

275)5(

300225250)5(

)5(60)5(9)5(2)5(

6092)(23

23

−=−−=

−−=−−=

P

P

P

xxxxP

Therefore, the local minimum is (5,

Maximum and Minimum Values: local maximum and

degree polynomial . This method has three

equal to zero and solve the

values in the original function to

Find the local maximum and minimum values of the function xP 2)( =

60186)('

6092)(2

23

−−=

−−=

xxxP

xxxxP

equal to zero and solve the quadratic equation

60

values in the original function to

Therefore, the local minimum is (5, -275) and the local maximum is (-2, 68).

xxx 6092 23 −−

Finding Function from Graph: The graph of a polynomial function can be used to find a function that describes the graph. Example: Given the graph of a polynomial function, determine a function which could describe it. Be sure to identify each of the following:

1. End Behavior 2. Roots & Multiplicity 3. # of Turning Points and n. 4. Factors of the Polynomial 5. Equation of Polynomial

Solution:

1. End Behavior: Starts low, ends high so according to the leading term test it has a positive, odd leading term.

2. Roots, Multiplicity & Turning Points:

2−=x , k = 1 0=x , k = 2 5=x , k = 2

3. Turning Points: There are 4 turning points, therefore, n = 5.

4. Using the roots, determine the factors and multiply to find the function.

05058

0)5058(

0)502022510(

0)2510)(2(

0)5)(5)(2)()((

2345

232

2232

22

=++−=++−

=+−++−=+−+

=−−+

xxxx

xxxx

xxxxxx

xxxx

xxxxx

5. Therefore, a function is 2345 5058)( xxxxxf ++−=

3.5-Applications

Example: A company’s weekly profit (in thousands of dollars) is given by the function

���� = �� − 3� + 2� + 3

Where x is the amount (in thousands of dollars) spent per week on advertising. Determine the local maximum and minimum value for the profit. Solution: To determine the local maximum and minimum values for profit, I will use the derivative.

2630

263)('

323)(

2

2

23

+−=

+−=

++−=

xx

xxxP

xxxxP

Solve the quadratic equation.

42.0

57.13

31

6

326

==

±=

±=

x

x

x

x

To find the y-values which represent revenue, evaluate these two points in the original polynomial function.

615.2)57.1(

385.3)42.0(

323)( 23

==

++−=

P

P

xxxxP

The local maximum profit is $3,385.00 when $420.00 is spent on advertising. The local minimum is $2,615.00 when $1,570.00 is spent on advertising.

Example: An open-top box is to be made from a 6 in. by 7 in. piece of copper by cutting equal squares from each corner and folding up the sides. Write the volume of the box as a function of x. Determine the maximum possible volume of the box. Solution: The formula for the volume of a rectangular box is V=LWH. Let x = length of square cut from the box. We can then create the following equation.

425212)('

42264)(

)26)(27()(

2

23

+−=+−=−−=

xxV

xxxxV

xxxxV

Using the quadratic equation we obtain x = 1.07 and x = 3.26. However, since the width of the copper sheet is only 6 in. it would be impossible to cut two 3.26 in sections out of the corners. Therefore the only legitimate solution is x = 1.07. The actual volume is:

1.20)(

)07.1(42)07.1(26)07.1(4)07.1( 23

=+−=

xV

V

The volume of the box is 20.1 in.cu.

Example: Find a function that represents the given graph: Solution:

1. End Behavior: Starts low, ends high so according to the leading term test it has a positive, odd leading term.

2. There are 4 turning points; therefore, the

function always has one less turning point than the degree of the function.

3. Roots & Multiplicity:

3−=x , k = 2 0=x , k = 2 1=x , k = 1

4. Using the roots and the value of

Therefore, the function is f

Find a function that represents the given graph:

: Starts low, ends high so according to the leading term test it has a positive, odd

There are 4 turning points; therefore, the function will be a 5th degree polynomial. A polynomial function always has one less turning point than the degree of the function.

Using the roots and the value of a, determine the factors and multiply to find the function.

2345

232

22

935)(

)935()(

)96)(1()(

)1)(3)(3)()(()(

xxxxxf

xxxxxf

xxxxxf

xxxxxxf

−++=−++=++−=

−++=

2345 935)( xxxxxf −++=

: Starts low, ends high so according to the leading term test it has a positive, odd

degree polynomial. A polynomial function always has one less turning point than the degree of the function.

, determine the factors and multiply to find the function.