Math 3201 Notes
Chapter 5: Polynomial Functions
Learning Goals: See p. 271 text.
§5.1 Exploring Graphs of Polynomial Functions (2 classes)
Read Goal p. 274 text.
1.
2.
3.
4.
Define and give examples and non-examples of a polynomial function. pp. 273, 698
Define a scatter plot. pp. 273, 699
Identify and label the four quadrants of the Cartesian plane. p. 275
Identify the characteristics of the graph of a polynomial function including the: pp. 274-276
i.
general shape of the graph
ii.
degree of the function
iii. number of x-intercepts
iv.
number of y-intercepts
v.
end behaviour
vi.
domain of the function
vii.
range of the function
viii. number of turning points
Def n : A function that contains only the operations of multiplication and addition with real-number
coefficients, whole-number exponents, and two variables is called a polynomial function. Recall that
the coefficients are the numbers in front of the variable in each term, and that the constant term is the
coefficient that has no variable.
E.g.:
Polynomial
f ( x)  2 x  1
f ( x)  x 2  1
Coefficient(s)
2
1
Constant Term
1
1
f ( x)  x 2  4 x  4
1,4
4
f ( x )   x  3  1  x 2  6 x  8
1,6
8
f ( x)  2 x 3
2
0
f ( x)  x  2 x  x  2
1
f ( x)   x 2  2 x  5
2
1,2,-1
1
 ,  2
2
-2
2
3
2
3
Non E.g.: f ( x)  x 2  1 because the exponent is NOT a whole number.
Non E.g.: f ( x, y)  xy  1 because there are too many variables.
1
-5
Polynomial functions are named according to their degree.
Def n : The degree of a function is the highest exponent on the variable in the polynomial.
E.g.:
Polynomial
Degree
Name of Polynomial
f ( x)  8  8x0
0
Constant
f ( x)  2 x  3
1
Linear
f ( x)  x 2  1
2
Quadratic
f ( x)  x 2  4 x  4
2
Quadratic
f ( x )   x  3  1  x 2  6 x  8
2
Quadratic
f ( x)  2 x 3
3
Cubic
f ( x)  x 3  2 x 2  x  2
3
Cubic
1
f ( x)   x 2  2 x  5
2
2
f ( x)  x 3  2 x 4  x 5  2
5
2
Quadratic
Quintic
Special Types of Polynomials
We will study 4 special types of polynomials in this chapter.
Type of
Polynomial
Standard Form
Example
Degree
Graph
Constant
f ( x)  a
f ( x)  6  6 x0
0
Horizontal Line
Linear (M1201)
f ( x)  ax  b
f ( x)  2 x  3
1
Oblique line with
slope a and yintercept b.
Quadratic (M2201)
f ( x)  ax2  bx  c
f ( x)  x 2  4 x  4
2
Parabola
(U-shape)
Cubic (M3201)
f ( x)  ax3  bx 2  cx  d
f ( x)  x 3  2 x 2  x  2
3
Sideways “S”
2
Note that a polynomial function is usually written in descending order. This means that the term with
the highest exponent is written first, the term with the second highest degree is written second, and so
on.
E.g.: f ( x)  x3  2 x 2  x  2 is in descending order.
E.g.: f ( x)  x3  2 x 4  x5  2 is NOT in descending order.
Also note that a vertical line does NOT represent a polynomial function because it does not pass the
vertical line test (VLT). The VLT was covered in Math 1201.
Def n : When the polynomial is written in standard form, the coefficient (the number) in front of the
term with the highest exponent is called the leading coefficient. In standard form, it is denoted as a.
E.g.: For f ( x)  2 x3  3x 2  4 x  1 , the leading coefficient is -2.
The Cartesian Plane and its Quadrants
The Cartesian plane is divided into 4 regions called quadrants. They are identified as in the diagram
below.
Def n : A description of the behaviour or shape of the graph, from left to right, on the coordinate plane
is called the graph’s end behaviour. It is a description of what happens to the y-values as x becomes
large in the positive or negative direction.
E.g.: In the graph to the right, as the x-values become very large
in the negative direction, the y-values become very large in the
negative direction (falls to the left) and as the x-values become
very large, the y-values become very large (rises to the right).
Note that we also say that the graph extends from quadrant 3 to
quadrant 1.
3
E.g.: In the graph to the right, as the x-values become very
large in the negative direction, the y-values become very large
in the positive direction (rises to the left) and as the x-values
become very large in the positive direction, the y-values
become very large in the negative direction (falls to the right).
Note that we also say that the graph extends from quadrant 2 to
quadrant 4.
Def n : Any point on the graph where the graph changes from increasing to decreasing or from
decreasing to increasing is called a turning point.
This curve has two turning points.
This curve does not have any!
4
E.g.: Complete the table below. The first one is done for you.
Graph
No. Turning Points
0
5
Polynomial Graphs in Detail
Now let`s put all this together and look at our 3 of our 4 polynomial graphs in detail.
Graph of a Constant Function f ( x)  a
Possible Graphs
End behaviour
No. x-intercepts
No. y-intercepts
Horizontal Line
Horizontal Line
Extends from Q2 to Q1
Extends from Q3 to Q4
0, except when the line falls on the x-axis
1  y  a
No. Turning Points
Domain
0
x x 

 y y  a
Range
Graph of a Linear Function f ( x)  ax  b
Possible Graphs
End behaviour
Oblique Line
Extends from Q3 to Q1. Rises
to the right.
No. x-intercepts
No. y-intercepts
6
Oblique Line
Extends from Q2 to Q4. Falls to the
right.
1
1  y  b
No. Turning Points
Domain
0
x x 
Range
y y 


Graph of a Quadratic Function f ( x)  ax2  bx  c
Possible Graphs
End behaviour
Parabola
Parabola
Extends from Q2 to Q1. Rises to Extends from Q3 to Q4. Falls to the left
the left and right.
and right.
No. x-intercepts
No. y-intercepts
0, 1, or 2
1  y  c
No. Turning Points
Domain
1
x x 
Range

 y y  k OR  y y  k where V  h, k 
7
Def n : A polynomial function of degree 3 is called a cubic function.
E.g.: f ( x)  x3 ,
f ( x)  2 x3  x2  4 x 10,
f ( x)  2 x3  4 x
Characteristics of the Graph of a Cubic Function
Use graphing technology to sketch the graph of each cubic function and complete the table.
Function
Leading
3
2
y  ax  bx  cx  d Coeff.
(a)
Const.
Term
(d)
Sketch
a)
y  x3
b)
y  2 x3  1
c)
y  x3  4 x
d)
y   x 3  3x 2  x  3
e)
y  x3  x 2  x  1
f)
y   x3  3x  2
8
End
Behav.
No.
No.
Turn. x-ints.
Pts.
y-int.
Investigation Questions:
1. How are the sign of the leading coefficient and the end behavior related?
2. How are the degree of the function and the number of x-intercepts related?
3. How is the y-intercept related to the equation?
4. How many turning points can a cubic function have?
5. In general, how are the number of turning points and the degree related?
6. What is the domain and range for cubic functions?
7. Explain why quadratic functions have maximum or minimum values, but cubic polynomial
functions have only turning points?
9
Function
Leading
3
2
y  ax  bx  cx  d Coeff.
(a)
Const.
Term
(d)
Sketch
End
Behav.
No.
Turn.
Pts.
No.
xints.
y-int.
a)
y  x3
1
0
Q3 to Q1
0
1
y=0
-2
1
Q2 to Q4
0
1
y=1
1
0
Q3 to Q1
2
3
y=0
-1
-3
Q2 to Q4
2
3
y = -3
1
1
Q3 to Q1
2
2
y=1
-1
-2
Q2 to Q4
2
2
y = -2
b)
y  2 x3  1
c)
y  x3  4 x
d)
y   x 3  3x 2  x  3
e)
y  x3  x 2  x  1
f)
y   x3  3x  2
10
Graph of a Cubic Function f ( x)  ax3  bx 2  cx  d
Possible Graphs
End behaviour
Sideways S-shape
Sideways S-shape
Extends from Q3 to Q1. Increases to Extends from Q2 to Q4. Decreases
the right and decreases to the left.
to the right and increases to the left.
No. x-intercepts
No. y-intercepts
1, 2, or 3
1 y  d
No. Turning Points
Domain
0 or 2
x x  
Range
y y  
Summary for Polynomial Functions of Degree 3 or Less






The graphs are continuous. This means there are no holes or breaks in the graph.
The degree of the polynomial determines the shape of the graph.
There is only 1 y-intercept and it is the constant term of the polynomial function.
The maximum number of turning points is 1 less than the degree of thje polynomial.
For linear and cubic functions, the end behaviour is opposite as you go from left to right.
For quadratic functions, the end behaviour is the same as you go from left to right.
Read In Summary & Need to Know p. 276 text
Do #’s 1-4, p. 277 text in your homework booklet.
11
§5.2 Characteristics of the Equations of Polynomial Functions (1 class)
Read Goal p. 274 text.
1. Describe the characteristics of a polynomial function by analyzing its equation. pp. 278-286
2. Match a polynomial function to its corresponding graph. p. 282
If you are given the equation of a polynomial function with degree 3 or less, you need to be able to
describe the characteristics of the corresponding graph without first graphing the polynomial.
Describing the graph of f ( x)  ax  b


Slope is a. If the value of a is positive the graph increases to the right. If the value of a is
negative the graph decreases to the right (increases to the left).
y-intercept is b or  0,b  .
Polynomial
Equation in
Standard Form
f ( x)  3x  2
 a  3
 b  2
 degree = 1
Corresponding Graph


1
x4
2
1
a
2
b4
degree = 1
f ( x) 



Characteristics of Graph
12


Shape: Linear (line)
Slope is negative. Graph decreases
to the right.
End behavior: Q2 to Q4
y-intercept: -2 or  0, 2 


# x-intercepts: 1
Domain:  x x 

Range:  y y 





# Turning Points: 0
Shape: Linear (line)
Slope is positive. Graph increases to
the right.
End behavior: Q3 to Q1
y-intercept: 4 or  0, 4 


# x-intercepts: 1
Domain:  x x 

Range:  y y 

# Turning Points: 0




Describing the graph of f ( x)  ax2  bx  c


If the value of a is positive the graph opens up. If the value of a is negative the graph opens
down.
y-intercept is c or  0, c  .
Polynomial Equation
in Standard Form
f ( x)  2 x 2  4 x  6
 a2
 b  4
 c  6
 degree = 2
Corresponding Graph
Characteristics of Graph




Shape: U-shape (parabola)
a is positive. Graph opens
upward.
End behavior: Q2 to Q1
y-intercept: -6 or  0, 6 

V 1, 8


# x-intercepts: 2
Domain:  x x 




f ( x)  2 x  12 x  18
2




a  2
b  12
c  18
degree = 2
f ( x)  2 x  8x  11
 a  2
 b 8
 c  11
 degree = 2
2
13

Range:  y y  8


# Turning Points: 1
Shape: U-shape (parabola)
a is negative. Graph opens
downward.
End behavior: Q3 to Q4
y-intercept: -18 or  0, 18

V  3, 0 


# x-intercepts: 1
Domain:  x x 

Range:  y y  0





# Turning Points: 1
Shape: U-shape (parabola)
a is negative. Graph opens
downward.
End behavior: Q3 to Q4
y-intercept: -11 or  0, 11

V  2, 3


# x-intercepts: 0
Domain:  x x 



Range:  y y  3

# Turning Points: 1
Describing the graph of f ( x)  ax3  bx 2  cx  d


If the value of a is positive the graph increases to the right and decreases to the left. If the value
of a is negative the graph decreases to the right and increases to the left.
y-intercept is d or  0, d  .
Polynomial Equation in
Standard Form
f ( x)  x3  2 x 2  15x  36
 a 1
 d  36
 degree = 3
Corresponding Graph


1
f ( x)   x 3  5
2
1
 a
2
 d  5
 degree = 3
f ( x)   x  2 x  11x  12
3



Characteristics of Graph
2
a  1
d  12
degree = 3
Read In Summary & Need to Know p. 286 text
Do #’s 1-7, 9, 10, 13 pp. 287-289 text in your homework booklet.
14


Shape: Sideways S
a is positive. Graph increases
to the right and decreases to
the left.
End behavior: Q3 to Q1
y-intercept: -36 or  0, 36 


# x-intercepts: 2
Domain:  x x 

Range:  y y 





# Turning Points: 2
Shape: Sideways S
a is negative. Graph
decreases to the right and
increases to the left.
End behavior: Q2 to Q4
y-intercept: -5 or  0, 5


# x-intercepts: 1
Domain:  x x 

Range:  y y 





# Turning Points: 0
Shape: Sideways S
a is negative. Graph
decreases to the right and
increases to the left.
End behavior: Q2 to Q4
y-intercept: 12 or  0,12 


# x-intercepts: 3
Domain:  x x 

Range:  y y 

# Turning Points: 2






§5.3 Modelling Data with a Line of Best Fit (2 classes)
Read Goal p. 295 text.
1.
2.
3.
4.
5.
6.
7.
8.
Define a scatter plot. p. 273
Create a scatter plot using technology. pp. 295-300
Define the line of best fit. p. 296
Determine the line of best fit for a set of data using technology. p. 296
Define interpolation. p. 296
Use interpolation in problem solving. pp. 296-299
Define extrapolation. p. 299
Use extrapolation in problem solving. pp. 299-300
If you are given a set of data, you need to be able to draw a scatter plot of the data and determine the
equation of the line of best fit.
Def n : A scatter plot is a set of points on a grid that is used to visualize a relationship or trend in data.
E.g.: Look at the scatter plot below and determine a relationship or trend in the data.
Drawing a Scatter Plot using the TI 83 Plus
E.g.: Nathan wonders if a person’s height can be predicted from that person’s hand span. He measured
the hand span and height of 15 of his classmates and recorded the data in tha table below.
Hand
Span
(cm)
Height
(cm)
20.0
21.1
17.6
16.5
Hand
Span
(cm)
Height
(cm)
20.7
16.0
21.2
165.0
165.0
175.0
17.5
19.0
20.8
22.5
25.0
23.0
20.2
21.1
165.0 172.5 172.5 153.8 157.5 170.0 168.8 177.5 182.5 172.5 180.0 177.5
15
Create a scatter plot of this data using the TI 83 Plus.
Step 1:
Press the STAT button.
Step 2:
Select EDIT by pressing ENTER or 1.
Step 3:
If L1 contains data, press the up
arrow key until L1 is
highlighted, then press CLEAR
and the down arrow key to
erase the data.
Step 4:
If L2 contains data, press the
right arrow key to move to L2,
press the up arrow key until L2
is highlighted, then press
CLEAR and the down arrow
key to erase the data.
Step 5:
Move to L1 and enter the data for
the independent variable (hand
span). Type in each data value and
press ENTER to add it to L1.
16
Step 6:
Move to L2 and enter the data for the
dependent variable (height). Type in
each data value and press ENTER to
add it to L2.
Step 7:
Press 2nd & then press STATPLOT (Y=).
You should see the screen to the far right.
Step 8:
Press ENTER or 1 to select PLOT 1.
Step 9:
Highlight ON & Press ENTER to turn on Plot 1.
Step
10:
Use the arrow keys to go to TYPE and highlight the
scatterplot icon.
Step
11:
Press ENTER to select scatterplot.
Step
12:
If necessary, use the down arrow key to set XList to L1 and YList to L2.
17
Step
13:
Press ZOOM and then 9. Your scatter plot should be similar to the one
below.
Finding the Line of Best Fit (LOBF) for Continuous Data using the TI-83 PLUS
Def n : The line of best fit is the line that best approximates the trend in a scatter plot.
E.g.: Determine the LOBF for the scatter plot of height vs hand span.
Step 1:
Press 2nd & then QUIT (MODE) to go to the home screen.
Step 2:
Press CLEAR to clear the home screen.
Step 3:
Press STAT.
Step 4:
Use the right arrow key to go to CALC.
18
Step 5:
Step 6:
Press 4 or use the down arrow key to go to LinReg (ax+b). You should see the screen below.
Note that L1 and L2 are the default lists for the independent and dependent variables
respectively. If you use other lists, you have to enter the list names after LinReg(ax+b).
Now you have to tell the calculator that you want to put the equation of the LOBF into Y1 so
you can graph it.
Press the VARS key. You should
see the screen to the far right.
Step 7:
Use the right arrow key to move to Y – VARS. You
should see the screen to the right.
Step 8:
Press ENTER or 1 to select FUNCTION. You
should see the screen to the right.
19
Step 9:
Press ENTER or 1 to select Y1. You should see the
screen to the right. This tells the calculator to copy
the equation of the LOBF to Y1.
Step 10:
Press ENTER. You should see the screen to the
right.
Step 11:
If you press Y= you should see the screen to the
right.
Step 12:
Press GRAPH and you should see the screen to the
right.
E.g.: Use interpolation to estimate a person’s height if their hand span is 22.0cm.
Method 1: Using the graph
First let’s go to WINDOW and enter some new values.
20
Old Window Settings
New Window Settings
Now press GRAPH to redraw the scatter plot and LOBF.
From the graph, if hand span is 22.0cm then height is about
175 cm.
Method 2: Using the equation
Press 2nd and QUIT to go to the home screen.
Press 2nd and CALC.
Press ENTER or 1 to select value
Type in 22.0 after X = and press ENTER. You should see the screen below.
The height of a person with a hand span of 22.0cm is about 174.7cm
E.g.: Use extrapolation to estimate a person’s height if their hand span is 26.0cm.
Method 1: Using the graph
Press CLEAR to show only the scatter plot and LOBF.
From the graph, if hand span is 26.0cm then height is about 185cm.
21
Method 2: Using the equation
Press 2nd and QUIT to go to the home screen.
Press 2nd and CALC.
Press ENTER or 1 to select value
Type in 26.0 after X = and press ENTER. You should see the screen below.
The height of a person with a hand span of 26.0cm is about 184.2cm
Determining How Well the Line of Best Fit (LOBF) Fits the Data
When we find the equation of the line of best fit using technology, we can also find a number that tells
us how well the line fits the data.
Step 1:
Step 2:
Step 3:
Press 2nd & Quit to go the the home screen. Press CLEAR to clear the screen
Press 2nd and Catalog. You should see the screen to
the right. Note the A (alpha) in the top right corner.
This tells you that the letter keys can be used.
Press D  x 1 key  to go the beginning of the catalogue of commands that
begin with D.
22
Step 4:
Press the down arrow button until the arrow head
points to “Diagnostics On”
Step 5:
Press ENTER & ENTER to turn on diagnostics.
Step 6:
Run the linear regression again (see previous Steps 110). You should see the screen to the right.
Look for the value of r 2 . The closer this number is to 1 the better the line fits the data. Our fit
is only moderate.
Do #’s 4, 7, 10, 12, pp. 302-305 text in your homework booklet.
23
§5.4 Modelling Data with a Curve of Best Fit (2 classes)
Read Goal p. 307 text.
1.
2.
3.
4.
Define the curve of best fit. p. 308
Determine the curve of best fit for a set of data using technology. pp. 308-312
Use interpolation in problem solving. pp. 308-312
Use extrapolation in problem solving. pp. 308-312
Sometimes a line of best fit does not show the trend in the data as well as another type of curve. In this
section, we will investigate whether a linear, quadratic, or cubic polynomial function best approximates
the trend in the data. We will determine the function that fits best using the value of r 2 .
Def n : A curve of best fit (COBF) is the curve that best approximates the trend in the data.
E.g.: Determine if a linear, quadratic, or cubic polynomial best fits the data given in the tabel below.
x
y
0
0
1
55
2
120
3
365
4
1070
We enter the data into L1 and L2 and run a linear regression, a quadratic regression, and a cubic
regression. The results are presented in the screens below.
Note that the value of r 2 that is closest to 1 occurs with the cubic regression. This means that a cubic
polynomial best fits the data. This fit is excellent and means that the curve go through nearly all the
points on the scatter plot.
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E.g.: The following table shows the average price of gasoline, in cents per litre, for a selection of years
since 1979.
Years after 1979
Price of Gas  cents/L 
Years after 1979
Price of Gas  cents/L 
0
1
2
3
4
7
8
9
12
14
21.98
26.18
35.63
43.26
45.92
45.78
47.95
47.53
57.05
54.18
17
20
22
23
24
25
26
27
28
29
58.52
59.43
70.56
70.00
74.48
82.32
92.82
97.86
102.27
115.29
a) Use technology to create a scatter plot of the data and determine if the a linear, quadratic, or cubic
polynomial best fits the data.
b) Use the equation of the line/curve of best ft to determine the price of gas 18 years after 1979
(interpolation).
c) Use the equation of the line/curve of best ft to determine the price of gas 30 years after 1979
(extrapolation).
a) Years after 1979 is the independent variable so those values go in L1. Price of Gas is the
dependent variable so those values go in L2. The value of r 2 that is closest to 1 occurs for the
cubic regression so the cubic polynomial best fits the data.
b) Press 2nd & CALC and select value. Type in 18 and press ENTER.
The price of gas in 1997 was 57.90 cents/L.
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OR
y  0.0122836118 18  0.46451659 18  6.295033755 18  23.45161524
3
2
y  57.90
The price of gas in 1997 was 57.90 cents/L.
c) Press 2nd & CALC and select value. Type in 30 and press ENTER.
The price of gas in 2009 would be 125.90 cents/L.
OR
y  0.0122836118  30   0.46451659  30   6.295033755  30   23.45161524
3
2
y  125.90
The price of gas in 2009 would be 125.90 cents/L.
E.g.: The scatterplot and the curve of best fit for a set of data are shown below. Determine the value for
y when x  3.5 using:
a) the line of best fit.
b) the equation.
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a)
When x  3.5 , the value of y is approximately -4.5
b) When x  3.5 ,
y  0.194  3.5  0.1785714286  3.5   0.5396825397  3.5   4.285714286
3
2
y  4.35
The average price of gasoline in Canada from 2005 to 2009 is shown in the table below. Use a cubuc
regression to predict the price of gasoline in 2020. Does your prediction make sense?
No. years since 2005
Price in cents/L
0
92.3
1
87.7
2
101.8
3
114.1
Enter the year values in L1 and the price values in L2 and do a cubic regression.
Press 2nd & CALC and select value. Type in 15 and press ENTER.
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4
94.5
This error occurs because the Xmax in the Window setting does not go to 15.
Press WINDOW and change Xmax to 16.
Press 2nd & CALC and select value. Type in 15 and press ENTER.
The price of gasoline in 2020 will be -9318.85 cents/L, which does not make sense in this context. This
example shows why you need to be careful when extrapolating.
Do #’s 1, 6, 9, 10, pp. 313-316 text in your homework booklet.
Do #’s 1, 2, 3 a, b, 4-6, 8, 9, pp. 321-322 text in your homework booklet.
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