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Copyright © 2007 Pearson Education, Inc.
Slide 3-1
Chapter 3: Polynomial Functions
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
Complex Numbers
Quadratic Functions and Graphs
Quadratic Equations and Inequalities
Further Applications of Quadratic Functions and Models
Higher Degree Polynomial Functions and Graphs
Topics in the Theory of Polynomial Functions (I)
Topics in the Theory of Polynomial Functions (II)
Polynomial Equations and Inequalities; Further
Applications and Models
Copyright © 2007 Pearson Education, Inc.
Slide 3-2
3.2 Quadratic Functions and Graphs
• Quadratic Functions are polynomial functions, discussed later.
• P is often used to represent a polynomial function.
• A function of the form
P( x)  ax  bx  c
2
with a  0 is called a quadratic function.
• Recall
g ( x )  a ( x  h)  k
2
is the graph of f ( x)  x stretched or shrunk and shifted horizontally
and vertically.
• Example
2
Figure 9 pg 3-13
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Slide 3-3
3.2 Completing the Square
• Rewrite P( x)  2 x  4 x  16 in the form g ( x)  a( x  h)  k .
2
2
Completing The Square
1. Divide both sides of the equation by a, so that the
coefficient of x 2 is 1.
2. Add  ac to both sides.
3. Add to both sides the square of half the coefficient of
b 2
x,  2a
.
4. Factor the right side as the square of a binomial and
combine terms on the left.
5. Isolate the term involving P(x) on the left.
6. Multiply both sides by a.
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Slide 3-4
3.2 Example of Completing the Square
P( x)  2 x  4 x  16
2
P( x)
 x  2x  8
2
2
P( x)
 8  x  2x
2
2
Add 8 to both sides.
P( x)
 8 1  x  2x  1
2
2
P( x)
 9  ( x  1)
2
P( x)
 ( x  1)  9
2
P( x)  2( x  1)  18
2
2
2
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Divide by 2 to make the coefficient
of x2 equal to 1.
Add [½·2]2 to both sides to complete
the square on the right.
Combine terms on the left; factor
on the right.
Subtract 9 from both sides.
Multiply both sides by 2.
Slide 3-5
3.2 Example of Completing the Square
• From P( x)  2( x 1) 2 18, we can determine several
components of the graph of P( x)  2 x 2  4 x 16.
- a  2, h  1, k  18
- vertex : (h, k )  (1,18)
- axis of symmetry : x  1
- domain : (, ), range : [18, )
- decreasing on (,1]
- increasing on [1, )
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Slide 3-6
3.2 Graphs of Quadratic Functions
Transform P( x)   x 2  6 x  8 into P( x)  a( x  h) 2  k.
P( x)   x 2  6 x  8
 P( x)  x 2  6 x  8
 P( x)  8  x 2  6 x
 P( x )  8  9  x 2  6 x  9
 P( x)  1  ( x  3) 2
 P( x)  ( x  3) 2  1
P( x)  ( x  3) 2  1
Copyright © 2007 Pearson Education, Inc.
- P has vertex (-3,1), so the
graph of f (x) = x2 is shifted left
3 and up 1.
- The coefficient of (x+3)2 is –1,
so the graph opens downward.
- y-intercept: (0,–8)
- Axis of symmetry: line x = -3
- Domain: (-,); Range: (-,1]
- increasing: (-,-3]; decreasing: [-3,)
Slide 3-7
3.2 Graph of P(x) = a(x-h)2 + k
The graph of P( x)  a( x  h) 2  k , a  0,
(a) is a parabola with vertex (h,k), and the vertical line
x = h as axis of symmetry;
(a) opens upward if a > 0 and downward if a < 0;
(b) is broader than y  x 2 if 0  a  1 and narrower than
y  x 2 if a  1.
•
•
One method to determine the coordinates of the vertex is
to complete the square.
Rather than go through the procedure for each individual
function, we generalize the result for P(x) = ax² + bx + c.
Copyright © 2007 Pearson Education, Inc.
Slide 3-8
3.2 Vertex Formula for Parabola
P(x) = ax² + bx + c (a  0)
P ( x)  ax 2  bx  c (a  0)
y  ax 2  bx  c (a  0)
y
b
c
 x2  x 
a
a
a
y c
b
  x2  x
a a
a
y c b2
b
b2
2
  2  x  x 2
a a 4a
a
4a
2
y b 2  4ac 
b 

 x 
2
a
2a 
4a

2
y 
b  b 2  4ac
 x  
a 
2a 
4a 2
2
2
  b  4ac  b
P( x)  a  x     
2a   
4a
 



h
k
Copyright © 2007 Pearson Education, Inc.
Standard form
Replace P(x) with y to simplify
notation.
Divide by a.
Subtract
Add
c
a
.
 1  b 
 2  a 
2

b2
4a
.
2
Combine terms on the left;
factor on the right.
Get y-term alone on the left.
Multiply by a and write in the
2
form P ( x )  a ( x  h )  k .
Slide 3-9
3.2 Vertex Formula
The vertex of the graph of P( x)  ax 2  bx  c (a  0) is the
point
 b  b 
  , P   .
 2a  2a  
Example Use the vertex formula to find the coordinates of the vertex of
the graph of P ( x )  .65 x 2  2x  4.
Analytic Solution – exact solution
b
2
2


a  .65, b  2, so x  
2a
2 ( .65) 2 (.65)
and
Approximation
Copyright © 2007 Pearson Education, Inc.
b 
 2 

y  P 

  .65
 2a 
 2(.65) 
2

 2 
2
  4.
 2(.65) 
Using a calculator, we find x  1.09 and y  4.77.
Slide 3-10
3.2 Extreme Values
• The vertex of the graph of P( x)  ax 2  bx  c is the
– lowest point on the graph if a > 0, or
– highest point on the graph if a < 0.
• Such points are called extreme points (also extrema,
singular: extremum).
For the quadratic function defined by P( x)  ax 2  bx  c,
(a) if a > 0, the vertex (h,k) is called the minimum point
of the graph. The minimum value of the function is
P(h) = k.
(b) if a < 0, the vertex (h,k) is called the maximum point
of the graph. The maximum value of the function is
P(h) = k.
Copyright © 2007 Pearson Education, Inc.
Slide 3-11
3.2 Identifying Extreme Points and Extreme
Values
Example Give the coordinates of the extreme point and the
corresponding maximum or minimum value for each function.
(a) P( x)  2 x  4 x  16
(b) P( x)   x  6 x  8
2
The vertex of the graph
is (–1,–18). Since a > 0,
the minimum point is
(–1,–18), and the
minimum value is –18.
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2
The vertex of the graph
is (–3,1). Since a < 0,
the maximum point is
(–3,1), and the maximum
value is 1.
Slide 3-12
3.2 Finding Extrema with the Graphing
Calculator
Let P( x)  2 x 2  4 x  16.
• One technique is to use the fmin function. We get the
x-value where the minimum occurs. The y-value is found
by substitution.
Figure 14 pg 3-20b
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Slide 3-13
3.2 Applications and Modeling
Example The table gives data for the percent increase (y) on
hospital services in the years 1994 – 2001, where x is the
number of years since 1990. The data are plotted in the
scatter diagram.
Year x
Percent
increase y
Year
Percent
increase y
4
1.8
8
3.4
5
0.8
9
5.8
6
0.5
10
7.1
7
1.3
11
12.0
A good model for the data is the function defined by f ( x)  .37 x  4.1x  12
2
(a) Use f (x) to approximate the year when the percent increase was a minimum.
The x-value of the minimum point is
b
4.1


 5.54  6
2a
2(0.37)
(b) Find the minimum percent increase.
f (5.54)  .37(5.54)  4.1(5.54)  12  .64
The minimum value is .64 differing slightly from the data value of .5 in the table.
Copyright © 2007 Pearson Education, Inc.
Slide 3-14
3.2 Height of a Propelled Object
Height of a Propelled Object
If air resistance is neglected, the height s (in feet) of an
object propelled directly upward from an initial height
s0 feet with initial velocity v0 feet per second is
s(t )  16t 2  v0t  s0
where t is the number of seconds after the object is
propelled.
• The coefficient of t², 16, is a constant based on
gravitational force and thus varies on different surfaces.
• Note that s(t) is a parabola, and the variable x will be used
for time t in graphing-calculator-assisted problems.
Copyright © 2007 Pearson Education, Inc.
Slide 3-15
3.2 Solving a Problem Involving Projectile
Motion
A ball is thrown directly upward from an initial height of 100 feet with
an initial velocity of 80 feet per second.
(a) Give the function that describes height in terms of time t.
s (t )  16t  80t  100
2
(b)
Graph this function.
(c)
The cursor in part (b) is at the point (4.8,115.36). What does this
mean?
After 4.8 seconds, the object will be at a height of 115.36 feet.
Copyright © 2007 Pearson Education, Inc.
Slide 3-16
3.2 Solving a Problem Involving Projectile
Motion
(d) After how many seconds does the projectile reach its maximum
height?
The maximum occurs at the vertex
b
80
x

 2.5
2a
2( 16)
and
2
y  16( 2.5)  80( 2.5)  100  200.
After 2.5 seconds, the projectile reaches
a height of 200 ft.
(e)
Figure 19 pg 3-24
For what interval of time is the height of the ball greater than
160 feet?
Using the graphs, t must
be between .92 and 4.08
seconds.
Copyright © 2007 Pearson Education, Inc.
Slide 3-17
3.2 Solving a Problem Involving Projectile
Motion
(f)
After how many seconds will the ball fall to the ground?
Figure 21 pg 3-25
When the ball hits the ground, its height will be 0,
so we need to find the positive x-intercept. From the
graph, the x-intercept is about 6.04, so the ball will
reach the ground 6.04 seconds after it is projected.
Copyright © 2007 Pearson Education, Inc.
Slide 3-18