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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
3.3 Quadratic Equations and Inequalities
Quadratic Equation in One Variable
An equation that can be written in the form
ax2 + bx + c = 0
where a, b, and c are real numbers with a  0, is a
quadratic equation in standard form.
• Consider the
– zeros of P( x)  2 x 2  4 x  16
– x-intercepts of P( x)  2 x 2  4 x  16
2
– solution set of 2 x  4 x  16  0.
• Each is solved by finding the numbers that make
2
2 x  4 x  16  0 true.
Copyright © 2007 Pearson Education, Inc.
Slide 3-3
3.3 Zero-Product Property
• If a and b are complex numbers and ab = 0, then a = 0 or
b = 0 or both equal zero.
Example Solve 2 x 2  4 x  16  0.
Solution
2 x 2  4 x  16  0
x2  2x  8  0
( x  4)( x  2)  0
x40
x  4
or
or
x20
x2
The solution set of the equation is {– 4,2}.
Note that the zeros of P(x) = 2x2 + 4x – 16
are –4 and 2, and the x-intercepts are (– 4,0)
and (2,0).
Copyright © 2007 Pearson Education, Inc.
Slide 3-4
3.3 Example of a Quadratic with One
Distinct Zero
Example
2
Solve x  6 x  9  0.
Solution
x2  6x  9  0
( x  3) 2  0
x  3  0 or
x  3 or
x 3  0
x3
There is one distinct zero, 3. It is sometimes called a double
zero, or double solution (root) of the equation.
Graphing Calculator Solution
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Slide 3-5
3.3 Solving x2 = k
x2  k
x2  k  0
( x  k )( x  k )  0  x  k  0
x k
or x  k  0
or
x k
Square Root Property
The solution of x 2  k is
(a) { k} if k  0 (b) {0} if k  0 (c) {i k } if k  0.
Figure 26 pg 3-38
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Slide 3-6
3.3 Examples Using the Square Root
Property
Solve each equation.
(a) 2 x 2  10
x 2  5
x  i 5
(b) 2( x  1) 2  14
( x  1) 2  7
x 1   7  x  1  7
Graphing Calculator Solution
(a)
(b)
This solution is an approximation.
Copyright © 2007 Pearson Education, Inc.
Slide 3-7
3.3 The Quadratic Formula
• Complete the square on P( x)  ax 2  bx  c and rewrite P in the form
P( x)  a( x  h) 2  k. To find the zeros of P, use the square root property
to solve for x.
ax 2  bx  c  0
b
c
x2  x   0
a
a
b
c
2
x  x
a
a
b
b2
b2 c
2
x  x 2  2 
a
4a
4a a
2
b 
b 2  4ac

x  
2a 
4a 2

b
b 2  4ac
x

2a
4a 2
b
b 2  4ac  b  b 2  4ac
x 

2a
2a
2a
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Slide 3-8
3.3 The Quadratic Formula and the
Discriminant
The Quadratic Formula
The solutions of the equation ax 2  bx  c  0, where a  0 ,
are
b  b 2  4ac
x
.
2a
• The expression under the radical, b 2  4ac, is called the
discriminant.
• The discriminant determines whether the quadratic
equation has
2
– two real solutions if b  4ac  0,
2
– one real solution if b  4ac  0,
2
– no real solutions if b  4ac  0.
Copyright © 2007 Pearson Education, Inc.
Slide 3-9
3.3 Solving Equations with the Quadratic
Formula
Solve x( x  2)  2 x  2.
Analytic Solution
x( x  2)  2 x  2
x2  2x  2x  2
x2  4x  2  0
 (4)  (4) 2  4(1)(2)
x
2(1)
4  16  8
x
2
4 8

2
42 2

 2 2
2
Copyright © 2007 Pearson Education, Inc.
Rewrite in the form
ax2 + bx + c = 0
Substitute into the
quadratic formula a = 1,
b = – 4, and c = 2, and
simplify.
Slide 3-10
3.3 Solving Equations with the Quadratic
Formula
Graphical Solution
With the calculator, we get an approximation for
2  2 and 2  2.
Copyright © 2007 Pearson Education, Inc.
Slide 3-11
3.3 Solving a Quadratic Equation with
No Solution
Solve 2 x 2  x  4  0.
 (1)  (1) 2  4(2)(4)
x
Analytic Solution
2 ( 2)
1  1  32

4
1   31 1  i 31 1
31


 i
4
4
4
4
Graphing Calculator Solution
Example
Figure 30 pg 3-43
Copyright © 2007 Pearson Education, Inc.
Slide 3-12
3.3 Possible Orientations for Quadratic
Function Graphs (a > 0)
P(x) a>0
x-axis
(a,0)
(b,0)
P(x) a>0
x-axis
(a,0)
P(x) a>0
x-axis
a
Solution Set of
Is
P(x)=0
P(x)<0
P(x)>0
{a,b}
interval (a,b)
(  , a )  ( b ,  )
Solution Set of
Is
P(x)=0
P(x)<0
P(x)>0
{a}

(  , a )  ( a ,  )
Solution Set of
Is
P(x)=0
P(x)<0
P(x)>0


(  ,  )
• Similar for a quadratic function P(x) with a < 0.
Copyright © 2007 Pearson Education, Inc.
Slide 3-13
3.3 Solving a Quadratic Inequality
Analytically
Solve x 2  x  12  0.
x 2  x  12  0
( x  3)( x  4)  0

x  3 or
x4
The function is 0 when x = – 3 or x = 4. It will be greater than 0 or
less than zero for any other value of x. So we use a sign graph to
determine the sign of the product (x + 3)(x – 4).
For the interval (–3,4), one factor is positive and the other is
negative, giving a negative product. The product is positive
elsewhere since (+)(+) and (–)(–) is positive.
Copyright © 2007 Pearson Education, Inc.
Slide 3-14
3.3 Solving a Quadratic Inequality
Graphically
2
Verify the analytic solution to x  x  12  0 : Y1 < 0 in the
interval (–3,4).
Solving a Quadratic Inequality
1.
2.
3.
4.
Solve the corresponding quadratic equation.
Identify the intervals determined by its solutions.
Use a sign graph to determine the solution interval(s).
Decide whether or not the endpoints are included.
Copyright © 2007 Pearson Education, Inc.
Slide 3-15
3.3 Literal Equations Involving Quadratics
Example Solve (a) A 
 d2
4
for d (b) rt 2  st  k (r  0) for t.
Solution 2
d
 4A   d 2
(a) A 
4
4A
d2 

d 
4A

 2
A

(b) rt 2  st  k  rt 2  st  k  0
 b  b 2  4ac
t
where a  r , b   s, c  k
2a
 ( s)  ( s) 2  4(r )(k ) s  s 2  4rk
t

2r
2r
Copyright © 2007 Pearson Education, Inc.
Slide 3-16
3.3 Modeling Length of Life
The survival rate after age 65 is approximated by S ( x)  1  .058x  .076x 2 ,
where x is measured in decades. This function gives the probability that an
individual who reaches age 65 will live at least x decades (10x years)
longer. Find the age for which the survival rate is .5 for people who reach
the age of 65. Interpret the result.
Solution Let Y1 = S(x), Y2 = .5, and find the positive x-value of an
intersection point.
Discard x = – 3. For x  2.2, half the people who reach age
65 will live about 2.2 decades or 22 years longer to age 87.
Copyright © 2007 Pearson Education, Inc.
Slide 3-17