Transcript Slide 1

Objectives
Students will learn how to:
Graph functions (x) = axn
Polynomial Functions: Graphs,
Graph Applications,
General Polynomial
and Functions
Models
Find Turning Points and End Behavior
Intermediate Value Theorems
Approximate Real Zeros Using Graphing
Calculator
Use Graphing Calculator to Determine
Polynomial Models and Curve Fitting
Example 1
GRAPHING FUNCTIONS OF THE
FORM (x) = axn
Graph the function.
a. f ( x )  x 3
Solution Choose several values for x, and
find the corresponding values of (x), or y.
f (x)  x3
x
–2
–1
0
1
2
(x)
–8
–1
0
1
8
Example 1
GRAPHING FUNCTIONS OF THE
FORM (x) = axn
Graph the function.
b. g ( x )  x 5
Solution The graphs of (x) = x3 and g(x) = x5
are both symmetric with respect to the origin.
g( x )  x 5
x
–1.5
–1
0
1
1.5
g(x)
–7.6
–1
0
1
7.6
Example 1
GRAPHING FUNCTIONS OF THE
FORM (x) = axn
Graph the function.
c. f ( x )  x 4 , g ( x )  x 6
Solution The graphs of (x) = x3 and g(x) = x5
are both symmetric with respect to the origin.
f (x)  x
x
–2
–1
0
1
2
4
(x)
16
1
0
1
16
g( x )  x 6
x
–1.5
–1
0
1
1.5
g(x)
11.4
1
0
1
11.4
Graphs of General Polynomial
Functions
As with quadratic functions, the absolute
value of a in (x) = axn determines the width
of the graph. When a> 1, the graph is
stretched vertically, making it narrower, while
when 0 < a < 1, the graph is shrunk or
compressed vertically, so the graph is
broader. The graph of (x) = –axn is
reflected across the x-axis compared to the
graph of (x) = axn.
Graphs of General Polynomial
Functions
n
Compared with the graph of f ( x )  ax , the
n
f
(
x
)

ax
 k is translated (shifted) k
graph of
units up if k > 0 andkunits down if k < 0.
n
f
(
x
)

ax
,
Also, when compared with the graph of
the graph of (x) = a(x – h)n is translated h
units to the right if h > 0 and hunits to the left
if h < 0.
n
The graph of f ( x )  ax  k shows a
combination of these translations. The effects
here are the same as those we saw earlier with
quadratic functions.
Example 2
EXAMINING VERTICAL AND
HORIZONTAL TRANSLATIONS
Graph the function.
5
a. f ( x )  x  2
Solution The graph
will be the same as that
of (x) = x5, but
translated 2 units down.
Example 2
EXAMINING VERTICAL AND
HORIZONTAL TRANSLATIONS
Graph the function.
6
b. f ( x )  ( x  1)
Solution
In (x) = (x + 1)6,
function  has a graph
like that of
(x) = x6, but since
x + 1 = x – (–1), it is
translated 1 unit to the
left.
Example 2
EXAMINING VERTICAL AND
HORIZONTAL TRANSLATIONS
Graph the function.
3
c. f ( x )  2( x  1)  3
Solution
The negative sign in –2 causes
the graph of the function to be
reflected across the x-axis when
compared with the graph of (x)
= x3. Because  – 2> 1, the
graph is stretched vertically as
compared to the graph of
(x) = x3. It is also translated 1
unit to the right and 3 units up.
Unless otherwise restricted, the domain of a
polynomial function is the set of all real numbers.
Polynomial functions are smooth, continuous curves
on the interval (–, ). The range of a polynomial
function of odd degree is also the set of all real
numbers. Typical graphs of polynomial functions of
odd degree are shown in next slide. These graphs
suggest that for every polynomial function  of odd
degree there is at least one real value of x that
makes (x) = 0 . The real zeros are the x-intercepts
of the graph.
Odd Degree
A polynomial function of even degree has a
range of the form (–, k] or [k, ) for some
real number k. Here are two typical graphs
of polynomial functions of even degree.
Even Degree
Recall that a zero c of a polynomial function
has as multiplicity the exponent of the factor
x – c. Determining the multiplicity of a zero
aids in sketching the graph near that zero. If
the zero has multiplicity one, the graph
crosses the x-axis at the corresponding xintercept as seen here.
If the zero has even multiplicity, the graph is
tangent to the x-axis at the corresponding
x-intercept (that is, it touches but does not
cross the x-axis there).
If the zero has odd multiplicity greater than
one, the graph crosses the x-axis and is
tangent to the x-axis at the corresponding
x-intercept. This causes a change in
concavity, or shape, at the x-intercept and
the graph wiggles there.
Turning Points and End Behavior
The previous graphs show that
polynomial functions often have turning
points where the function changes from
increasing to decreasing or from
decreasing to increasing.
Turning Points
A polynomial function of degree n has
at most n – 1 turning points, with at
least one turning point between each
pair of successive zeros.
End Behavior
The end behavior of a polynomial graph is
determined by the dominating term, that is,
the term of greatest degree. A polynomial of
the form
f ( x )  an x n  an 1x n 1   a0
n
f
(
x
)

a
x
has the same end behavior as
.
n
End Behavior
For instance,
f ( x )  2x  8 x  9
3
2
has the same end behavior as f ( x )  2 x .
It is large and positive for large positive
values of x and large and negative for
negative values of x with large absolute
value.
3
End Behavior
The arrows at the ends of the graph look like those
of the graph shown here; the right arrow points up
and the left arrow points down.
The graph shows that as x takes on larger and
larger positive values, y does also. This is
symbolized as x  , y  ,
read “as x approaches infinity, y
approaches infinity.”
End Behavior
For the same graph, as x takes on negative values
of larger and larger absolute value, y does also:
as x  ,
y  ,
End Behavior
For this graph, we have
as
x  ,
y  ,
and as x   , y  
End Behavior of Polynomials
Suppose that axn is the dominating term of a
polynomial function  of odd degree.
1. If a > 0, then as x  , f ( x )  , and as
x  , f ( x )  .
Therefore, the end behavior of the graph is
of the type that looks like the figure shown
here.
We symbolize it as
.
End Behavior of Polynomials
Suppose that axn is the dominating term of a
polynomial function  of odd degree.
2. If a < 0, then as x  , f ( x )  , and as
x  , f ( x )  .
Therefore, the end behavior of the graph
looks like the graph shown here.
We symbolize it as
.
End Behavior of Polynomials
Suppose that axn is the dominating term of a
polynomial function  of even degree.
1. If a > 0, then as x  , f ( x )  .
Therefore, the end behavior of the
graph looks like the graph shown here.
We symbolize it as
.
End Behavior of Polynomials
Suppose that is the dominating term of a
polynomial function  of even degree.
2. If a < 0, then as x  , f ( x )  .
Therefore, the end behavior of the graph
looks like the graph shown here.
We symbolize it as
.
Example 3
DETERMINING END BEHAVIOR GIVEN
THE DEFINING POLYNOMIAL
Match each function with its graph.
f ( x )  x  x  5x  4
4
A.
B.
2
C.
D.
Solution Because  is of even degree with
positive leading coefficient, its graph is C.
Example 3
DETERMINING END BEHAVIOR GIVEN
THE DEFINING POLYNOMIAL
Match each function with its graph.
g( x )   x  x  3x  4
6
A.
B.
2
C.
D.
Solution Because g is of even degree with
negative leading coefficient, its graph is A.
Example 3
DETERMINING END BEHAVIOR GIVEN
THE DEFINING POLYNOMIAL
Match each function with its graph.
h( x )  3 x  x  2 x  4
3
A.
B.
2
C.
D.
Solution Because function h has odd
degree and the dominating term is positive,
its graph is in B.
Example 3
DETERMINING END BEHAVIOR GIVEN
THE DEFINING POLYNOMIAL
Match each function with its graph.
k( x )  x  x  4
7
A.
B.
C.
D.
Solution Because function k has odd
degree and a negative dominating term, its
graph is in D.
Graphing Techniques
We have discussed several characteristics of the graphs of
polynomial functions that are useful for graphing the
function by hand. A comprehensive graph of a polynomial
function will show the following characteristics:
1. all x-intercepts (zeros)
2. the y-intercept
3. the sign of (x) within the intervals formed by the xintercepts, and all turning points
4. enough of the domain to show the end behavior.
In Example 4, we sketch the graph of a polynomial function
by hand. While there are several ways to approach this,
here are some guidelines.
Graphing a Polynomial
Function
n
n 1
f
(
x
)

a
x

a
x
  a1x  a0 , an  0,
Let
n
n 1
be a polynomial function of degree n. To
sketch its graph, follow these steps.
Step 1 Find the real zeros of . Plot them as
x-intercepts.
Step 2 Find (0). Plot this as the y-intercept.
Graphing a Polynomial
Function
Step 3 Use test points within the intervals
formed by the x-intercepts to
determine the sign of (x) in the
interval. This will determine whether
the graph is above or below the xaxis in that interval.
Graphing a Polynomial
Function
Use end behavior, whether the graph
crosses, bounces on, or wiggles through the
x-axis at the x-intercepts, and selected points
as necessary to complete the graph.
Objectives
Students will learn how to:
Synthetic Division
Solve Equations of Higher Order Than
Quadratics by Factoring
Use Synthetic Division to Divide Polynomials
Evaluate Polynomial Functions Using the
Remainder Theorem (synthetic substitution)
Test Potential Zeros (Roots, Solutions)
Solving Higher Degree Equations by Factoring
The equation x3 + 8 = 0 that follows is called
a cubic equation because of the degree 3
term. Some higher-degree equations can
be solved using factoring and/or the
quadratic formula.
SOLVING A CUBIC EQUATION
Solve x 3  8  0.
Solution
3
x 8 0
 x  2  x  2x  4   0
2
2
or
x
 2x  4  0
x20
Factor as a sum of
cubes.
Zero-factor property
( 2)  ( 2)  4(1)(4)
x  2 or x 
2(1)
2
Quadratic formula; a = 1, b = –2, c = 4
SOLVING A CUBIC EQUATION
Solve x 3  8  0.
Solution
2  12
x
2
Simplify.
2  2i 3
x
2
Simplify the radical.
x

2 1 i 3
2

Factor out 2 in the
numerator.
SOLVING A CUBIC EQUATION
Solve x 3  8  0.
Solution
x  1 i 3
Lowest terms


The solution set is 2,1  i 3 .
SOLVING A CUBIC EQUATION
Solve x3  2 x 2  48x  0
Factor out common factor
x( x  2x  48)  0
2
Factor completely if possible or use other
methods.
x( x  8)(x  6)  0
Set each factor equal to zero and
solve
SOLVING A CUBIC EQUATION
x0
x 8  0
x 88  08
x 8
x60
x66  06
x  6
Solution set x={0,8,-6}
If a polynomial is not easily factored than we
must use division to determine the solutions
Division Algorithm
Let (x) and g(x) be polynomials with g(x) of
lower degree than (x) and g(x) of degree one
or more. There exists unique polynomials q(x)
and r(x) such that
f  x   g  x  q  x   r  x ,
where either r(x) = 0 or the degree of r(x) is
less than the degree of g(x).
Synthetic Division
Synthetic division is a shortcut method of
performing long division with polynomials.
It is used only when a polynomial is divided
by a first-degree binomial of the form x – k,
where the coefficient of x is 1.
Divide
by
Traditional Division
3x  10x  40
3
2
x  4 3 x  2 x  0 x  150
3 x 3  12 x 2
10 x 2  0 x
2
10 x  40 x
40x  150
40x  160
10
Remainder
10
2
3 x  10 x  40 
x4
2
Quotient
Synthetic Division
2
0  150
12 40
160
10
3 10 40
43
Remainder
10
Quotient
3 x  10 x  40 
x4
Caution To avoid errors, use 0 as
the coefficient for any missing terms,
including a missing constant, when
setting up the division.
2
USING SYNTHETIC DIVISION
Example 1
Use synthetic division to divide
5 x  6 x  28 x  2
.
x2
Solution Express x + 2 in the form x – k by
writing it as x – (–2). Use this and the
coefficients of the polynomial to obtain
x + 2 leads
to – 2
2 5
3
2
6
 28
 2.
Coefficients
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
5 x  6 x  28 x  2
.
x2
Solution Bring down the 5, and multiply:
–2(5) = –10
2 5  6  28  2
10
5
3
2
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
5 x  6 x  28 x  2
.
x2
Solution Add –6 and –10 to obtain –16.
Multiply –2(–16) = 32.
3
2 5
2
 6  28
10
32
5 16
2
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
5 x  6 x  28 x  2
.
x2
Solution Add –28 and 32 to obtain 4.
Finally, –2(4) = – 8.
3
2 5
2
 6  28
10
32
5 16
4
2
8
Add
columns.
Watch your
signs.
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
5 x  6 x  28 x  2
.
x2
Solution Add –2 and –8 to obtain –10.
3
2
2 5
 6  28  2
10
32 8
5 16
4 10
Quotient
Remainder
Example 1
USING SYNTHETIC DIVISION
Use synthetic division to divide
5 x  6 x  28 x  2
.
x2
Since the divisor x – k has degree 1, the
degree of the quotient will always be written
one less than the degree of the polynomial to
be divided. Thus,
3
2
5 x  6 x  28 x  2
10
2
 5 x  16 x  4 
.
x2
x2
Remember to
3
2
add remainder
divisor
.
Special Case of the Division
Algorithm
For any polynomial (x) and any
complex number k, there exists a
unique polynomial q(x) and number r
such that
f ( x )  ( x  k )q( x )  r .
For Example
In the synthetic division in Example 1,
5 x 3  6 x 2  28 x  2  ( x  2)(5 x 2  16 x  4)  ( 10).
f (x)
 (x  k )
q( x )

r
Here g(x) is the first-degree polynomial x – k.
Remainder Theorem
If the polynomial (x) is divided by
x – k, the remainder is equal to (k).
Remainder Theorem
In Example 1, when (x) = 5x3 – 6x2 – 28x – 2
was divided by x + 2 or x –(–2), the
remainder was –10. Substituting –2 for x in
(x) gives
f ( 2)  5( 2)3  6( 2)2  28( 2)  2
 40  24  56  2
 10
Use parentheses
around substituted
values to avoid errors.
Remainder Theorem
A simpler way to find the value of a polynomial
is often by using synthetic division. By the
remainder theorem, instead of replacing x by –2
to find (–2), divide (x) by x + 2 using synthetic
division as in Example 1. Then (–2) is the
remainder, –10.
2 5  6  28  2
10
32 8
 (–2)
5 16
4 10
Example 2
APPLYING THE REMAINDER
THEOREM
Let (x) = –x4 + 3x2 – 4x – 5. Use the
remainder theorem to find (–3).
Solution Use synthetic division with k = –3.
3 1 0 3  4  5
3  9 18  42
1 3  6 14  47
By this result, (–3) = –47.
Remainder
Testing Potential Zeros
A zero of a polynomial function  is a number k
such that (k) = 0. The real number zeros are
the x-intercepts of the graph of the function.
The remainder theorem gives a quick way to
decide if a number k is a zero of the polynomial
function defined by (x). Use synthetic division to
find (k); if the remainder is 0, then (k) = 0 and k
is a zero of (x). A zero of (x) is called a root or
solution of the equation (x) = 0.
Example 3
DECIDING WHETHER A NUMBER
IS A ZERO
Decide whether the given number k is a zero
of (x).
3
2
f
(
x
)

x

4
x
 9 x  6; k  1
a.
Solution
Proposed
zero
11 4 9 6
1 3
6
1 3 6
0
f ( x )  x 3  4x 2  9x  6
Remainder
Since the remainder is 0, (1) = 0, and 1 is a zero of the
polynomial function defined by (x) = x3 – 4x2 + 9x – 6. An
x-intercept of the graph (x) is 1, so the graph includes the
point (1, 0).
Example 3
DECIDING WHETHER A NUMBER
IS A ZERO
Decide whether the given number k is a zero
of (x).
4
2
f
(
x
)

x

x
 3 x  1; k   4
b.
Solution Remember to use 0 as coefficient for
the missing x3-term in the synthetic division.
Proposed
zero
4 1 0
1 3
1
4 16  68 284
1  4 17  71 285
Remainder
The remainder is not 0, so – 4 is not a zero of
(x) = x4 +x2 – 3x + 1. In fact, (– 4) = 285, indicating that
(– 4, 285) is on the graph of (x).
Example 3
DECIDING WHETHER A NUMBER
IS A ZERO
Decide whether the given number k is a zero
of (x).
4
3
2
c. f ( x )  x  2 x  4 x  2 x  5; k  1  2i
Solution Use synthetic division and operations
with complex numbers to determine whether 1
+ 2i is a zero of (x) = x4 – 2x3 + 4x2 + 2x – 5.
1  2i 1  2
4
1  2i  5
(1  2i )( 1  2i )
 1  4 i 2
 5
1  1  2i  1
2
5
 1  2i
5
1  2i
0
Remainder
Example 3
DECIDING WHETHER A NUMBER
IS A ZERO
Decide whether the given number k is a zero
of (x).
4
3
2
c. f ( x )  x  2 x  4 x  2 x  5; k  1  2i
Since the remainder is 0, 1 + 2i is a zero of the given
polynomial function. Notice that 1 + 2i is not a real
number zero. Therefore, it would not appear as an
x-intercept in the graph of (x).
1  2i 1  2
4
1  2i  5
(1  2i )( 1  2i )
 1  4 i 2
 5
1  1  2i  1
2
5
 1  2i
5
1  2i
0
Remainder
Objectives
Students will learn how to:
Use the Factor Theorem to find Solutions
Use Rational
Theorem to find
Zero Zeros
of Polynomial
Functions
Possible Solutions
Use Number of Zeros Theorum to
determine the Maximum Number of Zeros
Use Conjugate Zeros Theorem to find
additional solutions
Find All Zeros(solutions) of a Polynomial
Function
Descartes’ Rule of Signs
Factor Theorem
The polynomial x – k is a factor of the
polynomial (x) if and only if (k) = 0.
DECIDING WHETHER x – k IS A
FACTOR OF (x)
Example 1
Determine whether x – 1 is a factor of (x).
a. f ( x )  2 x 4  3 x 2  5 x  7
Solution By the factor theorem, x – 1 will be a
factor of (x) if and only if (1) = 0. Use synthetic
division and the remainder theorem to decide.
0 3 5 7
2 2
5 0
2 2 5
0 7
12
Use a zero
coefficient for
the missing
term.
(1) = 7
Since the remainder is 7 and not 0,
x – 1 is not a factor of (x).
Example 1
DECIDING WHETHER x – k IS A
FACTOR OF (x)
Determine whether x – 1 is a factor of (x).
b. f ( x )  3 x 5  2 x 4  x 3  8 x 2  5 x  1
Solution
2 1 8 5 1
3 1
2  6 1
3
1 2  6 1 0
13
(1) = 0
Because the remainder is 0, x – 1 is a factor.
Additionally, we can determine from the coefficients in
the bottom row that the other factor is
DECIDING WHETHER x – k IS A
FACTOR OF (x)
Example 1
Determine whether x – k is a factor of (x).
b. f ( x )  3 x 5  2 x 4  x 3  8 x 2  5 x  1
Solution
2 1 8 5 1
3 1
2  6 1
1 2  6 1 0
1 3
3
(1) = 0
3 x  x  2x  6 x  1.
4
3
2
Thus, f ( x )  ( x  1)(3 x  x  2 x  6 x  1).
4
3
2
Example 2
FACTORING A POLYNOMIAL
GIVEN A ZERO
Factor the following into linear factors if –3 is
3
2
a zero of . f ( x )  6 x  19 x  2 x  3
Solution Since –3 is a zero of ,
x – (–3) = x + 3 is a factor.
3 6 19 2
18  3
6
1 1
3
3
0
Use synthetic division to
divide (x) by x + 3.
The quotient is 6x2 + x – 1.
Example 2
FACTORING A POLYNOMIAL
GIVEN A ZERO
Factor the following into linear factors if –3 is
3
2
a zero of . f ( x )  6 x  19 x  2 x  3
Solution x – (–3) = x + 3 is a factor.
The quotient is 6x2 + x – 1, so
f ( x )  ( x  3)(6 x  x  1)
2
f ( x )  ( x  3)(2x  1)(3 x  1).
Factor 6x2 + x – 1.
These factors are all linear.
Example 2
Solve:
FACTORING A POLYNOMIAL
GIVEN A ZERO
f ( x )  ( x  3)(2x  1)(3 x  1).
0  ( x  3)(2 x  1)(3x  1)
Set:
x3 0
2x 1  0
3x  1  0
x  3
x  1 / 2
x  1/ 3
If no zeros are given than must use trial and error. The
following theorum helps limit the possibilities.
Rational Zeros Theorem
p
q
If
is a rational number written in
p
lowest terms, and if q is a zero of , a
polynomial function with integer
coefficients, then p is a factor of the
constant term and q is a factor of the
leading coefficient.
Example 3
USING THE RATIONAL ZERO
THEOREM
Do the following for the polynomial function
4
3
2
defined by f ( x )  6 x  7 x  12 x  3 x  2.
a. List all possible rational zeros.
p
q
Solution For a rational number to be
zero, p must be a factor of a0 = 2 and q
must be a factor of a4 = 6. Thus, p can be
1 or 2, and q can be 1, 2, 3, or 6.
p
The possible rational zeros, q are,
possible p
possible q
1 1 1 2
1,  2,  ,  ,  ,  .
2 3 6 3
Example 3
USING THE RATIONAL ZERO
THEOREM
Do the following for the polynomial function
4
3
2
defined by f ( x )  6 x  7 x  12 x  3 x  2.
b. Find all rational zeros and factor (x) into
linear factors.
Solution Use the remainder theorem to show
that 1 is a zero.
1 6 7  12  3 2
Use “trial and
error” to find
6
13
1 2
zeros.
(1) = 0
6 13
1 2 0
The 0 remainder shows that 1 is a zero. The quotient is
6x3 +13x2 + x – 4, so (x) = (x – 1)(6x3 +13x2 + x – 2).
Example 3
USING THE RATIONAL ZERO
THEOREM
Do the following for the polynomial function
4
3
2
defined by f ( x )  6 x  7 x  12 x  3 x  2.
b. Find all rational zeros and factor (x) into
linear equations.
Solution Now, use the quotient polynomial
and synthetic division to find that –2 is a zero.
2 6 13 1  2
12  2
2
(–2 ) = 0
6 1 1
0
The new quotient polynomial is 6x2 + x – 1.
Therefore, (x) can now be factored.
Example 3
USING THE RATIONAL ZERO
THEOREM
Do the following for the polynomial function
4
3
2
defined by f ( x )  6 x  7 x  12 x  3 x  2.
b. Find all rational zeros and factor (x) into
linear equations.
Solution
f ( x )  ( x  1)( x  2)(6 x  x  1)
2
 ( x  1)( x  2)(3 x  1)(2x  1).
Example 3
USING THE RATIONAL ZERO
THEOREM
Do the following for the polynomial function
4
3
2
defined by f ( x )  6 x  7 x  12 x  3 x  2.
b. Find all rational zeros and factor (x) into
linear equations.
Solution Setting 3x – 1 = 0 and 2x + 1 = 0
yields the zeros ⅓ and –½. In summary the
rational zeros are 1, –2, ⅓, –½, and the linear
factorization of (x) is
f ( x )  6 x 4  7 x 3  12 x 2  3 x  2
Check by
 ( x  1)( x  2)(3 x  1)(2x  1).
multiplying
these factors.
Note In Example 3, once we obtained
the quadratic factor of 6x2 + x – 1, we were
able to complete the work by factoring it
directly. Had it not been easily factorable,
we could have used the quadratic formula
to find the other two zeros (and factors).
Caution The rational zeros theorem gives only
possible rational zeros; it does not tell us whether
these rational numbers are actual zeros. We must rely
on other methods to determine whether or not they are
indeed zeros. Furthermore, the function must have
integer coefficients. To apply the rational zeros theorem
to a polynomial with fractional coefficients, multiply
through by the least common denominator of all fractions.
For example, any rational zeros of p(x) defined below will
also be rational zeros of q(x).
1 3 2 2 1
1
p( x )  x  x  x  x 
6
3
6
3
4
q( x )  6 x 4  x 3  4 x 2  x  2
Multiply the terms of
p(x) by 6.
Fundamental Theorem of
Algebra
Every function defined by a polynomial
of degree 1 or more has at least one
complex zero.
Number of Zeros Theorem
A function defined by a polynomial of
degree n has at most n distinct zeros.
Example 4
FINDING A POLYNOMIAL FUNCTION
THAT SATISFIES GIVEN CONDITIONS
(REAL ZEROS)
Find a function  defined by a polynomial of
degree 3 that satisfies the given conditions.
a. Zeros of –1, 2, and 4; (1) = 3
Solution These three zeros give x – (–1) =
x + 1, x – 2, and x – 4 as factors of (x).
Since (x) is to be of degree 3, these are
the only possible factors by the number of
zeros theorem. Therefore, (x) has the form
f ( x )  a( x  1)( x  2)( x  4)
for some real number a.
Example 4
FINDING A POLYNOMIAL FUNCTION
THAT SATISFIES GIVEN CONDITIONS
(REAL ZEROS)
Find a function  defined by a polynomial of
degree 3 that satisfies the given conditions.
a. Zeros of –1, 2, and 4; (1) = 3
Solution To find a, use the fact that (1) = 3.
f (1)  a(1  1)(1  2)(1  4)
3  a(2)( 1)( 3)
3  6a
1
a
2
Let x = 1.
(1) = 3
Solve for a.
Example 4
FINDING A POLYNOMIAL FUNCTION
THAT SATISFIES GIVEN CONDITIONS
(REAL ZEROS)
Find a function  defined by a polynomial of
degree 3 that satisfies the given conditions.
a. Zeros of –1, 2, and 4; (1) = 3
Solution Thus,
1
f ( x )  ( x  1)( x  2)( x  4),
2
1 3 5 2
f ( x )  x  x  x  4. Multiply.
or
2
2
Note In Example 4a, we cannot clear the denominators in (x)
by multiplying both sides by 2 because the result would equal 2 • (x),
not (x).
Example 4
FINDING A POLYNOMIAL FUNCTION
THAT SATISFIES GIVEN CONDITIONS
(REAL ZEROS)
Find a function  defined by a polynomial of
degree 3 that satisfies the given conditions.
b. –2 is a zero of multiplicity 3; (–1) = 4
Solution The polynomial function defined
by (x) has the form
f ( x )  a( x  2)( x  2)( x  2)
 a( x  2) .
3
Example 4
FINDING A POLYNOMIAL FUNCTION
THAT SATISFIES GIVEN CONDITIONS
(REAL ZEROS)
Find a function  defined by a polynomial of
degree 3 that satisfies the given conditions.
b. –2 is a zero of multiplicity 3; (–1) = 4
Solution Since (– 1) = 4,
f ( 1)  a( 1  2)
3
Remember:
(x + 2)3 ≠ x3 + 23
4  a(1)
3
a  4,
3
3
2
f
(
x
)

4(
x

2)

4
x

24
x
 48 x  32.
and
Conjugate Zeros Theorem
If (x) defines a polynomial function
having only real coefficients and if
z = a + bi is a zero of (x), where a
and b are real numbers,
Then its conjugate
z  a  bi is also a zero of f ( x ).
Example 5
FINDING A POLYNOMIAL FUNCTION THAT
SATISFIES GIVEN CONDITIONS
(COMPLEX ZEROS)
Find a polynomial function of least degree
having only real coefficients and zeros 3 and
2 + i.
Solution The complex number 2 – i must
also be a zero, so the polynomial has at
least three zeros, 3, 2 + i, and 2 – i. For the
polynomial to be of least degree, these must
be the only zeros. By the factor theorem
there must be three factors, x – 3, x – (2 + i),
and x – (2 – i), so
Example 5
FINDING A POLYNOMIAL FINCTION THAT
SATISFIES GIVEN CONDITIONS
(COMPLEX ZEROS)
Find a polynomial function of least degree
having only real coefficients and zeros 3 and
2 + i.
Solution
f ( x )  ( x  3)  x  (2  i ) x  (2  i )
 ( x  3)( x  2  i )( x  2  i )
 ( x  3)( x  4 x  5)
2
 x 3  7 x 2  17 x  15.
Remember:
i2 = – 1
Example 5
FINDING A POLYNOMIAL FINCTION THAT
SATISFIES GIVEN CONDITIONS
(COMPLEX ZEROS)
Find a polynomial function of least degree
having only real coefficients and zeros 3 and
2 + i.
Solution Any nonzero multiple of
x3 – 7x2 + 17x – 15 also satisfies the given
conditions on zeros. The information on
zeros given in the problem is not enough to
give a specific value for the leading
coefficient.
Example 6
FINDING ALL ZEROS OF A
POLYNOMIAL FUNCTION GIVEN ONE
ZERO
Find all zeros of
(x) = x4 – 7x3 + 18x2 – 22x + 12,
given that 1 – i is a zero.
Solution Since the polynomial function has
only real coefficients and since 1 – i is a
zero, by the conjugate zeros theorem 1 + i is
also a zero. To find the remaining zeros,
first use synthetic division to divide the
original polynomial by x – (1 – i).
Example 6
FINDING ALL ZEROS OF A
POLYNOMIAL FUNCTION GIVEN ONE
ZERO
Find all zeros of
(x) = x4 – 7x3 + 18x2 – 22x + 12,
given that 1 – i is a zero.
Solution
1 i 1  7
18
 22
12
1  i  7  5i 16  6i  12
1 6i
11  5i
 6  6i
0
Example 6
FINDING ALL ZEROS OF A
POLYNOMIAL FUNCTION GIVEN ONE
ZERO
Find all zeros of
(x) = x4 – 7x3 + 18x2 – 22x + 12,
given that 1 – i is a zero.
Solution By the factor theorem, since
x = 1 – i is a zero of (x), x – (1 – i) is a
factor, and (x) can be written as
f ( x )   x  (1  i )  x  ( 6  i )x  (11  5i ) x  ( 6  6i )  .
3
2
We know that x = 1 + i is also a zero of (x), so
f ( x )   x  (1  i ) x  (1  i ) q( x ),
for some polynomial q(x).
Example 6
FINDING ALL ZEROS OF A
POLYNOMIAL FUNCTION GIVEN ONE
ZERO
Find all zeros of
(x) = x4 – 7x3 + 18x2 – 22x + 12,
given that 1 – i is a zero.
Solution Thus,
x  ( 6  i )x  (11  5i )x  ( 6  6i )   x  (1  i ) q( x ).
3
2
Use synthetic division to find q(x).
1  i 1  6  i 11  5i
1  i  5  5i
1 5
6
 6  6i
6  6i
0
Example 6
FINDING ALL ZEROS OF A
POLYNOMIAL FUNCTION GIVEN ONE
ZERO
Find all zeros of
(x) = x4 – 7x3 + 18x2 – 22x + 12,
given that 1 – i is a zero.
Solution Since q(x) = x2 – 5x + 6, (x) can
be written as
f ( x )   x  (1  i ) x  (1  i )( x 2  5 x  6).
Factoring x2 – 5x + 6 as (x – 2)(x – 3), we see
that the remaining zeros are 2 and 3. The
four zeros of (x) are 1 – i, 1 + i, 2, and 3.
Descartes’ Rule of Signs
Let (x) define a polynomial function with real coefficients
and a nonzero constant term, with terms in descending
powers of x.
a. The number of positive real zeros of  either equals the
number of variations in sign occurring in the coefficients
of (x), or is less than the number of variations by a
positive even integer.
b. The number of negative real zeros of  either equals
the number of variations in sign occurring in the
coefficients of (– x), or is less than the number of
variations by a positive even integer.
APPLYING DESCARTES’ RULE OF
SIGNS
Example 7
Determine the possible number of positive
real zeros and negative real zeros of
f ( x )  x 4  6 x 3  8 x 2  2 x  1.
Solution We first consider the possible
number of positive zeros by observing that
(x) has three variations in signs:
 x 4  6 x 3  8 x 2  2x  1
1
2
3
Example 7
APPLYING DESCARTES’ RULE OF
SIGNS
Determine the possible number of positive
real zeros and negative real zeros of
f ( x )  x 4  6 x 3  8 x 2  2 x  1.
Solution Thus, by Descartes’ rule of signs,
 has either 3 or 3 – 2 = 1 positive real zeros.
For negative zeros, consider the variations in
signs for (–x):
4
3
2
f (  x )  (  x )  6(  x )  8(  x )  2(  x )  1
 x 4  6 x 3  8 x 2  2x  1.
1
APPLYING DESCARTES’ RULE OF
SIGNS
Example 7
Determine the possible number of positive
real zeros and negative real zeros of
f ( x )  x 4  6 x 3  8 x 2  2 x  1.
Solution
 x  6 x  8 x  2x  1.
4
3
2
1
Since there is only one variation in sign, (x)
has only 1 negative real zero.
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 1 The possible rational zeros are 1,
2, 3, 6, ½, and 3/2. Use synthetic
division to show that 1 is a zero.
12
2
5 1  6
2
7
6
7
6
0
(1) = 0
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 1 Thus,
f ( x )  ( x  1)(2 x 2  7 x  6)
 ( x  1)(2x  3)( x  2).
Factor 2x2 + 7x + 6.
Set each linear factor equal to 0, then solve for x to
find real zeros. The three real zeros of  are 1, – 3/2,
and – 2.
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 2 f (0)  6, so plot (0, 6).
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
The x-intercepts divide the x-axis into four
intervals: (–, –2), (–2, –3/2), (–3/2, 1), and
(1, ). Because the graph of a polynomial
function has no breaks, gaps, or sudden
jumps, the values of (x) are either always
positive or always negative in any given
interval.
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
To find the sign of (x) in each interval, select
an x-value in each interval and substitute it
into the equation for (x) to determine
whether the values of the function are
positive or negative in that interval.
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
When the values of the function are negative,
the graph is below the x-axis, and when (x)
has positive values, the graph is above the xaxis.
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
A typical selection of test points and the
results of the tests are shown in the table on
the next slide. (As a bonus, this procedure
also locates points that lie on the graph.)
GRAPHING A POLYNOMIAL
FUNCTION
Example 4
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
Graph Above
Value
or Below xof (x) Sign of (x)
Axis
Interval
Test
Point
(–, –2)
–3
–12
Negative
Below
(–2, –3/2)
–7/4
11/32
Positive
Above
(–3/2, 1)
0
–6
Negative
Below
(1, )
2
28
Positive
Above
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
Plot the test points and
join the x-intercepts, yintercept, and test
points with a smooth
curve to get the graph.
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
Because each zero has
odd multiplicity (1), the
graph crosses the x-axis
each time. The graph
has two turning points,
the maximum number for
a third-degree polynomial
function.
Example 4
GRAPHING A POLYNOMIAL
FUNCTION
Graph f ( x )  2 x  5 x  x  6.
3
Solution
Step 3
The sketch could be improved
by plotting the points found in
each interval in the table. Notice
that the left arrow points down
and the right arrow points up.
This end behavior is correct
since the dominating term of the
polynomial is 2x3.
Graphing Polynomial Functions
Note If a polynomial function is given in
factored form, such as
2
f ( x )  (  x  1)( x  3)( x  2) ,
Step 1 of the guidelines is easier to perform,
since real zeros can be determined by
inspection. For this function, we see that 1
and 3 are zeros of multiplicity1, and –2 is a
zero of multiplicity 2.
Graphing Polynomial Functions
Note Since the
dominating term is
 x( x )( x 2 )   x 4 , the
end behavior of the
graph is
.
The y-intercept is
f (0)  1( 3)(2)  12.
2
Graphing Polynomial Functions
Note
The graph intersects the
x-axis at 1 and 3 but
bounces at –2.
This information is
sufficient to quickly
sketch the graph of (x).
Important Relationships
We emphasize the important relationships
among the following concepts.
1. the x-intercepts of the graph of y = (x)
2. the zeros of the function 
3. the solutions of the equation (x) = 0
4. the factors of (x)
x-Intercepts, Zeros, Solutions,
and Factors
If a is an x-intercept of the graph of
y  f ( x ), then a is a zero of , a is a
solution of (x) = 0, and x – a is a
factor of (x).
Intermediate Value Theorem
for Polynomials
If (x) defines a polynomial function
with only real coefficients, and if for
real numbers a and b, the values (a)
and (b) are opposite in sign, then
there exists at least one real zero
between a and b.
Example 5
LOCATING A ZERO
Use synthetic division and a graph to show
that (x) = x3 – 2x2 – x + 1 has a real zero
between 2 and 3.
Solution Since
(2) is negative and
(3) is positive, by
the intermediate
value theorem there
must be a real zero
between 2 and 3.
2 1  2 1 1
2
02
1
0  1  1  f (2)
3 1  2 1 1
3
3 6
1
1
2 7  f (3 )
Caution Be careful how you interpret the
intermediate value theorem.
If (a) and (b) are not opposite in sign, it does not
necessarily mean that there is no zero between a
and b.
In the graph shown here, for
example, (a) and (b) are both
negative, but –3 and –1, which
are between a and b, are zeros
of (x).
Example 7
APPROXIMATING REAL ZEROS
OF A POLYNOMIAL FUNCTION
Approximate the real zeros of
(x) = x4 – 6x3 + 8x2 + 2x – 1.
Solution The greatest degree term is x4, so
the graph will have end behavior similar to
the graph of (x) = x4, which is positive for all
values of x with large absolute values. That
is, the end behavior is up at the left and the
right,
.
There are at most four real zeros, since the
polynomial is fourth-degree.
Example 7
APPROXIMATING REAL ZEROS
OF A POLYNOMIAL FUNCTION
Approximate the real zeros of
(x) = x4 – 6x3 + 8x2 + 2x – 1.
Solution Since (0) = –1, the y-intercept is –1.
Because the end behavior is positive
on the left and the right, by the intermediate
value theorem  has at least one zero on either
side of x = 0. To approximate the zeros, we use
a graphing calculator.
Example 7
APPROXIMATING REAL ZEROS
OF A POLYNOMIAL FUNCTION
Approximate the real zeros of
(x) = x4 – 6x3 + 8x2 + 2x – 1.
Solution The graph below shows that there are four
real zeros, and the table indicates that they are
between –1 and 0, 0 and 1, 2 and 3, and 3 and 4
because there is a sign change in (x) in each case.
Example 7
APPROXIMATING REAL ZEROS
OF A POLYNOMIAL FUNCTION
Solution Using the capability of the calculator,
we can find the zeros to a great degree of
accuracy. The graph shown here shows that the
negative zero is approximately – .4142136.
Similarly, we find that the other three zeros are
approximately
.26794919, 2.4142136,
and 3.7320508.
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Year
The table shows the
number of
transactions, in
millions, by users of
bank debit cards for
selected years.
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
a. Using x = 0 to
Year
represent 1995, x = 3
to represent 1998, and 1995
so on, use the
1998
regression feature of a 2000
calculator to determine 2004
the quadratic function
2009
that best fits the data.
Plot the data and
graph.
Transactions
(in millions)
829
3765
6797
14,106
22,120
Example 8
Solution
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Year
a. The best-fitting
quadratic function
for the data is
defined by
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
y  25.53 x 2  120 x  453.1
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Solution
a. The regression
coordinates screen
is shown below.
Year
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
Example 8
Solution
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
a. The graph is shown
below.
Year
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
b. Repeat part (a) for a
cubic function.
Year
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Example 8
Solution
Year
b. The best-fitting cubic
function is shown
below and is defined
by
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
y  6.735 x  164.1x  543.1x  831.0
3
2
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Solution
b. The graph of the bestfitting cubic function is
shown below.
Year
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
c. Repeat part (a) for a
quartic function.
Year
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Solution
Year
c. The best-fitting quartic
function is defined by
1995
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
y  .0576 x 4  5.198 x 3  151.9 x 2  571.4 x  829
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Solution
c. The graph of the best- Year
fitting quartic function is
1995
shown below.
Transactions
(in millions)
829
1998
3765
2000
6797
2004
14,106
2009
22,120
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
d. The correlation coefficient,
R, is a measure of the
Year
Transactions
strength of the relationship
(in millions)
between two variables. The
1995
829
2
values of R and R are used
1998
3765
to determine how well a
2000
6797
regression model fits a set
of data. The closer the
2004
14,106
value of R2 is to 1, the better
2009
22,120
2
the fit. Compare R for the
three functions to decide
which function fits the data.
Example 8
Example 8
EXAMINING A POLYNOMIAL
MODEL FOR DEBIT CARD USE
Solution
d. Find the correlation
values R2. See the graph
for the quadratic function.
The others are
.999999265 for the cubic
function and 1 for the
quartic function.
Therefore, the quartic
function provides the best
fit.