Transcript No Slide Title

Copyright © 2004 Pearson Education, Inc.
Chapter 3
Polynomial and
Rational Functions
Copyright © 2004 Pearson Education, Inc.
3.1
Quadratic Functions and Models
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Polynomial Function

A polynomial function of degree n, where n is
a nonnegative integer, is a function defined by
an expression of the form
f ( x)  an xn  an1 xn1  ...  a1 x  a0 ,

where an, an  1, …a1, and a0 are real numbers,
with an  0.
Examples
f ( x)  6 x  2
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g ( x)  x  3 x  x
4
3
2
Slide 3-4
Quadratic Functions




A function f is a quadratic function if
f(x) = ax2 + bx + c, where a, b, and c are real
numbers with a  0.
The simplest quadratic function is f(x) = x2. This
is the graph of a parabola.
The line of symmetry for a parabola is called the
axis of the parabola.
The vertex is the point where the axis intersects
the parabola.
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Slide 3-5
Examples


Graph each function. Give the domain and
range.
a) f ( x)  x 2  6 x  2

1 2
b) g ( x)   x
4

c) h( x)   1 ( x  2)2  3
4
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Slide 3-6
Solutions

a) f ( x)  x  6 x  2
2


x
y
0
2
1
3
2
6
3
7
5
3
6
2
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
The domain is (, ).
The range is [7, )
The vertex is (3, 7)
Slide 3-7
Solutions continued





1 2
b) g ( x)   x
4
The graph is broader than
y = x2 it is reflected
across the x-axis.
The vertex is (0, 0)
The domain is (, )
The range is (, 0].
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y  x2
y  14 x2
1
g ( x)   x 2
4
Slide 3-8
Solutions continued





c) h( x)   1 ( x  2)2  3
4
The graph is translated 2
units to the left and 3
units down.
The vertex is (2, 3).
The domain is (, )
The range is (, 3].
g ( x)   14 x 2
1
h( x)   ( x  2)2  3
4
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Slide 3-9
Completing the Square


Graph f(x) = x2  4x + 7 by completing the
square and locating the vertex.
Solution:
f ( x)  ( x 2  4 x
)7
f ( x)  ( x 2  4 x  4  4)  7
f ( x)  ( x 2  4 x  4)  4  7
f ( x)  ( x  2) 2  3

The vertex is (2, 3) and the axis is the line x = 2.
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Slide 3-10
Completing the Square continued


Find additional ordered
pairs that satisfy the
equation.
Plot the points and
connect in a smooth
curve.
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x
y
0
7
1
4
2
3
3
4
4
7
Slide 3-11
Example

Graph f(x) = 3x2 + 6x  1 by completing the
square and locating the vertex.
f ( x)  3( x 2  2 x
) 1
f ( x)  3( x 2  2 x  1  1)  1
 3( x 2  2 x  1)  3(1)  1
 3( x  1) 2  2
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Slide 3-12
Example continued

The vertex is (1, 2). Find
additional points by
substituting x-values into
the original equation.
x
y
0
1
1
2
2
1
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Slide 3-13
Graph of a Quadratic Function

The quadratic function defined by
f(x) = ax2 + bx + c can be written in the form
y = f(x) = a(x  h)2 + k, a  0, where

The graph of f has the following characteristics.
1. It is a parabola with vertex (h, k) and the
vertical line x = h as axis.

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Slide 3-14
Graph of a Quadratic Function
continued




2. It opens up if a > 0 and down if a < 0.
3. It is broader than the graph of y = x2 if |a| < 1
and narrower if |a| > 1.
4. The y-intercept is f(0) = c.
5. If b2  4ac  0, the x-intercepts are
b  b 2  4ac
x
2a
If b2  4ac < 0, there are no x-intercepts.
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Slide 3-15
Example

Find the axis and vertex of the parabola having
equation f(x) = 4x2 + 8x + 3 using the formula.

Solution: Here a = 4, b = 8, and c = 3. The axis
of the parabola is the vertical line
b
8
xh

 1
2a 2(4)

The vertex is (1, 1).
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Slide 3-16
3.2
Synthetic Division
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Division Algorithm
Let f(x) and g(x) be polynomials with g(x) of
lower degree than f(x) and g(x) of degree of 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).
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Slide 3-18
Synthetic Division

A shortcut method of performing long division with
certain polynomials, called synthetic division, 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.

In the following example, the example on the right will
show how the division process is simplified by omitting
all variables and writing only coefficients, with 0 used to
represent the coefficient of any missing terms. Since the
coefficient of x in the divisor is always 1 in these
divisions, it too can be omitted. These omissions simplify
the problem.
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Slide 3-19
Example
2x2  x  2
x  2 2 x3  3x 2
 32
2x3  4x2
x2
x2  2x
2x  32
2x  4
 28
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2
1
2
2 2  3  0 - 32
24
1
12
2  32
24
 28
Slide 3-20
Example continued

The numbers that are
repetitions of the numbers
directly above them can also
be omitted.
2 1
2
2 2  3 + 0  32
4
1
2
2  32
4
 28
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
The entire problem can now be
condensed vertically, and the
top row of numbers can be
omitted since it duplicates the
bottom row if the 2 is brought
down.
2 2  3  0  32
 4 2  4
2 1 2  28
The rest of the bottom row is
obtained by subtracting  4,
2, and  4 from the
corresponding terms above
them.
Slide 3-21
Example continued

With synthetic division it is useful to change the sign of
the divisor, so the  2 is changed to 2, which also
changes the sign of the numbers in the second row. To
compensate for this change, subtraction is changed to
addition. Doing this gives the final result.
2 2  3  0  32
2
Quotient
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4
1
2 4
2  28
2x2 + x + 2 
28
x2
Signs changed
Remainder
Slide 3-22
Caution

Note: To avoid errors, use 0 as coefficient for
any missing terms, including a missing constant,
when setting up the division.
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Slide 3-23
Example
Use synthetic division to divide
2x3  3x2  11x + 7 by x  3.
Solution: Since x  3 in the form x  k use this
and the coefficients of the polynomial to obtain
3 2  3  11  7
Bring down the 2 and multiply (3)(2) = 6
3 2  3  11  7
6
2
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Slide 3-24
Example continued
Add 3 and 6 to obtain 3. Multiply: 3(3) = 9
3 2  3  11  7
6 9
2 3
Add 11 and 9, obtaining 2. Finally 3(2) = 6
3 2  3  11  7
6 96
2 3 2
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Slide 3-25
Example continued
Add 7 and 6 to obtain 1.
3 2  3  11  7
6 96
2 321
Remainder
Quotient
Since the divisor x  k has degree 1, the degree of the
quotient will always be one less than the degree of the
polynomial to be divided. Thus
2 x3  3x 2  11x  7
1
2
 2 x  3x  2 
x 3
x 3
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Slide 3-26
Remainder Theorem

If a polynomial f(x) is divided by x  k, the
remainder is f(k).

For example: In the synthetic division problem given
previously, when f(x) = 2x3  3x2  11x + 7 was divided
by x  3, the remainder was 1. Substituting 3 for x, in
f(x) gives
f(3) = 2(3)3  3(3)2  11(3) + 7
= 54  27  33 + 7
=1
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Slide 3-27
Applying the Remainder Theorem
Example: Let f(x) = 3x4  7x3  4x + 5. Use the
remainder theorem to find f(2).
Solution: Use synthetic division with k = 2.
2 3 7 0 4 5
6 2 4 16
3 1 2 8 11
Remainder
By this result, f(2) = 11
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Slide 3-28
Testing Potential Zeros


A zero of a potential function f is a number k
such that f(k) = 0. The 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 a polynomial
function defined by f(x). Use synthetic division to
find f(k); if the remainder is 0, then f(k) = 0 and k
is a zero of f(x). A zero of f(x) is called a root or
solution of the equation f(x) = 0.
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Slide 3-29
Example
Decide whether the given number k is a zero of f(x).

f(x) = 2x3  2x2  34x  30 k =  1
Solution: Since the remainder is 0, f(1) = 0, and 1 is a
zero of the polynomial function defined by
f(x) = 2x3  2x2  34x  30
f(x) = 2x3 + 4x2  x + 5 k = 2
Solution: Since the remainder is 7, not 0, 2 is not a zero
of f(x) = 2x3 + 4x2  x + 5 . In fact, f(2 ) = 7.

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Slide 3-30
3.3
Zeros of Polynomial Functions
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Factor Theorem
By the remainder theorem, if f(k) = 0, then the
remainder when f(x) is divided by x  k is 0. This
means that x  k is a factor of f(x). Conversely, if
x  k is a factor of f(x), then f(k) must equal 0.
This is summarized in the following factor
theorem.
Factor Theorem
The polynomial x  k is a factor of the polynomial
f(x) if and only if f(k) = 0.
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Slide 3-32
Example
Determine whether x  3 is a factor of f(x) for
f(x) = x3  7x2 + 11x + 3.
Solution: By the factor theorem, x  3 will be the
factor of f(x) only if f(3) = 0. Use synthetic
division and the remainder theorem to decide.
3 1  7 11 3
3 12 3
1 4 1 0
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Slide 3-33
Example continued
Because the remainder is 0, x  3 is a factor.
Additionally, we can determine from the
coefficients in the bottom row that the other
factor is x2  4x  1,
and
f(x) = (x  3) (x2  4x  1).
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Slide 3-34
Rational Zeros Theorem


The rational zeros theorem gives a method to
determine all possible candidates for rational
zeros of a polynomial function with integers.
Rational Zeros Theorem
p
If q is a rational number written in lowest terms,
p
and if q is a zero of f, a polynomial function with
integer coefficients, then p is a factor of the
constant term and q is a factor of the leading
coefficient.
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Slide 3-35
Using the Rational Zeros Theorem
Example: Do each of the following for the
polynomial function defined by
f(x) 6x3  5x2  7x + 4.

List all possible rational zeros.

Find all rational zeros and factors f(x) into
linear factors.
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Slide 3-36
Using the Rational Zeros Theorem
continued
Solution:
p

For a rational number q to be a zero, p must be a factor
of a0 = 4 and q must be a factor of a4 = 6. Thus, p can
be 1, 2,  4, and q can be 1, 2, 3, or 6. The
possible rational zero p, are
q
1 1 1 2
4
 ,  ,  ,  , 1,  , 2, 4.
6 3 2 3
3

Use the remainder theorem to show that 1 is a zero.
1 6  5  7 4
6
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6 11 4
11 4 0
Slide 3-37
Using the Rational Zeros Theorem
continued
The new quotient for the polynomial is
6x2 11x + 4. This factors to (3x  4)(2x  1).
Setting 3x  4 = 0 and 2x  1= 0 yields zeros
4 and 1
.
3
2
1
4
Thus the rational zeros are 1, 2 , and 3 , and
the linear factors of f(x) are x + 1, 2x  1,
and 3x  4.
Therefore,
f(x) = (x + 1)(2x 1) (3x  4)
= 6x3  5x2 7x + 4.
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Slide 3-38
Number of Zeros

Fundamentals 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: f(x) = x3 + 3x2 + 3x + 1 = (x + 1)3 is of degree 3, but has
only one zero, 1. Actually, the zero 1 occurs three times, since
there are three factors of x + 1; this zero is called a zero of
multiplicity 3.
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Slide 3-39
Example

Find a function f defined by a polynomial of
degree 3 that satisfies the given conditions.
Zeros of  2, 3, and 4; f(1) = 3
Solution: These three zeros give x  (2) = x + 2,
x  3, x  4 as factors of f(x). Since f(x) is to be
of degree 3, these are the only possible factors
by the number of zeros theorem. Therefore, f(x)
has the form f(x) = a(x + 2)(x  3)(x  4) for
some real number a.
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Slide 3-40
Example continued
To find a, use the fact that f(1) = 3.
f(1) = a(1 + 2)(1  3)(1  4) = 3
a(3)(2)(3) = 3
18a = 3
1
a=6
1
Thus, f(x) = (x + 2)(x  3)(x  4),
6
or
f(x) = 1 x3  5 x 2  1 x  4
6
6
3
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Slide 3-41
Conjugate Zeros Theorem

The following Properties of complex conjugates are
needed to prove the conjugate zeros theorem.
Properties of Conjugates
For any complex numbers c and d,
c  d  c  d,

c d  c d,

n
c  c .
n
Conjugate Zeros Theorem
If f(x) is a polynomial having only real coefficients and if
z = a + bi is a zero of f(x), where a and b are real
numbers,then z  a  bi is also a zero of f(x).
Copyright © 2004 Pearson Education, Inc.
Slide 3-42
Example
Find a polynomial function of least degree having only
real coefficients and zeros 3 and 5i.
Solution: The complex number 5i also must be a zero, so
the polynomial has at least three zeros, 3, 5i, and  5i.
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  5i, and x + 5i, so
f(x) = (x  3)(x  5i)(x + 5i)
= (x  3)(x2 + 25)
= x3  3x2 + 25x  75.
 Any nonzero multiple also satisfies the given conditions.
Copyright © 2004 Pearson Education, Inc.
Slide 3-43
Descartes’ Rule of Signs


Let f(x) define a polynomial function with real
coefficients and a nonzero constant term, with terms in
descending powers of x.
The number of positive real zeros of f either equals the
number of variations in sign occurring in the
coefficients of f(x), or is less than the number of
variations by a positive even integer.
The number of negative real zeros of f either equals
the number of variations in sign occurring in the
coefficients of f(x), or is less than the number of
variations by a positive even integer.
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Slide 3-44
Example

Determine the possible number of positive real
number zeros and negative real number zeros of
f(x) = x6 + 7x4  x3  2x2 + 6x  5.

Solution: First consider the possible number of
positive zeros by observing that f(x) has three
variations in signs.
+ x6 + 7x4  x3  2x2 + 6x  5
1
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2
3
Slide 3-45
Example continued
For negative zeros, consider the variations in signs
for f(x):
f(x) = (x)6 + 7(x)4  (x)3  2(x)2 + 6(x)  5
= x6 + 7x4 + x3  2x2  6x  5.

Since there is only one variation in sign, f(x) has
only 1 negative real zero.
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Slide 3-46
3.4
Polynomial Functions: Graphs,
Applications, and Models
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Graphing Functions of the Form
f(x) = axn

Graph each function.

f ( x)  2 x ,
3
1 5
g ( x)   x
3
x
f(x)
x
g(x)
1
2
1
1/3
0
0
0
0
1
2
1
1/3
2
16
1.5
2.53
Copyright © 2004 Pearson Education, Inc.
Slide 3-48
Horizontal and Vertical Translations
Graph the function.


a) f(x) = x5 + 1
The graph of f(x) = x5 + 1
will be the same as that
of f(x) = x5, but translated
up 1 unit.
Copyright © 2004 Pearson Education, Inc.


b) g(x) = (x  1)6
The graph is translated 1
unit to the right.
Slide 3-49
Odd Degree

Typical graphs of polynomial functions of odd degree suggest
that for every polynomial function f of odd degree there is at
least one real value of x that make f(x) = 0. The zeros are the
x-intercepts of the graph.
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Slide 3-50
Even Degree

A polynomial of even degree has a range of the
form (, k] or [k, ) for some real number k.
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Slide 3-51
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.

The end behavior of a polynomial graph is
determined by the dominating term, that is, the
term of greatest degree.
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Slide 3-52
End Behavior of Graphs of Polynomial
Functions
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Slide 3-53
End Behavior of Graphs of Polynomial
Functions continued
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Slide 3-54
Graphing Techniques

A comprehensive graph of a polynomial function
will show the following characteristics.




all x-intercepts (zeros)
the y-intercept
all turning points
enough of the domain to show the end behavior.
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Slide 3-55
Graphing a Polynomial Function





n1
Let f ( x)  an x  an1 x  ...  a1 x  a0 , an  0, be a polynomial
of degree n. To sketch its graph, follow these steps.
n
Step 1:
Step 2:
Step 3:
Find the real zeros of f. Plot them as x-intercepts.
Find f(0) = a0. Plot this as the y-intercept.
Use test points within the intervals formed by the
x-intercepts to determine the sign of f(x) in the interval.
This will determine whether the graph is above or
below the x-axis in that interval.
Use end behavior, whether the graph crosses or is tangent to the xaxis at the x-intercept, and selected points as necessary to complete
the graph.
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Slide 3-56
Example




Graph f(x) = 3x3 + 4x2  5x  2.
Solution:
Step 1: The possible rational zeros are
1
2
1,  2,  , 
3
3
Use synthetic division to
1 3 4 5 2
show that 1 is a zero.
3
3 7
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7
2
2
0  f (1)  0
Slide 3-57
Example continued



Thus, f(x) = (x  1)(3x2 + 7x + 2)
= (x  1)(3x + 1)(x + 2)
The three zeros of f are 1, 2, and 1/3.
Step 2: f(0) = 2, so plot (0, 2)
Step 3: The x-intercepts divide the x-axis into 4
intervals: (, 2), (2, 1/3), (1/3, 1)
and (1, )
Copyright © 2004 Pearson Education, Inc.
Slide 3-58
Example continued

Select test points
Interval
Test Point
(, 2)
4
110
Negative
Below
(2, 1/3)
1
4
Positive
Above
(1/3, 1)
0
2
Negative
Below
(1, )
2
28
Positive
Above
Copyright © 2004 Pearson Education, Inc.
Value of f(x) Sign of f(x)
Graph above or
below x-axis
Slide 3-59
Example continued

Plot the test points and
join the x-intercepts,
y-intercepts, and test
points with a smooth
curve to get the graph.
Copyright © 2004 Pearson Education, Inc.
Slide 3-60
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 f, a is a solution of f(x) = 0, and
x  a is a factor of f(x).

Intermediate Value Theorem for Polynomials
If f(x) defines a polynomial function with only
real coefficients, and if for real numbers a and b,
the values f(a) and f(b) are opposite in sign,
then there exists at least one real zero between
a and b.
Copyright © 2004 Pearson Education, Inc.
Slide 3-61
Example: Locating a Zero


Use synthetic division and a graph to show that
f(x) = 2x3  8x2 + x + 15 has a zero between 1 and 2.
Solution: Use synthetic division to find f(1) and f(2).
1 2  8 1 15
 2 10  11
2  10 11 4 = f ( 1)

2 2  8 1 15
 4 24  50
2  12 25  35 = f ( 2)
Since f(1) is positive and f(2) is negative, by the
theorem there must be a real zero between 1 and 2.
Copyright © 2004 Pearson Education, Inc.
Slide 3-62
Example: Locating a Zero continued

Graph of
f(x) = 2x3  8x2 + x + 15.
Copyright © 2004 Pearson Education, Inc.
Slide 3-63
Boundedness Theorem



Let f(x) be a polynomial function of degree n  1 with real
coefficients and with a positive leading coefficient. If f(x)
is divided synthetically by x  c, and
a) if c > 0 and all numbers in the bottom row of the
synthetic division are nonnegative, then f(x) has no zero
greater than c;
b) if c < 0 and the numbers in the bottom row of the
synthetic division alternate in sign (with 0 considered
positive or negative, as needed), then f(x) has no zero
less than c.
Copyright © 2004 Pearson Education, Inc.
Slide 3-64
Example


Show that f(x) = 2x3  8x2 + x + 15 has no real zero greater
than 4.
Solution: Since f(x) has real coefficients and the leading
coefficient, 2, is positive, use the boundedness theorem.
Divide f(x) synthetically by x  4.
4 2  8 1 15
8 0 4
2 0 1 19

Since 4 > 0 and all numbers in the last row of the synthetic
division are nonnegative, f(x) has no real zero greater than 4.
Copyright © 2004 Pearson Education, Inc.
Slide 3-65
3.5
Rational Functions: Graphs,
Applications, and Models
Copyright © 2004 Pearson Education, Inc.
Rational Function

A function f of the form
p ( x)
f ( x) 
,
q( x)
where p(x) and q(x) are polynomials, with
q(x)  0, is called a rational function.

2
Examples: f ( x)  ,
x
Copyright © 2004 Pearson Education, Inc.
x2
f ( x)  2
3x  2 x  4
Slide 3-67
Reciprocal Function


The simplest rational function with a variable
denominator.
1
f ( x) 
x
Since x cannot equal 0, the graph will never
intersect the vertical line x = 0. This line is called
a vertical asymptote.
Copyright © 2004 Pearson Education, Inc.
Slide 3-68
Reciprocal Function
Copyright © 2004 Pearson Education, Inc.
Slide 3-69
Example

Graph the function and
give the domain and
range.
3
y
x


x and y axes are the
horizontal and vertical
asymptotes.
The domain and range
are (, 0)  (0, ).
Copyright © 2004 Pearson Education, Inc.
Slide 3-70
Example

Graph the function and
give the domain and
range.


Domain: (, 2)  (2, )
Range: (, 0)  (0, )
3
f ( x) 
x2


y = 0 is the horizontal
asymptote.
x = 2 is the vertical
asymptote.
Copyright © 2004 Pearson Education, Inc.
Slide 3-71
Rational Function
Copyright © 2004 Pearson Education, Inc.
Slide 3-72
Example

Graph the function and give
the domain and range.


Domain: (, 3)  (3, )
Range: (2, )
1
y
2
2
( x  3)



The graph will be shifted 3
units to the left and 2 units
down.
Vertical asymptote x = 3
Horizontal asymptote: y = 2
Copyright © 2004 Pearson Education, Inc.
Slide 3-73
Asymptotes

Let p(x) and q(x) define polynomials. For the
rational function defined by
p ( x)
f ( x) 
,
q( x)
written in lowest terms, for real numbers a and b:
1. If |f(x)|   as x  a, then the line x = a is a
vertical asymptote.
2. If |f(x)|  b as |x|  , then the line y = b is a
horizontal asymptote.
Copyright © 2004 Pearson Education, Inc.
Slide 3-74
Determining Asymptotes


To find the asymptotes of a rational function
defined by a rational expression in lowest terms,
use the following procedures.
1. Vertical Asymptotes
Find any vertical asymptotes by setting the
denominator equal to zero and solving for x.
If a is a zero of the denominator, then the line
x = a is a vertical asymptote.
Copyright © 2004 Pearson Education, Inc.
Slide 3-75
Determining Asymptotes continued

2. Horizontal Asymptotes
Determine any other asymptotes. Consider three
possibilities:
(a) If the numerator has a lower degree than the
denominator, then there is a horizontal
asymptote y = 0 (the x-axis).
(b) If the numerator and denominator have the
same degree, and the function is of the form
an x n  ...  a0
f ( x) 
,
n
bn x  ...b0
an
then the horizontal asymptote has equation y 
.
bn
Copyright © 2004 Pearson Education, Inc.
Slide 3-76
Determining Asymptotes continued

(c) If the numerator is of degree exactly one
more than the denominator, then there will be
an oblique (slanted) asymptote. To find it,
divide the numerator by the denominator and
disregard the remainder. Set the rest of the
quotient equal to y to obtain the equation of the
asymptote.
Copyright © 2004 Pearson Education, Inc.
Slide 3-77
Example

For the rational function f, find all asymptotes.
3x  1
f 
x2

Vertical: Set the denominator equal to 0 and
solve.
x20
x2
The equation of the vertical asymptote is x = 2.
Copyright © 2004 Pearson Education, Inc.
Slide 3-78
Example continued

Horizontal: Divide each term by the largest
power of x in the expression.
3x 1
1

3
x
f ( x)  x x 
x 2
2

1
x x
x

The line y = 3 is therefore the horizontal
asymptote.
Copyright © 2004 Pearson Education, Inc.
Slide 3-79
Steps for Graphing Rational Functions

A comprehensive graph of a rational function
exhibits these features:




all x- and y-intercepts;
all asymptotes: vertical, horizontal, and/or oblique;
the point at which the graph intersects its nonvertical
asymptote (if there is any such point);
enough of the graph to exhibit the correct end
behavior.
Copyright © 2004 Pearson Education, Inc.
Slide 3-80
Graphing a Rational Function





p ( x)
Let f ( x)  q( x) define a function where p(x) and
q(x) are polynomials and the rational expression is
written in lowest terms. To sketch its graph, follow
these steps:
1. Find any vertical asymptotes.
2. Find any horizontal or oblique asymptotes.
3. Find the y-intercept by evaluating f(0).
4. Find the x-intercepts, if any, by solving
f(x) = 0. These will be the zeros of the
numerator, p(x).
Copyright © 2004 Pearson Education, Inc.
Slide 3-81
Graphing a Rational Function
continued



5. Determine whether the graph will intersect its
nonvertical asymptote y = b or y = mx + b by
solving f(x) = b or f(x) = mx + b.
6. Plot selected points, as necessary.
Choose an x-value in each domain interval
determined by the vertical asymptotes
and x-intercepts.
7. Complete the sketch.
Copyright © 2004 Pearson Education, Inc.
Slide 3-82
Example

4x
Graph f ( x)  2
.
x  2x  3
Step 1: Set the denominator equal to zero.
( x  3)( x  1)  0
x3 0
x 1  0
x3
x  1

The vertical asymptotes are x = 3 and x = 1.
Step 2: The horizontal asymptote is the x-axis.
Copyright © 2004 Pearson Education, Inc.
Slide 3-83
Example continued

Step 3: The y-intercept is 0, since
4(0)
0
f (0) 

0
2
(0)  2(0)  3 3

Step 4: The x-intercept is found by solving f
(x) = 0.
4x
0
2
x  2x  3
4x  0
x0
Copyright © 2004 Pearson Education, Inc.
Slide 3-84
Example continued


Step 5 Solve f(x) = 0, this solution was found in
step (4). The graph intersects the
horizontal asymptote at (0, 0).
Step 6 Plot a point in each interval determined
the x-intercepts and vertical asymptotes
to get an idea of how the graph behaves
in each interval.
Copyright © 2004 Pearson Education, Inc.
Slide 3-85
Example continued

Test Points
Interval
f ( x) 
Test Point
(, 1)
4x
.
2
x  2x  3
Value of f(x)
Sign of f(x)
Graph above
or below
x-axis
2
8/5
Negative
Below
(1, 0)
1/2
1.143
Positive
Above
(0, 3)
1
1
Negative
Below
(3, )
4
16/5
Positive
Above
Copyright © 2004 Pearson Education, Inc.
Slide 3-86
Example continued

Step 7 Complete the
graph.
Copyright © 2004 Pearson Education, Inc.
Slide 3-87
Behavior of Graphs of Rational
Functions Near Vertical Asymptotes
Copyright © 2004 Pearson Education, Inc.
Slide 3-88
Example: Oblique

x2
Graph f ( x) 
x 1




Vertical Asymptote
x = 1
Oblique Asymptote
y = x 1
x-intercept
(0, 0)
y-intercept
(0, 0)
Copyright © 2004 Pearson Education, Inc.
Slide 3-89
3.6
Variation
Copyright © 2004 Pearson Education, Inc.
Direct Variation
Direct Variation
y varies directly as x, or y is directly
proportional to x, if a nonzero real number k,
called the constant of variation, such that
y = kx.
The phrase “directly proportional” is sometimes
abbreviated to just “proportional”.
Copyright © 2004 Pearson Education, Inc.
Slide 3-91
Solving Variation Problems




Step 1 Write the general relationship among the
variables as an equation. Use the
constant k.
Step 2 Substitute given values of the variables
and find the value of k.
Step 3 Substitute this value of k into the
equation from Step 1, obtaining a
specific formula.
Step 4 Substitute the remaining values and
solve for the required unknown.
Copyright © 2004 Pearson Education, Inc.
Slide 3-92
Example
At a distance of 6 feet from a 60-cd (candlepower) light
source, a photographer’s light meter registers an
illuminance of 9 units. Find the illuminance of a 45-cd
light source at the same distance.
Solution:
Step 1 Since the illuminance varies directly to the intensity
at a given distance,
E = kI,
where I represent intensity and E represent
illuminance, and k is a nonzero constant.
Copyright © 2004 Pearson Education, Inc.
Slide 3-93
Example continued
Step 2 Since E = 9 when I = 60, the equation E = kI
becomes
9 = k(60)
k = 0.15
Step 3 Using the value of k, we can express the
relationship
between the illuminance and intensity as
E = 0.15I
Step 4 To find the illuminance of the 45-cd source at this
distance, we replace I with 45.
E = 0.15(45) = 6.75
The illuminance is 6.75 units.
Copyright © 2004 Pearson Education, Inc.
Slide 3-94
Direct Variation as nth Power

Let n be a positive real number. Then y varies
directly as the nth power of x, or y is directly
proportional to the nth power of x if there
exists a nonzero real number k such that
y = kxn.
For example, the area of a square of side x is
given by the formula A = x2, so the area varies
directly as the square of the length of a side.
Here k = 1.
Copyright © 2004 Pearson Education, Inc.
Slide 3-95
Inverse Variation
Let n be a positive real number. Then y varies
inversely as the nth power of x, or y is inversely
proportional to the nth power of x, if there exists
a nonzero real number k such that
k
y n
x
If n = 1, then y  k , and y varies inversely as x.
x
Copyright © 2004 Pearson Education, Inc.
Slide 3-96
Example
In a certain manufacturing process, the cost of producing
a single item varies inversely as the square of the
number of items produced. If 50 items are produced,
each costs $3. Find the cost per item if 250 items are
produced.
Solution:
Step 1 Let x represent the number of items produced and
y represent the cost per item. Then for some
nonzero constant k,
k
y
Copyright © 2004 Pearson Education, Inc.
x
2
.
Slide 3-97
Example continued
Step 2 Since y = 3 when x = 50,
k
3 2
50
k  7500
Substitute.
Solve for k .
7500
Step 3 The relationship between x and y is y  2 .
x
Step 4 When 250 items are produced, the cost per item is
7500
y
 0.12, or 12 cents.
2
250
Copyright © 2004 Pearson Education, Inc.
Slide 3-98
Combined and Joint Variation



One variable may depend on more than one other
variable. Such variation is called combined variation.
When a variable depends on the product of two or more
other variables, it is referred to as joint variation.
Joint Variation
Let m and n be real numbers. Then y varies jointly as the
nth power of x and the mth power of z if there exists a
nonzero number k such that
y = kxnzm.
Copyright © 2004 Pearson Education, Inc.
Slide 3-99
Example
The area of a triangle varies jointly as the lengths of the
base and the height. A triangle with base 4 ft and a
height 5 ft has area 10 ft2. Find the area of a triangle with
base 6 ft and height 2 ft.
Solution:
Step 1 Let A represent the area, b the base, and h the
height of the triangle. Then for some number k,
A = kbh.
Step 2 Since A is 10 when b is 4 and h is 5,
10 = k(4)(5)
1
 k.
2
Copyright © 2004 Pearson Education, Inc.
Slide 3-100
Example continued
Step 3 The relationship among the variables is
1
A  bh,
2
the familiar formula for the area of a
triangle.
Step 4 When b = 6ft. and h = 2 ft,
1
A   6  2   6ft 2 .
2
Copyright © 2004 Pearson Education, Inc.
Slide 3-101
Combined Variation
Example:The current (I) in an electric circuit varies
directly with the voltage (E) and inversely with
resistance (R). When the voltage is 16 volts and
the resistance is 8 ohms, the current is 2 amps.
Give the current when the voltage is 10 volts and
the resistance is 20 ohms.
Copyright © 2004 Pearson Education, Inc.
Slide 3-102
Combined Variation continued
Solution: Using the formula I  k
E
and substituting the
R
values for I, E, and R, we are able to find k.
2k
16
8
k 1
Now use the second set of values for E and R, to find I.
I  1
I

10
20
1
2
1
The current is 2 amps when the voltage is 10 volts and
the resistance is 20 ohms.
Copyright © 2004 Pearson Education, Inc.
Slide 3-103