Download Polynomial and Rational Functions

Polynomial
and
Rational
Functions
Outline
Application
3-1 Polynomial Functions
A manufacturer of camping supplies wants to
produce an aluminum drinking cup in the shape of
a right circular cylinder with an open top and a
closed bottom. The cup will have a volume of 65
cubic inches. Find the dimensions of the cup that
will use the minimum amount of aluminum and
find the amount of aluminum used.
3-2 Finding Rational Zeros of
Polynomials
3-3 Approximating Real Zeros of
Polynomials
3-4 Rational Functions
Chapter 3 Group Activity:
Interpolating Polynomials
Chapter 3 Review
ecall that the zeros of a function f are the solutions or roots of
the equation f(x) ⫽ 0, if any exist. There are formulas that give
the exact values of the zeros, real or imaginary, of any linear or
quadratic function (see Table 1).
R
T A B L E
1 Zeros of Linear and Quadratic Functions
Function
Form
Equation
Zeros/Roots
Linear
f(x) ⫽ ax ⫹ b, a ⫽ 0
ax ⫹ b ⫽ 0
Quadratic
f(x) ⫽ ax2 ⫹ bx ⫹ c, a ⫽ 0
ax2 ⫹ bx ⫹ c ⫽ 0
b
a
⫺b ⫾ 兹b2 ⫺ 4ac
x⫽
2a
x⫽⫺
Linear and quadratic functions are also called first- and second-degree
polynomial functions, respectively. Thus, Table 1 contains formulas for
the zeros of any first- or second-degree polynomial function. What
about higher degree polynomial functions such as
p(x) ⫽ 4x3 ⫺ 2x2 ⫹ 3x ⫹ 5
Third degree (cubic)
q(x) ⫽ ⫺2x4 ⫹ 5x2 ⫺ 6
Fourth degree (quartic)
r(x) ⫽ x ⫺ x ⫹ x ⫺ 10
Fifth degree (quintic)
5
4
3
It turns out that there are direct, though complicated, methods for
finding formulas for the zeros of any third- or fourth-degree polynomial function. However, the Frenchman Evariste Galois (1811–1832)
proved at the age of 20 that for polynomial functions of degree
greater than 4 there is no formula or finite step-by-step process that
will always yield exact values for all zeros.* This does not mean that
we give up looking for zeros of higher-degree polynomial functions.
It just means that we will have to use a variety of specialized methods and sometimes we will have to approximate the zeros. The development of these methods is one of the primary objectives of this chapter. Throughout this chapter, we will always use the term zero to refer
to an exact value and will indicate clearly when approximate values
of the zero will suffice.
We begin in Section 3-1 by discussing the graphical properties of polynomial functions. In Section 3-2, we develop tools for finding all the
rational zeros of a polynomial equation with rational coefficients.
178
*Galois’s contribution, using the new concept of “group,” was of the highest mathematical significance and
originality. However, his contemporaries hardly read his papers, dismissing them as “almost unintelligible.” At the age of 21, involved in political agitation, Galois met an untimely death in a duel. A short but
fascinating account of Galois’s tragic life can be found in E. T. Bell’s Men of Mathematics (New York:
Simon & Schuster, 1937), pp. 362–377.
3-1 Polynomial Functions
179
In Section 3-3 we discuss methods for locating the real zeros of a polynomial with real coefficients. Once located, the real zeros are easily
approximated with a graphing utility. Section 3-4 deals with rational
functions and their graphs.
Preparing for This Chapter
Before getting started on this chapter, review the following concepts:
Polynomials (Appendix A, Sections 2 and 3)
Rational Expressions (Appendix A, Section 5)
Graphs of Functions (Chapter 1, Section 4)
Linear Regression (Chapter 1, Group Activity)
Linear Functions (Chapter 2, Section 1)
Quadratic Functions (Chapter 2, Section 3)
Complex Numbers (Chapter 2, Section 4)
Quadratic Formula (Chapter 2, Section 5)
Section 3-1 Polynomial Functions
Polynomial Functions
Polynomial Division
Division Algorithm
Remainder Theorem
Graphing Polynomial Functions
Polynomial Regression
Polynomial Functions
In Chapters 1 and 2 you were introduced to the basic functions
f(x) ⫽ b
f(x) ⫽ ax ⫹ b
Constant function
a⫽0
f(x) ⫽ ax2 ⫹ bx ⫹ c
Linear function
a⫽0
Quadratic function
as well as some special cases of more complicated functions such as
f(x) ⫽ ax3 ⫹ bx2 ⫹ cx ⫹ d
a⫽0
Cubic function
Notice the evolving pattern going from the constant function to the cubic function—the terms in each equation are of the form axn, where n is a nonnegative
integer and a is a real number. All these functions are special cases of the general class of functions called polynomial functions. The function
P(x) ⫽ anxn ⫹ an⫺1xn⫺1 ⫹ ⴢ ⴢ ⴢ ⫹ a1x ⫹ a0
an ⫽ 0
180
3 POLYNOMIAL AND RATIONAL FUNCTIONS
is called an nth-degree polynomial function. We will also refer to P(x) as a polynomial of degree n or, more simply, as a polynomial. The numbers an, an⫺1, . . . ,
a1, a0 are called the coefficients of the function. A nonzero constant function is
a zero-degree polynomial, a linear function is a first-degree polynomial, and a
quadratic function is a second-degree polynomial. The zero function
Q(x) ⫽ 0 is also considered to be a polynomial but is not assigned a degree. The
coefficients of a polynomial function may be complex numbers, or may be
restricted to real numbers, rational numbers, or integers, depending on our interests. The domain of a polynomial function can be the set of complex numbers,
the set of real numbers, or appropriate subsets of either, depending on our interests. In general, the context will dictate the choice of coefficients and domain.
The number r is said to be a zero of the function P, or a zero of the polynomial P(x), or a solution or root of the equation P(x) ⴝ 0, if
P(r) ⫽ 0
A zero of a polynomial may or may not be the number 0. A zero of a polynomial is any number that makes the value of the polynomial 0. If the coefficients
of a polynomial P(x) are real numbers, then a real zero is simply an x intercept
for the graph of y ⫽ P(x). Consider the polynomial
P(x) ⫽ x2 ⫺ 4x ⫹ 3
The graph of P is shown in Figure 1.
y
The x intercepts 1 and 3 are zeros of
P(x) ⫽ x2 ⫺ 4x ⫹ 3, since P(1) ⫽ 0
and P(3) ⫽ 0. The x intercepts 1 and
3 are also solutions or roots for the
equation x2 ⫺ 4x ⫹ 3 ⫽ 0.
FIGURE 1
Zeros, roots, and x intercepts.
5
5
x
y ⫽ P (x)
In general:
ZEROS AND ROOTS
If the coefficients of a polynomial P(x) are real, then the x intercepts of
the graph of y ⫽ P(x) are real zeros of P and P(x) and real solutions,
or roots, for the equation P(x) ⫽ 0.
Polynomial Division
We can find quotients of polynomials by a long-division process similar to that
used in arithmetic. An example will illustrate the process.
3-1 Polynomial Functions
EXAMPLE
181
Algebraic Long Division
1
Divide 5 ⫹ 4x3 ⫺ 3x by 2x ⫺ 3.
Solution
2x2 ⫹ 3x ⫹ 3
2x ⫺ 3 冄4x3 ⫹ 0x2 ⫺ 3x ⫹ 5
4x3 ⫺ 6x2
6x2 ⫺ 3x
6x2 ⫺ 9x
6x ⫹ 5
6x ⫺ 9
14 ⫽ R
Remainder
Arrange the dividend and the divisor
in descending powers of the variable.
Insert, with 0 coefficients, any
missing terms of degree less than 3.
Divide the first term of the divisor
into the first term of the dividend.
Multiply the divisor by 2x2, line up
like terms, subtract (change the sign
and add) as in arithmetic, and bring
down ⫺3x. Repeat the process until
the degree of the remainder is less
than that of the divisor.
Thus,
14
4x3 ⫺ 3x ⫹ 5
⫽ 2x2 ⫹ 3x ⫹ 3 ⫹
2x ⫺ 3
2x ⫺ 3
Check
冤
(2x ⫺ 3) (2x2 ⫹ 3x ⫹ 3) ⫹
冥
14
⫽ (2x ⫺ 3)(2x2 ⫹ 3x ⫹ 3) ⫹ 14
2x ⫺ 3
⫽ 4x3 ⫺ 3x ⫹ 5
MATCHED PROBLEM
Divide 6x2 ⫺ 30 ⫹ 9x3 by 3x ⫺ 4.
1
Being able to divide a polynomial P(x) by a linear polynomial of the form
x ⫺ r quickly and accurately will be of great help in the search for zeros of higherdegree polynomial functions. This kind of division can be carried out more efficiently by a method called synthetic division. The method is most easily understood through an example. Let’s start by dividing P(x) ⫽ 2x4 ⫹ 3x3 ⫺ x ⫺ 5 by
x ⫹ 2, using ordinary long division. The critical parts of the process are indicated
in color.
Divisor
2x3 ⴚ 1x2 ⴙ 2x ⴚ 5
x ⴙ 2冄2x4 ⫹ 3x3 ⫹ 0x2 ⴚ 1x ⴚ 5
2x4 ⫹ 4x3
ⴚ1x3 ⫹ 0x2
⫺1x3 ⴚ 2x2
2x2 ⫺ 1x
2x2 ⴙ 4x
ⴚ5x ⫺ 5
⫺5x ⴚ 10
5
Quotient
Dividend
Remainder
3 POLYNOMIAL AND RATIONAL FUNCTIONS
The numerals printed in color, which represent the essential part of the division process, are arranged more conveniently as follows:
Dividend coefficients
agfgddddgdbgdddfddggc
2
0 ⫺1 ⫺5
3
4 ⫺2
4 ⫺10
⫺1
2 ⫺5
5
2 2
agfgddddbddfddggc abc
Quotient Remainder
coefficients
Mechanically, we see that the second and third rows of numerals are generated as follows. The first coefficient, 2, of the dividend is brought down and multiplied by 2 from the divisor; and the product, 4, is placed under the second dividend coefficient, 3, and subtracted. The difference, ⫺1, is again multiplied by
the 2 from the divisor; and the product is placed under the third coefficient from
the dividend and subtracted. This process is repeated until the remainder is
reached. The process can be made a little faster, and less prone to sign errors, by
changing ⫹2 from the divisor to ⫺2 and adding instead of subtracting. Thus
Dividend coefficients
agfgddddgdbgdddfddggc
182
2
3
⫺2 2
⫺4
⫺1
0 ⫺1 ⫺5
2 ⫺4
2 ⫺5
10
5
agfgddddbddfddggc abc
Quotient Remainder
coefficients
KEY STEPS IN THE SYNTHETIC DIVISION PROCESS
To divide the polynomial P(x) by x ⫺ r:
Step 1. Arrange the coefficients of P(x) in order of descending powers
of x. Write 0 as the coefficient for each missing power.
Step 2. After writing the divisor in the form x ⫺ r, use r to generate the
second and third rows of numbers as follows. Bring down the
first coefficient of the dividend and multiply it by r; then add
the product to the second coefficient of the dividend. Multiply
this sum by r, and add the product to the third coefficient of the
dividend. Repeat the process until a product is added to the
constant term of P(x).
Step 3. The last number to the right in the third row of numbers is the
remainder. The other numbers in the third row are the coefficients of the quotient, which is of degree 1 less than P(x).
EXAMPLE
2
Synthetic Division
Use synthetic division to find the quotient and remainder resulting from dividing P(x) ⫽ 4x5 ⫺ 30x3 ⫺ 50x ⫺ 2 by x ⫹ 3. Write the answer in the form
Q(x) ⫹ R/(x ⫺ r), where R is a constant.
3-1 Polynomial Functions
Solution
183
Since x ⫹ 3 ⫽ x ⫺ (⫺3), we have r ⫽ ⫺3, and
4
0
⫺30
0
⫺50
⫺2
⫺3 4
⫺12
⫺12
36
6
⫺18
⫺18
54
4
⫺12
⫺14
The quotient is 4x4 ⫺ 12x3 ⫹ 6x2 ⫺ 18x ⫹ 4 with a remainder of ⫺14. Thus,
P(x)
⫺14
⫽ 4x4 ⫺ 12x3 ⫹ 6x2 ⫺ 18x ⫹ 4 ⫹
x⫹3
x⫹3
MATCHED PROBLEM
Repeat Example 2 with P(x) ⫽ 3x4 ⫺ 11x3 ⫺ 18x ⫹ 8 and divisor x ⫺ 4.
2
A calculator is a convenient tool for performing synthetic division. Any type
of calculator can be used, although one with a memory will save some keystrokes.
The flowchart in Figure 2 shows the repetitive steps in the synthetic division
process, and Figure 3 illustrates the results of applying this process to Example
2 on a graphing calculator.
Store r
in memory
Enter first
coefficient
Multiply
by r
Add to next
coefficient
Display
result
Yes
Are
there more
coefficients?
No
Stop
FIGURE 2 Synthetic division.
FIGURE 3
Division Algorithm
If we divide P(x) ⫽ 2x4 ⫺ 5x3 ⫺ 4x2 ⫹ 13 by x ⫺ 3, we obtain
2x4 ⫺ 5x3 ⫺ 4x2 ⫹ 13
4
⫽ 2x3 ⫹ x2 ⫺ x ⫺ 3 ⫹
x⫺3
x⫺3
x⫽3
184
3 POLYNOMIAL AND RATIONAL FUNCTIONS
If we multiply both sides of this equation by x ⫺ 3, then we get
2x4 ⫺ 5x3 ⫺ 4x2 ⫹ 13 ⫽ (x ⫺ 3)(2x3 ⫹ x2 ⫺ x ⫺ 3) ⫹ 4
This last equation is an identity in that the left side is equal to the right side for
all replacements of x by real or imaginary numbers, including x ⫽ 3. This example suggests the important division algorithm, which we state as Theorem 1 without proof.
THEOREM
1
DIVISION ALGORITHM
For each polynomial P(x) of degree greater than 0 and each number r,
there exists a unique polynomial Q(x) of degree 1 less than P(x) and a
unique number R such that
P(x) ⫽ (x ⫺ r)Q(x) ⫹ R
The polynomial Q(x) is called the quotient, x ⫺ r is the divisor, and R
is the remainder. Note that R may be 0.
Explore/Discuss
1
Let P(x) ⫽ x3 ⫺ 3x2 ⫺ 2x ⫹ 8.
(A) Evaluate P(x) for
(i) x ⫽ ⫺2
(ii) x ⫽ 1
(iii) x ⫽ 3
(B) Use synthetic division to find the remainder when P(x) is divided by
(i) x ⫹ 2
(ii) x ⫺ 1
(iii) x ⫺ 3
What conclusion does a comparison of the results in parts A and B
suggest?
Remainder Theorem
We now use the division algorithm in Theorem 1 to prove the remainder theorem. The equation in Theorem 1,
P(x) ⫽ (x ⫺ r)Q(x) ⫹ R
is an identity; that is, it is true for all real or imaginary replacements for x.
In particular, if we let x ⫽ r, then we observe a very interesting and useful
relationship:
P(r) ⫽ (r ⫺ r)Q(r) ⫹ R
⫽ 0 ⴢ Q(r) ⫹ R
⫽0⫹R
⫽R
3-1 Polynomial Functions
185
In words, the value of a polynomial P(x) at x ⫽ r is the same as the remainder
R obtained when we divide P(x) by x ⫺ r. We have proved the well-known
remainder theorem:
THEOREM
2
EXAMPLE
3
Solutions
REMAINDER THEOREM
If R is the remainder after dividing the polynomial P(x) by x ⫺ r, then
P(r) ⫽ R
Two Methods for Evaluating Polynomials
If P(x) ⫽ 4x4 ⫹ 10x3 ⫹ 19x ⫹ 5, find P(⫺3) by
(A) Using the remainder theorem and synthetic division
(B) Evaluating P(⫺3) directly
(A) Use synthetic division to divide P(x) by x ⫺ (⫺3).
4
10
0
19
⫺3 4
⫺12
⫺2
6
6
⫺18
1
5
⫺3
2 ⫽ R ⫽ P(⫺3)
(B) P(⫺3) ⫽ 4(⫺3)4 ⫹ 10(⫺3)3 ⫹ 19(⫺3) ⫹ 5
⫽2
MATCHED PROBLEM
Repeat Example 3 for P(x) ⫽ 3x4 ⫺ 16x2 ⫺ 3x ⫹ 7 and x ⫽ ⫺2.
3
You might think the remainder theorem is not a very effective tool for evaluating polynomials. But let’s consider the number of operations performed in
parts A and B of Example 3. Synthetic division requires only 4 multiplications
and 4 additions to find P(⫺3), while the direct evaluation requires 10 multiplications and 4 additions. [Note that evaluating 4(⫺3)4 actually requires 5 multiplications.] The difference becomes even larger as the degree of the polynomial
increases. Computer programs that involve numerous polynomial evaluations
often use synthetic division because of its efficiency. We will find synthetic division and the remainder theorem to be useful tools later in this chapter.
Graphing Polynomial Functions
The shape of the graph of a polynomial function is connected to the degree of
the polynomial. The shapes of odd-degree polynomial functions have something
in common, and the shapes of even-degree polynomial functions have something
186
3 POLYNOMIAL AND RATIONAL FUNCTIONS
in common. Figure 4 shows graphs of representative polynomial functions from
degrees 1 to 6 and suggests some general properties of graphs of polynomial
functions.
FIGURE 4
y
y
y
Graphs of polynomial functions.
5
5
x
⫺5
5
5
x
⫺5
⫺5
5
x
⫺5
5
⫺5
⫺5
(a) f(x) ⫽ x ⫺ 2
(b) g(x) ⫽ x3 ⫹ 5x
y
y
5
(c) h(x) ⫽ x5 ⫺ 6x3 ⫹ 8x ⫹ 1
y
5
x
⫺5
5
⫺5
5
x
⫺5
5
x
⫺5
5
⫺5
(d) F (x) ⫽ x ⫺ x ⫹ 1
2
⫺5
(e) G(x) ⫽ 2x ⫺ 7x ⫹ x ⫹ 3
4
2
(f) H(x) ⫽ x ⫺ 7x4 ⫹ 12x2 ⫺ x ⫺ 2
6
Notice that the odd-degree polynomial graphs start negative, end positive, and
cross the x axis at least once. The even-degree polynomial graphs start positive,
end positive, and may not cross the x axis at all. In all cases in Figure 4, the coefficient of the highest-degree term was chosen positive. If any leading coefficient
had been chosen negative, then we would have a similar graph but reflected in
the x axis.
Explore/Discuss
2
Using a graphing utility, discuss the shape of the graph of each of the
following functions.
(A) y1 ⫽ x2 ⫹ x
(B) y2 ⫽ x3 ⫹ x2
(C) y3 ⫽ x4 ⫹ x3
(D) y4 ⫽ x5 ⫹ x4
In each case, which term seems to most influence the shape of the graph?
The shape of the graph of a polynomial is also related to the shape of the
graph of the term with highest degree or leading term of the polynomial. Figure
5 compares the graph of one of the polynomials from Figure 4 with the graph of
3-1 Polynomial Functions
187
its leading term. Although quite dissimilar for points close to the origin, as we
“zoom out” to points distant from the origin, the graphs become quite similar.
The leading term in the polynomial dominates all other terms combined.
FIGURE 5
y ph
p(x) ⫽ x5,
h(x) ⫽ x5 ⫺ 6x3 ⫹ 8x ⫹ 1.
y
5
ph
500
ZOOM OUT
x
⫺5
x
⫺5
5
⫺5
5
⫺500
In general, the behavior of the graph of a polynomial function as x decreases
without bound to the left or as x increases without bound to the right is determined by its leading term. We often use the symbols ⫺⬁ and ⬁ to help describe
this left and right behavior.* The various possibilities are summarized in Theorem 3.
THEOREM
3
LEFT AND RIGHT BEHAVIOR OF A POLYNOMIAL
P(x) ⫽ anxn ⫹ an⫺1xn⫺1 ⫹ ⭈ ⭈ ⭈ ⫹ a1x ⫹ a0
an ⫽ 0
2. an ⬎ 0 and n odd
1. an ⬎ 0 and n even
Graph of P(x) increases without
Graph of P(x) decreases without
bound as x decreases to the left
bound as x decreases to the left
and as x increases to the right.
and increases without bound as
x increases to the right.
y ⫽ P (x)
y ⫽ P (x)
x
P(x) ®
⬁
冦⬁
as x ® ⫺⬁
as x ® ⬁
x
P(x) ®
⫺⬁
冦⬁
as x ® ⫺⬁
as x ® ⬁
*Remember, the symbol ⬁ does not represent a real number. Earlier, we used ⬁ to denote unbounded intervals, such as [0, ⬁). Now we are using it to describe quantities that are growing with no upper limit on
their size.
188
3 POLYNOMIAL AND RATIONAL FUNCTIONS
THEOREM
3
3. an ⬍ 0 and n even
4. an ⬍ 0 and n odd
Graph of P(x) decreases without
Graph of P(x) increases without
bound as x decreases to the left
bound as x decreases to the left
and as x increases to the right.
and decreases without bound as
x increases to the right.
continued
y ⫽ P (x)
y ⫽ P (x)
x
P(x) ®
⫺⬁
冦 ⫺⬁
as x ® ⫺⬁
as x ® ⬁
x
P(x) ®
⬁
冦 ⫺⬁
as x ® ⫺⬁
as x ® ⬁
Figure 4 shows examples of polynomial functions with graphs containing the
maximum number of local extrema. The points on a continuous graph where the
local extrema occur are sometimes referred to as turning points. Listed in Theorem 4 are useful properties of polynomial functions we accept without proof.
Property 3 is discussed in detail later in this chapter. The other properties are
established in calculus.
THEOREM
4
Explore/Discuss
3
GRAPH PROPERTIES OF POLYNOMIAL FUNCTIONS
Let P be an nth-degree polynomial function with real coefficients.
1. P is continuous for all real numbers.
2. The graph of P is a smooth curve.
3. The graph of P has at most n x intercepts.
4. P has at most n ⫺ 1 local extrema or turning points.
(A) What is the least number of local extrema an odd-degree polynomial
function can have? An even-degree polynomial function?
(B) What is the maximum number of x intercepts the graph of a polynomial function of degree n can have?
(C) What is the maximum number of real solutions an nth-degree
polynomial equation can have?
(D) What is the least number of x intercepts the graph of a polynomial
function of odd degree can have? Of even degree?
(E) What is the least number of real zeros a polynomial function of odd
degree can have? Of even degree?
3-1 Polynomial Functions
EXAMPLE
189
Analyzing the Graph of a Polynomial
4
Approximate to two decimal places the zeros and local extrema for
P(x) ⫽ x3 ⫺ 14x2 ⫹ 27x ⫺ 12
Solution
Examining the graph of P in a standard viewing window [Fig. 6(a)], we see
two zeros and a local maximum near x ⫽ 1. Zooming in shows these points
more clearly [Fig. 6(b)]. Using built-in routines (details omitted), we find that
P(x) ⫽ 0 for x ⬇ 0.66 and x ⬇ 1.54, and that P(1.09) ⬇ 2.09 is a local maximum value.
FIGURE 6
(1.09, 2.09)
P(x) ⫽ x3 ⫺ 14x2 ⫹ 27x ⫺ 12.
10
⫺10
3
10
⫺1
3
⫺10
⫺3
0.66
(a)
1.54
(b)
Have we found all the zeros and local extrema? The graph in Figure 6(a) seems
to indicate that P(x) is decreasing as x decreases to the left and as x increases to
the right. However, the leading term for P(x) is x3. Since x3 increases without
bound as x increases to the right without bound, P(x) must change direction at
some point and become increasing. Thus, there must exist a local minimum and
another zero that are not visible in this viewing window. Tracing along the graph
in Figure 6(a) to the right and observing the coordinates on the screen we discover a local minimum near x ⫽ 8 and a zero near x ⫽ 12. Adjusting the window variables produces the graph in Figure 7. Using built-in routines (details
omitted) we find that P(x) ⫽ 0 for x ⬇ 11.80 and that P(8.24) ⬇ ⫺180.61 is
a local minimum. Since a third-degree polynomial can have at most three zeros
and two local extrema, we have found all the zeros and local extrema for this
polynomial.
FIGURE 7
50
P(x) ⫽ x3 ⫺ 14x2 ⫹ 27x ⫺ 12.
⫺10
15
⫺200
MATCHED PROBLEM
4
Approximate to two decimal places the zeros and the coordinates of the local
extrema for
P(x) ⫽ ⫺x3 ⫺ 14x2 ⫺ 15x ⫹ 5
190
3 POLYNOMIAL AND RATIONAL FUNCTIONS
Polynomial Regression
In the first two chapters, we saw that regression techniques can be used to construct a linear or quadratic model for a set of data. Most graphing utilities have
the ability to use a variety of functions for modeling data. We discuss polynomial
regression models in this section and other types of regression models in later
sections.
EXAMPLE
5
Estimating the Weight of Fish
Using the length of a fish to estimate its weight is of interest to both scientists and sport anglers. The data in Table 1 give the average weight of North
American sturgeon for certain lengths. Use these data and regression techniques to find a cubic polynomial model that can be used to estimate the
weight of a sturgeon for any length. Estimate (to the nearest ounce) the
weights of sturgeon of lengths 45, 46, 47, 48, 49, and 50 inches, respectively.
T A B L E
1 Sturgeon
Length (in.)
x
Weight (oz.)
y
18
22
26
30
34
38
44
52
60
13
26
46
75
115
166
282
492
796
Source: www.thefishernet.com
Solution
The graph of the data in Table 1 [Fig. 8(a)] indicates that a linear regression model
would not be appropriate for this data. And, in fact, we would not expect a linear relationship between length and weight. Instead, since weight is associated
with volume, which involves three dimensions, it is more likely that the weight
would be related to the cube of the length. We use a cubic regression polynomial
to model this data [Fig. 8(b)]. (Consult your manual for the details of calculating
regression polynomials on your graphing utility.) Figure 8(c) adds the graph of
the polynomial model to the graph of the data. The graph in Figure 8(c) shows
that this cubic polynomial does provide a good fit for the data. (We will have
more to say about the choice of functions and the accuracy of the fit provided by
regression analysis later in the text.) Figure 8(d) shows the estimated weights for
the requested lengths.
3-1 Polynomial Functions
191
FIGURE 8
1,000
1,000
0
70
0
70
0
0
(a)
(b)
MATCHED PROBLEM
5
(c)
(d)
Find a quadratic regression model for the data in Table 1 and compare it with the
cubic regression model found in Example 5. Which model appears to provide a
better fit for this data? Use numerical and/or graphical comparisons to support
your choice.
Answers to Matched Problems
2
P(x)
0
2.
⫽ 3x3 ⫹ x2 ⫹ 4x ⫺ 2 ⫹
⫽ 3x3 ⫹ x2 ⫹ 4x ⫺ 2
3x ⫺ 4
x⫺4
x⫺4
3. P(⫺2) ⫽ ⫺3 for both parts, as it should
4. Zeros: ⫺12.80, ⫺1.47, 0.27; local maximum: P(⫺0.57) ⬇ 9.19; local minimum: P(⫺8.76) ⬇ ⫺265.71
5. The cubic regression model provides a better model for this data, especially for 18 ⱕ x ⱕ 26.
1. 3x2 ⫹ 6x ⫹ 8 ⫹
EXERCISE 3-1
y
y
A
In Problems 1–4, a is a positive real number. Match each
function with one of graphs (a)–(d).
1. f(x) ⫽ ⫺ax3
2. g(x) ⫽ ⫺ax4
3. h(x) ⫽ ax6
4. k(x) ⫽ ⫺ax5
y
(c)
y
x
(a)
x
x
(b)
x
(d)
192
3 POLYNOMIAL AND RATIONAL FUNCTIONS
21. (2x3 ⫺ 3x ⫹ 1) ⫼ (x ⫺ 2)
Problems 5–8 refer to the graphs of functions f, g, h, and k
shown below.
f (x)
22. (x3 ⫹ 2x2 ⫺ 3x ⫺ 4) ⫼ (x ⫹ 2)
g (x)
B
x
x
Use synthetic division and the remainder theorem in Problems
23–28.
23. Find P(⫺2), given P(x) ⫽ 3x2 ⫺ x ⫺ 10.
24. Find P(⫺3), given P(x) ⫽ 4x2 ⫹ 10x ⫺ 8.
25. Find P(2), given P(x) ⫽ 2x3 ⫺ 5x2 ⫹ 7x ⫺ 7.
h (x)
26. Find P(5), given P(x) ⫽ 2x3 ⫺ 12x2 ⫺ x ⫹ 30.
k (x)
27. Find P(⫺4), given P(x) ⫽ x4 ⫺ 10x2 ⫹ 25x ⫺ 2.
28. Find P(⫺7), given P(x) ⫽ x4 ⫹ 5x3 ⫺ 13x2 ⫺ 30.
x
x
In Problems 29–44, divide, using synthetic division. Write the
quotient, and indicate the remainder. As coefficients get more
involved, a calculator should prove helpful. Do not round
off—all quantities are exact.
29. (3x4 ⫺ x ⫺ 4) ⫼ (x ⫹ 1)
5. Which of these functions could be a second-degree
polynomial?
30. (5x4 ⫺ 2x2 ⫺ 3) ⫼ (x ⫺ 1)
31. (x5 ⫹ 1) ⫼ (x ⫹ 1)
32. (x4 ⫺ 16) ⫼ (x ⫺ 2)
6. Which of these functions could be a third-degree
polynomial?
33. (3x4 ⫹ 2x3 ⫺ 4x ⫺ 1) ⫼ (x ⫹ 3)
7. Which of these functions could be a fourth-degree
polynomial?
35. (2x6 ⫺ 13x5 ⫹ 75x3 ⫹ 2x2 ⫺ 50) ⫼ (x ⫺ 5)
8. Which of these functions is not a polynomial?
36. (4x6 ⫹ 20x5 ⫺ 24x4 ⫺ 3x2 ⫺ 13x ⫹ 30) ⫼ (x ⫹ 6)
In Problems 9–16, divide, using algebraic long division. Write
the quotient, and indicate the remainder.
9. (4m2 ⫺ 1) ⫼ (2m ⫺ 1)
10. (y2 ⫺ 9) ⫼ (y ⫹ 3)
12. (11x ⫺ 2 ⫹ 12x2) ⫼ (3x ⫹ 2)
14. (a3 ⫹ 27) ⫼ (a ⫹ 3)
15. (3y ⫺ y2 ⫹ 2y3 ⫺ 1) ⫼ (y ⫹ 2)
38. (2x3 ⫺ 5x2 ⫹ 6x ⫹ 3) ⫼ (x ⫺ 12)
39. (4x3 ⫹ 4x2 ⫺ 7x ⫺ 6) ⫼ (x ⫹ 23)
41. (3x4 ⫺ 2x3 ⫹ 2x2 ⫺ 3x ⫹ 1) ⫼ (x ⫺ 0.4)
42. (4x4 ⫺ 3x3 ⫹ 5x2 ⫹ 7x ⫺ 6) ⫼ (x ⫺ 0.7)
43. (3x5 ⫹ 2x4 ⫹ 5x3 ⫺ 7x ⫺ 3) ⫼ (x ⫹ 0.8)
44. (7x5 ⫺ x4 ⫹ 3x3 ⫺ 2x2 ⫺ 5) ⫼ (x ⫹ 0.9)
16. (3 ⫹ x3 ⫺ x) ⫼ (x ⫺ 3)
In Problems 17–22, use synthetic division to write the quotient
P(x) ⫼ (x ⫺ r) in the form P(x)/(x ⫺ r) ⫽ Q(x) ⫹ R/(x ⫺ r),
where R is a constant.
17. (x2 ⫹ 3x ⫺ 7) ⫼ (x ⫺ 2)
37. (4x4 ⫹ 2x3 ⫺ 6x2 ⫺ 5x ⫹ 1) ⫼ (x ⫹ 21)
40. (3x3 ⫺ x2 ⫹ x ⫹ 2) ⫼ (x ⫹ 23)
11. (6 ⫺ 6x ⫹ 8x2) ⫼ (2x ⫹ 1)
13. (x3 ⫺ 1) ⫼ (x ⫺ 1)
34. (x4 ⫺ 3x3 ⫺ 5x2 ⫹ 6x ⫺ 3) ⫼ (x ⫺ 4)
18. (x2 ⫹ 3x ⫺ 3) ⫼ (x ⫺ 3)
For each polynomial function in Problems 45–50:
(A) State the left and right behavior, the maximum number of x
intercepts, and the maximum number of local extrema.
(B) Approximate (to two decimal places) the x intercepts and
the local extrema.
19. (4x2 ⫹ 10x ⫺ 9) ⫼ (x ⫹ 3)
45. P(x) ⫽ x3 ⫺ 5x2 ⫹ 2x ⫹ 6
20. (2x2 ⫹ 7x ⫺ 5) ⫼ (x ⫹ 4)
46. P(x) ⫽ x3 ⫹ 2x2 ⫺ 5x ⫺ 3
3-1 Polynomial Functions
47. P(x) ⫽ ⫺x3 ⫹ 4x2 ⫹ x ⫹ 5
48. P(x) ⫽ ⫺x3 ⫺ 3x2 ⫹ 4x ⫺ 4
49. P(x) ⫽ x4 ⫹ x3 ⫺ 5x2 ⫺ 3x ⫹ 12
50. P(x) ⫽ ⫺x4 ⫹ 6x2 ⫺ 3x ⫺ 16
In Problems 51–54, either give an example of a polynomial
with real coefficients that satisfies the given conditions or
explain why such a polynomial cannot exist.
51. P(x) is a third-degree polynomial with one x intercept.
52. P(x) is a fourth-degree polynomial with no x intercepts.
53. P(x) is a third-degree polynomial with no x intercepts.
54. P(x) is a fourth-degree polynomial with no turning points.
193
69. (A) Divide P(x) ⫽ a2x2 ⫹ a1x ⫹ a0 by x ⫺ r, using both
synthetic division and the long-division process, and
compare the coefficients of the quotient and the remainder produced by each method.
(B) Expand the expression representing the remainder.
What do you observe?
70. Repeat Problem 69 for
P(x) ⫽ a3x3 ⫹ a2x2 ⫹ a1x ⫹ a0
71. Polynomials also can be evaluated conveniently using a
“nested factoring” scheme. For example, the polynomial
P(x) ⫽ 2x4 ⫺ 3x3 ⫹ 2x2 ⫺ 5x ⫹ 7 can be written in a
nested factored form as follows:
P(x) ⫽ 2x4 ⫺ 3x3 ⫹ 2x2 ⫺ 5x ⫹ 7
⫽ (2x ⫺ 3)x3 ⫹ 2x2 ⫺ 5x ⫹ 7
⫽ [(2x ⫺ 3)x ⫹ 2]x2 ⫺ 5x ⫹ 7
C
⫽ {[(2x ⫺ 3)x ⫹ 2]x ⫺ 5}x ⫹ 7
In Problems 55 and 56, divide, using algebraic long division.
Write the quotient, and indicate the remainder.
55. (16x ⫺ 5x3 ⫺ 8 ⫹ 6x4 ⫺ 8x2) ⫼ (2x ⫺ 4 ⫹ 3x2)
56. (8x2 ⫺ 7 ⫺ 13x ⫹ 24x4) ⫼ (3x ⫹ 5 ⫹ 6x2)
In Problems 57 and 58, divide, using synthetic division. Do not
use a calculator.
Use the nested factored form to find P(⫺2) and P(1.7).
[Hint: To evaluate P(⫺2), store ⫺2 in your calculator’s
memory and proceed from left to right recalling ⫺2 as
needed.]
72. Find P(⫺2) and P(1.3) for P(x) ⫽ 3x4 ⫹ x3 ⫺ 10x2 ⫹ 5x
⫺ 2 using the nested factoring scheme presented in Problem 71.
57. (x3 ⫺ 3x2 ⫹ x ⫺ 3) ⫼ (x ⫺ i)
58. (x3 ⫺ 2x2 ⫹ x ⫺ 2) ⫼ (x ⫹ i)
59. Let P(x) ⫽ x2 ⫹ 2ix ⫺ 10. Find
(A) P(2 ⫺ i)
(B) P(5 ⫺ 5i)
(C) P(3 ⫺ i)
(D) P(⫺3 ⫺ i)
60. Let P(x) ⫽ x ⫺ 4ix ⫺ 13. Find
2
APPLICATIONS
73. Revenue. The price-demand equation for 8,000 BTU window air conditioners is given by
p ⫽ 0.0004x2 ⫺ x ⫹ 569
0 ⱕ x ⱕ 800
(A) P(5 ⫹ 6i)
(B) P(1 ⫹ 2i)
where x is the number of air conditioners that can be sold
at a price of p dollars each.
(C) P(3 ⫹ 2i)
(D) P(⫺3 ⫹ 2i)
(A) Find the revenue function.
In Problems 61–68, approximate (to two decimal places) the x
intercepts and the local extrema.
61. P(x) ⫽ 40 ⫹ 50x ⫺ 9x2 ⫺ x3
62. P(x) ⫽ 40 ⫹ 70x ⫹ 18x2 ⫹ x3
64. P(x) ⫽ ⫺0.01x3 ⫹ 2.8x ⫺ 3
65. P(x) ⫽ 0.1x4 ⫹ 0.3x3 ⫺ 23x2 ⫺ 23x ⫹ 90
66. P(x) ⫽ 0.1x4 ⫹ 0.2x3 ⫺ 19x2 ⫹ 17x ⫹ 100
67. P(x) ⫽ x ⫺ 24x ⫹ 167x ⫺ 275x ⫹ 131
3
74. Profit. Refer to Problem 73. The cost of manufacturing
8,000 BTU window air conditioners is given by
C(x) ⫽ 10,000 ⫹ 90x
63. P(x) ⫽ 0.04x3 ⫺ 10x ⫹ 5
4
(B) Find the number of air conditioners that must be sold
to maximize the revenue, the corresponding price to
the nearest dollar, and the maximum revenue to the
nearest dollar.
2
68. P(x) ⫽ x4 ⫹ 20x3 ⫹ 118x2 ⫹ 178x ⫹ 79
where C(x) is the total cost in dollars of producing x air
conditioners.
(A) Find the profit function.
(B) Find the number of air conditioners that must be sold
to maximize the profit, the corresponding price to the
nearest dollar, and the maximum profit to the nearest
dollar.
194
3 POLYNOMIAL AND RATIONAL FUNCTIONS
75. Construction. A rectangular container measuring 1 foot
by 2 feet by 4 feet is covered with a layer of lead shielding
of uniform thickness (see the figure).
(A) Let x represent the number of years since 1960 and
find a cubic regression polynomial for the total
national expenditures.
(B) Use the polynomial model from part A to estimate the
total national expenditures (to the nearest tenth of a
billion) for 1995.
4
1
78. Health Care. Refer to Table 2.
(A) Let x represent the number of years since 1960 and
find a cubic regression polynomial for the per capita
expenditures.
2
Lead shielding
(B) Use the polynomial model from part A to estimate the
per capita expenditures (to the nearest dollar) for
1995.
79. Marriage. Table 3 shows the marriage and divorce rates
per 1,000 population for selected years since 1950.
(A) Find the volume of lead shielding V as a function of
the thickness x (in feet) of the shielding.
(B) Find the thickness of the lead shielding to three
decimal places if the volume of the shielding is 3
cubic feet.
76. Manufacturing. A rectangular storage container measuring 2 feet by 2 feet by 3 feet is coated with a protective
coating of plastic of uniform thickness.
(A) Find the volume of plastic V as a function of the
thickness x (in feet) of the coating.
(B) Find the thickness of the plastic coating to four
decimal places if the volume of the shielding is 0.1
cubic feet.
77. Health Care. Table 2 shows the total national expenditures (in billion dollars) and the per capita expenditures (in
dollars) for selected years since 1960.
T A B L E
2 National Health Expenditures
Per Capita
Expenditures ($)
Year
Total Expenditures
(billion $)
1960
27.1
143
1965
41.6
204
1970
74.4
346
1975
132.9
592
1980
247.2
1,002
1985
428.2
1,666
1990
697.5
Source: U.S. Census Bureau.
2,588
T A B L E
3 Marriages and Divorces
(per 1,000 population)
Year
Marriages
Divorces
1950
11.1
2.6
1955
9.3
2.3
1960
8.5
2.2
1965
9.3
2.5
1970
10.6
3.5
1975
10.0
4.8
1980
10.6
5.2
1985
10.1
5.0
1990
9.8
4.7
Source: U.S. Census Bureau.
(A) Let x represent the number of years since 1950 and
find a cubic regression polynomial for the marriage
rate.
(B) Use the polynomial model from part A to estimate the
marriage rate (to one decimal place) for 1995.
80. Divorce. Refer to Table 3.
(A) Let x represent the number of years since 1950 and
find a cubic regression polynomial for the divorce
rate.
(B) Use the polynomial model from part A to estimate the
divorce rate (to one decimal place) for 1995.