polynomial and rational functions
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Polynomial and Rational Functions. Chapter 3. 3.1 Polynomial Functions and Modeling. Determine the behavior of the graph of a polynomial function using the leading-term test. Factor polynomial functions and find the zeros of their multiplicities. - PowerPoint PPT PresentationTRANSCRIPT
Slide 3-1Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Polynomial andRational Functions
Chapter 3
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
3.1 Polynomial Functions
and Modeling
Determine the behavior of the graph of a polynomial function using the leading-term test.
Factor polynomial functions and find the zeros of their multiplicities. Use a graphing calculator to graph a polynomial
function and find its real-number zeros, relative maximum and minimum values, and domain and range.
Solve applied problems using polynomial models; fit linear, quadratic, power, cubic, and quartic polynomial functions to data.
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Polynomial Function
A polynomial function P is given by
where the coefficients an, an - 1, …, a1, a0 are real numbers and the exponents are whole numbers.
1 21 2 1 0( ) ... ,n n
n nP x a x a x a x a x a
f(x) = x4 3.2x3 + 0.1x4Quartic
f(x) = x3 +2x2 x + 113Cubic
f(x) = 4x2 x + 92Quadratic
f(x) = 3x + 11Linear
f(x) = 40Constant
ExampleDegreePolynomial Function
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Quadratic Function
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Cubic Function
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Examples of Polynomial Functions
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Examples of Nonpolynomial Functions
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The Leading-Term Test
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Example
Using the leading term-test, match each of the following functions with one of the graphs AD, which follow.
a) b)
c) d)
4 3( ) 3 2 3f x x x 3 2( ) 5 4 2f x x x x
5 14( ) 1f x x x 6 5 3( ) 4f x x x x
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Graphs
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Solution
CNegativeEvenx6
APositiveOddx5
BNegativeOdd5x3
DPositiveEven3x4
GraphSign of Leading Coeff.
Degree of Leading Term
Leading Term
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Finding Zeros of Factored Polynomial Functions
If c is a real zero of a function (that is, f(c) = 0), then (c, 0) is an x-intercept of the graph of the function.
Example: Find the zeros of
3
( ) 5( 1)( 1)( 1)( 2)
5( 1) ( 2).
f x x x x x
x x
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Finding Zeros of Factored Polynomial Functions continued
Solution:
To solve the equation f(x) = 0, we use the principle of zero products, solving x 1 = 0
and x + 2 = 0.
The zeros of f(x) are 1 and 2.
See graph on right.
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Even and Odd Multiplicity
If (x c)k, k 1, is a factor of a polynomial function P(x) and:
k is odd, then the graph crosses the x-axis at (c, 0);
k is even, then the graph is tangent to the x-axis at (c, 0).
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Finding Real Zeros on a Calculator
Find the real zeros of the function f given by f(x) = 0.2x3 1.1x2 0.3x + 3
Look for points where the graph crosses the x-axis. It appears that there are three zeros, one between 2 and 1, one near 2, and one near 5. We use the ZERO feature to find them.
The zeros are approximately
x = 1.56689, y = 0
x = 1.82693, y = 0
x = 5.2399, y = 0
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3.2 Graphing
Polynomial Functions
Graph polynomial functions. Use the intermediate value theorem to determine
whether a function has a real zero between two given real numbers.
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Graphing Polynomial Functions
If P(x) is a polynomial function of degree n, the graph of the function has:
at most n real zeros, and thus at most n x-intercepts;
at most n 1 turning points.
(Turning points on a graph, also called relative maxima and minima, occur when the function changes from decreasing to increasing or from increasing to decreasing.)
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Steps to Graph a Polynomial Function
1. Use the leading-term test to determine the end behavior.2. Find the zeros of the function by solving f(x) = 0. Any real zeros
are the first coordinates of the x-intercepts.3. Use the x-intercepts (zeros) to divide the x-axis into intervals and
choose a test point in each interval to determine the sign of all function values in that interval.
4. Find f(0). This gives the y-intercept of the function.5. If necessary, find additional function values to determine the
general shape of the graph and then draw the graph.6. As a partial check, use the facts that the graph has at most n
x-intercepts and at most n 1 turning points. Multiplicity of zeros can also be considered in order to check where the graph crosses or is tangent to the x-axis.
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Example
Graph the polynomial function f(x) = 2x3 + x2 8x 4.
Solution:
1. The leading term is 2x3. The degree, 3, is odd, the coefficient, 2, is positive. Thus the end behavior of the graph will appear as:
2. To find the zero, we solve f(x) = 0. Here we can use factoring by grouping.
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Example continued
Factor:
The zeros are 1/2, 2, and 2. The x-intercepts are (2, 0), (1/2, 0), and (2, 0).
3. The zeros divide the x-axis into four intervals:
(, 2), (2, 1/2), (1/2, 2), and (2, ).
We choose a test value for x from each interval and find f(x).
3 2
2
2
2 8 4 0
(2 1) 4(2 1) 0
(2 1)( 4) 0
(2 1)( 2)( 2) 0
x x x
x x x
x x
x x x
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Example continued
Above x-axis+353(2, )
Below x-axis91(1/2, 2)
Above x-axis+31(2, 1/2)
Below x-axis253(, 2)
Location of points on graph
Sign of f(x)Function value, f(x)
Test Value, x
Interval
4. To determine the y-intercept, we find f(0):
The y-intercept is (0, 4).
3 2
3 2
( ) 2 8 4
( ) 2( ) 8( )0 0 0 0 4 4
f x x x x
f
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Example continued
5. We find a few additional points and complete the graph.
6. The degree of f is 3. The graph of f can have at most 3 x-intercepts and at most 2 turning points. It has 3 x-intercepts and 2 turning points. Each zero has a multiplicity of 1; thus the graph crosses the x-axis at 2, 1/2, and 2. The graph has the end behavior described in step (1). The graph appears to be correct.
71.5
3.51.5
92.5
f(x)x
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Intermediate Value Theorem
For any polynomial function P(x) with real coefficients, suppose that for a b, P(a) and P(b) are of opposite signs. Then the function has a real zero between a and b.
Example: Using the intermediate value theorem, determine, if possible, whether the function has a real zero between a and b.
a) f(x) = x3 + x2 8x; a = 4 b = 1 b) f(x) = x3 + x2 8x; a = 1 b = 3
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Solution
We find f(a) and f(b) and determine where they differ in sign. The graph of f(x) provides a visual check.
f(4) = (4)3 + (4)2 8(4) = 16
f(1) = (1)3 + (1)2 8(1) = 8
By the intermediate value theorem, since f(4) and f(1) have opposite signs, then f(x) has a zero between 4 and 1.
f(1) = (1)3 + (1)2 8(1) = 6
f(3) = (3)3 + (3)2 8(3) = 12
By the intermediate value theorem, since f(1) and f(3) have opposite signs, then f(x) has a zero between 1 and 3.
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3.3 Polynomial Division; The
Remainder and Factor Theorems
Perform long division with polynomials and determine whether one polynomial is a factor of another. Use synthetic division to divide a polynomial by x c. Use the remainder theorem to find a function value f(c). Use the factor theorem to determine whether x c is a factor of f(x).
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Division and Factors
When we divide one polynomial by another, we obtain a quotient and a remainder. If the remainder is 0, then the divisor is a factor of the dividend.
Example: Divide to determine whether x + 3 and x 1
are factors of 3 22 5 4.x x x
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Division and Factors continued
Divide:
Since the remainder is –64, we know that x + 3 is not a factor.
2
3 2
3 2
2
2
5 20
3 2 5 4
3
5 5
5 15
2
rema
0 4
20 60
64 inder
x x
x x x x
x x
x x
x x
x
x
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Division and Factors continued
Divide:
Since the remainder is 0, we know that x 1 is a factor.
2
3 2
3 2
2
2
4
1 2 5 4
5
4 4
4 4
0 remainder
x x
x x x x
x x
x x
x x
x
x
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The Remainder Theorem
If a number c is substituted for x in a polynomial f(x), then the result f(c) is the remainder that would be obtained by dividing f(x) by x c. That is, if f(x) = (x c) • Q(x) + R, then f(c) = R.
Synthetic division is a “collapsed” version of long division; only the coefficients of the terms are written.
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Example
Use synthetic division to find the quotient and remainder.
The quotient is – 4x4 – 7x3 – 8x2 – 14x – 28 and the remainder is –6.
5 4 3 24 6 2 50 ( 2)x x x x x
–6–28–14–8–7–4
–56–28–16–14–8
500261–42 Note: We must write a 0 for the missing term.
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Example
Determine whether 4 is a zero of f(x), where f(x) = x3 6x2 + 11x 6.
We use synthetic division and the remainder theorem to find f(4).
Since f(4) 0, the number is not a zero of f(x).
63–21
12–84
–611–614
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The Factor Theorem
For a polynomial f(x), if f(c) = 0, then x c is a factor of f(x).
Example: Let f(x) = x3 7x + 6. Factor f(x) and solve the equation f(x) = 0.
Solution: We look for linear factors of the form x c. Let’s try x 1.
Since f(1) = 0, we know that x 1 is one factor and the quotient x2 + x 6 is another.So, f(x) = (x 1)(x + 3)(x 2).For f(x) = 0, x = 3, 1, 2.
0–611
–611
6–701 1
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3.4 Theorems about Zeros of
Polynomial Functions
Find a polynomial with specified zeros. For a polynomial function with integer coefficients, find
the rational zeros and the other zeros, if possible. Use Descartes’ rule of signs to find information about
the number of real zeros of a polynomial function with real coefficients.
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The Fundamental Theorem of Algebra
Every polynomial function of degree n, with n 1, has at least one zero in the system of complex numbers.
Example: Find a polynomial function of degree 4 having zeros 1, 2, 4i, and 4i.
Solution: Such a polynomial has factors (x 1),(x 2), (x 4i), and (x + 4i), so we have:
Let an = 1.
( ) ( 1)( 2)( 4 )( 4 )nf x a x x x i x i
2 2
4 3 2 2
4 3 2
( ) ( 1)( 2)( 4 )( 4 )
( 3 2)( 16)
3 2 16 48 32
3 18 48 32
f x x x x i x i
x x x
x x x x x
x x x x
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Zeros of Polynomial Functions with Real Coefficients
If a complex number a + bi, b 0, is a zero of a polynomial function f(x) with real coefficients, then its conjugate, a bi, is also a zero. (Nonreal zeros occur in conjugate pairs.)
Rational Coefficients If where a and b are rational and c is not a perfect square, is a zero of a polynomial function f(x) with rational coefficients, then is also a zero.
,a b c
a b c
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Example
Suppose that a polynomial function of degree 6 with rational coefficients has 3 + 2i, 6i, and as three of its zeros. Find the other zeros.
Solution: The other zeros are the conjugates of the given zeros, 3 2i, 6i, and There are no other zeros because the polynomial of degree 6 can have at most 6 zeros.
1 2
1 2.
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Rational Zeros Theorem
Let
where all the coefficients are integers. Consider a rational number denoted by p/q, where p and q are relatively prime (having no common factor besides 1 and 1). If p/q is a zero of P(x), then p is a factor of a0 and q is a factor of an.
1 21 2 1 0( ) ... ,n n
n nP x a x a x a x a x a
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Example
Given f(x) = 2x3 3x2 11x + 6:
a) Find the rational zeros and then the other zeros.
b) Factor f(x) into linear factors.
Solution:
a) Because the degree of f(x) is 3, there are at most 3 distinct
zeros. The possibilities for p/q are:
3 31 12 2 2 2
1, 2, 3, 6:
1, 2
/ : 1, 1,2, 2,3, 3,6, 6, , , ,
Possibilities for p
Possibilities for q
Possibilities for p q
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Example continued
Use synthetic division to help determine the zeros. It is easier to consider the integers before the fractions.
We try 1: We try 1:
Since f(1) = 6, 1 is not a zero. Since f(1) = 12, 1 is not zero.
–6–12–12
–12–12
6–11–321
12–6–52
65–2
6–11–32–1
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Example continued
We try 3:
.
We can further factor 2x2 + 3x 2 as (2x 1)(x + 2).
0–232
–696
6–11–323
Since f(3) = 0, 3 is a zero. Thus x 3 is a factor. Using the results of the division above, we can express f(x) as
2( ) ( 3)(2 3 2)f x x x x
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Example continued
The rational zeros are 2, 3 and
The complete factorization of f(x) is:
1.
2
( ) (2 1)( 3)( 2)f x x x x
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Descartes’ Rule of Signs
Let P(x) be a polynomial function with real coefficients and a nonzeroconstant term. The number of positive real zeros of P(x) is either:
1. The same as the number of variations of sign in P(x), or2. Less than the number of variations of sign in P(x) by a positive even
integer.
The number of negative real zeros of P(x) is either:
3. The same as the number of variations of sign in P(x), or4. Less than the number of variations of sign in P(x) by a positive even
integer.
A zero of multiplicity m must be counted m times.
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Example
What does Descartes’ rule of signs tell us about the number of positive real zeros and the number of negative real zeros?
There are two variations of sign, so there are either two or zero positive real zeros to the equation.
4 3 2
4 3 2
( ) 3 6 7
3 2
2
6 7
P x x x x x
x x x x
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Example continued
The number of negative real zeros is either two or zero.
Total Number of Zeros4
Positive 2 2 0 0
Negative 2 0 2 0
Nonreal 0 2 2 4
4 3 2
4 3 2
( ) 3( ) 6( ) ( ) 7( )
6 7 2
2
3
P x x x x x
x x xx
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3.5 Rational Functions
For a rational function, find the domain and graph the function, identifying all of the asymptotes.
Solve applied problems involving rational functions.
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Rational Function
A rational function is a function f that is a quotient of two polynomials, that is,
where p(x) and q(x) are polynomials and where q(x) is not the zero polynomial. The domain of f consists of all inputs x for which q(x) 0.
( )( ) ,
( )p x
f xq x
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Example
Consider .
Find the domain and graph f.
Solution:
When the denominator x + 4 = 0, we have x = 4, so the only input that results in a denominator of 0 is 4. Thus the domain is {x|x 4} or (, 4) (4, ).
The graph of the function is the graph of y = 1/x translated to the left 4 units.
1( )
4f x
x
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Vertical Asymptotes
The vertical asymptotes of a rational function f(x) = p(x)/q(x) are found by determining the zeros of q(x) that are not also zeros of p(x). If p(x) and q(x) are polynomials with no common factors other than constants, we need to determine only the zeros of the denominator q(x).
If a is a zero of the denominator, then the line x = a is a vertical asymptote for the graph of the function.
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Example
Determine the vertical asymptotes of the function.
Factor to find the zeros of the denominator:x2 4 = (x + 2)(x 2)
Thus the vertical asymptotes are the lines x = 2 and x = 2.
2
2 3( )
4
xf x
x
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Horizontal Asymptote
The line y = b is a horizontal asymptote.
When the numerator and the denominator of a rational function have the same degree, the line y = a/b is the horizontal asymptote, where a and b are the leading coefficients of the numerator and the denominator, respectively.
Example: Find the horizontal asymptote:
.
The numerator and denominator have the same degree. The ratio of the leading coefficients is 6/9, so the line y = 2/3 is the horizontal asymptote.
4 2
4
6 3 1( )
9 3 2
x xf x
x x
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Determining a Horizontal Asymptote
When the numerator and the denominator of a rational function have the same degree, the line y = a/b is the horizontal asymptote, where a and b are the leading coefficients of the numerator and the denominator, respectively.
When the degree of the numerator of a rational function is less than the degree of the denominator, the x-axis, or y = 0, is the horizontal asymptote.
When the degree of the numerator of a rational function is greater than the degree of the denominator, there is no horizontal asymptote.
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True Statements
The graph of a rational function never crosses a vertical asymptote.
The graph of a rational function might cross a horizontal asymptote but does not necessarily do so.
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Example
Graph .
Vertical asymptotes: x + 3 = 0, so x = 3.
The degree of the numerator and denominator is the same. Thus y = 2, is the horizontal asymptote.
1. Draw the asymptotes with dashed lines.
2. Compute and plot some ordered pairs and draw the curve.
2( )
3
xh x
x
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Example continued
4/52
00
42
84
55
3.57
h(x)x
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Oblique or Slant Asymptote
Find all the asymptotes of .
The line x = 2 is a vertical asymptote.
There is no horizontal asymptote because the degree of the numerator is greater than the degree of the denominator.
22 3 5( )
2
x xh x
x
22 3 5 3( ) (2 1)
2 2
x xh x x
x x
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Oblique or Slant Asymptote continued
Divide to find an equivalent expression.
The line y = 2x 1 is the oblique asymptote.
22 3 5 3( ) (2 1)
2 2
x xh x x
x x
2
2
2 12 2 3 5
2 4
5
2
3
xx x x
x x
x
x
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Occurrence of Lines as Asymptotes of Rational Functions
For a rational function f(x) = p(x)/q(x), where p(x) and q(x) have no common factors other than constants:
Vertical asymptotes occur at any x-values that make the denominator 0.
The x-axis is the horizontal asymptote when the degree of the numerator is less than the degree of the denominator.
A horizontal asymptote other than the x-axis occurs when the numerator and the denominator have the same degree.
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Occurrence of Lines as Asymptotes of Rational Functions continued
An oblique asymptote occurs when the degree of the numerator is 1 greater than the degree of the denominator.
There can be only one horizontal asymptote or one oblique asymptote and never both.
An asymptote is not part of the graph of the function.
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Graphing Rational Functions
1. Find the real zeros of the denominator. Determine the domain of the function and sketch any vertical asymptotes.
2. Find the horizontal or oblique asymptote, if there is one, and sketch it.
3. Find the zeros of the function. The zeros are found by determining the zeros of the numerator. These are the first coordinates of the x-intercepts of the graph.
4. Find f(0). This gives the y-intercept (0, f(0)), of the function.
5. Find other function values to determine the general shape. Then draw the graph.
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Example
Graph .
1. Find the zeros by solving:
The zeros are 1/2 and 3, thus the domain excludes these values.
The graph has vertical asymptotes at x = 3 and x = 1/2. We sketch these with dashed lines.
2. Because the degree of the numerator is less than the degree of the denominator, the x-axis, y = 0, is the horizontal asymptote.
2
3( )
2 5 3
xf x
x x
2
2
2 5 3 0
2 5 3 (2 1)( 3)
x x
x x x x
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Example continued
3. To find the zeros of the numerator, we solve x + 3 = 0 and get x = 3. Thus, 3 is the zero of the function, and the pair (3, 0) is the x-intercept.
4. We find f(0):
Thus (0, 1) is the y-intercept.
2
00
0
3( )
2( ) 5( ) 3
31
3
0f
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Example continued
5. We find other function values to determine the general shape of the graph and then draw the graph.
7/94
12
2/31
1/21
f(x)x
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More Examples
Graph the following functions.
a)
b)
c)
3( )
2
xf x
x
2
2
8( )
9
xf x
x
2
( )1
xf x
x
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Graph a:
Vertical Asymptote
x = 2 Horizontal Asymptote
y = 1 x-intercept
(3, 0) y-intercept
(0, 3/2)
3( )
2
xf x
x
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Graph b:
Vertical Asymptote
x = 3, x = 3 Horizontal Asymptote
y = 1 x-intercepts
(2.828, 0) y-intercept
(0, 8/9)
2
2
8( )
9
xf x
x
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Graph c:
Vertical Asymptote
x = 1 Oblique Asymptote
y = x 1 x-intercept
(0, 0) y-intercept
(0, 0)
2
( )1
xf x
x
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3.6 Polynomial and
Rational Inequalities
Solve polynomial and rational inequalities.
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Polynomial Inequalities
A quadratic inequality can be written in the form ax2 + bc + c > 0, where the symbol could be replaced with either <, , or .
A quadratic inequality is one type of polynomial inequality.
Examples:
4 2 3 232 2 4 6 0 8 2 6 5
4x x x x x x
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Example
Solve: 4x3 7x2 15x.
We need to find all the zeros of the function so we solve the related equation.
The zeros are 0, 3 and 5/4. Thus the x-intercepts of the graph are (0, 0), (3, 0) and (5/4, 0).
3 2
3 2
2
4 7 15
4 7 15 0
(4 7 15) 0
(4 5)( 3) 0
x x x
x x x
x x x
x x x
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Example continued
The zeros divide the x-axis into four intervals. For all x-values within a given interval, the sign of 4x3 7x2 15x 0 must be either positive or negative. To determine which, we choose a test value for x from each interval and find f(x).
Since we are solving 4x3 7x2 15x 0, the solution set consists of only two of the four intervals, those in which the sign of f(x) is negative. {x| < x < 5/4 or 0 < x < 3}.
Positivef(4) = 84(3, )
Negativef(1) = 18(0, 3)
Positivef(1) = 4(5/4, 0)
Negativef(2) = 30(, 5/4)
Sign of f(x)Test ValueInterval
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To Solve a Polynomial Inequality
1. Find an equivalent inequality with 0 on one side.2. Solve the related polynomial equation.3. Use the solutions to divide the x-axis into intervals.
Then select a test value from each interval and determine the polynomial’s sign on the interval.
4. Determine the intervals for which the inequality is satisfied and write interval notation or set-builder notation for the solution set. Include the endpoints of the intervals in the solution set if the inequality symbol is or .
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To Solve a Rational Inequality
1. Find an equivalent inequality with 0 on one side.2. Change the inequality symbol to an equals sign and solve the
related equation.3. Find the values of the variable for which the related rational
function is not defined.3. The numbers found in steps (2) and (3) are called critical values.
Use the critical values to divide the x-axis into intervals. Then test an x-value from each interval to determine the function’s sign in that interval.
5. Select the intervals for which the inequality is satisfied and write interval notation or set-builder notation for the solution set. If the inequality symbol is or , then the solutions to step (2) should be included in the solution set. The x-values found in step (3) are never included in the solution set.
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Example
Solve .
The denominator tells us that f(x) is not defined for
x = 1 and x = 1.
Next, solve f(x) = 0.2
30
1
x
x
2
2
2 22
30
13
01
3( 1) 0( 1)
1
3 0
3
x
xx
xx
x xx
x
x
Slide 3-75Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example continued
The critical values are 3, 1, and 1. These values divide the x-axis into four intervals. We use a test value to determine the sign of f(x) in each interval.
+f(2) = 5/3(1, )
f(0) = 3(1, 1)
+f(2) = 1/3(3, 1)
f(4) = 1/15(, 3)
Sign of f(x)Test ValueInterval
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Example continued
Function values are positive in the intervals (3, 1) and (1, ). Since neither 1 or 3 is in the domain of f, they cannot be part of the solution set.
The solution set is
(3, 1) (1, ).
Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
3.7 Variation and Applications
Find equations of direct, inverse, and combined variation given values of the variables.
Solve applied problems involving variation.
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Direct Variation
If a situation gives rise to a linear function f(x) = kx, or y = kx, where k is a positive constant, we say that we have direct variation, or that y varies directly as x, or that y is directly proportional to x. The number k is called the variation constant, or constant of proportionality.
The graph of y = kx, k > 0, always goes through the origin and rises from left to right. As x increases, y increases; that is, the function is increasing on the interval (0,). The constant k is also the slope of the line.
Slide 3-79Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Direct Variation
Example: Find the variation constant and an equation of variation in which y varies directly as x, and y = 42 when x = 3.
Solution: We know that (3, 42) is a solution of y = kx. y = kx
42 = k 3
= k
14 = k The variation constant 14, is the rate of change of y with respect to
x. The equation of variation is y = 14x.
42
3
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Application
Example: Wages. A cashier earns an hourly wage. If the cashier worked 18 hours and earned $168.30, how much will the cashier earn if she works 33 hours?
Solution: We can express the amount of money earned as a function of the amount of hours worked.
I(h) = kh I(18) = k 18 $168.30 = k 18 $9.35 = k The hourly wage is the variation constant.
Next, we use the equation to find how much the cashier will earn if she works 33 hours.
I(33) = $9.35(33) = $308.55
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Inverse Variation
If a situation gives rise to a function f(x) = k/x, or y = k/x, where k is a positive constant, we say that we have inverse variation, or that y varies inversely as x, or that y is inversely proportional to x. The number k is called the variation constant, or constant of proportionality.
For the graph y = k/x, k 0, as x increases, y decreases; that is, the function is decreasing on the interval (0, ).
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Inverse Variation
Example: Find the variation constant and an equation of variation in which y varies inversely as x, and y = 22 when x = 0.4.
Solution:
The variation constant is 8.8. The equation of variation is y = 8.8/x.
220.4
(0.4)22
8.8
ky
xk
k
k
Slide 3-83Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Application
Example: Road Construction. The time t required to do a job varies inversely as the number of people P who work on the job (assuming that all work at the same rate). If it takes 180 days for 12 workers to complete a job, how long will it take 15 workers to complete the same job?
Solution: We can express the amount of time required, in days, as a
function of the number of people working.
t varies inversely as P
This is the variation constant.
( )
(12)12
18012
2160
kt P
Pk
t
k
k
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Application continued
The equation of variation is t(P) = 2160/P.
Next we compute t(15).
It would take 144 days for 15 people to complete the same job.
2160( )
2160(15)
15144
t PP
t
t
Slide 3-85Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Combined Variation
Other kinds of variation: y varies directly as the nth power of x if there is some
positive constant k such that .
y varies inversely as the nth power of x if there is some positive constant k such that .
y varies jointly as x and z if there is some positive constant k such that y = kxz.
ny kx
n
ky
x
Slide 3-86Copyright © 2006 Pearson Education, Inc. Publishing as Pearson Addison-Wesley
Example
The illuminance of a light (E) varies directly with the intensity (I) of the light and inversely with the square distance (D) from the light. At a distance of 10 feet, a light meter reads 3 units for a 50-cd lamp. Find the illuminance of a 27-cd lamp at a distance of 9 feet.
Solve for k.
Substitute the second set of data into the equation.
The lamp gives an illuminance reading of 2 units.
2
2
2
503
106
6 27
92
IE k
Dk
k
E
E