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Copyright © 2014, 2010 Pearson Education, Inc.
Chapter 2
Polynomials and Rational
Functions
Copyright © 2014, 2010 Pearson Education, Inc.
Copyright © 2014, 2010 Pearson Education, Inc.
Section 2.1 Quadratic Functions
1. Graph a quadratic function in standard form.2. Graph any quadratic function.3. Solve problems modeled by quadratic functions.
SECTION 1.1
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QUADRATIC FUNCTION
A function of the form
where a, b, and c, are real numbers with a ≠ 0, is called a quadratic function.
f x ax2 bx c ,
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THE STANDARD FORM OFA QUADRATIC FUNCTION
The quadratic function
is in standard form. The graph of f is a parabola with vertex (h, k). The parabola is symmetric with respect to the line x = h, called the axis of the parabola. If a > 0, the parabola opens up, and if a < 0, the parabola opens down.
f x a x h 2 k , a 0
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Find the standard form of the quadratic function whose graph has vertex (–3, 4) and passes through the point (–4, 7).
Let y = f (x) be the quadratic function.
2
2
2
3 4
3 4
y a x
y a x
a x
kh
y
2
2
7 3 4
3
Hence,
3 3 4
–4a
a
y x
Example: Writing the Equation of a Quadratic Function
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Example: Graphing a Quadratic Function in Standard Form
Sketch the graph of f (x) = a(x – h)2 + k
Step 1 The graph is a parabola because it has the form f(x) = a(x – h)2 + k Identify a, h, and k.
Sketch the graph of
1. The graph of f(x) = –3(x + 2)2 + 12
is the parabola: a = –3, h= – 2, and k = 12.
23 2 12.f x x
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Example: Graphing a Quadratic Function in Standard Form
Step 2 Determine how the parabola opens. If a > 0, the parabola opens up. If a < 0, the parabola opens down.
Step 3 Find the vertex. The vertex is (h, k). If a > 0 (or a < 0), the function f has a minimum (or a maximum) value k at x = h.
2. Since a = –3, a < 0, the parabola opens down.
3. The vertex (h,k) = (–2, 12). Since the parabola opens down, the function f has a maximum value of 12 at x = –2.
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Example: Graphing a Quadratic Function in Standard Form
Step 4 Find the x-intercepts (if any). Set f (x) = 0 and solving the equation a(x – h)2 + k = 0 for x. If the solutions are real numbers, they are the x-intercepts. If not, the parabola lies above the x-axis (when a > 0) or below the x-axis (when a < 0).
4.
2
2
2
0 3 2 12
12 3 2
4 2
x
x
x
2 2
0 or 4
-intercepts: 0 and 4
x
x x
x
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Example: Graphing a Quadratic Function in Standard Form
Step 5
Find the y-intercept. Replace x with 0. Thenf (0) = ah2 + k is the y-intercept.
5.
20 3 0 2 12
3 4 12 0
-intercept is 0 .
f
y
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Example: Graphing a Quadratic Function in Standard Form
Step 6
Sketch the graph. Plot the points found in Steps 3–5 and join them by a parabola. Show the axis x = h of the parabola by drawing a dashed line.
6. axis: x = –2
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Example: Graphing a Quadratic Function
Graph f(x) = ax2 + bx + c, a ≠ 0.
Step 1 Identify a, b, and c.
Sketch the graph of
1. In the equation y = f(x) = 2x2 + 8x – 10,
a = 2, b = 8, and c = – 10.
22 8 10.f x x x
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Example: Graphing a Quadratic Function
Step 2 Determine how the parabola opens. If a > 0, the parabola opens up. If a < 0, the parabola opens down.
Step 3 Find the vertex.
2. Since a = 2 > 0, the parabola opens up.
3.
2
2
bh
ab
k fa
2
82
2 2 2
2
2 2 8 2 10 18
, 2, 18
bh
a
k f
h k
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Example: Graphing a Quadratic Function
Step 4 Find the x-intercepts (if any). Set f (x) = 0 and solving the equation a(x – h)2 + k = 0 for x. If the solutions are real numbers, they are the x-intercepts. If not, the parabola lies above the x-axis (when a > 0) or below the x-axis (when a < 0).
4. 22 8 10 0x x
22 4 5 0
2 5 1 0
x x
x x
5 0 or 1 0
5 1
x x
x x
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Example: Graphing a Quadratic Function
Step 5
Find the y-intercept. Let x = 0. The result,f (0) = c, is the y-intercept.
5. Let x = 0.
20 2 0 8 0 10
-intercept is 10 .
f
y
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Example: Graphing a Quadratic Function
Step 6
The parabola is symmetric with respect to its axis,
Use this symmetry to find additional points.
6.
Axis of symmetry is x = –2. The symmetric image of (0, –10) with respect to the axis x = –2 is (–4, –10).
.2
bx
a
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Example: Graphing a Quadratic Function
Step 7
Draw a parabola through the points found in Steps 3–6.
7. Sketch the parabola passing through the points found in Steps 3–6.
.2
bx
a
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The graph of f (x) = –2x2 +8x – 5 is shown.a. Find the domain and range of f.b. Solve the inequality –2x2 +8x – 5 > 0.
b. The graph is above the x-axis in the interval
4 6 4 6, .
2 2
a. The domain is (–∞, ∞). The range is (–∞, 3].
Example: Identifying the Characteristics of a Quadratic Function from Its Graph
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Section 2.2 Polynomial Functions
1. Learn properties of the graphs of polynomial functions.2. Determine the end behavior of polynomial functions.3. Find the zeros of a polynomial function by factoring.4. Identify the relationship between degrees, real zeros, and
turning points.5. Graph polynomial functions.
SECTION 1.1
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Definitions
A polynomial function of degree n is a function of the form
where n is a nonnegative integer and the coefficients an, an–1, …, a2, a1, a0 are real numbers with an ≠ 0.
f x an xn an 1xn 1 ... a2 x2 a1x a0 ,
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Definitions
The term anxn is called the leading term.
The number an is called the leading coefficient, and a0 is the constant term.
A constant function f (x) = a, (a ≠ 0) which may be written as f (x) = ax0, is a polynomial of degree 0.
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Definitions
The zero function f (x) = 0 has no degree assigned to it.
Polynomials of degree 3, 4, and 5 are called cubic, quartic, and quintic polynomials.
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Definitions
an xn an 1xn 1 ... a2 x2 a1x a0
The expression
is a polynomial, the function f defined by
is a polynomial function, and the equation
is a polynomial equation.
f x an xn an 1xn 1 ... a2 x2 a1x a0
an xn an 1xn 1 ... a2 x2 a1x a0 0
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COMMON PROPERTIES OFPOLYNOMIAL FUNCTIONS
1. The domain of a polynomial function is the set of all real numbers.
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2. The graph of a polynomial function is a continuous curve.
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3. The graph of a polynomial function is a smooth curve.
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State which functions are polynomial functions. For each polynomial function, find its degree, the leading term, and the leading coefficient.a. f (x) = 5x4 – 2x + 7b. g(x) = 7x2 – x + 1, 1 x 5
a. f (x) is a polynomial function. Its degree is 4, the leading term is 5x4, and the leading coefficient is 5.b. g(x) is not a polynomial function because its domain is not (–, ).
Example: Polynomial Functions
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POWER FUNCTION
A function of the form
is called a power function of degree n, where a is a nonzero real number and n is a positive integer.
f x axn
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POWER FUNCTIONS OF EVEN DEGREE
The graph is symmetric with respect to the y-axis.
The graph of y = xn (n is even) is similar to the graph of y = x2.
Let f x axn . If n is even, then x n xn .
Then f x a x n axn f x .
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POWER FUNCTIONS OF ODD DEGREE
The graph is symmetric with respect to the origin.
The graph of y = xn (n is odd) is similar to the graph of y = x3.
Let f x axn . If n is odd, then x n xn .
Then f x a x n axn f x .
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END BEHAVIOR OF POLYNOMIAL FUNCTIONS
Case 1a > 0
The graph rises to the left and right, similar to y = x2.
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END BEHAVIOR OF POLYNOMIAL FUNCTIONS
Case 2a < 0
The graph falls to the left and right, similar to y = –x2.
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END BEHAVIOR OF POLYNOMIAL FUNCTIONS
Case 3a > 0
The graph rises to the right and falls to the left, similar to y = x3.
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END BEHAVIOR OF POLYNOMIAL FUNCTIONS
Case 4a < 0
The graph rises to the left and falls to the right, similar to y = –x3.
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Let be a polynomialfunction of degree 3. Show thatwhen |x| is very large.
P x 2x3 5x2 7x 11P x 2x3
P x x3 2 5
x
7
x2 11
x3
When |x| is very large are close to 0.5
x,
7
x2 and 11
x3
P x x3 2 0 0 0 2x3.Therefore,
Example: Understanding the End Behavior of a Polynomial Function
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THE LEADING–TERM TEST
Its leading term is anxn.
The behavior of the graph of f as x → ∞ or as x → –∞ is similar to one of the following four graphs and is described as shown in each case.
The middle portion of each graph, indicated by the dashed lines, is not determined by this test.
Let 11 1 0... 0n n
nn nf x a x ax ax aa
be a polynomial function.
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Case 1
an > 0
THE LEADING–TERM TEST
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Case 2
an < 0
THE LEADING–TERM TEST
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Case 3
an > 0
THE LEADING–TERM TEST
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Case 4
an < 0
THE LEADING–TERM TEST
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Use the leading-term test to determine the end behavior of the graph of
y f x 2x3 3x2 4.
Here n = 3 (odd) and an = –2 < 0. Thus, Case 4 applies. The graph of f (x) rises to the left and falls to the right. This behavior is described as y → ∞ as x → –∞ and y → –∞ as x → ∞.
Example: Using the Leading-Term Test
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REAL ZEROS OF POLYNOMIAL FUNCTIONS
1. c is a zero of f .2. c is a solution (or root) of the equation f (x) = 0.3. c is an x-intercept of the graph of f . The point
(c, 0) is on the graph of f .
If f is a polynomial function and c is a real number, then the following statements are equivalent.
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Find all zeros of each polynomial function.
4 3 2
3 2
a. 2 3
b. 2 2
f x x x x
g x x x x
Factor f (x) and then solve f (x) = 0.
4 3 2
2
2
2
2
a. 2 3
3
0, 1 0, 3 0
0,
2 3
1
1, 3
f x x x x
x
x x
x x
x
x
x
x
x
x
x
Example: Finding the Zeros of a Polynomial Function
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3 2
2
2
2
2
b. 2 2
1
1
0 2 1
2 0 or
2 2
2
1 0
2
f x x x x
x
x
x x
x
x
x
x
x
x
The only zero of g(x) is 2, since x2 + 1 > 0 for all real numbers x.
Example: Finding the Zeros of a Polynomial Function
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REAL ZEROS OF POLYNOMIAL FUNCTIONS
A polynomial function of degree n with real coefficients has at most n real zeros.
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Find the number of distinct real zeros of the following polynomial functions of degree 3.
a. Solve f (x) = 0.
22
a. 1 2 3
b. 1 1 c. 3 1
f x x x x
g x x x h x x x
1 2 3 0
1 0 or 2 0 or 3 0
1 or 2 or 3
f x x x x
x x x
x x x
f (x) has three real zeros: –2, 1, and 3.
Example: Finding the Number of Real Zeros
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2
2
b. 1 1 0
1 0 or 1 0
1 No Real Solutions
g x x x
x x
x
h(x) has two distinct real zeros: –1 and 3.
2c. 3 1
3 0 or 1 0
3 or 1
h x x x
x x
x x
g(x) has only one real zero: –1.
Example: Finding the Number of Real Zeros
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MULTIPLICITY OF A ZERO
If c is a zero of a polynomial function f (x) and the corresponding factor (x – c) occurs exactly m times when f (x) is factored, then c is called a zero of multiplicity m.
1.If m is odd, the graph of f crosses the x-axis at x = c.
2. If m is even, the graph of f touches but does not cross the x-axis at x = c.
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MULTIPLICITY OF A ZERO
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MULTIPLICITY OF A ZERO
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Find the zeros of the polynomial function f (x) = x2(x + 1)(x – 2), and give the multiplicity of each zero.
f (x) is already in factored form.
f (x) = x2(x + 1)(x – 2) = 0 x2 = 0, or x + 1 = 0, or x – 2 = 0 x = 0 or x = –1 or x = 2
The zero x = 0 has multiplicity 2, while each of the zeros –1 and 2 have multiplicity 1.
Example: Finding the Zeros and Their Multiplicities
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TURNING POINTS
A local (or relative) maximum value of f is higher than any nearby point on the graph.
A local (or relative) minimum value of f is lower than any nearby point on the graph.
The graph points corresponding to the local maximum and local minimum values are called turning points. At each turning point the graph changes from increasing to decreasing or vice versa.
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TURNING POINTS
The graph of f has turning points at (–1, 12) and at (2, –15).
f x 2x3 3x2 12x 5
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NUMBER OF TURNING POINTS
If f (x) is a polynomial of degree n, then the graph of f has at most (n – 1) turning points.
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Use a graphing calculator and the window –10 x 10; –30 y 30 to find the number of turning points of the graph of each polynomial.
4 2
3 2
3 2
a. 7 18
b. 12
c. 3 3 1
f x x x
g x x x x
h x x x x
Example: Finding the Number of Turning Points
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f has three total turning points; two local minimum and one local maximum.
a. f x x4 7x2 18
Example: Finding the Number of Turning Points
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g has two total turning points; one local maximum and one local minimum.
b. g x x3 x2 12x
Example: Finding the Number of Turning Points
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h has no turning points, it is increasing on the interval (–∞, ∞).
c. h x x3 3x2 3x 1
Example: Finding the Number of Turning Points
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Example: Graphing a Polynomial Function
Sketch the graph of a polynomial function
Step 1 Determine the end behavior. Apply the leading-term test.
Sketch the graph of
1. Degree = 3 Leading coefficient = –1End behavior shown.
3 24 4 16.f x x x x
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Example: Graphing a Polynomial Function
Step 2 Find the zeros of the polynomial function.
Set f (x) = 0 and solve. The zeros give the x-intercepts.
2.
Each zero has multiplicity 1, the graph crosses the x-axis at each zero.
3 24 4 16 0
4 2 2 0
4, 2, or 2
x x x
x x x
x x x
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Example: Graphing a Polynomial Function
Step 3 Find the y-intercept by computing f (0).
Step 4 Use symmetry to check whether the function is odd, even, or neither.
3. The y-intercept is f(0) = 16. the graph passes through the point (0,16).
4.
There is no symmetry in the y-axis nor with respect to the origin.
3
3 2
4 16
4 4 16
f x x x
x x x
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Example: Graphing a Polynomial Function
Step 5 Determine the sign of f(x) by using arbitrarily chosen “test numbers” in the intervals defined by the x-intercepts. Find the intervals on which the graph lies above or below the x-axis.
5.
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Example: Graphing a Polynomial Function
Step 6 Draw the graph. Use the fact that the number of turning points is less than the degree of the polynomial to check whether the graph is drawn correctly.
6. Draw the graph.
The number of turning points is 2, which is less than 3, the degree of f.
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Section 2.3 Dividing Polynomials and the Rational Zeros Test
1. Learn the Division Algorithm.2. Use the Remainder and Factor Theorems.3. Use the Rational Zeros Test.
SECTION 1.1
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POLYNOMIAL FACTOR
A polynomial D(x) is a factor of a polynomial F(x) if
there is a polynomial Q(x) such that F(x) = D(x) ∙
Q(x).
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THE DIVISION ALGORITHM
If a polynomial F(x) is divided by a polynomial D(x), with
D(x) ≠ 0, there are unique polynomials Q(x) and R(x)
such that
Either R(x) is the zero polynomial, or the degree of
R(x) is less than the degree of D(x).
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Use long division and synthetic division to find the quotient and remainder when
2x4 + x3 − 16x2 + 18 is divided by x + 2.
We will start by performing long division.
Example: Using Long Division and Synthetic Division
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Example: Using Long Division and Synthetic Division
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quotient: 2x3 – 3x2 – 10x + 20 remainder: –22
The result is
Example: Using Long Division and Synthetic Division
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THE REMAINDER THEOREM
If a polynomial F(x) is divided by x – a, then the remainder R is given by
.R F a
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Find the remainder when the polynomialF x 2x5 4x3 5x2 7x 2
is divided by x 1.
5 3 21 1 1 1 12 4 5 7 2
2 4 5 7 2 2
F
By the Remainder Theorem, F(1) is the remainder.
The remainder is –2.
Example: Using the Remainder Theorem
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Let 4 3 23 5 8 75.f x x x x x Find f 3 .
One way is to evaluate f (x) when x = –3.
Another way is to use synthetic division.
4 3 23 53 3 3 3 638 75f
Either method yields a value of 6.
1 3 5 8 75
3 0 15 69
1 0 5 23
3
6
Example: Using the Remainder Theorem
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THE FACTOR THEOREM
A polynomial F(x) has (x – a) as a factor if and only if F(a) = 0.
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Given that 2 is a zero of the function
solve the polynomial equation
Since 2 is a zero of f (x), f (2) = 0 and (x – 2) is a factor of f (x). Perform synthetic division by 2.
f x 3x2 2x2 19x 6,
3x2 2x2 19x 6 0.
2 3 2 19 6
6 16 6
3 8 3 0
Example: Using the Factor Theorem
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2 2 23 2 19 6 2 3 8 3f x x x x x x x
Since the remainder is 0,
To find other zeros, solve the depressed equation.
23 8 3 0
3 1 3 0
3 1 0 or 3 0
1 or 3
3
x x
x x
x x
x x
Example: Using the Factor Theorem
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Including the original zero of 2, the solution set is
3,1
3, 2
.
Example: Using the Factor Theorem
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THE RATIONAL ZEROS TEST
1. p is a factor of the constant term a0;
2. q is a factor of the leading coefficient an.
is a polynomial function with integer coefficients (an ≠ 0, a0 ≠ 0) and
is a rational number in lowest terms that is a zero of F(x), then
1 21 2 1 0If ...n n
n nF x a x a x a x a x a
p
q
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Find all the rational zeros of
3 25 4 .2 3F x x x x List all possible zeros
Factors of the constant term,
Factors of the leading coefficien
3
t, 2
p
q
Factors of : 1, 3
Factors 2 1
3
of : , 2
Possible rational zeros are: 1, 1
2,
3
2, 3.
Example: Using the Rational Zeros Test
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Begin testing with 1. if it is not a rational zero, then try another possible zero.
1 2 5 4 3
2 7 3
2 7 3 0
The remainder of 0 tells us that (x – 1) is a factor of F(x). The other factor is 2x2 + 7x + 3. To find the other zeros, solve 2x2 + 7x + 3 = 0.
Example: Using the Rational Zeros Test
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22 7 3 0
2 1 3 0
2 1 0 or 3 0
1 or 3
2
x x
x x
x x
x x
The solution set is 1, 1
2, 3
.
The rational zeros of F are 3, 1
2, and 1.
Example: Using the Rational Zeros Test
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Section 2.4 Zeros of a Polynomial Function
1. Learn basic facts about the complex zeros of polynomials.2. Use the Conjugate Pairs Theorem to find zeros of
polynomials.
SECTION 1.1
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If we extend our number system to allow the coefficients of polynomials and variables to represent complex numbers, we call the polynomial a complex polynomial. If P(z) = 0 for a complex number z we say that z is a zero or a complex zero of P(x).
In the complex number system, every nth-degree polynomial equation has exactly n roots and every nth-degree polynomial can be factored into exactly n linear factors.
Definitions
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FUNDAMENTAL THEOREM OF ALGEBRA
Every polynomial
with complex coefficients an, an – 1, …, a1, a0 has at least one complex zero.
11 1 0... 1, 0n n
n n nP x a x a x a x a n a
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FACTORIZATION THEOREM FOR POLYNOMIALS
If P(x) is a complex polynomial of degree n ≥ 1, it can be factored into n (not necessarily distinct) linear factors of the form
where a, r1, r2, … , rn are complex numbers.
1 2 ... ,nP x a x r x r x r
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Find a polynomial P(x) of degree 4 with a leading coefficient of 2 and zeros –1, 3, i, and –i. Write P(x)
a. Since P(x) has degree 4, we write
1 2 3 4
1
2 1 3
32
P x x x x x
x x x x
x x x i x i
r r r
i i
a r
a. in completely factored form;b. by expanding the product found in part a.
Example: Constructing a Polynomial Whose Zeros Are Given
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b. Expand the product found in part a.
P x 2 x 1 x 3 x i x i 2 x 1 x 3 x2 1 2 x 1 x3 3x2 x 3 2 x4 2x3 2x2 2x 3 2x4 4x3 4x2 4x 6
Example: Constructing a Polynomial Whose Zeros Are Given
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CONJUGATE PAIRS THEOREM
If P(x) is a polynomial function whose coefficients are real numbers and if z = a + bi is a zero of P, then its conjugate, is also a zero of P.,z a bi
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ODD–DEGREE POLYNOMIALSWITH REAL ZEROS
Any polynomial P(x) of odd degree with real coefficients must have at least one real zero.
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A polynomial P(x) of degree 9 with real coefficients has the following zeros: 2, of multiplicity 3; 4 + 5i, of multiplicity 2; and 3 – 7i. Write all nine zeros of P(x).
Since complex zeros occur in conjugate pairs, the conjugate 4 – 5i of 4 + 5i is a zero of multiplicity 2, and the conjugate 3 + 7i of 3 – 7i is a zero of P(x). The nine zeros of P(x) are:
2, 2, 2, 4 + 5i, 4 – 5i, 4 + 5i, 4 – 5i, 3 + 7i, 3 – 7i
Example: Using the Conjugate Pairs Theorem
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FACTORIZATION THEOREM FOR A POLYNOMIAL WITH REAL COEFFICIENTS
Every polynomial with real coefficients can be uniquely factored over the real numbers as a product of linear factors and/or irreducible quadratic factors.
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Given that 2 – i is a zero of
find the remaining zeros.
The conjugate of 2 – i, 2 + i is also a zero.
P x x4 6x3 14x2 14x 5,
x 2 i x 2 i x 2 i x 2 i
x 2 2 i2 x2 4x 5
So P(x) has linear factors: x 2 i x 2 i
Example: Finding the Complex Zeros of a Polynomial
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Divide P(x) by x2 – 4x + 5.
x2 4x 5 x4 6x3 14x2 14x 5
x4 4x3 5x2
2x3 9x2 14x
2x3 8x2 10x
x2 4x 5
x2 4x 5
0
x2 2x 1
Example: Finding the Complex Zeros of a Polynomial
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Therefore
The zeros of P(x) are 1 (of multiplicity 2), 2 – i, and 2 + i.
2 2
2
2 1 4 5
1 1 4 5
1 1 2 2
P x x x x x
x x x x
x x x i x i
Example: Finding the Complex Zeros of a Polynomial
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Find all zeros of the polynomial P(x) = x4 – x3 + 7x2 – 9x – 18.
Possible zeros are: ±1, ±2, ±3, ±6, ±9, ±18Use synthetic division to find that 2 is a zero.
2 1 1 7 9 18
2 2 18 18
1 1 9 9 0
(x – 2) is a factor of P(x). Solve x3 x2 9x 9 0
Example: Finding the Zeros of a Polynomial
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3 2
2
2
2
2
9 9 0
9 0
9 0
1 0 or 9 0
1 or 9
1 or
1
3
1
1
x x x
x
x
x
x x
x
x
x x
x x i
The four zeros of P(x) are –1, 2, –3i, and 3i.
Example: Finding the Zeros of a Polynomial
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Section 2.5 Rational Functions
1. Define a rational function.2. Define vertical and horizontal asymptotes.3. Graph translations of 4. Find vertical and horizontal asymptotes (if any).5. Graph rational functions.6. Graph rational functions with oblique asymptotes.7. Graph a revenue curve.
SECTION 1.1
1( ) .f x
x
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RATIONAL FUNCTION
A function f that can be expressed in the form
where the numerator N(x) and the denominator D(x) are polynomials and D(x) is not the zero polynomial, is called a rational function. The domain of f consists of all real numbers for which D(x) ≠ 0.
f x N x D x ,
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Find the domain of each rational function.
a. f x 3x2 12
x 1
a. The domain of f (x) is the set of all real numbers for which x – 1 ≠ 0; that is, x ≠ 1 .In interval notation:
b. g x x
x2 6x 8
c. h x x2 4
x 2
,1 U 1,
Example: Finding the Domain of a Rational Function
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b. Find the values of x for which the denominator x2 – 6x + 8 = 0, then exclude those values from the domain.
The domain of g (x) is the set of all real numbers such that x ≠ 2 and x ≠ 4 .
In interval notation: x 2 x 4 0
x 2 0 or x 4 0
x 2 or x 4
,2 U 2, 4 U 4,
Example: Finding the Domain of a Rational Function
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,2 U 2,
Example: Finding the Domain of a Rational Function
c. The domain of
is the set of all real numbers for which x – 2 ≠ 0; that is, x ≠ 2 .
The domain of g (x) is the set of all real numbers such that x ≠ 2.
In interval notation:
2 4
2
xh x
x
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VERTICAL ASYMPTOTES
The line with equation x = a is called a vertical asymptote of the graph of a function f if
as or as
or if as or as .
f x x a x a
f x x a x a
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VERTICAL ASYMPTOTES
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VERTICAL ASYMPTOTES
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LOCATING VERTICAL ASYMPTOTES OF RATIONAL FUNCTIONS
If is a rational function,
where the N(x) and D(x) do not have a common factor and a is a real zero of D(x), then the line with equation x = a is a vertical asymptote of the graph of f.
f x N x D x
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Find all vertical asymptotes of the graph of each rational function.
a. No common factors, zero of the denominator is x = 1. The line with equation x = 1 is a vertical asymptote of f (x).
a. f x 1
x 1
b. g x 1
x2 9
c. h x 1
x2 1
Example: Finding Vertical Asymptotes
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b. No common factors. Factoring x2 – 9 = (x + 3)(x – 3), we see the zeros of the denominator are –3 and 3. The lines with equations x = – 3 and x = 3 are the two vertical asymptotes of g (x).
c. The denominator x2 + 1 has no real zeros. Hence, the graph of the rational function h (x) has no vertical asymptotes.
Example: Finding Vertical Asymptotes
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Find all vertical asymptotes of the graph of each rational function.
The graph is the line with equation y = x + 3, with a gap (hole) corresponding to x = 3.
2 9
a. 3
xh x
x
b. g x x 2
x2 4
2 3 39a.
3 33, 3
x xxh x
x xx x
Example: Rational Function Whose Graph Has a Hole
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Example: Rational Function Whose Graph Has a Hole
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The graph has a hole at x = –2. However, the graph of g(x) also has a vertical asymptote at x = 2.
b. g x x 2
x2 4
x 2
x 2 x 2
1
x 2, x 2
Example: Rational Function Whose Graph Has a Hole
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Example: Rational Function Whose Graph Has a Hole
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HORIZONTAL ASYMPTOTES
The line with equation y = k is called a horizontal asymptote of the graph of a function f if
f x k as x or x .
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RULES FOR LOCATINGHORIZONTAL ASYMPTOTES
Let f be a rational function given by
To find whether the graph of f has one horizontal asymptote or no horizontal asymptote, we compare the degree of the numerator, n, with that of the denominator, m:
an xn an 1x
n 1 ... a2 x2 a1x a0
bm xm bm 1xm 1 ... b2 x2 b1x b0
,an 0,bn 0
f x N x D x
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1. If n < m, then the x-axis (y = 0) is the horizontal asymptote.
2. If n = m, then the line with equationis the horizontal asymptote.
3. If n > m, then the graph of f has no horizontal asymptote.
y an
bm
RULES FOR LOCATINGHORIZONTAL ASYMPTOTES
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Find the horizontal asymptote (if any) of the graph of each rational function.
a. f x 5x 2
1 3x
a. Numerator and denominator have degree 1.
is the horizontal asymptote.
b. g x 2x
x2 1
c. h x 3x2 1
x 2
5
3
5
3y
Example: Finding the Horizontal Asymptote
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degree of denominator > degree of numerator y = 0 (the x-axis) is the horizontal asymptote.
b. g x 2x
x2 1
c. h x 3x2 1
x 2
degree of numerator > degree of denominator The graph has no horizontal asymptote.
Example: Finding the Horizontal Asymptote
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Example: Graphing a Rational Function
Graph where f(x) is in lowest terms.
Step 1 Find the intercepts. Since f is in lowest terms, the x-intercepts are found by solving the equation N(x) = 0. The y-intercept is, if there is one, is f (0).
Sketch the graph of
1.
( )
,( )
N xf x
D x
2
2
2 2.
9
xf x
x
22 2 0
2 1 1 0
1 0 or 1 0
1 or 1
x
x x
x x
x x
The x-intercepts are 1 and 1, so the graph passes through (1, 0) and (1, 0).
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Example: Graphing a Rational Function
Step 1 Find the intercepts. Since f is in lowest terms, the x-intercepts are found by solving the equation N(x) = 0. The y-intercept is, if there is one, is f (0).
1.
The y-intercept is , so the graph of f passes through the point (0, ).
2
2
2 0 2 20
90 9f
2
9
2
9
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Example: Graphing a Rational Function
Step 2 Find the vertical asymptotes (if any). Solve D(x) = 0 to find the vertical asymptotes of the graph. Sketch the vertical asymptotes.
2.
The vertical asymptotes are x = 3 and x = –3.
2Solve 9 0, 3x x
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Example: Graphing a Rational Function
Step 3 Find the horizontal asymptotes (if any). Use the rules from the previous slide.
3. Since n = m = 2 the horizontal asymptote is
Leading coefficient of 22.
Leasing coefficient of ( ) 1
N xy
D x
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Example: Graphing a Rational Function
Step 4 Test for symmetry. If f (–x) = f (x), then f is symmetric with respect to the y-axis. If f (–x) = –f (x), then f is symmetric with respect to the origin.
4.
The graph of f is symmetric in the y-axis. This is only one symmetry.
2
2
2
2
2 2
9
2 2.
9
xxf
xf x
x
x
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Example: Graphing a Rational Function
Step 5 Locate the graph relative to the horizontal asymptote (if any). Use the sign graphs and test numbers associated with the zeros of N(x) and D(x), to determine where the graph of f is above the x-axis and where it is below the x-axis.
5.
R(x) = 16 has no zeros and D(x) has zeros –3 and 3. These zeros divide the x-axis into three intervals. We choose test points –4, 0, and 4 to test the sign of
2
2 2
2 2 162
9 9
xf
x xx
2
16.
9x
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Example: Graphing a Rational Function
Step 6 Sketch the graph. Plot some points and graph the asymptotes found in Steps 1-5; use symmetry to sketch the graph of f.
6.
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Sketch the graph of
Step 1 Since x2 + 2 > 0, no x-intercepts.
; y-intercept is –1.
Step 2 Solve (x + 2)(x – 1) = 0; x = –2, x = 1Vertical asymptotes are x = –2 and x = 1.
f x x2 2
x 2 x 1 .
20
00
21
2 10f
Example: Graphing a Rational Function
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Step 3 degree of den = degree of numy = 1 is the horizontal asymptote
Step 4 Symmetry. None
Step 5 The zeros of the denominator –2 and 1 yield the following figure:
Example: Graphing a Rational Function
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Example: Graphing a Rational Function
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Step 6 Sketch the graph.
Example: Graphing a Rational Function
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Sketch a graph of f x x2
x2 1.
Step 1 Since f (0) = 0 and setting f (x) = 0 yields 0, x-intercept and y-intercept are 0.
Step 2 Because x2 +1 > 0 for all x, the domain is the set of all real numbers. Since there are no zeros for the denominator, there are no vertical asymptotes.
Example: Graphing a Rational Function
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Step 3 degree of den = degree of numy = 1 is the horizontal asymptote.
Step 4 Symmetry.
Symmetric with respect to the y-axis.
Step 5 The graph is always above the x-axis, except at x = 0.
2 2
2 2 11
xx
xfx
xf
x
Example: Graphing a Rational Function
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Step 6 Sketch the graph.
f x x2
x2 1
Example: Graphing a Rational Function
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OBLIQUE ASUMPTOTES
Suppose f x N x D x ,
is greater than the degree of D(x). Then
and the degree of N(x)
f x N x D x Q x R x
D x .
Thus, as x , f x Q x 0 Q x .
That is, the graph of f approaches the graph of the oblique asymptote defined by Q(x).
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Sketch the graph of f x x2 4
x 1.
Step 2 Solve x + 1 = 0; x = –1; domain is set of all real numbers except –1.
Vertical asymptote is x = –1.
f 0 0 4
0 1 4 ; y-intercept: –4.
Step 1 Solve x2 – 4 = 0; x-intercepts: –2, 2
Example: Graphing a Rational Function with an Oblique Asymptote
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y = x – 1 is an oblique asymptote.
Step 3 degree of numerator > degree of denominator
Step 4 Symmetry. None
Step 5 Use the intervals determined by the zeros of the denominator.
f x x2 4
x 1x 1
3
x 1
Example: Graphing a Rational Function with an Oblique Asymptote
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Example: Graphing a Rational Function with an Oblique Asymptote
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Step 6 Sketch the graph.
2 4
1
xf x
x
Example: Graphing a Rational Function with an Oblique Asymptote
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