1. Complex Numbers

2. The imaginary unit i is defined as
Complex Numbers
The imaginary unit i is defined as

3. Example

4. Complex Numbers
The set of all numbers in the form a+bi with real numbers a and b, and i, the imaginary unit, is called the set of complex numbers. The real number a is called the real part, and the real number b is called the imaginary part, of the complex number a+bi.

5. Equality of Complex Numbers
a+bi = c+di if and only if a = c and b = d

6. Adding and Subtracting Complex Numbers
(a+bi) + (c+di) = (a+b) + (c+d)i
(a+bi) - (c+di) = (a-c) + (b-d)i

7. Text Example
Perform the indicated operation, writing the result in standard form.
(-5 + 7i) - ( i)
Combine the real and imaginary parts: (-5-(-11)) + (7-(-6))i = (-5+11) + (7+6)i = 6+13i

8. Example

9. Example

10. Conjugate of a Complex Number
The complex conjugate of the number a+bi is a-bi, and the conjugate of a-bi is a+bi.

11. Example

12. Principal Square Root of a Negative Number
For any positive real number b, the principal square root of the negative number -b is defined by
(-b) = ib

13. Example

14. Quadratic Functions

15. Graphs of Quadratic Functions
The graph of any quadratic function is called a parabola. Parabolas are shaped like cups, as shown in the graph below. If the coefficient of x2 is positive, the parabola opens upward; otherwise, the parabola opens downward. The vertex (or turning point) is the minimum or maximum point.

16. The Standard Form of a Quadratic Function
The quadratic function
f (x) = a(x - h)2 + k, a  0
is in standard form. The graph of f is a parabola whose vertex is the point (h, k). The parabola is symmetric to the line x = h. If a > 0, the parabola opens upward; if a < 0, the parabola opens downward.

17. Graphing Parabolas With Equations in Standard Form
To graph f (x) = a(x - h)2 + k:
Determine whether the parabola opens upward or downward. If a > 0, it opens upward. If a < 0, it opens downward.
Determine the vertex of the parabola. The vertex is (h, k).
Find any x-intercepts by replacing f (x) with 0. Solve the resulting quadratic equation for x.
Find the y-intercept by replacing x with zero.
Plot the intercepts and vertex. Connect these points with a smooth curve that is shaped like a cup.

18. Text Example Graph the quadratic function f (x) = -2(x - 3)2 + 8.
Standard form f (x) = a(x - h)2 + k  
a = h = 3 k = 8
Given equation f (x) = -2(x - 3)2 + 8
Solution We can graph this function by following the steps in the preceding box. We begin by identifying values for a, h, and k.
Step Determine how the parabola opens. Note that a, the coefficient of x 2, is -2. Thus, a < 0; this negative value tells us that the parabola opens downward.

19. Text Example cont.
Step Find the vertex. The vertex of the parabola is at (h, k). Because h = 3 and k = 8, the parabola has its vertex at (3, 8).
Step Find the x-intercepts. Replace f (x) with 0 in f (x) = -2(x - 3)2 + 8.
0 = -2(x - 3) Find x-intercepts, setting f (x) equal to zero.
2(x - 3)2 = 8
Solve for x. Add 2(x - 3)2 to both sides of the equation.
(x - 3)2 = Divide both sides by 2.
(x - 3) = ±2 Apply the square root method.
x - 3 = or x - 3 = Express as two separate equations.
x = or x = Add 3 to both sides in each equation.
The x-intercepts are 1 and 5. The parabola passes through (1, 0) and (5, 0).

20. Text Example cont.
f (0) = -2(0 - 3) = -2(-3) = -2(9) + 8 = -10
Step Find the y-intercept. Replace x with 0 in f (x) = -2(x - 3)2 + 8.
The y-intercept is –10. The parabola passes through (0, -10).
Step Graph the parabola. With a vertex at (3, 8), x-intercepts at 1 and 5, and a y-intercept at –10, the axis of symmetry is the vertical line whose equation is x = 3.

21. The Vertex of a Parabola Whose Equation Is f (x) = ax 2 + bx + c
Consider the parabola defined by the quadratic function
f (x) = ax 2 + bx + c. The parabola's vertex is at

22. Example Graph the quadratic function f (x) = -x2 + 6x -. Solution:
Step Determine how the parabola opens. Note that a, the coefficient of x 2, is -1. Thus, a < 0; this negative value tells us that the parabola opens downward.
Step Find the vertex. We know the x-coordinate of the vertex is x = -b/(2a). We identify a, b, and c to substitute the values into the equation for the x-coordinate:
x = -b/(2a) = -6/2(-1)=3.
The x-coordinate of the vertex is 3. We substitute 3 for x in the equation of the function to find the y-coordinate:
y=f(3) = -(3)^2+6(3)-2= =7, the parabola has its vertex at (3,7).

23. Example Graph the quadratic function f (x) = -x2 + 6x -.
Step Find the x-intercepts. Replace f (x) with 0 in f (x) = -x2 + 6x - 2.
0 = -x2 + 6x - 2

24. Example Graph the quadratic function f (x) = -x2 + 6x -.
Step Find the y-intercept. Replace x with 0 in f (x) = -x2 + 6x - 2.
f (0) = • = -
The y-intercept is –2. The parabola passes through (0, -2).
Step Graph the parabola.

25. Minimum and Maximum: Quadratic Functions
Consider f(x) = ax2 + bx +c.
If a > 0, then f has a minimum that occurs at x = -b/(2a). This minimum value is f(-b/(2a)).
If a < 0, the f has a maximum that occurs at x = -b/(2a). This maximum value is f(-b/(2a)).

26. Strategy for Solving Problems Involving Maximizing or Minimizing Quadratic Functions
Read the problem carefully and decide which quantity is to be maximized or minimized.
Use the conditions of the problem to express the quantity as a function in one variable.
Rewrite the function in the form f(x) = ax2 + bx +c.
Calculate -b/(2a). If a > 0, then f has a minimum that occurs at x = -b/(2a). This minimum value is f(-b/(2a)). If a < 0, the f has a maximum that occurs at x = -b/(2a). This maximum value is f(-b/(2a)).
Answer the question posed in the problem.

27. Polynomial Functions and Their Graphs

28. Definition of a Polynomial Function
Let n be a nonnegative integer and let an, an-1,…, a2, a1, a0, be real numbers with an ¹ 0. The function defined by
f (x) = anxn + an-1xn-1 +…+ a2x2 + a1x + a0
is called a polynomial function of x of degree n. The number an, the coefficient of the variable to the highest power, is called the leading coefficient.

29. Smooth, Continuous Graphs
Two important features of the graphs of polynomial functions are that they are smooth and continuous. By smooth, we mean that the graph contains only rounded curves with no
sharp corners. By continuous, we mean that the graph has no breaks and can be drawn without lifting your pencil from the rectangular coordinate system. These ideas are illustrated in the figure.
Smooth rounded corner x y

30. The Leading Coefficient Test
As x increases or decreases without bound, the graph of the polynomial function
f (x) = anxn + an-1xn-1 + an-2xn-2 +…+ a1x + a0 (an ¹ 0)
eventually rises or falls. In particular,
1. For n odd: an > an < 0
If the leading coefficient is positive, the graph falls to the left and rises to the right.
If the leading coefficient is negative, the graph rises to the left and falls to the right.
Rises right
Falls left
Falls right
Rises left

31. The Leading Coefficient Test
As x increases or decreases without bound, the graph of the polynomial function
f (x) = anxn + an-1xn-1 + an-2xn-2 +…+ a1x + a0 (an ¹ 0)
eventually rises or falls. In particular,
1. For n even: an > an < 0
If the leading coefficient is positive, the graph rises to the left and to the right.
If the leading coefficient is negative, the graph falls to the left and to the right.
Rises right
Rises left
Falls left
Falls right

32. Text Example
Use the Leading Coefficient Test to determine the end behavior of the graph of Graph the quadratic function f (x) = x3 + 3x2 - x - 3.
Falls left y
Rises right x
Solution Because the degree is odd (n = 3) and the leading coefficient, 1, is positive, the graph falls to the left and rises to the right, as shown in the figure.

33. Text Example Find all zeros of f (x) = -x4 + 4x3 - 4x2.
Solution We find the zeros of f by setting f (x) equal to 0.
-x4 + 4x3 - 4x2 = We now have a polynomial equation.
x4 - 4x3 + 4x2 = Multiply both sides by -1. (optional step)
x2(x2 - 4x + 4) = Factor out x2.
x2(x - 2)2 = Factor completely.
x2 = or (x - 2)2 = Set each factor equal to zero.
x = x = Solve for x.

34. Multiplicity and x-Intercepts
If r is a zero of even multiplicity, then the graph touches the x-axis and turns around at r. If r is a zero of odd multiplicity, then the graph crosses the x-axis at r. Regardless of whether a zero is even or odd, graphs tend to flatten out at zeros with multiplicity greater than one.

35. Example Find the x-intercepts and multiplicity of f(x) = 2(x+2)2(x-3)
Solution:
x=-2 is a zero of multiplicity 2 or even
x=3 is a zero of multiplicity 1 or odd

36. Graphing a Polynomial Function
f (x) = anxn + an-1xn-1 + an-2xn-2 + ¼ + a1x + a0 (an ¹ 0)
Use the Leading Coefficient Test to determine the graph's end behavior.
Find x-intercepts by setting f (x) = 0 and solving the resulting polynomial equation. If there is an x-intercept at r as a result of (x - r)k in the complete factorization of f (x), then:
a. If k is even, the graph touches the x-axis at r and turns around.
b. If k is odd, the graph crosses the x-axis at r.
c. If k > 1, the graph flattens out at (r, 0).
3. Find the y-intercept by setting x equal to 0 and computing f (0).

37. Graphing a Polynomial Function
f (x) = anxn + an-1xn-1 + an-2xn-2 + ¼ + a1x + a0 (an ¹ 0)
Use symmetry, if applicable, to help draw the graph:
a. y-axis symmetry: f (-x) = f (x)
b. Origin symmetry: f (-x) = - f (x).
5. Use the fact that the maximum number of turning points of the graph is n - 1 to check whether it is drawn correctly.

38. Text Example Graph: f (x) = x4 - 2x2 + 1. Solution
Step Determine end behavior. Because the degree is even (n = 4) and the leading coefficient, 1, is positive, the graph rises to the left and the right:
Solution y x
Rises left
Rises right

39. Text Example cont. Graph: f (x) = x4 - 2x2 + 1. Solution
Step Find the x-intercepts (zeros of the function) by setting f (x) = 0.
x4 - 2x2 + 1 = 0
(x2 - 1)(x2 - 1) = Factor.
(x + 1)(x - 1)(x + 1)(x - 1) = Factor completely.
(x + 1)2(x - 1)2 = Express the factoring in more compact notation.
(x + 1)2 = or (x - 1)2 = Set each factor equal to zero.
x = x = Solve for x.

40. Text Example cont. Graph: f (x) = x4 - 2x2 + 1. Solution
Step We see that -1 and 1 are both repeated zeros with multiplicity 2. Because of the even multiplicity, the graph touches the x-axis at -1 and 1 and turns around. Furthermore, the graph tends to flatten out at these zeros with multiplicity greater than one:
Rises left
Rises right x y
1 1

41. Text Example cont. Graph: f (x) = x4 - 2x2 + 1. Solution
Step Find the y-intercept. Replace x with 0 in f (x) = -x + 4x - 1.
f (0) = • = 1
There is a y-intercept at 1, so the graph passes through (0, 1). 1
Rises left
Rises right x y
1 1

42. f (-x) = (-x)4 - 2(-x)2 + 1 = x4 - 2x2 + 1
Text Example cont.
Graph: f (x) = x4 - 2x2 + 1.
Solution
Step Use possible symmetry to help draw the graph. Our partial graph suggests y-axis symmetry. Let's verify this by finding f (-x).
f (x) = x x
Replace x with -x.
f (-x) = (-x)4 - 2(-x)2 + 1 = x4 - 2x2 + 1
Because f (-x) = f (x), the graph of f is symmetric with respect to the y-axis. The following figure shows the graph.

43. Text Example cont. Graph: f (x) = x4 - 2x2 + 1. Solution
Step Use the fact that the maximum number of turning points of the graph is n - 1 to check whether it is drawn correctly. Because n = 4, the maximum number of turning points is 4 - 1, or 3. Because our graph has three turning points, we have not violated the maximum number possible. y x

44. Dividing Polynomials

45. Long Division of Polynomials
Arrange the terms of both the dividend and the divisor in descending powers of any variable.
Divide the first term in the dividend by the first term in the divisor. The result is the first term of the quotient.
Multiply every term in the divisor by the first term in the quotient. Write the resulting product beneath the dividend with like terms lined up.
Subtract the product from the dividend.
Bring down the next term in the original dividend and write it next to the remainder to form a new dividend.
Use this new expression as the dividend and repeat this process until the remainder can no longer be divided. This will occur when the degree of the remainder (the highest exponent on a variable in the remainder) is less than the degree of the divisor.

46. Text Example Divide 4 – 5x – x2 + 6x3 by 3x – 2.
Solution We begin by writing the divisor and dividend in descending powers of x. Next, we consider how many times 3x divides into 6x3.
3x – 2
6x3 – x2 – 5x + 4
6x3 – 4x2
Multiply.
Multiply: 2x2(3x – 2) = 6x3 – 4x2.
2x2
Divide: 6x3/3x = 2x2.
– 5x
Subtract 6x3 – 4x2 from 6x3 – x2 and bring down –5x.
3x2
Now we divide 3x2 by 3x to obtain x, multiply x and the divisor, and subtract.
3x – 2
6x3 – x2 – 5x + 4
6x3 – 4x2
2x x
3x2 – 5x
Multiply.
Multiply: x(3x – 2) = 3x2 – 2x.
3x2 – 2x
Divide: 3x2/3x = x.
Subtract 3x2 – 2x from 3x2 – 5x and bring down 4.
+ 4
-3x

47. Text Example cont. Divide 4 – 5x – x2 + 6x3 by 3x – 2.
Solution Now we divide –3x by 3x to obtain –1, multiply –1 and the divisor, and subtract.
3x – 2
6x3 – x2 – 5x + 4
6x3 – 4x2
2x x – 1
3x2 – 5x
+ 4
3x2 – 2x
-3x
Multiply.
Multiply: -1(3x – 2) = -3x + 2.
-3x + 2
Divide: -3x/3x = -1.
Subtract -3x + 2 from -3x + 4, leaving a remainder of 2. 2

48. The Division Algorithm
If f (x) and d(x) are polynomials, with d(x) = 0, and the degree of d(x) is less than or equal to the degree of f (x), then there exist unique polynomials q(x) and r(x) such that
f (x) = d(x) • q(x) + r(x).
The remainder, r(x), equals 0 or its is of degree less than the degree of d(x). If r(x) = 0, we say that d(x) divides evenly in to f (x) and that d(x) and q(x) are factors of f (x).
Dividend
Divisor
Quotient
Remainder

49. Example
Divide:

50. Example cont.

51. Example cont. 3x

52. Example cont. 3x
3x3 + 9x2 + 9x

53. Example cont. 3x
3x3 + 9x2 + 9x
-11x2 - 5x - 3

54. Example cont.
3x -11
-(3x3 + 9x2 + 9x)
-11x2 - 5x - 3

55. Example cont.
3x -11
3x3 + 9x2 + 9x
-11x2 - 5x - 3
-11x2 - 33x - 33

56. Example cont.
3x -11
3x3 + 9x2 + 9x
-11x2 - 5x - 3
-11x2 - 33x - 33
28x+30

57. Example cont.

58. Synthetic Division To divide a polynomial by x – c Example
1. Arrange polynomials in descending powers,
with a 0 coefficient for any missing terms. x – 3 x3 + 4x2 – 5x + 5
2. Write c for the divisor, x – c. To the right,
write the coefficients of the dividend.
3. Write the leading coefficient of the dividend
on the bottom row. Bring down 1. 1
4. Multiply c (in this case, 3) times the value
just written on the bottom row. Write the 3
product in the next column in the 2nd row
Multiply by 3.

59. Synthetic Division 5. Add the values in this new column, writing
the sum in the bottom row.
6. Repeat this series of multiplications and
additions until all columns are filled in.
7. Use the numbers in the last row to write the
quotient and remainder in fractional form.
The degree of the first term of the quotient is
one less than the degree of the first term of
the dividend. The final value in the row is the
remainder. 3
1 7
Add.
3 21
1 7 16
Add.
Multiply by 3.
Add.
Multiply by 3.
1x2 + 7x 53
x – 3

60. Use the coefficients of the dividend in descending powers of x.
Text Example
Use synthetic division to divide 5x3 + 6x + 8 by x + 2.
Solution The divisor must be in the form x – c. Thus, we write x + 2 as x – (-2). This means that c = -2. Writing a 0 coefficient for the missing x2-term in the dividend, we can express the division as follows:
x – (-2)
5x3 + 0x x
Now we are ready to set up the problem so that we can use synthetic division.
Use the coefficients of the dividend in descending powers of x.
This is c in x-(-2).

61. Text Example cont.
Solution We begin the synthetic division process by bringing down 5. This is following by a series of multiplications and additions. 5
1. Bring down 5.
2. Multiply: -2(5) = -10.
-10 5
3. Add: 0 + (-10) = -10.
-10
5 -10
Add.
4. Multiply: -2(-10) = 20.
-10 20
5 -10
5. Add: = 26.
-10 20
Add.
6. Multiply: -2(26) = -52.
7. Add: 8 + (-52) = -44.
Add.

62. Text Example cont. Solution
The numbers in the last row represent the coefficients of the quotient and the remainder. The degree of the first term of the quotient is one less than that of the dividend. Because the degree of the dividend is 3, the degree of the quotient is 2. This means that the 5 in the last row represents 5x2.
Thus,
x + 2
5x3 + 0x x + 8
5x2 – 10x + 26 – 44

63. The Remainder Theorem
If the polynomial f (x) is divided by x – c, then the remainder is f (c).

64. The Factor Theorem Let f (x) be a polynomial.
If f (c ) = 0, then x – c is a factor of f (x).
If x – c is a factor of f (x), then f ( c) = 0.

65. Text Example
Solve the equation 2x3 – 3x2 – 11x + 6 = 0 given that 3 is a zero of f (x) = 2x3 – 3x2 – 11x + 6.
Solution We are given that f (3) = 0. The Factor Theorem tells us that x – 3 is a factor of f (x). We’ll use synthetic division to divide f (x) by x – 3.
Equivalently,
2x3 – 3x2 – 11x + 6 = (x – 3)(2x2 + 3x – 2)
6 9 -6
2x3 – 3x2 – 11x + 6
x – 3
2x2 + 3x – 2

66. Text Example cont. Solution Now we can solve the polynomial equation.
2x3 – 3x2 – 11x + 6 = 0
This is the given equation.
(x – 3)(2x2 + 3x – 2) = 0
Factor using the result from the synthetic division.
(x – 3)(2x – 1)(x + 2) = 0
Factor the trinomial.
x – 3 = 0 or 2x – 1 = 0 or x + 2 = 0
Set each factor equal to 0.
x = x = 1/ x = -2
Solve for x.
The solution set is {-2, 1/2 , 3}.

67. Zeros of Polynomial Functions

68. The Rational Zero Theorem
If f (x) = anxn + an-1xn-1 +…+ a1x + a0 has integer coefficients and p/q (where p/q is reduced) is a rational zero, then p is a factor of the constant term a0 and q is a factor of the leading coefficient an.

69. Example
Find all of the possible real, rational roots of f(x) = 2x3-3x2+5
Solution:
p is a factor of 5 = 1, 5
q is a factor of 2 = 1, 2
p/q = 1, 1/2, 5, 5/2

70. Properties of Polynomial Equations
If a polynomial equation is of degree n, then counting multiple roots separately, the equation has n roots
If a+bi is a root of the equation, then a-bi is also a root.

71. Example Find all zeros of f(x) = x3+12x2+21x+10 p/q = 1, 2, 5, 10
Divide out -1 to get x2+11x-10
Use the quadratic formula to find the last 2 zeros. x= and .844
The solutions are -1, , and .844

72. Text Example Solve: x4 - 6x2 - 8x + 24 = 0.
Solution The graph of f (x) = x4 - 6x2 - 8x + 24 is shown the figure below. Because the x-intercept is 2, we will test 2 by synthetic division and show that it is a root of the given equation.
x-intercept: 2 2
The zero remainder indicates that 2 is a root of x4 - 6x2 - 8x + 24 = 0.

73. Text Example cont. Solve: x4 - 6x2 - 8x + 24 = 0.
Solution Now we can rewrite the given equation in factored form.
x4 - 6x2 + 8x + 24 = This is the given equation.
(x – 2)(x3 + 2x2 - 2x - 12) = 0
This is the result obtained from the
synthetic division.
x – 2 = or x3 + 2x2 - 2x - 12 = Set each factor equal to zero.

74. Text Example cont. Solve: x4 - 6x2 - 8x + 24 = 0.
Solution We can use the same approach to look for rational roots of the polynomial equation x3 + 2x2 - 2x - 12 = 0, listing all possible rational roots. However, take a second look at the figure of the graph of x4 - 6x2 - 8x + 24 = 0. Because the graph turns around at 2, this means that 2 is a root of even multiplicity. Thus, 2 must also be a root of x3 + 2x2 - 2x - 12 = 0, confirmed by the following synthetic division.
x-intercept: 2
These are the coefficients of x3 + 2x2 - 2x - 12 = 0.
The zero remainder indicates that 2 is a root of x3 + 2x2 - 2x - 12 = 0.

75. Text Example cont. Solve: x4 - 6x2 - 8x + 24 = 0.
Solution Now we can solve the original equation as follows.
x4 - 6x2 + 8x + 24 = This is the given equation.
(x – 2)(x3 + 2x2 - 2x - 12) = 0
This was obtained from the first
synthetic division.
(x – 2)(x – 2)(x2 + 4x + 6) = 0
This was obtained from the second
synthetic division.
x – 2 = 0 or x – 2 = 0 or x2 + 4x + 6 = Set each factor equal to zero.
x = x = x2 + 4x + 6 = Solve.

76. Text Example cont. Solve: x4 - 6x2 - 8x + 24 = 0.
Solution We can use the quadratic formula to solve x2 + 4x + 6 = 0.
The solution set of the original equation is: {2, -2 - i2, -2+i2}

77. Descartes’s Rule of Signs
If f (x) = anxn + an-1xn-1 + … + a2x2 + a1x + a0 be a polynomial with real coefficients.
1. The number of positive real zeros of f is either equal to the number of sign changes of f (x) or is less than that number by an even integer. If there is only one variation in sign, there is exactly one positive real zero.
2. The number of negative real zeros of f is either equal to the number of sign changes of f (-x) or is less than that number by an even integer. If f (-x) has only one variation in sign, then f has exactly one negative real zero.

78. Text Example
Determine the possible number of positive and negative real zeros of
f (x) = x3 + 2x2 + 5x + 4.
Solution
1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f (x). Because all the terms are positive, there
are no variations in sign. Thus, there are no positive real zeros.
2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f (-x). We obtain this equation by replacing x with -x in the given function.
f (-x) = (-x)3 + 2(-x)2 + 5(-x) + 4
f (x) = x x x This is the given polynomial function.
Replace x with -x.
= -x3 + 2x2 - 5x + 4

79. Text Example cont.
Determine the possible number of positive and negative real zeros of
f (x) = x3 + 2x2 + 5x + 4.
Solution
Now count the sign changes.
f (-x) = -x3 + 2x2 - 5x + 4
There are three variations in sign. The number of negative real zeros of f is either equal to the number of sign changes, 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros or = 1 negative real zero.

80. Rational Functions and Their Graphs

81. Example Find the Domain of this Function. Solution:
The domain of this function is the set of all real numbers not equal to 3.

82. Arrow Notation Symbol Meaning x  a + x approaches a from the right.
x  a - x approaches a from the left.
x   x approaches infinity; that is, x increases without bound.
x  -  x approaches negative infinity; that is, x decreases without bound.

83. Definition of a Vertical Asymptote
The line x = a is a vertical asymptote of the graph of a function f if f (x) increases or decreases without bound as x approaches a.
f (x)   as x  a f (x)   as x  a - f a y x
x = a f a y x
x = a
Thus, f (x)   or f(x)  -  as x approaches a from either the left or the right.

84. Definition of a Vertical Asymptote
The line x = a is a vertical asymptote of the graph of a function f if f (x) increases or decreases without bound as x approaches a.
f (x)  as x  a + f (x)  -  as x  a - f a y x
x = a
x = a f a y x
Thus, f (x)   or f(x)  -  as x approaches a from either the left or the right.

85. Locating Vertical Asymptotes
If f(x) = p(x) / q(x) is a rational function in which p(x) and q(x) have no common factors and a is a zero of q(x), the denominator, then x = a is a vertical a vertical asymptote of the graph of f.

86. Definition of a Horizontal Asymptote
The line y = b is a horizontal asymptote of the graph of a function f if f (x) approaches b as x increases or decreases without bound. f y x
y = b x y f
y = b f y x
y = b
f (x)  b as x   f (x)  b as x   f (x)  b as x 

87. Locating Horizontal Asymptotes
Let f be the rational function given by
The degree of the numerator is n. The degree of the denominator is m.
If n<m, the x-axis, or y=0, is the horizontal asymptote of the graph of f.
If n=m, the line y = an/bm is the horizontal asymptote of the graph of f.
If n>m,t he graph of f has no horizontal asymptote.

88. Strategy for Graphing a Rational Function
Suppose that f(x) = p(x) / q(x),
where p(x) and q(x) are polynomial functions with no common factors.
1. Determine whether the graph of f has symmetry.
f (-x) = f (x): y-axis symmetry f (-x) = -f (x): origin symmetry
Find the y-intercept (if there is one) by evaluating f (0).
Find the x-intercepts (if there are any) by solving the equation p(x) = 0.
Find any vertical asymptote(s) by solving the equation q (x) = 0.
Find the horizontal asymptote (if there is one) using the rule for determining the horizontal asymptote of a rational function.
Plot at least one point between and beyond each x-intercept and vertical asymptote.
Use the information obtained previously to graph the function between and beyond the vertical asymptotes.

89. Sketch the graph of

90. the graph has no symmetry

91. the graph has no symmetry
The y-intercept is (0,-3/10)

92. the graph has no symmetry
The y-intercept is (0,-3/10)
The x-intercept is (3/2, 0)

93. the graph has no symmetry
The y-intercept is (0,-3/10)
The x-intercept is (3/2, 0)
The vertical asymptote is x = -2

94. the graph has no symmetry
The y-intercept is (0,-3/10)
The x-intercept is (3/2, 0)
The vertical asymptote is x = -2
The horizontal asymptote is y = 2/5

95. the graph has no symmetry
The y-intercept is (0,-3/10)
The x-intercept is (3/2, 0)
The vertical asymptote is x = -2
The horizontal asymptote is y = 2/5
Test points include (-3, 9/5), (0, -3/10), (2, 1/20)

97. Text Example Find the slant asymptote of
Solution Because the degree of the numerator, 2, is exactly one more than the degree of the denominator, 1, the graph of f has a slant asymptote. To find the equation of the slant asymptote, divide x - 3 into x2 - 4x - 5.
The equation of the slant asymptote is y = x - 1. Using our strategy for graphing rational functions, the graph is shown. -2 -1 4 5 6 7 8 3 2 1 -3
Vertical asymptote:
x = 3
Slant asymptote:
y = x - 1

98. Polynomial and Rational Inequalities

99. Definition of a Polynomial Inequality
A polynomial inequality is any inequality that can be put in one of the forms
f(x) < 0, f(x) > 0, f(x) < 0, or f(x)> 0
where f is a polynomial function.

100. Procedure for Solving Polynomial Inequalities
Express the inequality in the standard form
f(x) < 0 or f(x) > 0.
Find the zeros of f. The real zeros are the boundary points.
Locate these boundary points on a number line, thereby dividing the number line into intervals.
Choose one representative number within each interval and evaluate f at that number. If the value of f is positive, then f(x)>0 for all numbers, x, in the interval. If the value of f is negative, then f(x)<0 for all numbers, x, in the interval.
Write the solution set; selecting the interval(s) that satisfy the given inequality.

101. Example
Solve and graph the solution set on a real number line: 2x2 – 3x > 2.
Solution
Step 1 Write the inequality in standard form. We can write by subtracting 2 from both sides to get zero on the right.
2x2 – 3x – 2 > 2 – 2
2x2 – 3x – 2 > 0
Step 2 Solve the related quadratic equation. Replace the inequality sign with an equal sign. Thus, we will solve.
2x2 – 3x – 2 = 0 This is the related quadratic equation.
(2x + 1)(x – 2) = 0 Factor.
2x + 1 = 0 or x – 2 = 0 Set each factor equal to 0.
x = -1/2 or x = 2 Solve for x.
The boundary points are –1/2 and 2.

102. Example cont.
Solve and graph the solution set on a real number line: 2x2 – 3x > 2.
Solution
Step 3 Locate the boundary points on a number line. The number line with the boundary points is shown as follows:
-1/2 2 x
The boundary points divide the number line into three test intervals. Including the boundary points (because of the given greater than or equal to sign), the intervals are (-ºº, -1/2], [-1/2, 2], [2, -ºº).

103. Example cont.
Solve and graph the solution set on a real number line: 2x2 – 3x > 2.
Solution
Step 4 Take one representative number within each test interval and substitute that number into the original inequality.
[2, ºº) belongs to the solution set.
2(3)2 – 3(3) > 2
9 > 2, True 3
[2, ºº)
[-1/2, 2] does not belong to the solution set.
2(0) 2 – 3(0) > 2
0 > 2, False
[-1/2, 2]
(-ºº, -1/2] belongs to the solution set.
2(-1) 2 – 3(-1) > 2
5 > 2, True -1
(-ºº, -1/2]
Conclusion
Substitute into
2x∆ – 3x > 2
Representative Number
Test Interval

104. Example cont.
Solve and graph the solution set on a real number line: 2x2 – 3x > 2.
Solution
Step 5 The solution set are the intervals that produced a true statement. Our analysis shows that the solution set is
(-ºº, -1/2] or [2, ºº).
The graph of the solution set on a number line is shown as follows:
-1/2 2 ) ( x

105. Text Example Solve and graph the solution set: Solution
Step 1 Express the inequality so that one side is zero and the other side is a single quotient. We subtract 2 from both sides to obtain zero on the right.

106. Text Example cont. Solve and graph the solution set: Solution
Step 2 Find boundary points by setting the numerator and the denominator equal to zero.
Set the numerator and denominator equal to 0. These are the values that make the previous quotient zero or undefined. =
x + 3
-x - 5
Solve for x. -3 = x
x = -5
The boundary points are -5 and -3. Because equality is included in the given less-than-or-equal-to symbol, we include the value of x that causes the quotient to be zero. Thus, -5 is included in the solution set. By contrast, we do not include 1 in the solution set because -3 makes the denominator zero.

107. Text Example cont. Solve and graph the solution set:
Step 3 Locate boundary points on a number line.
The boundary points divide the number line into three test intervals, namely
(-ºº, -5], [-5, -3), (-3, ºº).

108. Text Example cont. Solve and graph the solution set:
Step 4 Take one representative number within each test interval and substitute that number into the original equality.
f(x) < 0 for all x in (-3, ºº)
(-3, ºº)
f(x) > 0 for all x in (-5,-3) -4
(-5,-3)
f(x )< 0 for all x in (-ºº, -5) -6
(-ºº, -5)
Conclusion
Substitute into
Representative Number
Test Interval

109. Text Example cont. Solve and graph the solution set:
Step 5 The solution set are the intervals that produced a true statement. Our analysis shows that the solution set is [-5, -3).

110. The Position Formula for a Free-Falling Object Near Earth’s Surface
An object that is falling or vertically projected into the air has its height in feet above the ground given by
s = -16 t 2 + v0 t + s0
where s is the height in feet, v0 is the original velocity (initial velocity) of the object in feet per second, t is the time that the object is in motion in seconds, and s0 is the original height (initial height) of the object in feet.

111. Example
An object is propelled straight up from ground level with an initial velocity of 80 fps. Its height at time t is described by s = -16 t t where the height, s, is measured in feet and the time, t, is measured in seconds. In which time interval will the object be more than 64 feet above the ground?
-16 t t > 64
This is the inequality implied by the problem’s question. We must find t.
Solution
-16 t t – 64 > 0
Subtract 64 from both sides.
-16 t t – 64 = 0
Solve the related quadratic equation.
-16 (t – 1)(t – 4) = 0
Factor.
(t – 1)(t – 4) = 0
Divide each side by -16.
t – 1 = 0 or t – 4 = 0
Set each factor equal to 0.
t = 4
Solve for t. The boundary points are 1 and 4.
t = 1

112. Example cont.
An object is propelled straight up from ground level with an initial velocity of 80 fps. Its height at time t is described by s = -16 t ∆ + 80 t where the height, s, is measured in feet and the time, t, is measured in seconds. In which time interval will the object be more than 64 feet above the ground?
Solution
-16 t t > 64
This is the inequality implied by the problem’s question. We must find t.
t = 4
The boundary points are 1 and 4.
t = 1
Since neither boundary point satisfy the inequality, 1 and 4 are not part of the solution. 1 4 x
With test intervals (-ºº, 1), (1, 4), and (4, ºº), we could use 0, 2, and 5 as test points for our analysis.

113. Example cont.
Solution
An object is propelled straight up from ground level with an initial velocity of 80 fps. Its height at time t is described by s = -16 t t where the height, s, is measured in feet and the time, t, is measured in seconds. In which time interval will the object be more than 64 feet above the ground?
(4, ºº) does not belong to the solution set.
(5 – 1)(5 – 4) < 0
4 < 0, False 5
(4, ºº)
(1, 4) belongs to the solution set.
(2 – 1)(2 – 4) < 0
-2 < 0, True 2
(1, 4)
(-ºº, 1) does not belong to the solution set.
(0 – 1)(0 – 4) < 0
(-ºº, 1)
Conclusion
Substitute into
(x – 1)(x – 4) < 0
Representative Number
Test Interval
The object will be above 64 feet between 1 and 4 seconds.

114. Modeling Using Variation

115. Direct Variation
If a situation is described by an equation in the form
y = kx
where k is a constant, we say that y varies directly as x. The number k is called the constant of variation.

116. Solving Variation Problems
Write an equation that describes the given English statement.
Substitute the given pair of values into the equation in step 1 and find the value of k.
Substitute the value of k into the equation in step 1.
Use the equation from step 3 to answer the problem's question.

117. Text Example
The amount of garbage, G, varies directly as the population, P. Allegheny County, Pennsylvania, has a population of 1.3 million and creates 26 million pounds of garbage each week. Find the weekly garbage produced by New York City with a population of 7.3 million.
Solution
Step Write an equation We know that y varies directly as x is expressed as
y = kx.
By changing letters, we can write an equation that describes the following English statement: Garbage production, G, varies directly as the population, P.
G = kP

118. Text Example cont.
The amount of garbage, G, varies directly as the population, P. Allegheny County, Pennsylvania, has a population of 1.3 million and creates 26 million pounds of garbage each week. Find the weekly garbage produced by New York City with a population of 7.3 million.
Solution
Step 2 Use the given values to find k. Allegheny County has a population of 1.3 million and creates 26 million pounds of garbage weekly. Substitute 26 for G and 1.3 for P in the direct variation equation. Then solve for k.
26 = 1.3k
26/1.3 = k
20 = k

119. Text Example cont.
The amount of garbage, G, varies directly as the population, P. Allegheny County, Pennsylvania, has a population of 1.3 million and creates 26 million pounds of garbage each week. Find the weekly garbage produced by New York City with a population of 7.3 million.
Solution
Step 3 Substitute the value of k into the equation.
G = kP Use the equation from step 1.
G = 20P Replace k, the constant of variation, with 20.

120. Text Example cont.
The amount of garbage, G, varies directly as the population, P. Allegheny County, Pennsylvania, has a population of 1.3 million and creates 26 million pounds of garbage each week. Find the weekly garbage produced by New York City with a population of 7.3 million.
Solution
Step 4 Answer the problem's question New York City has a population of 7.3 million. To find its weekly garbage production, substitute 7.3 for P in
G = 20P and solve for G.
G = 20P Use the equation from step 3.
G = 20(7.3) Substitute 7.3 for P.
G = 146
The weekly garbage produced by New York City weighs approximately 146 million pounds.

121. Direct Variation with Powers
y varies directly as the nth power of x if there exists some nonzero constant k such that
y = kx n.
We also say that y is directly proportional to the nth power of x.

122. Inverse Variation
If a situation is described by an equation in the form
y = k/x
where k is a constant, we say that y varies inversely as x. The number k is called the constant of variation.

123. Example Describe in words the variation shown by the given equation:
Solution: H varies directly as T and inversely as Q

124. Joint Variation
Joint variation is a variation in which a variable varies directly as the product of two or more other variables. Thus, the equation y = kxz is read "y varies jointly as x and z."

125. Example
An object’s weight on the moon, M, varies directly as its weight on Earth, E. A person who weighs 75 kilograms on Earth weighs 12 kilograms on the moon. What is the moon weight of a person who weighs 80 kilograms on Earth?
Solution
M = kE
12 = k*75
k = .16
M = .16E
M = .16 * 80
M = 12.8 kilograms