1. 1 Preliminaries Precalculus Review I Precalculus Review II
The Cartesian Coordinate System
Straight Lines

2. 1.1
Precalculus Review I

3. The Real Number Line
We can represent real numbers geometrically by points on a real number, or coordinate, line:
This line includes all real numbers.
Exactly one point on the line is associated with each real number, and vice-versa.
Origin
Negative Direction
Positive Direction
– 4 – 3 – 2 – p

4. Finite Intervals Open Intervals
The set of real numbers that lie strictly between two fixed numbers a and b is called an open interval (a, b).
It consists of all the real numbers that satisfy the inequalities a < x < b.
It is called “open” because neither of its endpoints is included in the interval.

5. Finite Intervals Closed Intervals
The set of real numbers that lie between two fixed numbers a and b, that includes a and b, is called a closed interval [a, b].
It consists of all the real numbers that satisfy the inequalities a  x  b.
It is called “closed” because both of its endpoints are included in the interval.

6. Finite Intervals Half-Open Intervals
The set of real numbers that between two fixed numbers a and b, that contains only one of its endpoints a or b, is called a half-open interval (a, b] or [a, b).
It consists of all the real numbers that satisfy the inequalities a < x  b or a  x < b.

7. Infinite Intervals Examples of infinite intervals include:
(a, ), [a, ), (–, a), and (–, a].
The above are defined, respectively, by the set of real numbers that satisfy x > a, x  a, x < a, x  a.

8. Exponents and Radicals
If b is any real number and n is a positive integer, then the expression bn is defined as the number
bn = b · b b · · … · b
The number b is called the base, and the superscript n is called the power of the exponential expression bn.
For example:
If b ≠ 0, we define b0 = 1.
20 = 1 and (–p)0 = 1, but 00 is undefined.
n factors

9. Exponents and Radicals
If n is a positive integer, then the expression b1/n is defined to be the number that, when raised to the nth power, is equal to b, thus
Such a number is called the nth root of b, also written as
Similarly, the expression bp/q is defined as the number
(b1/q)p or
Examples:
(b1/n)n = b

10. Laws of Exponents Law Example
1. am · an = am + n x2 · x3 = x2 + 3 = x5 2.
3. (am)n = am · n (x4)3 = x4 · 3 = x12
4. (ab)n = an · bn (2x)4 = 24 · x 4 = 16x4 5.

11. Examples
Simplify the expressions
Example 1, page 6

12. Examples Simplify the expressions (assume x, y, m, and n are positive)
Example 2, page 7

13. Examples Rationalize the denominator of the expression
Example 3, page 7

14. Examples
Rationalize the numerator of the expression
Example 4, page 7

15. Operations With Algebraic Expressions
An algebraic expression of the form axn, where the coefficient a is a real number and n is a nonnegative integer, is called a monomial, meaning it consists of one term.
Examples:
7x2 2xy 12x3y4
A polynomial is a monomial or the sum of two or more monomials.
x2 + 4x + 4 x4 + 3x2 – 3 x2y – xy + y

16. Operations With Algebraic Expressions
Constant terms, or terms containing the same variable factors are called like, or similar, terms.
Like terms may be combined by adding or subtracting their numerical coefficients.
Examples:
3x + 7x = 10x 12xy – 7xy = 5xy

17. Examples
Simplify the expression
Example 5, page 8

18. Examples
Simplify the expression
Example 5, page 8

19. Examples Perform the operation and simplify the expression
Example 6, page 8

20. Examples Perform the operation and simplify the expression
Example 6, page 9

21. Factoring
Factoring is the process of expressing an algebraic expression as a product of other algebraic expressions.
Example:

22. Factoring
To factor an algebraic expression, first check to see if it contains any common terms.
If so, factor out the greatest common term.
For example, the greatest common factor for the expression
is 2a, because

23. Examples Factor out the greatest common factor in each expression
Example 7, page 9

24. Examples Factor out the greatest common factor in each expression
Example 8, page 10

25. Factoring Second Degree Polynomials
The factors of the second-degree polynomial with integral coefficients
px2 + qx + r
are (ax + b)(cx + d), where ac = p, ad + bc = q, and bd = r.
Since only a limited number of choices are possible, we use a trial-and-error method to factor polynomials having this form.

26. Examples Find the correct factorization for x2 – 2x – 3 Solution
The x2 coefficient is 1, so the only possible first degree terms are
(x )(x )
The product of the constant term is –3, which gives us the following possible factors
(x – 1)(x + 3)
(x + 1)(x – 3)
We check to see which set of factors yields –2 for the x coefficient:
(–1)(1) + (1)(3) = or (1)(1) + (1)(–3) = –2
and conclude that the correct factorization is
x2 – 2x – 3 = (x + 1)(x –3)

27. Examples Find the correct factorization for the expressions
Example 9, page 11

28. Roots of Polynomial Expressions
A polynomial equation of degree n in the variable x is an equation of the form
where n is a nonnegative integer and a0, a1, … , an are real numbers with an ≠ 0.
For example, the equation
is a polynomial equation of degree 5.

29. Roots of Polynomial Expressions
The roots of a polynomial equation are the values of x that satisfy the equation.
One way to factor the roots of a polynomial equation is to factor the polynomial and then solve the equation.
For example, the polynomial equation
may be written in the form
For the product to be zero, at least one of the factors must be zero, therefore, we have
x = 0 x – 1 = 0 x – 2 = 0
So, the roots of the equation are x = 0, 1, and 2.

30. The Quadratic Formula The solutions of the equation
ax2 + bx + c = (a ≠ 0)
are given by

31. Examples Solve the equation using the quadratic formula: Solution
For this equation, a = 2, b = 5, and c = –12, so
Example 10, page 12

32. Examples Solve the equation using the quadratic formula: Solution
First, rewrite the equation in the standard form
For this equation, a = 1, b = 3, and c = – 8, so
Example 10, page 12

33. 1.2
Precalculus Review II

34. Rational Expressions
Quotients of polynomials are called rational expressions.
For example

35. Rational Expressions
The properties of real numbers apply to rational expressions.
Examples
Using the properties of number we may write
where a, b, and c are any real numbers and b and c are not zero.
Similarly, we may write

36. Examples
Simplify the expression
Example 1, page 16

37. Examples
Simplify the expression
Example 1, page 16

38. Rules of Multiplication and Division
If P, Q, R, and S are polynomials, then
Multiplication
Division

39. Example Perform the indicated operation and simplify
Example 2, page 16

40. Rules of Addition and Subtraction
If P, Q, R, and S are polynomials, then
Addition
Subtraction

41. Example Perform the indicated operation and simplify
Example 3b, page 17

42. Other Algebraic Fractions
The techniques used to simplify rational expressions may also be used to simplify algebraic fractions in which the numerator and denominator are not polynomials.

43. Examples
Simplify
Example 4a, page 18

44. Examples
Simplify
Example 4b, page 18

45. Rationalizing Algebraic Fractions
When the denominator of an algebraic fraction contains sums or differences involving radicals, we may rationalize the denominator.
To do so we make use of the fact that

46. Examples
Rationalize the denominator
Example 6, page 19

47. Examples
Rationalize the numerator
Example 7, page 19

48. Properties of Inequalities
If a, b, and c, are any real numbers, then
Property 1 If a < b and b < c, then a < c.
Property 2 If a < b, then a + c < b + c.
Property 3 If a < b and c > 0, then ac < bc.
Property 4 If a < b and c < 0, then ac > bc.

49. Examples Find the set of real numbers that satisfy –1  2x – 5 < 7
Solution
Add 5 to each member of the given double inequality
4  2x < 12
Multiply each member of the inequality by ½
2  x < 6
So, the solution is the set of all values of x lying in the interval [2, 6).
Example 8, page 20

50. Examples Solve the inequality x2 + 2x – 8 < 0. Solution
Factorizing we get (x + 4)(x – 2) < 0.
For the product to be negative, the factors must have opposite signs, so we have two possibilities to consider:
The inequality holds if (x + 4) < 0 and (x – 2) > 0, which means x < – 4, and x > 2, but this is impossible: x cannot meet these two conditions simultaneously.
The inequality also holds if (x + 4) > 0 and (x – 2) < 0, which means x > – 4, and x < 2, or – 4 < x < 2.
So, the solution is the set of all values of x lying in the interval (– 4, 2).
Example 9, page 20

51. Examples Solve the inequality Solution
For the quotient to be positive, the numerator and denominator must have the same sign, so we have two possibilities to consider:
The inequality holds if (x + 1)  0 and (x – 1) < 0, which means x  – 1, and x < 1, both of these conditions are met only when x  – 1.
The inequality also holds if (x + 1)  0 and (x – 1) > 0, which means x  – 1, and x > 1, both of these conditions are met only when x > 1.
So, the solution is the set of all values of x lying in the intervals (– , – 1] and (1, ).
Example 10, page 21

52. Absolute Value
The absolute value of a number a is denoted |a| and is defined by

53. Absolute Value Properties
If a, b, and c, are any real numbers, then
Property 5 |– a| = |a|
Property 6 |ab| = |a| |b|
Property 7 (b ≠ 0)
Property 8 |a + b| ≤ |a| + |b|

54. Examples Evaluate the expression |p – 5| + 3 Solution
Since p – 5 < 0, we see that
|p – 5| = –(p – 5).
Therefore
|p – 5| + 3 = – (p – 5) +3
= 8 – p ≈
Example 12a, page 22

55. Examples Evaluate the expression Solution Since , we see that
Similarly, , so
Therefore,
Example 12b, page 22

56. Examples Evaluate the inequality |x|  5. Solution
If x  0, then |x| = x, so |x|  5 implies x  5.
If x < 0, then |x| = – x , so |x|  5 implies – x  5 or x  – 5.
So, |x|  5 means – 5  x ≤ 5, and the solution is [– 5, 5].
Example 13, page 22

57. Examples Evaluate the inequality |2x – 3|  1. Solution
From our last example, we know that |2x – 3|  1 is equivalent to –1  2x – 3  1.
Adding 3 throughout we get 2  2x  4.
Dividing by 2 throughout we get 1  x  2, so the solution is [1, 2].
Example 14, page 22

58. The Cartesian Coordinate System
1.3
The Cartesian Coordinate System

59. The Cartesian Coordinate System
At the beginning of the chapter we saw a one-to-one correspondence between the set of real numbers and the points on a straight line (one dimensional space).

60. The Cartesian Coordinate System
The Cartesian coordinate system extends this concept to a plane (two dimensional space) by adding a vertical axis. 4 3 2 1
– 1
– 2
– 3
– 4

61. The Cartesian Coordinate System
The horizontal line is called the x-axis, and the vertical line is called the y-axis. y 4 3 2 1
– 1
– 2
– 3
– 4 x

62. The Cartesian Coordinate System
The point where these two lines intersect is called the origin. y 4 3 2 1
– 1
– 2
– 3
– 4
Origin x

63. The Cartesian Coordinate System
In the x-axis, positive numbers are to the right and negative numbers are to the left of the origin. y 4 3 2 1
– 1
– 2
– 3
– 4
Negative Direction
Positive Direction x

64. The Cartesian Coordinate System
In the y-axis, positive numbers are above and negative numbers are below the origin. y 4 3 2 1
– 1
– 2
– 3
– 4
Positive Direction x
Negative Direction

65. The Cartesian Coordinate System
A point in the plane can now be represented uniquely in this coordinate system by an ordered pair of numbers (x, y). y
(– 2, 4) 4 3 2 1
– 1
– 2
– 3
– 4
(4, 3) x
(3,–1)
(– 1, – 2)

66. The Cartesian Coordinate System
The axes divide the plane into four quadrants as shown below. y 4 3 2 1
– 1
– 2
– 3
– 4
Quadrant II
(–, +)
Quadrant I
(+, +) x
Quadrant III
(–, –)
Quadrant IV
(+, –)

67. The Distance Formula
The distance between any two points in the plane may be expressed in terms of their coordinates.
Distance formula
The distance d between two points P1(x1, y1) and P2(x2, y2) in the plane is given by

68. Examples Find the distance between the points (– 4, 3) and (2, 6).
Solution
Let P1(– 4, 3) and P2(2, 6) be points in the plane.
We have
x1 = – 4 y1 = 3 x2 = 2 y2 = 6
Using the distance formula, we have
Example 1, page 26

69. Examples
Let P(x, y) denote a point lying on the circle with radius r and center C(h, k). Find a relationship between x and y.
Solution
By definition in a circle, the distance between P(x, y) and C(h, k) is r.
With distance formula we get
Squaring both sides gives y
P(x, y)
C(h, k) r k x h
Example 2, page 27

70. Equation of a Circle
An equation of a circle with center C(h, k) and radius r is given by

71. Examples
Find an equation of the circle with radius 2 and center (–1, 3).
Solution
We use the circle formula with r = 2, h = –1, and k = 3: y
(–1, 3) 3 2 x –1
Example 3, page 27

72. Examples
Find an equation of the circle with radius 3 and center located at the origin.
Solution
We use the circle formula with r = 3, h = 0, and k = 0: y 3 x
Example 3, page 27

73. 1.4
Straight Lines

74. Slope of a Vertical Line
Let L denote the unique straight line that passes through the two distinct points (x1, y1) and (x2, y2).
If x1 = x2, then L is a vertical line, and the slope is undefined. y L
(x1, y1)
(x2, y2) x

75. Slope of a Nonvertical Line
If (x1, y1) and (x2, y2) are two distinct points on a nonvertical line L, then the slope m of L is given by y L
(x2, y2)
y2 – y1 = y
(x1, y1)
x2 – x1 = x x

76. Slope of a Nonvertical Line
If m > 0, the line slants upward from left to right. y L
m = 2
y = 2
x = 1 x

77. Slope of a Nonvertical Line
If m < 0, the line slants downward from left to right. y
m = –1
x = 1
y = –1 x L

78. Examples
Sketch the straight line that passes through the point (2, 5) and has slope – 4/3.
Solution
Plot the point (2, 5).
A slope of – 4/3 means that if x increases by 3, y decreases by 4.
Plot the point (5, 1).
Draw a line across the two points. y 6 5 4 3 2 1
x = 3
(2, 5)
y = – 4
(5, 1) x L
Example 1, page 34

79. Examples
Find the slope m of the line that goes through the points (–1, 1) and (5, 3).
Solution
Choose (x1, y1) to be (–1, 1) and (x2, y2) to be (5, 3).
With x1 = –1, y1 = 1, x2 = 5, y2 = 3, we find
Example 2, page 35

80. Equations of Lines Let L be a straight line parallel to the y-axis.
Then L crosses the x-axis at some point (a, 0) , with the x-coordinate given by x = a, where a is a real number.
Any other point on L has the form (a, ), where is an appropriate number.
The vertical line L can therefore be described as
x = a y L
(a, )
(a, 0) x

81. Equations of Lines Let L be a nonvertical line with a slope m.
Let (x1, y1) be a fixed point lying on L and (x, y) be variable point on L distinct from (x1, y1).
Using the slope formula by letting (x, y) = (x1, y1) we get
Multiplying both sides by x – x2 we get

82. Point-Slope Form
An equation of the line that has slope m and passes through point (x1, y1) is given by

83. Examples
Find an equation of the line that passes through the point (1, 3) and has slope 2.
Solution
Use the point-slope form
Substituting for point (1, 3) and slope m = 2, we obtain
Simplifying we get
Example 5, page 36

84. Examples
Find an equation of the line that passes through the points (–3, 2) and (4, –1).
Solution
The slope is given by
Substituting in the point-slope form for point (4, –1) and slope m = – 3/7, we obtain
Example 6, page 36

85. Perpendicular Lines
If L1 and L2 are two distinct nonvertical lines that have slopes m1 and m2, respectively, then L1 is perpendicular to L2 (written L1 ┴ L2) if and only if

86. Example
Find the equation of the line L1 that passes through the point (3, 1) and is perpendicular to the line L2 described by
Solution
L2 is described in point-slope form, so its slope is m2 = 2.
Since the lines are perpendicular, the slope of L1 must be
m1 = –1/2
Using the point-slope form of the equation for L1 we obtain
Example 7, page 37

87. Crossing the Axis
A straight line L that is neither horizontal nor vertical cuts the x-axis and the y-axis at , say, points (a, 0) and (0, b), respectively.
The numbers a and b are called the x-intercept and y-intercept, respectively, of L. y
y-intercept
(0, b)
x-intercept x
(a, 0) L

88. Slope Intercept Form
An equation of the line that has slope m and intersects the y-axis at the point (0, b) is given by
y = mx + b

89. Examples
Find the equation of the line that has slope 3 and y-intercept of – 4.
Solution
We substitute m = 3 and b = – 4 into y = mx + b, and get
y = 3x – 4
Example 8, page 38

90. Examples
Determine the slope and y-intercept of the line whose equation is 3x – 4y = 8.
Solution
Rewrite the given equation in the slope-intercept form.
Thus,
Comparing to y = mx + b we find that m = ¾ , and b = – 2.
So, the slope is ¾ and the y-intercept is – 2.
Example 9, page 38

91. Applied Example
An art object purchased for $50,000 is expected to appreciate in value at a constant rate of $5000 per year for the next 5 years.
Write an equation predicting the value of the art object for any given year.
What will be its value 3 years after the purchase?
Solution
Let x = time (in years) since the object was purchased
y = value of object (in dollars)
Then, y = 50,000 when x = 0, so the y-intercept is b = 50,000.
Every year the value rises by 5000, so the slope is m = 5000.
Thus, the equation must be y = 5000x + 50,000.
After 3 years the value of the object will be $65,000:
y = 5000(3) + 50,000 = 65,000
Applied Example 11, page 39

92. General Form of an Linear Equation
The equation
Ax + By + C = 0
where A, B and C are constants and A and B are not both zero, is called the general form of a linear equation in the variables x and y.

93. Theorem 1
An equation of a straight line is a linear equation; conversely, every linear equation represents a straight line.

94. Example Sketch the straight line represented by the equation
3x – 4y – 12 = 0
Solution
Since every straight line is uniquely determined by two distinct points, we need find only two such points through which the line passes in order to sketch it.
For convenience, let’s compute the x- and y-intercepts:
Setting y = 0, we find x = 4; so the x-intercept is 4.
Setting x = 0, we find y = –3; so the y-intercept is –3.
Thus, the line goes through the points (4, 0) and (0, –3).
Example 12, page 40

95. Example Sketch the straight line represented by the equation
3x – 4y – 12 = 0
Solution
Graph the line going through the points (4, 0) and (0, –3). y L 1
– 1
– 2
– 3
– 4
(4, 0) x
(0, – 3)
Example 12, page 40

96. End of Chapter