3 The Real Number LineWe 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.OriginNegative DirectionPositive 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 numberbn = b · b b · · … · bThe 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, thusSuch a number is called the nth root of b, also written asSimilarly, the expression bp/q is defined as the number(b1/q)p orExamples:(b1/n)n = b
10 Laws of Exponents Law Example 1. am · an = am + n x2 · x3 = x2 + 3 = x52.3. (am)n = am · n (x4)3 = x4 · 3 = x124. (ab)n = an · bn (2x)4 = 24 · x 4 = 16x45.
11 ExamplesSimplify the expressionsExample 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 ExamplesRationalize the numerator of the expressionExample 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 12x3y4A 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 ExamplesSimplify the expressionExample 5, page 8
18 ExamplesSimplify the expressionExample 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 FactoringFactoring is the process of expressing an algebraic expression as a product of other algebraic expressions.Example:
22 FactoringTo 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 expressionis 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 coefficientspx2 + qx + rare (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) = –2and conclude that the correct factorization isx2 – 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 formwhere n is a nonnegative integer and a0, a1, … , an are real numbers with an ≠ 0.For example, the equationis 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 equationmay be written in the formFor the product to be zero, at least one of the factors must be zero, therefore, we havex = 0 x – 1 = 0 x – 2 = 0So, 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, soExample 10, page 12
32 Examples Solve the equation using the quadratic formula: Solution First, rewrite the equation in the standard formFor this equation, a = 1, b = 3, and c = – 8, soExample 10, page 12
34 Rational ExpressionsQuotients of polynomials are called rational expressions.For example
35 Rational ExpressionsThe properties of real numbers apply to rational expressions.ExamplesUsing the properties of number we may writewhere a, b, and c are any real numbers and b and c are not zero.Similarly, we may write
36 ExamplesSimplify the expressionExample 1, page 16
37 ExamplesSimplify the expressionExample 1, page 16
38 Rules of Multiplication and Division If P, Q, R, and S are polynomials, thenMultiplicationDivision
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, thenAdditionSubtraction
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.
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 ExamplesRationalize the denominatorExample 6, page 19
47 ExamplesRationalize the numeratorExample 7, page 19
48 Properties of Inequalities If a, b, and c, are any real numbers, thenProperty 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 SolutionAdd 5 to each member of the given double inequality4 2x < 12Multiply each member of the inequality by ½2 x < 6So, 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 ValueThe 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, thenProperty 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, , soTherefore,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 1.3The 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.4321– 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.y4321– 1– 2– 3– 4x
62 The Cartesian Coordinate System The point where these two lines intersect is called the origin.y4321– 1– 2– 3– 4Originx
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.y4321– 1– 2– 3– 4Negative DirectionPositive Directionx
64 The Cartesian Coordinate System In the y-axis, positive numbers are above and negative numbers are below the origin.y4321– 1– 2– 3– 4Positive DirectionxNegative 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)4321– 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.y4321– 1– 2– 3– 4Quadrant II(–, +)Quadrant I(+, +)xQuadrant III(–, –)Quadrant IV(+, –)
67 The Distance FormulaThe distance between any two points in the plane may be expressed in terms of their coordinates.Distance formulaThe 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). SolutionLet P1(– 4, 3) and P2(2, 6) be points in the plane.We havex1 = – 4 y1 = 3 x2 = 2 y2 = 6Using the distance formula, we haveExample 1, page 26
69 ExamplesLet 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.SolutionBy definition in a circle, the distance between P(x, y) and C(h, k) is r.With distance formula we getSquaring both sides givesyP(x, y)C(h, k)rkxhExample 2, page 27
70 Equation of a CircleAn equation of a circle with center C(h, k) and radius r is given by
71 ExamplesFind an equation of the circle with radius 2 and center (–1, 3).SolutionWe use the circle formula with r = 2, h = –1, and k = 3:y(–1, 3)32x–1Example 3, page 27
72 ExamplesFind an equation of the circle with radius 3 and center located at the origin.SolutionWe use the circle formula with r = 3, h = 0, and k = 0:y3xExample 3, page 27
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.yL(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 byyL(x2, y2)y2 – y1 = y(x1, y1)x2 – x1 = xx
76 Slope of a Nonvertical Line If m > 0, the line slants upward from left to right.yLm = 2y = 2x = 1x
77 Slope of a Nonvertical Line If m < 0, the line slants downward from left to right.ym = –1x = 1y = –1xL
78 ExamplesSketch the straight line that passes through the point (2, 5) and has slope – 4/3.SolutionPlot 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.y654321x = 3(2, 5)y = – 4(5, 1)xLExample 1, page 34
79 ExamplesFind the slope m of the line that goes through the points (–1, 1) and (5, 3).SolutionChoose (x1, y1) to be (–1, 1) and (x2, y2) to be (5, 3).With x1 = –1, y1 = 1, x2 = 5, y2 = 3, we findExample 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 asx = ayL(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 getMultiplying both sides by x – x2 we get
82 Point-Slope FormAn equation of the line that has slope m and passes through point (x1, y1) is given by
83 ExamplesFind an equation of the line that passes through the point (1, 3) and has slope 2.SolutionUse the point-slope formSubstituting for point (1, 3) and slope m = 2, we obtainSimplifying we getExample 5, page 36
84 ExamplesFind an equation of the line that passes through the points (–3, 2) and (4, –1).SolutionThe slope is given bySubstituting in the point-slope form for point (4, –1) and slope m = – 3/7, we obtainExample 6, page 36
85 Perpendicular LinesIf 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 ExampleFind the equation of the line L1 that passes through the point (3, 1) and is perpendicular to the line L2 described bySolutionL2 is described in point-slope form, so its slope is m2 = 2.Since the lines are perpendicular, the slope of L1 must bem1 = –1/2Using the point-slope form of the equation for L1 we obtainExample 7, page 37
87 Crossing the AxisA 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.yy-intercept(0, b)x-interceptx(a, 0)L
88 Slope Intercept FormAn equation of the line that has slope m and intersects the y-axis at the point (0, b) is given byy = mx + b
89 ExamplesFind the equation of the line that has slope 3 and y-intercept of – 4.SolutionWe substitute m = 3 and b = – 4 into y = mx + b, and gety = 3x – 4Example 8, page 38
90 ExamplesDetermine the slope and y-intercept of the line whose equation is 3x – 4y = 8.SolutionRewrite 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 ExampleAn 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?SolutionLet x = time (in years) since the object was purchasedy = 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 = 65,000Applied Example 11, page 39
92 General Form of an Linear Equation The equationAx + By + C = 0where 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 1An 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 = 0SolutionSince 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 = 0SolutionGraph the line going through the points (4, 0) and (0, –3).yL1– 1– 2– 3– 4(4, 0)x(0, – 3)Example 12, page 40