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1
Preliminaries
 Precalculus Review I
 Precalculus Review II
 The Cartesian Coordinate System
 Straight Lines
1.1
Precalculus Review I
Origin
Negative Direction
–4
–3
–2
–1
– 2
Positive Direction
0
1
2
3

4
The Real Number Line
 We can represent real numbers geometrically by points on
a real number, or coordinate, line:
Origin
Negative Direction
–4
–3
–2
–1
Positive Direction
0
1
 2
2
3
4

 This line includes all real numbers.
 Exactly one point on the line is associated with each real
number, and vice-versa.
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.
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.
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.
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.
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
n factors
 The number b is called the base, and the superscript n is
called the power of the exponential expression bn.
 For example:
3
2 2 2 2 8
       
25  2  2  2  2  2  32
 3   3   3   3  27
 If b ≠ 0, we define b0 = 1.
 For example:
20 = 1 and (–)0 = 1, but 00 is undefined.
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
(b1/n)n = b
 Such a number is called the nth root of b, also written as
n
b
 Similarly, the expression bp/q is defined as the number
(b1/q)p
 Examples:
2
3/2
4
 2
5/2
or

1/2 3
q
bp
 1.41412   2.8283
3
1
1
1
1
 5/2 
 5
5
1/2
4
 4  2 32
Laws of Exponents
Law
Example
1. am · an = am + n
x2 · x3 = x 2 + 3 = x5
am
2. n  a m n
a
x7
74
3

x

x
x4
(a  0)
3. (am)n = am · n
(x4)3 = x4 · 3 = x12
4. (ab)n = an · bn
(2x)4 = 24 · x 4 = 16x4
n
a
a
5.    n
b
b
n
3
x3 x3
 x
   3
8
2 2
Examples
 Simplify the expressions
a.
 3x  4 x 
2
3
 12 x 23  12 x 5
165 / 4
3/4
1/ 2  16

16
b.
6 
2/3 3
c.
d.
x y
e.
y 
 1/ 4 
x 
3
3/ 2
Example 1, page 6

2  2
 
4
16
3
 23  8
 6(2/3)(3)  66/3  62  36
 x
 y 
3 2
2 2
x
(3)( 2)
2
y (3/2)( 2)
y 3 x1/2
 (1/4)( 2)  1/2  3
x
x
y
y
( 2)( 2)
y4
x y  6
x
6
4
Examples
 Simplify the expressions (assume x, y, m, and n are positive)
4
a.
16 x y  16 x y
4
8
4

8 1/4
 161/4 x 4/4 y 8/4  2 xy 2
b. 12m n  3m n  36m n   36m n
3
5
3
c.
27 x
3
Example 2, page 7
8y
3
8 2
6
27 x 


8 y 
6 1/3
3 1/3

8 2 1/2
 361/2 m8/2n2/2  6m4n
271/3 x 6/3
3x 2
 1/3 3/3  
8 y
2y
Examples
 Rationalize the denominator of the expression
3x
x
3x


x
2 x 2 x

3x x
2 x2
3x x
2x
3

x
2

Example 3, page 7
Examples
 Rationalize the numerator of the expression
3 x 3 x
x


2x
2x
x
3 x2

2x x
3x

2x x
3

2 x
Example 4, page 7
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.
Examples:
x2 + 4x + 4
x4 + 3x2 – 3
x2y – xy + y
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
Examples
 Simplify the expression
(2 x 4  3x 3  4 x  6)  (3x 4  9 x 3  3x 2 )
 2 x 4  3x 3  4 x  6  3x 4  9 x 3  3x 2
 2 x 4  3x 4  3x 3  9 x 3  3x 2  4 x  6
  x 4  6 x 3  3x 2  4 x  6
Example 5, page 8
Examples
 Simplify the expression
2t 3  {t 2  [t  (2t  1)]  4}
 2t 3  {t 2  [t  2t  1]  4}
 2t 3  {t 2  [t  1]  4}
 2t 3  {t 2  t  1  4}
 2t 3  t 2  t  1  4
 2t 3  t 2  t  3
Example 5, page 8
Examples
 Perform the operation and simplify the expression
( x 2  1)(3x 2  10 x  3)
 x 2 (3x 2  10 x  3)  1(3x 2  10 x  3)
 3x 4  10 x 3  3x 2  3x 2  10 x  3
 3x 4  10 x 3  6 x 2  10 x  3
Example 6, page 8
Examples
 Perform the operation and simplify the expression
( e t  e  t )e t  e t ( e t  e  t )  e 2 t  e 0  e 2 t  e 0
 11
2
Example 6, page 9
Factoring
 Factoring is the process of expressing an algebraic
expression as a product of other algebraic expressions.
 Example:
3x 2  x  x(3x  1)
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
2a 2 x  4ax  6a
is 2a, because
2a 2 x  4ax  6a  2a  ax  2a  2 x  2a  3
 2a (ax  2 x  3)
Examples
 Factor out the greatest common factor in each expression
a.
b.
c. 2 ye
Example 7, page 9
 0.3t 2  3t  0.3t (t  10)
2 x 3/ 2  3 x1/ 2  x1/2 (2 x  3)
xy 2
 2 xy e
3 xy 2
 2 ye (1  xy 2 )
xy 2
Examples
 Factor out the greatest common factor in each expression
a.
2ax  2ay  bx  by  2a ( x  y )  b( x  y )
 (2a  b)( x  y )
b. 3x y  4  2 y  6 x  3x y  2 y  6 x  4

y (3x  2)  2(3x  2)
 (3x  2)( y  2)
Example 8, page 10
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.
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) = 2 or (1)(1) + (1)(–3) = –2
and conclude that the correct factorization is
x2 – 2x – 3 = (x + 1)(x –3)
Examples
 Find the correct factorization for the expressions
3x 2  4 x  4  (3x  2)( x  2)
3x 2  6 x  24  3( x 2  2 x  8)
 3( x  4)( x  2)
Example 9, page 11
Roots of Polynomial Expressions
 A polynomial equation of degree n in the variable x is an
equation of the form
an x n  an 1 x n1    a0  0
where n is a nonnegative integer and a0, a1, … , an are real
numbers with an ≠ 0.
 For example, the equation
2 x 5  8 x 3  6 x 2  3x  1  0
is a polynomial equation of degree 5.
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
x 3  3x 2  2 x  0
may be written in the form
x( x 2  3x  2)  0
or
x( x  1)( x  2)  0
 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.
The Quadratic Formula
 The solutions of the equation
ax2 + bx + c = 0
(a ≠ 0)
are given by
b  b2  4ac
x
2a
Examples
 Solve the equation using the quadratic formula:
2 x 2  5 x  12  0
Solution
 For this equation, a = 2, b = 5, and c = –12, so
b  b2  4ac 5  52  4(2)(12)
x

2a
2(2)
5  121 5  11


4
4
3
 4 or
2
Example 10, page 12
Examples
 Solve the equation using the quadratic formula:
x 2  3x  8
Solution
 First, rewrite the equation in the standard form
x 2  3x  8  0
 For this equation, a = 1, b = 3, and c = – 8, so
b  b2  4ac 3  32  4(1)( 8)
x

2a
2(1)
3  41

2
3  41
3  41
so, x 
 1.7 or x 
 4.7
2
2
Example 10, page 12
1.2
Precalculus Review II


1  h  1
22
1 h 1 1 h 1
1 h 1



h
h
1 h 1 h 1 h 1



1 h 1
h

h 1  h  1 h 1  hh 11

1
1 h 1
 

22

Rational Expressions
 Quotients of polynomials are called rational
expressions.
 For example
3 x  8
2x  3
5 x 2 y 3  2 xy
4x
2
5ab
Rational Expressions
 The properties of real numbers apply to rational
expressions.
Examples
 Using the properties of number we may write
ac a c a
a
   1 
bc b c b
b
where a, b, and c are any real numbers and b and c
are not zero.
 Similarly, we may write
( x  2)( x  3) ( x  2)

( x  2)( x  3) ( x  2)
( x  2,3)
Examples
 Simplify the expression
x 2  2 x  3 ( x  3)( x  1)

2
x  4 x  3 ( x  3)( x  1)
x 1

x 1
Example 1, page 16
Examples
 Simplify the expression
2
2
( x 2  1)2 ( 2)  (2 x)(2)( x 2  1)(2 x) ( x  1) ( x  1)( 2)  (2 x )(2)(2 x) 

2
4
( x  1)
( x 2  1)4
( x 2  1)( 2 x 2  2  8 x 2 )

( x 2  1)4
( x 2  1)(6 x 2  2)

( x 2  1) 4
(6 x 2  2)
 2
( x  1)3
2(3x 2  1)

( x 2  1)3
Example 1, page 16
Rules of Multiplication and Division
If P, Q, R, and S are polynomials, then
 Multiplication
P R PR
 
Q S QS
(Q, S  0)
 Division
P R P S PS
   
Q S Q R QR
(Q, R, S  0)
Example
 Perform the indicated operation and simplify
( x  2) 2
2 x  8 x 2  4 x  4 2( x  4)



2
x2
x  16
x  2 ( x  4)( x  4)
2( x  4)( x  2)( x  2)

( x  2)( x  4)( x  4)
2( x  2)

x4
Example 2, page 16
Rules of Addition and Subtraction
If P, Q, R, and S are polynomials, then
 Addition
P Q PQ
 
R S
R
( R  0)
 Subtraction
P Q P Q
 
R S
R
( R  0)
Example
 Perform the indicated operation and simplify
1
1 x  ( x  h)
 
xh x
x( x  h)
x xh

x( x  h)
h

x( x  h)
Example 3b, page 17
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.
Examples
 Simplify
x 11
1
x 1  x 1  x  2  x
4
x2  4
x  1 x2  4
x
x
x
1
x2
x


x  1 ( x  2)( x  2)
x

( x  1)( x  2)
Example 4a, page 18
Examples
 Simplify
x 1  y 1
x 2  y 2
1 1
yx

x y
xy

 2
2
1
1
y

x
 2
2
x
y
x2 y2
y  x x2 y2
yx
( xy )2

 2


2
xy y  x
xy ( y  x )( y  x )
xy

yx
Example 4b, page 18
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

a b

  a    b
a b 
 ab
2
2
Examples
 Rationalize the denominator
1
1
1 x


1 x 1 x 1 x

1 x
1
2

 x
1 x

1 x
Example 6, page 19
2
Examples
 Rationalize the numerator


1  h  1
1 h 1 1 h 1
1 h 1



h
h
1 h 1
h 1 h 1

h

1 h 1
1

1 h 1
Example 7, page 19

1 h 1

h

h


1 h 1
2

2
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.
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
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:
1. 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.
2. 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
Examples
 Solve the inequality
Solution
x 1
0
x 1
 For the quotient to be positive, the numerator and
denominator must have the same sign, so we have two
possibilities to consider:
1. 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.
2. 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
Absolute Value
 The absolute value of a number a is
denoted |a| and is defined by
 a if a  0
a 
 a if a  0
Absolute Value Properties
 If a, b, and c, are any real numbers, then
Property 5
|–a| = |a|
Property 6
|ab| = |a| |b|
Property 7
Property 8
a
a

b
b
(b ≠ 0)
|a + b| ≤ |a| + |b|
Examples
 Evaluate the expression
| – 5| + 3
Solution
 Since  – 5 < 0, we see that
| – 5| = –( – 5).
 Therefore
| – 5| + 3 = – ( – 5) +3
=8–
≈ 4.8584
Example 12a, page 22
Examples
 Evaluate the expression
32  2 3
Solution
 Since
3  2  0 , we see that
32  

32
 Similarly, 2  3  0 , so 2  3  2  3
 Therefore,
32  2 3  

42 3

2 2 3
 0.5359
Example 12b, page 22
 
32  2 3



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
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
1.3
The Cartesian Coordinate System
P(x, y)
C(h, k)
k
h
r
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).
–4
–3
–2
–1
0
1
2
3
4
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
–4
–3
–2
–1
–1
–2
–3
–4
1
2
3
4
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
–4
–3
–2
–1
–1
–2
–3
–4
1
2
3
4
x
The Cartesian Coordinate System
 The point where these two lines intersect is called the origin.
y
4
3
2
1
–4
–3
–2
–1
–1
–2
–3
–4
Origin
1
2
3
4
x
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
Negative Direction
–4
–3
–2
–1
1
–1
–2
–3
–4
Positive Direction
1
2
3
4
x
The Cartesian Coordinate System
 In the y-axis, positive numbers are above and negative
numbers are below the origin.
3
2
1
–4
–3
–2
–1
–1
–2
–3
–4
1
Negative Direction
4
Positive Direction
y
2
3
4
x
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
(4, 3)
3
2
1
–4
–3
–2
–1
(– 1, – 2)
–1
–2
–3
–4
1
2
3
4
(3,–1)
x
The Cartesian Coordinate System
 The axes divide the plane into four quadrants as shown below.
y
4
Quadrant II
(–, +)
Quadrant I
(+, +)
3
2
1
–4
–3
–2
Quadrant III
(–, –)
–1
–1
–2
–3
–4
1
2
3
4
Quadrant IV
(+, –)
x
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
d
 x2  x1    y2  y1 
2
2
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
d

 x2  x1 
2
  y2  y1 
2  (4)   6  3
2
2
 62  32  45  3 5
Example 1, page 26
2
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.
y
 With distance formula we get
 x  h   y  k   r
2
 Squaring both sides gives
 x  h
2
P(x, y)
2
k
C(h, k)
  y  k   r2
2
h
Example 2, page 27
r
x
Equation of a Circle
 An equation of a circle with center C(h, k)
and radius r is given by
 x  h
2
  y  k   r2
2
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:
 x  h   y  k   r2
2
2
 x  ( 1)   y  3
2
2
 x  1
2
 22
  y  3  4
2
y
(–1, 3)
2
–1
Example 3, page 27
3
x
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:
 x  h   y  k   r2
2
2
2
x

0

y

0

3

 

2
2
x  y 9
2
Example 3, page 27
y
2
3
x
1.4
Straight Lines
y
6
5
4
x = 3
(2, 5)
y = – 4
3
2
(5, 1)
1
1
2
3
4
5
x
6 L
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
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
m
y y2  y1

x x2  x1
y
L
(x2, y2)
y2 – y1 = y
(x1, y1)
x2 – x1 = x
x
Slope of a Nonvertical Line
 If m > 0, the line slants upward from left to right.
L
y
m=2
y = 2
x = 1
x
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
Examples
 Sketch the straight line that passes through the point
(2, 5) and has slope –4/3.
Solution
1. Plot the point (2, 5).
2. A slope of –4/3 means
that if x increases by 3,
y decreases by 4.
3. Plot the point (5, 1).
4. Draw a line across the
two points.
y
6
5
4
x = 3
(2, 5)
y = –4
3
2
(5, 1)
1
x
1
Example 1, page 34
2
3
4
5
6
L
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
y2  y1
3 1
2 1
m

 
x2  x1 5  ( 1) 6 3
Example 2, page 35
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, y ), where y is
an appropriate number.
 The vertical line L can
therefore be described as
x=a
y
L
(a, y)
(a, 0)
x
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
y  y1
m
x  x1
 Multiplying both sides by x – x2 we get
y  y1  m( x  x1 )
Point-Slope Form
 An equation of the line that has slope m and
passes through point (x1, y1) is given by
y  y1  m( x  x1 )
Examples
 Find an equation of the line that passes through the point
(1, 3) and has slope 2.
Solution
 Use the point-slope form
y  y1  m( x  x1 )
 Substituting for point (1, 3) and slope m = 2, we obtain
y  3  2( x  1)
 Simplifying we get
2x  y  1  0
Example 5, page 36
Examples
 Find an equation of the line that passes through the points
(–3, 2) and (4, –1).
Solution
 The slope is given by
y  y1
1  2
3
m


x  x1 4  ( 3)
7
 Substituting in the point-slope form for point (4, –1) and
slope m = – 3/7, we obtain
3
y  1   ( x  4)
7
7 y  7  3x  12
3x  7 y  5  0
Example 6, page 36
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
1
m1  
m2
Example
 Find the equation of the line L1 that passes through the
point (3, 1) and is perpendicular to the line L2 described by
y  3  2( x  1)
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
1
y  1   ( x  3)
2
2 y  2  x  3
x  2y 5  0
Example 7, page 37
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
(a, 0)
x
L
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
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
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,
3x  4 y  8
4 y  8  3x
3
y  x2
4
 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
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
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.
Theorem 1
 An equation of a straight line is a linear
equation; conversely, every linear equation
represents a straight line.
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
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
(4, 0)
x
–1
1
–2
–3
–4
Example 12, page 40
(0, – 3)
2
3
4
5
6
End of
Chapter