Title: Precalculus%20Review%20I
11
Preliminaries
- Precalculus Review I
- Precalculus Review II
- The Cartesian Coordinate System
- Straight Lines
21.1
3The 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
Positive Direction
Negative Direction
4 3 2 1 0 1 2 3 4
p
4Finite 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 lt x lt b. - It is called open because neither of its
endpoints is included in the interval.
5Finite 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.
6Finite 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 lt x ? b or a ? x lt b.
7Infinite 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 gt a, x ? a, x lt a,
x ? a.
8Exponents 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.
- For example
- 20 1 and (p)0 1, but 00 is undefined.
n factors
9Exponents 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
10Laws 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.
11Examples
Example 1, page 6
12Examples
(assume x, y, m, and n are positive)
Example 2, page 7
13Examples
- Rationalize the denominator of the expression
Example 3, page 7
14Examples
- Rationalize the numerator of the expression
Example 4, page 7
15Operations 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
16Operations 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
17Examples
Example 5, page 8
18Examples
Example 5, page 8
19Examples
- Perform the operation and simplify the expression
Example 6, page 8
20Examples
- Perform the operation and simplify the expression
Example 6, page 9
21Factoring
- Factoring is the process of expressing an
algebraic expression as a product of other
algebraic expressions. - Example
22Factoring
- 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
23Examples
- Factor out the greatest common factor in each
expression
Example 7, page 9
24Examples
- Factor out the greatest common factor in each
expression
Example 8, page 10
25Factoring 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.
26Examples
- 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)
27Examples
- Find the correct factorization for the expressions
Example 9, page 11
28Roots 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.
29Roots 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.
30The Quadratic Formula
- The solutions of the equation
- ax2 bx c 0 (a ? 0)
- are given by
31Examples
- Solve the equation using the quadratic formula
- Solution
- For this equation, a 2, b 5, and c 12, so
Example 10, page 12
32Examples
- 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
331.2
34Rational Expressions
- Quotients of polynomials are called rational
expressions. - For example
35Rational 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
36Examples
Example 1, page 16
37Examples
Example 1, page 16
38Rules of Multiplication and Division
- If P, Q, R, and S are polynomials, then
- Multiplication
- Division
39Example
- Perform the indicated operation and simplify
Example 2, page 16
40Rules of Addition and Subtraction
- If P, Q, R, and S are polynomials, then
- Addition
- Subtraction
41Example
- Perform the indicated operation and simplify
Example 3b, page 17
42Other 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.
43Examples
Example 4a, page 18
44Examples
Example 4b, page 18
45Rationalizing 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
46Examples
- Rationalize the denominator
Example 6, page 19
47Examples
- Rationalize the numerator
Example 7, page 19
48Properties of Inequalities
- If a, b, and c, are any real numbers, then
- Property 1 If a lt b and b lt c, then a lt c.
- Property 2 If a lt b, then a c lt b c.
- Property 3 If a lt b and c gt 0, then ac lt bc.
- Property 4 If a lt b and c lt 0, then ac gt bc.
49Examples
- Find the set of real numbers that satisfy
- 1 ? 2x 5 lt 7
- Solution
- Add 5 to each member of the given double
inequality - 4 ? 2x lt 12
- Multiply each member of the inequality by ½
- 2 ? x lt 6
- So, the solution is the set of all values of x
lying in the interval 2, 6).
Example 8, page 20
50Examples
- Solve the inequality x2 2x 8 lt 0.
- Solution
- Factorizing we get (x 4)(x 2) lt 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) lt 0 and (x 2) gt
0, which means x lt 4, and x gt 2, but this
is impossible x cannot meet these two
conditions simultaneously. - The inequality also holds if (x 4) gt 0 and (x
2) lt 0, which means x gt 4, and x lt
2, or 4 lt x lt 2. - So, the solution is the set of all values of x
lying in the interval ( 4, 2).
Example 9, page 20
51Examples
- 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) lt
0, which means x ? 1, and x lt 1, both of
these conditions are met only when x ? 1. - The inequality also holds if (x 1) ? 0 and (x
1) gt 0, which means x ? 1, and x gt 1, both
of these conditions are met only when x gt 1. - So, the solution is the set of all values of x
lying in the intervals ( ?, 1 and (1, ?).
Example 10, page 21
52Absolute Value
- The absolute value of a number a is denoted a
and is defined by -
53Absolute 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
54Examples
- Evaluate the expression
- p 5 3
- Solution
- Since p 5 lt 0, we see that
- p 5 (p 5).
- Therefore
- p 5 3 (p 5) 3
- 8 p
- 4.8584
Example 12a, page 22
55Examples
- Evaluate the expression
- Solution
- Since , we see that
-
- Similarly, , so
- Therefore,
Example 12b, page 22
56Examples
- Evaluate the inequality x ? 5.
- Solution
- If x ? 0, then x x, so x ? 5 implies x ? 5.
- If x lt 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
57Examples
- 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
581.3
- The Cartesian Coordinate System
59The 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).
60The 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
61The 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
62The 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
63The 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
Positive Direction
Negative Direction
x
64The 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
65The 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)
66The Cartesian Coordinate System
- The axes divide the plane into four quadrants as
shown below.
y
4 3 2 1 1 2 3 4
Quadrant I (, )
Quadrant II (, )
x
Quadrant IV (, )
Quadrant III (, )
67The 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
68Examples
- 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
69Examples
- 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
70Equation of a Circle
- An equation of a circle with center C(h, k) and
radius r is given by
71Examples
- 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
72Examples
- 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
731.4
74Slope 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
75Slope 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
76Slope of a Nonvertical Line
- If m gt 0, the line slants upward from left to
right.
y
L
m 2
?y 2
?x 1
x
77Slope of a Nonvertical Line
- If m lt 0, the line slants downward from left to
right.
y
m 1
?x 1
?y 1
x
L
78Examples
- 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
1 2 3 4 5 6
L
Example 1, page 34
79Examples
- 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
80Equations 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
81Equations 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
82Point-Slope Form
- An equation of the line that has slope m and
passes through point (x1, y1) is given by
83Examples
- 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
84Examples
- 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
85Perpendicular 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
86Example
- 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
87Crossing 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
88Slope 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
89Examples
- 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
90Examples
- 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
91Applied 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
92General 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.
93Theorem 1
- An equation of a straight line is a linear
equation conversely, every linear equation
represents a straight line.
94Example
- 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, lets 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
95Example
- 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
1 2 3 4 5 6
(0, 3)
Example 12, page 40
96End of Chapter