The Cartesian Coordinate System and Straight lines - PowerPoint PPT Presentation

View by Category
About This Presentation
Title:

The Cartesian Coordinate System and Straight lines

Description:

2 Functions and Their Graphs The Cartesian Coordinate System and Straight lines Equations of Lines Functions and Their Graphs The Algebra of Functions – PowerPoint PPT presentation

Number of Views:883
Avg rating:3.0/5.0
Slides: 149
Provided by: PeterT154
Learn more at: http://www.shsu.edu
Category:

less

Write a Comment
User Comments (0)
Transcript and Presenter's Notes

Title: The Cartesian Coordinate System and Straight lines


1
2
Functions and Their Graphs
  • The Cartesian Coordinate System and
    Straight lines
  • Equations of Lines
  • Functions and Their Graphs
  • The Algebra of Functions
  • Linear Functions
  • Quadratic Functions
  • Functions and Mathematical Models

2
2.1
  • The Cartesian Coordinate System and Straight lines

3
The Cartesian Coordinate System
  • We can represent real numbers geometrically by
    points on a real number, or coordinate, line

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 1 2 3 4
5
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
6
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
7
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
Positive Direction
Negative Direction
x
8
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
9
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)
10
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 I (, )
Quadrant II (, )
x
Quadrant IV (, )
Quadrant III (, )
11
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
12
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
13
Slope of a Nonvertical Line
  • If m gt 0, the line slants upward from left to
    right.

y
L
m 1
?y 1
?x 1
x
14
Slope 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
15
Slope 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
16
Slope of a Nonvertical Line
  • If m lt 0, the line slants downward from left to
    right.

y
m 2
?x 1
?y 2
x
L
17
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 resulting point (5, 1).
  • Draw a line through 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
18
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

19
Examples
  • Find the slope m of the line that goes through
    the points (2, 5) and (3, 5).
  • Solution
  • Choose (x1, y1) to be (2, 5) and (x2, y2) to be
    (3, 5).
  • With x1 2, y1 5, x2 3, y2 5, we find

20
Examples
  • Find the slope m of the line that goes through
    the points (2, 5) and (3, 5).
  • Solution
  • The slope of a horizontal line is zero

y
6 4 3 2 1
(3, 5)
(2, 5)
L
m 0
x
2 1 1 2 3 4
21
Parallel Lines
  • Two distinct lines are parallel if and only if
    their slopes are equal or their slopes are
    undefined.

22
Example
  • Let L1 be a line that passes through the points
    (2, 9) and (1, 3), and let L2 be the line that
    passes through the points ( 4, 10) and (3, 4).
  • Determine whether L1 and L2 are parallel.
  • Solution
  • The slope m1 of L1 is given by
  • The slope m2 of L2 is given by
  • Since m1 m2, the lines L1 and L2 are in fact
    parallel.

23
2.2
  • Equations of Lines

24
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
25
Equations of Lines
  • Let L be a nonvertical line with a slope m.
  • Let (x1, y1) be a fixed point lying on L, and let
    (x, y) be a variable point on L distinct from
    (x1, y1).
  • Using the slope formula by letting (x, y) (x2,
    y2), we get
  • Multiplying both sides by x x1 we get

26
Point-Slope Form
  • An equation of the line that has slope m and
    passes through point (x1, y1) is given by

27
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

28
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

29
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

30
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

31
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
32
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

33
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

34
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.
  • Comparing to y mx b, we find that m ¾ and
    b 2.
  • So, the slope is ¾ and the y-intercept is 2.

35
Applied Example
  • Suppose 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

36
General Form of a 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.

37
General Form of a Linear Equation
  • An equation of a straight line is a linear
    equation conversely, every linear equation
    represents a straight line.

38
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, 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).

39
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
1 2 3 4 5 6
(0, 3)
40
Equations of Straight Lines
  • Vertical line x a
  • Horizontal line y b
  • Point-slope form y y1 m(x x1)
  • Slope-intercept form y mx b
  • General Form Ax By C 0

41
2.3
  • Functions and Their Graphs

42
Functions
  • A function f is a rule that assigns to each
    element in a set A one and only one element in a
    set B.
  • The set A is called the domain of the function.
  • It is customary to denote a function by a letter
    of the alphabet, such as the letter f.
  • If x is an element in the domain of a function f,
    then the element in B that f associates with x is
    written f(x) (read f of x) and is called the
    value of f at x.
  • The set B comprising all the values assumed by y
    f(x) as x takes on all possible values in its
    domain is called the range of the function f.

43
Example
  • Let the function f be defined by the rule
  • Find f(1)
  • Solution

44
Example
  • Let the function f be defined by the rule
  • Find f( 2)
  • Solution

45
Example
  • Let the function f be defined by the rule
  • Find f(a)
  • Solution

46
Example
  • Let the function f be defined by the rule
  • Find f(a h)
  • Solution

47
Applied Example
  • ThermoMaster manufactures an indoor-outdoor
    thermometer at its Mexican subsidiary.
  • Management estimates that the profit (in dollars)
    realizable by ThermoMaster in the manufacture and
    sale of x thermometers per week is
  • Find ThermoMasters weekly profit if its level of
    production is
  • 1000 thermometers per week.
  • 2000 thermometers per week.

48
Applied Example
  • Solution
  • We have
  • The weekly profit by producing 1000 thermometers
    is
  • or 2,000.
  • The weekly profit by producing 2000 thermometers
    is
  • or 7,000.

49
Determining the Domain of a Function
  • Suppose we are given the function y f(x).
  • Then, the variable x is called the independent
    variable.
  • The variable y, whose value depends on x, is
    called the dependent variable.
  • To determine the domain of a function, we need to
    find what restrictions, if any, are to be placed
    on the independent variable x.
  • In many practical problems, the domain of a
    function is dictated by the nature of the problem.

50
Applied Example Packaging
  • An open box is to be made from a rectangular
    piece of cardboard 16 inches wide by cutting away
    identical squares (x inches by x inches) from
    each corner and folding up the resulting flaps.

x
10 10 2x
x
16 2x
x
x
16
51
Applied Example Packaging
  • An open box is to be made from a rectangular
    piece of cardboard 16 inches wide by cutting away
    identical squares (x inches by x inches) from
    each corner and folding up the resulting flaps.
  • Find the expression that gives the volume V of
    the box as a function of x.
  • What is the domain of the function?
  • The dimensions of the resulting box are

x
10 2x
16 2x
52
Applied Example Packaging
  • Solution
  • a. The volume of the box is given by multiplying
    its dimensions (length ? width ? height), so

x
10 2x
16 2x
53
Applied Example Packaging
  • Solution
  • b. Since the length of each side of the box must
    be greater than or equal to zero, we see that
  • must be satisfied simultaneously. Simplified
  • All three are satisfied simultaneously provided
    that
  • Thus, the domain of the function f is the
    interval 0, 5.

54
More Examples
  • Find the domain of the function
  • Solution
  • Since the square root of a negative number is
    undefined, it is necessary that x 1 ? 0.
  • Thus the domain of the function is 1,?).

55
More Examples
  • Find the domain of the function
  • Solution
  • Our only constraint is that you cannot divide by
    zero, so
  • Which means that
  • Or more specifically x ? 2 and x ? 2.
  • Thus the domain of f consists of the intervals (
    ?, 2), (2, 2), (2, ?).

56
More Examples
  • Find the domain of the function
  • Solution
  • Here, any real number satisfies the equation, so
    the domain of f is the set of all real numbers.

57
Graphs of Functions
  • If f is a function with domain A, then
    corresponding to each real number x in A there is
    precisely one real number f(x).
  • Thus, a function f with domain A can also be
    defined as the set of all ordered pairs (x, f(x))
    where x belongs to A.
  • The graph of a function f is the set of all
    points (x, y) in the xy-plane such that x is in
    the domain of f and y f(x).

58
Example
  • The graph of a function f is shown below

y
y
(x, y)
Range
x
x
Domain
59
Example
  • The graph of a function f is shown below
  • What is the value of f(2)?

y
4 3 2 1 1 2
x
1 2 3 4 5 6 7 8
(2, 2)
60
Example
  • The graph of a function f is shown below
  • What is the value of f(5)?

y
4 3 2 1 1 2
(5, 3)
x
1 2 3 4 5 6 7 8
61
Example
  • The graph of a function f is shown below
  • What is the domain of f(x)?

y
4 3 2 1 1 2
x
1 2 3 4 5 6 7 8
Domain 1,8
62
Example
  • The graph of a function f is shown below
  • What is the range of f(x)?

y
4 3 2 1 1 2
Range 2,4
x
1 2 3 4 5 6 7 8
63
Example Sketching a Graph
  • Sketch the graph of the function defined by the
    equation
  • y x2 1
  • Solution
  • The domain of the function is the set of all real
    numbers.
  • Assign several values to the variable x and
    compute the corresponding values for y

x y
3 10
2 5
1 2
0 1
1 2
2 5
3 10
64
Example Sketching a Graph
  • Sketch the graph of the function defined by the
    equation
  • y x2 1
  • Solution
  • The domain of the function is the set of all real
    numbers.
  • Then plot these values in a graph

y
10 8 6 4 2
x y
3 10
2 5
1 2
0 1
1 2
2 5
3 10
x
3 2 1 1 2 3
65
Example Sketching a Graph
  • Sketch the graph of the function defined by the
    equation
  • y x2 1
  • Solution
  • The domain of the function is the set of all real
    numbers.
  • And finally, connect the dots

y
10 8 6 4 2
x y
3 10
2 5
1 2
0 1
1 2
2 5
3 10
x
3 2 1 1 2 3
66
Example Sketching a Graph
  • Sketch the graph of the function defined by the
    equation
  • Solution
  • The function f is defined in a piecewise fashion
    on the set of all real numbers.
  • In the subdomain ( ?, 0), the rule for f is
    given by
  • In the subdomain 0, ?), the rule for f is given
    by

67
Example Sketching a Graph
  • Sketch the graph of the function defined by the
    equation
  • Solution
  • Substituting negative values for x into
    , while
  • substituting zero and positive values into
    we get

x y
3 3
2 2
1 1
0 0
1 1
2 1.41
3 1.73
68
Example Sketching a Graph
  • Sketch the graph of the function defined by the
    equation
  • Solution
  • Plotting these data and graphing we get

y
x y
3 3
2 2
1 1
0 0
1 1
2 1.41
3 1.73
3 2 1
x
3 2 1 1 2 3
69
The Vertical Line Test
  • A curve in the xy-plane is the graph of a
    function y f(x) if and only if each vertical
    line intersects it in at most one point.

70
Examples
  • Determine if the curve in the graph is a function
    of x
  • Solution
  • The curve is indeed a function of x, because
    there is one and only one value of y for any
    given value of x.

y
x
71
Examples
  • Determine if the curve in the graph is a function
    of x
  • Solution
  • The curve is not a function of x, because there
    is more than one value of y for some values of x.

y
x
72
Examples
  • Determine if the curve in the graph is a function
    of x
  • Solution
  • The curve is indeed a function of x, because
    there is one and only one value of y for any
    given value of x.

y
x
73
2.4
  • The Algebra of Functions

74
The Sum, Difference, Product and Quotient
of Functions
  • Consider the graph below
  • R(t) denotes the federal government revenue at
    any time t.
  • S(t) denotes the federal government spending at
    any time t.

y
2000 1800 1600 1400 1200 1000
y R(t)
y S(t)
S(t)
Billions of Dollars
R(t)
t
1990 1992 1994 1996 1998 2000
t
Year
75
The Sum, Difference, Product and Quotient
of Functions
  • Consider the graph below
  • The difference R(t) S(t) gives the budget
    deficit (if negative) or surplus (if positive) in
    billions of dollars at any time t.

y
2000 1800 1600 1400 1200 1000
y R(t)
y S(t)
S(t)
Billions of Dollars
D(t) R(t) S(t)
R(t)
t
1990 1992 1994 1996 1998 2000
t
Year
76
The Sum, Difference, Product and Quotient
of Functions
  • The budget balance D(t) is shown below
  • D(t) is also a function that denotes the federal
    government deficit (surplus) at any time t.
  • This function is the difference of the two
    functions R and S.
  • D(t) has the same domain as R(t) and S(t).

y
400 200 0 200 400
y D(t)
t
Billions of Dollars
t
1992 1994 1996 1998 2000
D(t)
Year
77
The Sum, Difference, Product and Quotient
of Functions
  • Most functions are built up from other, generally
    simpler functions.
  • For example, we may view the function f(x) 2x
    4 as the sum of the two functions g(x) 2x and
    h(x) 4.

78
The Sum, Difference, Product and Quotient of
Functions
  • Let f and g be functions with domains A and B,
    respectively.
  • The sum f g, the difference f g, and the
    product fg of f and g are functions with domain A
    n B and rule given by
  • (f g)(x) f(x) g(x) Sum
  • (f g)(x) f(x) g(x) Difference
  • (fg)(x) f(x)g(x) Product
  • The quotient f/g of f and g has domain A n B
    excluding all numbers x such that g(x) 0 and
    rule given by
  • Quotient

79
Example
  • Let and g(x) 2x 1.
  • Find the sum s, the difference d, the product p,
    and the quotient q of the functions f and g.
  • Solution
  • Since the domain of f is A 1,?) and the
    domain of g is B ( ?, ?), we see
    that the domain of s, d, and p is A
    n B 1,?).
  • The rules are as follows

80
Example
  • Let and g(x) 2x 1.
  • Find the sum s, the difference d, the product p,
    and the quotient q of the functions f and g.
  • Solution
  • The domain of the quotient function is 1,?)
    together with the restriction x ? ½ .
  • Thus, the domain is 1, ½) U ( ½,?).
  • The rule is as follows

81
Applied Example Cost Functions
  • Suppose Puritron, a manufacturer of water
    filters, has a monthly fixed cost of 10,000 and
    a variable cost of
  • 0.0001x2 10x (0 ? x ? 40,000)
  • dollars, where x denotes the number of filters
    manufactured per month.
  • Find a function C that gives the total monthly
    cost incurred by Puritron in the manufacture of x
    filters.

82
Applied Example Cost Functions
  • Solution
  • Puritrons monthly fixed cost is always 10,000,
    so it can be described by the constant function
  • F(x) 10,000
  • The variable cost can be described by the
    function
  • V(x) 0.0001x2 10x
  • The total cost is the sum of the fixed cost F and
    the variable cost V
  • C(x) V(x) F(x)
  • 0.0001x2 10x 10,000 (0 ? x ? 40,000)

83
Applied Example Cost Functions
  • Lets now consider profits
  • Suppose that the total revenue R realized by
    Puritron from the sale of x water filters is
    given by
  • R(x) 0.0005x2 20x (0 x 40,000)
  • Find
  • The total profit function for Puritron.
  • The total profit when Puritron produces 10,000
    filters per month.

84
Applied Example Cost Functions
  • Solution
  • The total profit P realized by the firm is the
    difference between the total revenue R and the
    total cost C
  • P(x) R(x) C(x)
  • ( 0.0005x2 20x) ( 0.0001x2 10x
    10,000)
  • 0.0004x2 10x 10,000
  • The total profit realized by Puritron when
    producing 10,000 filters per month is
  • P(x) 0.0004(10,000)2 10(10,000) 10,000
  • 50,000
  • or 50,000 per month.

85
The Composition of Two Functions
  • Another way to build a function from other
    functions is through a process known as the
    composition of functions.
  • Consider the functions f and g
  • Evaluating the function g at the point f(x), we
    find that
  • This is an entirely new function, which we could
    call h

86
The Composition of Two Functions
  • Let f and g be functions.
  • Then the composition of g and f is the function
    ggf (read g circle f ) defined by
  • (ggf )(x) g(f(x))
  • The domain of ggf is the set of all x in the
    domain of f such that f(x) lies in the domain of
    g.

87
Example
  • Let
  • Find
  • The rule for the composite function ggf.
  • The rule for the composite function fgg.
  • Solution
  • To find ggf, evaluate the function g at f(x)
  • To find fgg, evaluate the function f at g(x)

88
Applied Example Automobile Pollution
  • An environmental impact study conducted for the
    city of Oxnard indicates that, under existing
    environmental protection laws, the level of
    carbon monoxide (CO) present in the air due to
    pollution from automobile exhaust will be
    0.01x2/3 parts per million when the number of
    motor vehicles is x thousand.
  • A separate study conducted by a state government
    agency estimates that t years from now the number
    of motor vehicles in Oxnard will be 0.2t2 4t
    64 thousand.
  • Find
  • An expression for the concentration of CO in the
    air due to automobile exhaust t years from now.
  • The level of concentration 5 years from now.

89
Applied Example Automobile Pollution
  • Solution
  • Part (a)
  • The level of CO is described by the function
  • g(x) 0.01x2/3
  • where x is the number (in thousands) of motor
    vehicles.
  • In turn, the number (in thousands) of motor
    vehicles is described by the function
  • f(t) 0.2t2 4t 64
  • where t is the number of years from now.
  • Therefore, the concentration of CO due to
    automobile exhaust t years from now is given by
  • (ggf )(t) g(f(t)) 0.01(0.2t2 4t 64)2/3

90
Applied Example Automobile Pollution
  • Solution
  • Part (b)
  • The level of CO five years from now is
  • (ggf )(5) g(f(5)) 0.010.2(5)2 4(5)
    642/3
  • (0.01)892/3 0.20
  • or approximately 0.20 parts per million.

91
2.5
  • Linear Functions

92
Linear Function
  • The function f defined by
  • where m and b are constants, is called a linear
    function.

93
Applied Example Linear Depreciation
  • A Web server has an original value of 10,000 and
    is to be depreciated linearly over 5 years with a
    3000 scrap value.
  • Find an expression giving the book value at the
    end of year t.
  • What will be the book value of the server at the
    end of the second year?
  • What is the rate of depreciation of the server?

94
Applied Example Linear Depreciation
  • Solution
  • Let V(t) denote the Web servers book value at
    the end of the tth year. V is a linear function
    of t.
  • To find an equation of the straight line that
    represents the depreciation, observe that V
    10,000 when t 0 this tells us that the line
    passes through the point (0, 10,000).
  • Similarly, the condition that V 3000 when t 5
    says that the line also passes through the point
    (5, 3000).
  • Thus, the slope of the line is given by

95
Applied Example Linear Depreciation
  • Solution
  • Using the point-slope form of the equation of a
    line with point (0, 10,000) and slope m 1400,
    we obtain the required expression
  • The book value at the end of the second year is
    given by
  • or 7200.
  • The rate of depreciation of the server is given
    by the negative slope of the depreciation line m
    1400, so the rate of depreciation is 1400 per
    year.

96
Applied Example Linear Depreciation
  • Solution
  • The graph of V is

V
(0, 10,000)
10,000
(5, 3000)
3000
t
1 2 3 4 5 6
97
Cost, Revenue, and Profit Functions
  • Let x denote the number of units of a product
    manufactured or sold.
  • Then, the total cost function is
  • C(x) Total cost of manufacturing x units of
    the product
  • The revenue function is
  • R(x) Total revenue realized from the sale of x
    units of the product
  • The profit function is
  • P(x) Total profit realized from manufacturing
    and selling x units of the product

98
Applied Example Profit Function
  • Puritron, a manufacturer of water filters, has a
    monthly fixed cost of 20,000, a production cost
    of 20 per unit, and a selling price of 30 per
    unit.
  • Find the cost function, the revenue function, and
    the profit function for Puritron.
  • Solution
  • Let x denote the number of units produced and
    sold.
  • Then,

99
Finding the Point of Intersection
  • Suppose we are given two straight lines L1 and L2
    with equations
  • y m1x b1 and y m2x b2
  • (where m1, b1, m2, and b2 are constants) that
    intersect at the point P(x0, y0).
  • The point P(x0, y0) lies on the line L1 and so
    satisfies the equation y m1x b1.
  • The point P(x0, y0) also lies on the line L2 and
    so satisfies y m2x
    b2 as well.
  • Therefore, to find the point of intersection
    P(x0, y0) of the lines L1 and L2, we solve for x
    and y the system composed of the two equations
  • y m1x b1 and y m2x b2

100
Example
  • Find the point of intersection of the straight
    lines that have equations
  • y x 1 and y 2x 4
  • Solution
  • Substituting the value y as given in the first
    equation into the second equation, we obtain
  • Substituting this value of x into either one of
    the given equations yields y 2.
  • Therefore, the required point of intersection is
    (1, 2).

101
Example
  • Find the point of intersection of the straight
    lines that have equations
  • y x 1 and y 2x 4
  • Solution
  • The graph shows the point of intersection (1, 2)
    of the two lines

y
5 4 3 2 1
L1
(1, 2)
x
1 1 2 3 4 5
L2
102
Applied Example Break-Even Level
  • Prescott manufactures its products at a cost of
    4 per unit and sells them for 10 per unit.
  • If the firms fixed cost is 12,000 per month,
    determine the firms break-even point.
  • Solution
  • The revenue function R and the cost function C
    are given respectively by
  • Setting R(x) C(x), we obtain

103
Applied Example Break-Even Level
  • Prescott manufactures its products at a cost of
    4 per unit and sells them for 10 per unit.
  • If the firms fixed cost is 12,000 per month,
    determine the firms break-even point.
  • Solution
  • Substituting x 2000 into R(x) 10x gives
  • So, Prescotts break-even point is 2000 units of
    the product, resulting in a break-even revenue of
    20,000 per month.

104
2.6
  • Quadratic Functions

105
Quadratic Functions
  • A quadratic function is one of the form
  • where a, b, and c are constants and a ? 0.
  • For example, the function
  • is quadratic, with a 2, b 4, and c 3.

106
Quadratic Functions
  • Below is the graph of the quadratic function
  • The graph of a quadratic function is a curve
    called a parabola that opens upward or downward.

y
10 8 6 4 2
Parabola
x
2 1 1 2 3 4
107
Quadratic Functions
  • The parabola is symmetric with respect to a
    vertical line called the axis of symmetry.
  • The axis of symmetry also passes through the
    lowest or highest point of the parabola, which is
    called the vertex of the parabola.

Axis of symmetry
y
10 8 6 4 2
Parabola
Vertex (1, 1)
x
2 1 1 2 3 4
108
Quadratic Functions
  • We can use these properties to help us sketch the
    graph of a quadratic function.
  • Suppose we want to sketch the graph of
  • If we complete the square in x, we obtain
  • Note that (x 1)2 is nonnegative it equals to
    zero when x 1 and is greater than zero if x ?
    1.
  • Thus, we see that f(x) ? 2 for all values of
    x.
  • This tells us the vertex of the parabola is the
    point (1, 2).

109
Quadratic Functions
  • We know the vertex of the parabola is the point
    (1, 2) and that it is the minimum point of the
    graph, since f(x) ? 2 for all values of x.
  • Thus, the graph of f(x) 3x2 6x 1 looks as
    follows

y
4 2 2
x
2 2 4
Vertex (1, 2)
110
Properties of Quadratic Functions
  • Given f(x) ax2 bx c (a ? 0)
  • The domain of f is the set of all real numbers.
  • If a gt 0, the parabola opens upward, and if a lt
    0, it opens downward.
  • The vertex of the parabola is
  • The axis of symmetry of the parabola is
  • The x-intercepts (if any) are found by solving
    f(x) 0. The y-intercept is f(0)
    c.

111
Example
  • Given the quadratic function f(x) 2x2 5x
    2
  • Find the vertex of the parabola.
  • Find the x-intercepts (if any) of the parabola.
  • Sketch the parabola.
  • Solution
  • Here a 2, b 5, and c 2. therefore, the
    x-coordinate of the vertex of the parabola is
  • The y-coordinate of the vertex is therefore
    given by
  • Thus, the vertex of the parabola is the point

112
Example
  • Given the quadratic function f(x) 2x2 5x
    2
  • Find the vertex of the parabola.
  • Find the x-intercepts (if any) of the parabola.
  • Sketch the parabola.
  • Solution
  • For the x-intercepts of the parabola, we solve
    the equation
  • using the quadratic formula with a 2, b 5,
    and c 2.
  • We find
  • Thus, the x-intercepts of the parabola are 1/2
    and 2.

113
Example
  • Given the quadratic function f(x) 2x2 5x
    2
  • Find the vertex of the parabola.
  • Find the x-intercepts (if any) of the parabola.
  • Sketch the parabola.
  • Solution
  • The sketch

y
Vertex
1 1 2
x-intercepts
x
1 1 2
y-intercept
114
Some Economic Models
  • Peoples decision on how much to demand or
    purchase of a given product depends on the price
    of the product
  • The higher the price the less they want to buy of
    it.
  • A demand function p d(x) can be used to
    describe this.

115
Some Economic Models
  • Similarly, firms decision on how much to supply
    or produce of a product depends on the price of
    the product
  • The higher the price, the more they want to
    produce of it.
  • A supply function p s(x) can be used to
    describe this.

116
Some Economic Models
  • The interaction between demand and supply will
    ensure the market settles to a market
    equilibrium
  • This is the situation at which quantity demanded
    equals quantity supplied.
  • Graphically, this situation occurs when the
    demand curve and the supply curve intersect
    where d(x) s(x).

117
Applied Example Supply and Demand
  • The demand function for a certain brand of
    bluetooth wireless headset is given by
  • The corresponding supply function is given by
  • where p is the expressed in dollars and x is
    measured in units of a thousand.
  • Find the equilibrium quantity and price.

118
Applied Example Supply and Demand
  • Solution
  • We solve the following system of equations
  • Substituting the second equation into the first
    yields
  • Thus, either x 400/9 (but this is not
    possible), or x 20.
  • So, the equilibrium quantity must be 20,000
    headsets.

119
Applied Example Supply and Demand
  • Solution
  • The equilibrium price is given by
  • or 40 per headset.

120
2.7
Functions and Mathematical Models
121
Mathematical Models
  • As we have seen, mathematics can be used to solve
    real-world problems.
  • We will now discuss a few more examples of
    real-world phenomena, such as
  • The solvency of the U.S. Social Security
    trust fund
  • Global warming

122
Mathematical Modeling
  • Regardless of the field from which the real-world
    problem is drawn, the problem is analyzed using a
    process called mathematical modeling.
  • The four steps in this process are

Real-world problem
Mathematical model
Formulate
Solve
Test
Solution of real-world problem
Solution of mathematical model
Interpret
123
Modeling With Polynomial Functions
  • A polynomial function of degree n is a function
    of the form
  • where n is a nonnegative integer and the numbers
    a0, a1, . an are
    constants called the coefficients of the
    polynomial function.
  • Examples
  • The function below is polynomial function of
    degree 5

124
Modeling With Polynomial Functions
  • A polynomial function of degree n is a function
    of the form
  • where n is a nonnegative integer and the numbers
    a0, a1, . an are
    constants called the coefficients of the
    polynomial function.
  • Examples
  • The function below is polynomial function of
    degree 3

125
Applied Example Global Warming
  • The increase in carbon dioxide (CO2) in the
    atmosphere is a major cause of global warming.
  • Below is a table showing the average amount of
    CO2, measured in parts per million volume (ppmv)
    for various years from 1958 through 2007

Year 1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount 315 325 330 335 345 355 365 375 380
126
Applied Example Global Warming
Year 1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount 315 325 330 335 345 355 365 375 380
  • Below is a scatter plot associated with these
    data

y (ppmv)
380 360 340 320
t (years)
10 20 30 40 50
127
Applied Example Global Warming
Year 1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount 315 325 330 335 345 355 365 375 380
  • A mathematical model giving the approximate
    amount of CO2 is given by

y (ppmv)
380 360 340 320
t (years)
10 20 30 40 50
128
Applied Example Global Warming
Year 1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount 315 325 330 335 345 355 365 375 380
  1. Use the model to estimate the average amount of
    atmospheric CO2 in 1980 (t 23).
  2. Assume that the trend continued and use the model
    to predict the average amount of atmospheric CO2
    in 2010.

129
Applied Example Global Warming
Year 1958 1970 1974 1978 1985 1991 1998 2003 2007
Amount 315 325 330 335 345 355 365 375 380
  • Solution
  • The average amount of atmospheric CO2 in 1980 is
    given by
  • or approximately 338 ppmv.
  • Assuming that the trend will continue, the
    average amount of atmospheric CO2 in 2010 will be

130
Applied Example Social Security Trust Fund
  • The projected assets of the Social Security trust
    fund (in trillions of dollars) from 2008 through
    2040 are given by
  • The scatter plot associated with these data is

Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0
y (trillion)
6 4 2
t (years)
5 10 15 20 25 30
131
Applied Example Social Security Trust Fund
  • The projected assets of the Social Security trust
    fund (in trillions of dollars) from 2008 through
    2040 are given by
  • A mathematical model giving the approximate value
    of assets in the trust fund (in trillions of
    dollars) is

Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0
y (trillion)
6 4 2
t (years)
5 10 15 20 25 30
132
Applied Example Social Security Trust Fund
Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0
  1. The first baby boomers will turn 65 in 2011. What
    will be the assets of the Social Security trust
    fund at that time?
  2. The last of the baby boomers will turn 65 in
    2029. What will the assets of the trust fund be
    at the time?
  3. Use the graph of function A(t) to estimate the
    year in which the current Social Security system
    will go broke.

133
Applied Example Social Security Trust Fund
Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0
  • Solution
  • The assets of the Social Security fund in 2011 (t
    3) will be
  • or approximately 3.18 trillion.
  • The assets of the Social Security fund in 2029
    (t 21) will be
  • or approximately 5.59 trillion.

134
Applied Example Social Security Trust Fund
Year 2008 2011 2014 2017 2020 2023 2026 2029 2032 2035 2038 2040
Assets 2.4 3.2 4.0 4.7 5.3 5.7 5.9 5.6 4.9 3.6 1.7 0
  • Solution
  • The graph shows that function A crosses the
    t-axis at about t 32, suggesting the system
    will go broke by 2040

y (trillion)
6 4 2
t (years)
5 10 15 20 25 30
135
Rational and Power Functions
  • A rational function is simply the quotient of two
    polynomials.
  • In general, a rational function has the form
  • where f(x) and g(x) are polynomial functions.
  • Since the division by zero is not allowed, we
    conclude that the domain of a rational function
    is the set of all real numbers except the zeros
    of g (the roots of the equation g(x) 0)

136
Rational and Power Functions
  • Examples of rational functions

137
Rational and Power Functions
  • Functions of the form
  • where r is any real number, are called power
    functions.
  • We encountered examples of power functions
    earlier in our work.
  • Examples of power functions

138
Rational and Power Functions
  • Many functions involve combinations of rational
    and power functions.
  • Examples

139
Applied Example Driving Costs
  • A study of driving costs based on a 2007
    medium-sized sedan found the following average
    costs (car payments, gas, insurance, upkeep, and
    depreciation), measured in cents per mile
  • A mathematical model giving the average cost in
    cents per mile is
  • where x (in thousands) denotes the number of
    miles the car is driven in 1 year.

Miles/year, x 5000 10,000 15,000 20,000
Cost/mile, y () 83.8 62.9 52.2 47.1
140
Applied Example Driving Costs
Miles/year, x 5000 10,000 15,000 20,000
Cost/mile, y () 83.8 62.9 52.2 47.1
  • Below is the scatter plot associated with this
    data

y ()
140 120 100 80 60 40 20
C(x)
x (?1000 miles/year)
5 10 15 20 25
141
Applied Example Driving Costs
Miles/year, x 5000 10,000 15,000 20,000
Cost/mile, y () 83.8 62.9 52.2 47.1
  • Using this model, estimate the average cost of
    driving a 2007 medium-sized sedan 8,000 miles per
    year and 18,000 miles per year.
  • Solution
  • The average cost for driving a car 8,000 miles
    per year is
  • or approximately 68.8/mile.

142
Applied Example Driving Costs
Miles/year, x 5000 10,000 15,000 20,000
Cost/mile, y () 83.8 62.9 52.2 47.1
  • Using this model, estimate the average cost of
    driving a 2007 medium-sized sedan 8,000 miles per
    year and 18,000 miles per year.
  • Solution
  • The average cost for driving a car 18,000 miles
    per year is
  • or approximately 48.95/mile.

143
Constructing Mathematical Models
  • Some mathematical models can be constructed using
    elementary geometric and algebraic arguments.
  • Guidelines for constructing mathematical
    models
  • Assign a letter to each variable mentioned in the
    problem. If appropriate, draw and label a figure.
  • Find an expression for the quantity sought.
  • Use the conditions given in the problem to write
    the quantity sought as a function f of
    one variable.
  • Note any restrictions to be placed on the domain
    of f by the nature of the
    problem.

144
Applied Example Enclosing an Area
  • The owner of the Rancho Los Feliz has 3000 yards
    of fencing with which to enclose a rectangular
    piece of grazing land along the straight portion
    of a river.
  • Fencing is not required along the river.
  • Letting x denote the width of the rectangle, find
    a function f in the variable x giving the area of
    the grazing land if she uses all of the fencing.

145
Applied Example Enclosing an Area
  • Solution
  • This information was given
  • The area of the rectangular grazing land is A
    xy.
  • The amount of fencing is 2x y which must equal
    3000 (to use all the fencing), so
  • 2x y 3000
  • Solving for y we get
  • y 3000 2x
  • Substituting this value of y into the expression
    for A gives
  • A x(3000 2x) 3000x 2x2
  • Finally, x and y represent distances, so they
    must be nonnegative, so x ? 0 and y 3000 2x ?
    0 (or x ? 1500).
  • Thus, the required function is
  • f(x) 3000x 2x2 (0 ? x ? 1500)

146
Applied Example Charter-Flight Revenue
  • If exactly 200 people sign up for a charter
    flight, Leisure World Travel Agency charges 300
    per person.
  • However, if more than 200 people sign up for the
    flight (assume this is the case), then each fare
    is reduced by 1 for each additional person.
  • Letting x denote the number of passengers above
    200, find a function giving the revenue realized
    by the company.

147
Applied Example Charter-Flight Revenue
  • Solution
  • This information was given.
  • If there are x passengers above 200, then the
    number of passengers signing up for the flight is
    200 x.
  • The fare will be (300 x) dollars per passenger.
  • The revenue will be
  • R (200 x)(300 x)
  • x2 100x 60,000
  • The quantities must be positive, so x ? 0 and 300
    x ? 0 (or x ? 300).
  • So the required function is
  • f(x) x2 100x 60,000 (0 ? x ? 300)

148
End of Chapter
About PowerShow.com