1 The Cartesian Coordinate System Straight Lines Linear Functions and Mathematical Models Intersection of Straight Lines The Method of Least

11

The Cartesian Coordinate SystemThe Cartesian Coordinate System Straight LinesStraight Lines Linear Functions and Mathematical Linear Functions and Mathematical

ModelsModels Intersection of Straight LinesIntersection of Straight Lines The Method of Least SquaresThe Method of Least Squares

Straight Lines and Linear Functions Straight Lines and Linear Functions

1.11.1The Cartesian Coordinate SystemThe Cartesian Coordinate System

C(h, k)

h

k r

P(x, y)

y

x

– 4 – 3 – 2 – 1 0 1 2 3 4

We can represent real numbers We can represent real numbers geometricallygeometrically by points on by points on a a real numberreal number, or, or coordinate coordinate,, line line::

This line includes This line includes allall real numbersreal numbers.. Exactly Exactly oneone point on the line point on the line is associated with is associated with eacheach real real

numbernumber, and vice-versa , and vice-versa (one dimensional space)(one dimensional space)..

OriginOrigin

PositivePositive Direction DirectionNegativeNegative Direction Direction

The Cartesian Coordinate SystemThe Cartesian Coordinate System

2 3


The The Cartesian coordinate systemCartesian coordinate system extends this concept to a extends this concept to a plane plane (two dimensional space)(two dimensional space) by adding a by adding a vertical axisvertical axis..

– 4 – 3 – 2 – 1 1 2 3 4

44

33

22

11

– – 11

–– 22

–– 33

–– 44

The Cartesian Coordinate System

The horizontal line is called the x-axis, and the vertical line is called the y-axis.

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4

x

y


The point where these two lines The point where these two lines intersectintersect is called the is called the originorigin..

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4

xx

yy

OriginOrigin


In theIn the xx-axis-axis, , positive numberspositive numbers are to the are to the rightright and negative and negative numbersnumbers are to the are to the leftleft of the origin. of the origin.

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4

xx

yy

PositivePositive Direction DirectionNegativeNegative Direction Direction


In theIn the yy-axis-axis, , positive numberspositive numbers are are aboveabove and and negativenegative numbersnumbers are are belowbelow the origin. the origin.

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4

xx

yy

Pos

itive

Pos

itive

Dir

ecti

on D

irec

tion

Neg

ativ

eN

egat

ive

Dir

ecti

on D

irec

tion

(–(– 2, 4)2, 4)

(–1, (–1, –– 2)2)

(4, 3)(4, 3)


A A pointpoint in the in the planeplane can now be represented uniquely in this can now be represented uniquely in this coordinate system by ancoordinate system by an ordered pair of numbers ordered pair of numbers ((xx, , yy))..

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4

xx

yy

(3, –1)(3, –1)


The axes divide the plane into The axes divide the plane into four quadrantsfour quadrants as shown below. as shown below.

– 4 – 3 – 2 – 1 1 2 3 4

4

3

2

1

– 1

– 2

– 3

– 4

xx

yy

Quadrant IQuadrant I(+, +)(+, +)

Quadrant IIQuadrant II(–, +)(–, +)

Quadrant IVQuadrant IV(+, –)(+, –)

Quadrant IIIQuadrant III(–, –)(–, –)

The Distance FormulaThe Distance Formula

The The distancedistance between any two pointsbetween any two points in the plane can be in the plane can be expressed in terms of the coordinates of the pointsexpressed in terms of the coordinates of the points..

Distance formulaDistance formula The distance The distance dd between two points between two points PP11((xx11, , yy11))

and and PP22((xx22, , yy22)) in the plane is given by in the plane is given by

2 2

2 1 2 1d x x y y 2 2

2 1 2 1d x x y y

Examples Examples

Find the Find the distancedistance between the points between the points (–(– 4, 3)4, 3) and and (2, 6)(2, 6)..

SolutionSolution Let Let P1(–(– 4, 34, 3) and and P2(2, 6(2, 6) be points in the plane. be points in the plane.

We have We have

xx11 = = –– 44 yy11 = = 33 xx22 = = 22 yy22 = = 66

Using the Using the distance formuladistance formula, we have, we have

2 2

2 1 2 1

22

2 2

2 ( 4) 6 3

6 3 45 3 5

d x x y y

Example 1, page 4

ExamplesExamples

Let Let PP((xx, , yy)) denote a point lying on the denote a point lying on the circlecircle with with radiusradius rr and and centercenter CC((hh, , kk)). Find a relationship between . Find a relationship between xx and and yy..

SolutionSolution By the definition of a circle, the distance between By the definition of a circle, the distance between PP((xx, , yy))

and and CC((hh, , kk)) is is rr.. With the distance formula With the distance formula

we getwe get

Squaring both sides givesSquaring both sides gives

2 2x h y k r

2 2 2x h y k r

CC((hh, , kk))

hh

kk rr

PP((xx, , yy))

yy

xx

Example 3, page 4

Equation of a CircleEquation of a Circle

An equation of a circle with An equation of a circle with centercenter CC((hh, , kk)) and and radiusradius rr is given by is given by

2 2 2x h y k r 2 2 2x h y k r

ExamplesExamples

Find an Find an equation of the circle equation of the circle withwith radiusradius 22 and and centercenter (–1, 3)(–1, 3). .

SolutionSolution We use the We use the circle formulacircle formula with with rr = 2 = 2, , hh = = –1–1, and, and kk = 3 = 3::

2 2 2

22 2

2 2

( 1) 3 2

1 3 4

x h y k r

x y

x y

(–1, 3)(–1, 3)

––11

33

22

yy

xx

Example 4, page 5

ExamplesExamples

Find an Find an equation of the circle equation of the circle withwith radiusradius 33 and and centercenter located at the origin. located at the origin.

SolutionSolution We use the We use the circle formulacircle formula with with rr = 3 = 3, , hh = = 00, and, and kk = 0 = 0::

2 2 2

2 2 2

2 2

0 0 3

9

x h y k r

x y

x y

33

yy

xx

Example 4, page 5

1.21.2Straight LinesStraight Lines

11 22 33 44 55 66

(2, 5)(2, 5)

yy

xx

LL

y y = = ––44

x x = 3= 366

55

44

33

22

11 (5, 1)(5, 1)

Slope of a Vertical LineSlope of a Vertical Line

Let Let LL denote the unique straight line that passes through denote the unique straight line that passes through the two distinct points the two distinct points ((xx11, , yy11)) and and ((xx22, , yy22))..

If If xx11 == xx22, then , then LL is a is a vertical linevertical line, and the , and the slopeslope is is

undefinedundefined..

((xx11, , yy11))

((xx22, , yy22))

yy

xx

LL

Slope of a Nonvertical LineSlope of a Nonvertical Line

If If ((xx11, , yy11)) and and ((xx22, , yy22)) are two distinct points on a are two distinct points on a

nonvertical linenonvertical line LL, then the , then the slopeslope mm of of LL is given by is given by

((xx11, , yy11))

((xx22, , yy22))

yy

xx

2 1

2 1

y y ym

x x x

LL

yy22 – – yy11 = = yy

xx22 – – xx11 = = xx


If If mm > 0 > 0, the line , the line slants slants upwardupward fromfrom left to rightleft to right..

xx

LL

y y = 1= 1

x x = 1= 1

m m = 1= 1

yy


If If mm > 0 > 0, the line , the line slants slants upwardupward fromfrom left to rightleft to right..

yy

xx

LL

y y = 2= 2

x x = 1= 1

m m = 2= 2

m m = –1= –1


If If mm < 0 < 0, the line , the line slants slants downwarddownward fromfrom left to rightleft to right..

xx

LL

y y = –1= –1

x x = 1= 1

yy

m m = –2= –2


If If mm < 0 < 0, the line , the line slants slants downwarddownward fromfrom left to rightleft to right..

yy

xx

LL

y y = –2= –2

x x = 1= 1

11 22 33 44 55 66

(2, 5)(2, 5)

ExamplesExamples

Sketch the straight line that passes through the point Sketch the straight line that passes through the point (2, 5)(2, 5) and has slope and has slope –– 4/34/3..

SolutionSolution

1.1. Plot the Plot the pointpoint (2, 5)(2, 5)..

2.2. A A slopeslope of of –– 4/34/3 means means that if that if xx increasesincreases by by 33, , yy decreasesdecreases by by 44..

3.3. Plot the Plot the resulting resulting pointpoint (5, 1)(5, 1)..

4.4. Draw a Draw a lineline through through the two points.the two points.

yy

xx

LL

y y = –= – 44

x x = 3= 366

55

44

33

22

11 (5, 1)(5, 1)

ExamplesExamples

Find the Find the slopeslope m m of the line that goes through the of the line that goes through the pointspoints (–1, 1)(–1, 1) and and (5, 3)(5, 3). .

SolutionSolution Choose Choose ((xx11, , yy11)) to be to be (–1, 1)(–1, 1) and and ((xx22, , yy22)) to be to be (5, 3)(5, 3). .

With With xx11 = = –1–1, , yy11 = 1 = 1, , xx22 = = 55, , yy22 = = 33, we find, we find

2 1

2 1

3 1 2 1

5 ( 1) 6 3

y ym

x x

2 1

2 1

3 1 2 1

5 ( 1) 6 3

y ym

x x

Example 2, page 11

ExamplesExamples


SolutionSolution Choose Choose ((xx11, , yy11)) to be to be (–2, 5)(–2, 5) and and ((xx22, , yy22)) to be to be (3, 5)(3, 5). .

With With xx11 = = –2–2, , yy11 = 5 = 5, , xx22 = = 33, , yy22 = = 55, we find, we find

2 1

2 1

5 5 00

3 ( 2) 5

y ym

x x

2 1

2 1

5 5 00

3 ( 2) 5

y ym

x x

Example 3, page 11

––2 2 –1 –1 11 22 33 44

ExamplesExamples


SolutionSolution The slope of a The slope of a horizontal linehorizontal line is is zerozero::

yy

xx

LL

66

44

33

22

11

(–2, 5)(–2, 5) (3, 5)(3, 5)

m m = 0= 0

Example 3, page 11

Parallel LinesParallel Lines

Two distinct lines are Two distinct lines are parallelparallel if and only if their if and only if their slopes are equalslopes are equal or their or their slopes are undefinedslopes are undefined..

ExampleExample

Let Let LL11 be a line that passes through the points be a line that passes through the points (–2, 9)(–2, 9) and and

(1, 3)(1, 3), and let , and let LL22 be the line that passes through the points be the line that passes through the points

(–(– 4, 10)4, 10) and and (3, –(3, – 4)4). . Determine whether Determine whether LL11 and and LL22 are parallel. are parallel.

SolutionSolution The The slopeslope mm11 of of LL11 is given by is given by

The The slopeslope mm22 of of LL22 is given by is given by

Since Since mm11 == mm22, the lines , the lines LL11 and and LL22 are in fact are in fact parallelparallel..

1

3 92

1 ( 2)m

1

3 92

1 ( 2)m

2

4 102

3 ( 4)m

2

4 102

3 ( 4)m

Example 4, page 12

Equations of LinesEquations of Lines

Let Let LL be a be a straight linestraight line parallelparallel to the to the yy-axis-axis..

Then Then LL crossescrosses the the xx-axis-axis at at some some pointpoint ((aa, 0), 0) , with the , with the xx-coordinate-coordinate given by given by x = ax = a, , where where a a is a real number.is a real number.

Any other point on Any other point on LL has has the form the form ((aa, , )), where , where is an appropriate number.is an appropriate number.

The The vertical linevertical line LL can can therefore be described astherefore be described as

x = ax = a

((aa, , ))

yy

xx

LL

((aa, 0), 0)

y y

y

Equations of LinesEquations of Lines

Let Let LL be a be a nonvertical linenonvertical line with a slope with a slope mm.. Let Let ((xx11, , yy11)) be a be a fixed pointfixed point lying on lying on LL, and let , and let ((xx, , yy)) be a be a

variable point on variable point on LL distinct from distinct from ((xx11, , yy11))..

Using the slope formula by letting Using the slope formula by letting ((xx, , yy) =) = ((xx22, , yy22)), we get, we get

Multiplying both sides by Multiplying both sides by xx – – x x11 we get we get

1

1

y ym

x x

1 1( )y y m x x

Point-Slope FormPoint-Slope Form

1 1( )y y m x x 1 1( )y y m x x

An equation of the line that has slope An equation of the line that has slope mm and and passes through point passes through point ((xx11, , yy11)) is given by is given by

ExamplesExamples

Find an equation of the line that passes through the point Find an equation of the line that passes through the point (1, 3)(1, 3) and has slope and has slope 22..

SolutionSolution Use the Use the point-slope formpoint-slope form

Substituting for Substituting for pointpoint (1, 3)(1, 3) and and slopeslope mm = 2 = 2, we obtain, we obtain

SimplifyingSimplifying we get we get

1 1( )y y m x x

3 2( 1)y x

2 1 0x y

Example 5, page 13

ExamplesExamples

Find an equation of the line that passes through the points Find an equation of the line that passes through the points (–3, 2)(–3, 2) and and (4, –1)(4, –1)..

SolutionSolution The slope is given byThe slope is given by

Substituting in the Substituting in the point-slope formpoint-slope form for point for point (4, –1)(4, –1) and and slope slope mm = – 3/7 = – 3/7, we obtain, we obtain

31 ( 4)

7y x

3 7 5 0x y

2 1

2 1

1 2 3

4 ( 3) 7

y ym

x x

7 7 3 12y x

Example 6, page 14

Perpendicular LinesPerpendicular Lines

If If LL11 and and LL22 are two distinct nonvertical lines that are two distinct nonvertical lines that

have slopes have slopes mm11 and and mm22, respectively, then , respectively, then LL11 is is

perpendicularperpendicular to to LL22 (written (written LL1 1 ┴┴ LL22) if and only if ) if and only if

12

1m

m1

2

1m

m

ExampleExample

Find the equation of the line Find the equation of the line LL11 that passes through the that passes through the point point (3, 1)(3, 1) and is perpendicular to the line and is perpendicular to the line LL22 described by described by

SolutionSolution LL22 is described in is described in point-slope formpoint-slope form, so its , so its slopeslope is is mm22 = 2 = 2.. Since the lines are Since the lines are perpendicularperpendicular, the , the slopeslope of of LL11 must be must be

mm11 = –1/2 = –1/2

Using the Using the point-slope formpoint-slope form of the equation for of the equation for LL11 we obtain we obtain

3 2( 1)y x

11 ( 3)

22 2 3

2 5 0

y x

y x

x y

Example 7, page 14

((aa, 0), 0)

(0, (0, bb))

Crossing the AxisCrossing the Axis

A straight line A straight line LL that is that is neither horizontal nor verticalneither horizontal nor vertical cuts the cuts the x-x-axisaxis and the and the y-y-axisaxis at, say, points at, say, points ((aa, 0), 0) and and (0, (0, bb)), respectively., respectively.

The numbers The numbers aa and and b b are called the are called the xx--intercept intercept andand y y--interceptintercept, respectively, of , respectively, of LL..

yy

xx

LL

yy-intercept-intercept

xx-intercept-intercept

Slope-Intercept FormSlope-Intercept Form

An equation of the line that has An equation of the line that has slopeslope mm and and intersectsintersects the the yy-axis-axis at the at the pointpoint (0, (0, bb)) is given by is given by

y y = = mxmx + + bb

ExamplesExamples

Find the equation of the line that has Find the equation of the line that has slopeslope 33 and and yy--interceptintercept of of –– 44..

SolutionSolution We substitute We substitute mm = 3 = 3 and and bb = – = – 44 into into y y = = mxmx + + bb and get and get

y y = 3= 3xx – 4 – 4

Example 8, page 15

ExamplesExamples

Determine the Determine the slopeslope and and yy-intercept -intercept of the line whose of the line whose equation is equation is 33xx – 4 – 4y y = 8= 8..

SolutionSolution Rewrite the given equation in the Rewrite the given equation in the slope-intercept formslope-intercept form..

Comparing to Comparing to y y = = mxmx + + bb, we find that , we find that mm = ¾ = ¾ and and bb = – = – 22.. So, the So, the slopeslope is is ¾¾ and the and the yy-intercept-intercept is is –– 22..

3 4 8

4 8 3

32

4

x y

y x

y x

Example 9, page 15

Applied ExampleApplied Example

Suppose an art object Suppose an art object purchasedpurchased for for $50,000$50,000 is expected to is expected to appreciate in valueappreciate in value at a at a constant rateconstant rate of of $5000$5000 per year for per year for the next the next 55 years. years.

Write an Write an equationequation predicting the value of the art object for predicting the value of the art object for any given year.any given year.

What will be its What will be its valuevalue 33 years after the purchase? years after the purchase?SolutionSolution Let Let xx == time (in years) since the object was purchasedtime (in years) since the object was purchased

yy == value of object (in dollars)value of object (in dollars)

Then, Then, yy = 50,000 = 50,000 when when xx = 0 = 0, so the , so the yy-intercept-intercept is is bb = = 50,00050,000.. Every year the value rises by Every year the value rises by 50005000, so the , so the slopeslope is is mm = 5000 = 5000.. Thus, the equation must be Thus, the equation must be yy = 5000 = 5000xx + 50,000 + 50,000.. After After 33 years the years the value of the objectvalue of the object will be will be $65,000$65,000::

yy = 5000(3) + 50,000 = 65,000 = 5000(3) + 50,000 = 65,000Applied Example 11, page 16

General Form of a Linear EquationGeneral Form of a Linear Equation

The equationThe equation

AxAx + + ByBy + + CC = 0 = 0

where where AA, , BB, and , and CC are are constantsconstants and and AA and and BB are not both zero, is called the are not both zero, is called the general form general form of a linear equationof a linear equation in the in the variablesvariables xx and and yy..

General Form of a Linear EquationGeneral Form of a Linear Equation

An equation of a An equation of a straight linestraight line is a is a linear linear equationequation; conversely, every ; conversely, every linear equationlinear equation represents a represents a straight linestraight line..

ExampleExample

Sketch the straight line represented by the equationSketch the straight line represented by the equation

33xx – 4 – 4yy – 12 = 0 – 12 = 0

SolutionSolution Since every straight line is Since every straight line is uniquely determineduniquely determined by by two two

distinct pointsdistinct points, we need find only two such points through , we need find only two such points through which the line passes in order to sketch it.which the line passes in order to sketch it.

For convenience, let’s compute the For convenience, let’s compute the xx-- and and yy-intercepts-intercepts::✦ Setting Setting yy = 0= 0, we find , we find xx = 4= 4; so the ; so the xx-intercept-intercept is is 44..

✦ Setting Setting xx = 0= 0, we find , we find yy = –3= –3; so the ; so the yy-intercept-intercept is is –3–3.. Thus, Thus, the line goes through the pointsthe line goes through the points (4, 0)(4, 0) and and (0, –3)(0, –3)..

Example 12, page 17

ExampleExample

Sketch the straight line represented by the equationSketch the straight line represented by the equation

33xx – 4 – 4yy – 12 = 0 – 12 = 0

SolutionSolution Graph the line going through the points Graph the line going through the points (4, 0)(4, 0) and and (0, –3)(0, –3)..

11 22 33 44 55 66

(0, –(0, – 3)3)

yy

xx

LL11

–– 11

–– 22

–– 33

–– 44

(4, 0)(4, 0)

Example 12, page 17

Equations of Straight LinesEquations of Straight Lines

Vertical line:Vertical line: xx = = aa

Horizontal line:Horizontal line: yy = = bb

Point-slope form:Point-slope form: yy – – yy1 1 = = mm((xx – – xx11))

Slope-intercept form:Slope-intercept form: yy = = mxmx + b + b

General Form:General Form: Ax Ax + + By By + + CC = 0 = 0

1.31.3Linear Functions and Mathematical ModelsLinear Functions and Mathematical Models

Real - world problem

Mathematical model

Solution of real-world Problem

Solution of mathematical model

Formulate

Interpret

SolveTest

Mathematical ModelingMathematical Modeling

Mathematics can be used to solve Mathematics can be used to solve real-world problemsreal-world problems.. Regardless of the field from which the real-world problem Regardless of the field from which the real-world problem

is drawn, the problem is analyzed using a is drawn, the problem is analyzed using a processprocess called called mathematical modelingmathematical modeling..

The The four stepsfour steps in this process are: in this process are:

Real-world Real-world problemproblem Mathematical modelMathematical model

Solution of real- Solution of real- world Problemworld Problem

Solution of Solution of mathematical modelmathematical model

FormulateFormulate

InterpretInterpret

SolveSolveTestTest

FunctionsFunctions

A function A function ff is a is a rulerule that assigns to each value of that assigns to each value of xx one and one and only one value of only one value of yy..

The value The value yy is normally denoted by is normally denoted by ff((xx)), emphasizing the , emphasizing the dependencydependency of of yy on on xx..

ExampleExample

Let Let xx and and yy denote the denote the radiusradius and and areaarea of a of a circlecircle, , respectively. respectively.

From From elementaryelementary geometrygeometry we have we have

yy = = xx22

This equation defines This equation defines yy as a as a functionfunction of of xx, since for each , since for each admissible value of admissible value of xx there corresponds there corresponds precisely oneprecisely one number number yy = = xx22 giving the giving the areaarea of the circle. of the circle.

The The area functionarea function may be written as may be written as

ff((xx) = ) = xx22

To To computecompute the the areaarea of a circle with a of a circle with a radiusradius of of 55 inches, we inches, we simply simply replacereplace x x in the equation by the number in the equation by the number 55::

ff(5) = (5) = 5522)= 25)= 25

Domain and RangeDomain and Range

Suppose we are given the function Suppose we are given the function y y == ff((xx)).. The variable The variable xx is referred to as the is referred to as the independent variableindependent variable, ,

and the variable and the variable y y is called the is called the dependent variabledependent variable.. The set of all the possible values of The set of all the possible values of xx is called the is called the domaindomain of of

the function the function ff.. The set of all the values of The set of all the values of ff((xx)) resulting from all the possible resulting from all the possible

values of values of xx in its domain is called the in its domain is called the rangerange of of ff.. The output The output ff((xx)) associated with an input associated with an input xx is is uniqueunique: :

✦ EachEach xx must correspond to must correspond to one and onlyone and only oneone value of value of ff((xx))..

Linear FunctionLinear Function

The function The function ff defined by defined by

where where mm and and bb are are constantsconstants, is called a , is called a linear linear functionfunction..

( )f x mx b ( )f x mx b

Applied Example:Applied Example: U.S. Health-Care Expenditures U.S. Health-Care Expenditures

Because the over-65 population will be Because the over-65 population will be growing more rapidlygrowing more rapidly in the next few decades, in the next few decades, health-care spendinghealth-care spending is expected to is expected to increaseincrease significantly in the coming decades. significantly in the coming decades.

The following table gives the The following table gives the projected U.S. health-care projected U.S. health-care expendituresexpenditures (in trillions of dollars) from (in trillions of dollars) from 20052005 through through 20102010::

A A mathematical modelmathematical model giving the giving the approximate U.S. health-approximate U.S. health-care expenditurescare expenditures over the period in question is given by over the period in question is given by

wherewhere t t is measured in years, with is measured in years, with t t = 0= 0 corresponding to corresponding to 20052005..

( ) 0.178 1.989S t t ( ) 0.178 1.989S t t

YearYear 20052005 20062006 20072007 20082008 20092009 20102010

ExpenditureExpenditure 2.002.00 2.172.17 2.342.34 2.502.50 2.692.69 2.902.90

Applied Example 1, page 29


We haveWe have

a.a. Sketch the graphSketch the graph of the of the functionfunction SS and the given and the given datadata on on the same set of axes.the same set of axes.

b.b. Assuming that the trend continues, how much will Assuming that the trend continues, how much will U.S. U.S. health-care expenditureshealth-care expenditures be in be in 20112011??

c.c. What is the projected What is the projected rate of increaserate of increase of of U.S. health-U.S. health-care expenditurescare expenditures over the period in question? over the period in question?

( ) 0.178 1.989S t t ( ) 0.178 1.989S t t

YearYear 20052005 20062006 20072007 20082008 20092009 20102010




We haveWe have

SolutionSolutiona.a. TheThe graph graph of the given of the given datadata and of the and of the functionfunction SS is: is:

( ) 0.178 1.989S t t ( ) 0.178 1.989S t t

YearYear 20052005 20062006 20072007 20082008 20092009 20102010


S(t)S(t)

tt

3.03.0

2.82.8

2.62.6

2.42.4

2.22.2

2.02.0

11 22 33 44 55

( ) 0.178 1.989S t t ( ) 0.178 1.989S t t



We haveWe have

SolutionSolutionb.b. The The projected U.S. health-care expendituresprojected U.S. health-care expenditures in in 20112011 is is

or approximately or approximately $3.06$3.06 trillion. trillion.

( ) 0.178 1.989S t t ( ) 0.178 1.989S t t

YearYear 20052005 20062006 20072007 20082008 20092009 20102010


(6) 0.178(6) 1.989 3.057S (6) 0.178(6) 1.989 3.057S



We haveWe have

SolutionSolutionc.c. The function The function SS is is linearlinear, so the , so the rate of increaserate of increase of the of the

U.S. health-care expendituresU.S. health-care expenditures is given by the is given by the slope slope of of thethe straight line straight line represented by represented by SS, which is , which is approximately approximately $0.18$0.18 trillion per year. trillion per year.

( ) 0.178 1.989S t t ( ) 0.178 1.989S t t

YearYear 20052005 20062006 20072007 20082008 20092009 20102010



Cost, Revenue, and Profit FunctionsCost, Revenue, and Profit Functions

Let Let xx denote the number of denote the number of units of a productunits of a product manufactured or sold.manufactured or sold.

Then, the Then, the total cost functiontotal cost function is is

CC((xx)) == Total cost of manufacturingTotal cost of manufacturing xx units of the productunits of the product

The The revenue functionrevenue function is is

RR((xx)) == Total revenue realized from Total revenue realized from the sale ofthe sale of xx units of the productunits of the product

The The profit functionprofit function is is

PP((xx)) == Total profit realized from Total profit realized from manufacturing and selling manufacturing and selling xx units of the units of the productproduct

Applied Example:Applied Example: Profit Function Profit Function

Puritron, a manufacturer of water filters, has a monthly Puritron, a manufacturer of water filters, has a monthly fixed costfixed cost of of $20,000$20,000, a , a production costproduction cost of of $20$20 per unit, per unit, and a and a selling priceselling price of of $30$30 per unit. per unit.

Find the Find the cost functioncost function, the , the revenue functionrevenue function, and the , and the profit functionprofit function for Puritron. for Puritron.

SolutionSolution Let Let xx denote the number of denote the number of units produced and soldunits produced and sold.. Then,Then,

( ) 20 20,000C x x ( ) 20 20,000C x x

( ) 30R x x( ) 30R x x

( ) ( ) ( )

30 (20 20,000)

10 20,000

P x R x C x

x x

x

( ) ( ) ( )

30 (20 20,000)

10 20,000

P x R x C x

x x

x


1.41.4Intersections of Straight LinesIntersections of Straight Lines

––11 11 22 33 44 55

55

44

33

22

11

yy

xx

(1, 2)(1, 2)

LL11

LL22

Finding the Point of IntersectionFinding the Point of Intersection

Suppose we are given Suppose we are given two straight linestwo straight lines LL11 and and LL22 with with

equations equations yy = = mm11xx + + bb11 and and yy = = mm22xx + + bb22

(where (where mm11, , bb11, , mm22, and , and bb22 are are constantsconstants) that ) that intersectintersect at the at the

point point PP((xx00, , yy00))..

The point The point PP((xx00, , yy00)) lies on the line lies on the line LL11 and so and so satisfies the satisfies the

equationequation yy = = mm11xx + + bb11..

The point The point PP((xx00, , yy00)) also lies on the line also lies on the line LL22 and so and so satisfiessatisfies

yy = = mm22xx + + bb22 as well. as well.

Therefore, to find the Therefore, to find the point of intersectionpoint of intersection PP((xx00, , yy00)) of the of the

lines lines LL11 and and LL22, we , we solve solve for for xx and and yy thethe system system composed of composed of

the the two equationstwo equations

yy = = mm11xx + + bb11 and and yy = = mm22xx + + bb22

ExampleExample Find the Find the point of intersectionpoint of intersection of the of the straight linesstraight lines that have that have

equationsequationsyy = = xx + 1 + 1 and and yy = – = – 22xx + 4 + 4

SolutionSolution SubstitutingSubstituting the value the value yy as given in the as given in the first equationfirst equation into into

the the second equationsecond equation, we obtain, we obtain

SubstitutingSubstituting this value of this value of xx into into either oneeither one of the given of the given equationsequations yields yields yy = 2 = 2..

Therefore, the requiredTherefore, the required point of intersection point of intersection is is (1, 2)(1, 2)..

1 2 4x x 1 2 4x x

3 3x 3 3x 1x 1x

Example 1, page 40

––11 11 22 33 44 55

55

44

33

22

11

ExampleExample Find the Find the point of intersectionpoint of intersection of the of the straight linesstraight lines that have that have

equationsequationsyy = = xx + 1 + 1 and and yy = – = – 22xx + 4 + 4

SolutionSolution The graph shows theThe graph shows the point of intersection point of intersection (1, 2)(1, 2) of the two of the two

lines:lines:yy

xx

(1, 2)(1, 2)

LL11

LL22

Example 1, page 40

Applied Example:Applied Example: Break-Even Level Break-Even Level

Prescott manufactures its products at a Prescott manufactures its products at a costcost of of $4$4 per unit per unit and and sellssells them for them for $10$10 per unit. per unit.

If the firm’s If the firm’s fixed costfixed cost is is $12,000$12,000 per month, determine the per month, determine the firm’s firm’s break-even pointbreak-even point..

SolutionSolution The The revenue functionrevenue function RR and the and the cost functioncost function CC are given are given

respectively byrespectively by

Setting Setting RR((xx) =) = C C((xx)), we obtain, we obtain

( ) 10 ( ) 4 12,000 d an .R x x C x x ( ) 10 ( ) 4 12,000 d an .R x x C x x

10 4 12,000x x 10 4 12,000x x 6 12,000x 6 12,000x

2000x 2000x


Applied Example:Applied Example: Break-Even Level Break-Even Level

Prescott manufactures its products at a Prescott manufactures its products at a costcost of of $4$4 per unit per unit and and sellssells them for them for $10$10 per unit. per unit.

If the firm’s If the firm’s fixed costfixed cost is is $12,000$12,000 per month, determine the per month, determine the firm’s firm’s break-even pointbreak-even point..

SolutionSolution Substituting Substituting x x = 2000= 2000 into into RR((xx) =) = 1010x x givesgives

So, Prescott’s So, Prescott’s break-even pointbreak-even point is is 20002000 units of the product, units of the product, resulting in a resulting in a break-even revenuebreak-even revenue of of $20,000 $20,000 per month.per month.

(2000) 10(2000) 20,000R (2000) 10(2000) 20,000R


Applied Example:Applied Example: Market Equilibrium Market Equilibrium

The management of ThermoMaster, which manufactures The management of ThermoMaster, which manufactures an indoor-outdoor thermometer at its Mexico subsidiary, an indoor-outdoor thermometer at its Mexico subsidiary, has determined that the has determined that the demand equationdemand equation for its product is for its product is

where where pp is the price of a thermometer in dollars and is the price of a thermometer in dollars and xx is the is the quantity demandedquantity demanded in units of a thousand. in units of a thousand.

The The supply equationsupply equation of these thermometers is of these thermometers is

where where xx (in thousands) is the (in thousands) is the quantityquantity that ThermoMaster that ThermoMaster will will make availablemake available in the market at in the market at pp dollars each. dollars each.

Find the Find the equilibrium quantity and priceequilibrium quantity and price..

5 3 30 0x p 5 3 30 0x p

52 30 45 0x p 52 30 45 0x p



SolutionSolution We need to We need to solve the system of equationssolve the system of equations

for for xx and and pp.. Let’s Let’s solvesolve the the first equationfirst equation for for pp in terms of in terms of xx::

5 3 30 0x p 5 3 30 0x p 52 30 45 0x p 52 30 45 0x p

3 5 30p x 3 5 30p x 5 3 30 0x p 5 3 30 0x p

510

3p x

510

3p x




for for xx and and pp.. Now we Now we substitutesubstitute the value of the value of pp into the into the second equationsecond equation::

5 3 30 0x p 5 3 30 0x p 52 30 45 0x p 52 30 45 0x p

552 30 10 45 0

3x x

552 30 10 45 0

3x x

52 50 300 45 0x x 52 50 300 45 0x x

102 255 0x 102 255 0x

102 255x 102 255x 255 5

102 2x

255 5

102 2x




for for xx and and pp.. Finally, we Finally, we substitutesubstitute the value the value x x = 5/2= 5/2 into the into the firstfirst equation equation

thatthat we alreadywe already solved solved::

5 3 30 0x p 5 3 30 0x p 52 30 45 0x p 52 30 45 0x p

5 510

3 2p

5 510

3 2p

2510

6

2510

6

355.83

6

355.83

6



SolutionSolution We conclude that the We conclude that the equilibrium quantityequilibrium quantity is is 25002500 units and units and

the the equilibrium priceequilibrium price is is $5.83$5.83 per thermometer. per thermometer.


1.51.5The Method of Least SquaresThe Method of Least Squares

55 1010xx

yy LL

d4

d5

d3

d2

d1

1010

55

In this section, we describe a general method known as the In this section, we describe a general method known as the method for least squaresmethod for least squares for determining a straight line for determining a straight line that, in a sense, best fits a set of data points when the that, in a sense, best fits a set of data points when the points are scattered about a straight line.points are scattered about a straight line.

The Method of Least SquaresThe Method of Least Squares

55 1010


Suppose we are given five Suppose we are given five data pointsdata points

PP11((xx11, , yy11)), , PP22((xx22, , yy22)), , PP33((xx33, , yy33)), , PP44((xx44, , yy44)), and , and PP55((xx55, , yy55))

describing describing thethe relationshiprelationship between between two variables two variables xx and and yy.. By By plotting these data pointsplotting these data points, we obtain a , we obtain a scatter diagramscatter diagram: :

xx

yy

PP11

PP22

PP33

PP44

PP55

1010

55

1010

55

55 1010


Suppose we try to fit a Suppose we try to fit a straight linestraight line LL to the data points to the data points PP11, , PP22, , PP33, , PP44, and , and PP55..

The line will The line will miss these pointsmiss these points by the by the amountsamounts dd11, , dd22, , dd33, , dd44, and , and dd55 respectively. respectively.

xx

yy LL

d4

d5

d3

d2

d1

55 1010


The principle of The principle of least squaresleast squares states that the states that the straight linestraight line L L that that fits the data points fits the data points bestbest is the one chosen by requiring is the one chosen by requiring that the that the sum of the squaressum of the squares of of dd11, , dd22, , dd33, , dd44, and , and dd55, that is, that is

be made be made as small as possibleas small as possible..

xx

yy

2 2 2 2 21 2 3 4 5d d d d d 2 2 2 2 21 2 3 4 5d d d d d

LL

d4

d5

d3

d2

d1

1010

55


Suppose we are given Suppose we are given nn data pointsdata points::

PP11((xx11, , yy11)), , PP22((xx22, , yy22)), , PP33((xx33, , yy33)), . . ., . . . , , PPnn((xxnn, , yynn))

Then, the Then, the least-squares (regression) lineleast-squares (regression) line for the data for the data is given by the is given by the linear equationlinear equation

yy = = ff((xx) = ) = mxmx + + bb

where the where the constantsconstants mm and and bb satisfy the equations satisfy the equations

andand

simultaneously.simultaneously. These last two equations are called These last two equations are called normal equationsnormal equations..

1 2 3 1 2 3( )n nx x x x m nb y y y y 1 2 3 1 2 3( )n nx x x x m nb y y y y

2 2 2 21 2 3 1 2 3

1 1 2 2 3 3

( ) ( )n n

n n

x x x x m x x x x b

y x y x y x y x

2 2 2 21 2 3 1 2 3

1 1 2 2 3 3

( ) ( )n n

n n

x x x x m x x x x b

y x y x y x y x

ExampleExample

Find the equation of the Find the equation of the least-squares lineleast-squares line for the data for the data

PP11(1, 1)(1, 1), , PP22(2, 3)(2, 3), , PP33(3, 4)(3, 4), , PP44(4, 3)(4, 3), and , and PP55(5, 6)(5, 6)

SolutionSolution Here, we have Here, we have nn = 5 = 5 and and

xx11 = 1 = 1 xx22 = 2 = 2 xx33 = 3 = 3 xx44 = 4 = 4 xx55 = 5 = 5

yy11 = 1 = 1 yy22 = 3 = 3 yy33 = 4 = 4 yy44 = 3 = 3 yy55 = 6 = 6

xx yy xx22 xyxy

11 11 11 11

22 33 44 66

33 44 99 1212

44 33 1616 1212

55 66 2525 3030

1515 1717 5555 6161

Before using the Before using the equations it is convenient equations it is convenient to to summarize these datasummarize these data in the form of a in the form of a tabletable::

Example 1, page 53

ExampleExample




xx11 = 1 = 1 xx22 = 2 = 2 xx33 = 3 = 3 xx44 = 4 = 4 xx55 = 5 = 5

yy11 = 1 = 1 yy22 = 3 = 3 yy33 = 4 = 4 yy44 = 3 = 3 yy55 = 6 = 6

Using theUsing the table table toto substitute substitute in the in the second equationsecond equation we get we get2 2 2 21 2 3 1 2 3

1 1 2 2 3 3

( ... ) ( ... )

... n n

n n

x x x x m x x x x b

y x y x y x y x

2 2 2 21 2 3 1 2 3

1 1 2 2 3 3

( ... ) ( ... )

... n n

n n

x x x x m x x x x b

y x y x y x y x

55 15 61m b 55 15 61m b

Example 1, page 53

ExampleExample




xx11 = 1 = 1 xx22 = 2 = 2 xx33 = 3 = 3 xx44 = 4 = 4 xx55 = 5 = 5

yy11 = 1 = 1 yy22 = 3 = 3 yy33 = 4 = 4 yy44 = 3 = 3 yy55 = 6 = 6

Using theUsing the table table toto substitute substitute in the in the first equationfirst equation we get we get

15 5 17m b 15 5 17m b

1 2 3 1 2 3( ... ) ...n nx x x x m nb y y y y 1 2 3 1 2 3( ... ) ...n nx x x x m nb y y y y

Example 1, page 53

ExampleExample



SolutionSolution Now we need to Now we need to solvesolve the the simultaneous equationssimultaneous equations

SolvingSolving the the first equationfirst equation for for bb gives gives

15 5 1755 15 61

m bm b

15 5 1755 15 61

m bm b

15 5 17m b 15 5 17m b

5 15 17b m 5 15 17b m

173

5b m

173

5b m

Example 1, page 53

ExampleExample




Substituting Substituting bb into the into the second equationsecond equation gives gives

1755 15 3 61

5m m

1755 15 3 61

5m m

55 15 61m b 55 15 61m b

55 45 51 61m m 55 45 51 61m m 10 10m 10 10m

1m 1m

15 5 1755 15 61

m bm b

15 5 1755 15 61

m bm b

Example 1, page 53

ExampleExample




Finally, Finally, substitutingsubstituting the value the value m m = 1= 1 into the into the firstfirst equation equation thatthat we alreadywe already solved solved gives gives

173

5b m

173

5b m

173(1)

5

173(1)

5

0.40.4

15 5 1755 15 61

m bm b

15 5 1755 15 61

m bm b

Example 1, page 53

ExampleExample




Thus,Thus, we find that we find that mm = 1 = 1 and and bb = 0.4 = 0.4.. Therefore, the required Therefore, the required least-squares lineleast-squares line is is

0.4

y mx b

y x

0.4

y mx b

y x

15 5 1755 15 61

m bm b

15 5 1755 15 61

m bm b

Example 1, page 53

ExampleExample



SolutionSolution Below is the Below is the graphgraph of the required of the required least-squares lineleast-squares line

yy = = xx + 0.4 + 0.4

66

55

44

33

22

11

11 22 33 44 55xx

yyLL

Example 1, page 53


Because the Because the over-65 populationover-65 population will be will be growinggrowing more more rapidly in the next few decades, rapidly in the next few decades, health-care spendinghealth-care spending is is expected to expected to increase significantlyincrease significantly in the coming decades. in the coming decades.

The following table gives the The following table gives the U.S. health expendituresU.S. health expenditures (in trillions of dollars) from (in trillions of dollars) from 20052005 through through 20102010::

Find a function giving the Find a function giving the U.S. health-care spendingU.S. health-care spending between between 20052005 and and 20102010, using the , using the least-squares techniqueleast-squares technique..

Year,Year, tt 00 11 22 33 44 55

Expenditure,Expenditure, yy 2.002.00 2.172.17 2.342.34 2.502.50 2.692.69 2.902.90



SolutionSolution The The calculationscalculations required for obtaining the required for obtaining the normal normal

equationsequations are are summarizedsummarized in the following in the following tabletable::

Use theUse the table table toto obtainobtain the the second normal equationsecond normal equation::2 2 2 21 2 3 1 2 3

1 1 2 2 3 3

( ... ) ( ... )...

n n

n n

t t t t m t t t t by t y t y t y t

2 2 2 21 2 3 1 2 3

1 1 2 2 3 3

( ... ) ( ... )...

n n

n n

t t t t m t t t t by t y t y t y t

55 15 39.61m b 55 15 39.61m b

tt yy tt22 tyty00 2.002.00 00 0011 2.172.17 11 2.172.1722 2.342.34 44 4.684.6833 2.502.50 99 7.507.5044 2.692.69 1616 10.7610.7655 2.902.90 2525 14.5014.50

1515 14.6014.60 5555 39.6139.61



SolutionSolution The The calculationscalculations required for obtaining the required for obtaining the normal normal

equationsequations are are summarizedsummarized in the following in the following tabletable::

Use theUse the table table toto obtainobtain the the first normal equationfirst normal equation::

15 6 14.60m b 15 6 14.60m b 1 2 3 1 2 3( ... ) ...n nt t t t m nb y y y y 1 2 3 1 2 3( ... ) ...n nt t t t m nb y y y y

tt yy tt22 tyty00 2.002.00 00 0011 2.172.17 11 2.172.1722 2.342.34 44 4.684.6833 2.502.50 99 7.507.5044 2.692.69 1616 10.7610.7655 2.902.90 2525 14.5014.50

1515 14.6014.60 5555 39.6139.61



SolutionSolution Now we Now we solvesolve the the simultaneous equationssimultaneous equations

SolvingSolving the the first equationfirst equation for for bb gives gives

15 6 14.6055 15 39.61

m bm b

15 6 14.6055 15 39.61

m bm b

15 6 14.60m b 15 6 14.60m b

6 15 14.60b m 6 15 14.60b m

2.5 2.4333b m 2.5 2.4333b m




Substituting Substituting bb into the into the second equationsecond equation gives gives

55 15 39.61m b 55 15 39.61m b

55 15 2.5 2.4333 39.61m m 55 15 2.5 2.4333 39.61m m

55 37.5 36.4995 39.61m m 55 37.5 36.4995 39.61m m

17.5 3.1105m 17.5 3.1105m

0.1777m 0.1777m

15 6 14.6055 15 39.61

m bm b

15 6 14.6055 15 39.61

m bm b




Finally, Finally, substitutingsubstituting the value the value m m ≈≈ 0.1777 0.1777 into the into the firstfirst equation equation thatthat we alreadywe already solved solved gives gives

2.5 2.4333b m 2.5 2.4333b m

2.5(0.1777) 2.4333 2.5(0.1777) 2.4333

1.98911.9891

15 6 14.6055 15 39.61

m bm b

15 6 14.6055 15 39.61

m bm b




Thus,Thus, we find that we find that mm ≈ 0.1777≈ 0.1777 and and bb ≈≈ 1.9891 1.9891.. Therefore, the required Therefore, the required least-squares functionleast-squares function is is

( ) 1.178 1.989S t y t ( ) 1.178 1.989S t y t

15 6 14.6055 15 39.61

m bm b

15 6 14.6055 15 39.61

m bm b


End of End of Chapter Chapter

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1 The Cartesian Coordinate System Straight Lines Linear Functions and Mathematical Models Intersection of Straight Lines The Method of Least