The Cartesian Coordinate System and Straight lines Equations of Lines

22 The Cartesian Coordinate System The Cartesian Coordinate System

and Straight lines and Straight lines Equations of LinesEquations of Lines Functions and Their GraphsFunctions and Their Graphs The Algebra of FunctionsThe Algebra of Functions Linear FunctionsLinear Functions Quadratic FunctionsQuadratic Functions Functions and Mathematical ModelsFunctions and Mathematical Models

Functions and Their Graphs Functions and Their Graphs

2.12.1The Cartesian Coordinate System The Cartesian Coordinate System and Straight linesand Straight lines

11 22 33 44 55 66

(2, 5)(2, 5)

y y = = ––44

x x = 3= 366

11 (5, 1)(5, 1)

– 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::

The Cartesian Coordinate SystemThe Cartesian Coordinate System

The Cartesian Coordinate SystemThe Cartesian Coordinate System 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

– – 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

The Cartesian Coordinate SystemThe Cartesian Coordinate System 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

OriginOrigin

The Cartesian Coordinate SystemThe Cartesian Coordinate System 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

PositivePositive Direction DirectionNegativeNegative Direction Direction

The Cartesian Coordinate SystemThe Cartesian Coordinate System 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

tive D

(–(– 2, 4)2, 4)

(–1, (–1, –– 2)2)

(4, 3)(4, 3)

The Cartesian Coordinate SystemThe Cartesian Coordinate System 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

(3, –1)(3, –1)

The Cartesian Coordinate SystemThe Cartesian Coordinate System 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

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

Quadrant IIQuadrant II(–, +)(–, +)

Quadrant IVQuadrant IV(+, –)(+, –)

Quadrant IIIQuadrant III(–, –)(–, –)

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))

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))

y y ymx x x

yy22 – – yy11 = = yy

xx22 – – xx11 = = xx

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

y y = 1= 1

x x = 1= 1

m m = 1= 1

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

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..

y y = –1= –1

x x = 1= 1

m m = –2= –2

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

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..

SolutionSolution1.1. 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.

y y = –= – 44

x x = 3= 366

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

3 1 2 15 ( 1) 6 3

y ymx x

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

5 5 0 03 ( 2) 5

y ymx x

––2 2 –1 –1 11 22 33 44

ExamplesExamples

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

(–2, 5)(–2, 5) (3, 5)(3, 5)

m m = 0= 0

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..

13 9 2

1 ( 2)m

24 10 2

3 ( 4)m

2.22.2Equations of LinesEquations of Lines

11 22 33 44 55 66

(0, (0, –– 3)3)

–– 11

–– 22

–– 33

–– 44

(4, 0)(4, 0)

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, , ))

((aa, 0), 0)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

y ymx x

1 1( )y y m x x

Point-Slope FormPoint-Slope Form

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

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)7

3 7 5 0x y

1 2 34 ( 3) 7

y ymx x

7 7 3 12y x

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

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)2

2 2 32 5 0

y xx y

((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-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

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 84 8 3

x yy x

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 = 65,000

General Form of a Linear EquationGeneral Form of a Linear Equation

The equationThe equationAxAx + + 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 equation33xx – 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)..

ExampleExample

Sketch the straight line represented by the equationSketch the straight line represented by the equation33xx – 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)

–– 11

–– 22

–– 33

–– 44

(4, 0)(4, 0)

Equations of Straight LinesEquations of Straight Lines

Vertical line:Vertical line: xx = = aaHorizontal 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 + bGeneral Form:General Form: Ax Ax + + By By + + CC = 0 = 0

2.32.3Functions and Their GraphsFunctions and Their Graphs

f x x f x x

–– 33 –– 22 –– 11 11 22 33

FunctionsFunctions

A function A function ff is a is a rulerule that assigns to each element in a that assigns to each element in a setset AA one and only one element in a one and only one element in a setset BB..

The The setset AA is called the is called the domaindomain of the function. of the function. It is customary to denote a function by a letter of the It is customary to denote a function by a letter of the

alphabet, such as the letter alphabet, such as the letter ff.. If If xx is an is an elementelement in the in the domaindomain of a function of a function ff, then the , then the

elementelement in in BB that that ff associates with associates with xx is written is written ff((xx)) (read (read ““f f of of xx”) and is called the ”) and is called the valuevalue of of ff at at xx..

The The setset BB comprising all the values assumed by comprising all the values assumed by y =y = ff((xx)) as as xx takes on all possible values in its domain is called takes on all possible values in its domain is called the the rangerange of the function of the function ff..

ExampleExample

Let the function Let the function ff be defined by the rulebe defined by the rule

Find: Find: ff(1)(1)Solution:Solution:

22 1f x x x

21 2 1 1 1 2 1 1 2f

ExampleExample

Find: Find: ff(( – 2)2)Solution:Solution:

22 1f x x x

22 2 2 2 1 8 2 1 11f

ExampleExample

Find: Find: ff((a))Solution:Solution:

22 1f x x x

2 22 1 2 1f a a a a a

ExampleExample

Find: Find: ff((a + h))Solution:Solution:

22 1f x x x

2 2 22 1 2 4 2 1f a h a h a h a ah h a h

Applied ExampleApplied Example

ThermoMaster manufactures an indoor-outdoor ThermoMaster manufactures an indoor-outdoor thermometer at its Mexican subsidiary.thermometer at its Mexican subsidiary.

Management estimates that the Management estimates that the profitprofit (in dollars) (in dollars) realizable by ThermoMaster in the manufacture and realizable by ThermoMaster in the manufacture and salesale of of xx thermometers per week is thermometers per week is

Find ThermoMaster’s weekly Find ThermoMaster’s weekly profitprofit if its if its level of level of productionproduction is: is:

a.a. 10001000 thermometers per week. thermometers per week. b.b. 20002000 thermometers per week. thermometers per week.

20.001 8 5000P x x x

Applied ExampleApplied Example

SolutionSolution We haveWe have

a.a. The weekly The weekly profitprofit by by producingproducing 10001000 thermometers is thermometers is

or or $2,000$2,000..

b.b. The weekly The weekly profitprofit by by producingproducing 20002000 thermometers is thermometers is

or or $7,000$7,000..

21000 0.001 1000 8 1000 5000 2000P =

22000 0.001 2000 8 2000 5000 7000P =

20.001 8 5000P x x x

Determining the Domain of a FunctionDetermining the Domain of a Function

Suppose we are given the function Suppose we are given the function y y = = ff((xx)).. Then, the variable Then, the variable xx is called the is called the independent variableindependent variable.. The variable The variable yy, whose value depends on , whose value depends on xx, is called the , is called the

dependent variabledependent variable.. To determine the To determine the domaindomain of a function, we need to find of a function, we need to find

what what restrictionsrestrictions, if any, are to be placed on the , if any, are to be placed on the independent variable independent variable xx..

In many practical problems, the domain of a function is In many practical problems, the domain of a function is dictated by the dictated by the nature of the problemnature of the problem..

16 – 216 – 2xx

Applied Example:Applied Example: Packaging Packaging An open box is to be made from a rectangular piece of An open box is to be made from a rectangular piece of

cardboard cardboard 1616 inches wide by cutting away identical inches wide by cutting away identical squares (squares (xx inches by inches by xx inches) from each corner and inches) from each corner and folding up the resulting flaps.folding up the resulting flaps.

10 10 – 210 10 – 2xx

Applied Example:Applied Example: Packaging Packaging An open box is to be made from a rectangular piece of An open box is to be made from a rectangular piece of

cardboard cardboard 1616 inches wide by cutting away identical inches wide by cutting away identical squares (squares (xx inches by inches by xx inches) from each corner and inches) from each corner and folding up the resulting flaps.folding up the resulting flaps.

a.a. Find the expression that gives the Find the expression that gives the volumevolume VV of the box as of the box as a function of a function of xx. .

b.b. What is the What is the domaindomain of the function? of the function?

The The dimensionsdimensions of the of the resulting box are:resulting box are:

16 – 216 – 2xx10 – 210 – 2xx

16 – 216 – 2xx

Applied Example:Applied Example: Packaging Packaging

SolutionSolutiona.a. The The volumevolume of the box is given by of the box is given by multiplying its multiplying its

dimensionsdimensions (length (length ☓☓ width width ☓☓ height) height), so:, so:

10 – 210 – 2xx

16 2 10 2

160 52 4

4 52 160

V f x x x x

Applied Example:Applied Example: Packaging Packaging

SolutionSolutionb.b. Since theSince the length length of each of each sideside of the box must be of the box must be greatergreater

than or than or equalequal to to zerozero, we see that, we see that

must be satisfied simultaneously. Simplified:must be satisfied simultaneously. Simplified:

All three are satisfied simultaneously provided that:All three are satisfied simultaneously provided that:

Thus, the Thus, the domaindomain of the function of the function ff is the interval is the interval [0, 5][0, 5]..

8 5 0x x x =

16 2 0 10 2 0 0x x x

More ExamplesMore Examples

Find the domain of the function:Find the domain of the function:

SolutionSolution Since the square root of a negative number is Since the square root of a negative number is undefinedundefined, it , it

is necessary that is necessary that xx – 1 – 1 0 0.. Thus the Thus the domaindomain of the function is of the function is [1,[1,)). .

1f x x

SolutionSolution Our only constraint is that you Our only constraint is that you cannot divide by zerocannot divide by zero, so , so

Which means that Which means that

Or more specifically Or more specifically x x ≠≠ –2–2 and and x x ≠ ≠ 22.. Thus the Thus the domaindomain of of ff consists of the intervals consists of the intervals (–(– , –2), –2), ,

(–2, 2)(–2, 2), , (2, (2, )). .

2 4 0x

2 4 2 2 0x x x

SolutionSolution Here, any real number satisfies the equation, so the Here, any real number satisfies the equation, so the

domaindomain of of ff is the set of is the set of all real numbersall real numbers..

2 3f x x

Graphs of FunctionsGraphs of Functions

If If f f is a function with domain is a function with domain AA, then corresponding to , then corresponding to each real number each real number xx in in AA there is there is precisely oneprecisely one real real number number ff((xx))..

Thus, a function Thus, a function ff with domain with domain AA can also be defined as can also be defined as the set of all the set of all ordered pairsordered pairs ((x, fx, f((xx)))) where where xx belongs to belongs to AA..

The The graph of a functiongraph of a function ff is the set of all points is the set of all points ((xx, , yy)) in in the the xyxy-plane such that -plane such that xx is in the domain of is in the domain of ff and and y = fy = f((xx))..

ExampleExample

The graph of a function The graph of a function ff is shown below: is shown below:

Domain

y (x, y)

ExampleExample

The graph of a function The graph of a function ff is shown below: is shown below:✦ What is the value of What is the value of ff(2)(2)??

–2 (2, –2)

1 2 3 4 5 6 7 8

ExampleExample

The graph of a function The graph of a function ff is shown below: is shown below:✦ What is the value of What is the value of ff(5)(5)??

(5, 3)

1 2 3 4 5 6 7 8

ExampleExample

The graph of a function The graph of a function ff is shown below: is shown below:✦ What is the What is the domaindomain of of ff((xx))??

Domain: [1,8]

1 2 3 4 5 6 7 8

ExampleExample

The graph of a function The graph of a function ff is shown below: is shown below:✦ What is the What is the rangerange of of ff((xx))??

Range: [–2,4]

1 2 3 4 5 6 7 8

Example:Example: Sketching a Graph Sketching a Graph

Sketch the graph of the function defined by the equation Sketch the graph of the function defined by the equation yy = = xx22 + 1 + 1

SolutionSolution The The domaindomain of the function is the set of of the function is the set of all real numbersall real numbers.. Assign several valuesAssign several values to the variable to the variable x x and and compute the compute the

corresponding valuescorresponding values for for yy: :

xx yy––33 1010––22 55––11 2200 1111 2222 5533 1010

– – 33 – 2 – 2 – 1 – 1 11 22 33

SolutionSolution The The domaindomain of the function is the set of of the function is the set of all real numbersall real numbers.. Then Then plotplot these values in a these values in a graphgraph: :

yyxx yy

––33 1010––22 55––11 2200 1111 2222 5533 1010

– – 33 – 2 – 2 – 1 – 1 11 22 33

SolutionSolution The The domaindomain of the function is the set of of the function is the set of all real numbersall real numbers.. And finally, And finally, connect the dotsconnect the dots: :

yyxx yy

––33 1010––22 55––11 2200 1111 2222 5533 1010

Example:Example: Sketching a Graph Sketching a Graph Sketch the graph of the function defined by the equation Sketch the graph of the function defined by the equation

SolutionSolution The function The function ff is defined in a piecewise fashion on the set is defined in a piecewise fashion on the set

of of all real numbersall real numbers.. In the In the subdomainsubdomain (–(– , 0), 0), the , the rulerule for for ff is given by is given by

In the In the subdomainsubdomain [0, [0, )), the , the rulerule for for ff is given by is given by

x xf x

SolutionSolution Substituting Substituting negative valuesnegative values for for xx into , while into , while

substituting substituting zero and positive valueszero and positive values into into we get: we get:

x xf x

f x x f x x

xx yy––33 33––22 22––11 1100 0011 1122 1.411.4133 1.731.73

SolutionSolution PlottingPlotting these data and these data and graphinggraphing we get: we get:

x xf x

– – 33 – 2 – 2 – 1 – 1 11 22 33

yyxx yy––33 33––22 22––11 1100 0011 1122 1.411.4133 1.731.73

The Vertical Line TestThe Vertical Line Test

A curve in the A curve in the xyxy-plane-plane is the graph of a is the graph of a functionfunction yy = = ff((xx)) if and only ifif and only if each vertical line intersects it each vertical line intersects it in in at mostat most one pointone point..

ExamplesExamples

Determine if the curve in the graph is a Determine if the curve in the graph is a functionfunction of of xx::

SolutionSolution The curve The curve is indeed a functionis indeed a function of of xx, because there is , because there is one one

and only oneand only one value of value of yy for any given value of for any given value of xx..

ExamplesExamples

SolutionSolution The curve The curve is is notnot a function a function of of xx, because there is , because there is more more

than onethan one value of value of yy for some values of for some values of xx..

ExamplesExamples

SolutionSolution The curve The curve is indeed a functionis indeed a function of of xx, because there is , because there is one one

and only oneand only one value of value of yy for any given value of for any given value of xx..

2.42.4The Algebra of FunctionsThe Algebra of Functions

19901990 19921992 19941994 19961996 19981998 20002000

20002000

18001800

16001600

14001400

12001200

10001000

YearYear

y y = = SS((tt))

y y = = RR((tt))

RR((tt))

SS((tt))

The Sum, Difference, Product and Quotient The Sum, Difference, Product and Quotient of Functions of Functions

Consider the graph below:Consider the graph below:✦ R(t)R(t) denotes the federal government denotes the federal government revenuerevenue at any time at any time tt..✦ S(t)S(t) denotes the federal government denotes the federal government spendingspending at any time at any time tt..

19901990 19921992 19941994 19961996 19981998 20002000

20002000

18001800

16001600

14001400

12001200

10001000

YearYear

y y = = SS((tt))

y y = = RR((tt))

RR((tt))

SS((tt))

DD((tt) = ) = RR((tt) ) – – SS((tt))

Consider the graph below:Consider the graph below:✦ The difference The difference RR((tt) ) – S– S((tt)) gives the gives the budget deficitbudget deficit (if negative)(if negative)

or or surplussurplus (if positive)(if positive) in billions of dollars at any time in billions of dollars at any time tt..

19901990 19921992 19941994 19961996 19981998 20002000

20002000

18001800

16001600

14001400

12001200

10001000

YearYear

y y = = SS((tt))

y y = = RR((tt))

RR((tt))

SS((tt))

The budget balanceThe budget balance D D((tt)) is shown below: is shown below:✦ DD((tt)) is also a is also a function function that denotes the federal government that denotes the federal government

deficit (surplus)deficit (surplus) at any time at any time tt. . ✦ This function is the This function is the differencedifference of the two functions of the two functions R R and and SS..✦ DD((tt) ) has the has the same domainsame domain as as RR((tt) ) andand S S((tt))..

19921992 19941994 19961996 19981998 20002000

400400

200200

–– 200200

–– 400400

YearYear

tt DD((tt))

y y = = DD((tt))

Most functions are built up from other, generally Most functions are built up from other, generally simpler functions.simpler functions.

For example, we may view the function For example, we may view the function ff((xx) = 2) = 2xx + 4 + 4 as the as the sumsum of the two functions of the two functions gg((xx) = 2) = 2xx and and hh((xx) = 4) = 4..

( )( )

f f xxg g x

The Sum, Difference, Product and Quotient of FunctionsThe Sum, Difference, Product and Quotient of Functions Let Let ff and and gg be functions with domains be functions with domains AA and and BB, respectively., respectively. The The sumsum f + gf + g, the , the differencedifference f – gf – g, and the , and the productproduct fgfg of of ff

and and gg are are functionsfunctions with domain with domain A A ∩∩ B B and rule given byand rule given by

((ff + + gg)()(xx) = ) = ff((xx) + ) + gg((xx)) SumSum((ff –– gg)()(xx) = ) = ff((xx) ) –– gg((xx)) DifferenceDifference

((fgfg)()(xx) = ) = ff((xx))gg((xx)) ProductProduct

The The quotientquotient ff//gg of of f f and and gg has domain has domain A A ∩∩ B B excluding all excluding all numbers numbers xx such that such that gg((xx)) = 0= 0 and rule given by and rule given by

QuotientQuotient

ExampleExample

LetLet and and gg((xx) = 2) = 2xx + 1 + 1. . Find the Find the sumsum ss, the , the differencedifference dd, the , the productproduct pp, and the , and the

quotientquotient qq of the functions of the functions ff and and gg..SolutionSolution Since the domain of Since the domain of ff is is AA = [–1, = [–1,)) and the domain of and the domain of gg

is is BB = (– = (– , , )), we see that the , we see that the domaindomain of of ss, , dd, and , and pp is is A A ∩∩ B B = [–1,= [–1,))..

The rules are as follows:The rules are as follows:

( ) 1f x x

( ) ( )( ) ( ) ( ) 1 2 1

( ) ( )( ) ( ) ( ) 2 1 1

s x f g x f x g x x x

d x f g x f x g x x x

p x fg x f x g x x x

ExampleExample

LetLet and and gg((xx) = 2) = 2xx + 1 + 1. . Find the Find the sumsum ss, the , the differencedifference dd, the , the productproduct pp, and the , and the

quotientquotient qq of the functions of the functions ff and and gg..SolutionSolution The The domaindomain of the of the quotientquotient function is function is [–1,[–1,)) together together

with the restriction with the restriction xx ≠≠ –– ½½ ..

Thus, the domain is Thus, the domain is [–1, –[–1, – ½) ½) (– (– ½,½,)).. The rule is as follows:The rule is as follows:

( ) 1f x x

( ) 1( ) ( )( ) 2 1

f f x xq x xg g x x

Applied ExampleApplied Example: Cost Functions: Cost Functions

Suppose Puritron, a manufacturer of water filters, has a Suppose Puritron, a manufacturer of water filters, has a monthly monthly fixed costfixed cost of of $10,000$10,000 and a variable cost of and a variable cost of

–– 0.00010.0001xx22 + 10 + 10xx (0 (0 xx 40,000) 40,000)dollars, where dollars, where xx denotes the number of denotes the number of filters filters manufacturedmanufactured per month. per month.

Find a function Find a function CC that gives the that gives the total monthly costtotal monthly cost incurred by Puritron in the manufacture of incurred by Puritron in the manufacture of xx filters. filters.

Applied ExampleApplied Example: Cost Functions: Cost FunctionsSolutionSolution Puritron’s monthly Puritron’s monthly fixed costfixed cost is always is always $10,000$10,000, so it can , so it can

be described by the constant function: be described by the constant function: FF((xx)) = 10,000= 10,000

The The variable costvariable cost can be described by the function: can be described by the function:VV((xx)) = –= – 0.00010.0001xx22 + 10 + 10xx

The The total costtotal cost is the is the sumsum of the fixed cost of the fixed cost FF and the and the variable cost variable cost VV::

CC((xx)) = = VV((xx) + ) + FF((xx) ) = –= – 0.00010.0001xx22 + 10 + 10x x + 10,000+ 10,000 (0 (0 xx

40,000)40,000)

Applied ExampleApplied Example: Cost Functions: Cost FunctionsLet’s now consider Let’s now consider profitsprofits Suppose that the Suppose that the total revenuetotal revenue RR realized by Puritron realized by Puritron

from the sale of from the sale of xx water filters is given by water filters is given by RR((xx)) = –= – 0.00050.0005xx22 + 20 + 20x x (0 (0 ≤≤ xx ≤≤ 40,000) 40,000)

FindFinda.a. The total profit function for Puritron.The total profit function for Puritron.b.b. The total profit when Puritron produces The total profit when Puritron produces 10,00010,000 filters per filters per

month.month.

Applied ExampleApplied Example: Cost Functions: Cost FunctionsSolutionSolutiona.a. The The totaltotal profitprofit PP realized by the firm is the difference realized by the firm is the difference

between the between the total revenuetotal revenue R R and the and the total costtotal cost CC:: PP((xx)) = = RR((xx) – ) – CC((xx) )

= (–= (– 0.00050.0005xx22 + 20 + 20xx)) – (–– (– 0.00010.0001xx22 + 10 + 10x x + 10,000)+ 10,000)= –= – 0.00040.0004xx22 + 10 + 10x x – 10,000– 10,000

b.b. The The totaltotal profitprofit realized by Puritron when realized by Puritron when producingproducing 10,00010,000 filters per month isfilters per month is

PP((xx)) = –= – 0.0004(10,000)0.0004(10,000)22 + 10(10,000) + 10(10,000) – 10,000– 10,000 = 50,000= 50,000or or $50,000$50,000 per month. per month.

The Composition of Two FunctionsThe Composition of Two Functions

Another way to Another way to build a functionbuild a function from other functions is from other functions is through a process known as the composition of functions.through a process known as the composition of functions.

Consider the functions Consider the functions ff and and gg: :

Evaluating the function Evaluating the function gg at the pointat the point ff((xx)), we find that:, we find that:

This is an entirely This is an entirely new functionnew function, which we could call , which we could call hh::

2( ) 1f x x ( )g x x

2( ) ( ) 1g f x f x x

2( ) 1h x x

The Composition of Two FunctionsThe Composition of Two Functions

Let Let ff and and gg be functions. be functions. Then the Then the compositioncomposition of of gg and and f f is the function is the function

ggff (read “(read “gg circlecircle f f ”)”) defined by defined by

((ggff )()(xx) = ) = gg((ff((xx))))

The The domaindomain of of ggff is the set of all is the set of all xx in the domain in the domain of of ff such that such that ff((xx)) lies in the domain of lies in the domain of gg..

ExampleExample

LetLet Find:Find:

a.a. The rule for the composite function The rule for the composite function ggff..b.b. The rule for the composite function The rule for the composite function ffgg..

SolutionSolution To find To find ggff, evaluate the function , evaluate the function gg at at ff((xx))::

To find To find ffgg, evaluate the function , evaluate the function ff at at gg((xx))::

2( ) 1 ( ) 1an d .f x x g x x

2( )( ) ( ( )) ( ) 1 1 1g f x g f x f x x

2 2( )( ) ( ( )) ( ( )) 1 ( 1) 1

2 1 1 2

f g x f g x g x x

x x x x

Applied ExampleApplied Example: Automobile Pollution: Automobile Pollution An environmental impact study conducted for the city of An environmental impact study conducted for the city of

Oxnard indicates that, under existing environmental Oxnard indicates that, under existing environmental protection laws, protection laws, the level of carbon monoxidethe level of carbon monoxide (CO) (CO) present in the air due to pollution from automobile present in the air due to pollution from automobile exhaust will be exhaust will be 0.010.01xx2/32/3 parts per million when the parts per million when the number ofnumber of motor vehiclesmotor vehicles isis xx thousand. thousand.

A separate study conducted by a state government A separate study conducted by a state government agency estimates that agency estimates that tt years from now the years from now the number of number of motor vehiclesmotor vehicles in Oxnard will be in Oxnard will be 0.20.2tt22 + 4 + 4tt + 64 + 64 thousand. thousand.

Find:Find:a.a. An expression for the concentration of CO in the air due An expression for the concentration of CO in the air due

to automobile exhaust to automobile exhaust tt years from now. years from now.b.b. The level of concentration The level of concentration 55 years from now.years from now.

Applied ExampleApplied Example: Automobile Pollution: Automobile PollutionSolutionSolution Part Part (a)(a)::

✦ The level of CO is described by the function The level of CO is described by the function

gg((xx) =) = 0.010.01xx2/32/3

where where xx is the number (in thousands) of motor vehicles. is the number (in thousands) of motor vehicles.✦ In turn, the number (in thousands) of motor vehicles is In turn, the number (in thousands) of motor vehicles is

described by the functiondescribed by the function

ff((tt) = 0.2) = 0.2tt22 + 4 + 4tt + 64 + 64

where where tt is the number of years from now. is the number of years from now.✦ Therefore, the concentration of CO due to automobile Therefore, the concentration of CO due to automobile

exhaust exhaust tt years from now is given by years from now is given by

((ggff )()(tt) = ) = gg((ff((tt)) = 0.01(0.2)) = 0.01(0.2tt22 + 4 + 4tt + 64) + 64)2/32/3

Applied ExampleApplied Example: Automobile Pollution: Automobile PollutionSolutionSolution Part Part (b)(b)::

✦ The level of CO The level of CO five yearsfive years from now is: from now is:

((ggff )(5))(5) = = gg((ff(5)) = 0.01[0.2(5)(5)) = 0.01[0.2(5)22 + 4(5) + 64] + 4(5) + 64]2/32/3

= (0.01)89= (0.01)892/32/3 ≈≈ 0.20 0.20

or approximately or approximately 0.200.20 parts per million. parts per million.

2.52.5Linear FunctionsLinear Functions

––11 11 22 33 44 55

(1, 2)(1, 2)

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

Applied Example:Applied Example: Linear Depreciation Linear Depreciation

A Web server has an original value of A Web server has an original value of $10,000$10,000 and is to and is to be be depreciated depreciated linearlylinearly over over 55 years with a years with a $3000$3000 scrap scrap value.value.

Find an Find an expressionexpression giving the giving the book valuebook value at the end of at the end of year year tt..

What will be the What will be the book valuebook value of the server at the end of of the server at the end of the the second yearsecond year??

What is the What is the rate of depreciationrate of depreciation of the server? of the server?

Applied Example:Applied Example: Linear Depreciation Linear DepreciationSolutionSolution Let Let VV((tt)) denote the Web server’s denote the Web server’s book valuebook value at the end of at the end of

thethe ttth th year. year. VV is a linear function of is a linear function of tt.. To find an To find an equationequation of the of the straight linestraight line that represents the that represents the

depreciationdepreciation, observe that , observe that VV = 10,000= 10,000 when when t t = 0= 0; this tells ; this tells us that the line us that the line passes throughpasses through the point the point (0, 10,000)(0, 10,000)..

Similarly, the condition that Similarly, the condition that VV = 3000= 3000 when when t t = 5= 5 says that says that the line also the line also passes throughpasses through the point the point (5, 3000)(5, 3000)..

Thus, the Thus, the slopeslope of the line is given by of the line is given by

10,000 3000 7000 14000 5 5

Applied Example:Applied Example: Linear Depreciation Linear DepreciationSolutionSolution Using the Using the point-slope formpoint-slope form of the equation of a line with of the equation of a line with

pointpoint (0, 10,000)(0, 10,000) and and slopeslope mm = –1400 = –1400, we obtain the , we obtain the required expressionrequired expression

The The book valuebook value at the end of the at the end of the second yearsecond year is given by is given by

or or $7200$7200.. The The rate of depreciationrate of depreciation of the server is given by the of the server is given by the

negative negative slopeslope of the depreciation line of the depreciation line mm = –1400 = –1400, so the , so the rate of depreciation is rate of depreciation is $1400$1400 per year. per year.

10,000 1400( 0)1400 10,000

V tV t

(2) 1400(2) 10,000 7200V

Applied Example:Applied Example: Linear Depreciation Linear DepreciationSolutionSolution The graph of The graph of VV is: is:

1400 10,000V t

11 22 33 44 55 66

(0, 10,000)(0, 10,000)VV

(5, 3000)(5, 3000)

10,00010,000

30003000

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 isCC((xx)) ==Total cost of Total cost of manufacturingmanufacturing xx units of the productunits of the product

The The revenue functionrevenue function is isRR((xx)) ==Total revenue realized Total revenue realized from the sale offrom the sale of xx units of the units of the productproduct

The The profit functionprofit function is isPP((xx)) ==Total profit realized Total profit realized from manufacturing and selling from manufacturing and selling xx units of the productunits of the product

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 ( ) 30R x x

( ) ( ) ( )30 (20 20,000)10 20,000

P x R x C xx xx

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 3 3x

––11 11 22 33 44 55

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

(1, 2)(1, 2)

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 12,000x x 6 12,000x

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

2.62.6Quadratic FunctionsQuadratic Functions

–– 11 11 22

–– 11

–– 22

2( ) 2 5 2f x x x 2( ) 2 5 2f x x x

VertexVertex

xx--interceptsintercepts

5 9,4 8

yy--interceptintercept

Quadratic FunctionsQuadratic Functions

A A quadratic functionquadratic function is one of the form is one of the form

where where aa, , bb, and , and c c are are constantsconstants and and a a ≠≠ 00.. For example, the function For example, the function

is is quadraticquadratic, with , with aa = 2 = 2, , bb = – = – 44, and , and cc = 3 = 3..

2( )f x ax bx c

2( ) 2 4 3f x x x

Quadratic FunctionsQuadratic Functions Below is the Below is the graphgraph of the of the quadratic functionquadratic function

The The graphgraph of a quadratic function is a curve called a of a quadratic function is a curve called a parabolaparabola that opens upward or downward. that opens upward or downward.

2( ) 2 4 3f x x x

–– 22 – – 11 11 22 33 44

yy2( ) 2 4 3f x x x

ParabolaParabola

Quadratic FunctionsQuadratic Functions The parabola is The parabola is symmetricsymmetric with respect to a with respect to a vertical vertical

lineline called the called the axis of symmetryaxis of symmetry.. The The axis of symmetry axis of symmetry also passes through the also passes through the lowestlowest or or

highesthighest pointpoint of the parabola, which is called the of the parabola, which is called the vertexvertex of the parabola.of the parabola.

–– 22 – – 11 11 22 33 44

yy Axis of Axis of symmetrysymmetry

VertexVertex (1, 1)(1, 1)

2( ) 2 4 3f x x x

ParabolaParabola

Quadratic FunctionsQuadratic Functions We can We can use these propertiesuse these properties to help us to help us sketch the graphsketch the graph

of a of a quadratic functionquadratic function.. Suppose we want to sketch the graph of Suppose we want to sketch the graph of

If we If we complete the squarecomplete the square in in xx, we obtain, we obtain

Note that Note that ((xx – 1) – 1)22 is is nonnegativenonnegative: it equals to zero when : it equals to zero when x x == 1 1 and is greater than zero if and is greater than zero if x x ≠≠ 1 1. .

Thus, we see that Thus, we see that ff((xx)) – – 22 for all values of for all values of xx. . This tells us the This tells us the vertexvertex of the parabola is the point of the parabola is the point (1,(1, –– 2)2)..

2( ) 3 6 1f x x x

( ) 3( 2 ) 1

3[ 2 ( 1) ] 1 3

3( 1) 2

f x x x

Quadratic FunctionsQuadratic Functions We know the We know the vertexvertex of the parabola is the point of the parabola is the point (1,(1, –– 2)2) and and

that it is the that it is the minimumminimum point of the graph, since point of the graph, since ff((xx)) – – 22 for for all values of all values of xx..

Thus, the graph of Thus, the graph of ff((xx) = 3) = 3xx22 – 6 – 6xx +1 +1 looks as follows: looks as follows:

–– 22 22 44

–– 22

2( ) 3 6 1f x x x

VertexVertex (1, –2)(1, –2)

Properties of Quadratic FunctionsProperties of Quadratic Functions

Given Given ff((xx) = ) = axax22 + + bxbx + +c c ((a a ≠≠ 0)0)

1.1. The The domaindomain of of ff is the set of is the set of all real numbersall real numbers..2.2. If If aa > 0> 0, the , the parabolaparabola opens upwardopens upward, and if , and if aa < 0< 0, ,

it it opens downwardopens downward..

3.3. The The vertexvertex of the of the parabolaparabola is is

4.4. The The axis of symmetryaxis of symmetry of the of the parabolaparabola is is

5.5. The The xx-intercepts-intercepts (if any) are found by solving (if any) are found by solving ff((xx)) = 0= 0. The . The yy-intercept-intercept is is ff(0) =(0) = cc..

,2 2b bfa a

ExampleExample Given the quadratic function Given the quadratic function ff((xx) = –) = – 22xx22 + 5 + 5xx – 2 – 2

a.a. Find the Find the vertexvertex of the parabola. of the parabola.b.b. Find the Find the xx-intercepts-intercepts (if any) of the parabola. (if any) of the parabola.c.c. SketchSketch the parabola. the parabola.

SolutionSolutiona.a. Here Here aa = – = – 22, , bb = 5 = 5, and , and cc = – = – 22. therefore, the . therefore, the xx-coordinate -coordinate

of the of the vertexvertex of the parabola is of the parabola is

The The yy-coordinate -coordinate of the of the vertexvertex is therefore given by is therefore given by

Thus, the Thus, the vertexvertex of the parabola is the of the parabola is the pointpoint

5 52 2( 2) 4ba

25 5 5 92 5 24 4 4 8

5 9,4 8

SolutionSolutionb.b. For the For the xx-intercepts -intercepts of the parabola, we solve the equationof the parabola, we solve the equation

using the using the quadratic formulaquadratic formula with with aa = – = – 22, , bb = 5 = 5, and , and cc = – = – 22. . We findWe find

Thus, the Thus, the xx-intercepts -intercepts of the parabola are of the parabola are 1/2 1/2 and and 22..

22 5 2 0x x

2 5 25 4( 2)( 2)4 5 32 2( 2) 4

b b acxa

SolutionSolutionc.c. The The sketchsketch::

–– 11 11 22

–– 11

–– 22

2( ) 2 5 2f x x x

VertexVertex

xx-intercepts-intercepts

5 9,4 8

yy-intercept-intercept

Some Economic ModelsSome Economic Models

People’s decision on how much to People’s decision on how much to demanddemand or purchase of a or purchase of a given product depends on the given product depends on the priceprice of the product: of the product:

✦ TheThe higherhigher the the priceprice the the lessless they want to they want to buybuy of it. of it.

✦ A A demand functiondemand function pp = = dd((xx)) can be used to describe this. can be used to describe this.

Similarly, firms’ decision on how much to Similarly, firms’ decision on how much to supplysupply or or produce of a product depends on the produce of a product depends on the priceprice of the product: of the product:

✦ TheThe higherhigher the the priceprice, the, the moremore they want to they want to produceproduce of it. of it.

✦ A A supply functionsupply function pp = = ss((xx)) can be used to describe this. can be used to describe this.

The interaction between demand and supply will ensure The interaction between demand and supply will ensure the market settles to a the market settles to a market equilibriummarket equilibrium::

✦ This is the situation at which This is the situation at which quantity demanded quantity demanded equalsequals quantity suppliedquantity supplied..

✦ Graphically, this situation occurs when the Graphically, this situation occurs when the demand demand curvecurve and the and the supply curvesupply curve intersectintersect: where : where dd((xx)) = = ss((xx))..

Applied Example:Applied Example: Supply and Demand Supply and Demand

The The demanddemand functionfunction for a certain brand of bluetooth for a certain brand of bluetooth wireless headset is given bywireless headset is given by

The corresponding The corresponding supply functionsupply function is given by is given by

where where pp is the expressed in is the expressed in dollarsdollars and and xx is measured in is measured in units of a units of a thousandthousand. .

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

2( ) 0.025 0.5 60p d x x x

2( ) 0.02 0.6 20p s x x x

SolutionSolution We solve the following We solve the following system of equationssystem of equations::

SubstitutingSubstituting the the secondsecond equation into the equation into the firstfirst yields: yields:

Thus, either Thus, either xx = – = – 400/9400/9 (but this is (but this is not possiblenot possible), or ), or xx = 20 = 20. . So, the So, the equilibrium quantityequilibrium quantity must be must be 20,000 20,000 headsets.headsets.

0.025 0.5 600.02 0.6 20

p x xp x x

0.02 0.6 20 0.025 0.5 600.045 1.1 40 0

45 1100 40,000 09 220 8,000 09 400 20 0

x x x xx x

x xx xx x

SolutionSolution The The equilibrium priceequilibrium price is given by: is given by:

or or $40$40 per headset.per headset.

20.02 20 0.6 20 20 40p

2.72.7Functions and Mathematical ModelsFunctions and Mathematical Models

55 1010 1515 2020 2525 3030

t t (years)(years)

y y ($trillion)($trillion)

Mathematical ModelsMathematical Models

As we have seen, mathematics can be used to As we have seen, mathematics can be used to solve real-world problems.solve real-world problems.

We will now discuss a few more examples of We will now discuss a few more examples of real-world phenomena, such as:real-world phenomena, such as:✦ The solvency of the U.S. Social Security The solvency of the U.S. Social Security

trust fund trust fund✦ Global warmingGlobal warming

Mathematical ModelingMathematical Modeling

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

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

Real-world problemReal-world problem Mathematical Mathematical modelmodel

Solution of real-Solution of real-world problemworld problem

Solution of Solution of mathematical modelmathematical model

FormulateFormulate

InterpretInterpret

SolveSolveTestTest

Modeling With Polynomial FunctionsModeling With Polynomial Functions

A A polynomial functionpolynomial function of degree of degree nn is a function of the form is a function of the form

where where n n is a is a nonnegative integernonnegative integer and the numbers and the numbers aa00, , aa11, …. , …. aann are are constantsconstants called the called the coefficientscoefficients of of the polynomial function.the polynomial function.

Examples:Examples:✦ The function below is polynomial function of The function below is polynomial function of degree 5degree 5::

5 4 3 212( ) 2 3 2 6f x x x x x

1 21 2 1 0( ) ( 0)n n

n n nf x a x a x a x a x a a

Modeling With Polynomial FunctionsModeling With Polynomial Functions

A A polynomial functionpolynomial function of degree of degree nn is a function of the form is a function of the form

where where n n is a is a nonnegative integernonnegative integer and the numbers and the numbers aa00, , aa11, …. , …. aann are are constantsconstants called the called the coefficientscoefficients of of the polynomial function.the polynomial function.

Examples:Examples:✦ The function below is polynomial function of The function below is polynomial function of degree 3degree 3::

3 2( ) 0.001 0.2 10 200g x x x x

1 21 2 1 0( ) ( 0)n n

n n nf x a x a x a x a x a a

Applied ExampleApplied Example:: Global WarmingGlobal Warming

The increase in The increase in carbon dioxide (COcarbon dioxide (CO22)) in the atmosphere is in the atmosphere is a major cause of global warming.a major cause of global warming.

Below is a table showing the Below is a table showing the average amount of COaverage amount of CO22, , measured in measured in parts per million volume (ppmv)parts per million volume (ppmv) for various for various years from years from 19581958 through through 20072007::

YearYear 19581958 19701970 19741974 19781978 19851985 19911991 19981998 20032003 20072007

AmountAmount 315315 325325 330330 335335 345345 355355 365365 375375 380380

Below is a Below is a scatter plotscatter plot associated with these data: associated with these data:

YearYear 19581958 19701970 19741974 19781978 19851985 19911991 19981998 20032003 20072007

AmountAmount 315315 325325 330330 335335 345345 355355 365365 375375 380380

1010 2020 3030 4040 5050

380380

360360

340340

320320

t t (years)(years)

y y (ppmv)(ppmv)

A A mathematical modelmathematical model giving the approximate amount of giving the approximate amount of COCO22 is given by: is given by:

YearYear 19581958 19701970 19741974 19781978 19851985 19911991 19981998 20032003 20072007

AmountAmount 315315 325325 330330 335335 345345 355355 365365 375375 380380

1010 2020 3030 4040 5050

380380

360360

340340

320320

t t (years)(years)

y y (ppmv)(ppmv)2( ) 0.01076 0.8212 313.4A t t t

a.a. Use the model to Use the model to estimateestimate the average amount of atmospheric the average amount of atmospheric COCO22 in in 19801980 ((tt = 23= 23))..

b.b. Assume that the trend continued and use the model to Assume that the trend continued and use the model to predictpredict the average amount of atmospheric COthe average amount of atmospheric CO22 in in 20102010..

YearYear 19581958 19701970 19741974 19781978 19851985 19911991 19981998 20032003 20072007

AmountAmount 315315 325325 330330 335335 345345 355355 365365 375375 380380

2( ) 0.010716 0.8212 313.4 (1 50) A t t t t

SolutionSolutiona.a. The average amount of atmospheric COThe average amount of atmospheric CO22 in in 19801980 is given by is given by

or approximately 338 ppmv.or approximately 338 ppmv.b.b. Assuming that the trend will continue, the average amount Assuming that the trend will continue, the average amount

of atmospheric COof atmospheric CO22 in in 20102010 will be will be

YearYear 19581958 19701970 19741974 19781978 19851985 19911991 19981998 20032003 20072007

AmountAmount 315315 325325 330330 335335 345345 355355 365365 375375 380380

2(23) 0.010716 23 0.8212 23 313.4 337.96A

2(53) 0.010716 53 0.8212 53 313.4 387.03A

2( ) 0.010716 0.8212 313.4 (1 50) A t t t t

Applied Example: Applied Example: Social Security Trust FundSocial Security Trust Fund

The The projected assetsprojected assets of the Social Security trust fund of the Social Security trust fund (in (in trillions of dollars)trillions of dollars) from from 20082008 through through 20402040 are given by: are given by:

The scatter plot associated with these data is:The scatter plot associated with these data is:

YearYear 20082008 20112011 20142014 20172017 20202020 20232023 20262026 20292029 20322032 20352035 20382038 20402040

AssetsAssets 2.42.4 3.23.2 4.04.0 4.74.7 5.35.3 5.75.7 5.95.9 5.65.6 4.94.9 3.63.6 1.71.7 00

55 1010 1515 2020 2525 3030

t t (years)(years)

The The projected assetsprojected assets of the Social Security trust fund of the Social Security trust fund (in (in trillions of dollars)trillions of dollars) from from 20082008 through through 20402040 are given by: are given by:

A A mathematical modelmathematical model giving the approximate value of assets giving the approximate value of assets in the trust fund (in trillions of dollars) is:in the trust fund (in trillions of dollars) is:

55 1010 1515 2020 2525 3030

t t (years)(years)

y y ($trillion)($trillion)4 3 2( ) 0.00000324 0.000326 0.00342 0.254 2.4A t t t t t

YearYear 20082008 20112011 20142014 20172017 20202020 20232023 20262026 20292029 20322032 20352035 20382038 20402040

AssetsAssets 2.42.4 3.23.2 4.04.0 4.74.7 5.35.3 5.75.7 5.95.9 5.65.6 4.94.9 3.63.6 1.71.7 00

a.a. The first baby boomers will turn The first baby boomers will turn 6565 in in 20112011. What will be . What will be the the assetsassets of the Social Security trust fund of the Social Security trust fund at that timeat that time??

b.b. The last of the baby boomers will turn The last of the baby boomers will turn 6565 in in 20292029. What will . What will the the assetsassets of the trust fund be of the trust fund be at the timeat the time??

c.c. Use the graph of function Use the graph of function AA((tt)) to to estimateestimate the yearthe year in which in which the current Social Security system the current Social Security system will go brokewill go broke..

4 3 2( ) 0.00000324 0.000326 0.00342 0.254 2.4A t t t t t

YearYear 20082008 20112011 20142014 20172017 20202020 20232023 20262026 20292029 20322032 20352035 20382038 20402040

AssetsAssets 2.42.4 3.23.2 4.04.0 4.74.7 5.35.3 5.75.7 5.95.9 5.65.6 4.94.9 3.63.6 1.71.7 00

SolutionSolutiona.a. The assets of the Social Security fund in The assets of the Social Security fund in 20112011 ((tt = 3 = 3)) will be: will be:

or approximately or approximately $3.18$3.18 trillion. trillion.The assets of the Social Security fund in The assets of the Social Security fund in 20292029 ((tt = 21 = 21)) will be: will be:

or approximately or approximately $5.59$5.59 trillion. trillion.

4 3 2(3) 0.00000324 3 0.000326 3 0.00342 3 0.254 3 2.4 3.18A

4 3 2(21) 0.00000324 21 0.000326 21 0.00342 21 0.254 21 2.4 5.59A

YearYear 20082008 20112011 20142014 20172017 20202020 20232023 20262026 20292029 20322032 20352035 20382038 20402040

AssetsAssets 2.42.4 3.23.2 4.04.0 4.74.7 5.35.3 5.75.7 5.95.9 5.65.6 4.94.9 3.63.6 1.71.7 00

4 3 2( ) 0.00000324 0.000326 0.00342 0.254 2.4A t t t t t

SolutionSolutionb.b. The graph shows that function The graph shows that function AA crosses the crosses the tt-axis-axis at about at about

tt = 32 = 32, suggesting the system will go broke by , suggesting the system will go broke by 20402040::

55 1010 1515 2020 2525 3030

Trust runs out of funds

YearYear 20082008 20112011 20142014 20172017 20202020 20232023 20262026 20292029 20322032 20352035 20382038 20402040

AssetsAssets 2.42.4 3.23.2 4.04.0 4.74.7 5.35.3 5.75.7 5.95.9 5.65.6 4.94.9 3.63.6 1.71.7 00

4 3 2( ) 0.00000324 0.000326 0.00342 0.254 2.4A t t t t t

t t (years)(years)

Rational and Power FunctionsRational and Power Functions

A A rational functionrational function is simply is simply the quotient of two the quotient of two polynomialspolynomials..

In general, a rational function has the formIn general, a rational function has the form

where where ff((xx)) and and gg((xx)) are are polynomial functionspolynomial functions.. Since the Since the division by zero is not alloweddivision by zero is not allowed, we conclude that , we conclude that

the the domaindomain of a rational function is the set of all real of a rational function is the set of all real numbers except the numbers except the zeros zeros of of g g (the (the rootsroots of the equation of the equation gg((xx)) = 0= 0))

( )( )( )

f xR xg x

ExamplesExamples of rational functions: of rational functions:

3 23 1( )2

x x xF xx

Functions of the formFunctions of the form

where where rr is is any real numberany real number, are called , are called power functionspower functions.. We encountered examples of power functions earlier in We encountered examples of power functions earlier in

our work. our work. ExamplesExamples of power functions: of power functions:

1/2 22

1( ) ( ) and f x x x g x xx

( ) rf x x

Many functions involve Many functions involve combinationscombinations of of rationalrational and and power functionspower functions..

Examples:Examples:2

1/22 3/2

( ) 3 4

1( ) (1 2 )( 2)

g x x x

h x xx

Applied Example:Applied Example: Driving Costs Driving Costs

A study of driving costs based on a A study of driving costs based on a 20072007 medium-sized medium-sized sedan found the following average costs (car payments, sedan found the following average costs (car payments, gas, insurance, upkeep, and depreciation), measured in gas, insurance, upkeep, and depreciation), measured in cents per mile:cents per mile:

A A mathematical modelmathematical model giving the average cost in cents per giving the average cost in cents per mile is:mile is:

where where xx (in thousands)(in thousands) denotes the number of miles the denotes the number of miles the car is driven in car is driven in 11 year. year.

Miles/year, Miles/year, xx 50005000 10,00010,000 15,00015,000 20,00020,000

Cost/mile, Cost/mile, yy ((¢¢)) 83.883.8 62.962.9 52.252.2 47.147.1

164.8( )C xx

Below is the scatter plot associated with this data:Below is the scatter plot associated with this data:

164.8( )C xx

55 1010 1515 2020 2525

140140120120100100

8080606040402020

x x ((☓☓1000 miles/1000 miles/year)year)

y y ((¢¢))

CC((xx))

Using this model, estimate the Using this model, estimate the average costaverage cost of driving a of driving a 20072007 medium-sized sedan medium-sized sedan 8,0008,000 miles per year and miles per year and 18,00018,000 miles per year.miles per year.

SolutionSolution The The average costaverage cost for driving a car for driving a car 8,0008,000 miles per year is miles per year is

or approximately or approximately 68.868.8¢/mile¢/mile..

164.8( )C xx

0.42164.8(8) 68.818

Using this model, estimate the Using this model, estimate the average costaverage cost of driving a of driving a 20072007 medium-sized sedan medium-sized sedan 8,000 8,000 miles per year and miles per year and 18,00018,000 miles per year.miles per year.

SolutionSolution The The average costaverage cost for driving a car for driving a car 18,00018,000 miles per year is miles per year is

or approximately or approximately 48.9548.95¢/mile¢/mile..

164.8( )C xx

0.42164.8(18) 48.9518

Constructing Mathematical ModelsConstructing Mathematical Models Some mathematical models can be constructed using Some mathematical models can be constructed using

elementary geometricelementary geometric and and algebraic argumentsalgebraic arguments..

GuidelinesGuidelines for constructing mathematical models: for constructing mathematical models:1.1. Assign a letter to each variableAssign a letter to each variable mentioned in the mentioned in the

problem. If appropriate, problem. If appropriate, draw and labeldraw and label a figure. a figure.2.2. Find an expressionFind an expression for the quantity sought. for the quantity sought.3.3. Use the Use the conditionsconditions given in the problem to given in the problem to write write

the quantity sought as a function the quantity sought as a function ff of one variable. of one variable. Note any Note any restrictionsrestrictions to be placed on the to be placed on the domaindomain of of f f by the nature of the problem. by the nature of the problem.

Applied Example:Applied Example: Enclosing an Area Enclosing an Area

The owner of the Rancho Los Feliz has The owner of the Rancho Los Feliz has 30003000 yards of yards of fencingfencing with which to with which to enclose a rectangular pieceenclose a rectangular piece of of grazing land along the straight portion of a river.grazing land along the straight portion of a river.

Fencing is Fencing is not requirednot required along the along the riverriver.. Letting Letting xx denote the denote the widthwidth of the rectangle, find a of the rectangle, find a functionfunction

ff in the in the variablevariable xx giving the giving the areaarea of the grazing land if she of the grazing land if she uses alluses all of the fencing. of the fencing.

Applied Example:Applied Example: Enclosing an Area Enclosing an AreaSolutionSolution This information was This information was givengiven::

✦ The The areaarea of the rectangular grazing land is of the rectangular grazing land is AA = = xyxy..✦ The amount of The amount of fencingfencing is is 22xx + + yy which must equal which must equal 30003000

(to use all the fencing), so:(to use all the fencing), so:22xx + + y y = 3000 = 3000

Solving forSolving for y y we get: we get:y y = 3000 – 2= 3000 – 2xx

SubstitutingSubstituting this value of this value of yy into the expression for into the expression for AA gives: gives:A A = = xx(3000 – 2(3000 – 2xx) = 3000) = 3000xx – 2 – 2xx22

Finally,Finally, x x and and yy represent represent distancesdistances, so they must be , so they must be nonnegativenonnegative, so , so xx 0 0 and and yy = 3000 – 2= 3000 – 2xx 0 0 (or (or xx 1500 1500). ).

Thus, the Thus, the required functionrequired function is: is: ff((xx) =) = 30003000xx – 2 – 2xx22 (0 (0 xx 1500) 1500)

Applied Example:Applied Example: Charter-Flight Revenue Charter-Flight Revenue

If exactly If exactly 200200 people sign up for a people sign up for a charter flightcharter flight, Leisure , Leisure World Travel Agency charges World Travel Agency charges $300$300 per person. per person.

However, if However, if moremore than than 200200 people sign up for the flight people sign up for the flight (assume this is the case), then each fare is (assume this is the case), then each fare is reducedreduced by by $1$1 for each additional personfor each additional person..

Letting Letting xx denote the denote the number of passengers abovenumber of passengers above 200200, find , find a a functionfunction giving the giving the revenuerevenue realized by the company. realized by the company.

Applied Example:Applied Example: Charter-Flight Revenue Charter-Flight Revenue

SolutionSolution This information was given.This information was given.

✦ If there are If there are xx passengers above passengers above 200200, then the number of , then the number of passengers signing uppassengers signing up for the flight is for the flight is 200 + 200 + xx..

✦ The The farefare will be will be (300 – (300 – xx)) dollars dollars per passengerper passenger.. The The revenuerevenue will be will be

RR = (200 + = (200 + xx)(300 – )(300 – xx))= –= – xx22 + 100 + 100xx + 60,000 + 60,000

The quantities The quantities must be positivemust be positive, so , so xx 0 0 and and 300 – 300 – xx 0 0 (or (or xx 300 300). ).

So the So the required functionrequired function is: is: ff((xx) =) = –– xx22 + 100 + 100xx + 60,000 (0 + 60,000 (0 xx 300) 300)

End of End of Chapter Chapter

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