section 9b linear modeling

Section 9BSection 9BLinear ModelingLinear Modeling

Pages 542-553Pages 542-553

Linear FunctionsLinear Functions

A A Linear FunctionLinear Function changes by the changes by the same absolute amountsame absolute amount for each unit for each unit of change in the input (independent of change in the input (independent variable). variable).

A A Linear FunctionLinear Function has a constant has a constant rate of changerate of change..

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Examples:Straightown population as a function of time.

Postage cost as a function of weight.

Pineapple demand as a function of price.

First Class Mail – a linear functionFirst Class Mail – a linear function Weight Postage

cost

1 oz1 oz $0.37$0.37

2 oz2 oz $0.60$0.60

3 oz3 oz $0.83$0.83

4 oz4 oz $1.06 $1.06

5 oz5 oz $1.29$1.29

6 oz6 oz $1.52 $1.52

7 oz7 oz $1.75$1.75

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First Class Mail – a linear functionFirst Class Mail – a linear function Weight Postage

costDifferenc

e

1 oz1 oz $0.37$0.37

2 oz2 oz $0.60$0.60

3 oz3 oz $0.83$0.83

4 oz4 oz $1.06 $1.06

5 oz5 oz $1.29$1.29

6 oz6 oz $1.52 $1.52

7 oz7 oz $1.75$1.75

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First Class Mail – a linear functionFirst Class Mail – a linear function Weight Postage

costDifferenc

e

1 oz1 oz $0.37$0.37

2 oz2 oz $0.60$0.60 $0.23$0.23

3 oz3 oz $0.83$0.83

4 oz4 oz $1.06 $1.06

5 oz5 oz $1.29$1.29

6 oz6 oz $1.52 $1.52

7 oz7 oz $1.75$1.75

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First Class Mail – a linear functionFirst Class Mail – a linear function Weight Postage

costDifferenc

e

1 oz1 oz $0.37$0.37

2 oz2 oz $0.60$0.60 $0.23$0.23

3 oz3 oz $0.83$0.83 $0.23$0.23

4 oz4 oz $1.06 $1.06

5 oz5 oz $1.29$1.29

6 oz6 oz $1.52 $1.52

7 oz7 oz $1.75$1.75

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First Class Mail – a linear functionFirst Class Mail – a linear function Weight Postage

costDifferenc

e

1 oz1 oz $0.37$0.37

2 oz2 oz $0.60$0.60 $0.23$0.23

3 oz3 oz $0.83$0.83 $0.23$0.23

4 oz4 oz $1.06 $1.06 $0.23$0.23

5 oz5 oz $1.29$1.29

6 oz6 oz $1.52 $1.52

7 oz7 oz $1.75$1.75

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First Class Mail – a linear functionFirst Class Mail – a linear function Weight Postage

costDifferenc

e

1 oz1 oz $0.37$0.37

2 oz2 oz $0.60$0.60 $0.23$0.23

3 oz3 oz $0.83$0.83 $0.23$0.23

4 oz4 oz $1.06 $1.06 $0.23$0.23

5 oz5 oz $1.29$1.29 $0.23$0.23

6 oz6 oz $1.52 $1.52

7 oz7 oz $1.75$1.75

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First Class Mail – a linear functionFirst Class Mail – a linear function Weight Postage

costDifferenc

e

1 oz1 oz $0.37$0.37

2 oz2 oz $0.60$0.60 $0.23$0.23

3 oz3 oz $0.83$0.83 $0.23$0.23

4 oz4 oz $1.06 $1.06 $0.23$0.23

5 oz5 oz $1.29$1.29 $0.23$0.23

6 oz6 oz $1.52 $1.52 $0.23$0.23

7 oz7 oz $1.75$1.75

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First Class Mail – a linear functionFirst Class Mail – a linear function Weight Postage

costDifferenc

e

1 oz1 oz $0.37$0.37

2 oz2 oz $0.60$0.60 $0.23$0.23

3 oz3 oz $0.83$0.83 $0.23$0.23

4 oz4 oz $1.06 $1.06 $0.23$0.23

5 oz5 oz $1.29$1.29 $0.23$0.23

6 oz6 oz $1.52 $1.52 $0.23$0.23

7 oz7 oz $1.75$1.75 $0.23$0.23

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First Class PostageFirst Class Postage9-B

First Class Postage Cost as a function of Weight

$0.00

$0.50

$1.00

$1.50

$2.00

0 1 2 3 4 5 6 7 8

Weight (oz)

Po

stag

e

First class postage – a First class postage – a linear functionlinear function

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$0.23$0.23

1rateof change per ounce

oz

First class postage – a First class postage – a linear functionlinear function

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$0.46$0.23

2rateof change per ounce

oz

First class postage – a First class postage – a linear functionlinear function

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$1.15$0.23

5rateof change per ounce

oz

change in variable

change in

depen

ind

den

variablependent

t

erateof change

2

2 1

1(

( )

)yrateof ch

ya

xn

xge

We define ‘rate of change’ We define ‘rate of change’ of a linear function by:of a linear function by:

where (x1,y1) and (x2,y2) are any two ordered pairs of the function.

Slope = rate of changeSlope = rate of change9-B

Linear FunctionsLinear FunctionsA A linear functionlinear function has a has a constant rate of constant rate of

changechange and a and a straight line graphstraight line graph..

The The rate of change = slope of the rate of change = slope of the graphgraph..The greater the rate of change, the The greater the rate of change, the steeper the slope.steeper the slope.

positive slope positive slope negative negative slopeslope

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rise

runslope

Example: Price-Demand Example: Price-Demand FunctionFunction

A linear function is used to describe A linear function is used to describe how the demand for pineapples varies how the demand for pineapples varies with the price.with the price.

($2, 80 pineapples) and ($5, 50 ($2, 80 pineapples) and ($5, 50 pineapples).pineapples).

Find the rate of change (slope) for this Find the rate of change (slope) for this function and then graph the function.function and then graph the function.

independent variable = priceindependent variable = price

dependent variable = demand for dependent variable = demand for pineapplespineapples

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Example: Price-Demand Example: Price-Demand FunctionFunction

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change in demand

change in priceslope

($2, 80 pineapples) and ($5, 50 pineapples)($2, 80 pineapples) and ($5, 50 pineapples)

80 50

$2 $5

pineapples pineapples

30

$3

pineapples

10 /pineapple dollar

50 80

$5 $2

pineapples pineapplesor

($2, 80 pineapples) and ($5, 50 pineapples).($2, 80 pineapples) and ($5, 50 pineapples).

To graph a linear function you need 2 To graph a linear function you need 2 things:things:

• two pointstwo points or or• slope and one pointslope and one point

Example: Price-Demand Example: Price-Demand FunctionFunction

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change in demand10 $

change in priceslope pineapples per

Example: Price-Demand Example: Price-Demand FunctionFunction

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($2, 80 pineapples) and ($5, 50 pineapples).($2, 80 pineapples) and ($5, 50 pineapples).

Demand for Pineapples as a function of Price

0

20

40

60

80

100

$- $2.00 $4.00 $6.00 $8.00 $10.00

Price

Dem

and

Example: Price-Demand Example: Price-Demand FunctionFunction

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($2, 80 pineapples) and ($5, 50 pineapples).($2, 80 pineapples) and ($5, 50 pineapples).

Demand for Pineapples as a function of Price

0

20

40

60

80

100

120

$- $2.00 $4.00 $6.00 $8.00 $10.00

Price

Dem

and

General Equation for a General Equation for a Linear FunctionLinear Function

dependent = initial value + (slope)×independent yy = initial value + (slope)×xx(Initial value occurs when the independent variable =

0.)

y y = = mmxx + + b b or or

y y = b + mx

m = m = slope slope

bb = = yy-intercept -intercept

(The line goes through the point (0,b).)(The line goes through the point (0,b).)

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Example:Example:dep. variable = initial value + (slope)× indep. dep. variable = initial value + (slope)× indep.

variablevariable

slope = -10 pineapples/$slope = -10 pineapples/$ initial value = 100 pineapplesinitial value = 100 pineapples

Demand = 100 - 10×(price)Demand = 100 - 10×(price)DD = 100 – 10 = 100 – 10pp

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Demand for Pineapples as a function of Price

0

20

40

60

80

100

120

$- $2.00 $4.00 $6.00 $8.00 $10.00

Price

Dem

and

Example:Example:

Demand = 100 - 10×(price)Demand = 100 - 10×(price)DD = 100 – 10 = 100 – 10ppCheck: $2: 100 - 10×2 = 80 pineapplesCheck: $2: 100 - 10×2 = 80 pineapples

$5: 100 - 10×5 = 50 pineapples$5: 100 - 10×5 = 50 pineapples

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Demand for Pineapples as a function of Price

0

20

40

60

80

100

120

$- $2.00 $4.00 $6.00 $8.00 $10.00

Price

Dem

and

t P=f(t)

00 f(0)=10,00f(0)=10,0000

55 f(5)=12,50f(5)=12,5000

1010 f(10)=15,0f(10)=15,00000

1515 f(15)=17,5f(15)=17,50000

2020 f(20)=20,0f(20)=20,00000

4040 f(40)=30,0f(40)=30,00000

Growth of Straightown

20, 20000

5, 12500

0, 10000

40, 30000

15, 17500

10, 15000

0

5000

10000

15000

20000

25000

30000

35000

0 10 20 30 40 50

years

po

pu

lati

on

Data Table

Graph

old example: The initial population of Straightown is 10, 000 and increases by 500 people per year.

t P=f(t)

00 10,00010,000

55 12,50012,500

1010 15,00015,000

1515 17,50017,500

2020 20,00020,000

4040 30,00030,000

old example: The initial population of Straightown is 10, 000 and increases by 500 people per year.

15,000-10,000

10-0slope

12,500-10,000

5-0rateof change

20,000-12,500

20-5rateof change

= 500

= 500

= 500

Rate of change (slope) is ALWAYS 500 (people per year).

Initial population is 10,000 (people).

Linear Function: Population = 10,000 + 500×(year)

Example – First class postageExample – First class postage

Weight Postage cost

1 oz1 oz $0.37$0.37

2 oz2 oz $0.60$0.60

3 oz3 oz $0.83$0.83

4 oz4 oz $1.06 $1.06

5 oz5 oz $1.29$1.29

6 oz6 oz $1.52 $1.52

7 oz7 oz $1.75$1.75

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Slope = Slope = $.23/ounce$.23/ounce

initial value = initial value = $0.14$0.14

Example: First Class PostageExample: First Class Postage

Slope = $.23/ounceSlope = $.23/ounce

initial value = $0.14initial value = $0.14

Postage = $0.14 + $0.23×(weight)Postage = $0.14 + $0.23×(weight)

PP = $0.14+ $0.23 = $0.14+ $0.23ww

Check: 1 ounce: $0.14+ $0.23×1 = $0.37Check: 1 ounce: $0.14+ $0.23×1 = $0.37

6 ounces: $0.14 + $0.23×6 = 6 ounces: $0.14 + $0.23×6 = $1.52$1.52

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First class postage as a function of weight

$-

$0.50

$1.00

$1.50

$2.00

0 1 2 3 4 5 6 7 8

Weight (oz)

Po

stag

e

Example:Example:The world record time in the 100-meter butterfly The world record time in the 100-meter butterfly

was 53.0 seconds in 1988. Assume that the was 53.0 seconds in 1988. Assume that the record record fallsfalls at a constant rate of at a constant rate of 0.05 seconds 0.05 seconds per yearper year. What does the model predict for the . What does the model predict for the record in 2010?record in 2010?

dependent variable dependent variable = world record time (R)= world record time (R)independent variableindependent variable is time, is time, tt (years) after (years) after

1988.1988.SlopeSlope = 0.05 seconds; = 0.05 seconds; initial valueinitial value = 53.0 = 53.0

seconds;seconds;Record time = 53.0 – 0.05×(t years after 1988)

R = 53 – 0.05tRecord time in 2010 = 53 - .05×(22) = 51.9 Record time in 2010 = 53 - .05×(22) = 51.9

secondsseconds

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Example:Example:9-B

Suppose you were 20 inches long at birth Suppose you were 20 inches long at birth and 4 ft tall on your tenth birthday. and 4 ft tall on your tenth birthday. Create a Create a linear equationlinear equation that describes that describes how your height varies with age.how your height varies with age.

independent variable = age (years)independent variable = age (years)

dependent variable = height (inches)dependent variable = height (inches)

Two points: (0, 20) (10, 48)Two points: (0, 20) (10, 48)

Initial value = 20 inchesInitial value = 20 inches

Height = 20 + 2.8t t = years

48 202.8 /

10 0slope in yr

Example:Example:9-B

““Fines for Certain PrePayable Violations” – Fines for Certain PrePayable Violations” – Speeding other than residence zone, highway Speeding other than residence zone, highway work zone and school crosswalk: work zone and school crosswalk: $5.00 per $5.00 per MPH over speed limitMPH over speed limit

plus processing fee ($51.00) and local fees plus processing fee ($51.00) and local fees ($5.00)($5.00)

independent variable = miles over speed limitindependent variable = miles over speed limit

dependent variable = fine ($)dependent variable = fine ($)

Initial valueInitial value = $56.00 = $56.00 SlopeSlope = $5.00 = $5.00

Fine = $56 + $5(your speed-speed limit)

Example:Example:9-B

Mrs. M. was given a ticket for doing 52 Mrs. M. was given a ticket for doing 52 mph in a zone where the speed limit mph in a zone where the speed limit was 35 mph. How much was her fine?was 35 mph. How much was her fine?

Fine = $55 + $5(her speed-35)

Fine = $56 + $5(52-35) = $56 + $5(17)

= $141

Example:Example:9-B

““Fines for Certain PrePayable Violations” – Fines for Certain PrePayable Violations” – Speeding in a Speeding in a residence zoneresidence zone: $200 plus : $200 plus $7.00 per MPH over speed limit (25 mph), $7.00 per MPH over speed limit (25 mph), plus processing fee ($51.00) and local fees plus processing fee ($51.00) and local fees ($5.00)($5.00)

independent variable = miles over speed limitindependent variable = miles over speed limit

dependent variable = fine ($)dependent variable = fine ($)

Initial valueInitial value = $256.00 = $256.00 SlopeSlope = $7.00 = $7.00

Fine = $256 + $7(your speed-25)

Example:Example:9-B

The Psychology Club plans to pay a visitor $75 to The Psychology Club plans to pay a visitor $75 to speak at a fundraiser. Tickets will be sold for $2 speak at a fundraiser. Tickets will be sold for $2 apiece. Find a linear equation that gives the apiece. Find a linear equation that gives the profit/loss for the event as it varies with the profit/loss for the event as it varies with the number of tickets sold. number of tickets sold.

independent variable = number of tickets soldindependent variable = number of tickets sold

dependent variable = profit/loss ($)dependent variable = profit/loss ($)

(0, -$75) (0, -$75) slopeslope = +$2 (= rate of change in = +$2 (= rate of change in ticket price)ticket price)

Profit = -$75 +2×(number of tickets)Profit = -$75 +2×(number of tickets)

P = -$75 +2P = -$75 +2nn

Example:Example:9-B

How many people must attend for How many people must attend for the club to break even?the club to break even?

P = -$75 +2n

0 = -$75 + 20 = -$75 + 2nn

$75 = 2$75 = 2nn

37.5 = 37.5 = nn

Can’t sell half a ticket -- so we’ll Can’t sell half a ticket -- so we’ll need to sell need to sell 38 tickets38 tickets..

HomeworkHomework

Pages 553-555Pages 553-555

# 8, 12a-b, 14a-b, 18, 26, 28, 30, # 8, 12a-b, 14a-b, 18, 26, 28, 30, 3333

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