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### Section 9BLinear Modeling

Pages 542-553

Linear Functions

A Linear Functionchanges by the same absolute amount for each unit of change in the input (independent variable).

A Linear Function has a constant rate of change.

Examples:Straightown population as a function of time.

Postage cost as a function of weight.

Pineapple demand as a function of price.

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

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

Linear Functions

A linear function has a constant rate of change

and a straight line graph.

The rate of change = slope of the graph.

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

positive slope negative slope

Example: Price-Demand Function

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

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

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

independent variable = price

dependent variable = demand for pineapples

Example: Price-Demand Function

- ($2, 80 pineapples) and ($5, 50 pineapples).
- To graph a linear function you need 2 things:
- two points or
- slope and one point

General Equation for a Linear Function

dependent = initial value + (slope)×independent

y= initial value + (slope)×x

(Initial value occurs when the independent variable = 0.)

y = mx + b or

y = b + mx

m = slope

b = y-intercept

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

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

Graph

Data Table

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

= 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 Postage

Slope = $.23/ounce

initial value = $0.14

Postage = $0.14 + $0.23×(weight)

P = $0.14+ $0.23w

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

6 ounces: $0.14 + $0.23×6 = $1.52

Example:

The world record time in the 100-meter butterfly was 53.0 seconds in 1988. Assume that the record falls at a constant rate of 0.05 seconds per year. What does the model predict for the record in 2010?

dependent variable = world record time (R)

independent variable is time, t (years) after 1988.

Slope = 0.05 seconds; initial value = 53.0 seconds;

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

R = 53 – 0.05t

Record time in 2010 = 53 - .05×(22) = 51.9 seconds

Example:

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

independent variable = age (years)

dependent variable = height (inches)

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

Initial value = 20 inches

Height = 20 + 2.8t t = years

Example:

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

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

independent variable = miles over speed limit

dependent variable = fine ($)

Initial value = $56.00 Slope = $5.00

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

Example:

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

independent variable = miles over speed limit

dependent variable = fine ($)

Initial value = $256.00 Slope = $7.00

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

Example:

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

independent variable = number of tickets sold

dependent variable = profit/loss ($)

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

Profit = -$75 +2×(number of tickets)

P = -$75 +2n

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