Applications that Apply to Me!

Exponential Function

What do we know about exponents?

What do we know about functions?

Exponential Functions

Always involves the equation: bx

Example:23 = 2 · 2 · 2 = 8

Group investigation:Y = 2x

Create an x,y table.Use x values of -1, 0, 1, 2, 3, Graph the tableWhat do you observe.

The Table: ResultsX F(x) = 2x

-1 2-1 = ½

0 20 = 1

1 21 = 2

2 22 = 4

3 23 = 8

The Graph of y = 2x

ObservationsWhat did you notice?What is the pattern?What would happen if x= -2What would happen if x = 5What real-life applications are there?

Group: Money Doubling?You have a $100.00Your money doubles each year.How much do you have in 5 years?

Show work.

Money DoublingYear 1: $100 · 2 = $200Year 2: $200 · 2 = $400Year 3: $400 · 2 = $800Year 4: $800 · 2 = $1600Year 5: $1600 · 2 = $3200

Earning Interest onYou have $100.00.Each year you earn 10% interest.

How much $ do you have in 5 years?

Show Work.

Earning 10% resultsYear 1: $100 + 100·(.10) = $110Year 2: $110 + 110·(.10) = $121Year 3: $121 + 121·(.10) = $133.10Year 4: $133.10 + 133.10·(.10) = $146.41

Year 5: $146.41 + 1461.41·(.10) = $161.05

Growth Models: Investing

The Equation is:A = P (1+ r)t

P = Principalr = Annual Rate

t = Number of years

Using the Equation$100.0010% interest5 years100(1+ 100·(.10))5 = $161.05

What could we figure out now?

Comparing InvestmentsChoice 1

$10,000 5.5% interest 9 years

Choice 2$8,0006.5% interest10 years

Choice 1$10,000, 5.5% interest for 9 years.

Equation: $10,000 (1 + .055)9

 Balance after 9 years: $16,190.94

  

Choice 2$8,000 in an account that pays 6.5% interest for 10 years. Equation: $8,000 (1 + .065)10

 Balance after 10 years:$15,071.10

 

Which Investment?

The first one yields more money.

Choice 1: $16,190.94 Choice 2: $15,071.10

Exponential DecayInstead of increasing, it is

decreasing.

Formula: y = a (1 – r)t

a = initial amountr = percent decreaset = Number of years

Real-life ExamplesWhat is car depreciation?Car Value = $20,000Depreciates 10% a yearFigure out the following values:

After 2 yearsAfter 5 yearsAfter 8 yearsAfter 10 years

Exponential Decay: Car Depreciation

DepreciationRate

Value after 2 years

Value after 5 years

Value after 8 years

Value after 10 years

10% $16,200 $11,809.80 $8609.34 $6973.57

Assume the car was purchased for $20,000

Formula: y = a (1 – r)t

a = initial amountr = percent decreaset = Number of years

What Else?What happens when the depreciation rate changes.

What happens to the values after 20 or 30 years out – does it make sense?

What are the pros and cons of buying new or used cars.

Assignment 2 Worksheets:

Exponential Growth: Investing Worksheet (available at

ttp://www.uen.org/Lessonplan/preview.cgi?LPid=24626)

Exponential Decay: Car Depreciation