Applications that Apply to Me!
Exponential Function
What do we know about exponents?
What do we know about functions?
Exponential FunctionsAlways 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 = 11 21 = 22 22 = 43 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
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