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:
x
Y = 2
 Create an x,y table.
 Use x values of -1, 0, 1, 2, 3,
 Graph the table
 What do you observe.
The Table: Results
X
-1
0
1
2
3
F(x) = 2x
2-1 = ½
20 = 1
21 = 2
22 = 4
23 = 8
The Graph of y =
x
2
Observations
 What did you notice?
 What is the pattern?
 What would happen if x= -2
 What would happen if x = 5
 What real-life applications are there?
Group: Money Doubling?
 You have a $100.00
 Your money doubles each year.
 How much do you have in 5 years?
 Show work.
Money Doubling
Year 1: $100 · 2 = $200
Year 2: $200 · 2 = $400
Year 3: $400 · 2 = $800
Year 4: $800 · 2 = $1600
Year 5: $1600 · 2 = $3200
Earning Interest on
 You have $100.00.
 Each year you earn 10% interest.
 How much $ do you have in 5 years?
 Show Work.
Earning 10% results
Year 1: $100 + 100·(.10) = $110
Year 2: $110 + 110·(.10) = $121
Year 3: $121 + 121·(.10) = $133.10
Year 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+
t
r)
P = Principal
r = Annual Rate
t = Number of years
Using the Equation
 $100.00
 10% interest
 5 years
 100(1+ 100·(.10))5 = $161.05
 What could we figure out now?
Comparing Investments
 Choice 1
 $10,000
 5.5% interest
 9 years
 Choice 2
 $8,000
 6.5% interest
 10 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 Decay
Instead of increasing, it is decreasing.
Formula: y = a (1 – r)t
a = initial amount
r = percent decrease
t = Number of years
Real-life Examples
 What is car depreciation?
 Car Value = $20,000
 Depreciates 10% a year
 Figure out the following values:
 After 2 years
 After 5 years
 After 8 years
 After 10 years
Exponential Decay:
Car Depreciation
Assume the car was purchased for $20,000
Depreciation
Rate
10%
Value after
2 years
$16,200
Value after
5 years
$11,809.80
Value after
8 years
$8609.34
Formula: y = a (1 – r)t
a = initial amount
r = percent decrease
t = Number of years
Value after
10 years
$6973.57
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