functions? E XPONENTIAL F UNCTIONS Always involves the equation: b x - - PowerPoint PPT Presentation

DAY 107 – EXPONENTIAL GROWTH

AND DECAY

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:

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

x

y 2 

THE TABLE: RESULTS

x f(x) = 2x

• 1

2-1 = ½ 20 = 1 1 21 = 2 2 22 = 4 3 23 = 8

THE GRAPH OF

x

y 2 

OBSERVATIONS

What did you notice? What is the pattern? What would happen if What would happen if What real-life applications are there?

2   x

5  x

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*(0.10) = $110 Year 2: $110 + 110*(0.10) = $121 Year 3: $121 + 121*(0.10) = $133.10 Year 4: $133.10 + 133.10*(0.10) = $146.41 Year 5: $146.41 + 1461.41*(0.10) = $161.05

GROWTH MODELS: INVESTING

The Equation is:

P = Principal r = Annual Rate t = Number of years

t

r p a ) 1 (  

USING THE EQUATION

$100.00 10% interest 5 years 100(1+ (.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:

a = initial amount r = percent decrease t = Number of years

t

r a y ) 1 (  

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

Depreciation Rate 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:

a = initial amount r = percent decrease t = Number of years

t

a y ) 5 1 (  

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

 Exponential Decay: Car Depreciation