Algebra 1 Warm Up 9 April 2012State the recursive sequence (start?, how is it changing?), then find the next 3 terms.Also find the EQUATION for each. y = a b∙ x

1) 12000, 10800,9720, ___, ___, ___2) 100, 105.25,110.77, ___, ___, ___Rewrite as a fraction and decimal:3) a) 5% b) 50% c) 5.25%

Homework due TuesdayHomework due Tuesday: pg. 345: 1 – 5 ADV: 12: pg. 345: 1 – 5 ADV: 12

OBJECTIVEToday we will explore exponential

growth and decay patterns and write exponential equations.

Today we will take notes, work problems with our groups and present to the class.

Once upon a time, two merchants were trying to work out a deal. For the next month, the 1st merchant was going to give $10,000 to the 2nd merchant, and in return, he would receive 1 cent the first day, 2 cents the second, 4 cents in the third, and so on, each time doubling the amount.

After 1 month, who came out ahead?After 1 month, who came out ahead? THINK- PAIR- SHARETHINK- PAIR- SHARE

Group: Money Doubling?

• You have a $100.00• Your money doubles each year.• How much do you have in 5 years?• Show work. Use a table and/or equation!

Money Doubling

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

Earning Interest

• You have $100.00.• Each year you earn 10% interest.• How much $ do you have in 5 years?• Show Work.

• HINT…how much is 10% of $100? HINT…..can you find a constant multiplier?

Earning 10% results

Year 1: $100 + 100·(.10) = $110Year 2: $110 + 110·(.10) = $121Year 3: $121 + 121·(.10) = $133.10Year 4: $133.10 + 133.10·(.10) = $146.41Year 5: $146.41 + 1461.41·(.10) = $161.05Can you find an equation? start at 100, CM = 110/100 = 1.1 Equation?

y = 100(1.1)x y = 100(1.1)5=161.05

Growth Models: Investing

The equation for constant percent growth is

y = A (1+ )x

A = starting value (principal)r = rate of growth (÷100 to put in decimal form)

x = number of time periods elapsedy = final value

100

r

Using the Equation

• $100.00• 10% interest • 5 years• 100(1+ )5 = 100( 1 + 0.10)5

= 100 (1.1)5 = $161.05

10

100

10% as a fraction

Constant multiplier

10% as a decimal

Comparing Investmentswhich is better?

• 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: y =$10,000 (1 + )9

=10,000 (1 + 0.055)9

= 10,000(1.055)9

Balance after 9 years: $16,190.94

5.5

100

Choice 2

$8,000 in an account that pays 6.5% interest for 10 years. Equation: y=$8000 (1 + )10

=8,000 (1 + .065)10

=8,000(1 + 0.065)10

 Balance after 10 years: $15,071.10

 

6.5

100

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 – )x

A = starting valuer = rate of decrease (÷100 to put in decimal form)

x= number of time periods elapsedy = final value

100

r

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

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 – )t

a = initial amountr = percent decrease t = Number of years

100

r

debriefHow does the exponential growth differ from linear growth?

How does the difference show up in the table?

How does the difference show up on the graph?

Worksheet find then towards the end of page http://www.uen.org/Lessonplan/preview.cgi?LPid=24626

http://www.regentsprep.org/regents/math/algebra/AE7/ExpDecayL.htm