Download Algebra 1-9 April 2012 6 2 Exponential - Shope-Math

Algebra 1 Warm Up
9 April 2012
State the recursive sequence (start?, how is it
changing?), then find the next 3 terms.
Also find the EQUATION for each. y = a∙bx
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 Tuesday: pg. 345: 1 – 5 ADV: 12
OBJECTIVE
Today 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?
THINK- 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 = $200
Year 2: $200 · 2 = $400
Year 3: $400 · 2 = $800
Year 4: $800 · 2 = $1600
Year 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) = $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
Can you find an equation?
start at 100, CM = 110/100 = 1.1
y = 100(1.1)x
Equation?
y = 100(1.1)5=161.05
Growth Models: Investing
The equation for
constant percent growth is
y = A (1+
r
x
100
)
A = starting value (principal)
r = rate of growth (÷100 to put in decimal form)
x = number of time periods elapsed
y = final value
Using the Equation
•
•
•
•
$100.00
10% interest
10% as a fraction
5 years
10% as a decimal
10 5
100(1+ 100 )
= 100( 1 + 0.10)5
= 100 (1.1)5 = $161.05
Constant multiplier
Comparing Investments
which 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 +
5.5 9
100
)
=10,000 (1 + 0.055)9
= 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: y=$8000 (1
6.5
+
100
)10
=8,000 (1 + .065)10
=8,000(1 + 0.065)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
100
)x
A = starting value
r = rate of decrease (÷100 to put in decimal form)
x= number of time periods elapsed
y = final value
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
Formula: y = a (1 –
a = initial amount
r = percent decrease
t = Number of years
Value after
8 years
$8609.34
r
100
)t
Value after
10 years
$6973.57
debrief
How 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