1. C2 D6
Bellwork:
Solve for x. Be able to give the exact and approximate answers. 2) 3) 1) +1 +1 +1 +2 +1 +1 +1 +1 +1 +1
total: +1

2. ( ) ( ) Take out the worksheet
1) Find the annual multiplier for each situation.
a) increase of 8% compounded quarterly
b) increase of 15% compounded monthly
100%
100.00% 8%
15%
= 2% %
= 1.25% % 4 12
102%
101.25%
quarterly =
1.02
monthly =
1.0125
annually =
(1.02)(1.02)(1.02)(1.02)
annually =
(1.0125)(1.0125)…
(1.02)4
(1.0125)12
Or, you can do all of this in one step: ( ) 4 ( ) 12
0.08
0.15
1 +
1 + =
(1.02)4 =
(1.0125)12 4 12

3. ( ) Is there a faster way??? Take out the worksheet
1) Find the annual multiplier for each situation.
c) decrease of 12% compounded monthly
100%
12%
= 1%
– 1% 12
99%
monthly =
0.99
Or, you can do all of this in step:
annually =
(0.99)(0.99)(0.99)… ( ) 12
(0.99)12
0.12
1 – =
(0.99)12 12
Is there a faster way???

4. ( ) Compound Interest Formula Add to your notes: Example #1:
A(t) = amount (value)
P = initial value
r = rate (in decimal form)
n = number of compoundings
t = time
Esteban invests $1500 into an account earning 6% annual interest compounded quarterly. How much will he have in 10 years?
Example #1:
( ) 4
(10)
0.06
A(t) = amount (value)
A(10) =
1500
1 + 4
A0 =
$1500
r = 6%
= 0.06
n = 4
t = 10
Be careful! Change the rate to a decimal!

5. ( ) Compound Interest Formula Add to your notes: Example #2:
A(t) = amount (value)
P = initial value
r = rate (in decimal form)
n = number of compoundings
t = time
Keisha invests $800 into an account earning 5% annual interest compounded daily. How much will she have in 6 years?
Example #2:
( )
365
(6)
0.05
A(t) = amount (value)
A(6) =
800
1 +
365
A0 =
$800
r = 5%
= 0.05
n =
365
We cannot simplify this anymore without getting a long decimal, so leave it be and approximate at the very last step.
t = 6
Be careful! Change the rate to a decimal!

6. Complete the rest of the worksheet by tomorrow.
Practice:
1) Jim wants to invest $1000 at an 8% APR (annual percentage rate). Determine how much he will earn after 5 years given the various types of compounding. Use the formula:
a) quarterly
KEY:
A(t) = amount (value)
A0 = initial value
$1000
r = rate (in decimal form)
0.08
n = number of compoundings 4
t = time 5
Complete the rest of the worksheet by tomorrow.

7. Examples of Exponential Functions Not Exponential Functions
Add to your notes:
Exponential functions have the general form y = k(m)x, where k is the initial value and m is the multiplier.
The graph of an exponential function is continuous because x can assume all values (i.e., the domain is all real numbers).
The independent variable x has to be an exponent.
Examples of Exponential Functions
Not Exponential Functions

8. a) Calculate the predicted cost for the year 1997.
FX – 89
Suppose the annual fee for attending a public university were $1200 in 1986 and the cost was increasing 10% each year. (Round answers to the nearest dollar.)
a) Calculate the predicted cost for the year 1997.
Initial value = $1200
V(t) = 1200(1.10)t
1997
multiplier = 100% + 10%
–1986
V(11) = 1200(1.10)11
= 110%
11 years
V(11)  $3424
= 1.10
b) What was the cost in 1980? What did you have to assume?
1986
V(t) = 1200(1.10)t
We had to assume that the cost was increasing by 10% per year between 1980 and 1986.
–1980
V(–6) = 1200(1.10)–6
6 years
V(–6)  $677

9. c) Sketch a graph of this function.
FX – 89
c) Sketch a graph of this function.
d) By 1993 the annual cost was actually $ Find this point on your graph. How accurate was the model? What actually happened?
Year
1980
1985
1990
1995
2000
1000
3000
4000
Cost ($)
The fees are above the expected which implies the fees are growing faster than 10%.

10. There should be about 275.5 million people in 1990. –1980 10 years
FX – 90
According to the U.S. Census Bureau the population of the United States has been growing at an average rate of approximately 2% per year. The census is taken every 10 years and the population in 1980 was estimated at 226 million people.
a) How many people should the Census Bureau have expected to count in the 1990 census?
V(t) = 226,000,000(1.02)t
Initial value = 226,000,000
V(10) = 226,000,000(1.02)10
multiplier = 100% + 2%
= 102%
V(10)  275,492,739
= 1.02
1990
There should be about million people in 1990.
–1980
10 years

11. There should be about 336 million people in 1990.
FX – 90
b) How many people should the Census Bureau expect to count in census in the year 2000?
2000
V(t) = 226,000,000(1.02)t
–1980
V(20) = 226,000,000(1.02)20
20 years
V(20)  335,824,111
There should be about 336 million people in 1990.
c) If the rate of population growth in the U.S. were to continue at 2%, in about what year would the population in the United States reach and surpass one billion?
1980
V(t) = 226,000,000(1.02)t
1,000,000,000 = 226,000,000(1.02)t
2055
t  75 years
The population would be expected to reach 1 billion by the year 2055.

12. FX – 91
Solve each of the following equations for x. Notice the variety in the equations today! Each one requires some different thinking. a) b) 3)

13. FX – 91 e) d) f)

14. Finish today's assignment:FX
& Worksheet

15. Time Multiplier Initial value Equation a) b)
FX – 87
In your book, there are several situations that can be described by using exponential functions. They represent a small sampling of the situations where quantities grow or decay by a constant percentage over equal periods of time. For each situation (a) – (f):
Find an appropriate time unit (days, weeks, years, etc.).
State the multiplier that would be used for the situation.
Identify the initial value.
Write an equation, in exponential form: V(t) = k(m)t, that represents the growth or decay and V(t) represents the value at any given time t.
You can organize your work using a table such as the one below:
Time
Multiplier
Initial value
Equation a) b)

16. Time Multiplier Initial value Equation a) years 1.06 $120,000
FX – 87
Time
Multiplier
Initial value
Equation a)
years
1.06
$120,000
V(t)=120,000(1.06)t b)
hours
1.22
180
V(t)=180(1.22)t c)
weeks
0.94
100
V(t)=100(0.94)t d)
years
0.89
$12,250
V(t)=12250(0.89)t e)
months 6%
$1000
V(t)=1000(1.005)12t
= 0.5% 12
1.005 f)
days
5.5%
$2500
V(t)=2500( …)365t
365
= …% …
0.055
More accurately
365