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Quiz 7-1:
f ( x)  (1)ab
f ( x)  5(2)
( 1)( x c )
x2
d
3
1. Where does the graph cross the y-axis?
2. f(1) = ?
3. Horizontal asymptote = ?
4. How was the function g ( x)  5(2) x transformed
to get f(x) above?
5. Domain = ?
6. range = ?
7-2, and 7-3
Exponential Decay
And
The growth/decay factor (base) “e”
Putting it all together:
f ( x)  (1)ab
vertical shift
( 1)( x c )
d
If negative:
Reflect across x-axis
Horizontal shift
Initial value:
Crosses y-axis here
If negative:
Reflect across y-axis
Growth factor:
f ( x)  10(0.4) x  2
f ( x)  3(5) x 2
f ( x)  4(0.7) 2 x
f ( x)  6(1.1) x  2  5
Population Growth
The population at the end of one period of time equals the
initial population plus the growth/decay of the population.
P1  P0  Pchange
We can rewrite the change in the population
as some percentage of the original population.
P1  P0  P0 % change 
If “r” is the % change in population (decimal equivalent),
then we can rewrite that as:
P1  P0  P0 r
Factoring out the common factor results in:
P1  P0 (1  r )
Population Growth
over “t” time periods:
Percent rate of change
(in decimal form)
P(t )  P0 (1  r )
Population (as a
function of time)
Initial
population
t
time
Growth
rate
It’s just a formula!!!
The initial population of a colony of bacteria
is 1000. The population increases by 50%
every hour. What is the population after 5 hours?
P(5)  1000(1  0.50)5
P(5)  1000(1.5)
5
P(5)  7593
Interest (savings account)
Interest rate
kt
r
A(t )  P(1  )
k
Amount (as a
function of time)
Initial amount
(“principle”)
time
Number of times the
interest is paid per year
(compounded)
In problems like this, the words in the problem will give
you all but one of the quantities in the equation. The
quantities are:
A(t)
P
r
k
t
It will be up to you to solve for the missing quantity.
kt
r
A(t )  P(1  )
k
A bank account pays 3.5% interest per year compounded
monthly. If you initially invest $200, how much money
will you have after 5 years?
A(5)  $200(1  0.035 )12*5 A(5)  $238.19
12
What is the unknown quantity?
A(t)
A(5) = ?
P
r
k
t
kt
r
A(t )  P(1  )
k
A bank account pays quarterly compounded interest. If
you initially invested $1000 and after 10 years had $2200
in your account, what was the annual interest rate?
2200  1000(1  r ) 4*10
4
Divide left/right by 1000
What is the unknown quantity?
40
r
2.2  (1  )
4
1.019906  1  r
r  0.0796
4
40th root left/right
Subtract 1, multiply by 4
r  7.96%
Your turn:
1.
2.
kt
r
A(t )  P(1  )
k
A bank account pays 14% interest per year
compounded quarterly. If you initially invest $2500,
how much money will you have after 7 years?
Five years after you made a single deposit in an
earning 3% compounded monthly, it contains $580.81.
What was your initial deposit?
0 < ‘b’ < 1
Exponential Decay
f ( x)  4(0.5) x
20
18
16
f(x) (output values)
14
12
10
8
6
4
2
0
-3
-2
-1
0
1
2
3
4
f ( x)  ab  d
x
5
x (input value)
‘a’ is the initial value  f(0) = ‘a’
‘b’ is called the decay factor
‘d’ shifts everything up or down
Table of values
x 4(0.5) x f(x)
0
4
(
0
.
5
)
4
0
½
1
2
1 4(0.5)
½
2
2 4(0.5) 1
½
3
0.5
4
(
0
.
5
)
3
4 4(0.5) 4 0.25 ½
1
4
(
0
.
5
)
-1
8
2
4
(
0
.
5
)
16
-2
Exponential Growth and Decay
f ( x)  ab
x
exponential growth: growth factor > 1
exponential decay: growth factor 0 < b < 1
Your turn:
f ( x)  ab x  d
For each of the following what is the:
a. “initial value”?
b. “decay factor”?
c. “horizontal asymptote”
d. Any reflections (across x-axis or y-axis)
3.
f ( x)  2(0.3)
4.
g ( x)  10(5)
5.
f ( x)  0.5  4
x
x
x
Identifying the Parts of the function:
f ( x)  ab  d
x
‘a’ is the initial value  f(0) = ‘a’
‘b’ is called the decay factor (if 0 < b < 1)
‘d’ shifts graph up/down and is the horizontal asymptote
f ( x)  10(0.3) x  2
Initial value: 10
y-intercept: f(0) = 10 + 2 = 12
decay factor: 0.3
Horizontal asymptote: y = 2
Graphing Exponential Decay
Use the “power of the calculator” or:
f ( x)  6(0.5) x
10
9
1. f(0) = ? f(0) = 6
8
f(x) (output values)
7
6
2. Some other point
5
f(1) = ? f(1) = 3
4
3
2
3. Horizontal
asymptote
1
0
-5
-4
-3
-2
-1
0
1
2
3
x (input value)
Domain = ?
All real #’s
4
5
y=0
Range = ?
y>0
Where does the number ‘e’ come from?
Named after the Swiss mathematician
Leonard Euler (1707 – 1783).
The number ‘e’ can be found on your calculator
by “2nd” + “ln” + “1” = 2.718 281828459045 …
‘e’ is an irrational number like pi or the square
root of a prime number. The numbers after the
decimal point go on forever without any repetition
of number patterns.
The number ‘e’
(Named after Leonard Euler, a Swiss mathematician)
‘e’ is a very unique number
The slope of the tangent
line at any point on the
x
curve
is
f ( x)  e
e
x
The slope of the tangent
line at x = 0 is e 0  1
The slope of the tangent
line at x = 1 is e1  2.71828182845904523536......
Exponential Functions and the
number ‘e’
Any exponential function of the form:
Can be written in the form:
Exponential Growth: k > 0
Exponential Decay:
k<0
f ( x)  ab
f ( x)  aekx
x
Exponential Functions and ‘e’
What processes does ‘e’ have to do
with?
Think of a bacteria cell. Over
a certain period of time it splits
from one cell into two cells. The
number of bacteria doubles.
t=1
This type of growth occurs in “spurts”.
At one instant of time it is a single bacteria.
The next instant it is two bacteria.
The number of bacteria doubles.
A(t) = 2 ͭ
In “t” time periods it doubles ‘t’ times.
What processes does ‘e’ have to do with?
e= 2.718…
t=1
Instead of in “spurts,” this type of growth occurs continuously
 growth of money in an account that pays interest continuously
 the decay of radioactive material which occurs continuously
The number e (2.718…) represents the maximum compound rate of
growth from a process that grows at 100% for one time period. Sure,
you start out expecting to grow from 1 to 2. But with each tiny step
forward you create a little “dividend” that starts growing on its own.
When all is said and done, you end up with e (2.718…) at the end of
1 time period, not 2.
The “natural” number ‘e’ works
perfectly with natural processes
Exponential growth of populations
Exponential decay of radioactive material
(5e )(3e )  ?
x
2
15e
Your turn:
6. simplify
7.
simplify
x 2
e e
3x
2e 4e
5
x2
Exponential Function (base ‘e’)
What is the “initial value” ?
f ( x)  3e
2x
What is the horizontal asymptote ?
Is it growth or decay?
Describe the transformation:
g ( x)  3e  4
Where does g(x) cross the y-axis ?
2x
Putting it all together:
f ( x)  (1)ae  d
vertical shift
Horizontal asymptote
kx
If negative:
Reflect across x-axis
Initial value:
Crosses y-axis here
f ( x)  10e  2
x
If negative:
decay
Growth factor:
f ( x)  3e
0.2 x
f ( x)  4e
2x
Your turn:
kx
f ( x)  ae  d
For each of the following what is the:
a. “initial value”?
b. Growth or decay?
c. “horizontal asymptote”
d. Any reflections (across x-axis)
8.
f ( x)  2e  3
9.
x
f ( x)  10e
0.45 x
x
10.
f ( x)  e  4
Your turn:
P(t )  P0 (1  r ) t
11. The population of Detroit decreases by by 3% every
year. In 2000 the population was 1.5 million.
What was the population in 2009?
12. The population of a small town can be modeled by:
P(t )  P0 (1.05)
t
What is the % change in population for every time
period ‘t’ ?
Value of a depreciating asset.
A(t )  P(1  r )
t
According to tax law, the value of a piece of equipment can be
“depreciated” and the depreciation can be used as a “business
expense” to reduce the amount of taxes that you pay.
A company buys a car for $20,000. It can depreciate
the value of the car by 20% per year. What is the
value of the car after 2 years?
A(5)  $20,000(1  0.20)
A(2)  $20,000(0.8)
2
2
A(2)  $12,800
Your turn:
13.
A(t )  P(1  r )
t
A company is in debt for $300,000. It is reducing its
debt at the annual rate of 15%. What is the company’s
debt after 10 years?
$59,062.32
continuous compounding
Remember that base “e” is used for things that grow (or
decrease) continuously instead of in “spurts.”
The more frequently the interest is paid in a savings account
the faster the money will grow for a given interest rate.
The fastest possible rate of growth for a given interest rate
occurs with continuous compounding.
A(t )  P0 (1  r ) kt
k
Periodic compounding
A(t )  P0 e rt
continuous compounding
PERT
Your turn: A(t )  P e rt
0
14. You put $500 into an account earning 5% interest
which is compounded continuously. How much
money will be in the account after 6 years?
$675.93
15.
Radioactive material “decays” continuously
(to some other element). The “decay constant” of
Carbon-14 is k = -0.0001258. If 50 grams of C-14
decays for 2000 years, how much carbon-14 will
remain?
38.9 gms
HOMEWORK
 Section 7-1 (page 482)
2, 4, 16, 18, 20, 24, 28
Section 7-2 (page 489)
4, 6, 8, 16, 18, 30a
Section 7-3 (page 495)
6, 8, 20, 22, 32, 38, 40, 56
(21 total problems)