January 12, 2017
Sec 3.1 Exponential Functions and Their Graphs
Exponential Function - the independent variable
is in the exponent.
Model situations with constant percentage change
exponential growth
exponential decay
Domain:
Range:
asymptote:
x-intercept:
y-intercept:
For each function, state the domain, range,
asymptote and intercept. Then draw the
graph.
Answers are on next slide. Graphs follow.
January 12, 2017
For each function, state the domain, range,
asymptote and intercept. Then draw the
graph.
D: all real numbers
D: all real numbers
R: y < 0
R: y > 1
asymptote: y = 0
asymptote: y = 1
intercept: (0, -1)
intercept: (0, 2)
D: all real numbers
R: y > 0
asymptote: y = 0
intercept: (0, 1/8)
D: all real numbers
D: all real numbers
R: y > 0
R: y > 0
asymptote: y = 0
asymptote: y = 0
intercept: (0, 1)
intercept: (0, 1)
January 12, 2017
y = abx
Exponential functions model situations
involving constant percentage change.
a is the initial quantity
b is the growth(decay) factor
b = 1 + r where r is the growth rate
(r is negative for decay) If r = -.25
b = 1 - .25 = .75
y-intercept is (0, a)
Worksheet on application word problems.
January 12, 2017
Suppose you invest $400 at 3% annual
interest compounded annually. How much
money will you have after 5 years,
assuming you make no deposits or
withdrawals? Show work.
400(1 + .03)5 = 463.71
What if the interest is compounded monthly?
Then how much money would you have at the
end of one year? after five years?
A(5) = 400(1 + .0025)60 = $464.65
Can you write a formula for this type of
problem using the variables below?
P = principal (amount of money invested)
r = annual interest rate as a decimal
n = the number of times interest is compounded in a year
t = the number of years the money was invested.
January 12, 2017
The compound interest formula:
A(t) = P(1 +
)
nt
total # of times interest
is compounded
% interest earned each time interest is compounded (only a fraction of
the annual rate)
P = principal (amount of money invested)
r = annual interest rate as a decimal
n = the number of times interest is compounded in a year
t = the number of years the money was invested.
1.) Suppose you invest $5000 at 4.5% interest.
How much money will you have after 6 years,
assuming no deposits or withdrawals during this
time if interest is compounded
a.) semi-annually?
b.) daily
2.) Suppose you invest $15,000 at 6% interest
compounded quarterly. How much money would
you have after 10 years, assuming no deposits or
withdrawals?
January 12, 2017
Suppose you want to invest $1500 at an
annual interest rate of 5%. How much
money will you have after 10 years if
interest is compounded:
annually
monthly
daily
The more often you compound, the faster the money
grows.
Suppose you invest $3000 AT 5% APR
compounded monthly. Calculate the value
of the investment after
10 years
$4941.03
20 years
$8137.92
30 years
$13,403.23
40 years
$22,075.25
January 12, 2017
Find the value of $1 invested at 100%
interest for 1 year compounded:
annually
semi-annually
quarterly
daily
hourly (8760)
every second (31,536,000)
2
2.25
2.44140625
2.714568
2.718126
2.718282
The value does not increase without bound.
It approaches a limit. That limit is the number
we call e.
e is an irrational number
January 12, 2017
y = ex is called the natural exponential
function
It has many applications in science, population growth
radioactive decay, and continuous compounding of
interest.
Derivation of the continuous change formula
A(t) = P(1 +
)nt
Continuous Change Formula
A(t) = Pert
r = interest rate as a decimal (do not add 1)
P = principal
t = number of years of the investment
Invest $3000 at 5% interest compounded
continuously. How much money will you
have after 10 years?
A(10) = 3000e.05(10)
= $4946.16
January 12, 2017
You can use either y = abx or
y = Pert (contiuous change formula) to represent situations
involving constant percentage change.
1.) The population of a small city was 50,400 in 1995
and growing at a rate of 3% each year.
a.) Use y = abx to write a function for the population
x years after 1995. Then find the population in 2002.
f(x) = 50,400(1.03)x
f(7) = 50,400(1.03)7 = $61,986
b.) Use the continuous change formula to write a
function for the population x years after 1995.
Then find the population in 2002.
f(x) = 50,400e.03x
f(7) = 50,400e.03(7) = $62,177
Note that the continuous change formula gives a
slightly larger result.
We can convert our continuous change formula to
the form y = abx y = 50,400e.03x
e.03 = 1.0305 so,
y = abx form
y = 50,400(1.0305)x
Notice that the multiplier 1.0305 is larger than
1.03 in part (a). Therefore, the result is a little
bigger.
The annual percentage yield (APY) is the percentage
that your money effectively earns in one year due to
compounding.
Find the annual percentage yield on an account that pays
3% APR with monthly compounding.
Invest $1 for 1 year and see what it grows to.
represents the
percentage increase
Find the APY if interest is
compounded continuously.
January 12, 2017
(APY)
Find the annual percentage yield on
an account that pays 8% interest
compounded daily.
Find the annual percentage yield (APY) on
an account which pays 4.3% annual interest
compounded continuously.
APY = 4.39%