LESSON
13.4
Name
Compound Interest
Class
Date
13.4 Compound Interest
Essential Question: How do you model the value of an investment that earns compound
interest?
Common Core Math Standards
The student is expected to:
Resource
Locker
F-LE.2
Explore
Construct linear and exponential functions, including arithmetic and
geometric sequences, given a graph, a description of a relationship, or
two input-output pairs (include reading these from a table). Also F.IF.7e,
F-IF.7, A-REI.11, F-IF.8, F-BF.1
A bond is a type of investment that you buy with
cash and for which you receive interest either as the
bond matures or when it matures. A conventional
bond generates an interest payment, sometimes
called a coupon payment, on a regular basis, typically
twice a year. The interest payments end when the
bond matures and the amount you paid up-front is
returned to you.
Mathematical Practices
MP.4 Modeling
A zero-coupon bond, on the other hand, requires you to invest less money up-front and pays you its maturity value,
which includes all accumulated interest, when the bond matures. Over the life of the bond, the interest that is earned
each period itself earns interest until the bond matures.
Language Objective
Explain to a partner what simple interest is and what compound
interest is.
ENGAGE
The basic difference between a conventional bond and a zero-coupon bond is the type of interest earned.
A conventional bond pays simple interest, whereas a zero-coupon bond pays compound interest, which is interest
that earns interest.
Possible answer: You must know the principal
(amount invested) P, the annual interest rate r, and
the number of times per year, n, that the interest is
compounded. Then, the value V of the investment at
r nt
time t (in years) is V(t) = P 1 + n . This model
becomes V(t) = Pe rt when the interest is
compounded continuously.
(
_)
© Houghton Mifflin Harcourt Publishing Company • Image Credits: ©Treasury
Department/AP Images
A
Essential Question: How do you
model the value of an investment that
earns compound interest?
Comparing Simple and Compound Interest
For $1000, you can buy a conventional bond that has a maturity value of $1000, has a
maturity date of 4 years, and pays 5% annual interest. Calculate the interest that the
investment earns annually by completing the table. (Bear in mind that interest earned
is paid to you and not reinvested.)
Conventional Bond
Year
Value of Investment at
Beginning of Year
Interest Earned for Year
(Paid to the Investor)
0
$1000
1
$1000
$1000(0.05) = $ 50
$
1000
2
$1000
$1000(0.05) = $ 50
$
1000
3
$1000
$1000(0.05) = $ 50
$
1000
4
$1000
$1000(0.05) = $ 50
$
1000
PREVIEW: LESSON
PERFORMANCE TASK
View the online Engage. Discuss the photo and the
concept of investing to meet a changing goal amount.
Then preview the Lesson Performance Task.
Module 13
be
ges must
EDIT--Chan
DO NOT Key=NL-A;CA-A
Correction
Lesson 4
681
gh “File info”
made throu
Date
Class
Name
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13.4 Comp
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Explore
A2_MNLESE385900_U6M13L4.indd 681
HARDCOVER PAGES 495504
Sim
with
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
Turn to these pages to
find this lesson in the
hardcover student
edition.
ntional
st. Calculate
interest earned
buy a conve
annual intere
in mind that
, you can
pays 5%
For $1000
table. (Bear
years, and
eting the
date of 4
lly by compl
maturity
earns annua
investment
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t at
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is paid to
Investmen
nal Bond
Value of of Year
Conventio
Year
End
Earned for
Interest the Investor)
$1000
t at
(Paid to
Investmen
of
Value ning of Year
Begin
$ 1000
Year
) = $ 50
0
$1000(0.05
$1000
$ 1000
1
) = $ 50
$1000(0.05
$1000
$ 1000
2
) = $ 50
$1000(0.05
$1000
$ 1000
3
) = $ 50
$1000(0.05
$1000
4
Lesson 4
681
Module 13
Lesson 13.4
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A2_MNLE
681
Value of Investment at
End of Year
681
24/03/14
11:33 PM
24/03/14 11:31 PM
B
For $822.70, you can buy a zero-coupon bond that has a maturity value of $1000, has
a maturity date of 4 years, and pays 5% annual interest. Calculate the interest that the
investment earns annually by completing the table. (Bear in mind that interest earned
is reinvested and not paid to you until the bond matures.)
EXPLORE
Comparing Simple and
Compound Interest
Zero-Coupon Bond
Year
Value of
Investment
at Beginning
of Year
Interest Earned for Year
(Reinvested)
INTEGRATE TECHNOLOGY
Value of Investment at End of Year
0
$822.70
1
$822.70
$822.70(0.05) = $ 41.14
$822.70 + $ 41.14
= $ 863.84
2
$ 863.84
$ 863.84 (0.05 )= $ 43.19
$ 863.84 + $ 43.19
= $ 907.03
3
$ 907.03
$ 907.03 (0.05) = $ 45.35
$ 907.03 + $ 45.35
= $ 952.38
4
$ 952.38
$ 952.38 (0.05) = $ 47.62
$ 952.38 + $ 47.62
= $1000
Students have the option of completing the Explore
activity either in the book or online.
QUESTIONING STRATEGIES
Which type of bond generates the most
interest? How much more interest is generated
by this type of bond? The conventional bond
generates more interest than the zero-coupon
bond; $22.70
If you are an investor, what is the advantage of
investing in the zero-coupon bond? You may
invest less money.
Reflect
1.
Describe the difference between how simple interest is calculated and how compound interest is calculated.
Simple interest is paid back to the investor, and only calculated from the initial amount
It is first calculated from the initial amount invested. Then, the interest is added to the
initial amount invested and calculated again from this new value. So, the interest amount
grows exponentially over time.
2.
If V(t) represents the value of an investment at time t, in whole numbers of years starting with t = 0
and ending with t = 4, write a constant function for the last column in the first table and an exponential
function for the last column in the second table.
In the last column of the first table, the value of the investment is a constant $1000.
So, V(t) = 1000. In the last column of the second table, the value of the investment grows
© Houghton Mifflin Harcourt Publishing Company
invested. So, the interest amount is a constant value. Compound interest is reinvested.
exponentially with an initial investment of $822.70 and an annual growth rate of 5%.
So, V(t) = 822.70(1.05) .
t
Module 13
Lesson 4
682
PROFESSIONAL DEVELOPMENT
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Math Background
(
)
nt
r
n
The compound interest formula V(t) = P 1 + _
n gives the
2
total amount of money V(t) at the end of t years for an initial
5
investment P with an interest rate r that is compounded n
10
times per year. When $1 is invested at 100%n interest for 1
1
100
year, the formula becomes V(1) = 1 + _
n . The table gives
1000
the value of V(1) for various values of n. The table suggests
that as n grows larger, the value of V(1) approaches a limiting
value. This value, which is defined to be the constant e, is
about 2.718281828459045.
Written using the limit notation of calculus,
n
1
lim
_
e = n→∞ 1 + n .
(
(
)
)
V(1)
2.25
≈ 2.4883
≈ 2.5937
≈ 2.7048
≈ 2.7169
Compound Interest
682
Explain 1
EXPLAIN 1
Modeling Interest Compounded Annually
Recall that the general exponential growth function is ƒ(t) = a(1 + r) . When this function is applied to an
t
investment where interest is compounded annually at a rate r, the function is written as V(t) = P(1 + r)
where V(t) is the value V of the investment at time t and P is the principal (the amount invested).
t
Modeling Interest Compounded
Annually

A person invests $3500 in an account that earns 3% annual interest. Find when the value of
the investment reaches $10,000.
Let P = 3500 and r = 0.03. Then V(t) = P(1 + r) = 3500(1.03) .
t
QUESTIONING STRATEGIES
t
Graph y = 3500(1.03) on a graphing calculator. Also graph the line
y = 10,000 and find the point where the graphs intersect.
x
What is the difference between simple interest
and interest compounded annually? Simple
interest is interest earned on the principal. Interest
compounded annually is interest earned on the
principal and on the interest.
The value of the function is 10,000 at x ≈ 35.5. So, the investment reaches a value of $10,000 in
approximately 35.5 years.
When is the simple interest earned on an
investment equal to interest that is
compounded annually? At the end of the first year,
assuming the interest rates are the same.

A person invests $200 in an account that earns 6.25% annual interest. Find when the value
of the investment reaches $5000.
Let P = 200 and r = 0.0625 . Then V(t) = P(1 + r) = 200
t
Graph the function on a graphing calculator. Also graph the line
y=
5000
( 1.0625 ) .
t
and find the point of intersection.
The term compound has many meanings. It means
one thing in science, another in the everyday world
(for example, an army compound), and still another
in mathematics. Talk with students about the word’s
different meanings, specifying its particular meaning
in math class.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Discuss with students how compound
interest is an example of exponential growth. Help
them to see that the formula for compound interest is
an exponential function.
© Houghton Mifflin Harcourt Publishing Company
COMMUNICATE MATH
The value of the function is
approximately
53.1
5000
at x ≈
53.1
. So, the investment reaches a value of $5000 in
years.
Your Turn
3.
A person invests $1000 in an account that earns 5.5% annual interest. Find when the value of the
investment has doubled.
Let P = 1000 and r = 0.055. Then V(t) = P(1 + r) = 1000(1.055) . Graphing
t
t
y = 1000(1.055) along with the line y = 2000 shows that the value of the function is 2000
x
at x ≈12.9. So, the investment reaches a value of $2000 in approximately 12.9 years.
Module 13
683
Lesson 4
COLLABORATIVE LEARNING
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Small Group Activity
24/03/14 11:30 PM
Have students work in groups of 3 to 4. Instruct students to use the Internet to
research various types of investments and the interest programs they offer.
Have them create a scenario for each type of investment they identify, and show
how the investment grows over time. Have them create posters showing the results
of their research and present the posters to the class.
683
Lesson 13.4
Modeling Interest Compounded
More than Once a Year
Explain 2
EXPLAIN 2
Interest may be earned more frequently than once a year, such as semiannually (every 6 months), quarterly
(every 3 months), monthly, and even daily. If the number of times that interest is compounded in a year is n,
r
then the interest rate per compounding period is _
n (where r is the annual interest rate), and the number of
times that interestnt is compounded in t years is nt. So, the exponential growth function becomes
r
V(t) = P 1 + _
n .
(

Modeling Interest Compounded
More Than Once a Year
)
A person invests $1200 in an account that earns 2% annual interest compounded quarterly.
Find when the value of the investment reaches $1500.
(
)
(
)
AVOID COMMON ERRORS
4t
0.02 = 1200(1.005) 4t.
= 1200 1 + _
4
4x
Graph the function y = 1200(1.005) on a graphing calculator. Also graph the line y = 1500 and find
the point where the graphs intersect.
r
Let P = 1200, r = 0.02, and n = 4. Then V(t) = P 1 + _
n
nt
Some students may mistake the number of
months in a compounding period for the number
of compounding periods per year. For example, for
interest compounded quarterly, students may let
n = 3, since quarterly means every 3 months. Clarify
that n represents the number of times the interest is
compounded each year.
QUESTIONING STRATEGIES
The value of the function is 1500 at x ≈ 11.2. So, the investment reaches a value of $1500 in
approximately 11.2 years.

Why does the amount of interest earned
increase as the frequency of compounding
increases? The more often the interest is
compounded, the sooner the new interest is added
to the balance on which the interest is earned.
A person invests $600 in an account that earns 6.25% annual interest compounded
semiannually. Find when the value of the investment reaches $1700.
(
r
Then V(t) = P 1 + _
n
)
nt
)
0.0625
= 600 1 + _
2
2t
(
= 600 1.03125
Graph the function on a graphing calculator. Also graph the line y =
the graphs intersect.
The value of the function is
approximately
16.9
1700
)
1700
2t
.
and find the point where
© Houghton Mifflin Harcourt Publishing Company
(
Let P = 600 , r = 0.0625 , and n = 2 .
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Technology
MP.5 Use a graphing calculator to illustrate
how increasing the number of compounding
periods per year increases the rate of growth of an
investment. Have students graph and compare several
functions, each based on the same initial investment,
interest rate, and time period, but for varying
values of n.
at x ≈ 16.9 . So, the investment reaches a value of $1700 in
years.
Module 13
684
Lesson 4
DIFFERENTIATE INSTRUCTION
A2_MNLESE385900_U6M13L4.indd 684
Communicating Math
24/03/14 11:30 PM
Some students may feel overwhelmed by the complexity of the functions
presented in the lesson. These students may benefit from rewriting the formulas
using words in place of the various variables. Students can then use their “verbal
formulas” to solve the problems. After achieving some success, students will likely
feel less anxious about working with the formulas in their traditional forms.
Compound Interest
684
Your Turn
EXPLAIN 3
4.
A person invests $8000 in an account that earns 6.5% annual interest compounded daily. Find when the
value of the investment reaches $20,000.
365t
0.065
r nt
_
Let P = 8000, r = 0.065, and n = 365. Then V(t) = P 1 + _
n = 8000 1 + 365
365t
365x
≈ 8000(1.000178) . Graphing along y = 8000(1.000178) along with the y = 20,000
(
Modeling Interest Compounded
Continuously
(
)
)
shows that the value of the function is 20,000 at x = 14.1. So, the investment reaches a value
of $20,000 in approximately 14.1 years.
QUESTIONING STRATEGIES
Explain 3
How is interest that is compounded
continuously different from interest that is
compounded daily? Interest that is compounded
daily has 365 (or 366) compounding periods per
year. Interest that is compounded continuously has,
theoretically, infinitely many compounding periods
per year.
Modeling Interest Compounded Continuously
(
)
r
n , you can rewrite the model V(t) = P 1 + _
By letting m = _
n
r
(
)
nt
(
1
as V(t) = P 1 + _
m
(
)
)
mrt
because
m rt
⎡
r =_
1 and nt = mrt. Then, rewriting V(t) = P 1 + _
1
1 ⎥⎤ and
_
_
as
V(t)
=
P
1
+
⎢
m
m
m ⎦
n
⎣
m
1
letting n increase without bound, which causes m to increase without bound, you see that 1 + _
m
rt
approaches e, and the model simply becomes V(t) = Pe . This model gives the value of an investment
with principal P and annual interest rate r when interest is compounded continuously.

mrt
(
)
A person invests $5000 in an account that earns 3.5% annual interest compounded
continuously. Find when the value of the investment reaches $12,000.
Let P = 5000 and r = 0.035. Then V(t) = 5000e 0.035t.
AVOID COMMON ERRORS
Graph y = 5000e 0.035x. on a graphing calculator. Also graph the line y = 12,000 and find the point
where the graphs intersect.
© Houghton Mifflin Harcourt Publishing Company
Some students may, in error, raise the product of P
and e to the exponent rt. Reinforce that the order of
operations dictates that e be raised to the exponent,
with the result multiplied by P.
The value of the function is 12,000 at x ≈ 25. So, the investment reaches a value of $12,000 in
approximately 25 years.
Module 13
685
Lesson 4
LANGUAGE SUPPORT
A2_MNLESE385900_U6M13L4 685
Communicate Math
Have students work in pairs. Ask the first student to explain what simple interest
is and how to find it, while the second student takes notes. The students switch
roles; the second student explains what compound interest is and how to find it,
while the first student takes notes. The first student should then do the same for
constant compounding. Ask the pair to work together to read their notes and
make any edits needed to complete their explanations.
685
Lesson 13.4
6/16/15 9:31 AM

The principal amount, $350, earns 6% annual interest compounded continuously. Find
when the value reaches $1800.
0.06t
Let P = 350 and r = 0.06 . Then V(t) = 350 e .
EXPLAIN 4
Graph the function on a graphing calculator. Also graph the line y =
the graphs intersect.
Finding and Comparing Effective
Annual Interest Rates
1800
and find the point where
QUESTIONING STRATEGIES
Which is greater, the nominal annual interest
rate or its effective annual interest rate?
Explain. The effective annual interest rate is greater
because it takes into account the interest that is
generated on the interest.
The value of the function is 1800 at x ≈ 27.3 . So, the investment reaches a value of $1800 in
approximately
27.3 years.
Your Turn
5.
A person invests $1550 in an account that earns 4% annual interest compounded continuously. Find when
the value of the investment reaches $2000.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 The concept of compound interest provides
Let P = 1550 and r = 0.04. Then V(t) = 1550e 0.04t. Graphing y = 1550e 0.04x along with the
line y = 2000 shows that the value of the function is 2000 at x ≈ 6.4. So, the investment
reaches a value of $2000 in approximately 6.4 years.
Explain 4
Finding and Comparing Effective Annual
Interest Rates
)
(
)
r ⎤
= P⎡⎣ 1 + _
n ⎦^t, where interest is compounded
r n
n times per year, is an exponential function of the form ƒ(t) = ab t where a = P and b = 1 + _
n . When
nt
n
(
)
the base of an exponential function is greater than 1, the function is an exponential growth function where the
(
)
r
base is the growth factor and 1 less than the base is the growth rate. So, the growth rate for V(t) = P 1 + _
n
)
r n
b-1= 1+_
n - 1. This growth rate is called the investment’s effective annual interest rate R, whereas
nt
is
r is called the investment’s nominal annual interest rate.
t
Similarly, for the value-of-an-investment function V(t) = Pe rt =P⎡⎣e r⎤⎦ , where interest is compounded continuously,
the growth factor is e r, and the growth rate is e r - 1. So, in this case, the effective annual interest rate is R = e r - 1.
© Houghton Mifflin Harcourt Publishing Company
(
r
The value-of-an-investment function V(t) = P 1 + _
n
(
an opportunity to informally discuss the idea of a
limiting process. As the frequency of compounding
increases, the process gets closer and closer to
continuous compounding. Tell students that the
study of limiting processes is important in more
advanced mathematics, especially calculus.
For an account that earns interest compounded more than once a year, the effective annual interest rate is the rate that
would produce the same amount of interest if interest were compounded annually instead. The effective rate allows
you to compare two accounts that have different nominal rates and different compounding periods.
Module 13
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686
Lesson 4
24/03/14 11:30 PM
Compound Interest
686
A
Arturo plans to make a deposit in one of the accounts shown in the table. To maximize the
interest that the account earns, which account should he choose?
Account X
Account Y
Nominal Annual
Interest Rate
2.5%
2.48%
Compounding
Period
Quarterly
Monthly
For Account X, interest is compounded quarterly, so n = 4. The nominal rate is 2.5%, so r = 0.025.
(
)
n
r
RX = 1 + _
n -1
(
)
Use the formula for the effective rate.
4
0.025 - 1
= 1+_
4
≈ 0.02524
Substitute.
Simplify.
For Account Y, interest is compounded monthly, so n = 12. The nominal rate is 2.48%, so r = 0.0248.
(
)
n
r
RY = 1 + _
n -1
(
)
0.0248
= 1+_
12
≈ 0.02508
12
Use the formula for the effective rate.
-1
Substitute.
Simplify.
Account X has an effective rate of 2.524%, and Account Y has an effective rate of 2.508%, so Account X
has a greater effective rate, and Arturo should choose Account X.
B
Harriet plans to make a deposit in one of two accounts. Account A has a 3.24% nominal
rate with interest compounded continuously, and Account B has a 3.25% nominal rate with
interest compounded semiannually. To maximize the interest that the account earns, which
account should she choose?
© Houghton Mifflin Harcourt Publishing Company
For Account A, interest is compounded continuously. The nominal rate is 3.24%, so r = 0.0324 .
RA = er - 1
0.0324
= e -1
Substitute.
≈ 0.03293
Simplify.
Module 13
A2_MNLESE385900_U6M13L4.indd 687
687
Lesson 13.4
Use the formula for the effective rate.
687
Lesson 4
24/03/14 11:30 PM
For Account B, interest is compounded semiannually, so so n =
2 . The nominal rate is 3.25%,
so r = 0.0325
(
)
n
r
RB = 1 + _
n -1
(
)
0.0325
= 1+_
2
≈ 0.03276
Use the formula for the effective rate.
2
- 1 Substitute.
Simplify.
3.276 %,
Account A has an effective rate of 3.293 %, and Account B has an effective rate of
so Account
A
has a greater effective rate, and Harriet should choose Account
A
.
Your Turn
6.
Jaclyn plans to make a deposit in one of the accounts shown in the table. To maximize the
interest that the account earns, which account should she choose?
Account X
Account Y
Nominal Annual
Interest Rate
4.24%
4.18%
Compounding Period
Annually
Daily
For Account X, interest is compounded annually, so n = 1. The nominal rate is 4.24%,
(
(
)
)
For Account Y, interest is compounded daily, so n = 365. The nominal rate is 4.18%,
so r = 0.0418.
r n
RY = 1 + _
n -1
(
(
)
)
0.0418
= 1+_
365
≈ 0.04268
365
-1
Account X has an effective rate of 4.24%, and Account Y has an effective rate of 4.268%,
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
©David R. Frazier Photolibrary, Inc./Alamy
so r = 0.0424.
r n
RX = 1 + _
n -1
1
0.0424
= 1+_ -1
1
= 0.0424
so Account Y has a greater effective rate, and Jaclyn should choose Account Y.
Module 13
A2_MNLESE385900_U6M13L4.indd 688
688
Lesson 4
3/28/14 1:30 PM
Compound Interest
688
Elaborate
ELABORATE
7.
Explain the difference between an investment’s nominal annual interest rate and its effective annual
interest rate.
An investment’s nominal annual interest rate is the interest rate r that is used in the value
r nt
function V(t) = P 1 + n or V(t) = Pe rt. The investment’s effective annual interest rate is
t
the interest rate R that would need to be used in the value function V(t) = P(1 + R) , where
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Communication
MP.3 Ask students to describe how the three
( _)
interest is compounded annually, to produce the same amount of annual interest as the
investment actually earns.
functions used to model investments earning
compound interest are related. Students should point
out that the functions are all exponential growth
functions, but that the growth factors differ in each
case. Ensure that students use correct language in
describing the mathematical relationships that exist
among the functions.
8.
Essential Question Check-In List the three functions used to model an investment that earns
compound interest at an annual rate r. Identify when each function is used.
The three functions used to model an investment that earns compound interest at an annual
t
t
r nt
rate r are V(t) = P(1 + r) , V(t) = P 1 + n , and V(t) = Pe rt . The function V(t) = P(1 + r)
nt
r
is used when interest is compounded annually. The function V(t) = P 1 + n is used when
(
_)
(
_)
interest is compounded more than once a year but not continuously. The function V(t) = Pe rt is
used when interest is compounded continuously.
SUMMARIZE THE LESSON
What functions can you use to model
investments that earn compound
t
interest? You can use V(t) = P(1 + r) to model
interest that is compounded annually. You can use
r nt
V(t) = P 1 + n to model interest that is
compounded n times per year. You can use
V(t) = Pe rt to model interest that is compounded
continuously.
1.
_)
A person invests $2560 in an account that earns 5.2% annual interest. Find when the
value of the investment reaches $6000.
Let P = 2560 and r = 0.052. Then V(t) = P(1 + r) = 2560(1.052) t. Graphing
x
y = 2560(1.052) along with the line y = 6000 shows that the value of the
function is 6000 at x ≈ 16.8. So, the investment reaches a value of $6000 in
approximately 16.8 years.
2.
A person invests $1800 in an account that earns 2.46% annual interest. Find when the
value of the investment reaches $3500.
Let P = 1800 and r = 0.0246. Then V(t) = P(1 + r) = 1800(1.0246) .
x
Graphing y = 1800(1.0246) along with the line y = 3500 shows that the
value of the function is 3500 at x ≈ 27.4. So, the investment reaches a
value of $3500 in approximately 27.4 years.
t
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• Online Homework
• Hints and Help
• Extra Practice
t
© Houghton Mifflin Harcourt Publishing Company
(
Evaluate: Homework and Practice
689
t
Lesson 4
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3.
EVALUATE
Emmanuel invests $3600 and Kelsey invests $2400. Both investments earn 3.8%
annual interest. How much longer will it take Kelsey’s investment to reach $10,000
than Emmanuel’s investment?
Emmanuel’s investment:
Let P = 3600 and r = 0.038. Then V(t) = P(1 + r) = 3600(1.038) .
x
Graphing y = 3600(1.038) along with the line y = 10,000 shows that
the value of the function is 10,000 at x ≈ 27.4. So, the investment reaches
a value of $10,000 in approximately 27.4 years.
t
t
Kelsey’s investment:
Let P = 2400 and r = 0.038. Then V(t) = P(1 + r) = 2400(1.038) .
x
Graphing y = 2400(1.038) along with the line y = 10,000 shows that the
value of the function is 10,000 at x ≈ 38.3. So, the investment reaches a
value of $10,000 in approximately 38.3 years.
t
t
ASSIGNMENT GUIDE
Concepts and Skills
So, it takes Kelsey’s investment approximately 38.3 - 27.4 = 10.9 years
longer to reach a value of $10,000.
4.
Explore
Comparing Simple and
Compound Interest
Jocelyn invests $1200 in an account that earns 2.4% annual interest. Marcus invests
$400 in an account that earns 5.2% annual interest. Find when the value of Marcus’s
investment equals the value of Jocelyn’s investment and find the common value of the
investments at that time.
For Jocelyn’s investment, let P = 1200 and r = 0.024.
t
t
Then V(t) = P(1 + r) = 1200(1.024) .
For Marcus’s investment, let P = 400 and r = 0.052.
t
t
Then V(t) = P(1 + r) = 400(1.052) .
The graphs of y = 1200(1.024) and y = 400(1.052) intersect at
approximately the point (40.7, 3152.42).
x
x
So, the value of Marcus’s investment equals the value of Jocelyn’s
investment after approximately 40.7 years. The common value of the
investments is approximately $3152.42.
5.
(
)
r
Then V(t) = P 1 + _
n
nt
(
0.0365
= 350 1 + _____
2
)
2t
© Houghton Mifflin Harcourt Publishing Company
A person invests $350 in an account that earns 3.65% annual interest compounded
semiannually. Find when the value of the investment reaches $5675.
Let P = 350, r = 0.0365, and n = 2.
= 350(1.01825) .
2t
Graphing y = 350(1.01825) along with the line y = 5675, shows that the
2x
Practice
value of the function is 5675 at x ≈ 77. So, the investment reaches a value
of $5675 in approximately 77 years.
Example 1
Modeling Interest Compounded
Annually
Exercises 1–4, 18
Example 2
Modeling Interest Compounded
More Than Once a Year
Exercises 5–8,
15–17
Example 3
Modeling Interest Compounded
Continuously
Exercises 9–11
Example 4
Finding and Comparing Effective
Annual Interest Rates
Exercises 12–14
VISUAL CUES
Suggest that students circle the numbers in the
problem and, next to each number, write the variable
for which the number will be substituted.
AVOID COMMON ERRORS
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Exercise
Depth of Knowledge (D.O.K.)
Mathematical Practices
1–2
1 Recall of Information
MP.6 Precision
3–4
2 Skills/Concepts
MP.5 Using Tools
5–7
1 Recall of Information
MP.6 Precision
2 Skills/Concepts
MP.5 Using Tools
1 Recall of Information
MP.6 Precision
2 Skills/Concepts
MP.5 Using Tools
1 Recall of Information
MP.4 Modeling
8
9–10
11
12–13
24/03/14 11:30 PM
The formulas presented in this lesson each contain
numerous variables. Because of this, students may
mistake e, in the formula for investments that earn
interest that is compounded continuously, as a
variable, and look to make a substitution when
applying this formula to a problem. To avoid this,
suggest that students underline e when using this
formula, to remind them that e is a number, not a
variable, and that no number is substituted for e.
Compound Interest
690
6.
QUESTIONING STRATEGIES
If I(t) is a function that represents the amount
of interest earned on an account in which
interest is compounded continuously, what rule can
you write for I(t)? I(t) = Pe rt - P
Molly invests $8700 into her son’s college fund, which earns
2% annual interest compounded daily. Find when the value
of the fund reaches $12,000.
Let P = 8700, r = 0.02, and n = 365.
(
r
Then V(t) = P 1 + _
n
)
nt
≈ 8700(1.0000548)
(
0.02
= 8700 1 + ___
365
365t
.
Graphing y = 8700(1.0000548)
365x
)
365t
along with the line
y = 12,000 shows that the value of the function is
12,000 at x ≈ 16.1. So, the fund reaches a value of
$12,000 in approximately 16.1 years.
7.
A person invests $200 in an account that earns 1.98% annual interest compounded
quarterly. Find when the value of the investment reaches $500.
Let P = 200, r = 0.0198, and n = 4.
(
r
Then V(t) = P 1 + _
n
)
nt
(
0.0198
= 200 1 + _____
4
)
4t
≈ 200(1.00495) .
4t
Graphing y = 200(1.00495) along with the line y = 500, shows that
4x
the value of the function is 500 at x ≈ 46.4. So, the investment reaches
8.
a value of $500 in approximately 46.4 years.
Hector invests $800 in an account that earns 6.98% annual interest compounded
semiannually. Rebecca invests $1000 in an account that earns 5.43% annual interest
compounded monthly. Find when the value of Hector’s investment equals the value of
Rebecca’s investment and find the common value of the investments at that time.
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
©Rubberball/Alamy
For Hector’s investment, let P = 800, r = 0.0698, and n = 2.
(
)
0.0698
= 800 1 + _____
2
(
)
0.0543
= 1000 1 + _____
12
nt
r
Then V(t) = P 1 + _
n
≈ 800(1.0349) .
2t
nt
(
)
12t
= 1000(1.004525) .
12t
The graphs of y = 800(1.0349) and y = 1000(1.004525)
2x
12x
intersect at
approximately the point (15.5, 2310.98).
So, the value of Hector’s investment equals the value of Rebecca’s
investment after approximately 15.5 years. The common value of the
investments is approximately $2310.98.
Exercise
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Mathematical Practices
Depth of Knowledge (D.O.K.)
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2 Skills/Concepts
MP.5 Using Tools
16
3 Strategic Thinking
MP.5 Using Tools
17
3 Strategic Thinking
MP.3 Logic
18
3 Strategic Thinking
MP.2 Reasoning
14–15
Lesson 13.4
2t
For Rebecca’s investment, let P = 1000, r = 0.0543, and n = 12.
Module 13
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)
(
r
Then V(t) = P 1 + _
n
9.
A person invests $6750 in an account that earns 6.23% annual interest compounded
continuously. Find when the value of the investment reaches $15,000.
COGNITIVE STRATEGIES
Let P = 6750 and r = 0.0623. Then V(t) = Pe = 6570e
. Graphing y = 6750e
along
with the line y = 15,000 shows that the value of the function is 15,000 at x ≈ 12.8. So, the
investment reaches a value of $15,000 in approximately 12.8 years.
rt
0.0623t
As a memory device, students may find it helpful to
remember that the effective annual interest rate is a
result of the effect that the compounding of the
interest has on the nominal interest rate.
0.0623x
10. A person invests $465 in an account that earns 3.1% annual interest compounded
continuously. Find when the value of the investment reaches $2400.
Let P = 465 and r = 0.031. Then V(t) = Pe rt = 465e 0.031t. Graphing y = 465e 0.031x along
with the line y = 2400 shows that the value of the function is 2400 at x ≈ 52.9. So, the
investment reaches a value of $2400 in approximately 52.9 years.
11. Lucy invests $800 in an account that earns 6.12% annual interest compounded
continuously. Juan invests $1600 in an account that earns 3.9% annual interest
compounded continuously. Find when the value of Lucy’s investment equals the value
of Juan’s investment and find the common value of the investments at that time.
For Lucy’s investment, let P = 800 and r = 0.0612. Then V(t) = Pe rt = 800e 0.0612t.
For Juan’s investment, let P = 1600 and r = 0.039. Then V(t) = Pe rt = 1600e 0.039t.
The graphs of y = 800e 0.0612x and y = 1600e 0.039x intersect at approximately the point
(31.2, 5407.00).
So, the value of Lucy’s investment equals the value of Juan’s investment after approximately
31.2 years. The common value of the investments is approximately $5407.00.
12. Paula plans to make a deposit in one of the accounts shown in the table. To maximize
the interest that the account earns, which account should she choose?
Account Y
2.83%
2.86%
Compounding
Period
Continuously
Annually
© Houghton Mifflin Harcourt Publishing Company
Account X
Nominal Annual
Interest rate
For Account X, interest is compounded continuously. The nominal rate is 2.83%,
so r = 0.0283.
RX = er - 1
= e 0.0283 - 1
≈ 0.0287
For Account Y, interest is compounded annually, so n = 1. The nominal rate is 2.86%,
so r = 0.0286.
(
)
r
RY = 1 + _
n -1
(
n
)
1
0.0286
= 1 + ______
-1
1
≈ 0.0286
Account X has an effective rate of 2.87%, and Account Y has an effective rate of 2.86%,
so Account X has a greater effective rate, and Paula should choose Account X.
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Compound Interest
692
13. Tanika plans to make a deposit to one of two accounts. Account A has a 3.78%
nominal rate with interest compounded daily, and Account B has a 3.8% nominal rate
with interest compounded monthly. To maximize the interest that the account earns,
which account should she choose?
For Account A, interest is compounded daily, so n = 365. The nominal rate is 3.78%,
so r = 0.0378.
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Math Connections
MP.1 Students may be interested to learn that five
(
)
r
RA = 1 + _
n -1
of the most famous numbers in mathematics (e, i, π,
1, and 0) are related to each other by the seemingly
simple equation e iπ + 1 = 0. Students will learn the
derivation of this relationship in more advanced
courses.
n
(
0.0378
= 1 + _____
365
)
365
-1
≈ 0.0385
For Account B, interest is compounded monthly, so n = 12. The nominal rate is 3.8%,
so r = 0.038.
(
)
r
RB = 1 + _
n -1
n
(
0.038
= 1 + ____
12
≈ 0.0387
)
12
-1
Account A has an effective rate of 3.85%, and Account B has an effective rate of 3.87%,
so Account B has a greater effective rate, and Tanika should choose Account B.
14. Kylie plans to deposit $650 in one of the accounts shown in the table. She chooses the
account with the greater effective rate. How much money will she have in her account
after 10 years?
Account X
Account Y
Nominal Annual
Interest Rate
4.13%
4.12%
Compounding
Period
Semiannually
Quarterly
For Account X, interest is compounded semiannually, so n = 2. The nominal rate is 4.13%,
so r = 0.0413.
(
)
© Houghton Mifflin Harcourt Publishing Company
r
RX = 1 + _
n -1
n
(
)
0.0413
= 1 + _____
-1
2
≈ 0.0417
2
For Account Y, interest is compounded quarterly, so n = 4. The nominal rate is 4.12%,
so r = 0.0412.
(
)
r
RY = 1 + _
n -1
n
(
)
0.0412
= 1 + _____
-1
4
≈ 0.0418
4
Account Y has the greater effective rate, so Kylie chooses Account Y.
The value of Kylie’s investment after 10 years is:
(
)
0.0412
V(10) = 650 1 + _____
4
4(10)
≈ 979.33
So, Kylie will have $979.33 in her account after 10 years.
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Lesson 4
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15. A person invests $2860 for 15 years in an account that earns 4.6% annual interest.
Match each description of a difference in interest earned on the left with the actual
difference listed on the right.
C $14.89
A. Difference between compounding interest
______
semiannually and annually
B $21.98
______
B. Difference between compounding interest
quarterly and semiannually
E $0.25
______
C. Difference between compounding interest
monthly and quarterly
A $42.75
______
D. Difference between compounding interest
daily and monthly
E. Difference between compounding interest
continuously and daily
D $7.27
______
(
)
- 2860(1 + 0.046)
(
)
0.046
- 2860 1 + ____
2
(
)
(
)
0.046
A. V S(15) - V A(15) = 2860 1 + ____
2
2(15)
15
≈ 5657.64 - 5614.89 = 42.75
0.046
B. V Q(15) - V S(15) = 2860 1 + ____
4
4(15)
(
)
(
)
≈ 5679.62 - 5657.64 = 21.98
0.046
C. V M(15) - V Q(15) = 2860 1 + ____
12
12(15)
0.046
- 2860 1 + ____
4
≈ 5694.51 - 5679.62 = 14.89
0.046
D. V D(15) - V M(15) = 2860 1 + ____
365
365(15)
(
2(15)
4(15)
0.046
- 2860 1 + ____
12
≈ 5701.78 - 5694.51 = 7.27
(
0.046
E. V C(15) - V D(15) = 2860e0.046(15) - 2860 1 + ____
365
≈ 5702.03 - 5701.78 = 0.25
)
12(15)
)
365(15)
© Houghton Mifflin Harcourt Publishing Company • Image Credits:
©Visual Ideas/Camillo Morales/Blend Images/Corbis
H.O.T. Focus on Higher Order Thinking
16. Multi-Step Ingrid and Harry are saving to buy a house.
Ingrid invests $5000 in an account that earns 3.6% interest
compounded quarterly. Harry invests $7500 in an account
that earns 2.8% interest compounded semiannually.
a. Find a model for each investment.
Ingrid’s investment:
(
)
(
)
r
Let P = 5000, r = 0.036, and n = 4. Then V(t) = P 1 +_
n
(
0.036
= 5000 1 + ____
4
Harry’s investment:
)
4t
= 5000(1.009) .
4t
r
Let P = 7500, r = 0.028, and n = 2. Then V(t) = P 1 +_
n
(
)
0.028
= 7500 1 + _____
2
Module 13
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2t
= 7500(1.014) .
2t
694
nt
n
Lesson 4
8/20/14 3:32 PM
Compound Interest
694
b. Use a graphing calculator to find when the combined value of their investments
reaches $15,000.
PEER-TO-PEER DISCUSSION
The combined value of the investments is V(t) = 5000(1.009) + 7500(1.014) .
4x
2x
Graphing the function y = 5000(1.009) + 7500(1.014) along with the line
y = 15,000 shows that the value of the function is 15,000 at x ≈ 5.9. So, the
combined investment reaches a value of $15,000 after approximately 5.9 years.
4t
Ask students to work with a partner to make a
conjecture as to how much more money an
investment of $10,000 would make at a rate of 4%
compounded annually for 5 years versus 4%
compounded monthly for 5 years. Then have them
apply the formulas from this lesson to check their
conjectures. Students should find that when
the interest is compounded monthly, the
investment earns $43.44 more interest
($12,209.97 - $12,166.53).
2t
17. Explain the Error A student is asked to find when the value of an investment
of $5200 in an account that earns 4.2% annual interest compounded quarterly
3t
reaches $16,500. The student uses the model V(t) = 5200(1.014) and finds that the
investment reaches a value of $16,500 after approximately 27.7 years. Find and correct
the student’s error.
The interest is compounded quarterly. This means that the interest is
compounded every 3 months, or 4 times per year. The value of n in the
(
r
function V(t) = P 1 + _
n
)
nt
is the number of times the interest is compounded
per year, so the student should have used 4 instead of 3.
(
)
r
Let P = 5200, r = 0.042, and n = 4. Then V(t) = P 1 + _
n
nt
(
0.042
= 5200 1 + ____
4
)
4t
JOURNAL
= 5200(1.0105) . Graphing y = 5200(1.0105) along with the line y = 16,500
Have students describe the concept of compound
interest, citing the different types of compound
interest, and explaining why the functions that model
investments that earn compound interest are
exponential growth functions.
shows that the value of the function is 16,500 at x ≈ 27.6. So, the investment
4t
4x
reaches a value of $16,500 after approximately 27.6 years.
18. Communicate Mathematical Ideas For a certain price, you can buy a
zero-coupon bond that has a maturity value of $1000, has a maturity date of 4 years,
and pays 5% annual interest. How can the present value (the amount you pay for
the bond) be determined from the future value (the amount you get when the
bond matures)?
The future value of the investment is $1000. The interest is compounded
annually, so use the value function V(t) = P(1 + r) to find the present
t
© Houghton Mifflin Harcourt Publishing Company
value of the investment.
Let r = 0.05, t = 4, and V(4) = 1000. Solve for the present value P.
P(1 + 0.05)4 = 1000
P(1.05)4 = 1000
1000
P = _____
(1.05)4
P ≈ 822.70
So, the present value of the investment is $822.70.
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Lesson 13.4
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Lesson 4
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Lesson Performance Task
AVOID COMMON ERRORS
Students should compare the answers in Parts A and
B of the Lesson Performance Task and re-read the
original question in each part to determine whether
the answers are reasonable.
The grandparents of a newborn child decide to establish a college fund for her. They invest
$10,000 into a fund that pays 4.5% interest compounded continuously.
a. Write a model for the value of the investment over time and
find the value of the investment when the child enters college
at 18 years old.
b. In 2013, the average annual public in-state college tuition was
$8893, which was 2.9% above the 2012 cost. Use these figures
to write a model to project the amount of money needed
to pay for one year’s college tuition 18 years into the future.
What amount must be invested in the child’s college fund to
generate enough money in 18 years to pay for the first year’s
college tuition?
INTEGRATE MATHEMATICAL
PRACTICES
Focus on Modeling
MP.4 Students may consider alternative ways to
approach the problem. For example, the growth
during years 18–21 could also be considered, which
would require a larger investment. Encourage
students to pursue additional possibilities based on
different constraints. Note that the growth rate will
certainly change over time, so projections are unlikely
to be very accurate.
a. Use the formula for continuously compounded interest,
with P = 10,000, r = 0.045, and t = 18.
V(t) = Pe rt
= 10,000e (0.045)(18)
= 10,000e 0.81
≈ 22,479.08
The value of the investment when the child enters college is
$22,479.08.
b. To estimate the cost in 18 years, use the formula for compound annual
interest with P = 8893, r = 0.029, and t = 18.
t
V(t) = P(1 + r)
= 8893(1 + 0.029)
18
= 8893(1.029)
18
© Houghton Mifflin Harcourt Publishing Company
≈ 14,877.33
The model predicts that tuition will cost $14,877.33 eighteen years
into the future.
Now, use the continuously compounded interest formula to find
what initial principal P yields a value in 18 years of $14,877.33
when r = 0.045.
14,877.33 = Pe rt
14,877.33 = Pe(0.045)(18)
14,877.33 = Pe 0.81
14,877.33
_______
=P
e 0.81
≈ 6618.30
The amount that must be invested is $6618.30.
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Lesson 4
EXTENSION ACTIVITY
A2_MNLESE385900_U6M13L4.indd 696
Have students research the historical tuition costs of different four-year colleges
and predict the savings needed to cover the projected tuitions for 4 years
beginning 18–21 years from now. Have them consider different investment
amounts and interest rates that would meet those costs, and present their findings
to the class.
24/03/14 11:30 PM
Scoring Rubric
2 points: Student correctly solves the problem and explains his/her reasoning.
1 point: Student shows good understanding of the problem but does not fully
solve or explain his/her reasoning.
0 points: Student does not demonstrate understanding of the problem.
Compound Interest
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