User manual | EASY STEPS TO SUCCESS: A GRAPHING CALCULATOR GUIDE

EASY STEPS TO SUCCESS:
A GRAPHING CALCULATOR GUIDE
TO ACCOMPANY
MATHEMATICAL APPLICATIONS
FOR THE MANAGEMENT, LIFE, AND
SOCIAL SCIENCES
EIGHTH EDITION
HARSHBARGER/REYNOLDS
LISA S. YOCCO
Georgia Southern University
RONALD J. HARSHBARGER
University of South Carolina Beaufort
HOUGHTON MIFFLIN COMPANY
BOSTON
NEW YORK
Sponsoring Editor: Leonid Tunik
Development Editor: Jennifer King
Editorial Associate: Allison Carol Lewis
Marketing Manager: Danielle Potvin
Copyright © 2007 by Ronald J. Harshbarger and Lisa S.Yocco. All rights reserved.
No part of this work may be reproduced or transmitted in any form or by any means, electronic or
mechanical, including photocopying and recording, or by any information storage or retrieval system
without the prior written permission of Houghton Mifflin Company unless such copying is expressly
permitted by federal copyright law. Address inquiries to College Permissions, Houghton Mifflin Company,
222 Berkeley Street, Boston, MA 02116-3764.
Printed in the U.S.A.
ISBN 13: 978-0-618-67700-9
ISBN10: 0-618-67700-3
123456789–
– 10 09 08 07 06
PREFACE
Easy Steps To Success: A Graphing Calculator Guide For The TI-84 Plus, TI-83, TI-83 Plus, and TI-82 Graphing
Calculators gives step-by-step keystrokes and instructions for these calculators, along with examples using these
keystrokes to solve problems. The split screen format shows the menus and keystrokes needed to perform or to check
operations on the left with step-by-step examples of the use of these menus and keystrokes on the right.
The guide is written as a calculator supplement for Mathematical Applications for the Management, Life, and Social
Sciences, 8th edition, by Ronald J. Harshbarger and James J. Reynolds. The Mathematical Applications text is designed
to give flexibility in the use of technology, so this guide presents all steps for each topic presented. Because many of
the keystrokes and menus are used repeatedly throughout Mathematical Applications, the topics in this guide are not
matched to specific sections in the text, but are presented in a logical order consistent with the text. This permits easy
access to the calculator keystrokes and menus as they are needed, often repeatedly, in the text. The appropriate topics
can be found easily by referring to the Contents or Index of this guide.
This guide begins with an introduction of the keys and important menus of the TI-84 Plus, TI-83,
TI-83 Plus, and TI-82 graphing calculators, followed by the step-by-step procedures and examples for the topics of
Mathematical Applications. These topics include arithmetic calculations, graphing equations and functions, finding
intercepts of graphs, solving equations and systems of equations, evaluating algebraic expressions and functions,
finding domains and ranges of functions, finding vertices of parabolas, finding maxima and minima of polynomial
functions, modeling data, solving problems with matrices, solving inequalities, solving problems involving sequences,
finance, probability, and descriptive statistics, estimating limits, finding numerical derivatives, and evaluating definite
integrals.
Many of the keystrokes are identical on the TI-84 Plus, TI-83, TI-83 Plus, and TI-82, so the instructions given on the
pages of this guide are for the TI-84 Plus, TI-83, TI-83 Plus, and TI-82. In those cases where the instructions differ, the
special instructions will be stated clearly and identified as being for the TI-84 Plus, TI-83,
TI-83 Plus, or TI-82. This permits students with different calculators to work together collaboratively.
Lisa S. Yocco
Ronald J. Harshbarger
iii
CONTENTS
OPERATING THE TI-84 PLUS
1
OPERATING THE TI-83 PLUS
2
OPERATING THE TI-83
3
OPERATING THE TI-82
10
I. CALCULATIONS WITH THE TI-82, TI-83, TI-83 Plus AND TI-84 Plus
Calculations
Calculations with Radicals and Rational Exponents
15
15
16
II. EVALUATING ALGEBRAIC EXPRESSIONS
Evaluating Algebraic Expressions Containing One or More Variables
17
17
III. GRAPHING EQUATIONS
Using a Graphing Calculator to Graph an Equation
Viewing Windows
Finding y-Values for Specific Values of x
Graphing Equations on Paper
Using a Graphing Calculator to Graph Equations Containing y2
18
18
19
20
21
22
IV. EVALUATING FUNCTIONS
Evaluating Functions with TRACE
Evaluating Functions with TABLE
Evaluating Functions with y-VARS
Evaluating Functions of Several Variables
23
23
24
25
25
V. DOMAINS AND RANGES OF FUNCTIONS; COMBINATIONS OF FUNCTIONS
Finding or Verifying Domains and Ranges of Functions
Combinations of Functions
Composition of Functions
26
26
27
28
VI. FINDING INTERCEPTS OF GRAPHS
Finding or Approximating y- and x-Intercepts of a Graph Using TRACE
Using CALC, ZERO to Find the x-Intercepts of a Graph
29
29
30
VII. SOLVING EQUATIONS
Using TRACE to Find or Check Solutions of Equations
Solving Equations with the ZERO (or ROOT) Method
Solving Equations Using the Intersect Method
Solving Equations in One Variable with SOLVER
Solving Equations for One of Several Values with SOLVER
31
31
32
33
34
34
iv
VIII. SOLVING SYSTEMS OF EQUATIONS
Points of Intersection of Graphs - Solving a System of Two Linear
Equations Graphically
Solution of Systems of Equations - Using the Intersect Method
Solution of Systems of Equations - Finding or Approximating Using TABLE
35
IX. SPECIAL FUNCTIONS; QUADRATIC FUNCTIONS
Graphs of Special Functions
Approximating the Vertex of a Parabola with TRACE
Finding Vertices of Parabolas with CALC, MAXIMUM or MINIMUM
38
38
39
40
X. PIECEWISE-DEFINED FUNCTIONS
Graphing Piecewise-Defined Functions
41
41
XI. SCATTERPLOTS AND MODELING DATA
Scatterplots of Data
Modeling Data
42
42
43
XII. MATRICES
Entering Data into Matrices; the Identity Matrix
Operations with Matrices
Multiplying Two Matrices
Finding the Inverse of a Matrix
Determinant of a Matrix; Transpose of a Matrix
Solving Systems of Linear Equations with Unique Solutions
Solution of Systems of Three Linear Equations in Three Variables
Solution of Systems - Reduced Echelon Form on the TI-83, TI-83 Plus, and
TI-84 Plus
Solution of Systems of Equations: Non-Unique Solutions
44
44
45
46
47
48
49
50
51
XIII. SOLVING INEQUALITIES
Solving Linear Inequalities
Solving Systems of Linear Inequalities
Solving Quadratic Inequalities
Solving Quadratic Inequalities – Alternate Method
53
53
54
55
56
XIV. LINEAR PROGRAMMING
Graphical Solution of Linear Programming Problems
57
57
XV. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
Graphs of Exponential and Logarithmic Functions
Inverse Functions
Exponential Regression
Alternate Forms of Exponential Functions
Logarithmic Regression
58
58
59
60
61
62
XVI. SEQUENCES
Evaluating a Sequence
Arithmetic Sequences - nth Terms and Sums
Geometric Sequences - nth Terms and Sums
63
63
64
65
v
35
36
37
52
XVII. MATHEMATICS OF FINANCE
Future Value of an Investment
Future Values of Annuities and Payments into Sinking Funds
Present Value Formulas - Evaluating with TABLE
SOLVER and Finance Formulas on the TI-83, TI-83 Plus, or TI-84 Plus
Annuities and Loans - The FINANCE Key on the TI-83, TI-83 Plus, and
TI-84 Plus
66
66
67
68
69
70
XVIII. COUNTING AND PROBABILITY
Permutations and Combinations
Probability Using Permutations and Combinations
Evaluating Markov Chains; Finding Steady-State Vectors
71
71
72
73
XIX. STATISTICS
Histograms
Descriptive Statistics
Probability Distributions with the TI-83, TI-83 Plus, and TI-84 Plus
74
74
75
76
XX LIMITS
Limits
Limits with Piecewise-Defined Functions
Limits as x → ∞
77
77
78
79
XXI. NUMERICAL DERIVATIVES
Numerical Derivatives
Checking a Derivative
Finding and Testing Second Derivatives
80
80
81
82
XXII. CRITICAL VALUES
Critical Values
83
83
XXIII. RELATIVE MAXIMA AND MINIMA
Relative Maxima and Relative Minima Using The Derivative
Relative Maxima and Relative Minima Using Maximum or Minimum
Undefined Derivatives and Relative Extrema
84
84
85
86
XXIV. INDEFINITE INTEGRALS
Checking Indefinite Integrals
Families of Functions – Solving Initial Value Problems
87
87
88
XXV. DEFINITE INTEGRALS
Approximating a Definite Integral-Areas Under Curves
Approximating a Definite Integral-Alternate Method
Area Between Two Curves
89
89
89
90
vi
Operating the TI-84 and TI-84 Plus
TURNING THE
CALCULATOR
ON AND OFF
ADJUSTING THE
DISPLAY CONTRAST
É
Turns the calculator on
y É
y }
Turns the calculator off
Increases the display (darkens the screen)
y †
Decreases the contrast (lightens the screen)
Note: If the display begins to dim (especially during calculations),
and you must adjust the contrast to 8 or 9 in order to see the screen, then
batteries are low and you should replace them soon.
The TI-84/TI-84 Plus keyboard is divided into four zones: graphing keys, editing keys, advanced function
keys, and scientific calculator keys.
U|
|
V|
||
W
UV
W
UV
W
s
U|
||
|V
||
||
W
1
Homescreen
Graphing Keys
Editing keys (Allow you to edit
expressions and variables)
Advanced Function Keys (Display
menus that access advanced functions)
Scientific Calculator Keys
Operating the TI-83 Plus
TURNING THE
CALCULATOR
ON AND OFF
ADJUSTING THE
DISPLAY CONTRAST
É
Turns the calculator on
y É
y }
Turns the calculator off
Increases the display (darkens the screen)
y †
Decreases the contrast (lightens the screen)
Note: If the display begins to dim (especially during calculations),
and you must adjust the contrast to 8 or 9 in order to see the screen,
batteries are low and you should replace them soon.
The TI-83 Plus keyboard is divided into four zones: graphing keys, editing keys, advanced function keys,
and scientific calculator keys.
U|
||
V|
||
W
UV
W
UV
W
s
U|
||
|V
||
||
W
2
Homescreen
Graphing Keys
Editing keys (Allow you to edit
expressions and variables)
Advanced Function Keys (Display
menus that access advanced functions)
Scientific Calculator Keys
Operating the TI-83
U
TURNING THE
CALCULATOR
ON AND OFF
ADJUSTING THE
DISPLAY CONTRAST
É
Turns the calculator on
y É
y }
Turns the calculator off
Increases the display (darkens the screen)
y †
Decreases the contrast (lightens the screen)
Note: If the display begins to dim (especially during calculations),
and you must adjust the contrast to 8 or 9 in order to see the screen, then
batteries are low and you should replace them soon.
The TI-83 keyboard is divided into four zones: graphing keys, editing keys, advanced function keys, and
scientific calculator keys.
U|
||
V|
||
W
UV
W
UV
W
s
U|
||
|V
||
||
W
3
Homescreen
Graphing Keys
Editing keys (Allow you to edit
expressions and variables)
Advanced Function Keys (Display
menus that access advanced
functions)
Scientific Calculator Keys
Keystrokes on the TI-83, TI-83 Plus, TI-84, and TI-84 Plus
Í
Executes commands or performs a calculation
y
Pressing the y key before another key accesses the character
located above the key and printed in yellow
ƒ
Pressing the ƒ key before another key accesses the
character located above the key and printed in green
y [A-LOCK]
Locks in the ALPHA keyboard
‘
Pressing ‘ once clears the line
Pressing ‘ twice clears the screen
y [QUIT]
Returns to the homescreen
{
Deletes the character at the cursor
y [INS]
Inserts characters at the underline cursor
„
Enters an X in Function Mode, a T in Parametric Mode,
a θ in Polar Mode, or an n in Sequence Mode
¿
Stores a value to a variable
›
Raises to an exponent
y [π]
the number π
¹
Negative symbol
 ~ [1]
Computes the absolute value of a number or an expression in parentheses
y [ENTRY]
Recalls the last entry
y [:]
Used to enter more than one expression on a line
y [ANS]
Recalls the most recent answer to a calculation
¡
Squares a number or an expression
—
Inverse; can be used with a real number or a matrix
y
Computes the square root of number or an expression in parentheses
x
y [e ]
Returns the constant e raised to a power
ƒ [0]
Space
y|
Moves the cursor to the beginning of an expression
y~
Moves the cursor to the end of an expression
4
Special Features of theTI-83 Plus
U
The TI-83 Plus uses Flash technology, which lets you upgrade to future software versions and to download
helpful programs from the TI website, www.education.ti.com.
Special Features of theTI-84 and TI-84 Plus
U
The TI-84 Plus is an enhanced version of the TI-83 Plus that offers a built-in USB port, 3x the memory of
the TI-83 Plus, a high-contrast display, and many preloaded Apps.
Applications can be installed and accessed via the
APPS key. The Finance application is accessed via
this key rather than directly as it is with the TI-83.
Archiving on the TI-83/TI-84 Plus allows you to
store data, programs, or other variables to user data
archive, a protected area of memory separate from
RAM, where they cannot be edited or deleted
inadvertently.
Archived variables are indicated by asterisks (*) to
the left of variable names.
On the TI-83/TI-84 Plus, MATRX applications are accessed by pressing y [MATRX] rather than
directly as they are with the TI-83.
5
Setting Modes
Mode settings control how the TI-83, TI-84, TI-83 Plus, and TI-84 Plus display and interpret numbers and
graphs.
Numeric display format
Number of decimal places
Unit of angle measure
Type of graphing
Whether to connect graph points
Whether to plot simultaneously
Real, rectangular complex, or polar complex
Full, two split-screen modes
Clock (available on the TI-84)
MATH Operations
Displays the answer as a fraction
Displays the answer as a decimal
Calculates the cube
Calculates the cube root
Calculates the xth root
Finds the minimum of a function
Finds the maximum of a function
Computes the numerical derivative
Computes the function integral
Displays the equation solver
6
MATH NUM (Number) Operations
Absolute value
Round
Integer part
Fractional part
Greatest integer
Minimum value
Maximum value
Least common multiple
Greatest common divisor
MATH CPX (Complex) Operations
Returns the complex conjugate
Returns the real part
Returns the imaginary part
Returns the polar angle
Returns the magnitude (modulus)
Displays the result in rectangular form
Displays the result in polar form
7
MATH PRB (Probability) Operations
Random-number generator
Number of permutations
Number of combinations
Factorial
Random-integer generator
Random number from Normal distribution
Random number from Binomial distribution
Y= Editor
Up to 10 functions can be stored to the function
variables Y1 through Y9 , and Y0 . One or more
functions can be graphed at once.
VARS Menu
X/Y, T/θ, and U/V/W variables
ZX/ZY, ZT/Zθ, and ZU variables
GRAPH DATABASE variables
PICTURE variables
XY, Σ, EQ, TEST, and PTS variables
TABLE variables
STRING variables
8
VARS Y-VARS Menus
Yn functions
XnT, YnT functions
rn functions
Lets you select/deselect functions
TEST Menu
Returns 1 (true) if:
Equal
Not equal
Greater than
Greater than or equal to
Less than
Less than or equal to
TEST LOGIC Menu
Returns 1 (true) if:
Both values are nonzero (true)
At least one value is nonzero (true)
Only one value is zero (false)
The value is zero (true)
9
Operating the TI-82
TURNING THE
CALCULATOR
ON AND OFF
ADJUSTING THE
DISPLAY CONTRAST
É
Turns the calculator on
y É
y }
Turns the calculator off
Increases the display (darkens the screen)
y †
Decreases the contrast (lightens the screen)
Note: If the display begins to dim (especially during calculations),
and you must adjust the contrast to 8 or 9 in order to see the screen,
batteries are low and you should replace them soon.
The TI-82 keyboard is divided into four zones: graphing keys, editing keys, advanced function keys, and
scientific calculator keys.
Homescreen
Graphing Keys
Editing Keys (Allow you to edit
expressions and values)
Advanced Function Keys (Display
menus that access advanced functions)
Scientific Calculator Keys
10
Keystrokes on the TI-82
Í
Executes commands or performs a calculation
y
Pressing the y key before another key accesses the character
located above the key and printed in yellow
ƒ
Pressing the ƒ key before another key accesses the
character located above the key and printed in green
y [A-LOCK]
Locks in the ALPHA keyboard
‘
Pressing ‘ once clears the line
Pressing ‘ twice clears the screen
y [QUIT]
Returns to the homescreen
{
Deletes the character at the cursor
y [INS]
Inserts characters at the underline cursor
X,T,θ
Enters an X in Function Mode, a T in Parametric Mode,
or a θ in Polar Mode
¿
Stores a value to a variable
›
Raises to an exponent
y [π]
the number π
¹
Negative symbol
y [ABS]
Computes the absolute value of a number or an expression in parentheses
y [ENTRY]
Recalls the last entry
y [:]
Used to enter more than one expression on a line
y [ANS]
Recalls the most recent answer to a calculation
¡
Squares a number or an expression
—
Inverse; can be used with a real number or a matrix
y
Computes the square root of number or an expression in parentheses
x
y [e ]
Returns the constant e raised to a power
ƒ [0]
Space
y|
Moves the cursor to the beginning of an expression
y~
Moves the cursor to the end of an expression
11
Setting Modes
Mode settings control how the TI-82 displays and interprets numbers and graphs.
Numeric display format
Number of decimal places
Unit of angle measure
Type of graphing
Whether to connect graph points
Whether to plot simultaneously
Full or split-screen mode
MATH Operations
Displays the answer as a fraction
Displays the answer as a decimal
Calculates the cube
Calculates the cube root
Calculates the xth root
Finds the minimum of a function
Finds the maximum of a function
Computes the numerical derivative
Computes the function integral
Computes the solution of a function
12
MATH NUM (Number) Operations
Round
Integer part
Fractional part
Greatest integer
Minimum value
Maximum value
MATH PRB (Probability) Operations
Random-number generator
Number of permutations
Number of combinations
Factorial
Y= Editor
Up to 8 functions can be stored to the function variables
Y1 through Y8 . One or more functions can be graphed
at once.
13
VARS Menu
X/Y, T/θ, and U/V variables
ZX/ZY, ZT/Zθ, and ZU variables
GRAPH DATABASE variables
PICTURE variables
XY, Σ, EQ, BOX, and PTS variables
TABLE variables
2nd [Y-VARS] Menus
Yn functions
XnT, YnT functions
B
B
B
B
B
B
rn functions
B
B
Lets you select/deselect functions
TEST Menu
Returns 1 (true) if:
Equal
Not equal
Greater than
Greater than or equal to
Less than
Less than or equal to
14
I. CALCULATIONS WITH THE TI-84, TI-84 Plus, TI-83, TI-83 Plus, and TI-82
CAUTION!
The negative sign (-) and the subtraction sign − are different.
Use the − sign for subtraction and the (-) sign to write negative
numbers.
CALCULATIONS
EXAMPLES
Because the TI-84, TI-84 Plus, TI-83, TI-83 Plus,
and the TI-82 use standard algebraic order when
evaluating arithmetic expressions, the expression
can be entered as it appears. Working outwards
from inner parentheses, calculations are performed
from left to right. Powers and roots are evaluated
first, followed by multiplications and divisions, and
then additions and subtractions.
Calculate −4(9 − 8) + ( −7)(2) 3
If the numerator or denominator of a fraction
contains more than one operation, it should be
enclosed in parentheses when entering it into the
calculator.
Calculate
10 − 6
4 − 4(3)
Calculate
3 19
+
7 27
Note: To preserve the order of operations when
calculations involve fractions, enter the fractions in
parentheses.
Decimal answers will normally appear if the
answers are not integers. If an answer is a rational
number, its fractional form can be found by
pressing MATH 1: > Frac and pressing ENTER.
When entering an expression to be calculated, be
careful to enclose expressions in parentheses as
needed.
Calculate 3 ÷ 4 +
15
25
42
FG 5 IJ −3
H 4K
+ −
CALCULATIONS WITH RADICALS AND
RATIONAL EXPONENTS
Square roots can be evaluated by using the
EXAMPLES
Calculate
289
key,
when the expression is defined in the set of real
numbers. If the expression is undefined, an error
message appears.
−289
Cube roots can be evaluated by pressing MATH 4:
3 .
Calculate
3
−3375
Roots of other orders can be evaluated by entering
the index and then pressing MATH 5: x
Calculate
4
4096
Recall that roots can be converted to fractional
exponents using
n
a = a (1/ n ) and
n
am = a
bm/ ng .
Another way to calculate
b g
Calculate −64
On the TI-84, TI-84 Plus, TI-83 and TI-83 Plus,
expressions containing rational exponents can be
evaluated. On the TI-82, some expressions may
have to be rewritten using the property
e j
a m/ n = a 1/ n
m
.
If the result of a computation is an irrational
number, only the decimal approximation of this
irrational number will be shown. Pressing MATH,
1: > Frac will not give a fraction; it will give the
same decimal.
Calculate
16
18
3
2/3
4
4096
(On the TI-82):
II. EVALUATING ALGEBRAIC EXPRESSIONS
EVALUATING ALGEBRAIC EXPRESSIONS
CONTAINING ONE OR MORE VARIABLES
1. To evaluate an algebraic expression containing
one variable:
EXAMPLE
1. Evaluate
x+4
for x = −6.
5− x
Enter the x-value, press STO, X, ALPHA, : , enter
the expression, and press ENTER.
2. To evaluate an algebraic expression containing
two variables, x and y:
Enter the x-value, press STO, X, ALPHA, : , enter
the y-value, STO, ALPHA, Y, ALPHA, : , enter the
expression, and press ENTER.
3. When evaluating an algebraic expression
containing variables, any letter may be used as a
variable. Use the ALPHA key to enter the letters.
2. Evaluate 3x − 5 y for x = −2 and
y = −6.
3. Find the surface area of a right circular cylinder
with r = 5.2 and
h = 6.4.
The formula for the surface area is
S = 2πR 2 + 2πRH , so enter
5.2 → R:6.4 → H:2πR 2 + 2πRH and press ENTER.
To correct an entry or to evaluate the expression for
different values, press
2nd ENTER and edit the expression.
The output (surface area) is 379.002, to three
decimal places.
The surface area of a right circular cylinder for
different values of r and h can be found by pressing
2nd ENTER and entering the new values. For r =
1.3 and h = 2.7:
17
III. GRAPHING EQUATIONS
USING A GRAPHING CALCULATOR TO GRAPH
AN EQUATION
To graph an equation in the variables x and y:
EXAMPLES
1. Solve the equation for y in terms of x.
1. Solving for y gives −2y = −2x2 + 3, so
−2 x 2
3
3
y=
or y = x 2 − .
+
−2
−2
2
Graph 2x2−2y = 3 with a graphing calculator:
2
2. Press Y= key and enter y1 = x − 3 / 2 .
Both screens below will give the graph.
B
2. Press the Y= key to access the function entry screen
and enter the equation into y1. Use parentheses as
needed so that what you have entered agrees with the
order of operations.
To erase equations from the equation editor, press
CLEAR. To leave the equation editor and return to the
homescreen, press 2nd QUIT.
B
B
B
B
B
3. Determine an appropriate viewing window.
Frequently the standard window (ZOOM 6) is
appropriate, but often a decimal or integer viewing
window (ZOOM 4 or ZOOM 8) gives a better
representation of the graph.
Pressing GRAPH or a ZOOM key will activate the
graph.
ZOOM 8 must be followed by ENTER, but ZOOM 4
and ZOOM 6 do not.
2
3. The graphs of y = x − 3 / 2 using possible
windows are below.
ZOOM 6 (standard) ZOOM 4 (decimal)
B
B
ZOOM 8, ENTER (integer)
4. All equations in the equation editor that have their
“=“ signs highlighted (dark) will have their graphs
shown when GRAPH is pressed. If the “=“ sign of an
equation in the equation editor is not highlighted, the
equation will remain, but its graph is “turned off“ and
will not appear when GRAPH is pressed. The graph is
“turned on“ by repeating the process.)
18
4. Equation y1 is turned off.
Equation y2 and y3 are turned on. Only the graphs of
y2 and y3 are displayed.
VIEWING WINDOWS
EXAMPLE
With a TI-84, TI-84 Plus, TI-83, TI-83 Plus, and TI82, as with a graph plotted by hand, the appearance of
the graph is determined by the part of the graph we
are viewing. The viewing window determines how a
given graph appears in the same way that different
camera lenses show different views of an event.
The graph of y = x 3 − x looks somewhat like a line
in the region resulting from ZOOM 8.
The values that define the viewing window can be set
individually or by using ZOOM keys. The important
values are:
But its shape is defined better in the standard
viewing window, accessed by pressing ZOOM 6,
giving a window with x-values and y-values between
-10 and 10.
The graph of y = x 3 − x with ZOOM 6:
x-min: the smallest value on the x-axis
x-max: the largest value on the x-axis
x-scale: spacing for tics on the x-axis
y-min: the smallest value on the y-axis
y-max: the largest value on the y-axis
y-scale: spacing for tics on the y-axis
We can use a "friendly" window, which causes the
cursor to change by a "nice" value such as .1, .2, 1,
etc. when a right or left arrow is pressed. A window
will be "friendly" if xmax – xmin gives a "nice"
number when divided by 94.
ZOOM, 4 automatically gives
xmin = -4.7 and xmax = 4.7, so
xmax – xmin = 9.4. Thus each press of the right or
left arrow moves the cursor 9.4/94 = .1 units.
ZOOM, 8, ENTER gives a movement of 1 unit for
each press of an arrow.
The graph of y = x 3 − x with ZOOM 4:
The window should be set so that the important parts
of the graph are shown and the unseen parts are
suggested. Such a graph is called complete.
The values that define the viewing window can be set
individually. If necessary, using ZOOM, 3:Zoom Out
can help to determine the shape and important parts of
the graph.
The following window shows the complete graph
clearly.
Note: Using xmin = -9.4 and xmax = 9.4 with
ymin = -10 and ymax = 10 gives a window that is
"friendly" and close to the standard window.
19
FINDING y-VALUES FOR SPECIFIC VALUES
OF x
To find y-values at selected values of x by using
TRACE, VALUE:
1. Press the Y= key to access the function entry screen
and enter the right side of the equation. Use
parentheses as needed so that what you have entered
agrees with the order of operations.
EXAMPLE
Find y when x is 2, 6, and -1 for the equation y =
5x − 1.
1. Enter y1 = 5x − 1
B
B
2. Set the window with xmin = −10 and xmax= 10,
ymin = −10 and ymax = 10.
3.
2. Set the window so that it contains the x-value
whose y-value we seek.
3. Press GRAPH.
OR Do the following, which we will call TRACE,
VALUE:
4. ON THE TI-84, TI-84 PLUS, TI-83 and TI-83
PLUS:
Press TRACE and then enter the selected x-value
followed by ENTER.
The cursor will move to the selected value and give
the resulting y-value if the selected x-value is in the
window. If the selected x-value is not in the window,
Err: INVALID occurs.
If the x-value is in the window, the
y-value will occur even if it is not in the window.
ON THE TI-82 AND ON THE TI-84, TI-84 PLUS,
TI-83 AND TI-83 PLUS:
5. Press 2nd calculate, 1:(value), ENTER, enter the xvalue, and press ENTER. The corresponding y-value
will be displayed if the selected x-value is in the
window.
4. Press TRACE, enter the value
2, getting y = 9.
Enter 6, getting y = 29
Enter -1, getting y = -6
5. Press TRACE, enter 6, and press ENTER, getting
y = 29.
Enter 2, get y = 9.
Enter −1, get y = −6.
6. Press 2nd calculate, 1:(value), ENTER, enter −1,
and press ENTER, getting
y = −6.
20
GRAPHING EQUATIONS ON PAPER
EXAMPLE
To graph an equation in the variables x and y:
1. Solve the equation for y in terms of x.
Graph 2x − 3y = 12 .
1. Solving for y gives −3y = −2x + 12
−2 x
12
2x
y=
+
y=
−4
−3 −3 or
2. Press the Y= key to access the function entry
3
screen and enter the equation. Use parentheses as
needed so that what is entered agrees with the order 2. Enter y1= 2x/3 − 4
NOTE: Division is preserved without using
of operations.
parentheses in this case.
B
B
3. Determine an appropriate viewing that gives a
complete graph. Press GRAPH.
B
B
3. Graphing the equation with
xmin = -9.4 and xmax= 9.4, and with ymin = -10 and
ymax = 10 gives:
4. To sketch the graph on paper, use TABLE or
TRACE, VALUE to get x and y values of
representative points on the graph. Use these
points and the shape of the graph to sketch the
graph.
USING THE TABLE
1. a. To prepare the table, press 2nd Table Set,
enter an initial x-value in a table (Tblmin), and
enter the change ( ∆ Tbl) in the x-value we want in
the table.
b. If we change Independent variable to Ask, we
may enter any value of x we choose and get the
corresponding y-value.
2. Enter 2nd TABLE to get a list of x-values and
the corresponding y-values. The value of the
function at the given value of x can be read from
the table. OR simply enter the x-value if
Independent variable is set to Ask.
4. TRACE, VALUE gives coordinates of points.
1. Setting Tblmin = 0 and ∆ Tbl = 1
gives the table below.
2. The table shows a list of values of x and the
corresponding values of y.
The value of y when x is 3 is -2.
3. If Independent variable is at Auto, use the up or
down arrow to find the x-values where the function 3. Pressing the up arrow gives values less than 0 and
the corresponding y-values. The value corresponding
is to be evaluated.
to -3 is -6.
21
USING A GRAPHING CALCULATOR TO
EXAMPLE
2
GRAPH EQUATIONS CONTAINING y
To use a graphing calculator to graph an equation Graph the circle with equation
containing y 2 :
x 2 + y 2 = 49
1. Solve the equation for y using the property that 1. Solve for y.
y 2 = a if and only if y = ± a .
y 2 = 49 − x 2
y = ± 49 − x 2
2. Use the Y= key to enter each of the solutions for
y.
2. Enter y1 =
and y2 = −
3. Determine an appropriate viewing window and
graph the equation.
(49 − x 2 )
(49 − x 2 ) .
3. The standard window is appropriate but does
not produce a graph which appears to be a circle.
4. The graph will be in correct proportion (square) 4. Using a SQUARE window (ZOOM 5) will
if xmax − xmin is some multiple of 94 while
correct this dilemma.
ymax − ymin is that same multiple of 62. ZOOM 4
and ZOOM 8 are square windows, and ZOOM 5
changes the y-values of the window so that the
window is square.
ADDITIONAL EXAMPLE
Graph x 2 − 4 y 2 = 16 on a square window.
Solving for y gives two equations
y2 =
16 − x 2 x 2 − 16
=
4
−4
x 2 − 16
x 2 − 16
, y=−
2
2
Entering the equations as y1 and y2 gives the
graph of the relation.
y=
22
IV. EVALUATING FUNCTIONS
If y is a function of x, then the y-coordinate of the graph at a given value of x is the functional value.
EVALUATING FUNCTIONS WITH TRACE,
VALUE
To evaluate functions at selected values of x by using
TRACE:
Evaluate f(2), f(6), and f(−1) if
f(x) = 5x − 1.
1. Use the Y= key to store y1= f(x).
1. Enter y1 = 5x − 1
2. Graph using an appropriate viewing window that
gives a complete graph.
2.
3. Use TRACE with one of the following methods.
The resulting y-value is the function value.
3. a. Press TRACE, enter 2, getting y = 9. Thus
f(2) = 9.
B
B
EXAMPLE
B
B
B
B
a. ON THE TI-84, TI-84 PLUS, TI-83 and TI-83
PLUS:
Press TRACE and then enter the selected x-value
followed by ENTER. The corresponding y-value will
be displayed if the selected x-value is in the window.
ON THE TI-82 OR THE TI-84 OR TI-84 PLUS OR
THE TI-83 OR TI-83 PLUS:
b. Press 2nd calculate, 1:(value), ENTER, enter the xvalue, and press ENTER. The corresponding y-value
will be displayed if the selected x-value is in the
window.
23
b. Press 2nd calculate, 1:(value), ENTER, enter −1,
and press ENTER, getting y = −6. Thus f(−1) = −6.
We can also evaluate functions by means other than TRACE. Some alternate ways follow.
EVALUATING A FUNCTION WITH TABLE
EXAMPLE
To evaluate a function with a table:
Evaluate y = −x2 + 8x + 9 when x = 3 and when x =
−5.
1. Enter the function with the Y= key.
1. Enter y1 = −x2 + 8x + 9.
2. To find f(x) for specific values of x in the table,
press 2nd Table Set, move the cursor to Ask
opposite Indpnt:, and press ENTER. Then press
2nd TABLE and enter the specific values.
2.
B
B
ALTERNATE METHOD:
3. Press 2nd Table Set, enter an initial x-value in a
table (Tblmin), and enter the desired change
( ∆ Tbl) in the x-value in the table.
3. Setting Tblmin = 0 and ∆ Tbl = 1
gives the table below.
4. Enter 2nd TABLE to get a list of x-values and
the corresponding y-values. The value of the
function at the given value of x can be read from
the table.
4. The table shows a list of values of x and the
corresponding values of y.
The value of y when x is 3 is 24.
5. Use the up or down arrow to find the x-values
where the function is to be evaluated.
5. Pressing the up arrow gives values less than 0
and the corresponding y-values. The value
corresponding to −5 is −56.
24
EVALUATING FUNCTIONS WITH
y-VARS
To evaluate the function f at one or more values of x:
EXAMPLE
Given the function
f(x) = −16x2 + 20x – 2, find
f(4), f(1.45), f(−2), f(−8.4).
1. Enter y1 = −16x2 + 20x − 2 in y1:
1. Use the Y= key to store y1 = f(x).
Press 2nd QUIT.
B
B
B
B
B
2. Press VARS, Y-VARS 1,1
(2nd, Y-VARS, 1,1 on the TI-82) to get y1.
Enter the x-values needed as follows:
y1 ({value 1, value 2, etc.}) ENTER.
Values of the function will be displayed.
B
B
B
2.
Enter y1 ({4,1.45,−2,84})
B
B
B
B
The display gives the values
{−178, −6.64, −106, −111,218}.
EVALUATING FUNCTIONS OF SEVERAL
VARIABLES
To evaluate a function of two variables:
EXAMPLES
If z = f ( x , y ) = 2 x + 5 y , find f(3, 5).
1. Enter 3 for x and 5 for y, using STO:
1. Enter the x-value, press STO, X, ENTER. Enter
the y-value, press STO,
ALPHA Y (above 1), ENTER.
2. Enter the functional expression and press
ENTER. The functional value will be displayed.
2.
3. To evaluate the function for different values of
the variables, enter new values of the variables and
press 2nd ENTER to repeat the functional
expression.
3. To evaluate the same function at
x = −3, y = 4, enter the new values for x and y, and
use 2nd ENTER to find the functional value.
25
Entering 2x + 5y and pressing ENTER gives
f(3, 5)
V. DOMAINS AND RANGES OF FUNCTIONS; COMBINATIONS OF FUNCTIONS
The graphs of functions can be used to determine or to verify their domains and ranges.
Domain: set of all x-values for which a function is defined
Range: set of all y-values resulting from these x-values
FINDING OR VERIFYING DOMAINS
AND RANGES OF FUNCTIONS
To visually find the domain and range of a function:
EXAMPLES
Find the domains and ranges of the following
functions.
1. Graph the function with a window that shows all
the important parts and suggests where unseen parts
are located. Such a graph is called complete.
f(x) = x2+2
f(x) = 3x−8
2. Visually determine if the graph is defined for the
set of all x-values (all real numbers) or for some
subset of the real numbers. This set is the domain
of the function.
3. Visually determine if the
y-values on the graph form the set of all real
numbers or some subset of the real numbers. This
set is the range of the function.
If a function contains a denominator, values of x
that make the denominator 0 are not in the domain
of the fraction.
Graphing the function with a window that contains
these values of x shows that the graph is not defined
for them.
domain: all reals
range: all reals
f(x) = −x3+5
domain: all reals
range: {y| y > 2 }
f(x) = −0.4x4 + 8
domain: all reals
range: all reals
domain: all reals
range: {y| y < 8}
f ( x) =
If a function contains a square root radical, values
of x that give negative values inside the radical are
not in the domain of the function.
4
x+2
domain: reals except−2
range: reals except 0
y = 2+ x
domain:{x| x> -2}
range: {y| y> 0}
26
g( x) = 4 − x 2
domain:{ x| -2 < x < 2}
range :{y| 0 < y < 2}
COMBINATIONS OF FUNCTIONS
EXAMPLES
To find the graphs of combinations of two functions
f(x) and g(x):
If f(x) = 4 x − 8 and g(x) = x 2 , find the following:
1.
1. Enter f(x) as y1 and g(x) as y2 under the Y=
menu.
2. Graph (f + g)(x)
2. To graph (f + g)(x), enter y1 + y2 as y3 under
the Y= menu. Place the cursor on the = sign beside
y1 and press ENTER to turn off the graph of y1 .
Repeat with y2 . Press GRAPH with an appropriate
window.
3. Graph (f − g)(x)
3. To graph (f – g)(x), enter y1 – y2 as y3 under
the Y= menu. Place the cursor on the = sign beside
y1 and press ENTER to turn off the graph of y1 .
Repeat with y2 . Press GRAPH.
4. To graph ( f ∗ g )(x), enter y1 ∗ y2 as y3 under
the Y= menu. Place the cursor on the = sign beside
y1 and press ENTER to turn off the graph of y1 .
Repeat with y2 . Press GRAPH.
4. Graph (f*g)(x)
5. Graph (f/g)(x)
5. To graph (f/g)(x), enter y1 / y2 as y3 under the
Y= menu. Place the cursor on the = sign beside y1
and press ENTER to turn off the graph of y1 .
Repeat with y2 . Press GRAPH with an appropriate
window.
6. Find (f + g)(3).
6. To evaluate f + g, f − g, f * g, or f/g at a
specified value of x, enter y1 , the value of x
enclosed in parentheses, the operation to be
performed, y2 , the value of x enclosed in
parentheses, and press ENTER. Or, if the
combination of functions is entered as y3 , enter y3
and the value of x enclosed in parentheses.
b
gb g
NOTE that entering Y1 + Y2 3 does not produce
the correct result.
27
COMPOSITION OF FUNCTIONS
EXAMPLES
To graph the composition of two functions f(x) and
g(x):
If f(x) = 4x − 8 and g(x) = x 2 , find the following:
1. Enter the two functions.
1. Enter f(x) as y1 and g(x) as y2 under the Y=
menu.
b
gb g
b
gb g b g
b
gb g b
2. Graph f D g x = f g ( x )
2. To graph f D g x , enter y1 ( y2 ) as y3 under
the Y= menu. Place the cursor on the = sign beside
y1 and press ENTER to turn off the graph of y1 .
Repeat with y2 . Press GRAPH with an appropriate
window.
b
gb g
3. Graph g D f x = g f ( x )
3. To graph g D f x , enter y2 ( y1 ) as y4 under
g
the Y= menu. Turn off the graphs of y1 , y2 , and
y3 . Press GRAPH with an appropriate window.
b
gb g
4. To evaluate f D g x at a specified value of x,
b
b
gb g c b gh
b
gb g c b gh
4. Evaluate f D g −5 = f g −5
enter y1 ( y2 ) and the value of x enclosed in
parentheses, and press ENTER. Or, if the
combination of functions is entered as y3 , enter y3
and the value of x enclosed in parentheses.
gb g
5. To evaluate g D f x at a specified value of x,
5. Evaluate g D f −5 = g f −5
enter y2 ( y1 ) and the value of x enclosed in
parentheses, and press ENTER. Or, if the
combination of functions is entered as y4 , enter y4
and the value of x enclosed in parentheses.
28
VI. FINDING INTERCEPTS OF GRAPHS
As we have seen, TRACE allows us to find a specific point on the graph. Thus TRACE can be used to
solve a number of important problems in algebra. For example, it can be used to find the x- and yintercepts of a graph.
FINDING OR APPROXIMATING Y- AND XINTERCEPTS OF A GRAPH USING TRACE
To find the y-intercepts and
x-intercepts of a graph:
EXAMPLE
1. Solve the equation for y. Enter the equation
with the Y= key.
1. Enter y = −x2 + 8x + 9.
2. Set the window so that the intercepts to be
located can be seen. The graph of an nth degree
equation (n is a positive integer) crosses the x-axis
at most n times.
2. Set the window with xmin = −9.4 and xmax = 9.4,
ymin = −10 and ymax = 10.
3. Graph the equation.
Change the y-values in the window so that the
graph is one curve with points where the curve
crosses the axes visible. Use ZOOM, Zoom Out if
necessary.
3. Graphs (before and after change).
4. Press TRACE and enter the value 0. The
resulting y value is the y-intercept of the graph.
4. Press TRACE, enter 0, giving the y-intercept
y = 9.
5. Press TRACE and use the right and left arrows
to move the cursor to values of x that give y = 0.
These are the x-intercepts of the graph.
Find the x- and y- intercepts of the graph of
y = −x2 + 8x + 9.
5. Press TRACE, move with arrow to
x = −1, which gives y = 0 (and so x = −1 is an
x-intercept).
Tracing to x = 9 gives y = 0 (and so
x = 9 is an x-intercept).
29
Methods other than TRACE can be used to find the x-intercepts of a graph.
USING CALC, ZERO TO FIND THE
X-INTERCEPTS OF A GRAPH
To find the x-intercepts of the graph of an equation by
using CALC, ZERO:
1. Solve the equation for y.
Using the Y= key, enter the equation.
EXAMPLE
Find the x-intercepts of the graph of
2 x 2 − 9 x − y = 11 .
1. y = 2 x 2 − 9 x − 11
2. Graph the equation with an appropriate window, and
note that the equation will intersect the x-axis where y =
0; that is, when x is a solution to the original equation.
Set the window so that all points where the graph
crosses the
x-axis are visible. (Using ZOOM OUT can help check
that all such points on the graph are present.)
2. Because this is a quadratic function, it could have
two x-intercepts. Pick x-values that give a friendly
window including the x-intercepts. In this case, using 9.4 to 9.4 shows the x-intercepts:
3. To find the point(s) where the graph crosses the
x-axis and the function has zeros, use 2nd CALC, 2
(ZERO) .
Answer the question "left bound?" with ENTER after
moving the cursor close to and to the left of an
x-intercept.
3. Using TRACE shows a solution near
x = 5.4. Use 2nd CALC, 2 (ZERO) .
Answer the question "right bound?" with ENTER after
moving the cursor close to and to the right of this
x-intercept.
To the question "guess?" press ENTER.
The coordinates of the x-intercept will be displayed.
Repeat to get all x-intercepts.
4. The steps on the TI-82 are identical except 2nd
CALC 2 shows the word ROOT, "left bound" is
written as "lower bound," and "right bound" is written
as "upper bound. "
30
The value of the x-intercept is 5.5. Using TRACE or
the steps above gives the second x-intercept as x = −1.
VII. SOLVING EQUATIONS
TRACE can be used to find solutions of equations. The equation can be solved by setting one side of the
equation equal to zero, graphing y = the nonzero side, and finding the x-intercepts of this graph.
USING TRACE TO FIND OR CHECK SOLUTIONS
OF EQUATIONS
To solve an equation (approximately) with a graphing
calculator:
EXAMPLE
1. Rewrite the equation with 0 on one side of the
equation.
1. x2 + 11x + 10 = 0
2. Using the Y= key, enter the non-zero side of the
equation in step 1.
2. y1 = x2 + 11x + 10
Solve x2 + 11x = -10.
3. Graph the equation with a friendly window such that 3. Because this is a quadratic function, it could have
two x-intercepts. Pick
the points where the graph crosses the x-axis are
x-values that give a friendly window including the xvisible.
intercepts. In this case, using x-values from -18.8 to 0
shows the two points where the graph crosses the xa. For a linear equation, set the window so the graph
axis.
crosses the x-axis in one point.
b. For a non-linear equation, set the window so that all
points where the graph crosses the x-axis are visible.
An nth degree equation will intersect the x-axis in at
most n points. (Using ZOOM, Zoom Out can help
check that all such points on the graph are present.)
4. Use TRACE to get the x-value(s) of point(s) on the
graph where the y-value(s) are zero. Using a friendly
window is frequently helpful in finding these values.
ZOOM, Zoom In may give a better approximation for
some solutions.
5. The graph will intersect the x-axis where y = 0; that
is, where x is a solution to the original equation
31
4. Using TRACE shows that y = 0 at x = -10 and at
x = -1.
5. Thus the solutions to
x2 + 11x = -10 are x = -10 and at x = -1.
Methods other than TRACE can be used to solve an equation. The word "zero" means a value of x that
makes an expression zero, so an equation can be solved for x by setting one side of the equation equal to
zero and using the "zero" method on the TI-83, TI-84, TI-83 Plus, or TI-84 Plus. The same commands
give “roots” (solutions) on the TI-82.
SOLVING EQUATIONS WITH THE
ZERO (OR ROOT) METHOD
To solve an equation (approximately) with the TI-83,
TI-84, TI-83 Plus, or TI-84 Plus:
EXAMPLE
Solve 2x2 − 9x = 11.
1. Rewrite the equation with 0 on one side of the
equation.
1. 2x2 − 9x − 11 = 0
2. Using the Y= key, enter the non-zero side of the
equation.
2. y = 2x2 − 9x − 11
B
B
B
3. Graph the equation with an appropriate window, and
note that the equation will intersect the x-axis where y =
0; that is, when x is a solution to the original equation.
Set the window so that all points where the graph
crosses the x-axis are visible. (Using ZOOM OUT can
help check that all such points on the graph are
present.)
4. To find the point(s) where the graph crosses the xaxis and the equation has solutions, use 2nd CALC,
2 :ZERO .
Answer the question "left bound?" with ENTER after
moving the cursor close to and to the left of an xintercept.
Answer the question "right bound?" with ENTER after
moving the cursor close to and to the right of this xintercept. To the question "guess?" press ENTER.
The coordinates of the x-intercept will be displayed.
The x-value is the solution to the original equation.
B
3. Because this is a quadratic function, it could have
two x-intercepts. Pick x-values that give a friendly
window including the x-intercepts. In this case, using 9.4 to 9.4 gives the graph:
4. Using TRACE shows a solution near
x = 5.4. Use 2nd CALC, 2 (ZERO) .
5. The same steps, with 2nd CALC, 2: Root gives the
solution on the TI-82.
6. Repeat to get all x-intercepts (and solutions).
The value of the x-intercept (zero) is 5.5, so a solution
is x = 5.5
6. Repeating the steps above near the second intercept
gives the second
x-intercept (and solution) as
x = -1.
32
SOLVING AN EQUATION USING THE
INTERSECT METHOD
To solve an equation (approximately) by the
intersect method:
EXAMPLE
Solve 2 x − 1 =
1
x + 2 for x.
3
1.
1. Under the Y= menu, assign the left side of the
equation to y1 and the right side of the equation to
y2.
B
B
B
B
2. Graph the equations using a friendly window
that contains the points of intersection of the
graphs. Using ZOOM OUT can be used to search
for all points of intersection.
3. Press 2nd CALC, 5 (INTERSECT) to find each
point of intersection of two curves.
2. Using ZOOM 4 gives the graphs:
3.
Answer the question "first curve?" with ENTER
and "second curve?" with ENTER. (Or press the
down arrow to move to one of the two curves.)
To the question "guess?" move the cursor close to
the desired point of intersection and press ENTER.
The coordinates of the point of intersection will be
displayed. Repeat to get all points of intersection.
4. The solution(s) to the equation will be the
values of x from the points of intersection found in
Step 3.
4. The solutions to the equation are
x = -.429 (approximately) and x = 1.8.
33
USING SOLVER ON THE TI-83, TI-84, TI-83 PLUS, or TI-84 PLUS
An equation involving one or more variables can be solved for one variable with the SOLVER function,
under the MATH menu of the TI-83, TI-84, TI-83 Plus, or TI-84 Plus.
SOLVING AN EQUATION IN ONE VARIABLE
WITH SOLVER ON THE TI-83, TI-84, TI-83
PLUS, or TI-84 PLUS
To solve an equation using SOLVER:
EXAMPLE
1. Rewrite the equation with 0 on one side.
1. Write the equation in the form
x2 − 7x + 12 = 0
Solve x2 − 7x = −12
2. Press MATH 0 (Solver).
Press the up arrow revealing EQUATION SOLVER
eqn: 0 =, and enter the nonzero side of the equation
to be solved.
2. Get the EQUATION SOLVER and enter x2 − 7x
+ 12.
3. Press the down arrow or ENTER and the variable
appears with a value (not the solution). Place the
cursor on the variable whose value is sought.
Press ALPHA SOLVE (ENTER).
The value of the variable changes to the solution of
the equation that is closest to that value.
4. To find additional solutions (if they exist), change
the value of the variable and press ALPHA SOLVE
(ENTER). The value of the variable changes to the
solution of the equation that is closest to that value.
SOLVING AN EQUATION FOR ONE OF
SEVERAL VARIABLES WITH SOLVER
3.
4.
Use I = PRT to find the rate R if an investment of
$1000 yields $180 in 3 years.
1. 0 = PRT − I
2.
1. Press MATH 0 (Solver).
Press the up arrow revealing EQUATION SOLVER
eqn: 0 =, and enter the nonzero side of the equation
to be solved.
2. Press the down arrow or ENTER and the variables
appear. Enter known values for the variables, and
place the cursor on the variable whose value is
sought.
3. Press ALPHA SOLVE (ENTER).
The value of the variable changes to the solution of
the equation.
The rate is 6%.
34
VIII. SOLVING SYSTEMS OF EQUATIONS
TRACE can be used to find the intersection of two graphs.
POINTS OF INTERSECTION OF GRAPHS SOLVING A SYSTEM OF TWO LINEAR
EQUATIONS GRAPHICALLY
To find the points of intersection of two graphs (or to
find the solution of a system of equations graphically).
1. Solve each equation for y and use the “Y=“ key with
y1 and y2 to enter the equations. Graph the equation
with a friendly window.
B
B
B
B
2.(a) If the two lines intersect in one point, the
coordinates give the x- and y-values of the solution.
To find or approximate the intersection, use TRACE
with a friendly window. Pressing the up and down
arrows moves the cursor from one line to the other. If
TRACE gives equal y-values on both lines, this y-value
and the x-value is the solution to the system.
EXAMPLES
Find the solution graphically:
4x + 3y = 11
(a)
2x − 5y = −1
RS
T
1.
e j
y1 = 11 3 − 4 3 x
y2 = 2 5 x + 15
e j
2. Using ZOOM 4 and TRACE gives:
Solution: x = 2, y = 1
(b)
(b) If the two lines are parallel, there is no solution;
the system of equations is inconsistent.
RS4x + 3y = 4
T8x + 6y = 25
If the lines are parallel, then when solving for y the
equations will show that the lines have the same slope
and different y-intercepts.
No solution; inconsistent system
(c)
(c) If the two graphs of the equations give only one
line, every point on the line gives a solution to the
system, and the system is dependent.
RS2x + 3y = 6
T4x + 6y = 12
The two graphs will be the same graph if, when solving
for y to use the graphing calculator, the equations are
identical.
Many solutions; dependent system
35
The graphing techniques discussed previously can also be used to find the intersection of non-linear
curves. There may be more than one solution to systems containing non-linear equations.
Methods other than TRACE can be used to find the intersection of two graphs.
SOLUTION OF SYSTEMS OF EQUATIONS USING THE INTERSECT METHOD
To solve two equations simultaneously using the
intersect method:
EXAMPLE
Solve the system
R| y = 2 x − 3x + 2 .
S| y = x + 2 x + 8
T
2
2
1. The graphs of the functions are:
1. Solve each equation for y and use the Y= key
with y1 and y2 to enter the equations. Graph the
equation with a friendly window.
Use the graphs to determine how many points of
intersection (solutions) there are and approximately
where they are.
B
B
B
B
2. Use 2nd CALC, 5 (INTERSECT) to find each
point of intersection of two curves.
Answer the question "first curve?" with ENTER
and "second curve?" with ENTER.
To the question "guess?" move the cursor close to
the desired point of intersection and press ENTER.
The coordinates of the point of intersection will be
displayed. Repeat to get all points of intersection.
2. Using 2nd CALC, 5 and answering the questions
gives a point of intersection at x = −1, y = 7.
Repeating the process near x = 5 gives the other point
of intersection at x = 6,
y = 56.
3. The coordinates of each point of intersection
give the x- and y-values of the solutions to the
system of equations.
3. The solutions to the system of equations are
x = 6, y = 56 and x = −1, y = 7.
36
SOLUTION OF SYSTEMS OF EQUATIONS –
FINDING OR APPROXIMATING USING
TABLE
To solve a system of equations using a table:
EXAMPLE
Solve the system
R| y = 2 x − 3x + 2 .
S| y = x + 2 x + 8
T
2
2
1. Solve each equation for y and use the Y= key
with y1 and y2 to enter the equations. Graph the
equations. Use the graphs to determine how many
points of intersection (solutions) there are and
approximately where they are.
B
B
B
B
1. The graph
2.
shows two solutions.
2. Use 2nd TABLE SET to build a table that
contains values near the x coordinate of a solution.
3.
3. The x-values resulting in EQUAL y-values are
the x-coordinates of the points of intersection and
the corresponding y-values are the y-coordinates of
the intersection points.
Use up or down arrows to move the table to all
necessary x-values.
The coordinates of the points of intersection are the
The solutions are x = -1, y = 7 and
solutions of the system.
x = 6, y = 56.
4. If specified values of x give y values that are
close to each other, but not equal, changing the
∆Tbl value or changing the Indpnt variable from
Auto to Ask may be useful in finding or
approximating the points of intersection.
4. To find the intersection of y = x + 5 and y = 7 - 2x
with TABLE SETUP set on
Indpnt: Auto, ∆Tbl must be set at 1/3.
TABLE set with Indpnt: Ask could be used to test
values that approach the intersection, until it is found.
The solution is x = 0.6667.
37
IX. SPECIAL FUNCTIONS; QUADRATIC FUNCTIONS
GRAPHS OF SPECIAL FUNCTIONS
EXAMPLES
1. Linear function:
Graph is a line.
1. y = 4x − 3
1(a). y = x
2. (a)i. y = 0.5 x 2
ii. y = 0.25x 3
y = ax + b
1. (a) Identity function: y = x
A special linear function.
2. Power Functions
y = ax b
(a) Power Functions with b > 1
i. When b is even, the graph is similar to the graph
of f ( x ) = x 2 .
ii. When b is odd, the graph is similar to the graph
of f ( x ) = x 3 .
The greater the value of n, the flatter the graph is on
the interval [−1, 1].
2. (b) y = 4 x1/ 2 = 4 x
y = 4 x 1/ 3 = 43 x
b
(b) Power Functions y = ax with 0 < b < 1
(Root functions)
3. y = 3x 2 + 2 x − 5
3. Polynomial Functions:
y = 2x3 + x − 3
y = an x n + an −1 x n −1 +.....+ a1 x + a0
All powers of x are positive integers
Highest power even ⇒
odd number of turns
Highest power odd ⇒
even number of turns
4. Rational Functions
Ratio of two polynomials
4(a). Rectangular hyperbola: y =
4. y =
1
x
A special rational function
38
x2
x −1
y=
1
x
TRACE can be used to approximate the vertex of the graph of a quadratic function.
APPROXIMATING THE VERTEX OF A
PARABOLA WITH TRACE
To approximate the vertex of a parabola by using
TRACE:
EXAMPLE
1. Solve the equation for y and enter the equation
under the Y= menu.
1. Enter y1 = x2 + 7x − 8.
2. Set the window with "friendly" values of x and
values of y that are large enough to show the graph
is a parabola, and graph the equation.
2. Using xmin = -9.4 and xmax = 9.4 and ymin = -25 and
ymax = 10 will show the parabola.
3. Press TRACE and use the right and left arrows
to move the cursor to the vertex (turning point) of
the parabola. The x-value that gives the lowest (or
highest) point on the parabola, and the
corresponding y-value, are the coordinates of the
vertex. Changing the window or using ZOOM
may be necessary to get more accurate values.
3. TRACE shows the vertex is between -3.4 and -3.6.
Changing the window by using ZOOM, Zoom In, or by
changing the window to xmin = -9.4 and xmax = 0, we
can trace to the vertex at x = − 3.5,
y = − 20.25.
ANOTHER EXAMPLE
Find the vertex of the graph of − x 2 + 8 x − y = 10 .
Using xmin = -9.4 and xmax = 9.4 and
ymin = -10 and ymax = 10 will show the parabola.
Find the vertex of the graph of
x2 + 7x – y = 8.
B
B
B
B
TRACE shows the vertex to be x = 4, y = 6.
39
The vertex of a parabola is a maximum or minimum, so we use 2nd CALC and Maximum (or Minimum).
FINDING VERTICES OF PARABOLAS
WITH CALC, MAXIMUM OR MINIMUM
To find the vertex of a parabola by using CALC,
minimum:
EXAMPLE
Find the vertex of the graph of
x2 +7x − y = 8.
1. Solve the equation for y and enter the equation
under the Y= menu.
1. Enter y1 = x2 + 7x − 8.
2. Set the window with values of x and values of y
that are large enough to show the graph is a
parabola, and graph the equation.
2. Using a standard window does not show the
vertex of the parabola. Changing ymin to −25 will
show the parabola.
3. If the parabola opens up, enter
2nd CALC, 3 (minimum) to locate the vertex. If
the parabola opens down, enter 2nd CALC, 4
(maximum).
3. The parabola opens up and TRACE shows the
vertex is between -3.4 and
-3.6. Using 2nd CALC, 3 (minimum), and answering
the questions locates the vertex at x = −3.5 y =
−20.25.
B
B
To the question "left bound?" ("lower bound" on
the TI-82) move the cursor close to and to the left
of the vertex and press ENTER
To the question "right bound?" ("upper bound" on
the TI-82) move the cursor close to and to the right
of the vertex and press ENTER.
To the question "guess?" press ENTER. The
coordinates of the vertex will be displayed.
The vertex is at x = −3.5, y = −20.25.
40
X. PIECEWISE-DEFINED FUNCTIONS
GRAPHING PIECEWISE-DEFINED
FUNCTIONS
To graph a piecewise-defined function
f ( x ) if x ≤ a
:
y=
g ( x ) if x > a
RS
T
EXAMPLE
RSx + 7 if x ≤ −5
T− x + 2 if x > −5
Graph y =
1. Enter y1 = (x + 7)/(x ≤ −5)
and
y2 = (−x + 2)/(x> −5)
B
B
1. Under the Y= key, set
y1 = f x / x ≤ a
c b gh b
B
B
g
B
B
where ≤ is found under 2nd TEST 6.
and set
y2 = g x / x > a
B
B
c b gh b
g
where > is found under 2nd TEST 3.
2. Use ZOOM 6 to see the graph of the function.
2. Use an appropriate window, and use ZOOM or
GRAPH to graph the function.
Evaluating a piecewise-defined function at a given
value of x requires that the correct equation
(“piece”) be selected.
To find f(-6), move the cursor on the graph to y1
and use TRACE, VALUE at -6.
To find f(3), move the cursor on the graph to y2
and use TRACE, VALUE at 3.
41
XI. SCATTERPLOTS AND MODELING DATA
SCATTERPLOTS OF DATA
EXAMPLE
To create a scatterplot of data points:
Use AT&T revenues with the number of years past 1980 (x)
and revenue in $billions(y) to model the revenue. The data
points are
(5, 63.1), (6, 69.9), (7, 60.5), (9, 61.1), (10, 62.2), (11, 63.1),
(12, 64.9), (13, 67.2) Create a scatterplot of this data.
1.
1. Press STAT and under EDIT press
1:Edit. Enter the x-values in the column headed L1
and the corresponding y-values in the column
headed L2.
2. Press 2nd STAT PLOT, 1:Plot 1. Highlight ON, 2.
and then highlight the first graph type, Enter
Xlist:L1, Ylist:L2, and pick the point plot mark you
want.
3. Press GRAPH with an appropriate window or
press ZOOM, 9:ZoomStat to plot the data points.
ADDITIONAL EXAMPLE
Once the STAT PLOT has been defined as you
want it, it can be turned on and off in the equation
editor (Y=) by placing the cursor on Plot1 and
pressing ENTER.
Remember to turn the Plot1 off when not plotting
points, as it may interfere with graphing other
equations.
3.
Create a scatterplot for the following data points.
The following table gives the average yearly income of male
householders with children under 18. Put the number of
years past 1960 in column L1 and the corresponding incomes
in L2, and create a scatterplot for the data.
1969
1972 1975 1978 1981
Year
Income $ 33,749 36,323 33,549 37,575 33,337
1984
1987
1990 1993 1996
Year
Income $ 36,002 34,747 33,769 29,320 31,020
42
MODELING DATA
EXAMPLE
To find an equation that models data points:
Use AT&T revenues with the number of years past 1980 (x)
and revenue in $billions(y) to model the revenue. The data
1. Press STAT and under EDIT press
points are
1:Edit.
(5, 63.1), (6, 69.9), (7, 60.5), (9, 61.1), (10, 62.2), (11, 63.1),
Enter the x-values in the column headed L1 and the (12, 64.9), (13, 67.2)
corresponding
1.
y-values in the column headed L2.
2. Press 2nd STAT PLOT, 1:Plot 1. Highlight ON,
and then highlight the first graph Type. Enter Xlist 2.
:L1, Ylist:L2, and pick the point plot mark you
want.
3.
3. Press GRAPH with an appropriate window or
ZOOM, 9:ZoomStat to plot the data points.
4. Observe the point plots to determine what type
function would best model the data.
5. Press STAT, move to CALC, and select the
function type to be used to model the data. Press
the number of this function type. Press ENTER to
obtain the equation form and coefficients of the
variables.
4. The graph looks like a parabola, so use the quadratic
model, with QuadReg.
5.
Changing the mode to 3 decimal places and repeating step 5
simplifies the equation.
6. Press the Y= key and place the cursor on y1.
Press the VARS key and press 5:Statistics, then
move the cursor to EQ and press 1:RegEQ. The
regression equation you have selected will appear
as y1.
7. To see how well the equation models the data,
press GRAPH. If the graph does not fit the points
well, another function may be used to model the
data.
6.
7.
43
XII. MATRICES *
ENTERING DATA INTO MATRICES;
THE IDENTITY MATRIX
EXAMPLE
To enter data into matrices:
Enter the matrix below as [A].
1 2
3
2 −2 1
3 1 −2
LM
MM
N
1.
OP
PP
Q
2.
1. Press the MATRX key [2nd MATRX on the TI83 Plus].
P
P
2. Move the cursor to EDIT. Enter the number of
the matrix into which the data is to be entered.
3. Enter 3’s to set the dimension, and enter the numbers.
3. Enter the dimensions of the matrix,
and enter the value for each entry of the matrix.
Press ENTER after each entry.
4. To perform operations with the matrix or leave
the editor, first press 2nd QUIT.
5. To view the matrix, press MATRX, the number
of the matrix, and ENTER.
5.
The n x n matrix consisting of 1’s on the main
diagonal and 0’s elsewhere is called the identity
matrix of order n and is denoted I n . To display an
identity matrix of order n:
Find the identity matrix of order 2.
1. Press MATRX, move to MATH, enter
5:identity(, and the order of the identity matrix
desired.
Find the identity matrix of order 3.
2. An identity matrix can also be created by
entering the numbers directly with MATRX, EDIT.
* Note that on the TI-83 Plus and TI-84 Plus, matrices are accessed by pressing 2nd MATRX.
P
44
P
OPERATIONS WITH MATRICES
EXAMPLES
To find the sum of two matrices,
[A] and [D]:
Find the sum
1 2
3
7 −3 2
2 −2 1 + 4 −5 3 .
3 1 −2
0 2 1
LM
MM
N
OP
PP
Q
LM
MM
N
OP
PP
Q
1.
1. Enter the values of the elements of
[A] using MATRX and EDIT.
Press 2nd QUIT.
Enter the values of the elements of
[D] using MATRX and EDIT.
Press 2nd QUIT.
2.
2. Use MATRX and NAME, to enter
[A] + [D], and press ENTER.
3. If the matrices have the same dimensions, they
can be added (or subtracted). If they do not have
the same dimensions, an error message will occur.
To find the difference of two matrices, [A] and [D]: Find the difference.
1 2
3
7 −3 2
2 −2 1 − 4 −5 3
4. If the matrices are not already entered, enter the
3 1 −2
0 2 1
values of the elements of [A] and [D] using
4.
MATRX and EDIT. After entering the elements
forming each matrix, press 2nd QUIT.
B
LM
MM
N
OP
PP
Q
LM
MM
N
OP
PP
Q
B
5. Use MATRIX and NAME, then enter [A] − [D]
and press ENTER.
5.
6. We can multiply a matrix [D] by a real number
(scalar) k by pressing
k [D]. (Or k*[D].)
6. Multiply the matrix [D] by 5.
45
MULTIPLYING TWO MATRICES
EXAMPLES
To find the product of matrices, [C][A]:
Compute the product
1 2
3
1 2 4
2 −2 1 .
−3 2 −1
3 1 −2
LM
N
OP LM
Q MMN
OP
PP
Q
1.
1. Press MATRX, move to EDIT,
enter 1: [A], enter the dimensions of [A], and enter
the elements of [A].
Press 2nd QUIT.
2.
2. Enter the elements in matrix [C].
Press 2nd QUIT.
3. Press MATRX, 3 [C], *, MATRX [A], and
ENTER. (Or press MATRX [C], MATRX [A],
and ENTER.)
4. Note that [A][C] does not always equal [C][A].
The product [A][C] may be the same as [C][A],
may be different from [C][A], or may not exist.
5. The product of a matrix [A] and the identity
matrix of the appropriate order is the matrix [A],
that is,
[A] [I] = [I] [A] = [A].
3.
4. [A][C] cannot be computed because their
dimensions do not match.
5. Show that the product of matrix
[A] and I 3 is matrix [A].
46
FINDING THE INVERSE OF A MATRIX
To find the inverse of a matrix:
EXAMPLES
LM 2
Find the inverse of E = M−1
MN 4
1.
1. Enter the elements of the matrix using MATRX
and EDIT.
Press 2nd QUIT.
2.
2. Press MATRX, the number of the matrix, and
ENTER, then press the x-1 key and ENTER.
3. To see the entries as fractions, press MATH and 3.
press 1:˛Frac, and press ENTER.
4. The product of a matrix and its inverse is the
identity matrix with the same dimension.
4.
5. Not all matrices have inverses. Matrices that do
not have inverses are called singular matrices.
5.
47
OP
P
0PQ
0 2
0 1 .
2
DETERMINANT OF A MATRIX;
TRANSPOSE OF A MATRIX
To find the determinant of a matrix [A]:
EXAMPLES
Find the determinant of the matrix A given below.
2 3 0 1 3
1 0 2 3 1
A= 0 1 0 2 3
1 0 2 2 1
1 0 0 0 3
LM
MM
MM
MN
OP
PP
PP
PQ
1.
1. Press MATRX, move to EDIT,
enter 1: [A], enter the dimensions of [A], and enter
the elements of [A]. Press 2nd QUIT when all
elements are entered.
To view matrix A, Press MATRX, 1:[A}.
2. Press MATRX, move to MATH, enter 1: det(,
press MATRX, 1:[A], and press ENTER.
This gives det([A]).
2.
3. If det([B]) = 0, the matrix is singular (it has no
inverse).
To find the transpose of a matrix [A]:
Find the transpose of the matrix A above.
1. Press MATRX, move to EDIT,
enter 1: [A], enter the dimensions of [A], and enter
the elements of [A]. Press 2nd QUIT when all
elements are entered.
To view matrix A, Press MATRX, 1:[A}.
1.
2. Press MATRX, press 1:[A], press MATRX,
move to MATH, enter 2: T , and press ENTER.
This gives transpose ([A]).
2.
48
SOLVING SYSTEMS OF LINEAR EQUATIONS EXAMPLES
WITH UNIQUE SOLUTIONS
To solve a system of equations that has a unique
x + 2 y + 3z = 0
solution:
Solve the system 2 x − 2 y + z = 7 .
3 x + y − 2 z = −1
1. Write each equation with the constant on one
R|
S|
T
side and the variables aligned on the other side.
Enter the coefficients of the variables into matrix
[A], with the coefficients of the x-variables as
column 1, the coefficients of the y-variables as
column 2, and the coefficients of the z-variables as
column 3. This is called the coefficient matrix.
1.
2.
2. Enter the constants into a second matrix [B].
3.
3. Multiply the inverse of the coefficient matrix
times the matrix of constants. The product is the
solution to the system.
4. If the solution to the system is not unique or
does not exist, an error statement will occur when
using this method.
The solution is x =1, y = -2, z = 1.
4. The system below has no solution.
x + 2 y + 3z = 0
2x − 2 y − z = 7
3 x + 2 z = −1
R|
S|
T
5. A system does not have a unique solution if the
inverse of the coefficient matrix does not exist. (Or
equivalently, the determinant of the matrix is 0.)
5.
49
SOLUTION OF SYSTEMS OF 3 LINEAR
EQUATIONS IN 3 VARIABLES
To solve a system of three equations in three
variables:
EXAMPLE
Solve the system:
R|2 x − 2 y = 6
S|x + 2 y + 3z = 9
T3x + 3z = 15
1. Create an augmented matrix with the coefficient 1.
matrix augmented by the constants.
2. Perform operations that make a 1 in row 1,
2. Interchange row 1 and row 2 of matrix A, using
column 1. Operations used to do this include
MATRX, MATH, C:rowswap([A], 1,2)
interchanging rows with MATRX, MATH,
C:rowswap(, with the first entry the matrix and the
next elements the rows to be interchanged and/or
multiplying row 1 with MATRX, MATH, E:*row(,
with the elements value, matrix, and row.
3. Use row 1 only to get zeros in the other entries
of column 1. The operation used is frequently
MATRX, MATH, F: *row+(, with elements value,
matrix, first row, and second row. Use 2nd ANS to
enter the required matrix.
3. Multiply row 1 of the matrix by -2, add it to row 2,
and place the sum into row 2, using
MATRX, MATH, F: *row+(-2,ANS,1,2). Repeat this
step with -3 and row 3.
4. Perform operations that make a 1 in row 2,
column 2. Use MATRX, MATH, E:*row(, with
the elements value, matrix, and row.
4. Use MATRX,MATH E:*row(-1/6,ANS,2)
to get a 1 in row 2, column 2.
5. Use row 2 only to get zeros as the other entries
in column 2.
6. If the bottom row contains all zeros except for
the entry in row 3,
column 4, there is no solution.
5. Use MATRIX,MATH, F: *row+
7. If the bottom row contains all zeros, the system
has many solutions.
The values for the first two variables are found as
functions of the third.
8. If there is a nonzero element in row 3, use row 3 7. The bottom row contains all zeros, so there are
to solve all equations by substitution.
many solutions.
From row 1, x + z = 5 or x = 5 – z and from row 2,
y + z = 2 or y = 2 – z for any z.
50
SOLUTION OF SYSTEMS - REDUCED
ECHELON FORM ON THE TI-83, TI-84, TI-83
PLUS, or TI-84 PLUS
To solve a system of three equations in three
variables by using rref under the MATRX MATH
menu:
1. Create an augmented matrix [A] with the
coefficient matrix augmented by the constants.
2. Use the MATRX menu to produce a reduced
row echelon form of Matrix A, as follows:
a. Press MATRX, move to the right to MATH
EXAMPLE
R|2 x − y + z = 6
Solve the system: S x + 2 y − 3z = 9
|T3x − 3z = 15
1.
2.
b. Scroll down to B: rref(, and press ENTER, or
press ALPHA B.
Press MATRX, 1:[A] to get
rref([A]). Press ENTER.
This gives the reduced echelon form.
3. If each row in the coefficient matrix (first 3
columns) contains a 1 with the other elements 0’s,
the solution is unique and the number in column 4
of a row is the value of the variable corresponding
to a 1 in that row.
4. If the bottom row contains all zeros, the system
has many solutions.
The values for the first two variables are found as
functions of the third.
3. The solution is unique.
x = 4, y = 1, and z = −1
5. If there is a nonzero element in the augment of
row 3 and zeros elsewhere in row 3, there is no
solution to the system.
51
SOLUTION OF SYSTEMS OF LINEAR
EQUATIONS: NON-UNIQUE SOLUTIONS
To solve a system of three equations in three
variables on the TI-83, TI-84, TI-83 Plus, or TI-84
Plus:
1. Create an augmented matrix [A] with the
coefficient matrix augmented by the constants.
2. Use the MATRX menu to produce a reduced
row echelon form of Matrix A, as follows:
a. Press MATRX, move to the right to MATH.
R|2 x − 2 y = 6
Solve the system: S x + 2 y + 3z = 9
|T3x + 3z = 15
1.
2.
b. Scroll down to B: rref(, and press ENTER, or
press ALPHA B.
Press MATRX, 1:[A] to get
rref([A]). Press ENTER.
This gives the reduced echelon form.
3. If each row in the coefficient matrix (first 3
columns) contains a 1 with the other elements 0’s,
the solution is unique and the number in column 4
of a row is the value of the variable corresponding
to a 1 in that row.
3. The bottom row does not contain a 1.
4. If the bottom row contains all zeros, the system
has many solutions.
The values for the first two variables are found as
functions of the third.
5. If there is a nonzero element in the augment of
row 3 and zeros elsewhere in row 3, there is no
solution to the system.
4. The bottom row contains all zeros, so there are many
solutions.
From row 1, x + z = 5 or x = 5 – z and from row 2, y
+ z = 2 or y = 2 – z,
for any z.
52
XIII. SOLVING INEQUALITIES
SOLVING LINEAR INEQUALITIES
EXAMPLE
To solve a linear inequality:
Solve 3x > 6 + 5x
1. Rewrite the inequality with 0 on the right side
and simplify.
1. 3x −5x −6 > 0
−2x −6 > 0
2. Under the Y= menu, assign the left side of the
inequality to y1 , so that
y1 = f(x), where f(x) is the left side.
2. Set y1 = −2x −6.
3. Graph this equation. Set the window so that the
point where the graph crosses the x-axis is visible.
Note that the graph will cross the axis in at most
one point because the graph is of degree 1. (Using
ZOOM OUT can help find this point.)
3. Using ZOOM 4, the graph is:
4. Use the ZERO command under the CALC menu
to find the x-value where the graph crosses the xaxis. This value can also be found by finding the
solution to 0 = f(x) algebraically.
4. The x-intercept is x = -3.
B
B
5. Observe the inequality in Step 1. If the
inequality is "<", the solution to the original
inequality is the interval (bounded by the xintercept) where the graph is below the x axis.
If the inequality is ">", the solution to the original
inequality is the interval (bounded by the xintercept) where the graph is above the x axis.
5. The inequality is ">", so the solution to the original
inequality is the interval (bounded by the x-intercept)
where the graph is above the x axis.
Thus the solution is x < -3.
6.
6. The region above the x-axis and under the graph
can be shaded with 2nd DRAW, 7 (Shade), and
entering Shade(0, y1) on the homescreen.
B
B
Using 2nd DRAW, 7 (Shade),and entering Shade(0,
y1 ) will shade the region above the x-axis and below
y1 . The x-interval where the shading occurs is the
solution.
53
SOLVING SYSTEMS OF LINEAR
INEQUALITIES
To solve a system of inequalities in two variables
(approximately):
EXAMPLE
1. Under the Y= menu, assign the right side of the
first inequality to y1 and the right side of the
second inequality to y2 . Graph the equations using
a friendly window that contains the points of
intersection of the graphs.
1.
R| y >|2 x − 1|
Solve S
|T y < 13 x + 2
2. Visually determine the interval over which the
graph of y2 is above y1 by using 2nd DRAW,
7 (Shade),and entering Shade( y1 , y2 ).
Using ZOOM OUT can be used to search for all
points of intersection.
for x.
Using ZOOM 4 gives:
2. We seek values of x above
B
B
y1 = |2 x − 1| and below y2 =
3. Use 2nd CALC, 5 (intersect) to find one point of
intersection of the two curves.
Answer the question "first curve?" with ENTER
and "second curve?" with ENTER.
To the question "guess?" move the cursor close to
the desired point of intersection and press ENTER.
The coordinates of the point of intersection will be
displayed.
1
x+2.
3
3.
4. Repeat to get all points of intersection.
4.
5. The solution to the system of inequalities will be
the interval determined by the values of x from the
points of intersection found in Steps 3 and 4.
5. The solutions to the equation are
x =-.429 (approximately) and x = 1.8
The solution is -.429 < x < 1.8 (approx.)
54
SOLVING QUADRATIC INEQUALITIES
EXAMPLE
To solve a quadratic inequality:
Solve x2 − 5x < 6
1. Rewrite the inequality with 0 on the right side.
1. x2 − 5x − 6 < 0
2. Under the Y= menu, assign the left side of the
inequality to y1 , so that
y1 = f(x), where f(x) is the left side.
2.
3. Graph this equation. Set the window so that all 3. Using ZOOM 6, the graph is:
points where the graph crosses the x-axis are
visible. Note that the graph will cross the axis in at
most two points because the equation is of degree
2. (Using ZOOM OUT can help find these points.)
4. Use the ZERO command under the CALC menu 4. The x-intercepts are x = -1 and x = 6.
to find the x-values (one at a time) where the graph
crosses the x-axis. These values can also be found
by finding the solution to
0 = f(x) algebraically.
5. Observe the inequality in Step 1. If the
inequality is "<", the solution to the original
inequality is the interval (bounded by the xintercepts) or union of intervals where the graph is
below the x-axis.
If the inequality is ">", the solution to the original
inequality is the interval (bounded by the xintercepts) or union of intervals where the graph is
above the x-axis.
5. The graph is below the x-axis between x = -1 and x
= 6. Thus the inequality has solution -1 < x < 6.
The region below the x-axis can be shaded with 2nd
DRAW, 7 (Shade),and entering Shade(y1, 0) on the
homescreen (with ymin = -15).
B
B
55
SOLVING QUADRATIC INEQUALITIES ALTERNATE METHOD
To solve a quadratic inequality:
EXAMPLES
Solve x 2 + 3x ≥ 10
1. Rewrite the inequality with 0 on the right side.
1. x 2 + 3x − 10 ≥ 0
2. Under the Y= menu, assign the left side of the
inequality to y1, so that
y1 = f(x), where f(x) is the left side.
2.
B
B
B
B
3. Graph this equation. Set the window so that all
points where the graph crosses the x-axis are
visible. Note that the graph will cross the x-axis in
at most two points because the equation is of degree
2. (Using ZOOM, Zoom Out can help find these
points.)
3. Using ZOOM 6, the graph is
4. Use the ZERO command under the CALC menu
to find the x-values (one at a time) where the graph
crosses the x-axis. These values can also be found
by finding the solution to
0 = f(x) algebraically.
4. The x-intercepts are x = −5 and x = 2.
5. Under the Y= menu, beside y2 =, enter the
inequality, with any side containing more than one
term enclosed in parentheses. “Turn off” y1 and
graph y2. The graph displayed resembles a “number
line” solution to the inequality, with the x-intercepts
as bounds. The zeros found in Step 4 are the
boundaries of the inequality.
5.
B
B
B
B
The solution to the inequality x 2 + 3x ≥ 10 is x <
−5 or x > 2.
56
XIV. LINEAR PROGRAMMING
GRAPHICAL SOLUTION OF LINEAR
PROGRAMMING PROBLEMS
To solve a linear programming problem involving
two constraints graphically:
EXAMPLE
Find the region defined by the inequalities
5x + 2 y ≤ 54
2 x + 4 y ≤ 60
x ≥ 0, y ≥ 0
1. y = 27 − 5x/2
y = 15 − x/2
2.
1. Write the inequalities as equations, solved for y.
2. Graph the equations. The inequalities
x ≥ 0, y ≥ 0 limit the graph to Quadrant I, so
choose a window with xmin = 0 and ymin = 0.
3.
3. Use TRACE or INTERSECT to find the corners
of the region, where the borders intersect.
The corners of the region
determined by the inequalities are (0, 15), (6, 12),
and (10.8, 0).
4.
4. Use SHADE to shade the region determined by
the inequalities. Shade under the border from x = 0
to a corner and shade under the second border from
the corner to the x-intercept.
5. Evaluating the objective function at the
coordinates of each of the corners determines where
the objective function is maximized or minimized.
5. At (0, 15), f = 165
At (6,12), f = 162
At (10.8, 0), f = 54
The maximum value of f is 165 at x = 0, y = 15.
57
XV. EXPONENTIAL AND LOGARITHMIC FUNCTIONS
GRAPHS OF EXPONENTIAL AND
LOGARITHMIC FUNCTIONS
To graph the exponential function
EXAMPLES
Graph y = 3 x .
y=a :
x
Press Y= and enter a^x. Press GRAPH with an
appropriate window.
To graph the exponential function y = e g ( x ) :
Graph y = e.5x .
Press Y= and press 2nd e x , then enter the exponent
g(x) to get e^(g(x)). Press GRAPH with an
appropriate window.
To graph the logarithmic function
y = log x = log10 x :
Graph y = log x.
Press Y= and press LOG, then enter x), getting
log(x). Press GRAPH with an appropriate window.
To graph the logarithmic function
y = ln x = log e x :
Graph y = ln x.
Press Y= and press 2nd LN, then enter x), getting
ln(x). Press GRAPH with an appropriate window.
To graph logarithmic functions to other bases, use
the change of base formula to convert the function
to base 10.
log x
logb x =
log b
Graph y = log5 x =
58
log x
.
log 5
INVERSE FUNCTIONS
EXAMPLES
To find the inverse of a function f(x):
Find the inverse of f(x) = x 3 − 3 and graph f(x) and
its inverse on the same set of axes to show that they
are inverses.
1. Write y = f(x).
1. y = x 3 − 3
2. Interchange x and y in the equation, and solve
the new equation for y. The new equation gives y
as the inverse of the original function f(x).
2. Interchanging x and y and solving for y:
x = y3 − 3 ⇒ x + 3 = y3 ⇒ y = 3 x + 3
3.
3. Under the Y= menu, enter f(x) as y1 and the
inverse function as y2 , and press ENTER with the
cursor to the left of y2 (to make the graph dark).
Press GRAPH with an appropriate window.
4. To show that the graphs are symmetrical about
the line y = x, enter y3 = x under the Y= menu and
graph using ZOOM,5: Zsquare.
4.
log x
and
log 5
5. To show that a logarithmic function and an
exponential function are inverse functions if they
have the same base, graph them on the same set of
axes, along with y = x. Use a square window.
5. Show that y = log5 x =
To graph a function f(x) and its inverse of a
function f(x):
1. Under the Y= menu, enter f(x) as y1 . Choose a
square window.
Graph the function f(x) = x 3 − 3 and its inverse on
the same set of axes.
1. Enter y1 = x^3 − 3
y = 5 x are inverse functions.
2.
2. Press 2nd DRAW, 8:DrawInv, press VARS,
move to Y-VARS, highlight Y-VARS and press
ENTER three times. The graph of f(x) and its
inverse will be displayed.
3. To clear the graph of the inverse, press 2nd
DRAW, 1:ClrDraw.
59
EXPONENTIAL REGRESSION
EXAMPLES
If a scatterplot for data appears to have an
exponential shape, the equation that models the data
can be created with STAT.
The following data gives the weekly sales for each of
10 weeks after the end of an advertising campaign,
with x representing the number of weeks and y
representing sales in thousands of dollars. Find the
equation that models this data.
x | 1 2 3 4 5 6 7 8 9 10
y | 43 38 33 29 25 22 19 15 14 12
To find an equation that models data points:
1.
1. Press STAT and under EDIT press
1:Edit. Enter the x-values in the column headed L1
and the corresponding y-values in the column
headed L2.
2. Press 2nd STAT PLOT, 1:Plot 1. Highlight ON,
and then highlight the first graph Type. Enter
Xlist:L1, Ylist:L2, and pick the point plot mark you
want.
2.
3.
3. Press GRAPH with an appropriate window or
ZOOM, 9:ZoomStat to plot the data points.
4. Observe the point plots to determine what type
function would best model the data.
4. The shape of the scatterplot appears to be
exponential decay.
5. Press STAT, move to CALC, and select the
function type to be used to model the data. Press
the number of this function type. Press ENTER to
obtain the equation form and coefficients of the
variables.
5. Use STAT, CALC, 0:ExpReg to get the model.
6. Press the Y= key and place the cursor on y1.
Press the VARS key and press 5:Statistics, then
move the cursor to EQ and press 1:RegEQ. The
regression equation you have selected will appear as
y1.
6.
7. To see how well the equation models the data,
press GRAPH. If the graph does not fit the points
well, another function may be used to model the
data.
60
7.
ALTERNATE FORMS OF EXPONENTIAL
FUNCTIONS
An exponential decay function is frequently written
with the base greater than 1 and with a negative
exponent. To convert an exponential equation
whose base is less than 1 to one whose base is
greater than 1:
EXAMPLES
Convert the equation y = 50.8424(0.8656) x
to an equation with a base greater than 1 and with a
negative exponent.
1. Enter the base, press the x −1 key, and press
enter.
1. Find the reciprocal of the base and change the
sign of the exponent, to get a base greater than 1.
2. Write the equation with the same coefficient, the
new base, and the negative of the original exponent.
2. The new form of the equation is
50.8424(11553
.
)−x
3.
3. To verify that the new form is equivalent to the
original, graph the two equations and note that the
second graph lies on top of the first.
To convert an exponential equation that does not
have base e to an exponential equation with base e:
Convert the original equation to an exponential
equation with base e:
1. Take the logarithm, base e, of the base.
1.
2. The new form of the equation uses this number
times the original exponent as the new exponent,
has the original coefficient, and has
base e.
2. The new exponent is -.1443 times x.
3. To verify that the new form is equivalent, graph
both equations on the same set of axes.
3.
The function is y = 50.8424e −0.1443x .
61
LOGARITHMIC REGRESSION
EXAMPLES
If a scatterplot for data appears to have a
logarithmic shape, the equation that models the data
can be created with STAT.
The following data gives the millions of hectares (y)
destroyed in selected years (x) from 1950. Find the
equation that models this data.
To find an equation that models data points:
x | 10
y | 2.21
1. Press STAT and under EDIT press
1:Edit. Enter the x-values in the column headed L1
and the corresponding y-values in the column
headed L2.
20
30
3.79 4.92
38
5.77
1.
2. Press 2nd STAT PLOT, 1:Plot 1. Highlight ON,
and then highlight the first graph Type. Enter
Xlist:L1, Ylist:L2, and pick the point plot mark you
want.
2.
3. Press GRAPH with an appropriate window or
ZOOM, 9:ZoomStat to plot the data points.
4. Observe the point plots to determine what type
function would best model the data.
3.
5. Press STAT, move to CALC, and select the
function type to be used to model the data. Press
the number of this function type. Press ENTER to
obtain the equation form and coefficients of the
variables.
4. The shape of the scatterplot appears to be
logarithmic.
5. Use STAT, CALC, 9:LnReg to get the model.
6. Press the Y= key and place the cursor on y1.
Press the VARS key and press 5:Statistics, then
move the cursor to EQ and press 1:RegEQ. The
regression equation you have selected will appear as
y1.
7. To see how well the equation models the data,
press GRAPH. If the graph does not fit the points
well, another function may be used to model the
data.
6.
62
7.
XVI. SEQUENCES
EVALUATING A SEQUENCE
EXAMPLES
To evaluate a sequence for different values of n:
Evaluate the sequence with nth term
n2 + 1 at n = 1, 3, 5, and 9.
1. Press MODE and highlight Seq. Press ENTER
and 2nd QUIT.
1.
2. Store the formula for the sequence (in quotes) in
u, using
“formula” STO u.
Press the X,T, θ ,n key to get n for the formula, and
2nd 7 to get u.
(On the TI-82, press 2nd n to get n for the formula,
and press 2nd VARS, 4: Sequence, 1:un to get un.)
2.
3. Write u({a,b,c,..}) to evaluate the sequence at a,
b, c, ..., and press ENTER. (On the TI-82, enter
un({a,b,c,..}) and press ENTER.
3.
4. To generate a sequence after the formula is
defined, enter
u(nstart, nstop, step), and press ENTER. (On the
TI-82, use
un(nstart, nstop, step).)
4. For this sequence, evaluate every third term
beginning with the second term and ending with the
eleventh term.
63
ARITHMETIC SEQUENCES
EXAMPLES
nth TERMS AND SUMS
To find the nth term of an arithmetic sequence with Find the 12th term of the arithmetic sequence with
first term a and common difference d:
first term 10 and common difference 5.
1.
1. Press MODE and highlight Seq. Press ENTER
and press 2nd QUIT.
2. Press Y=. At u(n) =, enter the formula for the
nth term of an arithmetic sequence, using the
x,T, θ ,n key to enter n. (Use 2nd n on the TI-82.)
The formula is
a + (n - 1) *d,
where a is the first term and d is the common
difference.
On the TI-82, u(n) is denoted un(n). (Press 2nd
VARS, 4: Sequence, 1:un.)
3. Press 2nd QUIT. To find the nth term of the
sequence, press 2nd u (above 7) followed by the
value of n, in parentheses, to get u(n), then press
ENTER (On the TI-82, use un(n).)
4. Additional terms can be found in the same
manner.
2. Substitute 10 for a and 5 for d.
3. The 12th term.
4. The 8th term.
To find the sum of the first n terms of an arithmetic
sequence:
1. Press MODE and highlight Seq,. Press ENTER
and press 2nd QUIT.
2. Press Y=. At v(n) =, enter the formula for the
sum of the first n terms of an arithmetic sequence,
using the x,T, θ ,n key to enter n. (Use 2nd n on
the TI-82.) The formula is
(n / 2)(a + (a + (n − 1)d )) , where a is the first term
and d is the common difference.
On the TI-82, v(n) is denoted vn(n).
3. Press 2nd QUIT. To find the sum of the first n
terms of the sequence, press 2nd v (above 8)
followed by the value of n, in parentheses, to get
v(n), then press ENTER.
On the TI-82, find vn(n).
4. Other sums can be found in the same manner.
Find the sum of the first 12 terms of the arithmetic
sequence with first term 10 and common difference 5.
1.
2.
3. The sum of the
first 12 terms.
64
4. The sum of the
first 8 terms.
GEOMETRIC SEQUENCES
nth TERMS AND SUMS
To find the nth term of a geometric sequence with
first term a and common ratio r:
1. Press MODE and highlight Seq. Press ENTER
and press 2nd QUIT.
EXAMPLE
Find the 8th term of the geometric sequence with first
term 40 and common ratio 1/2.
1.
2. Press Y=. At u(n) =, enter the formula for the
nth term of a geometric sequence, using the
x,T, θ ,n key to enter n. The formula is
ar n−1 , where a is the first term and r is the common
ratio.
On the TI-82, u(n) is denoted un(n).
2. Substitute 40 for a and (1/2) for r.
3. Press 2nd QUIT. To find the nth term of the
sequence, press 2nd u followed by the value of n, in
parentheses, to get u(n), then press ENTER (On the
TI-82, use un(n).)
To get a fractional answer, press MATH, 1˛: Frac.
4. Additional terms can be found in the same
manner.
3. The 8th term is:
4. The 12 term is:
To find the sum of the first n terms of a geometric
sequence:
Find the sum of the first 12 terms of the geometric
sequence with first term 40 and common ratio 1/2.
1.
1. Press MODE and highlight Seq. Press ENTER
and press 2nd QUIT.
2. Press Y=. At v(n) =, enter the formula for the
sum of the first n terms of a geometric sequence,
using the x,T, θ ,n key to enter n. The formula is
a (1 − r n ) / (1 − r ) , where a is the first term and r is
the common ratio.
On the TI-82, v(n) is denoted vn(n).
2. Substitute 40 for a and (1/2) for r.
3. Press 2nd QUIT. To find the sum of the first n
terms of the sequence, press 2nd v (above 8)
followed by the value of n, in parentheses, to get
v(n), then press ENTER.
On the TI-82, find vn(n).
To get a fractional answer, press MATH, 1˛: Frac.
3. The sum of the
first 12 terms
4. Other sums can be found in the same manner.
65
4. The sum of the
first 8 terms
XVII. MATHEMATICS OF FINANCE
FUTURE VALUE OF AN INVESTMENT
EXAMPLES
The future values of investments can be found for
different rates, times, and compounding periods by
entering the formula for S as y1 in the equation
editor, using STO (store) to enter different values
for the other variables, and then evaluating y1.
B
B
B
B
1. The future value of an investment of $P invested
for t years at a nominal interest rate, r, compounded
m times per year, can be found with the formula
FG
H
S = P 1+
r
m
IJ
K
1. To find the future value of $1000 invested at 8%,
compounded annually and compounded monthly for
10 years, enter the formula as y1, store the other
values, and evaluate y1. (Use 2nd ENTER to re-enter
values for the variables.)
mt
.
B
B
B
B
2. To find the future value of investments for
different numbers of years, enter the given values in
the formula of step 1, store the formulas as y1, and
read the values in TABLE.
2. To find the future value of $1000 invested at 8%,
compounded monthly for 10, 20, and 30 years,
respectively, enter
3. The future value of an investment of $P invested
for t years at a nominal interest rate, r, compounded
continuously, can be found with the formula
3. To find the future value of $1000 invested at 8%,
compounded continuously for 10, 20, and 30 years,
enter
B
FG
H
B
IJ 12 x
K
.08
12
and use TABLE with x = 10, 20, and 30.
y1 = 1000 1 +
S = Pe rt
y1 = 1000e.08 x
and use TABLE with x = 10, 20, and 30.
66
Finance Formulas and TABLE can be used to find future values of annuities for several values of another
variable, such as years.
FUTURE VALUES OF ANNUITIES AND
PAYMENTS INTO SINKING FUNDS
1. Ordinary Annuity: If $R is deposited at the end
of each period for n periods in an annuity that earns
interest at a rate of i per period, the future value of
the annuity is
S=R
EXAMPLES
1. To find the future value after 10, 20, and 30
years, respectively, of an ordinary annuity with $500
deposited at the end of each quarter at interest rate
8%, compounded quarterly, enter
LM (1 + i) − 1OP
MN i PQ
n
y1 = 500
LM (1+.02)
MN .02
4x
−1
OP
PQ
and use TABLE with x = 10, 20, and 30.
2. Annuity Due: If $R is deposited at the
beginning of each period for n periods in an annuity
that earns interest at a rate of i per period, the future
value of this annuity due is
Sdue = R
2. To find the future value after 10, 20, and 30 years,
respectively, of an ordinary annuity with $500
deposited at the beginning of each quarter at interest
rate 8%, compounded quarterly, enter
LM (1 + i) − 1OP(1 + i)
MN i PQ
n
y1 = 500
LM (1+.02)
MN .02
4x
−1
OP(1+.02)
PQ
and use TABLE with x = 10, 20, and 30.
3. Sinking Fund: If periodic payments are
deposited at the end of each of n periods into an
ordinary annuity (or sinking fund) earning interest
at a rate of i per period, such that at the end of n
periods its value
is $S, then the size of each required payment R is
L i
R = SM
MN (1 + i)
n
3. To find the size of deposits made at the end of
each month to accumulate money to discharge a
debt of $100,000 due in 10, 20, and 30 years,
respectively, with interest at 6%, compounded
monthly, enter
OP .
− 1 PQ
y1 = 100,000
LM .005
MN (1+.005)
12 x
OP
− 1 PQ
and use TABLE with x = 10, 20, and 30.
67
PRESENT VALUE FORMULAS
EVALUATING WITH TABLE
To find the present value of each of the following
types of investments for different numbers of years,
enter the given values in the formula, store the
formulas in y1, and read the values in TABLE.
B
EXAMPLES
1. The present value of an annuity paying $1000 at the end
of each month with interest at 12%, compounded monthly,
is
B
y1 = 1000
1. If a payment of $R is made at the end of each
period for n periods, into (or out of) an annuity that
earns interest at a rate of i per period, the present
value of the annuity is
An = R
−12 x
OP
PQ
where x is the number of years. To find the present value
for annuities lasting 10, 20, and 30 years, respectively,
enter and use TABLE with x = 10, 20, and 30.
LM1 − (1 + i) OP
MN i PQ
−n
2. If a payment of $R is to be made at the
beginning of each period for n periods from an
account that earns interest rate i per period, the
present value of this annuity due is
Adue = R
LM1 − (1+.01)
MN .01
2. If a payment of $1000 is to be made at the beginning of
each period for x years, from an account that earns interest
at rate 12%, compounded monthly, the present value of this
annuity due is found by entering
LM1 − (1 + i) OP(i + i)
MN i PQ
−n
y1 = 1000
LM1 − (1+.01)
MN .01
−12 x
OPb1+.01g .
PQ
The present value of annuities lasting 10, 20, and 30 years,
respectively and using TABLE WITH x = 10, 20, and 30.
3. The present value of a deferred annuity of $R
per period for n periods deferred for k periods with
interest rate i per period is given by
A( n ,k ) = R
LM1 − (1 + i) OP(1 + i)
MN i PQ
−n
3. The present value of a deferred annuity of $1000 per
month for x years, with interest rate 12%, compounded
monthly, after being deferred for 5 years, is
−k
y1 = 1000
LM1 − (1+.01)
MN .01
−12 x
OP(1+.01)
PQ
−5⋅12
To find the present value of annuties lasting 10, 20, and 30
years, respectively and use TABLE with x = 10, 20, and 30.
68
The SOLVER feature under the MATH menu makes it possible to solve the finance formulas for any
variable if the values of the other variables are entered.
SOLVER AND FINANCE FORMULAS
EXAMPLES
ON THE TI-83, TI-84, TI-83 PLUS, or TI-84
PLUS
To solve a finance formula for one of the variables: 1. Find the future value after 10 years, of an
ordinary annuity with $500 deposited at the end of
each quarter at interest rate 8%, compounded
quarterly.
1. Rewrite S = R
1. Rewrite the equation with 0 on one side.
LM (1 + i) − 1OP in a form with 0 on
MN i PQ
n
one side.
2. Press MATH 0 (Solver).
Press the up arrow revealing EQUATION SOLVER
eq: 0 =,
and enter the nonzero side of the equation to be
solved.
2. Press MATH 0 (Solver).
Press the up arrow revealing EQUATION
SOLVER and enter the equation.
3. Press the down arrow or ENTER and the
variables appear. Enter known values for the
variables, place the cursor on the variable whose
value is sought, and press ALPHA SOLVE
(ENTER). The value of the variable changes to the
solution of the equation.
3. Press the down arrow or ENTER and enter the
given values, place the cursor on S and press
ALPHA SOLVER.
The future value is $30,200.99.
4. To solve additional problems with this formula,
enter the values of the given variables, place the
cursor on the variable sought, and press ALPHA
SOLVE (ENTER).
4. To find the size of deposits made at the end of
each month to accumulate money to discharge a
debt of $100,000 due in 10 years, with interest at
6%, compounded monthly, use the same formula,
enter the values of the variables, and solve for R.
The required deposit is $610.21.
5. Other finance problems can be solved by
entering the appropriate formulas and using
SOLVER.
69
The FINANCE key on the TI-83, TI-84, TI-83 Plus, or TI-84 Plus can be used to solve many different
types of finance problems, including future and present values of annuities and loan payments. The
Finance Applications are found under APPS on the TI-83 Plus and TI-84 Plus.
ANNUITIES AND LOANS
THE FINANCE KEY ON THE TI-83, TI-84, TI-83
PLUS, AND TI-84 PLUS
ANNUITIES
EXAMPLES
Find the future value of an annuity with a payment
of $300 at the end of each of 36 months, with
interest at 8%, compounded monthly.
1.
To find the future value of an annuity:
1. Press 2nd FINANCE and from the CALC menu
choose 1: TVM Solver.
2. Enter N = 36, I% = 8, PV = 0, Payment PMT = 300 (negative because it is leaving you), and P/Y
and C/Y = 12 for compounding monthly.
2. Enter the number of periods, N, the interest
percent, I%, present value PV, Payment PMT, and
the number of compounding periods per year, C/Y.
3.
3. Place the cursor on FV, the future value, and
press ALPHA SOLVE. The future value of the
annuity will be displayed.
LOANS
To find the payment needed to repay a loan:
1. Press 2nd FINANCE and from the CALC menu
choose 1: TVM Solver.
Enter the number of periods, N, the interest percent,
I%, the amount of the loan (present value) PV, and
the number of payment periods P/Y (the
compounding periods per year, C/Y is usually the
same). Enter 0 for FV, the future value of the loan
(when it is repaid).
Find the monthly payment needed to anortize a loan
of $100,000 in 60 months if interest is 10% per year
on the unpaid balance.
1.
2.
2. Place the cursor on Payment PMT, and press
ALPHA SOLVE. The payment of the loan will be
displayed.
The negative means it is leaving you.
70
XVIII. COUNTING AND PROBABILITY
PERMUTATIONS AND COMBINATIONS
EXAMPLES
To compute the number of permutations of n things
taken r at a time:
Find the number of permutations of 8 things taken 4
at a time.
1. Enter n on the home screen.
1.
2.
3.
4.
2. Press MATH, move to PRB, and select 2: nPr.
3. Press ENTER, and enter r.
4. Press ENTER. The answer is displayed.
To compute the number of combinations of n things
taken r at a time:
Find the number of combinations of 8 things taken 4
at a time.
1.
2.
1. Enter n on the home screen.
2. Press MATH, move to PRB, and select 3: nCr.
3. Press ENTER, and enter r.
4. Press ENTER. The answer is displayed.
3
ADDITIONAL EXAMPLE
4.
Compare
10 P3
71
10 P3
and
10 C3 .
is 6 times larger than
10 C3 .
PROBABILITY USING PERMUTATIONS AND
COMBINATIONS
To solve a probability problem that involves
permutations or combinations:
EXAMPLES
1. Analyze the probability problem.
It usually involves counting the number of ways an
event can occur divided by the total number of
possible outcomes.
1. The probability that both are defective is given by
the number of ways 2 chips can be drawn from the 5
that are defective divided by the number of ways 2
can be drawn from the 10 in the box.
If a box contains 10 computer chips, of which 5 are
defective, what is the probability that two chips
drawn from the box are both defective?
2. Use the ratio of combinations because the order
in which the chips are drawn is not important. This
gives the probability as
5 C2
.
10 C2
2. Determine whether permutations or
combinations should be used to count the number of
ways an event can occur and the total number of
possible outcomes. (If order is not important, use
combinations.)
3.
3. Enter the ratios of permutations or combinations
to find the probability that the event will occur.
Using MATH ˛Frac gives the probability as a
fraction.
ADDITIONAL EXAMPLE
If a die is rolled four times, what is the probability
that a 5 will occur three times?
This is a binomial probability model, solved using n
= 5 trials and with the probability of success on each
trial
p = 1/6. The probability of 3 successes in 4 trials is
found using
FG 1 IJ FG 5 IJ .
H 6K H 6K
3
4 C3
This is evaluated as follows.
72
EVALUATING MARKOV CHAINS
FINDING STEADY-STATE VECTORS
To evaluate a Markov chain:
EXAMPLES
If the initial-probability vector and the transition
matrix for a Markov chain problem are
⎡.5 .4 .1⎤
[A] = [.4 .4 .2 ] and [B] = ⎢.4 .5 .1⎥ ,
⎢⎣.3 .3 .4⎥⎦
find the probabilities for the fourth state of the chain.
1.
1. Enter the initial-probability vector as matrix A
and the transition matrix as matrix B.
3
2. The fourth state is [ A][B] .
2. To find the probabilities for the (n+1)st state,
n
calculate [ A ][B] .
Find the steady state vector for the Markov chain
problem above.
If the transition matrix contains only positive
entries, the probabilities will approach a steadystate vector, which is found as follows:
1.
1. Calculate and store [ C] = [B] − [I ] , where [B]
is the regular transition matrix and [I] is the
appropriately sized identity matrix.
T
2. On the TI-83, solve [ C] = [0] by using
T
MATRX, MATH, B:rref([ C] ).
T
(Find [ C] using MATRX, MATH, 2: T .)
On the TI-82 use row operations to solve the
T
equation, because [ C] is a singular matrix (it has
no inverse.)
2.
⇒ x = 3z, y = 3z
3. 3z + 3z + z = 1 gives z = 1/7, and the
3. Choose the solutions that add to 1, because they
are probabilities.
probabilities
73
⎡3
⎣7
3
7
1⎤
7⎦
XIX. STATISTICS
HISTOGRAMS
EXAMPLES
To find a frequency histogram, or more simply, a
histogram, for a set of data:
Find the frequency histogram for the following
scores: 38, 37, 36, 40, 35, 40, 38, 37, 36, 37, 39,
38.
1. Press STAT, EDIT, 1:edit to enter each number
in the column headed by L1 and the corresponding
frequency of each number in L2.
1. Each number can be entered individually, with a
frequency of 1, or a frequency table can be used to
create the histogram.
OR
2. Press 2nd STAT PLOT, 1 (Plot 1). Highlight
ON, and then press ENTER with the cursor on the
histogram icon.
Enter xlist:L1, Freq:L2.
2.
3.
3. Press ZOOM, 9: ZoomStat or press GRAPH with
an appropriate window.
4. If the data is given in interval form, a histogram
can be created by using the steps above, with class
marks used to represent the intervals.
4. Create a histogram for the interval data below.
Interval
Frequency
1- 5
0
6 - 10
2
11 - 15
5
16 - 20
1
21 - 25
3
Creating a table with class marks and then using
ZOOMSTAT gives the histogram.
Interval Class Marks Frequency
1- 5
3
0
6 - 10
8
2
11 - 15
13
5
16 - 20
18
1
21 - 25
23
3
74
DESCRIPTIVE STATISTICS
EXAMPLES
To find descriptive statistics for a set of data:
Find descriptive statistics for the following salary
data.
Salary
$59,000
30,000
26,000
34,000
31,000
75,000
35,000
1. Press the STAT key and 1:Edit under EDIT. To
clear any elements from a list, place cursor at top of
the list and press CLEAR and ENTER. To enter
data in a list, enter each number and press ENTER.
1.
2. To find the mean and standard deviation of the
data in list L1, press STAT, move to CALC, and
press 1: 1-Var Stats, then ENTER.
2.
3. To arrange the data in L1 in descending order,
use STAT, EDIT, 3: SortD(L1). Press STAT,
EDIT 1:Edit to view the data in descending order.
3.
4. If L2 contains the frequencies of the data in L1,
and the data in L1 is in ascending or descending
order, the median and mode can easily be read.
4.
5. If L2 contains the frequencies of the data in L1,
the mean and standard deviation is found using
STAT, CALC 1:1-Vars Stats L1,L2, ENTER.
5.
75
Number Earning
1
2
7
2
1
1
1
Probability values of a random variable in probability distributions can be evaluated with the TI-83, TI-84,
TI-83 Plus, or TI-84 Plus calculator.
PROBABILITY DISTRIBUTIONS
EXAMPLES
WITH THE TI-83, TI-84, TI-84 PLUS, and TI-83
PLUS
BINOMIAL DISTRIBUTION
2nd DISTR 0:binompdf(n,p,x) computes the
probability at x for the binomial distribution with
number of trials n and probability of success p.
Using MATH 1:˛Frac gives the probabilities as
fractions.
The probability of 3 heads in 6 tosses of a coin is
found using
2nd DISTR 0:binompdf(6,.5,3)
The probabilities can be computed
for more than one number in one command, using
2nd DISTR 0:binompdf(n,p,{x1,, x2,..}).
Using MATH 1:˛Frac gives the probabilities as
fractions.
The probabilities of 4, 5, or 6 heads in 6 tosses of a
coin are:
2nd DISTR 0:binomcdf(n,p,x) computes the
probability that the number of successes is less
than or equal to x for the binomial distribution
with number of trials n and probability of success
p.
The probability of 4 or fewer heads in 6 tosses of a
coin is:
B
B
B
B
B
B
NORMAL DISTRIBUTION
Xmin = 29, Xmax = 41,
Ymin = 0, Ymax = .2
To graph the normal distribution, press Y= and
store 2nd DISTR 1:normalpdf(x, µ , σ ) into y1 .
Then set the window variables Xmin and Xmax so
that the mean µ falls between them. Press ZOOM
0: ZoomFit to graph the normal distribution.
The default values for mean µ and standard
deviation σ are 0 and 1.
The command 2nd DISTR
2:normalcdf(lowerbound, upperbound, µ , σ )gives
the normal distribution probability that x lies
between the lowerbound and the upperbound,
when the mean is µ and the standard deviation is
σ.
76
XX. LIMITS
The TI-82, TI-83, TI-84, TI-83 Plus, and TI-84 Plus calculators are not faultless in evaluating limits, but
they are useful in evaluating most limits that we encounter. We can also use them to confirm limits that are
evaluated analytically.
LIMITS
EXAMPLE
To find the limit lim f ( x ) for the function f(x):
x→c
1. Enter the function as y1, and graph the function
in a window that contains x = c and the graph of
the function where it appears to cross
x = c.
x2 − 9
.
x→3 x − 3
Evaluate lim
B
B
x2 − 9
, with a window
x−3
containing x = 3, follows.
1. The graph of y =
2. Evaluate the function for several values near x =
c and on each side of c, by one of the following
methods.
a. TRACE and ZOOM near x = c. If the values of
y approach the same number L as x approaches c
from the left and right, we have evidence that
the limit is L.
2. a.
b. Evaluate y1({c1, c2, ...}) for values that are
very close to, and to the left of, c. This indicates
the value of lim f ( x ) .
The y-values appear to approach 6.
b.
x→c −
Repeating this with values very close to, and to the
right of, c indicates the value of lim f ( x ) .
x→c +
If these two limits are the same, say L, then the
limit is L.
lim
x → 3+
x2 − 9
=6
x−3
lim
x → 3−
x2 − 9
= 6 , so
x−3
x −9
=6
x−3
2
c. Use TBLSET to start a table near
x = c with ∆ Tbl very small.
If the y-values approach L as the
x-values get very close to c from both sides of c,
we have evidence that the limit is L.
lim
x→3
c.
d. Use TBLSET with Indpnt: set to Ask, and enter
values very close to, and on both sides of, c. The
y-values will approach the same limit as above.
The limit as x approaches 3 appears to be 6. The
error at x=3 indicates that f(3) does not exist.
77
LIMITS WITH PIECEWISE-DEFINED
FUNCTIONS
To evaluate the limit as x → a for the piecewisef ( x ) if x ≤ a
defined function y =
:
g ( x ) if x > a
EXAMPLE
1. Graph the function, as follows:
Under the Y= key, set
y1 = f ( x ) x ≤ a
1. Enter y1 = (x + 7)/(x ≤ -5)
and
y2 = (-x + 2)/(x>-5)
RS
T
B
B
b
gb
Find lim y if y =
x →−5
RSx + 7 if x ≤ −5
T− x + 2 if x > −5
B
B
g
B
B
where ≤ is found under 2nd TEST 6.
and set
y2= g ( x ) x > a
B
B
b gb
g
where > is found under 2nd TEST 3.
Use GRAPH or ZOOM to graph the function.
2. Evaluate the function for several values near x =
c and on each side of c, by one of the following
2.a.
methods.
a. TRACE and ZOOM near x = c. If the values of
y approach the same number L as x approaches c
from the left and right, we have evidence that
the limit is L. Use the up or down arrows to
TRACE on the correct piece of the function.
The limit from the left does not equal the limit from
the right, so the limit does not exist.
b. Use TBLSET to start a table near
b.
x = c with ∆ Tbl very small.
If the y-values approach L as the
x-values get very close to c from both sides of c,
we have evidence that the limit is L. "ERROR"
will occur in the table where the piece y1 or y2 does
not exist.
B
B
B
B
lim y = 2 ,
x →−5−
c. Use TBLSET with Indpnt: set to Ask, and enter
values very close to, and on both sides of, c. The
y-values will approach the same limit as above.
lim y = 7 , so lim y = DNE .
x →−5+
Note that y(-5) = 2.
c.
78
x →−5
LIMITS AS x → ∞
EXAMPLE
To find the limit lim f(x) for the function f(x):
x→∞
1. Enter the function as y1, and graph the function
in a window that contains the graph for large
values of x.
B
B
3x − 2
x→∞ 1 − 5x
3x − 2
1. The graph of y =
follows.
1 − 5x
ZOOM 4
Xmin=-25, Xmax=25
Evaluate lim
2. Evaluate the function for several very large
positive values by one of the following methods.
a. TRACE on the graph toward very large values.
Holding the right arrow or repeatedly pressing it
will move the window to the right. If the
y-values approach a finite number, this number is
the limit.
b. Evaluate y1({c1, c2, ...}) for values that are
very large. If the y-values approach a finite
number, this number is lim f ( x ) .
a.
b.
x →∞
c. Use TBLSET to start a table with TblStart very
large and with ∆ Tbl large.
If the values approach L as the x-values get very
large, we have evidence that the limit is L.
d. Use TBLSET with Indpnt: set to Ask, and enter
very large values. The y-values will approach the
same limit as above.
lim
x→∞
3x − 2
= -.6
1 − 5x
c.
The values of the function round off to -.6 (to 4
decimal places) after 5700.
3x − 2
lim
= -.6
x →∞ 1 − 5x
In a similar fashion, the calculator can be used to investigate limits as x approaches - ∞ .
79
XXI. NUMERICAL DERIVATIVES
There are two ways to find the derivative of a function at a specified value of x. One uses the graph of the
function and one uses the numerical derivative operation of the MATH menu. Both of these methods
follow.
NUMERICAL DERIVATIVES
EXAMPLE
To find the (approximate) numerical derivative of
the function f(x) at the value x = c:
Find the numerical derivative of
Method 1:
1. Enter MATH, 8 nDeriv(
and then enter the function, x, and the value c,
giving the following:
nDeriv(f(x),x,c)
1.
2. The approximate derivative at the specified
value of x will be displayed. To get better
accuracy in the approximation, add an additional
entry with a ∆x less than .001.
Method 2:
1. Enter the function as y1, and graph the function
in a window that contains x = c and f(c). Press
GRAPH to graph the function.
f(x) = x 3 − 2 x 2 at x = 2.
2.
The numerical derivative is 4.
B
B
2. Press 2nd CALC and 6 (dy/dx).
Use an arrow to trace to the selected x-value, or
enter the x-value and press ENTER. The
approximate value of dy/dx will appear on the
screen if the x-value is in the window. An error
will occur if the x-value is not in the window or if
the derivative does not exist.
1. The graph using ZOOM 4:
2.
This approximates the numerical derivative, 4.
After you have found the derivative of a function, you can use its graph and the graph of the numerical
derivative of the function to check your work.
80
CHECKING A DERIVATIVE
EXAMPLE
To check the correctness of a derivative f ′( x ) of
the function f(x):
Verify that the derivative of
1. Enter the derivative f ′( x ) that you found as
y1, and graph this derivative function in a
convenient window.
f(x) = x 3 − 2 x 2 is f ′( x ) = 3x 2 − 4 x
1.
B
B
2. Press Y = and enter the following in y2:
nDeriv(f(x),x,x).
B
B
3. Graph using an appropriate window. Both
graphs will appear. If the second graph lies on top
of the first, then the derivatives agree, and your
solution checks.
2.
3.
4. On the TI-83, TI-84, TI-83 Plus, or TI-84 Plus,
move the cursor to the left of y2 = and press
ENTER to get a thicker \, which indicates the graph 4.
will be drawn thicker than normal. Press GRAPH.
The second graph will now be thicker as it graphs
over the first.
B
B
5. To further verify that the derivatives agree
(especially on the TI-82), move from one graph to
the other by using the up or down arrow, and use
TRACE to evaluate them both at several x-values.
The values may not be identical, but should agree
when rounded.
5.
81
FINDING AND TESTING SECOND
DERIVATIVES
To find the second derivative of a function at a
given value of x at x = c:
EXAMPLES
Find the second derivative of
f ( x ) = x 3 − 2 x 2 at x = 2.
1.
1. Enter the function as y1.
Enter nDeriv(y1,x,x) as y2.
B
B
B
B
B
B
2. The second derivative of the function given by
Y1 is approximated at x = c with nDeriv(Y2,x,c)
2.
To check the correctness of a derivative f ′′( x ) of
the function f(x):
Thus f ′′(2) = 8 .
1. Enter the function as y1.
Enter nDeriv(y1,x,x) as y2.
B
B
B
1.
B
B
B
2. Enter nDeriv(y2,x,x) as y3.
B
B
B
B
3. Enter the second derivative f ′′( x ) that you
found as y4. (On the TI-83, TI-84, TI-83 Plus, or
3.
TI-84 Plus, move the cursor to the left of Y4 =
and press ENTER to get a thicker \, which indicates
the graph will be drawn thicker than normal.)
B
B
4. Turn off the equations for y1 and y2. Graph y3
and y4 in the same window. If the second graph
lies on top of the first, then the derivatives agree,
and your solution checks.
B
B
B
B
B
B
B
B
4.
82
2.
XXII. CRITICAL VALUES
The values of x that make the derivative of a function 0 or undefined are critical values of the function.
We can find where the derivative is 0 by finding the zeros of the derivative function (that is, the xintercepts of the derivative function).
CRITICAL VALUES
EXAMPLE
To find or approximate values of x that make the
derivative of f(x) equal to 0:
Find the values that make the derivative of f(x) =
x3
− 4 x equal to 0.
3
I. Find the derivative of f(x).
II. Find the values of x that make the derivative 0:
I. The derivative is f ′( x ) = x 2 − 4 .
1.
2. a.
Method 1
1. Enter the equation of the derivative as y1.
B
B
2. Find where y1 = 0 by one of the following:
a. Finding the x-intercepts of y1 by using TRACE.
B
B
B
B
b. Finding the zeros of y1 by using
2nd CALC, 2 :zero (root on the TI-82).
TRACE gives one x-intercept to be 2.
b.
B
B
c. Use 2nd TBLSET and 2nd TABLE to find
values of x that give y = 0.
Thus 2nd CALC Zero gives the zero -2.
c.
Method 2 (On the TI-83, TI-84, TI-83 Plus, or TI84 Plus)
1. Press MATH, 0: Solver
1.
2.
2. Press ENTER and the up arrow, and then enter
the equation of the derivative on the right of 0=.
3. Press the down arrow to put the cursor on the
variable, and press
ALPHA, SOLVE, ENTER. A solution will appear
if one exists. Changing values set equal to the
variable and using ALPHA, SOLVE, ENTER will
give other solutions if they exist.
3.
The solutions (zeros) are 2 and -2.
83
XXIII. RELATIVE MAXIMA AND MINIMA
RELATIVE MAXIMA AND RELATIVE
MINIMA USING THE DERIVATIVE
To find the relative maximum and or relative
minimum of a polynomial function:
1. Find the derivative of the function, and enter
the equation of the derivative as y2.
B
EXAMPLE
Find the relative maximum of
x3
f ( x) =
− 4x
3
1. f '( x ) = x 2 − 4
B
2. Graph y2 and find the values of x that make the
derivative 0 or undefined, using TRACE, ZERO,
or TABLE. (No value of x will make the
derivative of a polynomial undefined.)
B
B
3. Enter the function as y1.
B
TBLSET -3, ∆ TBL 1
2.
B
4. Use TBLSET and TABLE to find derivatives
near, and to the left and right of the zeros. of the
derivative:
a. If the derivative, y2 is positive to the left of x =
c, 0 at c and negative to the right of c, the function
has a relative maximum at x = c.
The y-value, y1, where x = c is the relative
maximum that occurs there.
b. If the derivative, y2 is negative to the left of x =
c, 0 at c and positive to the right of c, the function
has a relative minimum at x = c.
The y-value, y1, where x = c is the relative
minimum that occurs there.
c. If the derivative does not have opposite signs on
opposite sides of
x = c, there is no maximum nor minimum, and a
horizontal point of inflection occurs.
The derivative in column y2 is 0 at
x = -2 and x = 2.
B
B
B
B
4. Using TABLE with the derivative in column Y2
gives
B
B
B
B
B
B
5. Graph the function to confirm the relative
maximum and relative minimum occur where you
have found them.
a. The values of y2 change from positive to 0 at x =
-2 to negative, so a relative maximum occurs at x = 2; and from y1, y = 5.3333 when x = -2, so a relative
maximum is at (-2, 16/3).
b. y2 changes from negative to 0 at
x = 2 to positive, so a relative minimum occurs at x
= 2; and from y1,y = -5.3333 when x = 2, so a
relative minimum is at (2, -16/3).
B
B
B
B
B
B
B
B
5. max @ (-2, 16/3), min @ (2, -16/3)
84
RELATIVE MAXIMA AND RELATIVE
MINIMA USING MAXIMUM OR MINIMUM
I. To find the relative maximum of a polynomial
function:
1. Enter the equation of the function as y1.
B
B
EXAMPLE
Find the relative maximum of
x3
f ( x) =
− 4x
3
1.
2.
2. Select a window that includes all possible
"turns" in the graph, using knowledge of the shapes
of polynomial functions. Graph y1, and use
ZOOM, Zoom Out to find all "turns."
B
B
B
B
3. If there appears to be a relative maximum,
locate it as follows:
3. a.
b.
c.
d.
a. Press 2nd CALC, 4 (maximum).
b. Move the cursor to a point on the left side of
where the maximum appears to occur.
c. Press ENTER and move the cursor to the right
side of where the maximum appears to occur.
d. Press ENTER twice. The resulting
point is an approximation of the observed relative
maximum.
II. To find the relative minimum, repeat the steps
above using
2nd CALC, 3 (minimum).
The relative maximum, found with calculus, is
really y = 16/3 at x = -2.
II.
The relative minimum is y = -16/3 at
x = 2.
85
UNDEFINED DERIVATIVES AND RELATIVE
EXTREMA
If the derivative of f(x) is undefined at x = c, and if
f(c) exists, then
find a relative maximum or minimum as follows:
1. Enter the equation of the derivative as y2.
EXAMPLE
Find the relative maximum and or relative minimum
of
f(x) = ( x − 1) 2 / 3 + 2 .
1.
B
B
2. Graph y2 and find the values of x that make the
derivative undefined. (Use TRACE, ZERO, or
TABLE.)
B
B
2.
3. Enter the equation of the function as y1.
B
B
4. If y1 exists where the derivative is undefined,
use TBLSET and TABLE to find derivatives near,
and to the left and right of this x-value.
a. If the derivative, y2 is positive to the left of x =
c, undefined at c and negative to the right of c, the
function has a relative maximum at
x = c. The y-value, y1, where x = c is the relative
maximum that occurs there.
b. If the derivative, y2 is negative to the left of x =
c, undefined at c and positive to the right of c, the
function has a relative minimum at x = c.
The y-value, y1, where x = c is the relative
minimum that occurs there.
c. If the derivative does not have opposite signs on
opposite sides of
x = c, there is no maximum nor minimum, and a
vertical point of inflection occurs.
B
B
The derivative is undefined at x = 1.
B
B
B
B
B
B
4. The values of y2 change from negative to
undefined at x = 1 to positive, so a relative
minimum occurs at x = 1.
Using the table that includes y1 shows that the
relative minimum is y = 2 where x = 1.
B
B
B
B
B
B
5. The graph of the function is
5. Graph the function to confirm the relative
maximum and relative minimum occur where you
have found them.
rel min at (1,2).
86
XXIV. INDEFINITE INTEGRALS
CHECKING INDEFINITE INTEGRALS
EXAMPLES
To check an indefinite integral with
fnInt:
Find the integral of f(x) = x 2 and check the result
with fnInt.
1. Enter the integral of f(x) (without the +C) as y1
under the Y= menu.
1. The integral is
2. Move the cursor to y2 , press
MATH, 9: fnInt(.
Enter f(x),x,0,x) so the equation is
y2 = fnInt(f(x),x,0,x).
2.
x3
+ C . Enter
3
y1 = x^3/3 under the Y= menu.
3. Press GRAPH with an appropriate window. If
the second graph lies on top of the first, the graphs
agree and the computed integral checks.
3. The second graph lies on top of the first, which
indicates that the integrals agree.
4.
4. On the TI-83, TI-84, TI-83 Plus, or TI-84 Plus,
pressing ENTER with the cursor to the left of y2
changes the thickness of the graph of the second
graph, making the fact that it lies on top of the first
more evident.
87
FAMILIES OF FUNCTIONS
SOLVING INITIAL VALUE PROBLEMS
The indefinite integral of the function f(x) has the
form F(x) + C,
where the derivative of F(x) is f(x).
Thus the indefinite integral gives a family of
functions, one for each value of C. Different values
of C give different functions. To graph some of
them:
EXAMPLES
Find the integral of f ( x ) = 2 x − 4
and graph the integrals for
C = 0, C = 1, C = -2, and C = 3.
1. The integral of f ( x ) = 2 x − 4 is
f ( x ) = x 2 − 4 x + C . The equation editor showing
these equations and the graphs of these equations
are shown below.
2.
3.
1. Integrate f(x).
2. Enter equations in the equation editor, using
different values for C.
3. Press GRAPH with an appropriate window. The
graphs will be graphs of y = F(x) shifted up or
down, depending on C.
If a value of x and a corresponding value of y are
given for the integral of a function, This “initial
value” can be used to solve for C and thus to find
the one function that satisfies the conditions.
To find this function:
1. The integral of f ( x ) = 2 x − 4 is
1. Integrate the function f(x).
y = x2 − 4x + C .
2.
3.
2. Press MATH, 0:Solver and press the up arrow to
see EQUATION SOLVER.
3. Set 0 equal the integral minus y, getting 0 = F(x)
+ C - y, and press the down arrow.
4. Enter the given values of x and y, place the
cursor on C, and press ALPHA, SOLVE (ENTER).
The value of C will appear. Replace C with this
value to find the function satisfying the conditions.
4.
The unique function is y = x 2 − 4 x + 3 .
Its graph is shown above.
88
XXV. DEFINITE INTEGRALS
APPROXIMATING A DEFINITE INTEGRAL AREAS UNDER CURVES
To find the area under the graph of
y = f(x) and above the x-axis:
EXAMPLES
EXAMPLE
Find the area under the graph of
f ( x ) = x 2 from x = 0 to x = 3.
1.
1. Enter f(x) under the Y= menu, and press
GRAPH with an appropriate window.
2. Press 2nd CALC and 7:
z f ( x)dx .
2.
3.
4.
5.
3. Press ENTER. Move the cursor to, or enter, the
lower limit (the left x-value).
4. Press ENTER. Move the cursor to, or enter the
upper limit (the right x-value).
5. Press ENTER. The area will be displayed.
APPROXIMATING A DEFINITE INTEGRALALTERNATE METHOD
To approximate the definite integral of y = f(x) in
the interval between
x = a and x = b:
EXAMPLE
Approximate the definite integral of
f(x) = 4 x 2 − 2 x from x = -1 to x = 3.
1.
1. Press MATH, 9: fnInt(.
Enter f(x),x,a,b) so the display shows
fnInt(f(x),x,a,b).
2. Press ENTER to find the approximation of the
integral.
2.
3.
3. The approximation may be made closer than that
in step 3 by adding a fifth argument with a number
(tolerance) smaller than 0.00001.
This approximation is not improved.
The exact integral is 88/3 = 29 1/3.
89
AREA BETWEEN TWO CURVES
EXAMPLES
To find the area enclosed by the graphs of two
functions:
Find the area enclosed by the graphs of y = 4 x 2 and
y = 8x.
1. Enter one equation as y1 and the second as y2 .
Press GRAPH using an appropriate window.
1.
2. Find the x-coordinates of the points of
intersection of the graphs.
Use 2nd CALC 5: intersect.
2. Graphs intersect at x = 0 and x = 2.
B
B
3. Determine visually which graph is above the
other over the interval between the points of
intersection.
3. y = 8x is above y = 4 x 2 in the interval from
0 to 2.
4.
4. Press MATH, 9: fnInt(.
Enter f(x),x,a,b) so the display shows
fnInt(f(x),x,a,b) where f(x) is y2 − y1 if the graph
of y2 is above the graph of y1 between a and b, or
y1 − y2 if y1 is
above y2.
B
B
B
B
B
B
B
B
The area between the graphs can also be found
using 2nd CALC, z f ( x )dx .
1. Enter y3 = y2 − y1 where y2
is above y1 .
1.
2.
3.
4.
5.
6.
B
B
2. Turn off the graphs of y1 and y2 and graph y3
with a window showing where y3 > 0.
3. Press 2nd CALC and 7: z f ( x )dx .
4. Press ENTER and select the lower limit (the left
x-intercept).
5. Press ENTER. Move the cursor to, or enter, the
upper limit (the right x-intercept).
6. Press ENTER. The area will be displayed.
90
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