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or
omissions
at
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earliest
Course
The
IB
Diploma
resource
suppor t
books
is
Companion
Programme
materials
through
will
help
expected
the
Course
to
gain
study
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provide
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IB
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aim
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help
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guidance
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action,
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draw
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conclusions
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Diploma
ser vice
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are
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book
with
IB
and
required
a
variety
of
fur ther
suggestions
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provided.
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on
requirements
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programmes
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world
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Companions
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honesty
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education
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range
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mindedness;
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A
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note
is
of
vital
on
academic
impor tance
appropriately
credit
to
the
acknowledge
owners
of
honesty
and
What
constitutes
is
Malpractice
when
that
information
is
used
in
your
work.
owners
of
ideas
(intellectual
proper ty)
rights.
To
have
an
authentic
in,
you
piece
it
must
be
based
on
your
in
individual
ideas
with
the
work
of
others
Therefore,
all
assignments,
oral,
completed
for
assessment
may
or
any
student
gaining
an
unfair
one
or
more
includes
assessment
plagiarism
and
component.
collusion.
is
defined
as
the
representation
of
the
must
or
work
of
another
person
as
your
own.
written
The
or
or
fully
ideas
acknowledged.
in,
and
Plagiarism
original
results
of
Malpractice
work,
that
have
advantage
proper ty
behavior
After
result
all,
malpractice?
information
use
following
are
some
of
the
ways
to
avoid
your
plagiarism:
own
language
and
expression.
Where
sources
are
●
used
or
referred
to,
whether
in
the
form
of
Words
and
suppor t
quotation
or
paraphrase,
such
sources
ideas
of
another
person
used
to
direct
must
one’s
arguments
must
be
be
acknowledged.
appropriately
acknowledged.
●
How
of
I
acknowledge
the
work
way
used
the
that
you
ideas
footnotes
that
are
enclosed
within
quoted
verbatim
quotation
marks
must
be
and
acknowledged.
others?
The
of
do
Passages
of
acknowledge
other
and
people
that
is
you
●
have
through
the
CD-ROMs,
Inter net,
use
be
bibliographies.
email
and
treated
in
any
the
messages,
other
same
web
sites
electronic
way
as
on
media
books
the
must
and
jour nals.
(placed
Footnotes
endnotes
be
(placed
provided
another
at
when
at
the
you
document,
or
information
provided
do
to
not
that
is
need
par t
of
do
are
the
par t
of
resources
“Formal”
in
need
means
you
that
a
or
of
a
page)
document)
paraphrase
summarize
for
be
are
to
as
used
you
in
a
formal
your
●
is,
they
should
of
work.
accepted
forms
of
use
one
presentation.
Works
involves
separating
the
into
different
categories
newspaper
ar ticles,
be
is
of
CDs
and
works
of
information
as
to
how
a
allowing
that
work
can
find
the
same
is
compulsor y
your
takes
the
place,
malpractice
by
includes:
work
assessment
duplicating
and
to
be
copied
or
submitted
by
another
student
work
for
different
assessment
and/or
diploma
requirements.
reader
or
viewer
information.
the
forms
of
malpractice
include
any
action
providing
gives
you
an
unfair
advantage
or
affects
the
of
of
another
student.
Examples
include,
A
extended
unauthorized
material
into
an
examination
essay
.
room,
misconduct
falsifying
iv
This
work
suppor ting
dance,
where
Inter net-based
ar t)
in
as
a
film,
work.
books,
taking
bibliography
of
par t
own
be
you
results
your
a
your
and
graphs,
must
the
that
full
music,
ar ts,
defined
student.
not
visual
acknowledged.
Other
resources,
are
whether
or
components
magazines,
they
data,
material
This
resources
(e.g.
if
maps,
programs,
similar
must
Collusion
●
use
and
use
for
usually
photographs,
computer
ar t,
ar ts,
all
creative
another
of
of
theatre
●
several
of
acknowledged
Y
ou
list
sources
audio-visual,
knowledge.
include
The
illustrations,
the
That
footnoted
●
from
information
knowledge”.
to
or
document.
footnote
of
should
that
of
another
a
assumed
Bi bliographies
the
end
closely
“body
not
bottom
quote
provide
a
definitions
the
a
CAS
during
record.
an
examination,
and
About
The
new
Level
is
syllabus
for
thoroughly
chapter
the
the
is
divided
following
book
Mathematics
covered
into
in
lesson
this
size
Standard
where
book.
designed
Each
sections
a
GDC
to
analytical
with
skills
applications
Exploration
and
concludes
Theory
of
you
Historical exploration
know?
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the
exam.
v
Contents
Chapter
Functions
1.1
Introducing
1.2
The
functions
domain
Cartesian
and
Function
1.4
Composite
1.5
Inverse
1.6
Transforming
range
4
of
a
relation
on
plane
1.3
Chapter
notation
functions
functions
functions
Quadratic
functions
a
Solving
quadratic
2.2
The
2.3
Roots
2.4
Graphs
2.5
Applications
quadratic
of
equations
Patter ns
6.2
Arithmetic
13
6.3
Geometric
14
6.4
Sigma
16
6.5
Arithmetic
21
6.6
Geometric
6.7
Convergent
6.8
Applications
6.9
Pascal’s
of
quadratic
of
and
(Σ)
and
sequences
162
sequences
164
sequences
notation
arithmetic
167
and
series
170
series
172
series
175
series
of
and
sums
geometric
to
infinity
178
and
patter ns
triangle
181
and
the
binomial
expansion
38
equations
sequences
6.1
34
formula
quadratic
Patterns,
series
2.1
8
and
equations
Chapter
184
41
functions
quadratics
43
Chapter
53
7.1
Limi ts
Limits
and
and
derivatives
convergence
196
n
Chapter
Probabi li ty
3.1
Definitions
64
3.2
Venn
68
3.3
Sample
diagrams
product
space
3.4
Conditional
3.5
Probability
Chapter
diagrams
and
the
r ule
probability
tree
diagrams
Exponential
and
Exponents
Solving
4.3
Exponential
4.4
Proper ties
4.5
Logarithmic
4.6
Laws
4.7
Exponential
4.8
Applications
of
exponential
logarithms
functions
logarithms
logarithmic
Chapter
equations
and
of
functions
Rational
5.1
Reciprocals
5.2
The
5.3
Rational
reciprocal
functions
function
functions
7.4
The
chain
line
for
and
derivative
of
x
200
derivatives
r ule
and
higher
208
order
derivatives
Rates
7.6
The
85
7.7
More
of
215
change
derivative
on
and
and
extrema
motion
in
a
line
221
graphing
and
230
optimization
problems
Chapter
240
Descriptive
8.1
Univariate
analysis
103
8.2
Presenting
data
107
8.3
Measures
of
109
8.4
Measures
of
115
8.5
Cumulative
118
8.6
Variance
equations 127
and
Chapter
statistics
256
257
central
tendency
260
dispersion
267
frequency
and
standard
271
deviation
276
Integration
9.1
Antiderivatives
9.2
More
131
and
the
indefinite
integral
291
on
and
indefinite
9.3
Area
142
9.4
Fundamental
143
9.5
Area
147
9.6
Volume
of
9.7
Definite
integrals
other
vi
r ules
122
logarithmic
exponential
tangent
7.5
logari thmic
functions
of
More
77
4.2
The
7.3
89
functions
4.1
7.2
definite
between
integrals
297
integrals
Theorem
two
of
302
Calculus
cur ves
313
revolution
problems
with
309
318
linear
motion
and
321
Chapter
Bivariate
10.1
Scatter
10.2
The
10.3
Least
10.4
Measuring
analysis
diagrams
line
of
best
squares
fit
regression
correlation
Chapter
334
15.1
Random
15.2
The
binomial
345
15.3
The
normal
Right-angled
triangle
11.2
Applications
of
trigonometr y
right-angled
Using
the
363
triangle
trigonometr y
11.3
369
coordinate
variables
axes
in
16.1
About
16.2
Inter nal
the
16.3
How
16.4
Academic
the
Record
16.6
Choosing
11.4
The
sine
380
16.7
Getting
11.5
The
cosine
11.6
Area
11.7
Radians,
Chapter
r ule
r ule
Vectors:
triangle
12.2
Addition
and
12.3
Scalar
12.4
Vector
12.5
Application
Chapter
and
basic
391
of
of
the
13.3
Trigonometric
13.4
Graphing
13.5
Translations
of
a
line
unit
using
the
circle
and
568
Chapter
Using
a
graphic
calculator
display
1
Functions
2
Differential
3
Integral
4
Vectors
5
Statistics
572
calculus
calculus
598
606
608
and
probability
612
stretches
sine
with
Prior
learning
Number
633
2
Algebra
657
448
3
Geometr y
673
454
4
Statistics
699
Chapter
Practice
papers
Practice
paper
1
708
469
Practice
paper
2
712
478
Answers
483
Index
sine
and
functions
Calculus
1
of
functions
with
Chapter
462
functions
transformations
star ted
456
functions
and
Chapter
unit
identities
Combined
Modeling
564
420
437
circle
circular
cosine
563
topic
430
functions
equations
trigonometric
cosine
vectors
vectors
Circular
Solving
keeping
a
562
562
426
equation
Using
marked
Honesty
557
407
subtraction
13.2
13.7
sectors
concepts
13.1
13.6
389
product
556
criteria
is
386
Vectors
12.1
538
exploration
exploration
16.5
arcs
527
Exploration
assessment
373
a
520
distribution
distribution
The
trigonometr y
of
distri butions
349
Trigonometry
11.1
Probabi li ty
339
Chapter
Chapter
wi th
trigonometric
functions
14.1
Derivatives
14.2
More
14.3
Integral
14.4
Revisiting
of
practice
of
trigonometric
functions
withderivatives
sine
and
linear
cosine
motion
496
500
505
510
vii
What's
The
as
on
material
well
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lear ning.
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you
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of
Full
material
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are
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provided.
CASIO 9860-GII
Practice
paper
2
Simultaneous and quadratic equations
1.5
Tolerance
Asian
Solving
simultaneous
linear
equations
European
Low
30
Medium
50
40
High
40
20
80
Practice paper 2
Practice
prepare
exam
for
solutions
papers
your
can
will
help
examinations.
be
found
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the
you
to
fully
Worked
website
www.oxfordsecondar y.co.uk/ibmathsl
Powerpoint
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in
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presentations
solutions
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cover
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9860-GII
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instructions for
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calculators, so
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understanding.
Functions
CHAPTER
OBJECTIVES:
2.1
Functions:
2.2
Graphs
of
domain,
range,
functions,
by
composite,
hand
and
identity
using
GDC,
and
inverse
their
functions
maxima
and
minima,
−1
asymptotes,
the
T
ransformations
2.3
graph
of
of
f
(x)
graphs,
translations,
reections,
stretches
and
composite
transformations
Before
Y
ou
1
you
should
Plot
e.g.
start
know
how
to:
coordinates.
Plot
Skills
1
y
a
2
the
check
Plot
these
A(1, 3),
points
B(5, −3),
on
a
coordinate
C(4, 4),
plane.
D(−3, 2),
D
C
1
points
A(4, 0),
B(0, −3),
E(2, −3),
F(0, 3).
y
A
C(−1, 1)
and
0
D(2, 1)
–2
x
1
–1
3
2
b
4
Write
down
2
the
A
–1
on
a
coordinate
plane.
coordinates
–2
1.5
of
E
H
B
–3
points
A
to
1
H
–4
0.5
2
Substitute
e.g.
values
into
an
expression.
D
C
B
0
Given x = 2, y = 3 and z = −5,
–2
x
–1
1
2
3
0.5
2
find
the
value
of
a
4x
+
2y
y
b
−
3z
–1
4x
a
+
2y
=
4(2)
+
2(3)
=
8
+
6
=
G
14
–0.5
2
y
b
2
−
3z
=
(3)
−3(−5)
=
9
+
15
=
24
F
–2
3
Solve
linear
equations.
2
e.g.
Solve
6
4x
6
−
4x
=
Given
that
x
=
4,
y
=
6
and
z
=
−10,
find
0
2
a
−
=
0
⇒
6
=
4x
+
3y
z
b
−
3y
y
c
−
z
d
4x
3
1.5
=
x
⇒
x
=
1.5
Solve
y
6
4
Use
your
GDC
to
a
3x
−
6
=
6
5x
b
+
7
=
−3
c
graph
4
a
function.
4
Graph
these
functions
on
your
GDC
2
e.g.
Graph
f (x)
=
within
0
–6
2x
−
1,
–3
≤
x
≤
–4
–2
the
given
domain.
Then
x
2
4
6
3
sketch
the
functions
on
paper.
–4
a
y
=
2x
b
y
=
10
−
3,
−4
≤
x
≤
7
–6
−
2x,
−2
≤
x
≤
5
–8
2
5
Expand
e.g.
linear
Expand
(x
binomials.
+
3)
(x
−
2)
c
5
y
=
x
–
3,
–3
≤
x
≤
3.
Expand
2
=
x
+
x
Functions
−
6
a
(x
+
4)
(x
+
5)
c
(x
+
5)
(x
−
4)
b
(x
−
1)
(x
−
3)
The
Inter national
Space
Station
(ISS)
has
been
orbiting
the
Ear th
[
International
Space
Station
over
15
have
times
actually
difficult
as
it
a
day
seen
for
it?
might
more
than
Spotting
seem
–
ten
the
years,
ISS
provided
with
you
yet
the
know
how
many
naked
in
eye
which
of
is
us
not
as
direction
–1
to
of
look.
the
heads.
‘stars’,
Although
lowest
the
orbits
Thanks
which
to
ISS
travels
possible,
its
makes
large
it
at
solar
fairly
at
a
speed
of
approximately
390
wings
of
easy
to
it
is
one
,
7.7 km s
distinguish
km
the
as
it
it
is
above
in
one
our
brightest
moves
across
One
the
night
of
the
rst
sky
.
mathematicians
to
The
relation
gives
the
speed
of
the
ISS,
where
t
is
the
time
study
the
concept
of
function
measured
in
hours
and
d
is
the
distance
traveled
in
was
kilometres.
philosopher
This
is
a
mathematical
relationship
called
a function
and
is
just
of
how
a
mathematical
function
can
be
used
to
describe
(1323–1382).
a
He
situation.
worked
this
with
independent
dependent
In
Nicole
one
Oresme
example
chapter
you
will
explore
functions
and
how
French
they
can
and
variable
be
quantities.
applied
to
a
wide
variety
of
mathematical
situations.
Chapter
.
Introducing
Investigation
In
some
hands
1
countries
with
How
b
Copy
–
is
ever ybody
handshake,
a
it
many
and
if
functions
there
handshakes
customar y
in
the
are
3
handshakes
complete
Number
people
are
this
at
business
meeting.
If
there
there
there
for
are
4
meetings
are
3
2
to
people
shake
there
handshakes
and
is
so
on.
people?
table.
Number
of
Y
ou
of
people
might
nd
it
helps
handshakes
to
tr y
this
out
with
a
2
group
3
in
of
your
friends
class.
4
5
6
7
8
Do
not
join
the
points
9
in
this
case
as
we
10
are
dealing
with
c
the
Plot
points
on
a
Car tesian
coordinate
plane
with
the
whole
number
numbers.
of
people
Write
d
a
the
formula
number
of
Relations
Distance
on
x-axis
for
people,
and
(m)
and
the
the
number
number
of
T ime
The
(s)
200
34
300
60
400
88
way
➔
A
There
of
and
is
showing
data
is
nothing
provided
In
other
that
Functions
in
a
a
table
time
within
set
of
it
shows
takes
words,
order.
brackets
all
any
numbers
cer tain
is
(400, 88).
ordered
at
r un
information
and
specific
special
these
this
(300, 60)
enclosed
relation
relation.
of
(200, 34),
pieces
comma
H,
in
the
terms
y-axis.
of
the
the
for
a
amount
student
15
to
two
handshakes,
on
n
of
(100, 15),
handshakes
functions
100
Another
of
as ordered pairs:
Each
They
in
distances.
the
are
ordered
separated
form
pair
by
a
(x, y).
pairs.
about
group
come
the
of
in
numbers
numbers
pairs.
that
is
a
are
in
relation
a
has
only
(discrete)
➔
The
is
domain
ordered
pairs.
domain
of
the
set
of
all
the
first
numbers
(x-values)
of
the
The
The
the
ordered
pairs
above
is
{100,
200,
300,
curly
mean
➔
The
range
is
the
set
of
the
second
numbers
brackets,
{
},
400}.
( y-values)
in
'the
set
of'.
each
pair.
The
range
of
Example
Find
the
the
ordered
pairs
above
is
{15,
34,
60,
88}.
domain
a
{(1, 4),
(2, 7),
b
{(−2, 4),
and
range
(3, 10),
(−1, 1),
of
these
relations.
(4, 13)}
(0, 0),
(1, 1),
(2, 4)}
Answers
a
b
The
domain
is
The
range
{4,
The
The
is
domain
range
is
is
{1,
2,
7,
10,
{−2,
{0,
1,
3,
−1,
4}
First
13}
0,
1,
2}
4}
elements
Second
elements
Do
repeat
not
there
are
ordered
➔
A
of
function
the
element
be
a
is
a
domain
of
the
function
mathematical
of
the
range
no
two
of
the
is
pairs
in
ordered
the
values
4s
such
and
that
associated
function.
ordered
two
the
pairs
ordered
even
two
pairs
though
1s
in
the
pairs.
relation
function
in
In
may
each
with
order
have
for
the
element
exactly
a
one
relation
same
to
first
element.
Example
Which
of
these
sets
of
ordered
pairs
a
{(1, 4),
(2, 6),
(3, 8),
(3, 9),
(4, 10)}
b
{(1, 3),
(2, 5),
(3, 7),
(4, 9),
(5, 11)}
c
{(−2, 1),
(−1, 1),
(0, 2),
(1, 4),
are
functions?
(2, 6)}
Answers
a
Not
a
function
number
3
because
occurs
twice
the
in
the
domain.
b
A
function;
elements
c
A
function;
elements
all
are
all
are
of
the
first
different.
of
the
first
different.
Note
some
that
of
it
the
doesn’t
matter
y-values
are
that
the
same.
Chapter
Exercise
1
2
Which
A
of
these
sets
a
{(5, 5),
b
{(−3, 4),
c
{(4, 1),
d
{(−1, 1),
(0, 3),
e
{(−4, 4),
(−4, 5),
f
{(1, 2),
For
each
whether
(4, 4),
(3, 3),
(−1, 6),
(4, 2),
(2, 2),
ordered
(2, 2),
(0, 5),
(4, 3),
(2, −1),
relation
is
a
(2, 8)}
(−2, 8)}
(5, 2)}
the
domain
and
range
and
say
function.
y
a
functions?
(3, −1)}
(−3, 7),
(4, 2),
identify
are
(4, 5)}
(1, 7),
(−3, 6),
(3, 2),
pairs
(1, 1)}
(4, 4),
(1, 6),
diagram,
the
of
y
b
2
2
Write
down
the
1
1
coordinates
x
0
1
2
3
as
x
0
1
–1
2
ordered
3
pairs.
4
–1
3
Look
it
back
takes
for
between
The
a
at
a
can
Y
ou
can
relation
student
line
represent
use
is
a
table
distance
vertical
Y
ou
the
the
on
to
r un
traveled
4
that
cer tain
and
shows
the
distances.
time
taken
a
Is
amount
the
of
time
relationship
function?
test
relations
ver tical
function
page
or
and
line
not,
test
by
functions
to
on
a
determine
drawing
Car tesian
whether
ver tical
lines
a
plane.
par ticular
across
the
Car tesian
graph.
and
the
plane
➔
A
relation
intersect
line
is
the
a
function
graph
if
more
any
than
ver tical
once.
line
This
drawn
is
called
will
not
after
the vertical
René
test
Example
Which
of
a
coordinates
Car tesian
are
named
Frenchman
Descar tes
(1596 – 1650).
these
relations
y
are
functions?
b
y
y
c
y
=
|x|
0
0
x
0
{
Functions
x
x
Continued
on
next
page
Answers
a
y
b
c
y
y
Crosses
0
a
A
function
Exercise
1
0
x
Which
A
b
x
x
0
function
Not
c
a
twice
function
B
of
these
relations
a
are
functions?
b
y
c
y
y
Draw,
or
3
imagine,
2
ver tical
1
lines
0
x
x
0
x
on
the
–1
graph.
d
e
y
f
y
If
the
a
‘solid
function
has
y
2
indicates
1
value
x
0
dot’
0
x
0
is
•,
that
this
the
included
in
x
2
the
function.
–1
If
the
function
has
a
–2
‘hollow
dot’
,
this
°
indicates
value
the
g
y
h
is
that
not
the
included
in
function.
i
y
y
3
2
1
0
x
1
–1
–2
2
3
4
5
2
2
1
1
0
–4
–3
–2
x
0
–1
x
1
–1
–1
–2
–2
Chapter
Use
2
your
GDC
to
sketch
these
straight
line
graphs.
Indicate
a
y
=
x
e
Are
f
Will
y
b
=
x
+
2
y
c
=
2x
−
3
y
d
=
x-
they
all
all
functions?
straight
lines
Explain
be
your
functions?
and /or
the
region
y
<
3x
−
2.
Is
this
a
function?
2
Use
not
an
a
algebraic
method
to
show
that x
of
view
y
=
on
a
and
Cartesian
range
on
your
sketch.
of
your
your
GDC,
graph
aim
near
to
the
have
the
corners
of
window.
4 is
T
ry
domain
the
2
+
function.
The
y-axis
using
substituting
values
.
crosses
Why?
the
4
line
Why?
ends
Sketch
the
answer.
When
3
where
4
a
of
positive
and
negative
x.
relation
plane
R
Y
ou
can
often
write
the
domain
and
range
of
a
relation
using
A
inter val
notation. This
is
another
method
of
writing
down
a
set
of
D O M A I
N
G
numbers.
For
example,
for
the
set
of
numbers
that
are
all
less
than
3,
E
you
can
write
the
inequality
x
<
3,
where
x
is
a
number
in
the
set.
[
In
inter val
Inter val
notation,
notation
this
uses
set
only
of
numbers
five
is
written
(–∞, 3)
A
function
domain
x-values)
symbols.
range
maps
the
(horizontal,
onto
the
(vertical,
y-values)
Brackets
(
)
Square
[
]
brackets
How
Innity
many
numbers
∞
are
Negative
innity
−∞
Union
there
in
sequence
3,
∪
4,…
if
the
0,
we
1,
2,
go
on
forever?
How
To
use
inter val
are
➔
Use
the
round
many
brackets
(
,
)
if
the
value
is
not
included
in
there
as
point
Use
(a
in
(–∞,
hole
the
or
square
3)
or
when
the
asymptote,
brackets
[
,
]
or
if
graph
a
is
undefined
at
in
the
the
sequence
graph
numbers
notation:
0,
0.5,
1,
that
1.5,
2,
4,…
if
2.5,
3,
3.5,
jump).
the
value
is
par t
of
the
we
go
on
graph.
forever?
Whenever
the
point.
there
Then
Put
a
For
example
If
graph
a
(−∞.
union
If
break
on
on
3)
in
the
another
between
(–∞,
goes
a
write
sign
goes
it
is
∪
values,
inter val
each
write
for
inter val
to
the
the
inter val
values
‘join’
after
them
up
to
that
point.
together.
(4, ∞)
forever
to
the
left,
forever
to
the
right
the
domain
then
the
(x-values)
domain
starts
ends
with
with
∞).
Why
If
a
graph
travels
downward
forever,
the
range
(y-values)
star ts
do
we
with
undened?
(−∞.
And
Usually
the
x-
or
we
a
graph
use
y-axis.
numbers.
if
For
Functions
goes
inter val
forever,
notation
However,
example,
up
in
you
to
can
inter val
then
the
describe
use
it
to
a
range
set
of
describe
notation x
≥
6
is
ends
with
values
any
along
group
[6, ∞).
∞).
of
call
innity
y
Asymptotes
8
Asymptotes
are
visible
on
your
GDC
for
some
functions.
An
1
6
y
asymptote
is
a
line
that
a
graph
approaches,
but
does
not
example,
in
the
graph
of
y
=
x
2
1
For
,
the
line
approaches
x
x
0
–6
the
x-axis
will
The
not
(y
=
0),
actually
x-axis
or
y
but
never
reach
=
0
is
y
=
touches
0,
callled
but
the
=
4
intersect.
it.
will
As
we
always
horizontal
go
get
to
infinity
closer
and
the
2
4
line
closer.
asymptote.
–8
The
y-axis
There
on
will
or
x
be
rational
=
a
0
is
the
more
ver tical
in-depth
asymptote
treatment
of
for
the
same
asymptotes
reasons.
in
the
chapter
functions.
F inding
asymptotes
looking
at
is
called
the
by
graph
locating
asymptotes
by
inspection.
Example
Identify
the
horizontal
and
ver tical
asymptotes
for
2x
x
these
functions
if
they
exist.
+ 2
x
a
y
=
2
y
b
=
y
c
x
=
(
+1
x
) (
+1
x
2
)
Answers
y
a
4
Horizontal
As
3
asymptote
y
=
we
go
along
the
x-axis
to
the
0
left
the
cur ve
gets
closer
but
never
2
actually
meets
the
x-axis.
1
0
–2.5
–2
–1.5
–1
–0.5
x
0.5
1
1.5
2
2.5
y
b
8
Horizontal
6
asymptote
y
=
2
4
Ver tical
2
x
=
asympote
–1
x
–3
–2
–1
1
2
4
–8
y
c
6
Horizontal
4
asymptote
2
Ver tical
x
0
–2
–1
=
–1
y
=
0
asympote
and
x
=
2
x
2
4
–4
Chapter
Exercise
Identify
they
C
the
horizontal
and
ver tical
asympotes
3
functions,
if
4
x
=
3
y
2
=
y
3
=
x
2x
y
4
these
exist.
y
1
for
x
2x
=
y
5
+1
+1
=
y
6
=
2
x
Set
In
+ 2
x
builder
set
builder
x
1
9
notation
notation
we
use
curly
brackets
{
}
and
variables
to
Y
ou
express
the
domain
and
range.
We
can
compile
sets
of
to
using
inequality
and
other
may
wish
inequalities
explore
the
symbols.
‘internationalism’
of
the
set
of
{
symbols
the
less
than
in
}
language
of
<
mathematics.
less
than
or
greater
than
greater
than
is
a
equal
to
≤
>
or
member
of
equal
the
to
set
≥
of
real
numbers
∈
➔
Set
notation:
{
x
:
x
>
6
}
Inter val
often
notation
considered
efcient
The
set
of
x-values
such
that
x
is
less
than
Set
Description
builder
Around
notation
notation
are
+∞)
x
is
greater
than
–2
{x : x
>
the
4]
x
is
less
x
lies
than
or
equal
to
4
{x : x
≤
3)
between
including
(–∞,
5)
∪
[6,
x
+∞)
is
less
than
or
−3
−3
but
than
5
equal
to
and
not
or
3
{x : −3
≤
x
<
3}
3
for
greater
{x : x
<
5,
x
≥
6}
+∞)
x
may
be
any
real
also
also
How
called
people
brackets’
than.
x
>
2,
For
to
use
show
‘backwards
greater
number
example:
and
]
Functions
∞,
−4[
]
2,
is
∞
square
than
[
is
or
less
equivalent
equivalent
to
x
<
Radicals
called
does
this
x
∈
nd
some
examples?
Some
same
surds.
affect
understanding?
you
(–∞,
the
Brackets
parentheses.
are
6
to
−4.
there
4}
are
[−3,
world
different
−2}
symbol.
(–∞,
more
set
notation
many
words
(–2,
than
6
builder
Inter val
is
Can
more
Example
Y
ou
Find
the
domain
and
range
of
this
may
wish
to
function.
explore
the
inuence
y
of
technology
on
2.5
notation
and
2
vice
versa.
1.5
1
0.5
0
–4
–3
–2
x
–1
1
2
3
4
Answer
The
{x :
domain
x
The
{ y :
≥
−4}
range
y
≥
of
or
of
0}
[–4,
the
or
Example
the
[0,
function
is
x
+∞)
only
equal
function
is
The
+∞)
takes
to
values
than
or
−4.
function
greater
greater
than
only
or
takes
equal
to
y-values
0.
What
Find
the
domain
and
range
of
each
values
included
y
a
are
function.
b
in
the
domain
y
0
≤
x
≤
1?
4
2
How
many
values
are
3
1
there?
2
0
x
1
–1
–2
x
0
1
–1
–3
–4
Do
we
all
use
the
Answers
same
a
The
domain
and
0
is
{x : −2
≤
x
<
notation
mathematics?
<
x
≤
[−2, −1)
∪
an
range
is
{y : −4
<
y
≤
(−4,
that
The
domain
∈
of
the
function
is
x
can
take
any
real
or
(–∞,
range
≥
−1
is
have
notations
of
−3}
to
+∞).
the
function
or
[–3,
the
same
is
thing.
{y : y
=
Different
value.
represent
The
to
1].
different
x
x
included.
countries
b
are
dot
1}
not
or
empty
(0, 3].
indicate
The
We
3}
using
or
in
−1
Fur thermore,
+∞).
different
teachers
from
same
the
countr y
use
different
notations!
Chapter
Exercise
1
Look
D
back
numbers
a
2
to
of
the
4
at
the
handshakes
function?
Find
page
If
so,
domain
a
what
and
graph
for
is
various
the
range
and
formula
numbers
domain
for
the
each
and
of
the
people.
Is
this
range?
these
b
y
of
for
relations.
c
y
E
4
4
3
3
2
2
y
1
0.5
1
1
F
x
0
–4
–3
–2
–1
1
2
3
0
x
0
4
1
2
3
–1
6
5
x
–0.5
0.5
1
y
d
e
f
6
y
4
4
2
2
–2
–4
0
x
0
–6
2
y
4
x
0
x
–2
–2
–2
–1
–4
–6
g
h
y
i
y
y
5
5
4
4
3
3
1
0
x
2
1
1
1
0
–2
–1
0
x
1
–5
2
–4
x
–3
1
3
4
5
–1
–2
–2
–3
–3
–4
Exam-Style
3
Use
your
Question
GDC
to
sketch
these
graphs.
Y
our
Write
down
the
domain
and
range
of
GDC
will
nd
the
x-
and
y-intercepts.
T
o
do
each.
this
algebraically,
use
the
fact
that
a
function
2
a
y
=
2x
−
3
b
y
=
x
crosses
2
c
y
=
x
x-axis
when
y
=
0
and
crosses
the
3
+
5x
+
6
d
y
=
f
x
–
4
y-axis
e
the
when
x
=
0.
For
example,
the
function
y
=
2x
−
4
crosses
the x-axis
where
2x − 4 = 0,
x
g
h
y
=
j
x = 2. It
e
crosses
the
y-axis
where
y
=
2(0)
−
i
3k
k
l
Functions
gives
a
most
unusual
answer
.
carefully
for
a
hole
where
x
=
−3.
Look
4
=
−4.
.
Function
Functions
equation
are
y
=
function
the
as
2x
f
f (x)
=
notation
often
2x
+
described
1
describes
symbol
+
➔
f (x)
is
(x)
can
also
1
‘f
and
read
as
be
‘f
’ we
so
y
of
x’
written
by
y
write
=
equations.
as
a
For
function
this
of
equation
example,
x.
in
By
the
giving
function
the
notation
f (x).
and
like
means
this:
f
:
the
x
→
value
2x
+
of
function
f
at
x.
1.
f
An
ordered
Finding
pair
f (x)
function
f
for
at
Example
(x, y)
a
can
be
par ticular
that
written
value
as
of
x
(x,
:
(x)
→
2x
+
1
means
evaluating
that
f
is
a
that
maps
function
the
x
to
f
(x)
1.
German
mathematician
Evaluate
If
+
the
function
f
(x)
=
2x
+
1
at
x
=
=
philosopher
x
+
4x
−
3,
find
f
i
(2)
f
ii
and
3.
2
b
2x
value.
The
a
means
f (x)).
(0)
iii
f
(−3)
iv
f
(x
+
Gottfried
1)
Leibniz
rst
used
mathematical
the
term
Answers
a
f
(3)
=
2(3)
+
1
=
‘function’
7
For
x,
substitute
in
1673.
3.
2
b
f
i
(2)
=
(2)
+
4(2)
–
3
=
4
+
8
−
3
=
9
2
ii
f
(0)
iii
f
(−3)
=
iv
f
(x
(0)
+
4(0)
–
3
=
0
+
0
−
3
=
−3
2
=
=
(−3)
9
−
+
12
4(−3)
−
3=
–
3
−6
2
+
1)
=
(x
+
1)
+
4(x
+
1)
–
3
+
4
−
2
=
x
=
x
+
2x
+
1
+
4x
3
2
Exercise
+
6x
+
2
E
1
Find
f (7)
i
ii
f (−3)
f
iii
(
)
iv
f (0)
f (a)
v
for
these
functions.
a
f (x)
=
x
−
2
f (x)
b
=
3x
c
f (x)
=
x
2
d
f (x)
=
2x
+
5
f (x)
e
=
x
+
2
2
2
If
a
f (x)
=
x
–
4,
find
f (−a)
b
f (a
+
5)
e
f (5
−
a)
c
f (a
−
1)
2
d
f (a
−
2)
Notice
Exam-Style
3
If
g (x)
=
that
we
do
not
Question
4x
−
5
and
h (x)
=
a
find
x
when
g (x)
=
3
b
find
x
when
h (x)
=
−15
c
find
x
when
g (x)
=
7
–
always
use
letter
for
the
2x
Here
g
h (x).
f
we
and
h.
a
have
of
used
When
considering
terms
function.
velocity
time
we
in
often
4
a
If
h (x)
=
find
b
Is
there
a
use
h (−3).
v(t).
value
where
h (x)
does
not
exist?
Explain.
Chapter
The
5
volume
of
a
cube
with
edges
of
length x
is
Y
ou
can
use
mathematical
functions
3
given
by
the
function
f (x)
=
x
to
a
Find
f (5).
b
Explain
represent
For
what
f (5)
represents.
example,
pizzas
the
number
from
suppose
family
of
watch.
the
eats
football
If
life.
number
depends
games
of
on
you
eat
3
pizzas
you
during
ever y
football
g (6)
i
g (−2)
ii
g (0)
iii
game,
the
function
‘number
of
pizzas’
(p)
‘number
of
football
=
3
would
be
times
g
iv
games’
(g)
or
p
=
3g.
Can
you
think
of
another
Evaluate
b
real-life
g (1)
i
g (1.5)
ii
g (1.9)
iii
g (1.999)
v
What
do
d
Is
e
Graph
there
when
you
a
x
=
Exam-Style
notice
value
the
of
x
about
for
function
2.
the
It
amount
could
of
on
your
which
your
answers
g (x)
GDC
does
and
to b?
not
look
the
number
talking
on
the
of
minutes
you
spend
phone.
what
happens
Explain.
The
of
a
par ticle
is
given
by v (t)
=
initial
a
Find
the
initial
b
Find
the
velocity
after
4
c
Find
the
velocity
after
10
d
At
−
9 m s
.
star t,
does
at
the
par ticle
when
t
=
0.
seconds.
come
to
par ticle
comes
when
v
=
0.
find
Extension
material
Worksheet
1
on
CD:
f
a
.
A
(2
+
h)
Composite
composi te
one
function
➔
f
b
The
the
written
as
is
a
result
composition
of
f (g (x)),
+
h)
functions
function
to
(3
combination
of
of
two
functions.
Y
ou
function
is
apply
another.
the
function
which
is
read
f
with
as
‘f
of
the
g
of
x’,
or
g
( f
g)(x),
°
which
When
you
another
For
is
read
as
evaluate
variable
example,
if
for
f (x)
‘f
a
composed
function
with
f (x),
you
2x
+
3
then
f (5)
2
Y
ou
can
find
2
➔
A
1)
=
2(x
+
1)
by
substituting
+
1)
and
+
3
=
2x
function
is
defined
+
Functions
substitute
a
number
or
=
2(5)
+
3
=
13
x
+
1
for
x
to
get
5
applies
by
( f
one
g)(x)
°
x’.
2
composi te
another
of
2
2
+
f (x
f (x
g
x.
=
to
rest?
rest
velocity
seconds.
The
time
the
velocity
.
the
what
velocity
−1
t
means
be
spend
Question
velocity
Given
you
exist?
2
The
perhaps
money
g (1.9999)
vi
or
c
function?
g (1.99)
iv
about
8
own
Evaluate
a
7
your
6
your
things
function
=
f ( g(x)).
to
the
result
of
-
Polynomials
Example
2
If
f
(x)
=
5
−
3x
and
g (x)
=
x
+
4,
find
(f
g)(x).
°
g (x)
goes
in
here
Answer
2
2
(f
g)(x)
=
5
–
3(x
=
5
–
3x
+
Substitute
4)
x
+
4
into
f
(x).
°
2
–
12
2
=
Y
ou
may
value
of
−3x
need
–
7
to
evaluate
a
composite
function
for
a
par ticular
x.
Example
2
f
(x)
=
5
−
3x
and
g (x)
=
x
+
4.
Find
(f
g)(3).
°
Answer
Method
1
2
(f
g)(x)
=
5
–
3(x
+
Work
4)
out
the
composite
function.
°
2
=
−3x
–
7
2
(f
g)(3)
=
–3(3)
–
=
−27
7
=
−34
7
Then
°
–
substitute
3
for
x.
Both
Method
methods
give
2
the
same
result
–
you
2
g (3)
f
=
(13)
(3)
=
5
+
–
Example
4
=
13
3(13)
=
−34
Substitute
3
into
Substitute
that
g (x).
value
can
into
f
(x).
use
the
one
you
prefer
.
2
Given
a
(f
f
(x)
=
2x
+
1
and
g)(x)
g (x)
(f
b
=
x
–
2,
find
g)(4)
°
°
Answers
2
2
a
(f
g)(x)
=
2(x
–
2)
+
Substitute
1
x
–
2
into
f
(x).
Or
°
use
Method
2:
2
=
2x
−
2
3
g (4)
and
=
(4)
–
2
=
14
then
2
b
(f
g)(4)
=
2(4)
–
3
=
29
Substitute
°
Exercise
4
for
x.
f (14)
=
2(14)
+
1
=
29
F
2
1
Given
a
( f
f (x)
=
g)(3)
3x,
g (x)
b
=
( f
°
e
( g
f
( f
)(4)
f
h (x)
c
=
j
n
)(5)
)(2)
find
g)(–6)
d
( f
g
( g
k
f
( f
o
)(–6)
h
( g
)(x)
l
(h
f
)(x)
°
h)(x)
°
f
°
h)(x)
( g
g)(x)
°
°
g)(3)
°
2,
°
f
(h
+
°
f
(h
x
( f
°
h)(3)
°
and
°
h)(2)
( g
1
g)(0)
( g
°
m
+
°
°
i
x
p
(h
(f
°
h)(2)
≠
(h
°
f)(2)
g)(x)
°
Chapter
2
Given
2
f (x)
( g
a
f
=
x
−
)(1)
1
and
( g
b
f
°
=
3
−
)(2)
x,
find
( g
c
f
°
( g
e
g (x)
f
)(3)
( f
f
g)(–4)
°
( f
g
( f
d
g)(3)
°
g)(x
°
Exam-Style
)(4)
°
+
1)
( f
h
°
g)(x
+
2)
°
Questions
2
Given
3
the
( f
a
functions
g)(x)
f (x)
=
( f
b
x
and
g (x)
=
x
+
2
find
g)(3)
°
°
2
Given
4
(
a
the
f
functions
g )(x)
f (x)
( g
b
°
=
f
5x
and
g (x)
=
x
+
1
find
)(x)
°
2
g (x)
5
=
x
+
Find
a
3
( g
and
h (x)
=
x
–
4
‘Hence’
h)(x).
means
‘Use
the
work
to
preceding
°
Find
b
(h
g)(x).
obtain
the
°
Hence
c
solve
the
equation
( g
h)(x)
=
(h
°
required
g)(x).
result’.
°
2
If
6
r (x)
=
x
–
4
and
s (x)
=
x
,
find
(r
s)(x)
and
state
its
domain
°
and
.
range.
Inverse
functions
–1
➔
The
inverse
of
that
function.
a
function
If
f (x)
=
3x
−
4
and
g (x)
f (x)
is
f
(x).
It
reverses
the
action
of
=
,
then
f (10)
=
3(10)
–
4
=
26
and
g ( 26)
=
=
10,
so
we
are
back
to
(f
°
g)
(10)
=
10
where
So
g (x)
Not
If
g
all
we
all
is
star ted.
is
the
inverse
functions
the
inverse
values
in
the
of
have
f (x).
an
inverse.
function
domain
of
of
f
f,
−1
Note
then
and
f
g
will
will
reverse
also
be
the
the
action
inverse
of
of
f
that
the
inverse
the
‘–1’
is
➔
and
g
are
Functions
(
f
inverse
f (x)
g)(x)
and
functions,
g (x)
are
we
write
inverses
of
g (x)
one
=
f
(x).
another
if:
=
x
for
all
of
the
x-values
in
the
domain
of
g
=
x
for
all
of
the
x-values
in
the
domain
of
f.
°
( g
f
)(x)
°
The
➔
horizontal
Y
ou
can
use
line
the
test
horizontal
line
test
to
identity
inverse
functions.
If
a
horizontal
once,
there
Functions
is
line
no
crosses
inverse
the
graph
function.
of
a
function
of
not
f;
an
g.
exponent
f
means
for
−1
When
f
more
than
(power).
Example
Which
of
these
functions
have
a
inverse
functions?
y
b
5
5
4
4
3
3
2
2
1
1
0
0
–3
–2
x
x
–1
1
2
1
3
3
–1
–2
–3
–4
–5
y
c
y
d
3
2
2
1
1
x
0
2
1
3
4
5
6
x
7
–1
–1
–2
–2
–3
–4
Answers
a
y
b
y
5
5
4
4
3
2
2
1
1
–3
–2
0
–1
2
1
3
x
–3
4
you
Wafa
0
x
–4
Did
know
Buzjani,
that
a
mathematician
Abul
Persian
from
1
–1
the
No
inverse
10th
centur y,
used
function
functions?
There
is
–3
a
crater
on
the
moon
–4
named
–5
Inverse
y
c
after
him.
function
y
d
3
3
2
2
1
1
0
x
1
2
3
4
5
6
x
7
1
–2
–3
Inverse
function
No
inverse
function
Chapter
The
➔
graphs
The
graph
function
Here
are
of
of
in
some
inverse
the
the
inverse
line
y
examples
=
of
functions
of
a
function
is
a
reflection
of
that
x
functions
and
their
inverse
y
=
functions.
x
y
y
y
1
f
(x)
y
=
y
x
f (x)
x
1
–1
f
=
(x)
f
(x)
x
f (x)
x
x
f (x)
−1
If
(x, y)
lies
function
in
the
on
in
line
the
y
Exercise
1
Use
the
have
the
=
line
line
x
y
f (x),
=
x
becomes
then
(y,
‘swaps’
point
x
x)
lies
and
y,
on
so
f
(x).
the
Reflecting
point
(1, 3)
the
reflected
(3, 1).
G
horizontal
inverse
line
test
to
determine
which
of
these
functions
functions.
y
y
a
b
Pioneering
7
Indian
6
6
5
5
in
the
scientist
6th
included
4
3
2
2
1
1
0
–5
–4
–3
–2
x
0
–1
–6
–3
–2
–1
x
1
2
3
4
–1
–2
a
–3
c
d
y
y
3
2
2
1
x
0
1
Functions
2
3
4
work
by
7
5
6
x
0
Panini
centur y
BCE
functions.
2
Copy
the
graphs
line
=
and
y
x
a
of
the
these
graph
functions.
of
the
For
inverse
each,
the
y
b
y
draw
function.
c
y
10
8
8
8
6
6
6
4
4
4
2
2
0
x
0
0
x
x
4
2
–2
–2
–2
–4
–4
–4
–6
–6
–6
–8
–8
–10
y
d
e
y
f
y
8
6
8
4
4
6
2
4
x
0
–4
2
–4
0
–1
x
1
2
3
4
–2
x
–3
–2
–1
0
1
2
3
–4
Finding
Look
x
on
at
inverse
how
the
the
functions
function
f (x)
=
3x
3x
form
inverse
–
2
is
made
up.
We
star t
with
left.
x
To
algebraically
the
inverse
function
we
–
2
reverse
the
process,
using
operations.
The
x
+
inverse
of
+2
is
2
2
x
The
3
So
inverse
of
×3
is
÷3
The
next
example
Example
shows
you
how
to
do
this
without
diagrams.
–1
If
f
(x)
=
3x
–
2,
find
the
inverse
function
f
(x).
Answer
y
=
3x
–
2
x
=
3y
–
2
Replace
f (x)
Replace
ever y
ever y
x
+
2
=
y
=
y
with
with
x
y.
with
y
and
x.
3y
Make
x
y
the
subject.
+ 2
3
x
1
f
(x )
+
2
1
Replace
=
y
with
f
(x).
3
Chapter
As
you
saw
function
in
the
line
swapped
➔
To
in
of
a
y
x
=
x,
and
find
and
the
and
functions
f
is
‘swaps’
then
inverse
for
of
function
which
y,
the
solve
Example
graphs
given
the
x
and
made
function
y
and
their
reflection
y.
the
So
in
inverses,
of
the
the
graph
Example
12
y
inverse
=
f
(x)
we
subject.
algebraically
,
replace f (x)
with
y
y
−1
If
f
(x)
=
4
−
3x,
find
f
(x)
Answer
x
−
y
=
4
−
3x
x
=
4
−
3y
=
−3y
4
x
Replace
f (x)
Replace
ever y
with
with
x
y.
with
y
and
ever y
y
x.
4
Make
=
y
the
subject.
y
3
4
y
x
=
3
4
x
1
f
(x )
–1
=
Replace
y
with
f
(x).
3
To
check
the
that
the
inverse
function
Example
13
is
correct,
combine
(f
f
–1
So
in
functions
−1
)(x)
=
x
and
f
and
f
are
inverses
of
each
other.
°
➔
The
It
function
leaves
x
I (x)
=
x
is
called
the
identity
function.
unchanged.
−1
So
f
=
f
I
°
Exercise
H
Exam-Style
1
If
f (x)
Question
=
and
g (x)
=
2x
–
4,
ii
f
find
a
g (1)
i
and
( f
g)(1)
(–3)
and
( g
f
°
( f
iii
g)(x)
iv
( g
°
b
2
What
Find
the
)(–3)
°
f
)(x)
°
does
this
inverse
tell
for
you
each
about
of
functions f
these
and
g?
functions.
3
a
f
(x)
=
3x
−
1
b
g (x)
=
x
–
2
c
h (x)
=
x
+
5
+
3
d
f
(x)
=
e
g (x)
=
Self-inverse
3
–
2
f
h (x)
=
2x
functions
are
g
such
h
function
Look
–1
is
f
(x)
f
(x)
=
1
–
x
b
f
(x)
=
x
c
f
(x)
Functions
for
functions
=
are
its
the
same.
self-inverse
if
a
and
inverse
What
a
3
that
in
question
3.
–1
Evaluate
4
f
(5)
where
f
a
(x)
=
6
–
x
f
b
(x)
=
f
c
(x)
=
Note
of
If
5
f
(x)
that
point
image
(a, –b)
after
a
–1
=
,
find
f
(x).
reection
the
in
the
line
y
Exam-Style
=
x
is
the
point
Question
(b, –a).
x
6
Draw
a
the
plotting
Draw
b
graph
several
the
line
of
f (x)
=
2
by
making
a
table
of
values
and
points.
y
=
x
on
the
same
graph.
–1
c
Draw
the
in
line
the
graph
y
=
of
f
by
reflecting
the
graph
of
f
x.
–1
State
d
the
domain
and
range
of
f
and
f
2
The
7
the
function
square
Find
By
this
f
(x)
root
=
x
has
no
inverse
function
function.
does
have
However,
an
inverse
function.
inverse.
comparing
the
range
and
domain
explain
why
the
inverse
2
of
Prove
8
never
.
that
be
is
the
not
the
graphs
a
as
f (x)
linear
=
x
function
and
its
inverse
can
per pendicular.
Investigation
1
of
Transforming
Y
ou
same
should
Sketch
use
y
=
your
x,
y
=
–
+
material
Worksheet
1
the
same
to
1,
y
sketch
=
x
−
all
4,
y
the
=
x
graphs
+
in
this
4
will
also
and
on
2
effect
the
contrast
y
=
−2x
y
3
=
+
Compare
What
x
3,
y
Compare
=
the
of
+
y
and
effect
Sketch
do
graphs
Sketch
your
y
3,
=
constant
=
y
x
=
+
2x
0.5x
+
contrast
does
|x|,
and
effect
y
=
|x
+
terms
does
graphs
of
y
=
+
3
3,
on
your
2|,
y
your
changing
2
4
Sketch
y
Compare
=
x
,
and
|x
+
=
−x
equation
written
as
have
y
=
mx
+
b
y
=
mx
+
c
or
y
=
the
3x
+
3,
same
The
axes.
is
functions.
the
=
x-coefcient
|x
−
3|
on
have?
the
same
coefcient
the
number
of
x
that
multiplies
the
x-value.
|x|
the
modulus
axes.
means
functions.
the
values
of
h
have
x.
See
chapter
18
on
more
explanation.
h|?
2
y
line
b?
changing
contrast
(number)
a
for
the
this
functions.
of
What
nd
axes.
of
What
CD:
investigation.
standard
Compare
on
Polynomials
functions
Y
ou
on
-
functions
GDC
x
Extension
2
,
contrast
y
=
2x
your
What
effect
does
the
What
effect
does
changing
,
2
y
=
0.5x
on
the
same
axes.
functions.
negative
sign
have
on
the
graph?
2
the
value
of
a
have
on
the
graphs
of
y
=
ax
?
Chapter
In
the
1,
2
investigation
and
3
changed.
by
were
The
all
you
the
graphs
should
same
in
have
shape
par t
4
found
but
should
the
that
your
position
have
been
graphs
of
the
in
par ts
graphs
reflected
or
changed
stretching.
These
look
are
at
examples
these
of
‘transformations’
transformations
in
of
graphs.
We
will
now
detail.
Translations
Shift
➔
f
upward
(x)
+
k
downward
translates
ver tically
units
or
a
f (x)
distance
➔
of
k
f
(x)
–
k
translates
ver tically
upward.
units
y
a
f (x)
distance
of
k
downward.
y
3
3
f (x)
+
1
2
2
f (x)
1
1
f (x)
x
–2
–1
x
–2
–1
1
2
1
3
–1
3
f (x)
Shift
➔
to
f (x
the
+
k)
right
or
k
left,
>
when
k
f (x)
units
to
➔
the
f (x
−
right,
y
translates
when
k
k
f (x)
units
>
to
the
0.
y
3
2
2
f (x)
f (x)
1
+
k)
horizontally
0.
3
f (x
1
left
translates
horizontally
–
1
2)
x
–1
1
2
x
3
–1
1
3
4
5
–1
f (x
–
2)
Translations
can
be
represented
by
vectors
in
the
form
is
the
horizontal
component
and b
is
the
vertical
where
a
T
ry
component.
transforming
some
functions
different
is
a
horizontal
shift
of
2
units
units
by
the
ver tical
shift
of
on
right,
Functions
and
a
denotes
vector
the
a
down.
to
is
ver tical
a
horizontal
shift
of
2
units
values
right.
Translation
3
down.
shift
of
3units
your
GDC.
with
of
k
Reflections
Reflection
➔
−f (x)
in
the
x-axis
reflects
f (x)
Reflection
in
the
➔
f (−x)
x-axis.
in
the y-axis
reflects
f (x)
in
the
y-axis.
y
y
3
3
f (x)
2
f (x)
1
x
0
–2
–1
1
2
x
3
–3
–2
–1
1
f (x)
2
3
f (–x)
Stretches
Horizontal
stretch
(or
compress)
Ver tical
stretch
(or
compress)
A
➔
f (qx)
stretches
or
➔
compresses
pf (x)
stretches
f
stretch
factor
f (x)
horizontally
with
ver tically
scale
with
0
factor
p
scale
<
p
<
1
will
actually
p.
compress
y
a
where
scale
factor
with
(x)
the
graph.
y
f (2x)
f (x)
3
3
2
2
1
1
2f (x)
0
–3
–2
x
–1
1
2
x
3
–2
–1
–1
1
3
–1
f (x)
Students
make
The
transformation
is
a
horizontal
The
transformation
is
often
mistakes
a vertical
with
stretches.
stretch
of
scale
stretch
factor
of
scale
factor
p
It
When
When
q
>
1
the
graph
towards
the
0
<
q
<
1
the
1
away
graph
Example
1
Given
sketch
a
f
(x
from
graph
the
the
impor tant
stretches
to
from
remember
0
<
p
<
the
x-axis.
different
1
the
graph
for
towards
the
effects
of,
is
is
compressed
stretched
the
y-axis
When
When
>
is
away
compressed
p
is
example,
2f (x)
x-axis.
and
y-axis.
f (2x).
the
the
+
graph
of
graphs
1)
b
f
the
function
f
(x)
shown
y
here,
of:
(x)
4
−
2
c
f
(−x)
d
−f
(x)
e
2f
(x)
3
2
f (x)
1
x
0
1
{
2
3
4
Continued
5
on
6
next
Chapter
page
Answers
y
a
b
y
c
y
4
4
2
3
3
1
f
f
(x)
–
(–x)
2
2
2
0
f (x
+
1)
x
1
2
3
4
5
6
1
1
–1
–2
x
–1
1
0
2
3
Translated
to
the
4
one
–6
unit
Translated
two
units
down
–5
–4
–3
Reflected
–2
in
–1
0
the y-axis
left
y
d
x
5
y
e
1
12
10
0
x
2
1
3
4
5
6
8
–1
6
–2
f (x)
2f (x)
–3
4
–4
2
x
0
1
Reflected
Supply
and
and
in
the
demand
economics
Supply
are
and
x-axis
cur ves
in
Ver tical
business
reections.
Radioactive
3
2
4
stretch
decay
5
of
6
scale
cur ves
factor
2
are
reections.
demand
y
6
Demand
Supply
y
Number
of
5
100
smota
P
Surplus
4
P
*
fo
ECIRP
3
Equilibrium
rebmuN
2
Shor tage
1
Q
*
75
50
25
x
0
10
20
30
40
QUANTITY
50
Number
parent
of
atoms
0
60
x
1
2
3
4
5
6
Q
Number
of
half
lives
y
Exercise
I
f (x)
4
Exam-Style
1
Copy
the
Question
graph.
Draw
2
these
functions
on
the
same
axes.
–6
a
f (x)
+
d
f (x
g
f (2x)
+
4
b
f (x)
3)
e
f (x
–
−
2
c
4)
f
–2
f (x)
x
0
2
4
6
–2
2f (x)
–4
y
g
f (x)
q
4
2
Functions
g,
h
and
q
are
transformations
of
f (x).
2
Write
each
transformation
in
terms
of
f (x).
h
0
–10
–8
–6
–4
–2
–2
–4
Functions
x
2
4
6
8
10
3
Functions
Write
q,
each
s
and
t
are
transformations
transformation
in
terms
of
of
y
f (x).
f (x).
6
t
f (x)
s
4
q
2
x
0
–10
–8
–6
2
4
6
8
–2
Exam-Style
4
Copy
the
Question
graph
of
f (x).
Sketch
the
graph
of
each
of
y
these
3
functions,
and
state
the
domain
and
range
for
each.
f (x)
2
a
2f (x
–
5)
1
f (2x)
b
+
3
0
x
2
–1
–2
5
The
graph
of
f (x)
is
shown.
A
is
the
point
y
(1, 1).
5
Make
separate
copies
of
the
graph
and
draw
the
4
function
after
each
On
graph,
transformation.
3
each
label
the
new
position
of
A
as
A
2
1
A
a
f (x
+
c
f (–x)
1)
b
f (x)
+
d
2f (x)
1
1
0
–4
e
f (x
−
2)
+
x
1
2
3
4
5
3
–4
–5
6
In
each
change
case,
the
describe
graph
of
the
f (x)
3
a
f (x)
=
x
transformation
into
f (x)
=
x
c
f (x)
=
x,
Exam-Style
7
Let
f (x)
=
graph
of
would
g (x).
3
,
g (x)
=
−(x
,
g (x)
=
(x
)
2
b
the
that
2
g (x)
=
−
−2x
3)
+
5
Question
2x
+
If
1.
in
a
Draw
b
Let
the
graph
of
f (x)
for
0
≤
x
≤
a
domain
the
=
f (x
+
3)
–
2.
On
the
same
graph
draw
−3
≤
x
≤
you
only
draw
the
g (x)
function
for
given
2.
must
g (x)
is
question,
for
that
−1.
domain.
Review
exercise
✗
1
a
If
g (a)
=
4a
−
5,
find
g (a
−
2).
b
If
h (x)
=
,
find
h (1
−
x).
2
2
a
Evaluate
f (x
−
3)
when
f (x)
=
2x
−
3x
+1.
2
b
For
f (x)
=
function
2x
+
7
defined
and
by
g (x)
( f
=
1
−
x
,
find
the
composite
g)(x).
°
Chapter
3
Find
the
inverses
a
f (x)
of
these
functions.
=
3
b
g (x)
=
2x
+
3
4
Find
the
inverse
of
f (x)
=
.
Then
graph
the
function
and
5
its
Find
inverse.
the
inverse
functions
for
a
6
f (x)
Copy
=
3x
each
a
+
5
b
graph
and
draw
the
inverse
of
each
function.
b
y
y
4
3
3
2
2
1
1
x
0
x
1
–2
–1
0
1
2
3
–1
–2
–3
–4
7
Find
the
domain
and
range
for
a
each
of
these
graphs.
b
y
y
10
7.5
5
2.5
5
x
–2
6
7
–2.5
0
x
–5
5
–1
–5
–7.5
Exam-Style
8
For
each
given
a
Question
function,
write
combination
f (x)
=
x,
of
reflected
a
single
equation
to
represent
the
transformations.
in
the
y-axis,
stretched
ver tically
by
a
factor
of
2,
horizontally
by
a
factor
and
of
translated
3
units
left
and
2
units
up.
2
b
f (x)
=
x
,
reflected
in
the
x-axis,
stretched
ver tically
by
a
factor
,
of
horizontally
by
a
factor
of
3,
translated
5
units
right
9
and
a
Explain
b
Graph
1
how
the
Exam-Style
unit
to
down.
draw
inverse
of
the
f (x)
inverse
=
2x
+
of
a
function
from
its
graph.
3.
Question
3
10
Let
f (x)
=
2x
+
3
and
g (x)
=
3x
–
2.
−1
a
Find
g (0).
b
Find
( f
g)(0).
°
Functions
c
Find
f
(x).
Exam-Style
y
Questions
4
11
The
graph
shows
the
function
f (x),
for
−2
≤
x
≤
4.
3
a
Let
h (x)
b
Let
g (x)
=
f (−x).
Sketch
the
graph
of
h (x).
=
f (x
−
1).
The
point
A(3,
2)
on
the
graph
of
f
is
1
transformed
to
the
point
P
on
the
graph
of
g.
0
–3
Find
12
The
and
a
the
coordinates
functions
g (x)
Find
=
x
an
f
+
and
g
of
are
defined
as
f
(x)
=
3x
The
2.
for
x
–1
instruction
‘Obtain
expression
–2
1
2
3
4
5
P
( f
g)
using
(x).
the
‘Show
required
information
that…’
result
given)
means
(possibly
without
the
°
−1
b
Show
that
f
−1
(12)
+
g
(12)
=
formality
14.
For
of
‘Show
proof ’.
that’
questions
you
do
not
13
Let
g (x)
=
2x
–
1,
h (x)
=
a
Find
an
expression
for
≠
usually
need
A
method
to
use
a
calculator
.
(h
g)
(x).
Simplify
good
is
to
cover
up
the
your
°
right-hand
side
then
out
of
the
equation
and
answer.
b
Solve
the
equation
(h
g)
(x)
=
work
the
left-hand
side
until
0.
°
your
answer
hand
Review
1
Use
your
and
the
same
as
the
right-
exercise
GDC
range
is
side.
of
to
sketch
2
Sketch
the
function
3
Sketch
the
function
the
function
and
state
the
domain
y
=
(x
+
1)(x
−
3)
and
state
its
domain
and
range.
1
y
and
=
x
Exam-Style
state
its
domain
and
range.
+ 2
Questions
4
The
function
f (x)
is
defined
as
a
Sketch
b
Use
the
your
and
the
cur ve
GDC
f (x)
to
for
help
−3
≤
you
x
≤
write
≠
−.
2.
down
the
value
of
the x-intercept
y-intercept.
5
a
Sketch
the
graph
of
b
For
what
c
State
the
value
of
domain
x
is
and
f (x)
undefined?
range
6
Given
the
function
write
b
sketch
down
c
write
the
the
Let
f (x)
down
=
a
Sketch
b
Solve
2
−
the
x
both
f (x)
equations
of
the
asymptotes
function
coordinates
2
7
f (x).
a
of
=
of
the
intercepts
with
both
axes.
2
and
g (x)
functions
=
x
on
−
2.
one
graph
with
−3
≤ x
≤
3.
g (x).
Chapter
Exam-Style
Questions
3
8
Let
f (x)
=
x
–
3.
−1
a
Find
the
inverse
function
f
(x).
−1
b
Sketch
c
Solve
both
f (x)
and
f
(x)
on
the
same
axes.
–1
f (x)
9
=
10
the
Consider
cur ve
the
≠
Sketch
(x).
f
of
f (x)
functions
for
f
−5
and
g
≤
x
≤
2,
where
including
f (x)
=
3x
–
any
2
asymptotes.
and
g (x)
=
x
–
3.
−1
a
Find
the
inverse
function,
f
−1
b
Given
that
g
−1
(x)
=
x
+
3,
find
( g
f
)(x).
°
When
−1
c
Show
that
(f
g)(x)
IB
exams
=
.
°
have
words
in
bold
–1
d
Solve
script,
–1
( f
g)(x)
=
( g
f
°
you
Let
it
means
that
)(x)
°
must
do
exactly
,
x
≠
2.
what
is
required.
For
example
d
Sketch
the
graph
of
h
for
−6
≤
x
≤
10
and
−4
≤
y
≤
any
Write
down
CHAPTER
the equations
1
Introducing
●
A
●
The
domain
●
The
range
●
A
function
●
A
relation
than
The
once.
Use
is
●
is
is
Set
set
the
the
is
is
a
a
of
ordered
set
set
of
of
all
the
relation
function
This
but
given
not
as
3.
asymptotes.
is
and
pairs.
the
if
called
any
the
range
first
second
where
round
brackets
square
at
that
( , )
if
point
brackets
[ , ]
numbers
numbers
ever y
vertical
of
x-value
ver tical
a
line
line
(x-values)
(y-values)
is
related
drawn
of
in
will
the
each
to
a
not
ordered
pairs.
pair.
unique
y-value.
intersect
the
graph
more
test
relation
on
a
Cartesian
plane
if
the
(a
value
hole
the
is
not
included
or asymptote,
value
is
included
or
in
in
a
the
graph
or
when
the
graph
jump).
the
graph.
notation:
:
The
set
of
x-values
x
<
6
}
such
that
x
is
less
than
6
Continued
as
just
notation:
undefined
Use
3
SUMMARY
a
domain
Interval
of the
=
functions
is
relation
be
asymptotes.
x
e
equation
10,
could
including
the
Functions
on
next
page
Function
●
f (x)
is
notation
read
as
Composite
●
The
as
‘ f
of
and
means
‘the
value
of
function
f
at
x’.
functions
composition
f (g (x)),
x’
which
of
is
the
function
read
as
‘f
of
f
g
with
of
the
x’,
or
function
( f
g)(x),
g
is
written
which
is
read
as
°
‘f
●
A
composed
with
composi te
another
and
g
of
x’.
applies
function
is
defined
by
( f
one
g)(x)
function
=
to
the
result
of
f ( g(x)).
°
Inverse
functions
−1
●
The
the
●
of
inverse
a
function
f (x)
is
f
(x).
It
reverses
the
action
of
function.
Functions
f (x)
( f
xfor
g)(x)
=
and
all
g (x)
of
are
the
inverses
x-values
of
in
one
the
another
domain
if:
of
gand
°
( g
f
)(x)
=
x
for
Y
ou
can
use
all
of
the
x-values
in
the
domain
of
f.
°
●
horizontal
inverse
The
●
in
line
graph
the
●
To
●
The
line
find
horizontal
crosses
a
line
test
function
to
identify
more
than
inverse
once,
functions.
there
is
If
a
no
function.
graphs
The
the
of
of
y
the
inverse
the
=
inverse
of
a
function
is
a
reflection
of
that
function
x.
inverse
function
functions
I(x)
function
=
x
is
algebraically
,
called
the
replace f (x)
identity
function.
with
It
y
and
leaves x
solve
for
y.
unchanged.
−1
So
f
f
=
I
°
Transformations
of
functions
●
f (x)
+
●
f (x)
–
●
f (x
+
k)
translates
f (x)
horizontally
k
units
to
the
left,
●
f (x
−
k)
translates
f (x)
horizontally
k
units
to
the
right,
●
−f (x)
●
f (−x)
●
f (qx)
stretches
f (x)
horizontally
with
scale
factor
●
pf (x)
stretches
f (x)
ver tically
ktranslates
k
translates
f (x)
f (x)
ver tically
ver tically
reflects
f (x)
in
the
x-axis.
reflects
f (x)
in
the
y-axis.
a
a
distance
distance
of
k
of
k
units
units
upward.
downward.
where
k
where
>
k
0.
>
0.
with
scale
factor
p.
Chapter
Theory
of
knowledge
Mathematical
Mathematics
lines,
and
When
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in
fatal
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shows
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Chapter
Quadratic
functions
and
equations
CHAPTER
OBJECTIVES:
2
the
2.4
quadratic
y-intercepts,
The
form
x
function
axis
↦
a(x
of
−
f (x)
=
ax
+
bx
+
c
=
0:
its
graph,
its
ver tex,
x-
and
symmetr y
p)(x
−
q),
x-intercepts
(p, 0)
and
(q, O)
2
The
form
x
↦
a(x
−
h)
+
k,
ver tex
(h, k)
2
2.7
solving
2.7
the
quadratic
2.7
the
discriminant
2.8
applications
Before
Y
ou
1
of
the
form
ax
+
bx
+
c
=
0
and
the
graphing
nature
skills
of
and
roots
solving
equations
to
real-life
situations
start
know
simple
equations
formula
of
you
should
Solve
quadratic
how
equations
to:
for
a
Skills
given
1
variable.
check
Solve
a
each
3a
–
5
equation.
=
a
+
7
2
e.g.
3b
Solve
−
2
=
for
b:
0
b
4x
+
c
3(n
–
1
=
4)
=
21
5(n
+2)
2
3b
=
2,
b
3
2
e.g.
Solve
the
equation
n
+
3
=
5.
2
n
+
3
=
=
2,
5
2
n
2
n
=
Factorize
±
2
mathematical
expressions.
2
Factorize
2
e.g.
Factorize
p
each
expression.
2
–
5p:
a
2k
−
b
14a
10k
3
p(p
–
5)
2
+
21a
−
49a
2
e.g.
Factorize
the
expression
c
2x
+
4xy
−
10a
+
3x
+
6y
ab
+
2b
2
ax
–
3x
+
2a
–
6:
d
5a
−
2
x(a
–
3)
+
2(a
–
3)
e
n
f
2x
+
g
m
4n
+
3
2
(x
+
2)(a
–
3)
−
x
−
3
2
2
e.g.
Factorize
the
expression
x
–
3x
–
−
36
10:
2
h
(x
+
2)(x
–
5)
2
e.g.
(2a
Factorize
+
5)(2a
–
the
expression
5)
Quadratic functions and equations
4a
–
25:
25x
2
−
81y
This
World
War
Washington
spray
This
water
of
DC.
in
picture
drinking
the
Memorial
The
fountain
parabolas,
cur ved
water
in
paths
a
opened
at
the
these
they
2004
in
Memorial
from
trajector y
.
streams
can
in
trajectories.
streaming
similar
of
and
was
fountains
beautiful
shows
cur ved
called
II
be
of
a
simple
The
shapes
water
modeled
are
by
2
mathematical
Functions
Other
In
like
these
situations
functions
the
functions
this
which
include
height
of
a
the
you
that
form
called
area
will
functions
the
can
dropped
chapter,
quadratic
are
of
modeled
a
object
study
are
(x)
=
quadratic
be
of
f
figure
over
how
given
+bx
ax
+c.
functions
by
and
quadratic
measuring
time.
to
in
graph
standard
2
form,
f
(x)
=
ax
+
bx
+
c;
tur ning
point
form,
2
y
f
=
a(x
(x)
in
=
−
h)
a(x
their
−
own
+
k;
p)(x
way
.
maximum
height
you
use
to
might
find
area,
the
the
and
−
q).
If
of
the
factorized
Each
you
a
of
these
wanted
spray
of
factorized
of
form
a
to
be
If
are
useful
the
from
form.
rectangle
would
forms
know
water
tur ning-point
dimensions
form,
a
fountain,
you
with
a
wanted
par ticular
helpful.
Chapter
.
Solving
quadratic
equations
Some
2
An
equation
that
can
be
written
in
the
form ax
+
bx
+
c
=
of
equations
0,
written
where
a
≠
0,
is
called
a
quadratic
equation .
These
are
these
in
are
the
not
form
all
2
ax
examples
of
quadratic
they
2
–
x
+
4x
+
7
=
bx
+
c
=
0,
but
equations:
can
be
rearranged
0
into
this
form.
In
quadratic
2
=
5x
2x(3x
(x
–
3x
–
–
7)
7)(2
2
=
–
0
5x)
=
a
trinomial
14x
2
2
ax
In
this
section,
you
will
begin
solving
quadratic
+
bx
Before
by
you
the
term,
bx
is
➔
xy
solve
This
=
quadratic
equations
by
term,
and
(x
to
0,
understand
then
proper ty
−
the
linear
x
=
0
can
or
be
y
an
=
a)(x
−
b)
=
0,
impor tant
x
−
This
a
0
proper ty
sometimes
to:
=
is
the
term.
proper ty:
0.
expanded
then
c
factorizing,
the
If
is
factorization
impor tant
If
ax
quadratic
is
constant
it
c,
equations.
called
Solving
+
or
x
−
b
=
zero
is
called
product
0.
property.
Example
Solve
these
equations
by
factorization.
2
a
x
2
−
5x
−
14
=
0
b
3x
2
+
2x
−
5
=
0
c
4x
+
4x
the
left-hand
+
1
=
0
Answers
Factorize
2
a
x
−
(x
−
5x
−
7)(x
x
−
7
x
=
7
=
x
=
−2
14
+
0
=
2)
=
or
or
the
expression
on
side
of
the
equation.
0
0
x
+
2
x
=
−2
=
0
Set
each
factor
equal
to
zero,
using
the
zero
product
proper ty.
7
2
b
3x
+
(3x
3x
+
+
2x
−
5)(x
5
=
5
–
=
1)
0
Factorize
0
=
Set
0
or
x
–
x
=
1
=
0
the
expression
factor
equal
to
on
the
left-hand
side
of
the
equation.
zero.
You can also find the solutions with your GDC. (See Chapter 17
Section 1.7.)
5
x
each
=
1
3
5
x
=
,
1
3
2
c
4x
+
4x
+
(2x
+
1)(2x
(2x
+
1)
1
=
+
0
1)
=
When
0
there
we
will
get
be
the
same
only
one
factor
twice,
solution.
2
=
0
equation
has
two
equal
roots.
1
2x +
1
=
0
x
=
−
GDC
help
on
CD:
Alternative
2
demonstrations
Plus
and
GDCs
Quadratic functions and equations
Casio
are
on
for
the
TI-84
FX-9860GII
the
CD.
We
it
is
a
‘perfect
sometimes
say
square’
that
and
this
Exercise
In
this
your
1
A
exercise,
answers
Solve
by
solve
with
all
your
the
equations
−
3x
+
2
=
0
−
Solve
+
25
by
=
0
x
e
+
−
56
=
0
m
c
−
2x
−
48
6c
−
=
0
b
f
+
2
5x
−
4
=
0
b
+
30
=
0
6b
+
9
=
0
2
5c
+
8
=
0
2h
c
2
4x
11m
factorization.
+
2
d
a
2
2
6x
a
check
2
a
b
2
x
then
factorization.
2
2
and
2
x
d
hand’
GDC.
2
a
‘by
−
16x
−
9
=
0
e
−
3h
+
x
−
5
=
0
2
3t
+
14t
+
8
=
0
6x
f
−
12
=
0
2
If
a
quadratic
you
will
shown
have
in
to
is
not
rearrange
Example
Example
Solve
equation
written
the
terms
in
the
before
form ax
you
can
+
bx
+
c
factorize
=
0,
as
2.
these
equations
by
factorization.
2
8x
a
−
5
=
10x
−
2
x (x
−
5
=
10x
−
2
Collect
−
10x
−
=
0
equation.
b
+
10)
=
4(x
−
2)
Answers
2
8x
a
like
ter ms
on
one
side
of
the
2
8x
(4x
+
4 x +
1)(2x
1
=
3
−
0
3)
or
=
2x
0
−
=
and
solve
for
x.
0
3
1
x
Factorize
3
=
x
=
2
4
Ancient
1
x
Babylonians
3
=
or
and
4
Egyptians
quadratic
x(x
b
+
10)
=
4(x
−
2)
Expand
the
brackets
and
collect
like
+
10x
=
4x
–
8
of
+
6x
+
8
=
these
0
Factorize
and
solve
for
years
4)(x
+
2)
=
ago
for
example,
0
solutions
x
+
4
=
0
or
x
+
2
=
to
=
−4
x
=
=
−4,
Exercise
1
Solve
factorization.
2
2x
−
7
=
13
+
x
2
b
2n
+
11n
d
2(a
−
5)(a
=
3n
−
n
−
4
Use
2
3z(z
c
+
4)
=
−(z
+
9)
+
5)
=
+ 5 =
f
2x
x
number
Exam-style
3
The
x
+
two
2
The
number
,
equation
and
to
write
solve
2x
and
its
square
differ
by
12.
Find
the
x
number.
question
per pendicular
and
represent
1
for
A
to
1
an
2
‘x’
21a
the
x
36
x
e
of
B
by
+
area
rectangle.
−2
2
x
a
the
−2
a
x
problems
0
concerning
x
to
x.
nd,
(x +
thousands
ter ms.
2
x
equations
like
2
x
studied
2
5x
–
sides
of
a
right-angled
triangle
have
lengths
3.
hypotenuse
has
length
4x
+
1.
Find
x.
Chapter
Investigation
Solve
these
–
equations
perfect
by
x
3
x
2
+
10x
+
25
=
0
2
x
+
14x
+
49
=
0
4
x
2
−
What
18x
do
original
a
are
0
by
Describe
+
9
8x
+
16
=
−
20x
0
=
0
using
any
+
patterns
100
you
=
0
see
in
the
equations.
these
with
are
equations
methods
your
three
called
completing
other
x
6
polynomial
think
quadratic
without
=
notice?
is
you
Solving
there
81
quadratic
do
Some
+
you
trinomial
Why
6x
−
2
x
A
+
2
2
5
trinomials
factorization.
2
1
square
‘perfect
the
cannot
you
can
terms.
square
trinomials’?
square
be
use
solved
to
by
solve
factorization,
a
quadratic
but
equation
GDC.
2
Consider
above.
the
The
equation
left
side
of
x
+
this
14x
+
49
equation
=
is
0
a
from
the
perfect
investigation
square,
because
it
2
has
two
identical
factors:
2
+
x
14x
+
49
=
(x
+
7)(x
+
7)
=
(x
+
7)
2
To
solve
the
equation
x
+
14x
+
49
=
0,
you
could
factorize,
which
2
would
give
the
equation
(x
+
7)
=
0,
and
lead
to
the
answer
x
=
−7.
2
What
If
if
you
you
were
collect
all
asked
the
to
terms
solve
on
the
one
equation x
side
of
the
+
14x
+
equation,
49
=
you
5?
get
2
x
+
get
14x
an
+
44
exact
Example
✗
Solve
=
0,
which
answer,
does
however,
not
as
in
easily
.
Y
ou
Example
could
still
3.
each
equation
without
using
the
2
a
factorize
shown
GDC.
2
x
+
14x
+
49
=
5
b
x
–
6x
+
9
=
6
Answers
2
a
x
+
14x
+
49
=
5
Factorize
the
perfect
square
trinomial
2
(x
+
7)
=
5
on
the
Take
x
+ 7
=
±
5
the
left
the
hand
side
of
the
equation.
square
root
of
both
sides
of
equation.
Leaving
x
has
two
solutions:
−7
+
5
in
x
=
−7 ±
5
−7
−
your
radical
(surd)
5
form
gives
2
b
x
–
6x
−
3)
+
9
=
6
Again,
we
see
that
the
left
side
of
2
(x
x
x
− 3
=
=
3 ±
=
±
6
the
is
a
perfect
trinomial,
so
we
method
in
par t
x
3
equation
can
square
use
the
same
6
6
Quadratic functions and equations
has
answers
and
as
two
6
a
solutions:
3
+
6
and
solutions.
the
exact
In
Example
Y
ou
can
equation,
➔
T
o
the
using
add
a
the
perfect
Example
Solve
equations
perfect
complete
and
a
3,
use
square
method
the
square
called
square,
result
to
take
both
trinomial
perfect
half
the
of
the
any
the
square
equation.
side
of
each
the
equation
by
completing
the
+
of
x,
This
square
step
it,
creates
equation.
square.
2
x
trinomials.
quadratic
coefficient
the
left
square
solve
completing
sides
on
to
2
a
involved
trinomials
10x
=
6
x
b
2
–
12x
=
3
x
c
–
3x
–
1
=
0
Answers
2
a
x
The
+
10x
=
6
+
10x
+
25
coefficient
of
x
is
10.
Halve
this
2
x
=
6
+
(5)
25
and
square
it
(25).
2
(x
x
+
5)
+ 5
=
=
±
Complete
31
both
31
Solve
x
=
−5 ±
x
–
12x
=
The
3
x
–
12x
+
36
=
3
+
12
36
2
for
adding
25
to
x.
coefficient
−
6)
=
÷
2
− 6
=
±
=
6,
Complete
39
Solve
x
by
of
x
is
12.
Over
one
thousand
2
2
(x
square
31
2
b
the
sides.
for
6
the
=
36
years
square.
Hindu
x.
were
ago,
=
and
mathematicians
developing
39
methods
x
Arab
6 ±
similar
to
39
completing
the
square
2
x
c
–
3x
–
1
=
Add
0
1
to
both
sides
of
the
equation.
to
2
solve
quadratic
2
x
–
3x
=
1
3
⎛
Half
9
of
3
is
,
⎝
− 3x
+
9
equations.
4
nding
They
were
is
⎜
2
x
⎞
and
2
9
3
⎟
2
⎠
solutions
= 1 +
4
4
9
Add
to
to
both
sides
of
the
mathematical
equation.
2
3 ⎞
⎛
4
13
x
problems
2
Solve
4
⎠
for
x.
‘What
must
square
3
±
x
=
Exercise
completing
the
its
This
amounts
is
written
to
as
+
10x
=
39.
square.
2
+
8x
=
3
x
2
−
5x
=
3
+
7x
−
4
+
x
2
−
6x
+
1
=
0
x
4
2
x
of
C
2
5
when
10
2
2
x
roots,
39?’
x
3
by
13
2
x
the
2
own
by
which,
increased
3 ±
x
be
13
=
2
1
as
⎟
⎝
Solve
such
=
⎜
=
0
2
−
2x
−
6
=
0
x
6
−
3
=
0
2
➔
In
order
to
complete
the
square,
the
coefficient
of
the x
term
2
must
be
1.
If
completing
divide
the
the
through
x
term
square,
by
the
has
you
a
coefficient
can
factor
out
other
the
than
1,
before
coefficient,
or
coefficient.
Chapter
Example
Abu
Solve
each
equation
by
completing
2
2x
a
the
+
8x
=
6
b
3x
Kamil
(c.850
–
Shuja
square.
2
15x
=
–
c.930),
also
2
known
Answers
as
al-Misri,
al-Hasib
meaning
'the
2
2x
a
+
8x
=
Divide
6
both
sides
of
the
equation
by
calculator
from
Egypt',
2
2
x
+
4x
=
3
x
+
4x
+
4
(x
+
2)
the
coefficient
of
x
,
which
is
2.
was
one
of
the
rst
to
2
=
3
+
Use
4
completing
the
square
to
introduce
symbols
2
=
solve
7
for
x.
for
indices,
m
x
+
2
=
x
=
–2
x
7
±
n
x
such
as
m+n
=
x
,
in
algebra.
7
±
2
4x
b
−
20x
=
5
2
4(x
−
5x)
=
Divide
5
through
by
the
coefficient
2
5
of
2
x
–
5x
x
,
which
is
4.
=
4
25
5
25
2
x
–
5x
=
+
+
2
4
4
4
5
Half
of
5
is
,
and
2
5 ⎞
⎛
30
x
15
=
⎜
2
⎞
⎜
⎟
⎝
2
25
is
.
4
⎠
=
⎟
2
⎝
4
⎠
5
2
15
x
=
±
2
2
This
5
x
⎛ 5
=
answer
could
also
be
written
as
15
±
x
2
±
30
=
2
2
Exercise
Solve
by
D
completing
the
square.
2
1
2x
3
5x
5
2x
2
+
12x
=
6
−
10x
+
2
−
x
2
3x
4
4x
6
10x
2
Y
ou
6x
=
3
+
6x
−
5
2
=
0
2
.
−
=
0
2
−
The
know
6
=
0
quadratic
that
a
quadratic
+
4x
−
5
=
0
formula
equation
can
be
written
in
the
form
2
ax
+
bx
+
equation
Y
ou
c
=
0.
using
would
Suppose
the
you
wanted
completing
the
to
solve
square
this
general
quadratic
method.
have:
Subtract
c
from
both
2
+
ax
bx
+
c
=
0
sides
of
the
equation.
2
ax
+
bx
=
–c
Divide
c
b
both
sides
of
2
x
+
x
=
–
a
a
the
2
b
⎛
2
x
+
x
b
=
⎜
a
⎝
⎠
⎛
b
⎞
+
−
⎟
2a
by
a
2
c
⎞
+
equation
⎜
a
⎝
⎟
2a
⎠
b
Half
of
b
is
a
.
2a
2
b
Squaring
this
gives
2
4a
Quadratic functions and equations
2
2
b
⎛
x
c
⎞
+
=
⎜
b
−
+
⎟
2a
⎝
2
a
⎠
4a
2
2
b
⎛
x
b
⎞
+
4 ac
=
⎜
⎟
2a
⎝
2
4a
⎠
2
b
x
2
b
+
=
4 ac
±
±
b
− 4 ac
=
2
2a
2a
4a
2
−b
x
±
b
− 4 ac
=
2a
This
gives
solve
➔
any
The
us
an
extremely
quadratic
quadratic
useful
formula
which
can
be
used
to
equation.
This
formula
2
For
any
equation
in
the
form
ax
+
bx
+
c
=
in
0,
for
2
−b
x
±
b
formula
the
is
Formula
the
IB
given
booklet
exam,
so
− 4 ac
=
you
do
not
have
to
2a
memorize
Example
Solve
each
equation
using
the
2
a
quadratic
formula.
2
x
+
it.
4x
−
6
−
6
=
0
=
0
b
2x
2
−
3x
=
7
3x
c
=
7x
+
6
Answers
2
a
x
+
4x
2
−4
x
±
4
− 4
2
4 ±
x
( 1) (
Use
the
a
1,
quadratic
for mula
with
6)
=
=
b
=
4,
and
c
=
−6.
(1)
40
This
=
answer
is
cor rect,
but
it
can
still
2
be
4 ± 2
x
simplified.
10
=
= − 2 ±
10
2
2
b
2x
−
3x
=
7
−
3x
−
7
(
3)
First
write
for m
ax
the
equation
in
standard
2
2x
=
2
0
+
bx
+
c
=
0
2
3 ±
x
4
2
3 ±
x
(
2)
(
7)
Use
the
a
2,
quadratic
for mula
with
=
(2)
=
b
=
−3,
and
c
=
−7.
65
=
4
2
c
3x
=
7x
+
First
6
write
the
equation
in
standard
2
2
3x
−
7x
−
6
=
for m
0
2
7 ±
x
(
7
4
)
2
7 ±
x
( 3) (
=
Use
the
a
3,
+
bx
+
c
quadratic
=
0
for mula
with
6)
=
b
=
−7,
and
c
=
−6.
(3)
121
=
ax
7 ± 11
=
6
6
2
x
=
−
, 3
3
Chapter
Exercise
Solve
each
E
equation
using
the
quadratic
2
1
4x
3
5x
5
x
7
2x
2
+
9x
−
7
=
0
2
3x
+
6x
+
1
=
0
4
x
6
3x
8
2x
2
+
2x
−
8
=
0
2
−
2
6x
=
−4
2
=
x
−
3
2
3x
=
1
x
–
2x
=
9
10x
=
9x
=
5
sum
Find
x
2
x
+ 1
of
the
4
=
5x
Example
+
+ 3
10
x
The
+
2
−
6
9
formula.
the
two
squares
of
two
consecutive
integers
is
613.
integers.
Answer
First,
2
+
(x
2
+
1)
=
need
to
write
an
equation.
613
2
x
+
x
+
2x
+
1
=
Let
613
2
2x
+
2x
−
612
=
x
0
like
+
x
−
306
=
be
the
consecutive
2
x
you
2
x
smaller
integer.
integer,
and
Expand
the
x
+
1
be
brackets
the
and
next
collect
ter ms.
0
Divide
by
2.
2
(1)
−1 ±
x
− 4
( 1) (
This
−1 ±
x
1225
=
factorization
(1)
equation
=
or
−18
The
or
two
Two
Find
Since
17
−18
and
−17,
or
17
and
18.
There
are
are
for
two
two
x
have
a
sum
of
50
and
a
product
of
576.
numbers.
rectangle
has
Find
the
length
Find
the
value
4x
x
+
–
a
perimeter
and
of
x
width
in
the
of
70
the
m
6
6
Quadratic functions and equations
and
an
rectangle.
diagram.
3x
of
area
of
264
m
+
be
solved
using
square.
values
.
for
x,
there
will
also
be
1.
possible
F
numbers
the
are
there
values
2
3
also
2
integers
Exercise
A
the
−1 ± 35
two
2
could
completing
=
2
1
quadratic
=
2
x
−306 )
pairs
of
consecutive
integers.
Exam-Style
4
Questions
A
rectangle
If
the
has
length
is
a
length
of
decreased
23 cm
by
x
and
cm,
a
and
width
the
of
width
16 cm.
is
increased
2
by
x
cm,
the
dimensions
area
of
of
the
the
new
new
rectangle
is
378 cm
.
Find
the
rectangle.
2
5
The
formula
of
ball
a
ball
.
in
t
h
=
2
+
seconds
the
14t
after
–
it
4.9t
is
gives
thrown.
the
For
height,
how
h
metres,
long
is
the
air?
Roots
of
quadratic
Extension
material
Worksheet
2
Solve
1
these
–
roots
equations
using
8x
+
these
16
=
0
4x
b
equations
Solve
3
5x
–
these
14
=
0
equations
What
4
3x
let’s
+
6
patterns
questions
Now
+
9
=
formula.
0
quadratic
25x
c
8x
the
+
2
=
,
take
=
0
did
2x
b
you
and
another
?
Why
look
at
10x
+
1
=
0
quadratic
5x
c
–
3x
–
4
=
0
formula.
2
–
notice
+
formula.
0
2
+
quadratics
2
–
using
2
x
a
quadratic
12x
the
3x
b
CD:
equations
2
+
on
more
2
–
using
2
x
a
quadratic
2
–
Solve
2
of
the
2
x
a
Two
equations
challenging
Investigation
-
in
do
4x
the
you
the
+
5
=
0
solutions
think
this
quadratic
4x
c
of
the
+
2x
+
1
equations
=
0
in
happened?
formula,
used
for
2
solving
and
c
equations
are
all
in
the
form
ax
+
bx
+
c
=
0,
where
a,
b,
constants.
2
−b
x
±
b
− 4 ac
=
2a
This
par t
formula
of
the
the
give
quadratic
information
us
will
about
solution.
all
the
formula,
the
The
us
roots
roots
the
of
an
discriminant
of
a
quadratic
discriminant ,
equation,
is
the
par t
will
without
of
the
equation.
give
One
us
actually
giving
quadratic
2
formula
the
under
symbol
‘△’
the
to
radical
(square
represent
the
root)
sign, b
–
4ac.
We
often
use
discriminant.
2
➔
For
a
quadratic
equation
ax
+
bx
+
c
=
0,
2
●
if
b
–
4ac
>
0,
the
equation
will
have
two
different
real
2
if
b
equal
–
4ac
=
0,
the
equation
will
have
two
equal
real
if
b
think
–
4ac
<
0,
the
equation
will
have
no
real
with
roots
of
an
two
as
having
roots
only
2
●
can
equation
roots
●
Y
ou
one
solution.
roots.
Chapter
Example
Use
the
discriminant
to
determine
the
nature
of
the
roots
of
each
equation
2
a
9x
+
6x
+
1
=
0
1
=
0
4
b
3x
–
5
=
x
Answers
2
a
9x
+
6x
+
This
a
=
is
9,
a
b
quadratic
=
6
and
c
equation
=
with
1.
2
△ = 6
The
–
4(9)(1)
equation
equal
=
36
will
−
36
have
=
Calculate
0
the
discriminant.
Discriminant
two
=
0
means
two
equal
roots.
roots.
4
b
3x
–
5
=
First,
get
for m.
Multiply
the
equation
into
standard
x
by
x
on
both
sides,
2
3x
–
5x
=
4
then
add
4.
2
3x
–
5x
–
4
=
0
2
Remember,
△
=
b
–
4ac.
2
△
=
(−5)
=
25
The
−
+
4(3)(−4)
48
=
equation
different
Example
real
73
will
△
have
>
0
means
two
dif ferent
real
roots.
two
roots.
2
Find
two
the
value(s)
different
of
real
k
for
which
the
equation
2x
–
kx
+
3
=
0
will
have
roots.
Answer
Solution:
For
the
equation
to
have
two
dif ferent
2
b
–
4ac
>
0
real
roots,
you
must
have
△
>
0.
2
(–k)
–
4(2)(3)
>
0
2
k
–
24
>
0
2
k
>
You
24
can
use
the
absolute
value
For
when
|k|
>
|k|
>
taking
the
square
root
in
more
on
value,
see
Chapter
inequality.
k
>
Exercise
1
Find
roots
6
2
6
2
or
Section
k
<
–2
6
G
the
for
value
each
of
the
discriminant,
x
c
4x
5x
–
3
=
0
b
2x
+
5
=
0
d
x
+
the
nature
4x
+
1
=
0
16
=
0
2
–
x
+
2
x
state
2
+
2
e
and
equation.
2
a
8x
+
2
–
absolute
an
24
3x
+
8
=
0
Quadratic functions and equations
f
12x
–
20x
+
25
=
0
of
the
2.7.
18,
EXAM-STYLE
Find
2
real
the
QUESTION
values
of
p
such
that
the
equation
+
4x
+
p
=
0
px
b
2
x
+
real
+
px
the
+
8
=
values
0
x
d
of
k
such
that
+
the
Find
10x
+
k
=
0
2x
b
–
2kx
the
+
5
values
=
0
of
x
d
m
such
that
+
1
=
0
has
two
equal
x
–
6x
3mx
+
+
k
=
0
–
4kx
–
3k
equation
=
0
has
no
real
roots.
m
=
0
x
b
+
5mx
+
25
=
0
2
–
8x
EXAM-STYLE
Find
3x
2
2
5
3px
–
the
2
c
0
2
3x
a
=
2
+
2
4
2
roots.
x
c
+
equation
2
a
5x
2
Find
3
different
2
x
c
two
roots.
2
a
has
+
1
=
0
x
d
+
6x
+
m
–
3
=
0
QUESTION
the
values
4qx
+
of
q
for
which
the
quadratic
equation
2
–
qx
5
–
q
=
0
Investigation
will
–
have
no
graphs
real
of
roots.
quadratic
functions
2
Each
For
of
these
each
functions
is
given
in
the
form
y
=
ax
+
bx
+
c.
function,
2
i
nd
the
ii
graph
value
of
b
–
4ac
For
the
function
on
your
2
a
y
=
x
c
y
=
x
3x
–
5
b
y
=
3x
d
y
=
4x
2
y
=
x
g
y
=
–
2x
+
7
do
–
x
6x
the
quadratics
on
a
–
6x
+
see
Chapter
17
Section
1.6.
4
+
3x
+
5
+
9
5x
+
f
y
=
2x
h
y
=
x
2
examples
between
graph
–
4x
+
2
2
+
these
relationship
and
graphing
2
2
What
with
2
+
2
e
GDC,
2
–
help
GDC.
of
a
the
suggest
value
quadratic
of
to
+
you
the
7x
+
about
3
the
discriminant
function?
y
.
Graphs
of
quadratic
functions
2
A
function
where
we
a
will
≠
of
0,
look
the
is
at
form
called
the
a
y
=
ax
of
=
x
2
+
bx
quadratic
graphs
2
y
+
c,
or
f
function.
quadratic
(x)
In
=
ax
this
+
bx
+
c,
section,
functions.
2
The
This
simplest
graph
quadratic
has
a
function
minimum
at
is
the
y
=
x
point
.
Its
(0,
graph
0),
and
is
it
shown.
is
symmetrical
0
about
the
x
y-axis.
Chapter
If
you
look
notice
at
some
the
graphs
of
other
2
y
=
quadratic
functions,
you
should
similarities.
2
x
+
2x
–
1
y
=
2
3x
–
y
4x
+
2
y
=
–2x
+
2x
+
3
y
y
0
0
x
x
0
Each
of
these
Each
graph
graphs
also
has
has
a
a
cur ved
minimum
x
shape
or
a
known
maximum
as
a parabola.
point
called
a vertex
2
If
the
with
coefficient
the
ver tex
of
as
x
is
the
positive,
minimum
the
parabola
point
on
the
will
open
upwards,
graph.
2
If
the
coefficient
downwards,
and
of
is
x
the
negative,
ver tex
will
the
be
a
parabola
will
maximum
open
point.
y
If
a
you
imagine
parabola,
and
is
right
called
in
red
you
sides
the
on
a
will
of
axis
this
ver tical
notice
this
of
line
r unning
the
ver tical
graph
line.
symmetry .
through
is
ver tex
symmetrical
This
This
the
imaginar y
axis
of
on
of
the
ver tical
symmetr y
is
left
line
shown
graph.
0
We
will
now
Consider
the
look
at
graphs
different
of
these
forms
of
quadratic
quadratic
x
functions.
functions
in
the
form
axis
2
y
=
ax
y
=
x
+
bx
+
c:
2
2
+
x
–
3
y
=
–
0.5x
2
–
2x
+
4
y
=
–
3x
+
3
1
x
=
1
y
y
y
x
x
–
=
2
2
(0, 4)
(0, 1)
0
0
x
x
0
x
(0, –3)
=
x
–2
2
➔
For
the
quadratic
graph
functions
crosses
the
in
standard
y-axis
at
(0,
form y
=
ax
c).
b
The
equation
of
the
axis
of
symmetr y
is
x
=
2a
Quadratic functions and equations
+
bx
+
c,
of
symmetr y
2
➔
When
the
basic
quadratic
function
y
=
x
undergoes
Y
ou
transformations,
the
resulting
functions
can
be
written
might
want
to
look
as
back
at
the
section
2
y
=
a(x
–
+
h)
k
about
Look
at
the
graphs
of
these
quadratic
functions
in
the
=
a (x
y
=
(x
–
h)
+
2
2)
–
of
this
in
Chapter
1
book.
k:
2
–
graphs
form
2
y
transformations
of
1
y
=
2(x
+
1)
y
2
–
4
y
=
– (x
y
–
3)
+
2
y
(3, 2)
0
x
0
0
x
x
(2, –1)
(–1, –4)
2
➔
For
quadratic
has
its
functions
in
the
form
y
=
a(x
–
h)
+
k,
the
graph
This
ver tex
at
(h,
form
quadratic
is
Example
of
a
k).
function
sometimes
‘turning-point
2
a
Write
b
Sketch
called
the
function
the
graph
y
of
=
x
the
form’.
2
–
6x
+
4
function,
in
the
form
labeling
the
y
=
(x
–
ver tex
h)
and
+
k
the
y-intercept.
Answers
2
a
y
=
x
–
6x
+
4
By
looking
standard
at
the
for m,
y-intercept
will
equation
you
be
know
in
the
(0, 4).
2
y
=
(x
–
6x
+
9)
+
4
–
9
Use
‘completing
the
square’
to
rewrite
2
y
=
(x
–
3)
–
5
the
equation.
subtracting
b
By
9,
adding
the
value
9,
of
then
the
right-
y
hand
side
of
the
equation
has
not
changed.
0
x
(3, –5)
Note:
of
The
equation
symmetr y
is
x
=
of
the
axis
3.
Chapter
Example
2
a
Write
the
function
f
(x)
=
2x
+
8x
+
11
in
the
form
labeling
the
ver tex
2
f
b
(x)
=
a(x
Sketch
and
–
the
the
h)
+
k.
graph
of
the
function,
y-intercept.
Answers
2
a
f
(x)
f
(x)
f
(x)
=
2x
+
8x
+
11
The
y-intercept
of
the
graph
2
=
2(x
+
4x
+
4)
+
11
–
8
is
(0,
11).
2
=
2(x
+
2)
+
3
Be
b
careful
when
completing
the
y
2
square
if
Factor
out
the
x
the
ter m
has
coefficient
a
coefficient!
from
the
12
(0, 11)
first
10
8
By
two
ter ms.
adding
2
×
4,
then
subtracting
8,
6
the
value
the
equation
of
the
right-hand
side
of
The
4
(–2, 3)
has
not
name
‘parabola’
changed.
was
2
introduced
Apollonius
0
–5
–4
–3
–2
axis
The
of
Exercise
1
For
1
2
equation
symmetr y
of
is
x
the
=
190
−2.
on
function,
write
the
equation
of
symmetr y
and
give
the
y-intercept
BCE)
conic
in
BCE–
his
wor k
sections.
can
nd
graph
of
each
the
ver tex
and
the
of
y-intercept
the
262
the
Y
ou
of
c.
H
each
axis
Perga
x
–1
(Greek,
Note:
of
by
function.
See
points
Chapter
17
using
your
Section
GDC.
1.8.
2
a
f
(x)
=
x
+
8x
+
5
2
b
f
(x)
=
x
c
f
(x)
=
5x
–
6x
–
3
2
+
10x
+
6
2
d
2
f
(x)
=
For
each
and
give
–
3x
+
10x
function,
+
9
write
the
coordinates
of
the
ver tex
It
the
coordinates
of
the
y-intercept
of
the
may
be
helpful
substitute
2
a
y
=
(x
–
–
2
b
y
=
(x
+
5)
+
=
4(x
–
1)
+
6
d
y
=
3(x
+
2)
–
the
the
each
graph
function
of
the
in
the
function,
form
f
(x)
labeling
=
the
a (x
f
(x)
=
f
(x)
=
h)
+
and
k.
Then
x
+
10x
–
6
b
f
(x)
=
3x
x
–
5x
+
+
8x
2
2
–
6x
+
7
Quadratic functions and equations
d
f
(x)
=
–2x
sketch
the y-intercept.
2
2
c
–
ver tex
2
a
or
the
function
in
7
2
Write
0,
2
standard
3
=
1
write
y
x
2
7)
2
c
to
graph.
–
3
form
to
y-intercept.
nd
W
e
will
For
now
consider
obvious
Look
at
y
=
a(x
y
=
(x
the
–
+
reasons,
graphs
p)(x
3)(x
–
–
quadratic
we
of
functions
sometimes
these
call
quadratic
in
the
this
form y
the
=
a(x
–
‘factorized
functions
in
the
p)(x
–
q).
form’.
form
q):
1)
y
=
–3(x
y
+
1)(x
–
4)
y
=
(x
+
2)(x
–
5)
y
y
–12
20
8
16
4
12
0
x
4
8
0
x
–8
4
–12
0
x
–16
–4
–8
➔
For
quadratic
graph
For
crosses
functions
the
quadratic
in
x-axis
at
functions
the
( p,
in
the
form
0)
y
and
form
=
a(x
at
y
(q,
=
of
symmetr y
will
have
the
equation x
p)(x
–
q),
the
–
q),
the
0).
a(x
p
–
–
p)(x
axis
+ q
=
2
Note:
tell
The
us
For
the
+
roots
example,
crosses
(x
x-intercepts
the
3)(x
–
of
in
the
x-axis
1)
Example
=
the
at
0
of
the
graph
quadratic
first
graph
(–3,
0)
has
x
=
a
quadratic
equation
above,
and
roots
of
at
–3
(1,
the
0).
and
x
in
the
form f
function y
The
=
function y
=
(x)
(x
=
+
=
f
(x)
0.
3)(x
–
1)
equation
1.
2
Write
Then
the
function
sketch
the
f
(x)
graph
=
of
x
+
the
3x
–
10
function,
in
the
form
labeling
the
f
(x)
x-
=
(x
and
–
p)(x
–q).
y-intercepts.
Answer
2
f
(x)
=
x
+
3x
f
(x)
=
(x
+
5)(x
–
10
–
The
2)
graph
Factorize
will
the
cross
the
right-hand
y-axis
side
of
at
(0, −10).
the
equation.
y
0
x
(0, –10)
p
Note:
The
equation
of
the
axis
of
symmetr y
use
x
+
q
=
2
(
is
x
5) +
=
2
3
=
2
2
Chapter
Example
2
Write
the
sketch
function
the
graph
y
of
=
2x
the
–
x
–
3
in
function,
the
form
labeling
y
the
=
x-
a(x
and
–
p)(x
–
q).
Then
y-intercepts.
Answer
2
y
=
2x
–
x
–
3
y
=
(2x
–
3)(x
y
=
2(x
–
1.5)(x
The
y-intercept
of
the
graph
will
be
(0, −3).
+
1)
Factorize
the
right-hand
side
of
the
equation.
+
1)
Factor
first
y
0
out
the
coefficient
of
x
in
the
factor.
x
(0, –3)
Note:
The
equation
of
the
axis
1
of
symmetr y
x
is
=
4
Exercise
1
Write
I
the
coordinates
of
the
x-
and
y-intercepts
of
the
graph
It
of
each
may
be
helpful
substitute
a
f
(x)
=
(x
c
f
(x)
=
–
+
3)(x
3(x
+
–
7)
2)(x
+
1)
b
f
(x)
=
2(x
–
4)(x
d
f
(x)
=
5(x
+
–
6)(x
–
5)
write
2)
Write
the
each
graph
function
of
the
in
the
function,
form
y
=
labeling
a(x
–
the x-
2
a
y
=
x
and
y
=
–
q).
Then
x
7x
–
8
y
b
=
x
–
–2x
+
form
sketch
15
2
+
3x
+
5
y
d
=
5x
+
6x
–
8
2
3
Write
form
each
y
of
the
a
y
=
function
a(x
–
p)(x
function,
–
in
the
q).
form
Then
labeling
the
y
=
a(x
make
ver tex
a
–
neat
and
x
y
=
k
and
sketch
of
and
in
the
the
graph
y-intercepts.
2
+
6x
–
16
y
b
=
–x
2
c
+
the x-
2
=
h)
–0.5x
Exam-Style
–
4x
+
–
18x
21
2
+
3.5x
–
3
y
d
=
4x
+
8
y
Question
2
4
Let
f
(x)
=
2x
A
–
12x.
Par t
of
the
graph
of
f
is
shown.
0
a
The
graph
crosses
x-coordinate
i
Write
c
The
x-axis
ii
down
ver tex
at
A
and
B.
Find
B
the
of
A
b
the
the
of
B
equation
the
graph
of
is
Quadratic functions and equations
at
the
C.
axis
Find
of
symmetr y
.
the
coordinates
of
C
0,
to
y-intercept.
y-intercepts.
8x
=
C
or
function
2
–
2
c
p)(x
the
standard
the
2
to
function.
x
in
nd
Exam-Style
Question
2
Let
5
f
(x)
=
Find
a
x
(f
+
g)
3,
and
let
g (x)
=
x
–
2.
(x).
°
Write
b
down
the
coordinates
of
the
vertex
of
the
graph
of
(f
g).
°
The
(f
graph
g)
by
5
of
the
units
function
in
the
h
is
formed
positive
by
translating
x-direction,
and
by
2
the
graph
units
in
of
the
°
negative
y-direction.
Write
c
the
equation
of
the
function
h (x)
in
the
form
2
h (x)
=
Finding
can
➔
of
a
lot
the
When
●
+
c
down
the
equation
from
tell
equation
bx
write
the
function
Y
ou
+
ax
Hence,
d
a
of
a
of
the
graph
of
h
quadratic
graph
about
the
function
the
y-intercept
in
equation
graph
of
a
different
is
function
by
looking
at
the
forms.
written
in
standard
form
2
f
(x)
=
ax
+
bx
+
c,
you
know
the
y-intercept
of
the
graph
is
b
(0,
c),
and
the
equation
of
the
axis
of
symmetr y
x
is
=
2a
2
When
●
the
known
●
When
the
f
=
a(x
at
(q,
(x)
and
Now
you
function
If
If
you
will
from
know
you
are
given
Example
write
the
the
Write
equation
–
in
p)(x
–
is
q),
the
form
form,
written
the
the
in
graph
f
(x)
=
a(x
ver tex
will
factorized
will
–
cross
h)
be
+
(h,
k,
also
k).
form
the
x-axis
at
(p,
0)
0).
look
the
is
tur ning-point
at
how
you
information
the x-intercepts,
tur ning-point
Using
as
equation
the
ver tex,
can
find
given
begin
you
in
with
can
the
its
the
star t
equation
of
a
quadratic
graph.
equation
with
the
in
factorized
equation
form.
in
form.
information
equation
your
final
of
provided
the
answer
in
the
quadratic
in
standard
graph,
y
function.
form
2
y
=
ax
+
bx
+
c
0
(–2, 0)
(4, 0)
x
(0, –16)
{
Continued
on
next
page
Chapter
Answer
y
=
a(x
+
2)(x
–
4)
Since
with
the
the
x-intercepts
equation
in
are
given,
factorized
star t
for m.
You know that y = −16 when x = 0.
–
16
=
a(0
–
8a
=
–16
+
a
=
2
y
=
2(x
2)(0
–
4)
Substitute
your
You
+
2)(x
–
these
equation
can
values
to
check
solve
this
into
for
a.
answer
by
4)
graphing
the
equation
on
your
GDC,
2
y
=
2x
–
4x
–
16
and
to
comparing
those
in
the
the
x-
given
and
y-intercepts
graph.
GDC
help
on
screenshots
Plus
and
GDCs
Example
CD:
for
Casio
are
on
Alternative
the
TI-84
FX-9860GII
the
CD.
y
Write
the
equation
in
graph.
of
the
quadratic
function
shown
(6, 3)
the
2
Write
your
final
answer
in
standard
form
y
=
ax
+
bx
+
c.
0
x
(0, –15)
Answer
2
y
=
a(x
–
6)
+
3
Since
the
ver tex
tur ning-point
2
–
15
=
a(0
–
36a
+
3
=
–
36a
=
–
18
6)
+
3
You
know
values
15
You
on
into
can
your
is
given,
star t
with
the
equation
in
for m.
that
y
your
check
GDC,
=
−15
when
equation
this
and
to
answer
by
checking
x
=
solve
0.
Substitute
for
graphing
the
these
a.
ver tex
the
equation
and
the
y-intercept.
1
a
=
–
2
1
2
y
=
–
(x
–
6)
+
3
2
GDC
help
on
screenshots
Plus
and
CD:
for
Casio
Alternative
the
TI-84
FX-9860GII
1
2
y
=
–
x
+
6x
–
15
2
Quadratic functions and equations
GDCs
are
on
the
CD.
Finally
,
or
the
three
let’s
axial
look
equations
Example
Write
the
at
what
intercepts
in
of
three
happens
the
if
graph.
variables
to
you
This
don’t
next
solve
know
the
example
using
a
ver tex
also
leads
to
GDC.
equation
of
the
quadratic
function
shown
in
the
graph.
y
(–2, 9)
(4, 3)
x
(2, –7)
Answer
For
the
point
(–2,
9),
In
this
case,
you
are
given
the
coordinates
of
2
9
=
a(–2)
9
=
4a
For
the
–
+
2b
point
b(–2)
+
(2,
+
c
three
points
on
the
graph
of
the
function.
c
Substitute
–7),
the
x-
and
y-coordinates
of
these
three
2
–7
=
a(2)
+
b(2)
–7
=
4a
2b
+
points
c
into
the
standard-for m
quadratic
equation
2
For
the
+
point
+
(4,
y
c
3),
=
ax
+
You
now
You
can
bx
+
have
c.
three
equations
with
three
variables.
2
3
=
a(4)
3
=
16a
GDC
help
solving
on
the
on
CD:
b(4)
4b
TI-84
Plus
GDCs
+
Help
simultaneous
FX-9860GII
the
+
+
c
use
your
GDC
to
solve
for
a,
b
and
c.
with
equations
and
is
+
c
Casio
given
on
CD.
To
find
see
Chapter
If
Using
the
GDC,
a
=
1.5,
b
=
−4,
and
c
=
−5.
you
will
these
17
graph
see
points
that
Section
this
it
on
the
1.5.
function
passes
graph,
on
through
your
all
GDC,
three
you
points,
2
y
=
1.5x
–
4x
–
5
as
described.
Chapter
Exercise
Use
the
J
information
provided
in
the
graphs
to
write
the
equation
2
of
each
function
in
standard
form
y
y
=
ax
+
bx
+
c
y
y
(–1, 8)
(0, 5)
(–2, 0)
0
(6, 0)
x
(0, 5)
(0, –12)
0
x
(2, 1)
0
x
y
y
y
(1, 13)
(5, 30)
0
(15, 30)
x
(–4, 8)
(0, 4)
(20, 0)
(4, –5)
x
0
y
x
y
(2, 25)
(1, 3)
(–3, 0)
(7, 0)
0
x
0
Quadratic functions and equations
(0.5, 0)
x
.
Applications
At
the
by
water
beginning
in
Quadratic
different
When
you
use
a
a
quadratics
chapter,
can
and
be
you
saw
modeled
their
graphs
that
by
can
a
the
shape
quadratic
be
used
to
formed
function.
model
many
situations.
solving
lear ned
your
problems
throughout
GDC
farmer
If
this
functions
Example
A
of
fountain
of
to
this
you
quadratics,
chapter.
answer
Y
ou
many
you
will
can
use
also
be
the
methods
expected
to
questions.
wishes
the
help
using
garden
to
is
enclose
x
metres
a
rectangular
wide,
find
garden
the
length
with
and
100
the
metres
area
of
of
fencing.
the
garden
in
terms
of
x
2
b
Find
the
width
c
Find
the
maximum
of
a
garden
area
with
the
an
garden
area
can
of
525
m
have.
Answers
a
If
of
50
–
the
far mer
the
has
rectangle
100 m
must
of
be
fencing,
100. The
the
sum
perimeter
of
the
length
x
and
width
will
therefore
be
50 m.
x
length
area
=
=
50
x(50
–
−
Area
x
Set
b
x(50
–
x)
=
=
width
×
length
x)
525
the
Write
area
as
a
equal
to
525.
quadratic
equation
in
standard
for m,
and
2
50x
–
x
=
525
+
525
solve
for
x.
2
x
–
50x
(x
–
15)(x
–
=
35)
0
=
0
You
the
x
=
15
m
or
35 m
could
also
square,
or
solve
by
this
using
equation
the
by
quadratic
using
your
If
the
width
is
15,
the
length
is
35.
If
the
width
is
35,
the
length
is
15.
completing
for mula,
or
by
GDC.
y
c
The
(25, 625)
easiest
way
to
find
the
maximum
area
is
to
600
graph
the
and
is
x
function
the
width.
y
=
x(50
You
can
–
x),
do
where
this
on
y
is
your
the
area
GDC.
400
See
200
Chapter
The
ver tex
graph,
and
17
(25,
Section
625)
tells
you
is
1.6.
the
the
highest
point
maximum
area
on
the
occurs
when
x
–20
the
width
of
the
garden
is
25
metres.
2
The
maximum
area
is
625 m
Chapter
Example
The
height
of
a
ball
t
seconds
after
it
is
thrown
is
modeled
by
the
2
function
a
Find
b
For
h
=
the
24t
–
4.9t
+
maximum
what
length
of
1,
where
height
time
h
is
the
reached
will
the
by
ball
height
the
be
of
the
ball
in
metres.
ball.
higher
than
20
metres?
Answers
a
Graph
y
the
function
(2.45, 30.4)
2
30
y
25
height
=
24x
–
of
4.9x
the
+
ball
1,
where
and
x
is
y
is
the
the
time
in
seconds.
20
The
ver tex
is
approximately
15
(2.45,
30.4). This
tells
you
the
10
maximum
height
occurs
when
the
ball
5
has
0
x
1
2
3
4
been
You
5
can
GDC.
The
maximum
height
the
find
See
air
the
for
2.45
ver tex
Chapter
seconds.
using
your
17
is
Section
30.4
in
1.8.
metres.
2
20
b
=
24t
–
4.9t
+
1
Let
h
+
=
0
Write
=
20.
2
4.9t
–
24t
19
as
a
standard
You
See
t
≈
0.9930
seconds
and
20
–
0.9930
=
ball
is
metres
and
2.912
solve
Chapter
The
3.905
seconds
3.905
can
quadratic
for m,
once
and
this
17,
at
a
using
Section
height
twice,
on
equation
solve
once
for
in
t.
your
GDC.
1.7.
of
on
the
way
up,
the
What
The
ball
will
be
higher
way
than
other
real-life
20
metres
for
be
Example
takes
Luisa
3
hours
to
ride
her
bicycle
up
a
hill
and
back
down.
−1
Her
average
average
the
top
speed
speed
of
the
riding
riding
hill
is
up
40
down
the
hill
the
is
the
hill.
If
km,
find
Luisa’s
35
km
distance
h
faster
from
average
the
uphill
than
her
bottom
and
to
downhill
speeds.
Answer
distance
Let
x
represent
Luisa’s
uphill
Remember
time
=
,
and
speed
riding
speed.
when
40
you
add
=
x
3
downhill
times,
uphill
the
total
and
is
+ 35
3
hours.
{
the
40
+
x
modeled
by
seconds.
quadratic
It
of
situations
about
might
2.91
kinds
down.
Quadratic functions and equations
Continued
on
next
page
functions?
You
can
multiply
through
by
x,
and
40 x
40 +
= 3x
then
x
by
(x
+
35),
to
get
rid
of
the
+ 35
denominators.
2
40x
+
1400
+
40x
=
3x
+
105x
Write
as
a
quadratic
equation
2
3x
+
25x
–
1400
=
0
in
standard
using
your
Section
for m
and
GDC.
See
solve
for
x
Chapter
17,
1.7.
−1
x
≈
17.8
km
h
−1
Luisa
averages
17.8
km
h
riding
−1
uphill,
and
52.8
km
h
riding
downhill.
Exercise
1
The
K
height
of
a
ball
t
seconds
after
it
is
thrown
is
modeled
by
the
2
function
h
=
15t
–
+
4.9t
3,
where
h
is
the
height
of
the
ball
in
metres.
a
Find
b
For
12
the
maximum
what
length
of
height
time
reached
will
the
by
ball
the
be
ball.
higher
than
metres?
2
2
The
area,
A
cm
,
of
a
rectangular
picture
is
given
by
the
2
formula
A
=
centimetres.
32x
–
Find
x
,
the
where
x
is
the
dimensions
width
of
the
of
the
picture
picture
if
the
in
area
2
is
3
252
A
cm
piece
are
If
a
of
the
side
40
into
side
that
cm
two
length
length
Show
b
wire
formed
of
the
the
long
is
cut
into
two
pieces.
The
two
pieces
squares.
of
one
other
of
the
squares
is x
cm,
what
is
the
square?
combined
area
of
the
two
squares
is
given
by
the
squares?
2
A
A
a
is
20x
the
same
The
of
the
length
width.
as
100.
por trait
measures
frame
the
area
combined
of
of
50
uniform
the
area
cm
of
by
width.
por trait,
70
If
what
two
cm.
the
is
It
is
area
the
surrounded
of
the
frame
approximate
frame?
of
Find
+
minimum
rectangular
the
width
5
–
rectangular
by
is
2x
What
c
4
=
a
the
rectangle
is
dimensions
five
of
less
the
than
three
rectangle
if
times
its
its
area
is
2
782
6
The
is
m
sum
251.
of
Find
the
the
squares
of
three
consecutive
positive
odd
integers
integers.
Chapter
7
A
‘golden
rectangle’
has
the
proper ty
that
if
it
is
divided
into
The
a
square
and
a
smaller
rectangle,
the
smaller
rectangle
will
length-to-width
be
ratio
similar
in
propor tion
rectangle
ABCD
rectangle
PBCQ,
A
to
below
,
as
the
PQ
original
forms
a
rectangle.
square
In
the
APQD
golden
and
a
rectangle
a
as
shown.
P
of
the
ratio.
B
to
golden
is
Y
ou
this
BC
may
investigate
situations
AB
known
golden
in
want
other
which
interesting
ratio
=
AD
D
Q
Given
8
A
a
house.
the
One
builder
has
area
the
of
Jaswinder
away
.
He
=
1,
find
wants
side
other
appears.
C
AD
homebuilder
and
9
that
PB
of
three
takes
a
travels
the
sides
enough
largest
to
build
deck
will
wood
deck
trip
360
AB
for
he
to
km
a
rectangular
will
15
by
a
wall
of
on
with
the
railing.
railing,
the
If
back
of
house,
the
what
is
the
build?
his
bus,
a
wooden
metres
could
visit
share
have
deck
sister,
and
who
140
lives
km
by
500
km
train.
−1
The
train
entire
the
10
bus
and
Working
than
the
averages
jour ney
the
in
2
John
If
faster
h
hours,
if
takes
they
hours
house
Review
km
8
find
than
the
the
bus.
average
If
the
speeds
of
train.
does.
house
clean
the
alone,
Jane
10
takes
work
24
he
two
is
more
hours
together,
minutes.
working
John
How
long
to
clean
and
the
Jane
does
it
house
can
take
clean
John
to
alone?
exercise
✗
Solve
each
equation.
2
a
(x
+
2)
=
16
2
b
x
c
3x
–16x
d
x
+
64
=
0
2
+
4x
–
7
=
0
+
12
=
0
–
12
=
0
=
0
2
–
7x
2
e
x
+
2x
2
f
3x
–
7x
+
3
y
EXAM-STYLE
QUESTION
2
Let
f
(x)
=
x
+
down
3x
4.
Par t
of
a
Write
b
Find
c
Write
down
the
equation
d
Write
down
the
x-coordinate
the
the
–
y-intercept
x-intercepts
of
Quadratic functions and equations
the
of
the
of
the
graph
graph
of
of
f
is
shown.
f
0
graph.
the
of
axis
the
of
symmetr y
.
ver tex
of
the
graph.
x
EXAM-STYLE
QUESTIONS
y
25
3
Let
f
(x)
=
a(x
–
p)(x
–
q).
Par t
of
the
graph
of
f
is
shown.
20
The
graph
a
Write
b
Find
passes
down
through
the
value
the
of
points
p
and
of
(–5,
0),
(1,
0)
and
(0,
10).
q
10
the
value
of
a
5
0
–2
–3
x
–1
2
4
Let
f
(x)
=
Write
a
a(x
+
down
b
Given
c
Hence
that
3)
–
the
f
6
Quadratic
coordinates
(1)
=
2,
find
of
the
the
ver tex
value
of
of
the
graph
of
are
f
closely
other
the
value
of
f
a
related
functions
‘conic
find
functions
to
called
sections’
(see
(3).
page
60).
How
are
2
5
The
equation
Find
the
Let
(x)
x
+
possible
2kx
+
3
values
=
of
0
has
two
equal
real
roots.
k
these
functions
in
real
the
used
world?
2
6
f
=
Write
a
2x
+
the
12x
+
function
5.
f,
giving
your
answer
in
the
form
2
f
(x)
The
b
=
a(x
graph
units
in
–
of
the
y-direction.
+
h)
g
is
k.
formed
positive
Find
by
translating
x-direction
the
and
coordinates
of
8
the
units
the
graph
in
the
ver tex
of
of
f
by
4
positive
the
graph
of
g
y
7
Write
the
function
Give
equation
shown
your
in
answer
of
the
the
in
quadratic
graph.
the
form
2
y
=
+
ax
bx
+
(–4, 0)
c.
(6, 0)
0
x
(2, –12)
Review
1
Solve
exercise
each
equation,
giving
your
2
3x
a
–
5x
–
7
=
0
2x
b
x
=
The
2x
–
1
to
3
significant
figures.
8x
=
3
1
+
d
+ 3
x
EXAM-STYLE
2
+
1
c
x
answers
2
= 5
x
+ 2
QUESTION
height,
h
metres
above
the
water,
of
a
stone
thrown
off
a
2
bridge
t
is
the
is
modeled
time
in
by
the
seconds
function
after
a
What
is
the
initial
b
What
is
the
maximum
c
For
d
How
what
length
long
does
of
it
the
height
take
stone
from
height
time
is
for
h(t)
is
15t
which
height
stone
+
20
–
,
4.9t
where
thrown.
the
reached
the
the
=
by
of
to
stone
the
the
hit
is
stone
the
thrown?
stone?
greater
water
than
below
the
20
m?
bridge?
Chapter
3
The
length
The
area
of
a
rectangle
is
5
the
rectangle
is
1428
cm
more
than
3
times
its
width.
2
width
of
of
the
EXAM-STYLE
cm
.
Find
the
length
and
rectangle.
QUESTION
2
The
4
of
f
function
is
The
f
is
given
by
f
(x)
=
ax
+
bx
+
c.
Par t
of
the
y
graph
shown.
graph
of
f
passes
through
the
points
P(−10,
12),
Q(−5,
−3)
R
and
R(5,
Find
the
27).
values
of
a,
b
and
c
P
Thomas
5
drives
his
car
120
km
to
work.
If
he
could
increase
0
x
−1
his
average
minutes
speed
faster.
CHAPTER
Solving
●
●
If
xy
0,
then
x
=
This
proper ty
can
To
a)(x
solve
equation.
of
−
an
coefficient
side
his
quadratic
is
−
is
km
b)
,
he
would
average
make
driving
it
to
work
Q
30
speed?
or
equations
y
=
0.
sometimes
be
=
x,
This
the
0
0,
by
square
step
called
expanded
then
equation
of
h
SUMMARY
proper ty
(x
20
What
This
If
●
=
by
x
−
a
the zero product property
to:
=
0
or
completing
it,
and
creates
a
add
x
the
the
perfect
−
b
=
0.
square,
result
square
to
take
both
half
the
sides
trinomial
on
of
the
the
left
equation.
2
●
In
order
to
complete
the
square,
thecoefficient
of
the x
term
2
must
be
factor
The
1.
out
If
the
the
x
term
has
coefficient,
quadratic
or
a
coefficient
divide
other
through
by
than
the
1,
you
can
coefficient.
formula
2
●
For
any
equation
in
the
form
ax
+
bx
+
c
=
0,
This
formula
is
given
2
−b
x
±
b
− 4 ac
in
=
the
IB
Formula
2a
booklet
Roots
of
quadratic
need
equations
to
so
you
do
memorize
not
it!
2
●
For
a
quadratic
equation
ax
+
bx
+
c
=
0,
2
■
if
b
Y
ou
–
4ac
>
0,
the
equation
will
have
two
different
can
think
of
an
real
equation
with
two
roots
equal
2
■
if
b
■
if
b
–
4ac
=
0,
the
equation
will
have
two
–
4ac
<
0,
the
equation
will
have
no
equal
real
roots
as
having
roots
only
one
solution.
2
real
roots.
Continued
Quadratic functions and equations
on
next
page
Graphs
of
quadratic
equations
2
●
For
the
quadratic
graph
functions
will
cross
the
in
standard
y-axis
at
(0,
form y
=
ax
+
bx
+
c
=
0,
c).
b
●
The
●
When
equation
of
the
axis
of
symmetr y
x
is
=
2a
2
the
basic
quadratic
function
y
=
x
undergoes
transformations,
2
the
resulting
functions
can
be
written
as
y
=
a(x
–
h)
+
k
2
●
For
the
●
quadratic
graph
For
the
quadratic
graph
functions
will
have
in
the
the
ver tex
functions
crosses
For quadratic
its
in
the
x-axis
functions
in
form
at
at
the
(h,
form
( p,
y
=
a(x
–
h)
+
=
a(x
–
p)(x
0)
form
y
and
y
=
at
a(x
(q,
–
–
of
symmetr y
will
have
the
p)(x
equation x
q),
0).
–
p
the axis
k,
k).
q),
+ q
=
2
2
●
When
the
known
●
as
When
the
f
=
a(x
at
(q,
(x)
and
equation
is
in
tur ning-point
equation
–
p)(x
–
is
q),
the
form
form,
written
the
the
in
graph
f (x)
=
a(x
ver tex
will
factorized
will
cross
–
h)
be
+
(h,
k,
also
k).
form
the
x-axis
at
(p,
0)
0).
Chapter
Theory
of
knowledge
Conic
sections:
shapes
The
graph
a
parabolas
air,
or
are
sections.
the
cones)
quadratic
in
the
one
of
of
conic
a
the
real
the
circle,
of
The
the
function
–
is
the
streaming
in
world
the
path
from
shape
of
a
a
of
a
parabola.
baseball
flying
We
also
through
fountain.
four
as
sections
plane.
real
world
water
known
intersection
and
are
the
shape
shapes
These
by
sections
the
just
mathematical
two
of
see
Parabolas
formed
in
mathematical
conic
are
a
cone
other
ellipse
(or
conic
and
the
hyperbola.
Circle
{
A
parabola
when
a
is
the
plane
parallel
to
shape
that
intersects
one
of
its
a
Ellipse
results
slanted
edges.
describe
ancient
sections,
(c.262
–
and
190
Hypatia
died
Greeks
415
Apollonius
BCE)
(bor n
CE)
studied
first
a
350
each
equations
of
these
and
of
the
Parabola:
y
=
Circle:
(x
ax²
+
bx
and
370,
h)²
+
(y
–
when
few
Alexandria,
women
Egypt,
received
–
h)²
work
on
They
were
Persian
Omar
Theory
of
She
conic
developed
at
a
fur ther
studied
(x
an
–
(y
Khayyám
knowledge:
by
the
poet
(c.1048–1131).
Conic
sections:
=
–
k)²
=
1
b²
h)²
mathematical
shapes
in
the
real
world
(y
–
k)²
–
a²
Apollonius’
and
k)²
time
sections.
mathematician
–
+
Hyperbola:
education.
to
c
Platonist
a²
in
+
and
Ellipse:
School
used
them.
mathematician
head
be
Perga
(x
astronomer
can
shapes.
conic
named
between
was
of
Hyperbola
cone
Mathematical
The
Parabola
=
b²
1
r²
the
ese
see
ever y
Why
is
Did
a
circle
you
This
Apollonius
How
shown
had
when
do
that
the
orbits
of
planets
are
shapes?
wasn’t
orbits
‘perfect’?
know
elliptical
day.
he
you
until
the
early
hypothesized
was
think
studying
this
17th
that
and
century
.
planets
naming
knowledge
had
such
conic
developed
sections.
over
time?
Nowadays,
suspension
bodies
Who
in
we
space,
knew
Look
see
help
that
Why
us
do
think
shapes
How
using
can
our
the
slicing
–
be
describe
and
and
a
what
people
of
shape
cone
and
the
tr y
and
spacecraft
of
satellite
could
would
and
in
other
dishes.
result
give
parabolas
in
us
such
useful
shapes
that
universe?
other
modeled
and
hyperbolas
paths
equations,
you
might
you
the
understand
around
that
ellipses,
bridges,
mathematical
can
see
by
to
patterns
mathematics
gures
and
shapes
mathematical
use
in
mathematics
the
help
us
world
do
you
equations?
to
around
understand
us?
our
world
universe?
Chapter
Probability
CHAPTER
OBJECTIVES:
Concepts
5.5a
of
trial,
outcome,
equally
likely
outcomes,
sample
space
(
U)
and
n( A)
event.
The
probability
of
an
event
A
P(A)
is
=
.
The
complementar y
events
n(U )
A
and
A′
(not
A).
The
use
of
Venn
diagrams,
tree
diagrams
and
tables
of
outcomes.
Combined
5.6
events,
the
formula
for
P(A
∪
B).
Mutually
exclusive
events:
P(A
∩
B)
=
0.
P( A B)
Conditional
probability;
the
denition
P(A | B)
.
Independent
events;
the
P(B)
denition
Before
Y
ou
1
you
should
Add,
+
3
=
5
P(A)
how
multiply
10
3
=
=
P(A | B′).
Probabilities
to:
and
divide
Skills
and
without
replacement.
fractions
1
check
Work
these
a
=
15
out
1
b
−
a
calculator.
5
1
c
+
5
7
15
without
2
3
13
+
15
with
start
know
subtract,
2
P(A|B)
7
2
×
5
3
3
2
1
2
9
–
=
7
–
=
⎛ 1
9
9
9
d
9
1
−
5 ⎞
⎝
3
3
20
e
×
⎜
⎟
3
9
7
⎠
9
3 3
20
×
=
4
7
4
7
=
4
×
7
Add,
20
4 5
3
÷
2
=
5
4
=
7
3
subtract
1
=
1
3
and
3
multiply
0
0.2
+
+
0.9
×
3
0.4
Work
these
out.
a
1
−
0.375
b
0.65
+
0.05
c
0.7
e
50% of
×
0.6
30
d
0.25
×
f
22%
of
g
12%
10%
0.64
0.62
0.75
2
×
0.2
of
34
×
Calculate
52%
2
1
0.22
0.38
of
of
0.8
0.34
Since
then
decimals
0.35
0.7
0.2
9
0.34
68
=
0.068
percentages
60
Probability
=
=
0.52
×
60
3
=
31.2
Check
using
your
a
answers
calculator.
to
questions
1
and
2
●
What
is
the
chance
of
it
raining
tomorrow?
[
According
US
●
What
is
the
likelihood
●
What
is
the
probability
that
I
will
pass
my
test?
winning
the
football
National
of
us
match
after noon?
Am
I
to
get
to
school
on
time
if
I
catch
the
bus
rather
struck
a
than
the
Weather
the
probabilit y
cer tain
the
this
Ser vice,
●
to
government ’s
by
given
of
being
lightning
year
in
is
train?
1
750 000
We
consider
‘chance’,
questions
‘likelihood’,
like
these
all
‘probability’
the
and
time.
We
‘cer tain’
use
in
the
words
ever yday
speech.
The
probabilit y
being
struck
lightning
These
words
also
describe
mathematical
probability
.
This
from
of
mathematics
spor ting
averages
helps
to
the
us
to
understand
weather
repor t
risk,
and
and
the
ever ything
chance
of
This
str uck
chapter
by
lifetime
6250
probabilities
lightning.
looks
at
the
language
of
probability
,
how
to
have
been
from
data
estimated
on
(give
it
a
numerical
value)
and
the
basic
tools
you
solve
problems
involving
probability
.
of
and
need
number
to
size
quantify
population
probability
is
1
These
being
an
impor tant
80-year
branch
in
of
by
struck
the
of
by
past
people
lightning
30
in
years.
Chapter
Investigation
During
de
the
Which
or
is
.
Antoine
rolling
dice
mathematicians
Gombaud
Blaise
puzzled
over
Pascal,
this
Pierre
simple
problem:
a
likely,
‘double
option
do
rolling
six’
you
on
a
‘six’
24
think
is
on
four
throws
throws
with
two
more
likely?
of
one
dice,
dice?
Why?
Definitions
An
event
An
experiment
A
Some
is
random
which
●
and
more
rolling
Which
➔
mid-1600s,
Fermat
gambling
–
an
event
a
●
tossing
●
picking
may
of
dice
a
is
the
is
an
by
one
experiment.
which
where
we
there
obtain
is
an
outcome.
uncer tainty
over
occur.
random
three
coin
from
process
experiment
examples
rolling
outcome
experiments
are:
The
times
on
once
of
two
cards
from
a
pack
of
52
playing
rst
recording
5-minute
We
can
Chance
0
and
represents
probabi li ty
number
of
cars
that
pass
the
and
Book
school
gate
in
written
by
Games,
Jerome
a
Cardan
(1501–75).
Cardan
was
period.
express
between
1
the
written
cards
was
●
book
probability
, The
the
1.
an
chance
On
event
that
the
this
that
event
of
an
scale,
is
occurring
represents
cer tain
will
impossible
0
event
to
an
happen.
using
a
number
impossible
This
is
event
called
Italian
and
gambler
.
chance
cer tain
physician,
His
and
book
contained
techniques
on
cheat
how
how
1
0
astrologer
,
philosopher
,
mathematician
the
happen.
even
an
to
to
catch
and
others
at
1
cheating.
2
We
write
P (A)
to
represent
the
probability
of
an
event A
occurring.
A
probability
greater
Hence
There
0
≤
are
P (A)
three
≤
than
cannot
be
1.
1.
ways
of
finding
the
value
of
the
probability
of
an
Y
ou
can
write
event:
probability
decimal,
●
theoretical
as
a
fraction
or
probability
percentage.
●
experimental
●
subjective
probability
probability
.
On
a
dice
Theoretical
fair
the
fair
occur.
dice
The
has
list
six
of
numbered
equally
sides,
likely
all
of
possible
which
are
outcomes
equally
is
1,
2,
3,
likely
4,
5,
to
6.
outcome
same.
dice,
are
On
Probability
a
some
more
others.
of
probability
each
A
(unbiased)
probability
is
the
biased
outcomes
likely
than
W
e
call
n(U)
Let
=
a
6
list
event
one
6.
The
of
possible
shows
A
be
n(A)
=
that
defined
1
as
shows
probability
of
outcomes
there
are
six
‘the
that
getting
number
there
a
the sample space,
members
6
is
6’.
one
when
of
In
6
you
the
this
in
sample
the
roll
U.
dice
notation
space.
space
sample
the
The
sample
there
is
space.
is
one
out
of
1
six,
or
.
In
probability
notation,
6
n( A)
P( A )
1
=
=
n (U )
6
n( A)
➔
The
theoretical
probability
of
an
event A
is
P( A )
n (U )
where
and
➔
If
the
the
n(A)
n(U)
is
is
the
the
number
total
probability
event
to
of
occur
an
n
of
ways
number
×
of
event
P
that
event
possible
is
P,
in
n
A
can
occur
outcomes.
trials
you
would
expect
times.
A
Example
20-sided
polyhedron
is
called
an
icosahedron
fair
to
20-sided
20
is
dice
rolled.
with
The
faces
event
A
numbered
is
defined
8
1
as
2
A
20
Processes
‘the
number
obtained
is
a
multiple
of
4’.
that
are
too
1
3
6
1
complicated
to
allow
41
P (A).
The
dice
is
rolled
100
exact
4
Determine
9
a
How
many
times
would
you
expect
a
9
1
of
solved
using
probability
1
multiple
may
times.
be
b
analysis
4?
that
methods
employ
large
the
numbers’.
‘law
of
These
Answers
methods,
a
n(A)
=
5
and
n(U )
=
Find
20
in
n( A)
P( A ) =
5
=
n (U )
developed
n(A)
1
There
4
these
are
20
possible
outcomes.
5
of
the
40s,
1930s
are
and
known
=
20
16
are
and
multiples
of
4
(4,
8,
12,
as
20).
Monte
methods
Carlo
after
the
1
×
b
100
=
Probability
25
×
number
of
trials
famous
casino.
They
4
are
used
variety
Experimental
Sometimes
experiment
(empirical)
outcomes
to
are
estimate
not
probability
equally
likely
but
you
from
can
use
is
example,
being
components.
conclude
not
be
to
find
produced
If
that
the
the
all
case.
by
of
If
probability
factor y
first
the
the
wide
estimating
of
a
the
game
the
hand
in
an
card
‘Bridge’
probabilities.
the
a
a
situations,
strength
to
For
in
of
is
that
faulty
component
we
components
second
a
we
par ticular
would
test
are
is
faulty
faulty
.
component
is
not
component
test
Y
ou
could
However,
this
then
statistics
chain
some
we
faulty
that
modeling
may
we
to
of
the
a
nuclear
reaction.
may
wish
explore
the
applications
of
Monte
method
the
could
Carlo
1
conclude
that
the
probability
of
a
component
being
faulty
since
is
2
half
of
all
components
so
far
are
fur ther
.
faulty
.
Chapter
Continuing
this
process
a
number
of
times
and
calculating
the
ratio
The
the number
of
faulty
US
National
components
Weather
the number
of
components
used
gives
the relative frequency
of
the
component
being
nd
faulty
.
of
As
the
number
of
Ser vice
tes
s ted
components
tested
increases,
the
relative
this
the
closer
and
closer
to
the
probability
that
a
component
being
is
struck
No.
Y
ou
can
The
is
use
larger
to
the
Example
relative
the
frequency
number
of
as
trials,
an
the
estimate
closer
the
of
of
of
people
people
relative
frequency
cars passing the school gate one morning are given in the table:
Frequency
Red
45
Black
16
Y
ellow
2
Green
14
Blue
17
Gray
23
Other
21
T
otal
These
138
numbers
estimates,
Estimate
the
probability
that
the
next
car
to
pass
the
school
are
because
using
the
red.
relative
The
b
are
gates
we
be
population
probability
.
Color
will
in
struck
probability
.
The colors of
a
by
using
faulty
.
No.
➔
to
frequency
lightning ,
gets
method
probability
next
mor ning
350
cars
pass
the
school
an
Estimate
the
number
of
red
cars
that
frequency
as
gates.
estimate
of
the
mor ning.
probability.
Answers
45
The
a
relative
frequency
of
red
cars
is
This
probability
is
138
given
as
a
fraction.
In
45
So
the
probability
of
a
red
car
is
IB
exams
you
need
to
answers
or
138
give
When
b
350
cars
pass
the
school
decimals
45
the
number
of
red
cars
will
be
approximately
×
350
=
Y
ou
can’t
these
cases
subjective
For
win?
last
few
performed
played
in,
in
but
Probability
look
games
the
experiment
the
are
due
What
at
is
past
each
par ticular
eventually
the
play
has
weather
will
an
and
soccer
the
and
conditions
need
to
make
event
times.
based
Arsenal
Liver pool
two
how
the
a
of
In
on
belief.
against
that
between
played
number
of
probability
matches
you
large
information
to
team
a
probability
experience,
Liver pool
could
an
estimate
Premiership.
Y
ou
the
can
judgment,
example
English
as
you
repeat
3 sf
probabilities.
probability
always
to
114.
138
Subjective
exact
gate,
as
teams
match
‘guess’.
the
will
teams
the
in
will
well
have
be
for
Exercise
1
An
A
octahedral
(eight-sided)
dice
is
thrown.
The
faces
are
numbered
In
1
2
to
8.
What
the
a
an
b
a
multiple
of
3
c
a
multiple
of
4
d
not
e
less
A
even
is
a
that
the
number
thrown
is:
multiple
car
probability
all
number
than
used
probability
‘fair’,
told
of
30
of
unless
dealer
the
has
cars
150
are
used
cars
defective.
on
One
his
of
lot.
the
The
150
dealer
cars
is
What
table
random’
is
the
probability
that
it
is
below
shows
the
relative
the
students
Age
(in
any
car
years)
frequencies
of
the
chance
13
0.15
14
0.31
15
0.21
16
0.19
17
0.14
A
an
of
being
defective
the
cars
is
school.
as
likely
as
one
that
is
to
of
be
the
not
chosen
cars
defective.
1
student
is
randomly
selected
the
student
is
15
years
old,
ii
the
student
is
16
years
of
are
1200
Calculate
students
the
The
sides
of
The
table
shows
Number
on
a
What
six-sided
the
b
Do
the
you
this
school.
the
age
or
Find
the
older.
school.
15-year-old
spinner
for
are
100
students.
numbered
2
3
4
5
6
27
18
17
15
16
7
frequency
spinner
is
from
1
to
6.
spins.
1
relative
think
this
of
results
spinner
is
at
number
Frequency
a
from
that
i
There
4
high
One
of
ages
frequency
probability
b
a
has
defective?
Relative
Total
a
at
means
at
30
of
are
knows
selected
selected.
The
you
are
otherwise.
equal
3
questions,
coins
4?
that
random.
and
4
‘At
that
dice
of
fair?
getting
Give
a
a
1?
reason
for
your
answer.
c
The
spinner
times
5
the
Each
letter
card.
The
at
the
will
word
cards
are
3000
be
a
times.
Estimate
the
number
of
4.
CONSECUTIVE
placed
face
is
written
downwards.
A
on
card
a
is
separate
drawn
random.
What is
a
spun
result
of
11
is
the
the
letter
probability
C
b
of
the
picking
letter
P
a
card
c
with
a
vowel?
Chapter
The
6
spinner
getting
green
blue
is
twice
that
red
Frequency
0.4
A
bag
is
contains
.
an
of
Find
even
Venn
biased.
in
The
the
getting
probabilities
table.
The
of
getting
probability
of
red
and
getting
yellow
.
yellow
blue
green
0.3
probability
random.
a
is
shown
Color
Find the
7
shown
are
40
the
of
getting
discs
green.
numbered
probability
number,
that
1
the
has
b
to
40.
A
disc
number
the
digit
on
1
in
is
selected
the
at
disc
it.
John
diagrams
Venn
England
There
are
100
students
in
a
year
was
in
of
them
do
Hull,
father
and
were
priests
and
John
archer y
.
was
This
in
His
group.
grandfather
38
born
1834.
information
can
be
shown
on
a Venn diagram
also
their
to
encouraged
footsteps.
Gonville
and
In
to
follow
1853
Caius
he
in
went
College,
U
Cambridge,
A
Set
A
is
The
becoming
students
represents
For
do
100
a
archer y.
Therefore
n(A)
n(U)
the
next
=
way
is
chosen
at
random.
The
probability
Venn
to
student
does
38
P( A )
archer y
is
written
years
and
in
1857
college.
he
went
returned
1862
into
to
teach
to
logic
theor y.
developed
look
at
sets.
a
graphical
This
graph
that
became
the
in
probability
John
student
the
100.
and
A
ve
priesthood
Cambridge
38.
of
students.
the
=
graduated
fellow
the
38
who
and
rectangle
known
as
a
Venn
diagram.
P (A).
19
n( A)
100
Remember P( A)
50
=
n(U )
Complementary
The
area
students
outside
that
do
set
not
event
A
(but
do
A′
still
in
archer y
.
the
This
sample
is A′,
space
the
U)
represents
complement
of
the
set
A
U
A
38
62
n(A′ )
From
The
=
n(U)
the
–
n(A)
Venn
probability
diagram
that
n ( A )
P( A )
a
we
see
student
62
that
does
n(A′ )
not
=
do
100
–
38
=
62
archer y
,
31
Ever y
n (U )
100
does
Note
student
archer y
or
that
doesn’t
31
P (A′ )
+
P(A)
=
Probability
19
1
50
either
50
50
do
archer y.
➔
As
an
event,
A,
P( A ) P( A )
P( A )
the
Of
those,
Y
ou
38
100
can
do
22
do
The
the
of
16
show
16
and
does
not
happen.
events
30
both
this
students
archer y
on
a
play
and
Venn
badminton.
badminton.
diagram
like
this:
do
30
do
and
38
–
=
students
badminton
archer y.
region
students
badminton.
16
16
U
so
30
do
just
–
do
and
16
=
archer y,
14
badminton
48
is
B.
This
There
represents
students
both
written
The
as
A
22
that
archer y
badminton.
and
it
intersection
those
The
or
students
shaded
A
do
so
just
region
is
do
badminton
archer y,
of
students,
students
archer y.
happens
1 P( A )
Intersection
Of
either
1
–
are
16
and
students
region
not
∩
14
that
–
=
48
do
archer y
or
badminton.
B
probability
badminton
do
100
−
that
is
a
student
written
P(A
chosen
∩
at
random
does
both
archer y
B).
n(A
in
n(A
∩
B)
=
the
B)
is
the
number
intersection
of
16
the
n( A B )
P( A B )
∩
16
n (U )
sets
A
and
B
4
100
25
U
A
If
a
student
is
chosen
at
random
then
B
the
A
probability
that
badminton
but
a
student
does
do
does
not
archer y
is
∩
B′
do
written
P(A ∩ B ′).
22
students
out
of
48
100
22
P( A B )
do
archer y
but
11
not
100
badminton.
50
U
A′ ∩ B ′
or
represents
the
students
who
do
not
do
badminton
archer y
.
48
A′
Chapter
∩
B′
Union
of
events
The
U
of
shaded
A
and
those
B,
region
The
students
archer y
region
or
is
is
the
region
that
represents
do
badminton
written
union
either
or
both.
The
A ∪ B
48
Notice
that
‘or’
mathematics
The
or
probability
badminton
that
is
a
student
written
chosen
at
random
does
either
archer y
the
–
we
call
‘inclusive’
From
the
diagram,
n(A
∪
B )
=
22
+
16
+
14
=
it
the
or
.
is
from
the
hence
denition
n( A B )
P( A B )
52
A ∪ B ′
represents
all
of
probability.
13
n (U )
do
of
52
This
and
includes
possibility
both
P(A ∪ B).
in
U
100
those
25
students
that
either
do
archer y or
do
not
badminton.
A
∪
B′
48
n(A
∪
B ′)
=
22
+
16
+
48
n ( A B )
P( A B )
In
a
Draw
Use
of
and
a
hence
43
9
30
100
students,
play
Venn
your
and
50
group
games
86
86
n (U )
Example
=
17
play
computer
games,
10
play
board
neither.
diagram
diagram
to
to
find
a
a
student
chosen
at
b
a
student
plays
both
c
a
student
plays
board
show
the
this
information.
probability
random
from
computer
games
the
games
but
not
that:
group
and
plays
board
computer
board
games,
games,
games.
Answers
Let
B
=
C
=
plays
plays
Let
x
n(C
∩
=
computer
board
n (C
B ′)
=
∩
games,
You
B )
17
First
−
x
∩
B )
=
10
−
don’t
computer
and
use
n(C ′
define
your
notation.
games.
x
for
know
games
this
how
many
AND
do
board
games;
value.
x
U
C
B
17
–
10
–
x
9
{
Probability
Continued
on
next
page
(17
36
−
−
x)
x
x
+
=
=
x
+
(10
−
x)
+
9
=
The
30
four
diagram
30
and
6
U
so
Solve
regions
make
must
for
add
the
Venn
universal
up
to
set
U
30.
x.
Substitute
each
of
the
x
=
section
of
6
to
the
get
the
number
in
diagram.
9
Use
10
a
the
Venn
P( B ) =
=
P ( A ) =
3
n (U
6
P(C
∩ B
=
5
4
2
P(C ′ ∩ B ) =
=
30
Exercise
1
In
a
and
Draw
a
A child
has
2
In
the
One
of
or
25
have
many
girl
is
to
of
that
in
taken
have
is
a
chosen
15
of
study
both
random.
she
has
taken
both
b
she
has
taken
gymnastics
7
the
play
One
32
have
brown
eyes.
situation.
that
the
child
study
French,
13
One
from
French
have
girl
of
them
language.
random
both
13
and
taken
has
the
class.
Malay?
aerobics
done
What
before
neither
before.
activities?
a
Of
14
probability
neither
at
group.
gymnastics.
at
the
them
studies
PE
done
brown
this
Find
chosen
he
and
hair,
eyes.
them
students
girls
hair
blonde
represent
brown
5
have
random.
students,
and
25
at
10
blonde
diagram
hair
Exam-Style
4
both
selected
these
are
17
How
children,
probability
There
and
35
have
Malay
One of
3
4
Venn
is
class
study
is
of
blonde
a
15
B
group
eyes,
)
1
) =
30
c
and
n ( A)
30
b
diagram
1
Find
the
probability
that:
activities,
but
not
18
play
aerobics.
Question
students
both.
student
How
is
in
a
many
chosen
a
he
plays
golf
b
he
plays
the
class,
but
at
not
piano
play
16
play
the
piano
and
neither?
random.
the
but
golf,
Find
the
probability
that:
piano,
not
golf.
Chapter
Exam-Style
5
The
universal
than
or
A
B
a
b
=
A
ii
B
6
In
the
is
both
ii
neither
town
subsets
are
the
A
set
and
multiples
are
factors
of
B
of
of
positive
are
integers
defined
less
as:
3}
30}
of
of
chosen
A
at
probability
a
as
and
B
in
the
appropriate
region
U
diagram.
i
a
that
that
elements
number
the
defined
The
elements
Venn
Find
is
15.
{integers
Place
a
to
U
{integers
the
i
A
set
equal
=
List
on
c
Question
multiple
a
40%
that
of
multiple
of
the
random
3
the
and
of
3
from U.
number
a
factor
nor
a
population
is
of
factor
read
30,
of
30.
newspaper
‘A
’,
For
30%
read
newspaper
‘B’,
10%
you
newspaper
It
is
found
both
‘A
’
Also,
Find
and
2%
the
random
a
c
of
‘C’;
the
from
read
and
reads
reads
both
3%
people
the
only
will
three
5%
probability
reads
b
‘C’.
that
this
read
that
‘A
’
read
a
question
read
and
both
all
‘B’
three
person
‘B’;
4%
and
read
Venn
‘C’.
to
circles
to
in
diagram
represent
use
the
–
one
each
newspaper
.
newspapers.
chosen
need
at
U
town
‘A
’,
only
‘B’,
none
of
the
three
newspapers.
C
The
addition
rule
Here
is
diagram
the
badminton
Venn
from
page
for
the
students
who
do
archer y
and
69.
U
The
probability
archer y
a
and
student
includes
that
the
does
the
a
student
probability
badminton
does
that
each
probabi li ty
that
a
48
student
does
badminton.
n(A
or
n(A
So,
➔
∪
B)
∪
P(A
For
any
Probability
B)
∪
P(A
=
38
=
B)
two
∪
+
30
n(A)
=
−
+
P(A)
events
B)
=
only
archery
wish
to
and
include
16,
n(B)
+
A
P(A)
both
We
−
P(B)
and
+
n(A
−
∩
P(A
∩
B)
B
P(B)
–
this
probability
one
of
once
so
we
B)
P(A
∩
B)
these
probabilities.
subtract
Playing
In
the
next
familiar
playing
four
example
with
suits
–
card
Find
is
the
a
clubs,
cards,
you
ordinar y
In
The
Example
A
an
cards.
diamonds.
black
red
cards
pack
need
to
be
in
pack
of
52
10,
there
spades,
clubs
hear ts
and
and
The
are
hear ts
spades
similar
2,
or
in
the
cards
4,
5,
6,
7,
8,
9,
King.
and
picture
cards
13
3,
and
Queen
the
to,
are
Ace,
Queen
Jack,
playing
are
There
suit:
Jack,
called
and
are
diamonds
cards.
each
King
cards.
your
are
Are
countr y
same
as
these?
drawn
at
random
probability
that
from
the
an
card
is
ordinar y
a
hear t
pack
or
a
of
52
playing
cards.
king.
Answer
We
require
P(H
∪
K).
U
H
Draw
a
Venn
There
are
13
There
are
4
There
is
Using
P( H
diagram.
K
A ♥
K ♣
Q ♥
J ♥
10 ♥
9 ♥
8 ♥
7 ♥
6 ♥
5 ♥
4 ♥
3 ♥
2 ♥
K ♥
K ♦
K ♠
hear ts
in
the
pack.
13
P(H )
=
52
4
P(K )
kings
in
the
pack.
=
52
1
P(H
∩
K )
1
card
that
is
both
a
king
and
a
hear t.
=
52
So
13
P(H
∪
K )
4
+
=
52
1
–
52
4
16
=
52
=
52
∪
K )
=
P( H )
+
P( K )
–
P( H
∩
K )
13
Chapter
Example
9
If
A
and
B
are
two
events
such
that
P(A)
3
=
and
P(B)
=
20
P(A
∪
B)
P(A
a
=
∪
2P(A
∩
B )
B)
find
P(A
b
and
10
∪
B )′
P(A
c
∩
B ′ ).
Answers
Let
a
P(A
∩
B )
9
=
x
Use
3
P (A ∪ B ) = P (A ) + P (B ) – P (A ∩ B )
2x
=
+
–
20
x
10
15
3x
=
20
3
x
=
÷
3
4
1
x
=
=
P(A
∩
B)
4
1
P(A
∪
B)
=
Since
P (A
∪
Since
P(A′)
B )
=
2P(A
∩
B).
2
1
If
b
P(A
∪
B )
=
then
2
1
1
P(A
∪
B )′
=
1
–
=
=
1
–
P(A).
2
2
1
If
c
P(A
∩
B )
Use
=
result
from
par t
a
4
P(A
∩
B ′ )
=
P(A)
–
P(A
1
9
–
=
20
∩
This
B )
is
the
diagram
1
region
that
is
A
on
the
Venn
without
its
=
intersection
5
4
with
B.
U
P(B)
P(A)
1
P(A
∩
B)
=
4
Exercise
1
Two
two
The
C
dice
are
numbers
shown
following
500
on
times.
the
frequencies
dice
are
For
is
each
throw
,
written
the
sum
of
the
down.
obtained:
Sum
2
3
4
5
6
7
8
9
10
11
12
Frequencies
6
8
21
34
65
80
63
77
68
36
42
Using these
thrown
frequencies,
calculate
a
the
sum
being
exactly
b
the
sum
being
an
c
the
sum
being
exactly
Probability
the
divisible
even
probability
by
5,
by
5
of
number,
divisible
or
being
an
even
number.
2
A
ten-sided
probability
3
dice,
numbered
1
to
10,
is
rolled.
a
the
number
scored
is
a
b
the
number
scored
is
either
a
prime
c
the
number
scored
is
either
a
multiple
In
a
22
group
are
of
80
females
picked
Calculate
the
that:
from
tourists
with
this
prime
40
have
cameras.
group
at
number,
number
of
cameras,
Find
random
the
is
4
50
or
or
a
are
a
multiple
multiple
female
probability
either
a
that
camera
a
of
of
4,
3.
and
tourist
owner
or
female.
4
A
letter
Find
5
is
the
chosen
at
probability
random
that
it
from
a
in
the
word
MA
THEMA
TICS
b
in
the
word
TRIGONOMETRY
c
in
the
word
MA
THEMA
TICS
and
d
in
the
word
MA
THEMA
TICS
or
A
a
student
work
fiction
What
the
is
librar y
.
0.40,
non-fiction
is
fiction,
b
to
fiction
and
What
a
goes
of
the
the
is
26-letter
English
alphabet.
The
work
in
in
the
the
word
word
probability
of
TRIGONOMETRY
TRIGONOMETRY
that
non-fiction
is
she
checks
0.30,
and
out
both
0.20.
probability
non-fiction,
is
a
the
is
or
that
the
student
checks
the
student
does
out
a
work
of
both?
probability
that
not
check
out
a
book?
Exam-Style
Questions
1
6
In
a
cer tain
road
of
the
houses
have
no
newspapers
delivered.
3
1
If
3
have
a
national
paper
delivered
and
4
delivered,
has
have
a
local
paper
5
what
is
the
probability
that
a
house
chosen
at
random
both?
1
7
If
X
and
Y
are
two
events
such
that
P(X)
=
1
and
P(Y)
4
=
and
8
1
P(X
∩
Y)
=
,
find
8
8
a
P(X
∪
Y)
b
P(X
∪
Y)′
=
0.2
If
a
P(A)
P(A
∪
b
P(A
c
P(A′
and
P(B)
=
0.5
and
P(A
∩
B)
=
0.1,
find
B)
∪
∪
B)′
B).
Chapter
U
Mutually
exclusive
In
sur vey
a
student
it
is
events
found
that
A
32
students
play
C
chess.
38
Chess
time
and
so
a
archer y
student
clubs
are
cannot
do
on
the
both
same
day
archer y
at
and
the
32
same
chess.
30
The
events
These
same
so
are
Now
∩
∪
Generally
follows
Hence
P(A
=
C )
if
∪
B)
=
=
Example
A
box
P(A)
∩
B)
adapt
P(A)
∪
two
see
hence
+
events
In general, if
P(A
called
can
and
P(A
can
are
we
two
that
we
0
C
where
Here
C )
P(A
and
events
time.
n(A
➔
A
that
P(A
and
the
∩
–
B
cannot
circles
C )
=
events.
exclusive
occur
do
not
at
the
overlap,
0.
0.
are
mutually
exclusive
then
it
0.
the
+
outcomes
P(C )
A
=
mutually
addition
r ule
in
these
cases
P(B).
A and B are mutually exclusive, then P(A ∩ B) = 0 and
B)
=
P(A)
+
P(B).
contains
board-pens
of
various
colors.
A
teacher
picks
out
a
1
pen
at
random.
The
probability
of
drawing
out
a
red
pen
is
the
probability
of
drawing
out
a
green
pen
,
and
5
3
is
7
What
is
the
probability
red
pen
of
drawing
neither
a
red
nor
a
green
pen?
Answer
Let
G
R
=
P(R
=
green
∪
G )
pen
=
drawn,
P(R)
1
+
P(G )
define
+
R
and
22
events.
35
The
3
=
First
your
notation.
drawn.
G
are
mutually
exclusive
=
5
7
pen
teacher
or
a
either
green
draws
pen,
but
out
not
a
red
both
colors.
13
22
P(R
∪
G)′
=
1
–
=
Since
Exercise
1
Here
A:
B:
C:
D:
E:
dice
total
there
a
A
e
B
is
two
both
Which
some
both
the
and
and
events
show
is
at
7
are
these
B
E
or
least
dice
dice
of
Probability
=
1
–
P(A).
D
are
the
P(A′)
35
35
a
to
throwing
two
dice:
4
more
one
show
6
the
same
number
odd
pairs
b
f
relating
of
A
C
events
and
and
C
D
are
mutually
c
g
A
B
and
and
exclusive?
D
C
d
A
and
E
Exam-Style
Question
1
1
Two
2
events
N
and
M
are
such
that
P(N)
=
and
P(M)
=
and
10
5
3
P(N
∪
M)
=
.
10
Are
In
3
N
a
and
and
M
group
of
27
are
a
that
In
an
students,
a
or
30
events?
are
freshmen
(second-year
student
freshman
Exam-Style
4
89
exclusive
sophomores
probability
either
mutually
picked
(first-year
students).
from
this
Find
students)
the
group
at
random
is
school
A
winning
sophomore.
Question
inter-school
quiz,
the
probability
of
competition
is
the
1
1
,
the
probability
of
school
B
winning
is
and
4
3
1
the
probability
of
school
C
winning
is
5
Find
the
a
A
c
none
.
or
probability
B
wins
of
these
Sample
the
the
that
competition,
wins
the
space
product
A,
b
B
or
C
wins,
competition.
diagrams
and
rule
A
Y
ou
can
list
all
the
possible
outcomes
of
an
experiment
if
there
question
you
not
too
may
tell
are
to
list
the
possible
many
.
outcomes.
Example
A
fair
as
spinner
shown
is
outcomes
Hence
is
from
find
greater
with
spun
the
than
the
three
this
numbers
times.
2
all
and
the
3
on
it
possible
experiment.
probability
the
1,
List
scores
that
on
the
the
score
first
two
on
the
last
spin
spins.
Answer
The
1
1
27
outcomes
1
1
2
are:
When
1
1
3
1
need
not
1
1
2
1
2
2
1
3
2
1
1
3
1
2
3
1
3
3
2
1
1
2
2
1
2
3
1
2
1
2
2
2
2
2
3
2
2
1
3
2
2
3
3
1
1
3
2
1
3
1
2
3
2
2
3
3
2
3
1
3
3
2
3
3
3
3
Of
these,
score
the
on
the
the
scores
five
last
on
2
3
in
miss
all
the
outcomes,
systematic
any
so
that
you
you
do
out.
3
have
greater
first
be
1
red
spin
the
3
3
listing
to
two
the
than
spins.
5
Hence
the
probability
is
27
Chapter
Sample
A
sample
space
space
outcomes
of
Example
Draw
Find
a
an
sample
the
diagram
is
Sample
another
way
of
showing
all
the
space
diagrams
possible
event.
are
also
called
probability
space
diagrams.
space
probability
obtaining
a
diagrams
a
diagram
to
represent
the
scores
when
two
dice
are
thrown.
of:
score
of
6
throwing
b
a
double
scoring
c
less
than
6.
Answers
DICE
1
2
1
3
4
5
6
2
ECID
1
(1,
1)
(2,
1)
(3,
1)
(4,
1)
(5,
1)
(6,
1)
(1, 1)
2
(1,
2)
(2,
2)
(3,
2)
(4,
2)
(5,
2)
(6,
2)
score
3
(1,
3)
(2,
3)
(3,
3)
(4,
3)
(5,
3)
(6,
3)
4
(1,
4)
(2,
4)
(3,
4)
(4,
4)
(5,
4)
(6,
4)
5
(1,
5)
(2,
5)
(3,
5)
(4,
5)
(5,
5)
(6,
5)
6
(1,
6)
(2,
6)
(3,
6)
(4,
6)
(5,
6)
(6,
6)
There
are
36
possible
outcomes
illustrated
on
gives
of
a
score
of
2,
(4, 6)
gives
a
10.
this
diagram.
The
5
a
P(6)
five
possible
ways
of
getting
highlighted
36
in
DICE
2
ECID
6
P(double)
=
of
6
are
2
4
5
6
1
(1,
1)
(2,
1)
(3,
1)
(4,
1)
(5,
1)
(6,
1)
2
(1,
2)
(2,
2)
(3,
2)
(4,
2)
(5,
2)
(6,
2)
3
(1,
3)
(2,
3)
(3,
3)
(4,
3)
(5,
3)
(6,
3)
4
(1,
4)
(2,
4)
(3,
4)
(4,
4)
(5,
4)
(6,
4)
5
(1,
5)
(2,
5)
(3,
5)
(4,
5)
(5,
5)
(6,
5)
6
(1,
6)
(2,
6)
(3,
6)
(4,
6)
(5,
6)
(6,
6)
1
The
six
possible
6
highlighted
in
ways
of
getting
a
double
are
red.
DICE
1
2
1
3
4
5
6
2
ECID
1
(1,
1)
(2,
1)
(3,
1)
(4,
1)
(5,
1)
(6,
1)
2
(1,
2)
(2,
2)
(3,
2)
(4,
2)
(5,
2)
(6,
2)
3
(1,
3)
(2,
3)
(3,
3)
(4,
3)
(5,
3)
(6,
3)
4
(1,
4)
(2,
4)
(3,
4)
(4,
4)
(5,
4)
(6,
4)
5
(1,
5)
(2,
5)
(3,
5)
(4,
5)
(5,
5)
(6,
5)
6
(1,
6)
(2,
6)
(3,
6)
(4,
6)
(5,
6)
(6,
6)
{
Probability
1
3
=
36
score
yellow.
1
b
a
=
Continued
on
next
page
5
10
P(score
c
<
6)
=
The
10
ways
of
getting
a
score
18
36
highlighted
in
DICE
2
ECID
In
an
Find
less
the
the
a
sample
coin
space
probability
than
6
are
2
1
3
4
5
6
1
(1,
1)
(2,
1)
(3,
1)
(4,
1)
(5,
1)
(6,
1)
2
(1,
2)
(2,
2)
(3,
2)
(4,
2)
(5,
2)
(6,
2)
3
(1,
3)
(2,
3)
(3,
3)
(4,
3)
(5,
3)
(6,
3)
4
(1,
4)
(2,
4)
(3,
4)
(4,
4)
(5,
4)
(6,
4)
5
(1,
5)
(2,
5)
(3,
5)
(4,
5)
(5,
5)
(6,
5)
6
(1,
6)
(2,
6)
(3,
6)
(4,
6)
(5,
6)
(6,
6)
experiment
Draw
than
green.
1
Example
less
=
3
on
the
is
tossed
diagram
that
in
a
and
for
a
dice
this
single
is
rolled.
experiment.
experiment
you
obtain
a
head
and
a
number
dice.
Answer
1
2
3
4
5
(1,
H)
(2,
H)
(3,
H)
(4,
H)
(5,
H)
T
(1,
T)
(2,
T)
(3,
T)
(4,
T)
(5,
T)
2
P(head
and
number
less
than
3)
Three
H)
(6,
T)
outcomes
than
3
are
that
give
a
head
and
=
One
6
coins
possible
are
tossed
outcome
is
one
that
at
a
all
time
the
and
coins
the
are
results
heads.
are
unbiased
HHH.
Another
is
that
the
first
two
coins
are
heads
one
is
a
tail.
This
is
written
that
the
Find
2
complete
the
a
the
b
at
c
heads
‘Two
probability
number
least
Draw
a
two
and
the
of
heads
tails
tetrahedral
the
the
1
to
number
blue
difference
c
the
red
the
for
just
as
heads
up
this
random
tails
up.
experiment.
greater
tossed
tossed
are
the
number
of
tails,
consecutively
,
alter nately
.
diagram
one
They
than
blue
for
the
random
and
the
other
rolled
and
the
experiment
red,
result
are
each
noted’.
that:
the
red
dice
is
greater
than
the
number
on
the
dice,
the
d
dice,
on
b
even
are
probability
is
are
space
4.
space
is
land
that:
heads
sample
numbered
Find
sample
to
HHT
.
as
List
is
and
likely
last
coin
This
one
written
the
less
=
An
is
number
Question
unbiased
noted.
a
highlighted.
E
Exam-Style
1
(6,
1
12
Exercise
The
6
H
dice
between
shows
an
the
odd
numbers
number
on
the
dice
and
the
blue
is
one,
dice
shows
an
number,
sum
of
the
numbers
on
the
dice
is
prime.
Chapter
Exam-Style
Question
Genetic
3
A
box
1,
2,
contains
three
cards
bearing
the
ngerprinting
numbers
Genetic
3.
A
second
box
contains
four
cards
in
the
numbers
2,
3,
4,
5.
A
card
is
chosen
at
1984
each
Draw
the
box.
of
sample
space
diagram
for
the
random
experiment.
Find
the
probability
a
the
cards
b
the
larger
have
that:
the
same
by
number,
University
us
has
the
two
numbers
drawn
is
the
sum
less
d
the
e
at
of
than
the
two
numbers
on
the
contained
is
inherited
The
DNA
and
body
of
one
can
cards
bag.
the
numbers
even
number
numbered
One
it
is
in
the
from
DNA
our
extracted
and
one
make-up
and
parents.
from
analyzed
characteristics
our
‘genetic
When
matching
usual
to
cells
to
(seen
ngerprint’.
‘ngerprints’
on
the
cards
is
at
least
is
compare
0,
1,
2,
is
drawn
at
replaced
in
it
is
3,
bag.
bands.
of
these
comparisons
have
chosen.
4
random,
the
these
8,
used
and
5,
are
placed
its
number
noted,
Then
a
second
as
but
evidence
the
eld
to
is
convict
under
in
due
to
its
reliance
on
and
probability.
then
Each
genetic
is
scrutiny
a
be
uids
the
–
criminals
cards,
at
3,
been
Six
Jeffreys
Leicester
.
is
Some
4
developed
Alec
7,
product
least
of
unique
which
below)
c
a
which
produce
of
was
Professor
random
the
from
ngerprinting
with
card
Usually
between
10
is
and
20
bands
are
examined
and
chosen.
compared.
Draw
the
sample
space
diagram
for
the
Experimental
evidence
random
has
suggested
one
band
that
the
probability
of
experiment.
Find
the
probability
that:
matching
(although
a
the
cards
have
b
the
larger
the
same
the
two
coincidence
numbers
this
gure
is
subject
drawn
is
The
probability
of
two
prime,
1
matching
the
c
sum
less
the
d
e
5
than
least
8,
at
least
one
plays
rolls
metre.
moves
metre.
the
If
If
at
exactly
c
more
the
2
2
or
numbers
number
with
the
metre.
6
he
on
1
but
If
the
cards
will
therefore
is
for
rolled
are
not
is
it
is
called
1
he
where
away
4
He
he
he
from
than
2
a
coin
he
the
up
If
moves
it
and
is
left
Go’.
one
3
he
one
is.
makes
two
steps.
is
star ted,
his
star ting
metres
is
independent .
influence
is
‘Come
moves
independent
and
cards
metre.
he
where
that
the
from
his
star ting
point?
events
tossed,
This
point,
away
is
outcome
as
in
Example
because
of
be
16
chosen.
dice
twice.
less
is
on
rightone
stays
dice
point
a
score
moves
metres
events
Probability
numbers
probability
rule
is
If
he
the
than
does
even
one
5
the
game
same
dice
the
coin
is
the
b
a
is
rolls
is
Product
it
a
When
a
of
dice.
down
Tilman
What
it
two
7,
at
He
page,
the
product
Tilman
the
of
is
to
4
number,
debate).
of
by
1
the
rolling
9
on
the
outcome
the
dice
and
of
previous
tossing
vice
versa.
bands
➔
Two
events
does
not
Here
is
the
A
the
(1,
Let
H
From
(2,
T)
stand
the
for
L
stand
dice
(3,
T)
4
H)
(3,
and
(4,
T)
event
‘coin
(5,
T)
lands
occurs.
6
H)
(5,
one
coin.
5
H)
(4,
a
other
of
(6,
T)
H)
(6,
T)
heads’.
2
for
the
event
‘dice
score
less
than
3’.
1
4
=
3
12
L)
There
1
2
∩
a
the
occurrence
1
=
P(H
for
that
the
=
=
P(L)
chance
if
diagram:
12
Let
independent
3
the
6
P(H)
space
H)
(2,
are
the
2
H)
(1,
T
B
affect
sample
1
H
and
=
are
outcomes
coin
we
can
also
note
∩
L)
P(H)
=
is
than
When
two
P(A
B)
∩
This
is
the
This
is
also
Sample
space
possible
One
1
bag
red
Find
and
a
both
c
at
A
×
3.
called
and
B
are
independent
P(B)
product
rule
the
diagrams
but
for
independent
multiplication
can
you
help
you
don’t
events
r ule.
visualize
always
need
the
to
number
draw
of
one.
contains
the
less
6
P(A)
outcomes,
Example
is
=
3
events
=
and
score
1
×
2
➔
heads
dice
P(L)
×
1
1
=
the
that
the
P(H
where
6
12
But
two
=
4
3
white
red
balls.
probability
the
least
balls
one
A
2
white
ball
is
balls,
selected
another
at
bag
random
contains
from
each
bag,
that
are
ball
and
red,
is
b
the
balls
are
different
colors,
white.
Answers
3
The
a
From
the
first
bag
P(R
)
events
‘picking
a
red
from
the
first
bag’
(R
=
)
1
1
5
and
‘picking
a
red
from
the
second
bag’
(R
the
second
bag
P(R
)
=
)
are
2
1
From
independent
events.
In
R
2
there
are
3
red
balls
in
5.
1
5
In
R
there
is
events
R
1
red
ball
in
5.
2
Therefore
P(R
1
∩
R
)
2
The
and
R
1
3
3
1
P( R
1
=
×
5
∩
R
2
)
=
are
independent,
so
2
P(R
1
)
×
P(R
).
2
=
5
25
{
Continued
on
next
Chapter
page
3
From
b
the
first
bag
P(R
)
=
If
the
balls
are
dif ferent
colors,
this
means
either
the
1
5
first
one
is
red
and
the
second
one
white
or
the
first
one
4
From
the
second
bag
P(W
)
=
is
white
and
the
second
one
red.
2
5
Therefore
P(R
∩
W
1
3
4
=
)
2
12
×
=
5
5
25
2
From
the
first
bag
P(W
)
=
1
5
1
From
the
second
bag
P(R
)
=
2
5
Therefore
P(W
∩
R
1
2
2
1
=
×
=
5
25
5
P(different
P(R
∩
W
1
2
)
25
=
1
=
1
P(W
least
–
one
exclusive
events.
)
2
–
P(R
∩
For
white)
probability
R
1
1
R
mutually
25
that
both
are
–
‘at
least
calculate
red
probability
)
one
the
of
the
balls
probability
that
the
first
is
that
is
white’
both
white
we
are
and
could
white,
the
the
second
red
2
and
22
3
=
∩
1
are
=
25
P (at
+
These
=
14
+
c
colors)
2
12
)
2
the
probability
that
the
first
is
red
and
the
second
=
is
25
25
white.
OR
If
at least one is white then it means that both cannot
be red.
This
is
involve
a
common
the
words
complement
Exercise
1
My
wardrobe
red,
and
choose
shir t
2
A
one
contains
white
a
shir t
another.
both
card
the
at
of
solving
least...’.
is
five
and
shir ts
one
without
What
is
with
black.
I
looking.
the
one
reach
I
blue,
into
replace
probability
that
one
the
choosing
large
this
I
shir t
will
and
choose
the
chosen
and
a
a
at
random
second
king
school
and
card
a
from
a
deck
of
52
cards.
It
is
then
is
chosen.
What
is
the
probability
questions
of
you
It
is
three
a
sur vey
found
Probability
are
chosen
students
may
need
to
ten?
conducts
cafeteria.
students
all
and
red
times?
of
the
food
provided
by
remind
yourself
playing
cards
the
that
of
the
students
like
pasta.
5
that
2
then
page
Three
the
brown,
4
school
that
–
wardrobe
8,
A
1
event.
For
replaced
3
problems
Calculate
F
one
choose
of
method
‘…
like
at
random.
pasta?
What
is
the
probability
73.
–
about
see
Exam-Style
4
Adam
is
Questions
playing
in
a
cricket
match
and
a
game
of
hockey
at
the
weekend.
The
and
probability
the
Assume
the
5
that
Three
events
exclusive
6
∪
C)
=
Determine
a
An
a
coin
head
will
the
team
and
=
in
C
0.2,
the
win
such
P (C )
cricket
hockey
matches
will
are
win
the
are
in
that
=
0.3,
is
is
What
is
matches?
and
P (A
0.75,
0.85.
independent.
both
A
match
match
B
∪
are
B)
=
mutually
0.4
and
0.34.
b
toss
his
P (A)
Calculate
I
B
team
winning
results
A,
and
his
of
that
a
get
7
the
probability
P (B
that
probability
P (B)
air-to-air
P (B
whether
and
on
and
roll
the
a
and
C ).
C
are
six-sided
coin,
missile
B
∩
and
has
independent.
dice.
don’t
get
Find
a
6
8
probability
the
on
of
probability
the
that
I
dice.
hitting
a
target.
If
five
9
missiles
not
8
with
a
cards
10
is
from
a
the
probability
chosen
What
is
standard
the
deck
probability
P (E ′ )
P (F )
=
0.6
and
P (E
explain
why
E
and
F
are
independent,
c
explain
why
E
and
F
are
not
d
find
P(E
bags
the
target
is
is
the
playing
choosing
4
cards
hear ts
and
B
=
are
∩
blue
dice
is
=
0.24
mutually
exclusive,
is
the
4
from
red
each
that
and
the
and
first
third
probability
blue
marbles.
One
marble
bag.
the
numbered:
8
1,
marble
marble
2,
that
2,
5,
the
will
be
red,
the
red?
6,
6.
It
scores
is
add
thrown
up
to
three
6?
Question
0.3.
P (A
contain
probability
What
F )
F ′ ).
chosen
marble
Exam-Style
P(B)
52
P(E ),
each
six-sided
times.
a
∪
randomly
second
A
of
of
∩
b
Three
down
=
write
A
that
Question
that
a
What
12
are
replacement.
Given
is
11
what
row?
Exam-Style
9
launched,
destroyed?
Four
in
are
independent
events
such
that
P(A)
=
0.9
and
Find:
B)
b
P (A
∩
B ′)
c
P (A
∪
B)′
Chapter
Exam-Style
13
Independent
and
P(G′
Draw
Let
15
I
a
∩
events
H)
Venn
P(G
Find
14
Question
∩
two
throw
=
and
=
possible
to
dice.
values
Find
four
dice
show
a
b
all
four
dice
show
the
more
such
represent
likely:
of
the
all
is
are
that
P (G
∩
H ′ )
=
0.12
the
events G
and
H
x.
a
Which
H
0.42.
diagram
H)
four
G
x
probability
that
6,
same
rolling
number.
a
‘six’
on
four
throws
of
one
dice,
This
or
rolling
a
‘double
six’
on
24
throws
with
two
you
16
A
program
produces
(independently)
three
is
to
9.
For
random
digits
from
or
309
or
088
a
Find
the
probability
that
none
b
Find
the
probability
that
at
Investigation
following
American
The
is
a
–
the
famous
television
name
comes
game
from
least
Monty
probability
show
the
of
the
one
digit
original
a
on
closed
opens
After
for
the
one
of
he
has
par ticipants
last
time
the
What
being.
two
opened
should
After
they
you
Monty
remaining
one
whether
remaining
are:
of
the
want
have
Hall,
doors
doors
to
stay
always
shown
with
their
the
the
main
other
three
prize
two
doors
prizes
the
are
doors
door
,
the
is
reveals
dud,
rst
door
behind
an
the
doors,
unwanted
Monty
choice
remains
or
Hall
to
then
prize.
asks
switch
the
to
the
door
.
you
do?
We
a
Stick
with
doors.
prizes.
unwanted
behind
show
of
show.
what
a
is
game
choice
unwanted
the
randomly
a
the
the
door
dud,
the
knows
and
on
behind
and
chosen
who
and
one
are
before
Hall.
given
and
car
placed
game
5.
5.
you’re
you’re
there
the
a
the
Monty
car)
The
a
is
Deal’.
Behind
of
64.
di lemma
host,
(a
digits
is
based
Make
and
rules
your
rst
will
revisit
Switch
c
It
to
the
other
remaining
closed
Probability
not
matter
.
Chances
are
at
the
door
.
this
does
this
choice.
problem
b
the
on
936
three
Hall
puzzle
‘Let’
s
show’
s
or
Suppose
The
in
Investigation
example
247
The
question
considered
dice
page
0
the
dice?
even.
chapter
.
end
of
.
Conditional
Here
is
the
Venn
probabi li ty
diagram
showing
students
who
do
archer y
and
badminton.
U
48
If
we
know
affect
the
do
write
30
par ticular
the
that
students
a
how
does
this
archer y?
of
student
these
does
16
also
archer y
do
archer y
.
given
that
they
that
A| B
(
)
16
=
is
known
dependent
8
=
as
15
conditional
the
on
=
30
n( B )
This
do
badminton,
)
n( A B )
P
also
does
badminton;
that
P ( A| B
as
student
they
do
probability
badminton
Note
a
probability
Altogether
We
that
outcome
since
probability
of
the
outcome
of
A
is
B.
16
P( A B )
It
also
follows
that
P ( A| B
)
100
=
=
30
P( B )
100
16
=
8
=
30
➔
In
general
occurring
for
two
given
events
that
B
A
has
and
15
B
the
occurred
probability
can
be
found
of
A
using
Recall
that
for
independent
events
P( A B )
P
(
A| B
)
=
P(A
∩
B)
=
P(A)
×
P(B).
P( B )
By
denition
independent
Rearranging
the
formula
of
∩
B)
=
P
(
A| B
)
×
A
B,
the
occurring
that
occurred
If
A
and
B
are
independent
A| B
)
=
P(A),
P
(
B|A
)
=
P(B),
P
(
A| B ′ )
=
P(A)
P
(
B | A′ )
=
will
has
equal
probability
occurring ,
not
and
B
events,
the
(
A
probability
P(B).
given
➔
events,
gives
and
P(A
for
of
since
affected
by
A
A
is
the
P(B).
occurrence
of
B.
Chapter
Example
Of
10
the
53
drink
How
a
One
staff
at
neither
many
member
school,
nor
staff
of
drinks
a
tea
drink
staff
tea
is
but
36
drink
tea,
18
drink
coffee,
and
coffee.
both
chosen
not
tea
at
and
coffee?
random.
b
he
c
if
he
is
a
tea
drinker
he
drinks
d
if
he
is
a
tea
drinker
he
does
Find
the
probability
that:
coffee,
coffee
not
as
drink
well,
coffee.
Answers
Draw
a
a
Venn
diagram
to
show
the
U
infor mation
36
–
18
–
x
10
n( T
Let
n (T
∩
C )
=
∩
C )
is
the
number
who
drink
x
both
tea
and
cof fee.
so
53
36
–
64
–
x
x
x
+
x
=
+
18
–
x
+
10
=
is
the
total
number
of
staf f
on
the
53
Venn
diagram.
Solve
for
Since
x
53
=
x.
11
11
Therefore
P(T
∩
C )
=
=
11
and
total
=
53.
53
25
P(T
b
∩
C ′)
=
36
–
11
=
25
33
11
P(C ∩ T)
P(C|T )
c
=
53
=
36
P(T)
53
11
53
×
=
53
11
=
56
36
25
P(C′ ∩ T)
P(C ′|T )
d
=
53
=
P (C ′
∩
T )
=
P (T
∩
C ′)
36
P(T)
53
25
53
×
=
53
Exercise
There
Four
One
are
do
36
G
Exam-Style
1
25
=
36
Questions
27
students
neither
person
is
in
subject.
chosen
at
a
class.
How
he
or
she
takes
Theater
b
he
or
she
takes
at
c
he
or
she
takes
Theater,
least
take
many
random.
a
Probability
15
but
one
given
and
students
Find
not
of
Ar t
the
do
20
take
both
Theater.
subjects?
probability
that
Ar t,
the
two
that
he
subjects,
or
she
takes
Ar t.
2
For
events
P(A)
48%
a
and
∩
of
A
B
P(B)
all
is
is
1
own
roller
blades
at
P(A′
∩
B ′)
=
0.35;
a
skateboard
that
What
the
random
4
P(B ′|A′).
c
blades.
given
chosen
2
that:
Find
P(A|B)
b
and
known
0.6.
teenagers
roller
number
it
=
B)
skateboard
owns
4
A
0.25;
P(A
a
3
=
7
and
is
teenager
from
this
11
39%
the
of
all
teenagers
probability
owns
list
a
of
16
that
a
own
teenager
skateboard?
eight
22
numbers:
29
Find:
5
a
P (it
is
even
b
P (it
is
less
than
15
c
P (it
is
less
than
5
d
P (it
lies
In
my
a
town
6
The
0.1.
is
The
the
U
P (V)
8
A
V
=
A
are
and
on
first
The
that
less
and
20
|
it
4)
than
than
5)
15)
lies
households
between
have
computer
household
a
student
that
that
of
draw
table
players
in
her
both
a
has
a
takes
student
student
exclusive
P (U
class
tests
those
a
and
a
5
and
desktop
a
laptop
25).
computer.
61%
computer.
What
laptop
Design
takes
takes
computer
given
Technology
Design
Spanish
events.
|V)
an
and
who
black
The
P (U)
and
IB
white
0.34,
and
is
0.47.
What
draw
,
Paper
passed
probability
the
given
is
P (U
c
52%
is
second
Exam-Style
10
a
b
gave
marble
the
is
of
greater
that
of
all
is
it
has
and
Spanish
Technology
given
that
the
is
0.6.
is
What
student
is
=
0.26;
Find
replacement.
the
multiple
is
desktop
that
V)
contains
white
it
10
mutually
0.37.
passed
jar
|
a
it
computer?
percentage
9
|
all
probability
teacher
class
not
Technology?
P (U
a
a
probability
and
is
of
have
probability
taking
7
95%
probability
desktop
it
between
households
the
|
of
1
the
the
and
first
test
selecting
the
a
of
first
also
Paper
the
of
chosen
black
a
was
the
What
second
and
selecting
drawn
of
test.
the
are
a
35%
first
marble
selecting
marble
2.
passed
marbles
black
probability
the
IB
passed
Two
probability
that
V).
an
class
marbles.
of
or
without
then
a
marble
white
test?
on
marble
black?
Question
below
a
shows
sample
of
the
50
number
males
of
and
left-
Left-handed
Right-handed
Total
5
32
37
Female
2
11
13
Total
7
43
50
probability
player
that
and
the
a
male
c
right-handed,
was
selected
player
left-handed,
given
at
right-handed
table-tennis
females.
Male
A table-tennis
and
random
from
the
group.
Find
the
is:
b
that
right-handed,
the
player
selected
is
female.
Chapter
J
11
and
P(K)
Y
our
12
K
=
are
independent
0.5,
find
neighbour
Sam.
What
is
P(J)
has
the
events.
two
children.
probability
Investigation
Given
that
P(J
|K)
=
0.3
and
.
–
the
that
Y
ou
lear n
Sam’s
Monty
that
sibling
Hall
he
is
a
has
a
son,
This
brother?
as
it
is
not
as
might
obvious
seem!
problem
revisited!
T
ake
a
typical
contestant
that
there
What
is
behind
Let
A
Let
the
Hall
the
chosen
an
the
Door
unwanted
game.
3
and
prize
Suppose
Monty
behind
probability
Hall
Door
that
the
the
reveals
2.
car
is
1?
for
Door
stand
Monty
in
conditional
stand
B
that
has
is
Door
behind
situation
1
the
condition
and
for
the
the
has
that
condition
revealed
contestant
there
contestant
has
is
a
car
chosen
Door
Analysis
that
that
there
has
chosen
and
B
is
Door
a
dud
behind
probability
of
A
(P(A
∩
B))
car
Monty
The
a
is
behind
Hall
has
problem
dud
is
behind
situation
can
Door
to
1
show
the
and
the
what
is
is
1
Door
2
arise
in
given
two
that
contestant
of
the
the
the
given
Door
=
3
has
choice
because
Door
3
2.
was
when
the
car
is
behind
Door
1
2
when
the
car
is
behind
Door
3.
of
being
Door
3.
shown
This
1
The
rst
way
has
a
probability
of
,
as
shown
above.
9
In
the
Door
second
1
or
way,
Door
2.
the
If
host
he
is
could
equally
reveal
likely
either
to
what
choose
is
behind
either
of
these
1
doors
then
the
probability
of
showing
what
is
behind
Door
2
is
1
×
2
Therefore
behind
1
1
+
9
We
Door
1
×
2
This
the
2
when
the
of
there
being
contestant
revealed
has
an
unwanted
chosen
Door
.
is
3
1
18
prize
is
=
18
P (B),
want
the
the
probability
conditional
of
B
probability,
P
(
A |
B
)
This
given
by
1
P( A ∩
P
(
A |
B
B)
) =
2
9
=
=
3
P(B)
3
18
This
means
Door
3
that
given
the
that
conditional
the
probability
contestant
has
that
chosen
the
Door
car
3
is
and
behind
has
1
been
shown
that
there
is
an
unwanted
prize
behind
Door
2
is
.
only
3
Therefore
Probability
it
is
wor thwhile
to
switch!
.
=
9
3
9
is
probability
Hall
using
probability
chosen
ways:
1
monty
if
9
probability
of
conditional
1
×
just
behind
computation
2
3.
3
the
Door
problem
1
The
3.
.
Tree
Probability
diagrams
occurs.
It
is
outcomes.
‘With
hits
is
the
the
problems
to
to
use
read
different
and
where
these
the
more
than
question
types
of
to
than
list
one
all
carefully
the
event
possible
and
problems.
repeated
events
that
bullseye
shot
Represent
Find
the
replacement’
probability
each
for
diagrams
easier
impor tant
between
Example
The
useful
sometimes
It
distinguish
are
tree
is
is
Samuel,
0.8.
a
hits
two
b
hits
only
c
hits
at
from
information
probability
keen
Samuel
independent
this
a
that
on
takes
the
a
member
two
the
shots.
previous
tree
of
school
Assume
Archer y
that
Club,
success
with
shot.
diagram.
Samuel
bullseyes
one
least
bullseye
one
bullseye.
The
Answers
first
section
of
the
tree
diagram
represents
Samuel's
first
HIT
shot.
He
will
either
hit
the
bullseye
or
miss
it. The
probability
0.8
that
he
The
outcome
beside
misses
the
is
is
1
–
on
0.8
the
=
0.2.
end
of
the
branch,
the
probability
is
branch.
0.2
MISS
The
0.8
HIT
second
There
are
a
hit
a
hit
a
miss
shot
will
therefore
followed
by
also
four
a
hit
either
hit
possible
(H
and
or
miss
outcomes
the
of
bullseye.
this
‘experiment’:
H),
HIT
0.8
0.2
followed
by
a
miss
(H
and
M),
MISS
followed
by
a
hit
(M
and
H),
a miss followed by a miss (M and M).
HIT
0.2
0.8
MISS
0.2
MISS
a
We
So
b
require
P(H
P(H
and
and
=
(0.8
=
0.32
P(H
×
H)
M)
+
0.2)
+
and
H).
=
0.8
=
0.64
P(M
(0.2
×
the
a
hit
second
product
and
×
Since
0.8
H)
0.8)
Only
rule).
one
second
These
add
c
P(at
least
one
=
1
–
(0.2
=
1
–
0.04
=
0.96
×
bullseye)
0.2)
two
could
events,
be
(H
them
1
1
–
–
shot
along
either
independent
and
both
the
the
a
other
hit
again
2
and
the
on
and
two
the
(M
at
events
outcomes
P(miss
getting
a
hit
together
with
(the
branches.
first
or
a
hit
on
the
one.
M)
(as
of
probabilities
top
happen
(as
P(M
is
multiply
the
can’t
branch
need
have
first
can
Multiply
they
each
we
we
the
we
missing
between
Here
So
hit
and
exclusive:
along
with
shot
and
the
are
are
bullseye
H)
same
are
mutually
time.
Multiply
independent)
mutually
both
and
then
exclusive).
times)
M)
Chapter
Exercise
H
Correct
1
Lizzie
is
attempting
two
exam
questions.
The
probability
2
that
she
gets
any
exam
question
correct
is
3
a
Copy
and
b
What
complete
the
diagram.
Correct
2
is
the
probability
that
she
will
get
3
Incorrect
only
one
of
them
correct?
What is the probability she will get at least
c
2
one correct?
2
When
Laura
3
and
Michelle
play
in
the
hockey
1
team
the
probability
that
Laura
scores
and
is
3
1
that
Michelle
scores
is
.
2
Draw a
it
to
tree
find
Exam-Style
3
There
diagram
the
are
to
illustrate
probability
that
this
neither
information
will
score
in
and
the
use
next
game.
Question
equal
numbers
of
boys
and
girls
in
a
school
and
it
is
In
1
known
of
that
the
boys
and
of
the
girls
walk
in
ever y
day
.
1
of
the
boys
and
of
3
come
will
in
the
girls
get
a
there
be
by
the
two
rst
branches
section
and
lift.
2
The rest
3
10
10
1
Also
question
1
three
branches
from
coach.
each
of
these
in
the
Determine
second
the
a
propor tion
come
the
b
4
by
the
school
population
that
are
girls
the
school
population
that
come
section.
who
coach,
propor tion
Determine
of
the
of
probability
of
getting
two
heads
in
by
three
coach.
tosses
of
2
a
biased
coin
for
which
P(head)
=
3
5
A
10-sided
twice.
Find
the
has
the
numbers
probability
1−10
exactly
one
prime
number
is
rolled,
b
at
one
prime
number
is
rolled.
least
written
on
it.
It
is
rolled
that:
a
Exam-Style
6
dice
Question
The probability of
Rain
a day being windy is 0.6. If
it’s windy the
Windy
probability of
of
rain is 0.4. If
it’s not windy the probability
rain is 0.2.
a
Copy
and
b
What
is
c
What
is
complete
the
tree
diagram.
Rain
Probability
the
the
probability
probability
of
of
a
given
two
day
being
successive
rainy?
days not
being
rainy?
‘Without
Example
A
bag
replacement’
and
conditional
probability
contains
5
green
and
6
red
balls.
If
two
balls
are
This
taken
out
successively
,
probability
least
without
replacement,
what
is
means
that
the
the
probability
the
second
of
that
a
at
one
green
is
b
red is picked on the first pick given that at least one green is chosen?
draw
is
chosen,
dependent
results
draw,
of
on
the
since
a
the
rst
ball
has
Answers
been
Since
a
red
ball
has
been
taken
Draw
a
tree
(and
5
there
green
will
be
balls)
5
red
balls
probabilities
left
section
what
5
of
on
the
branches
has
after
diagram. The
the
initially
removed
happened
rst
draw.
second
depend
in
the
on
first
R
10
section.
R
6
11
5
G
10
6
R
10
5
11
G
4
G
10
a
P(at
least
one
green)
It
=
1
–
P(both
is
quicker
probability
6
⎛
= 1
5
×
⎟
⎝ 11
10
8
3
⎞
⎜
work
11
11
ball
on
P(red
followed
by
out
is
green)
a
the
(
red
on 1st
calculate
and
at
this
way
the
than
probability
green
on
both
it
that
is
to
When
the
least
one green
on
the
second
first
the
pick,
pick
or
a
first,
the
or
a
green
picks.
red
is
probability
P
in
=
= 1
⎠
green
b
to
red)
picked
of
the second being
)
5
=
green
( at
P
least
one green
is
,
so
multiply
these
)
10
probabilities.
6
5
1
3
×
11
10
=
3
2
11
=
8
=
8
8
11
11
Some
have
tree
seen
diagrams
so
do
not
have
the
same
‘classic’
shape
as
the
we
far.
Chapter
Example
Toby
is
when
is
0.75.
him
in
a
he
rising
gets
When
winning
on
3
out
star
his
he
the
of
5
of
first
the
school
ser ve
uses
his
point.
in
the
second
He
occasions
is
Tennis
ser ve
there
successful
and
Club.
probability
his
at
second
He
that
is
a
that
that
chance
his
in
found
wins
0.45
getting
ser ve
has
he
first
on
3
point
of
ser ve
out
of
4
occasions.
a
b
Find
the
probability
wins
the
point.
Given
his
that
first
Toby
ser ve
that
wins
the
the
next
point,
time
what
it
is
is
Toby’s
the
tur n
probability
to
ser ve
that
he
he
got
in?
Answers
Win
0.75
On
this
tree
diagram,
not
necessar y
to
it
is
continue
the
In
3
branches
once
the
point
has
been
Win
5
0.45
0.25
won.
Lose
In
3
4
2
0.55
Lose
Out
5
1
4
a
P
(win)
win)
+
=
b
⎟
+
0.45
=
0.585
P(1st
P
+
⎝
and
in
×
5
and
get
win)
⎞
× 0.45
4
Multiply
⎟
along
the
branches.
⎠
0.135
ser ve
(1st
ser ve
ser ve,
3
⎜
⎠
=
in
⎛ 2
⎞
× 0.75
⎝ 5
first
first
ser ve
⎛ 3
⎜
(get
(miss
second
Out
in|win
serve in
and
point)
win
point
Both
)
of
these
values
have
been
=
P
(
⎛ 3
point
found
)
in
par t
a
⎞
×
⎜
win
0.75
⎝ 5
This
⎟
answer
has
been
given
⎠
=
= 0.769
(3
to
sf)
3
sf
as
the
exact
answer
0.585
(fractional)
Exercise
1
Three
Each
is
not
obvious.
I
cards
card
is
are
drawn
not
at
random
replaced.
Find
from
the
a
pack
of
probability
of
playing
cards.
obtaining
See
page
ordinar y
playing
a
three
Probability
picture
cards
b
two
picture
cards.
73
for
pack
cards.
of
the
52
Exam-Style
2
A
pencil
Question
case
contains
5
faulty
and
7
working
pens.
A
boy
and
Even
then
a
girl
each
need
to
take
a
does
a
What
is
the
probability
that
two
faulty
pens
are
What
c
If
In
is
a
the
exactly
the
3
is
girl
bag
one
chose
are
chosen
probability
4
at
faulty
that
pen
is
at
least
chosen,
one
faulty
what
is
pen
the
is
chosen?
probability
that
it?
red
not
may
balls,
3
green
and
not
question
balls
and
replaced.
2
A
yellow
second
ask
balls.
ball
is
A
a
nd
to
use
to
answer
these
random,
the
for
it
chosen?
you
b
if
pen.
tree
it
useful
diagram
some
of
questions.
ball
then
chosen.
4
a
Find
P(the
balls
are
both
b
Find
P(the
balls
are
the
c
Find
P(neither
ball
is
d
Find
P(at
one
ball
Four
balls
least
are
replacement,
5 red,
4
blue,
Find the
5
A
club
of
the
drawn
from
3
10
Find
b
Two
the
people
6
Billy
that
you
of
one
after
the
that
the
obtain
which
at
chosen
Find
yellow).
the
other
following
and
without
balls:
pur ple.
chosen
are
color).
containing
2
probability
competition.
are
is
is
random,
members,
members
same
red).
bag
orange,
probability
has
a
a
at
green).
6
one
are
random
the
at
random
probability
girls
to
chosen
ball
be
of
each
and
4
to
is
represent
that
are
President
President
one
color.
of
a
the
One
club.
boy
.
the
boy
boys.
club
and
one
in
a
girl
chosen.
answers
average
is
5
on
average
questions
5
out
problems
of
9.
They
correctly
both
out
of
attempt
7.
the
Natasha’s
same
problem.
a
b
What
is
the
probability
answers
the
question
If
the
that
c
If
the
that
d
If
question
Billy
Natasha
was
there
is
the
question
there
that
got
is
got
at
were
that
answered
correctly
,
correctly
,
correct
one
two?
one
of
the
students
what
is
the
probability
what
is
the
probability
answer?
answered
least
least
correctly?
correct
the
at
answer?
correct
answer,
what
is
the
probability
Extension
material
Worksheet
3
-
on
CD:
Conditional
probability
Chapter
Review
exercise
✗
1
2
A
two-digit
at
random.
is
divisible
c
is
greater
In
a
class
that
A
this
For
=
Find
Explain
probability
students,
is
18
both
a
at
cat
is
a
number?
a
D
∩
it
is
∩
known
What
a
is
cat
the
and
3
have
probability
dog?
D ′)
=
that:
0.25
P (D)
=
0.2.
D ′).
why
C
A
and
and
D
B
are
are
not
independent
such
that
the
probabilities
P(A)
=
events.
0.6,
P (B)
=
0.2
and
the
of
the
events
occur,
c
exactly
one
of
the
events
occur,
d
B
occurs
of
events
that:
one
given
100
programme,
the
18 watch
drama
22 watch
comedy
35 watch
drama
10 watch
of
than
comedy
only
than
write
and
as
an
watch
drama
Using
all
and
which
reality
,
of
three
they
types
watch
of
TV
regularly
.
information:
reality
reality
these
as
and
TV
asked
comedy;
and
times
reality
are
types;
and
none
occurred.
comedy
three
only
taken
has
following
all
three
A
students
15 watch
There are
occur,
that
drama,
They provide
only
,
a
have
3,
0.1.
of
is
20
random.
least
x
square
dog,
and
down
it:
d
at
If
written
by
both
a
that
is
divisible
b
group
inclusive
is
a
A
99
b
have
selected
has
and
events
=
and
50,
P (C ′
P (C
Calculate
5
C
b
P(A|B)
the
10
Questions
0.7
two
is
between
5,
student
a
The
30
student
events
P (C )
by
than
of
Exam-Style
4
What
a
neither.
3
number
TV;
TV;
programmes
many
two
students
times
as
regularly
.
who
many
watch
who
drama
watch
only
U
Drama
Comedy
comedy
only
.
the
number
expression
of
for
students
the
who
number
of
watch
reality
students
TV
who
only
.
x
b
the
above
information
copy
and
complete
the
Reality
Venn
c
diagram.
Calculate
Probability
the
value
of
x.
Review
1
Let
P(C )
a
Find
b
Are
exercise
=
0.4,
P(C
C
P(D)
and
and
=
0.5,
P(C
|D)
=
0.6.
D).
D
mutually
exclusive?
D
independent
Give
a
reason
for
your
answer.
c
Are
C
and
events?
Give
a
reason
for
your
answer.
d
Find
P(C
or
e
Find
P(D
D).
|C).
3
2
Jack
does
of
the
jobs
around
the
house
and
Jill
does
the
rest.
If
5
35%
of
Jack’s
finished
house
will
be
a
properly
,
b
by
Jill
if
Max
The
travels
for
a
c
What
that
he
he
writing
the
travels
travels
diagram
jour neys
in
day
by
by
the
bicycle,
bus
bicycle
which
on
by
on
on
shows
Monday
probabilities
probability
that
he
any
any
the
and
by
of
Jill’s
jobs
around
are
the
and
Tuesday
,
and
by
iii
by
the
traveled
is
the
on
is
school
probability
Wednesday
the
by
to
and
once
by
travel
that
on
he
in
bus
is
is
by
car.
0.6.
The
0.3.
outcomes
Label
the
tree
outcome.
on
does
any
or
on
Tuesday
,
Monday
and
Monday
not
Thursday
that
by
bus
bicycle
and
probability
car
of
day
or
travels
Monday
method
day
each
Monday
same
bus
Tuesday
.
on
What
of
done
possible
on
three
twice
and
travel
and
to
Tuesday
.
school
by
Friday?
days
by
Tuesday?
Max
bicycle
travels
and
once
car?
bag
contains
into
What
Without
is
What
passed
selects
is
is
is
red
bag,
apples
and
the
green
and
Tarish.
it
eats
bag,
probability
Janet
is
it.
she
that
Without
10
green
randomly
that
Maddy
into
and
Maddy
probability
red
the
to
two
What
the
the
is
6
looking
The apple
c
that
each
bicycle
Max
55%
job
properly
.
bicycle
The apple
is
done
by
looking
b
a
by
by
a
and
that
ii
twice
A
not
school
tree
is
bicycle
4
properly
probability
i
What
d
to
Max’s
clearly
,
b
was
probability
Draw
finished
the
Questions
probability
a
are
find
done:
it
Exam-Style
3
jobs
properly
,
it
apples.
selects
one
apple.
red?
Next
the
randomly
is
Without
bag
is
passed
selects
one
to
Janet.
apple.
green?
replaces
looking
it
in
into
the
the
bag.
bag,
Next
he
the
bag
randomly
apples.
the
probability
that
they
are
both
red?
Chapter
Exam-Style
5
On
a
walk
carrots
female
count
23
Venn
and
a
What
b
What
is
70
are
rabbits,
female
diagram
eating
the
is
eating
Is
I
and
Draw a
c
Question
and
and
are
not
hence
female,
eating
find
34
are
not
eating
carrots.
the
number
that
are
both
carrots.
probability
the
42
that
probability
a
that
rabbit
a
is
male
rabbit
is
and
female
not
eating
given
that
carrots?
it
is
carrots?
being
female
CHAPTER
3
independent
of
eating
carrots?
Justify
your
answer.
SUMMARY
Definitions
●
An
An
A
is
event
an
outcome
is
experiment
random
over
the
process
is
experiment
which
event
from
may
an
by
one
experiment.
which
where
we
there
obtain
is
an
outcome.
uncer tainty
occur.
n( A)
●
The
theoretical
probability
of
an
event A
is
P( A )
n (U )
where
●
n(A)
and
If
probability
the
Y
ou
The
the
of
to
the
ways
n
×
is
P
P,
of
in
event
A
possible
n
trials
can
outcomes.
you
would
times.
frequency
of
that
number
event
occur
number
to
of
total
an
relative
the
is
number
is
event
use
larger
Venn
As
n(U)
the
can
frequency
●
the
occur
expect
●
is
as
trials,
an
the
estimate
closer
the
of
probability
.
relative
probability
.
diagrams
an
event,
A,
either
happens
or
it
does
not
happen
U
A
P(A)
+
P (A ′)
●
For
P (A ′)
=
any
(A ∪
B)
1
−
1
P (A)
two
=
=
events
P (A)
+
A
and
P (B)
–
B
U
P (A
∩
B).
P(A)
P(B)
P(A
●
In
general,
if
A
and
B
are
mutually
exclusive,
∩
B)
then
A
P (A
∩
B)
=
0
and
P(A
∪
B)
=
P(A)
+
B
P(B).
P(A)
P(B)
Continued
Probability
on
next
page
Sample
space
product
●
Two
●
events
other
A
and
of
B
one
two
P(A
B)
events
=
This
is
the
This
is
also
P(A)
are
independent
does
A
×
not
the
if
affect
the
the
chance
that
called
For
two
has
occurred
and
B
are
independent
P(B)
product
Conditional
●
the
occurs.
When
∩
and
rule
occurrence
the
diagrams
rule
the
for
independent
multiplication
events
r ule.
probabi li ty
events
A
can
and
be
B
the
found
probability
of
A
occurring
given
that
B
using
P( A ∩ B )
P
( A| B ) =
P( B )
●
If
P
●
A
(
In
and
B
A| B ′ )
=
general
given
are
that
P (A)
for
B
independent
,
two
has
P
(
events,
B | A′ ) =
events
occurred
A
( A| B )
=
P(A),
P
( B|A)
=
P (B),
P (B)
and
can
P
B
be
the
probability
found
of
A
occurring
using
P( A B )
P
(
A| B
)
=
P( B )
Chapter
Theory
of
knowledge
Probability
Math
from
textbook
a
bag
sur prising
People
Why
the
do
people
chances
Asking
How
do
asks
a
of
–
rather
get
lotter y
winning
the
sensitive
such
tr uth
or
than
problems
is
as
that
finding
calculating
so
a
real
the
sur vey
headteacher
the
600
The
interested
wants
at
to
know
her
how
school
examinations.
in
probability
sensitive
by
whether
is
the
for
she
the
many
relies
a
She
of
a
winning
each
are
asking
student
doesn’t
just
wants
an
whole
knowing
know
a
sensitive
harmless
to
question
or
a
not
individual
student
in
If
you
got
a
head
with
overall
sur vey
ips
a
the
rst
ip,
question:
twice,
answer
‘Have
cheated
in
the
you
an
no-one
each
the
honestly.
result.
asking:
They
Have
they
one.
school.
questionnaire
that
whether
exam?’
student
your
response
ever
sends
their
of
showing
she
on
have
is
specic
on
headteacher
coin
If
some
lotter y?
the
the
estimate
relying
probability
cheated,
has
balls
method
Each
has
colored
questions.
randomized
perfectly
in
picking
question?
students
cheated
abuses
odds.
This
A
to
national
that
But
probability
,
What
small?
involve
life?
answers
when
questions
from
in
and
often
misunderstand
tickets
are
sensitive
you
useful
misuse
buy
uses
probability
how
uses
also
intuition
–
–
you
ever
cheated
in
school
If
you
got
a
tail
with
the
follow
rst
the
Yes
then
exams?
instructions
ip,
answer
question:
‘Did
the
the
No
on
this
second
card.
ip
give
a
tail?’
honestly.
she
is
unlikely
to
get
honest
answers.
Answer
P(Y
es
to
Q1)
‘yes’
1
p
p
×
=
p
=
2
Head
The
tree
–
Answer
question
1
diagram
1
helps
P(H)
=
Answer
2
1
to
estimate
the
2
fraction
p
p,
Probability
‘no’
=
of
P(Y
es
to
of
answer
Q1)
Head
1
students
P(H)
who
P(T)
have
cheated
2
=
T
ail
+
answer
=
1
p
=
2
Answer
in
an
question
exam.
1
T
ail
P(Y
es
to
Q2)
=
answer
2
1
‘yes’
×
=
2
Theory
of
knowledge:
Probability
–
uses
and
abuses
1
1
=
2
4
1
+
2
2
P(T)
P(Y
es
‘no’
–
4
to
Q2)
‘yes’
Suppose
220
‘yes’
of
out
students
600
answer
The
inter viewed.
estimated
students
who
number
have
of
cheated
in
7
p
1
220
+
an
exam
is
600
=
×
140
30
=
2
4
600
p
220
Provided
1
=
ever ybody
tells
the
–
2
600
p
4
tr uth
when
answering
their
7
question,
this
estimates
the
method
=
2
60
number
of
7
p
=
students
30
an
Would
students
answer
the
who
have
cheated
exam.
question
honestly?
he
Probability
the
and
birthday
in
intui tion
truth?
–
problem
people
as
share
the
same
bir thday?
10%?
the
rst
and
Let’s
do
the
math
and
work
it
means
there
are
253
possible
pairs
of
×
a
pair
for
pair
(Tim,
Jane)
is
students:
exactly
23
in
choices
second.
The
students
22
out.
the
23
person
then
the
same
as
the
22
=
pair
253
(Jane,
Tim),
so
halve
2
the
The
probability
of
two
people
having
different
bir thdays
total.
is
364
=
0.997260
365
Ignoring
So
for
253
possible
pairs,
the
probability
that
the
–
two
there
365
people
in
each
pair
have
different
bir thdays
leap
are
years
364
bir thdays
out
that
of
are
is
‘different’.
253
364
(
)
=
0.4995
365
So
the
have
1
–
probability
the
same
0.4995
Do
Are
What
you
=
bir thday
0.5005,
rely
there
that,
on
in
areas
other
the
253
pairs,
two
people
in
a
pair
is
50.05%
intuition
other
about
or
for
to
of
help
–
just
you
over
make
mathematics
areas
of
half !
decisions?
where
your
intuition
has
let
you
down?
knowledge?
Chapter
Exponential
and logarithmic
functions
CHAPTER
OBJECTIVES:
1.2
Elementar y
2.6
Exponential
Laws
of
treatment
exponents;
a
of
and
logarithms;
and
their
graphs
and
their
graphs
logarithms
change
of
base
x
,
a
> 0,
Logarithmic
x
exponents
laws
functions
x
x
of
log
x
e
functions
x,
x
> 0,
x
lnx ,
x
> 0
a
Relationship
x
between
x ln a
these
log
x
functions
x
a
a
=
e
,
log
a
=
x;
a
=
x,
x
> 0
a
x
2.7
Solving
2.8
Applications
Before
Y
ou
1
equations
you
should
Evaluate
of
the
graphing
form
a
skills
x
=
and
b,
a
y
=
b
solving
equations
to
real-life
situations
start
know
simple
of
how
positive
to:
Skills
exponents
1
check
Evaluate
4
e.g.
Evaluate
⎛
⎛
⎞
a
4
3
3
=
3
×
3
×
3
×
3
=
b
⎜
81
⎞
⎝
⎟
⎜
⎠
⎝
⎟
⎠
⎞
⎛
e.g.
Evaluate
3
⎜
⎟
⎝
c
0.001
⎠
2
Conver t
numbers
to
exponential
form
2
State
e.g.
Find
n
given
=
2
the
value
of
n
in
n
n
128
a
7
these
equations.
n
=
343
=
625
b
3
=
243
7
128
=
2
so
n
=
n
7
c
5
2
3
Transform
graphs
3
Transform
the
graph
2
2
e.g.
Given
the
graph
of
y
=
x
sketch
2
graph
of
y
=
x
+
3
y
2
y
=
x
+
3
8
6
4
2
2
y
=
x
x
–3
–2
–1
0
Exponential
1
2
3
and logarithmic functions
the
graph
of
y
=
(x
− 2)
of
y
=
x
to
give
the
Facebook,
the
social
y
media
Facebook
users
600
giant,
celebrated
with
more
up
million
a
Febr uar y
than
from
in
450
huge
200
100
from
0
when
90-ceD
80-ceD
70-ceD
million
60-ceD
one
300
50-ceD
there
400
40-ceD
2004
100
2008
increase
December
were
million
about
August
500
2010
stinU
users,
and
in
sixth
)snoillim(
bir thday
its
x
members.
Dates
This
graph
shows
how
the
(Source:
number
of
Facebook
http://www
.facebook.com/press/info.
users
php?timeline)
has
increased
Growth
growth.
growth
the
like
As
this
you
rate.
number
over
(cer tainly
move
The
of
time.
growth
users
until
along
at
the
rate
that
at
Febr uar y
cur ve
any
its
time
2010)
is exponential
gradient
is
increases
roughly
with
propor tional
the
to
time.
Chapter
A
good
model
for
the
data
on
Facebook
users
is
Y
ou
could
also
use
x
n
=
1.32
×
1.1
the
where
n
is
the
number
of
users
in
millions,
and x
is
the
number
model
to
predictions
months
after
December
2004.
the
could
use
the
equation
n
=
1.32
×
future
1.1
to
estimate
the
users
at
any
specified
date
or
to
find
the
date
at
which
number
of
users
was
‘extrapolation’.
are
and
as
will
its
you
come
across
opposite,
move
many
other
exponential
along
the
examples
decay
(where
of
exponential
the
gradient
growth
decreases
a
model
to
cur ve).
Malcolm
Imagine
again,
How
Gladwell
taking
until
high
1
Fold
2
For
a
you
do
the
have
sheet
number
can
of
this
piece
folded
think
fold,
folding
posed
large
you
each
Y
ou
a
the
paper
(any
of
layers
assume
a
size)
half
to
The
it
Tipping
over
as
of
many
paper
of
is
times
the
number
the
paper
about
0.1
is
The
1
rst
×
10
few
Number
of
Number
folds
of
have
been
done
Thickness
(km)
As
thick
as
a
layers
−7
0
1
1
×
10
Piece
of
paper
−7
1
2
2
×
10
2
4
4
×
10
3
8
4
16
−7
Credit
card
5
6
7
8
9
3
4
How
many
a
as
b
just
How
folds
thick
as
taller
thick
Exponential
would
the
than
would
you
height
the
the
need
of
a
height
paper
to
make
the
table?
of
be
and
a
man?
after
and logarithmic functions
50
folds?
paper
of
possible.
folds,
formed.
km.
entries
as
mm
−7
that
Point
again
be?
show
thickness
sheet
book
folding
would
in
table
the
his
and
times.
stack
this
and
that
in
paper
,
50
nal
complete
paper
problem
of
it
using
this
estimate
you
–
with
of
type
future
growth?
What
Investigation
the
reached.
problems
Y
ou
of
is
a
What
par ticular
This
number
called
of
about
growth
Facebook.
x
Y
ou
make
of
thick,
other
need
factors
to
do
consider?
Y
ou
can
probably
get
to
about
six
or
seven
folds
before
you
can’t
Does
fold
the
paper
any
more.
At
seven
folds
the
paper
is
already
about
how
thick
the
man.
a
50
thick.
of
a
table
After
folds
This
Paper
17
is
of
fact
and
folds
after
after
it
paper
about
is
the
an
paper
sequence
In
is
15
only
folds
roughly
13
it
folds
will
13 m
the
be
paper
much
thick,
is
taller
roughly
depend
roughly
to
big
the
on
star t
paper
with?
Tr y
is
it.
than
the
height
house!
the
folding
layers’
the
textbook.
two-storey
After
of
this
height
any
of
as
it
as
are
would
distance
example
for m
a
be
of
approximately
from
the
exponential
sequence .
a
function
of
Ear th
the
The
113
to
the
g rowth.
ter ms
number
million
of
km
Sun.
The
‘number
in
folds,
n,
where
n
f
(n)
f
(n)
In
=
2
is
an
this
their
exponential
chapter
inverses,
.
you
growth
will
which
lear n
are
function
more
called
about
exponential
functions
and
functions .
logari thmic
Exponents
Exponents
are
multiplication
a
shor thand
of
a
number
way
by
of
representing
the
repeated
itself.
5
The
expression
The
3
in
this
,
3
for
example,
expression
is
the
represents
base
3
number
×
3
and
×
3
the
×
5
3
is
×
3.
the
exponent.
Other
names
for
exponent
are
power
and
index
4
It
Y
ou
can
also
use
a
variable
as
the
base,
for
is
quicker
to
write
x
example,
than
x
×
x
×
x
×
x
4
x
=
x
×
Laws
x
×
of
x
×
x
exponents
Multiplication
5
5
x
3
x
×
x
=
(x
×
x
=
x
=
x
Simplify
3
×
x
×
x
×
×
x
x
×
×
x
x
×
×
x)
x
×
×
x
(x
×
×
x
x
×
×
x)
Remove
brackets.
x
8
5
So
x
3
×
x
(5 + 3)
=
m
➔
a
x
n
×
a
8
=
x
m+n
=
a
5
Notice
that
the
two
are
the
Y
ou
3
×
x
variables
simplify
3
×
y
using
5
x
base
x
same.
cannot
5
x
in
,
for
this
3
×
y
example,
law.
5
=
x
3
y
Chapter
Division
Cancel
5
Simplify
÷
x
5
factors.
3
x
÷
2
x
=
=
5
So
3
x
common
3
x
÷
=
x
×
x
=
x
(5−3)
x
=
x
Notice
that
you
2
=
x
5
can’t
m
➔
a
n
÷
m
a
=
because
n
to
a
5
Simplify
5
(x
(x
the
x
3
÷
y
bases
a
are
Raising
simplify
not
the
same.
power
3
)
3
)
=
(x
=
x
=
x
×
×
x
x
×
×
x
x
×
×
x
x
×
×
x)
x
×
×
x
(x
×
×
x
x
×
×
x
x
×
×
x
x
×
×
x
x)
×
×
x
(x
×
x
×
×
x
x
×
×
x
x
×
×
x
×
x)
x
15
5
So
=
(x
m
➔
(a
3
5×3
)
=
15
x
n
=
x
mn
)
=
a
Example
2
Expand
(2xy
3
)
Answer
2
(2xy
Don’t
3
)
2
=
(2xy
2
)
×
(2xy
)
(2xy
)
You
don’t
need
to
show
this
line
of
to
working.
3
=
3
2
×
x
2
×
(y
3
)
3
=
8x
in
raise
the
you
have
numbers
Apply
the
power
of
3
to
ever y
ter m
in
bracket
to
bracket.
power
as
x-
y-terms.
and
well
as
the
the
A
Simplify
1
3
2
x
a
×
2
x
b
4
3p
×
2p
2
q
c
×
3
d
(x
2
y
4
)(xy
)
Remember
to
multiply
the
constants
Simplify
2
5
2
x
a
÷
7
x
b
2a
3
÷
2a
7
c
2a
(2a)
well
Simplify
3
(x
a
The
4
2
)
b
power
3
)
3
c
3(x
zero
2
Simplify
(3t
2
x
÷
x
=
=
=
But
0
Therefore
x
=
Exponential
1
and logarithmic functions
together
as
d
3
(numbers)
3
÷
the
6
y
the
Exercise
forget
2
×
2
y
2
)
2
d
(−y
3
)
as
the
variables.
0
➔
a
=
‘Anything
1
Any
base
raised
to
the
power
of
zero
is
equal
to
zero
1.
is
‘Zero
Fractional
is
to
1.
’
power
0.
’
0
So
×
what
How
decide
Who
we
?
should
0
what
about
+
Law
any
But
equal
to
power
Using
the
exponents
Simplify
to
this
is
equal
should
to?
decide?
so
=
Similarly
and
and
so
Y
ou
can
assume
always
that
a
➔
=
is
positive
when
considering
Roots
of
2
=
2
x
a.
6
x
roots
Simplify
Since
even
×
2
x
×
=
x
2
=
x
6
3
=
x
➔
=
Example
(
)
=
(
)
=
‘Evaluate’
Without
using
a
calculator,
means
evaluate:
4
‘work
out
the
value
1
⎛
1
⎞
3
2
a
36
b
⎜
⎝
of ’.
⎟
27
⎠
Answers
1
1
2
a
36
=
36
=
n
6
Since
n
a
=
a
4
4
1
⎛
⎛
b
1
⎞
3
⎜
⎞
⎛
1
⎞
3
⎝
⎟
27
⎠
⎜ ⎜
⎝ ⎝
27
n
m
⎟
=
⎜
Since
⎟
⎟
⎠
⎠
(a
mn
)
=
a
4
1
⎛
⎜
⎟
3
⎝
⎞
27
⎠
4
⎛ 1 ⎞
⎜
⎟
⎝ 3 ⎠
1
81
Chapter
Negative
exponents
3
Simplify
5
x
÷
x
3
x
5
÷
x
=
=
×
=
3
Also
5
x
÷
x
3−5
=
−2
x
=
x
And
therefore
=
Y
ou
➔
must
learn
the
=
laws
as
for
they
exponents
are
Formula
Example
Without
using
a
calculator
evaluate
2
⎛
−2
a
6
3
⎞
b
⎜
⎝
⎟
4
⎠
Answers
1
1
1
n
2
a
6
=
=
Use
a
=
n
2
a
36
6
2
⎛
b
3
1
1
⎞
=
=
⎜
⎝
⎟
4
2
⎠
⎛
3
⎜
⎝
⎞
⎟
4
⎠
⎛
9
⎜
⎝
⎞
⎟
16
⎠
16
=
9
Exercise
✗
1
B
Evaluate
2
1
1
3
3
2
a
9
b
125
c
64
2
2
⎛
8
⎞
3
3
d
2
8
e
⎜
⎟
⎝
27 ⎠
Evaluate
−3
a
2
b
c
⎛
d
(
)
⎞
e
⎜
⎟
⎝ ⎠
Exponential
and logarithmic functions
not
in
booklet.
the
Example
Here
Simplify
these
expressions.
‘simplify’
1
2
0
−3
5d
a
2
6x
b
÷
(2x
3
3
)
⎛
6
27 a
c
d
⎞
9v
⎜
⎝
2
means
write
these
⎟
4
expressions
16w
using
only
⎠
positive
exponents.
Answers
0
0
5d
a
=
5
×
1
−3
2
6x
b
=
÷
(2 x
5
Use
3
−3
)
=
a
=
m
6
6x
÷
8x
6
1.
(a
Use
n
)
mn
= a
.
3
9
m
=
x
=
Use
a
n
÷
a
m
=
–
n
a
9
8
4 x
1
1
1
1
3
6
6
=
(
27 a
= 27
)
n
6
3
3
27 a
c
(
m
m
n
3
a
a
Use
)
= (a
)
.
2
= 3a
1
1
2
⎛
d
⎜
⎝
4
2
⎞
9v
4
16w
=
⎟
⎠
⎛ 16w
⎞
⎜
⎟
⎝
2
9v
2
1
n
a
Use
=
n
a
⎠
1
4
(16w
2
2
)
=
4w
=
1
3v
2
(9v
Exercise
C
Simplify
1
2
)
these
exponential
expressions.
In
this
exercise,
(
a
)
b
c
d
make
sure
your
e
answers
have
positive
exponents.
Simplify
2
these
expressions.
a
b
÷
c
.
Solving
Exponential
exponential
equations
are
equations
equations
involving
‘unknowns’
x
as
exponents,
for
example,
5
=
25.
y
x
Y
ou
can
write
Example
3
exponential
equation
in
the
form a
=
b
x –1
Solve
an
5x
=
3
Answer
x
1
3
x
5x
=
− 1 =
3
Both
5x
sides
powers
of
of
3
the
so
equation
the
two
are
exponents
are
equal.
−1 =
4 x
1
x
=
−
4
Chapter
Example
3x
Solve
+
1
For
3
=
this
many
of
the
questions
Answer
3 x +1
3
example
=
to
81
learn
3 x +1
2
4
=
Write
3
81
as
a
power
of
1
=
1
=
3
=
9
1
=
2
3
=
4
3
exponents.
2
=
3
2
3
2
x
powers.
4
2
3x
3
3.
2
Equate
+ 1 =
need
0
=
1
3x
following
you
these
0
3
and
81
3
=
8
3
=
16
3
=
32
3
=
64
=
128
=
1
7
=
5
7
4
= 1
2
5
2
=
27
=
81
=
243
=
1
=
7
4
5
6
2
Exercise
D
7
2
Solve
1
these
equations
for
x
0
5
x
✗
1−2x
2
a
=
32
3
b
1
=
243
5
1
2
2
x
2 x
5
2x−1
3
c
0
=
27
5
d
−
25
=
2
=
25
7
=
125
7
=
625
3
5
x
4
=
5
49
Solve
2
these
x−3
3
a
equations
for
2−x
=
x
3x
3
b
5
d
2
x−2
=
25
+
c
2−3x
=
x−1
=
4
EXAM-STYLE
QUESTION
+
Solve
3
Example
=
3
5
Solve
3x
= 24
Answer
3
Divide
both
sides
by
3.
5
3x
= 24
3
Multiply
5
x
the
exponent
a
reciprocal
5
since
−
5
x
)
3
= 8
5
3
Replace
3
x
=
x
=
(
)
2
3
−5
2
1
x =
32
its
Exponential
and logarithmic functions
8
with
2
b
×
b
5
3
(
by
=8
3
49
343
3
1
1
7
e
=
=
0
= 1
−
a
Exercise
Solve
1
E
these
equations
for
x
4
5
2x
a
=
162
x
b
−
−2
c
x
=
e
27x
16
d
8x
Solve
=
0
f
27x
3
=
−2
2
32
−3
(8x)
−3
=
81x
these
equations
for
=
64
x
a
=
b
c
=
=
d
=
e
=
f
=
Solve
3
these
equations
for
x
3
2
x
a
=
125
b
=
c
.
=
192
and
d
Exponential
Graphs
➔
216
=
16
functions
properties
of
An exponential function
is
a
exponential
function
of
the
functions
We
form
could
also
write
x
f
:
x
→
a
x
f
(x)
where
=
a
a
is
a
positive
Investigation
Using
a
GDC,
sketch
–
real
number
graphs
the
graphs
of
(that
of
is,
a
>
0)
and
exponential
these
exponential
a
≠
1.
functions
1
functions.
Think
about
the
x
a
y
=
3
b
y
=
5
domain,
range,
x
intercepts
on
the
x
y
c
=
axes,
10
asymptotes,
shape
Look
at
your
three
each
What
can
you
and
behavior
of
graphs.
deduce
about
the
exponential
graph
as
x
tends
function,
to
innity.
x
f (x)
=
a
,
Whatever
when
a
>
positive
1?
value
a
has
in
the
equation
y
x
f
(x)
=
x
, the graph will
a
always
have
the
same
f(x)
=
a
shape.
x
f
(x)
=
a
is
an
exponential
growth
function
1
(0, 1)
0
x
Chapter
x
➔
The
domain
The
range
The
cur ve
The
graph
value
of
x
of
is
f
the
does
(x)
=
set
a
of
not
is
all
the
of
positive
intercept
approaches
set
the
closer
all
real
real
numbers.
numbers.
x-axis.
and
closer
to
the x-axis
The
y-intercept
The
graph
is
of
f
passes
through
the
points
⎞
,
−
⎜
⎝
(1,
The
Now
between
at
0
increases
the
and
graphs
a
1)
⎠
GDC
exponential
continually
.
of
exponential
functions
when
the
base a
is
1.
Investigation
Using
(0,
⎟
a).
graph
look
the
1.
⎛
and
as
decreases.
sketch
-
the
graphs
graphs
of
of
exponential
functions
these
functions.
–x
y
a
=
−x
3
y
=
3
is
the
–x
b
y
=
5
c
y
=
10
1
equivalent
of
y
or
=
x
–x
y
–x
What
can
you
deduce
about
the
exponential
function,
f (x)
=
a
a
>
1,
from
these
three
graphs?
−x
Whatever
will
positive
always
have
–x
f(x)
=
value
this
a
has,
the
graph
of
shape.
y
a
(0, 1)
1
0
x
– x
f
(x)
=
a
is
an
Exponential
exponential
decay
and logarithmic functions
function
f
(x)
=
a
=
⎛ 1
⎞
⎜
⎟
⎝ 3
⎠
so
the
base
,
is
when
3
x
between
0
and
1.
The
The
natural
base
e
is
exponential
exponential
one
that
you
you
come
across
often
in
functions.
Investigation
When
will
function
invest
–
money
compound
it
earns
interest
interest.
n t
r
⎛
We
use
this
A = C
formula
⎜
r
is
of
the
is
the
nal
interest
happens
£1
a
is
if
in
you
invested
How
much
(capital
expressed
a
at
will
year
,
star t
an
and
as
t
+
a
is
interest),
decimal,
the
total
compounding
interest
you
calculate
the
interest,
n ⎠
amount
rate
compoundings
What
1
A
to
⎟
⎝
where
⎞
1 +
have
rate
if
this
of
is
is
the
the
number
more
100%
is
n
C
and
for
capital,
number
of
years.
more
1
frequently?
year
.
compounded
year ly?
100
P
=
1,
r
=
100%
=
=
1,
n
=
1,
t
=
1
and
n
=
100
1
1 ⎞
⎛
A
= C
⎜
1
+
b
How
C
=
⎟
1
⎝
much
1,
r
=
=
2
(since
r
=
1
1)
⎠
will
you
100%
=
have
1,
n
if
=
this
4,
t
is
=
compounded
quar terly?
1
4
1
⎛
A
=
⎜
1
+
2
Copy
and
⎟
4
⎝
⎞
=
2
44 140 625
⎠
complete
Compounding
the
table.
Calculation
F inal
all
amount
gures
on
(write
calculator)
1
1 ⎞
⎛
Y
early
⎜
1
+
2
⎟
1
⎝
⎠
2
1
⎛
Half-Y
early
⎜
1
+
⎟
2
⎝
⎞
2.25
⎠
4
1
⎛
Quar terly
⎜
⎝
1
+
⎞
⎟
4
2.44 140 625
⎠
Monthly
Weekly
Daily
Hourly
Ever y
minute
Ever y
second
Chapter
The
final
amount
compoundings
smaller
value
The
and
is
value
of
impor tant
subject
e
is
an
the
called
increases
decreases
final
as
but
amount
the
inter val
each
between
separate
converges
on
increase
a
value.
is
This
‘e’.
e
is
approximately
number
in
2.71828
mathematics
and
which
it
has
is
an
exceptionally
applications
in
many
areas.
number.
irrational
Jacob
Bernoulli
(1654–1705)
Mathematics
beautiful
sometimes
throws
out
some
surprising
one
and
is
one
results.
such
20
decimal
Swiss
is
no
places
e
=
2.718
281
828
459
045
235
obvious
pattern
to
this
chain
of
look
at
this
series,
which
gives
a
1
1
+
value
1
+
of
2 × 1
3 × 2 × 1
the
tried
to
nd
4 × 3 × 2 × 1
1 ⎞
⎛
+ ...
of
⎜
5 × 4 × 3 × 2 × 1
tends
might
[See
and
the
wonder
Theor y
discussion
about
of
on
the
connection
Knowledge
beauty
in
page
at
between
the
end
this
of
series
this
and
chapter
the
for
value
of
1 +
to
thoughts
used
of
the
exponential
function f
(x)
=
e
is
a
graph
growth
is
the
exponential
binomial
and
the
graph
of
f
(x)
=
is
e
limit
to
a
show
had
2
to
and
considered
that
lie
3.
to
This
be
graph
the
of
He
of
x
exponential
innity.
between
x
graph
n
⎠
mathematics.]
the
The
as
⎟
n
e.
theorem
➔
the
n
+
⎝
Y
ou
problem
interest,
e:
limit
1
at
compound
1
+
was
numbers.
he
1
e = 1 +
he
36…
of
However
of
Bernoulli
When
looking
There
great
result.
family.
T
o
the
mathematicians
the
Here
of
was
rst
approximation
decay
.
found
y
for
e.
y
x
f(x)
=
e
–x
y
=
e
An
(0, 1)
(0, 1)
x
0
Transformations
Now
you
function,
Chapter
know
you
1
to
the
can
help
Exponential
of
exponential
general
use
you
the
x
0
shape
r ules
sketch
of
for
graph
of
an
transformations
graphs
and logarithmic functions
the
functions
of
other
exponential
of
graphs
exponential
from
functions.
number
cannot
be
expressed
exactly
as
a
a
1
irrational
decimal.
fraction
or
➔
f
(x)
±
units
k
translates
ver tically
up
f
(x)
or
through
k
y
down
y
f
(x
±
units
or
k)
translates
horizontally
f
(x)
to
through
the
y
k
=
=
f(x
(x)
(−x)
pf
+
2
f(x)
+
2)
right
reflects
f
(x)
in
the
y
x-axis
y
f
f(x)
left
y
−f
=
(x)
reflects
f
stretches
scale
factor
f
stretches
(qx)
(x)
f
in
(x)
the
f(x)
f(x)
–f(x)
y
y-axis
ver tically
=
=
=
=
f(–x)
y
with
=
y
y
(x)
f(x)
2f(x)
p
f
=
=
f(x)
horizontally
y
=
f(2x)
with
scale
y
factor
=
f(x)
Example
x
The
diagram
shows
the
sketch
of
f
(x)
=
y
2
x−2
On
the
same
axes
sketch
the
graph
of
g (x)
=
8
2
6
4
2
x
0
–3
–1
1
3
Answer
y
8
You
find
f
through
(x)
g (x)
by
2
translating
units
to
the
right.
6
4
The
graph
of
2
the
(0, 1)
point
⎜
0,
⎝
–1
will
pass
through
⎞
⎟
4
⎠
x
0
–3
g (x)
1
⎛
1
3
4
5
Both
graphs
get
closer
and
closer
to
1
4
the
x-axis
as
the
value
of
x
decreases.
Chapter
Exercise
1
Given
the
F
the
graph
graph
of
intercepts
of
g (x)
on
the
on
f (x)
=
(x),
and
the
axes
x
a
f
without
same
and
set
any
of
using
axes
a
calculator,
showing
g (x)
=
x
2
+
3
b
f (x)
=
g (x)
8
8
6
6
4
4
2
2
x
–1
1
1
–4
–4
–6
–6
–8
–8
–10
–10
⎜
⎝
⎞
x
=
⎟
⎜
⎠
⎝
d
⎟
f (x)
=
x+1
e
g (x)
=
e
⎠
y
y
8
8
6
6
4
4
2
2
x
0
–3
3
⎛
⎞
=
x
–1
–2
3
0
–3
3
–2
⎛
=
y
0
c
any
–x
3
y
–3
clearly
asymptotes.
x
2
sketch
–1
x
0
3
–3
–1
1
–2
–2
–4
–4
–6
–6
–8
–8
–10
–10
3
2x
x
⎛ ⎞
e
=
⎜
⎝
⎟
=
f
⎜
⎠
⎝
f
⎟
⎛ 1 ⎞
⎛ 1 ⎞
⎛ ⎞
(
x
)
=
⎜
⎝
⎠
y
g
⎟
e
(
x
)
=
⎜
⎝
⎠
⎟
e
y
8
8
6
6
4
4
2
x
0
–3
2
State
–1
the
–3
–1
–2
–4
–4
–6
–6
–8
–8
–10
–10
domain
Exponential
and
range
x
0
3
–2
of
each g(x)
and logarithmic functions
function
1
in
3
question
1.
⎠
.
Properties
of
logarithms
3
Look
2
is
So
at
the
we
this
equation:
base
say
8
and
that
=
3
the
is
2
the
=
8
exponent
logari thm
of
8
or logari thm
to
the
base
2
is
3
and
write
this
3
as
log
In
general,
2
given
that
a
>
0:
x
➔
If
b
=
a
then
log
b
=
x
a
or,
Being
if
b
is
able
simplify
to
log
Example
Evaluate
a
to
the
change
power
x,
between
then
x
these
is
the
two
logarithm
forms
of
allows
b,
you
to
base
a
to
statements.
log
125
5
Answer
Write
x
=
log
‘x
=’
the
log
statement.
125
5
x
5
Change
=
125
=
5
=
3
x
5
x
Equate
3
Example
Evaluate
equation
to
exponent
for m.
exponent
for m.
exponents.
log
4
64
Answer
x
=
log
4
64
x
64
=
Change
4
equation
to
3
3
(4
x
)
3x
x
1
=
4
Write
64
=
1
Equate
1
and
as
the
solve
4
exponents
for
x.
=
3
Exercise
✗
1
G
Evaluate
a
these
expressions.
b
2
c
Evaluate
these
log
64
d
2
expressions.
1
a
log
b
3
c
d
81
Chapter
Example
Evaluate
log
4
4
Answer
Write
x
=
log
‘x
=’
log
statement.
4
4
Change
equation
to
exponent
x
4
=
4
=
1
for m.
x
1
Equate
In
general,
➔
log
a
the
=
log
to
base
a
of
any
exponents
number
a
=
(4
=
4
).
1.
1
a
Example
Evaluate
log
1
5
Answer
x
=
log
1
5
x
5
x
=
1
=
0
Any
any
Write
number
base
➔
log
is
1
raised
to
the
power
0
is
equation
equal
to
1
in
so
exponent
the
log
of
0.
=
0
a
Exercise
✗
1
H
Evaluate
log
a
6
log
b
6
log
d
1
log
e
8
Some
you
1
log
What
log
c
find
are
solutions
1
log
f
happens
when
1
b
undefined
for
n
n
2
expressions
can’t
10
10
–
this
means
that
them.
you
tr y
to
evaluate
the
expression
(−27)?
log
3
First
write
x
=
the
log
log
equation.
(−27)
3
Then
rewrite
the
equation
in
exponent
form.
x
3
This
−27
equation
Y
ou
➔
=
can
log
b
is
only
has
find
no
solution.
logarithms
undefined
for
any
of
base
a
Exponential
and logarithmic functions
posi tive
a
if
b
is
numbers.
negative.
for m.
1
in
What
2
is
the
value
of
log
0?
3
First
write
x
=
an
log
equation.
0
3
Rewrite
in
exponent
form.
x
3
This
➔
=
0
equation
log
0
is
has
no
solution.
undefined.
a
Example
13
Example
illustrates
another
proper ty
of
logarithms.
5
Evaluate
log
2
2
Answer
Write
5
x
=
log
log
equation.
2
2
Rewrite
x
2
in
exponent
for m.
5
=
2
=
5
Solve.
x
n
➔
log
(a
)
=
n
a
Summary
Given
that
a
of
>
properties
of
logarithms
0
b
●
If
x
=
a
then
log
x
=
b
a
●
log
a
=
1
a
●
log
●
log
1
=
0
a
b
is
undefined
if
b
is
negative
a
●
log
0
is
undefined
a
n
●
log
(a
)
=
n
a
Example
Find
the
value
of
x
if
log
x
=
5
2
Answer
log
x
=
5
=
x
Rewrite
=
32
Solve.
2
5
2
x
Exercise
1
Write
2
x
these
=
Write
a
x
equations
in
log
for m.
form.
5
2
b
these
=
exponent
I
9
a
in
log
2
8
x
=
equations
b
x
=
4
3
in
c
exponent
log
3
27
c
x
=
b
10
d
x
=
a
d
x
=
log
form.
x
=
log
10
1000
b
a
Chapter
Solve
3
these
log
a
x
=
equations.
3
log
b
4
x
=
4
log
c
3
64
=
2
x
log
d
6
=
log
e
x
x
=
−5
2
.
Logarithmic
functions
Investigation
–
What
would
kind
of
function
inverse
undo
functions
an
exponential
function
x
such
f
as
:
x
2
?
x
Copy
a
and
complete
this
table
of
values
for
the
function
y
=
2
x
x
−3
−2
1
0
1
2
f :
3
f
x
is
↦
a
2
means
function
that
under
1
which
x
is mapped
y
x
8
to
2
x
The
inverse
y-values
Copy
b
and
and
function
switch
complete
of
y
=
2
will
take
all
the
x-
and
for
the
inverse
them.
this
table
of
values
x
function
of
y
=
2
.
1
x
8
y
−3
x
Using
c
and
its
What
d
Now
➔
you
find
find
and
tables
inverse
do
let’s
To
these
then
values
function
on
sketch
the
a
same
graph
set
of
of
both
y
=
2
axes.
notice?
the
an
of
equation
inverse
of
rearrange
a
to
of
the
graph
function
make
y
of
the
inverse
algebraically
,
the
function.
switch
x
and
y
subject.
x
f
:
x
2
is
another
x
−1
To
get
the
inverse
function,
f
,
of
way
of
writing
y
=
2
:
x
Write
y
=
2
y
is
the
exponent
that
y
x
log
=
2
Switch
x
=
ylog
y
=
log
2
so
2
Take
x
and
logs
to
y
the
the
base
2
of
both
sides
by
base
in
2
order
Since
log
2
=
2
Log
is
shor t
logarithm.
➔
Generally
if
then
x
y
=
log
x
is
the
inverse
of
y
=
a
a
get
1
to
raised
2
x
2
So
is
Exponential
and logarithmic functions
for
x
x
The
graph
of
y
=
log
x
is
a
reflection
of
y
=
x
a
y
y
=
a
a
in
the
line
y
=
x
y
=
x
=
log
(0,1)
y
x
a
x
(1,0)
➔
A
logarithmic
function,
f
( x)
=
log
x,
has
these
John
proper ties:
Napier
(1550–
a
1617)
the
domain
is
the
set
of
all
positive
real
much
the
range
is
the
cur ve
does
the
set
of
all
real
intercept
of
the
y-axis
is
a
ver tical
the
x-intercept
on
logarithms.
you
graph
is
continually
Transformations
of
Again
the
once
you
logarithmic
to
consider
the
Exercise
1
Given
know
function
that
he
logarithms
or
1
discovered
the
say
asymptote
invented
is
with
early
y-axis
Would
the
credited
the
numbers
work
not
is
numbers
graphs
increasing.
logarithmic
you
general
can
of
them?
use
other
shape
what
functions
of
you
logarithmic
the
graph
lear nt
in
of
a
Chapter
1
functions.
J
the
function
f
( x)
=
log
x
y
describe
a
the
transformation
required
in
each
case
y
=
log
x
a
to
obtain
the
graph
a
g ( x)
=
log
b
g ( x)
=
log
(x)
−
of
g(x)
2
a
0
x
(1, 0)
(x
−
2)
a
c
g ( x)
=
2log
x
a
EXAM-STYLE
2
Sketch
the
QUESTION
graph
of
y
=
−2log(x
−
1)
without
using
a
calculator.
When
Include
on
your
graph
the
intercepts
with
the
two
the
(if
3
they
Sketch
exist).
the
of
y
=
log
(x
+
1)
+
2
clearly
base
logarithms
base
graph
no
is
given
axes
labeling
are
10.
any
2
asymptotes
4
The
sketch
on
the
shows
graph.
the
graph
of
y
=
log
x.
y
a
Find
the
value
of
(27, 3)
a
0
(1, 0)
x
−1
5
Given
that
f
(x)
=
log
x
find
f
(2)
3
Chapter
Logarithms
to
base
x
y
=
log
x
is
the
inverse
of
y
=
10
.
This
is
an
impor tant
logarithm
10
as
it
is
Base
and
one
10
of
logs
just
the
are
write
only
ones
called
log x
for
that
you
common
log
can
logs
use
and
the
you
calculator
can
omit
the
to
find.
base
x
10
There
is
a
‘log’
Example
Use
a
key
on
the
calculator.
calculator
to
evaluate
log 2
to
3 dp.
Answer
log
2
=
*Logarithms
1.1
0.301
to
3
log
dp.
0.30103
(2)
10
GDC
help
on
CD:
demonstrations
Plus
and
Casio
Alternative
for
the
TI-84
FX-9860GII
1/99
GDCs
Natural
The
are
on
the
CD.
logarithms
logari thm ,
natural
log
x
(log
to
the
base
e), is
the
other
e
impor tant
Y
ou
write
logarithm.
ln x
for
log
x.
There
is
an
‘ln’
key
on
the
calculator
e
Example
Make
ln 4
Use
a
calculator
to
sure
you
close
evaluate
the
ln 2
brackets
after
the
4
the
calculator
otherwise
Answer
⎛
calculate
=
2
will
*Logarithms
1.1
ln 4
In(4)
ln
4
⎞
⎜
⎟
⎝ In 2
⎠
2.
ln 2
In(2)
GDC
help
on
CD:
demonstrations
1/99
Plus
and
GDCs
Exercise
1
K
Use
a
to
significant
a
3
calculator
log 3
to
evaluate
4log 2
correct
c
e
f
(log 3)
Exponential
2
h
log
2
expressions
figures.
b
d
g
these
log 3
and logarithmic functions
Casio
are
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
➔
y
=
ln x
is
the
inverse
of
the
x
exponential
function
y
=
e
x
y
y
=
e
y
=
x
(0, 1)
y
=
In x
x
(1, 0)
This
relation
gives
us
x
➔
log
(a
three
log
)
=
x
and
results:
x
a
a
impor tant
=
x
a
x
ln(e
lnx
)
=
x
and
e
=
x
log (10
Solve
log x
)
Example
x
=
x
and
(10
)
=
x
these
equations,
giving
your
answers
x
e
a
to
3
significant
figures.
x
=
2.3
ln x
b
=
–1.5
c
10
=
0.75
d
log x
=
3
Answers
x
e
a
=
2.3
x
ln(e
)
=
ln2.3
x
=
0.833(3 sf)
Write
in
natural
log
for m.
x
b
ln x
=
–1.5
=
e
=
0.223(3 sf)
lnx
e
x
Use
ln
–1.5
(e
)
Use
(e
)
=
Use
log(10
Use
10
x
10
c
=
x
and
evaluate.
lnx
x
and
evaluate.
x
=
0.75
)
=
log 0.75
x
=
−0.125(3 sf)
)
=
x
and
evaluate.
x
log(10
log x
log x
d
=
3
log x
10
x
=
x
and
evaluate.
3
=
10
=
1000
Example
1
2x
Given
that
f (x)
=
e
,
−1
find
f
(x).
3
Answer
1
2x
f
(x)
=
e
3
1
2x
y
=
e
3
1
2y
x
=
e
Interchange
x
and
y.
3
{
Continued
on
next
page
Chapter
2y
3x
=
e
2y
ln(3x)
=
ln e
ln(3x)
=
2y
x
Use
ln(e
)
=
x.
1
ln(3x)
=
Solve
y
for
y.
2
1
–1
So
f
(x)
=
ln(3x),
x
>
0
2
Exercise
1
Solve
L
these
equations
giving
x
a
e
d
e
answers
to
3 sf
where
x
=
1.53
x
e
b
necessar y
.
=
0.003
e
c
=
1
x
x
=
5e
e
=
0.15
2
Solve
these
equations
giving
answers
to
3 sf
where
necessar y
.
x
x
10
a
=
2.33
x
10
b
=
0.6
10
c
x
=
1
d
10
=
3
Find
if
log x
a
4
x
=
Without
log
2
log x
b
using
a
5
−1
calculator
12
log
5
log x
c
evaluate
4
5
b
Without
using
a
0
d
evaluate
log x
=
−5.1
expressions.
ln4
c
calculator
=
these
5
5
a
=
d
these
e
expressions.
5
ln e
a
log 100
b
ln1
c
ln e
d
e
ln
EXAM-STYLE
QUESTIONS
2x−1
6
Given
that
f
(x)
=
e
7
Given
that
f
(x)
=
e
−1
find
f
(x)
and
state
its
domain.
0.25x
,
−2
≤
x
≤
4,
state
the
domain
and
−1
range
of
f
−1
8
Given
that
f
(x)
=
ln 3x,
9
Given
that
f
(x)
=
ln(x
x
>
0,
find
f
(x).
x
−
1),
x
>
1,
and
g(x)
=
2e
find
(g
f
°
.
We
Laws
can
of
deduce
logari thms
the
laws
of
logarithms
p
equations,
x
=
=
a
q
and
y
=
and
=
a
then
=
and
=
and
so
=
×
=
=
+
=
+
and
hence
Exponential
+
and logarithmic functions
from
the
exponential
)(x)
This
equation
is
tr ue
for
logarithms
in
any
base
so
Notice
➔
log
x
+
log
y
=
log
that
xy
log xy
≠
log x
×
log y
x
=
÷
=
and
that
log
log
x
log
y
≠
y
so
=
−
and
hence
=
−
x
➔
log
x
–
log
y
=
log
y
=
=
so
=
and
hence
=
n
➔
We
n
log
can
x
=
also
log
x
derive
this
➔
key
result
from
the
third
law
.
−1 ×
=
−
All
can
these
be
laws
are
omitted.
Formula
tr ue
Y
ou
for
must
logarithms
lear n
these
in
any
laws
base
as
and
they
are
so
the
not
in
bases
the
booklet.
Example
1
Express
log
5
+
log
2
36
2
log
10
as
a
single
logarithm.
2
2
Answer
1
log
5
+
log
2
36
log
2
10
2
2
1
n
2
=
log
5 + log
2
=
log
36
log
2
5 + log
2
10
n log
6
2
log
30
2
= log
x
a
10
2
log x
= log
x
a
2
log
+ log
y = log xy
10
2
x
=
log
3
log x
log y =
2
y
Chapter
Exercise
1
M
Express
as
single
logarithms:
a
b
log
24
e
3log
–
log
2
c
2log
f
log
8
–
4log
2
d
x
–
2log
y
x
–
log
y
–
log
g
2
Express
+
as
single
−
logarithms:
a
−
b
c
−
2ln3
d
–
ln18
e
3ln2
–
2
f
3
Find
the
a
value
of
each
expression
log
b
(each
24
–
answer
log
2
is
3
an
c
integer).
2
d
e
Example
Given
that
a
=
log
x,
b
=
log
5
⎛
log
write
5
in
⎟
2
y
and
c
=
log
z,
5
⎞
x
⎜
⎜
y
5
3
terms
of
a,
b
and
c
log
z
⎟
z
⎝
⎠
Answer
⎛
⎞
x
2
log
5
⎜
=
⎟
2
⎜
y
3
log
x
log
5
y
3
z
5
⎟
z
⎝
⎠
1
x
log
3
2
2
=
y
(log
5
+
5
)
5
1
=
log
x
− 2log
5
y
− 3log
5
z
5
2
1
=
a − 2b
− 3c
2
Exercise
N
EXAM-STYLE
1
Given
that
QUESTION
p
=
log
a
and
q
=
log
2
of
p
and/or
q
b,
find
an
expression
2
for:
3
a
log
ab
b
2
log
a
c
d
e
Exponential
2
and logarithmic functions
in
terms
z
Let
2
x
=
log P,
y
=
log Q
and
z
=
log R.
⎛
Express
Write
these
where
a
in
⎟
⎝
3
⎞
⎜
terms
expressions
and
b
of
x,
y
and
z
⎠
are
in
the
form
a
+
blog x
integers.
log10x
a
b
c
d
EXAM-STYLE
QUESTIONS
Given
4
that
write
y
in
the
form
y
=
pa
+
q
where
p
and
q
are
integers
to
be
found.
Write
5
in
the
form
a
+
blog
x
where
a
and
b
are
3
integers.
x
xln2
Show
6
Notice
that
that
e
=
2
question
6
in
Exercise
4 N
demonstrates
the
general
result
x
a
xlna
=
e
Change
of
Sometimes
there
is
a
Suppose
base
you
need
formula
y
=
log
to
that
a
change
enables
and
you
the
you
want
base
to
to
do
of
a
logarithm
and
this.
change
the
log
to
base c.
b
y
If
y
=
log
a
then
a
=
b
b
y
Star t
with
a
=
Take
logs
to
b
base
c
of
both
sides:
y
log
a
=
log
a
=
ylog
c
log
b
c
c
b
c
=
But
y
=
log
a
so
b
➔
Change
of
base
formula:
This
=
formula
is
useful
as
most
only
Y
ou
can
use
this
formula
to
evaluate
a
logarithm
or
to
logarithm
to
any
calculators
logs
to
base
change
10
a
give
or
e.
base.
Chapter
Example
Use
the
change
of
base
formula
to
evaluate
log
9
to
3
4
significant
figures.
Answer
log 9
log
9
For
=
Change
4
the
log
to
base
10
= 1.58 (3 sf)
Use
calculator
to
evaluate
answer.
Example
log
3
=
a
and
log
x
6
=
b.
x
Find
log
6
in
terms
of
a
and
b
3
Answer
log
Use
6
the
change
of
base
for mula.
x
log
6 =
3
log
3
x
b
=
a
Exercise
1
O
Use
the
to
significant
3
change
of
base
formula
to
evaluate
these
expressions
figures.
⎛ ⎞
a
log
7
b
⎜
c
⎟
log
2
(0.7)
3
⎝
⎠
7
d
log
e
log
e
7
2
Given
7
3
that
log
x
=
y,
express
log
3
EXAM-STYLE
3
If
log
2
log
=
d
log
x
and
log
6
a
y
log
24
e
log
y,
find
Given
terms
c
log
12
f
log
of
y
of
x
and
6
GDC
log
to
x
sketch
b
y
=
that
36
2
3
2
these
2log
4
5
in
2
6
your
=
=
b
a
Use
terms
a
6
2
4
in
QUESTION
a
a
x
9
graphs.
x
5
log
a
=
b
express
y
in
terms
of
b
4
2
a
y
=
log
a
b
y
=
log
4
a
16
c
=
Exponential
and logarithmic functions
d
=
y:
base
10
logs,
10.
log 4
is
omitted.
the
.
Exponential
Solving
Y
ou
In
can
exponential
use
Section
numbers
you
are
will
and
logarithms
4.2
you
were
the
lear n
to
solved
same
how
to
logari thmic
equations
solve
exponential
exponential
or
equations
could
solve
be
equations
made
equations
equations.
the
where
where
same.
the
In
base
the
this
base
section
numbers
different.
Example
x
Solve
5
=
9
Answer
x
5
=
9
=
log
9
=
log
9
Choose
base
10
or
x
log
x log
5
5
log 9
x
Take
logs
of
Now
bring
Rear range
both
down
the
sides.
the
natural
exponent.
you
logs
can
use
so
that
your
GDC.
equation.
=
log 5
x
=
1.3652…
x
=
1.37
(3
sf)
Check
an
Example
6
question
requires
ln a
x + 1
=
the
answer
x
Solve
whether
exact
3
giving
your
answer
in
the
form
ln b
where
a
and
b
are
integers.
Answer
x
x+1
6
=
3
x
ln 6
x
ln 6 −
x +1
=
x
ln 6
=
(x
x
ln 6
=
x
x
ln 3
x (ln 6 − ln 3)
=
=
Take
ln 3
+ 1) ln 3
ln 3 + ln 3
natural
Bring
down
Multiply
Collect
ln 3
the
out
of
both
sides.
exponents.
brackets.
x-ter ms
Factorize
ln 3
logs
and
together.
divide.
ln 3
x
=
(ln 6
ln 3 )
a
ln 3
x
ln a
=
ln 2
− ln b
=
ln
b
Chapter
Example
3x
Solve
1−x
e
=
5
,
giving
an
exact
answer.
Answer
x
3x
=
x (3
=
ln 5
3x
=
(1–
3x
=
ln 5
x ln 5
+
logs
since
ln
e
=
x
1 – x
ln e
+
natural
5
3x
3x
Use
1 – x
e
x)
=
ln 5
=
ln 5
ln 5)
–
ln 5
Bring
down
the
exponents.
x ln 5
Multiply
Collect
out
Leave
brackets.
x-ter ms
in
together.
log
your
form
answer
since
ln 5
Factorize
x
and
an
divide.
(3 + ln5)
Exercise
1
Solve
required.
P
these
equations
x
to
find
the
value
of
x
to
x
2
a
=
5
b
3
f
2
3
significant
x
=
50
c
5
g
e
figures.
x+1
=
17
7
d
=
16
2x−1
Solve
=
3.2
×
x
10
=
6
h
=
EXAM-STYLE
2
−3
e
QUESTION
these
equations
to
find
the
value
of
x
to
3
significant
figures.
x+2
x −3
2x −5
2−x
a
2
e
e
=
5
=
c
3
4e
f
7
d
=
x −1
=
(0.5)
−0.001x
3x −2
x
=
4
x
+
3
b
3x −1
Example
=
244
g
35e
=
95
ln a
x −1
x+2
Solve
3
×
6
=
2
×
3
,
giving
your
answer
in
the
form
x
,
=
ln b
where
a,
b
∈
Answer
x
ln (3
×
–
6
+
ln (6
1
x + 2
)
x
ln 3
=
ln (2
×
3
– 1
)
Take
+
ln 2
+
+
(x
–
1) ln 6
=
ln 2
ln(3
+
(x
both
+
+
x ln 6
x ln 6
x(ln 6
–
–
ln 6
xln 3
–
ln 3)
=
=
=
ln 2
ln 2
ln 2
+
+
x ln 3
ln 9
x-ter ms
2)ln 3
2ln 3
+
sides.
)
and
ln 3
logs
2
of
=
natural
)
x
Collect
ln 3
+
+
+
factorize.
2ln 3
ln 6
ln 6
–
–
ln 3
ln 3
⎛ 108 ⎞
ln
⎜
You
⎟
⎝
x
6
⎞
simplify
ln a
=
⎛
ln 2
fur ther
this
⎟
and logarithmic functions
a
–
ln b
⎜
⎝ 3 ⎠
Exponential
can’t
ln 36
⎠
3
=
ln
exact
=
ln
b
any
answer
is
Exercise
Q
EXAM-STYLE
1
Solve
QUESTIONS
these
equations
to
find
the
value
x
a
d
5
=
x – 1
2
×
Solve
b
4
e
3
x
2x
2
=
these
3
×
×
7
equations
to
find
3
of
x
the
3
significant
=
figures.
x
5
3
c
x – 1
4
to
2x – 1
×
2
x
=
4
×
5
x + 2
=
2
×
value
7
of
x
in
the
form
,
=
where
a,
b
∈
x + 2
a
2
c
5
x – 3
=
x
5
3
Solve
=
2
×
6
(6
b
4
−
Solving
=
x
–
8
×
7
1
)(2
x + 2
)
x
Some
d
x
3
=
2(4
)
x
a
×
3 – 2x
3
for
5
x
x + 1
×
b
=
logarithmic
logarithmic
x
–
3(2
)
=
0
equations
equations
can
be
solved
by
ensuring
that
both
The
sides
of
the
equation
contain
logarithms
written
to
the
same
argument
expression
Then
you
can
equate
the
is
the
base.
inside
the
arguments
brackets.
Example
2
Solve
log
(x
)
=
log
a
(3 x
+ 4)
a
Answer
2
(x
log
)
=
log
=
3x
4
=
0
1)
=
0
a
(3x
+
4)
a
2
x
+
4
Equate
the
arguments.
2
x
(x
−
−
3x
4)(x
x =
Y
ou
−
+
4
or
x
you
Substituting
the
log
Example
Solve
x
that
both
cannot
=
of
the
quadratic.
−1
check
must
Remember
gives
=
Solve
4
a
and
solutions
find
x
=
positive
the
−1
are
possible.
logarithm
into
number
both
so
of
sides
here
a
negative
of
both
the
number.
original
solutions
are
equation
possible.
ln(12 −
x )
=
ln x
+ ln( x
− 5)
Answer
ln(12
−
x)
=
ln x
ln(12
−
x)
=
ln x (x
+
ln(12
−
x)
=
ln(x
=
x
ln(x
−
−
5)
5)
2
−
5x)
2
12
−
x
−
5x
Equate
arguments.
2
x
(x
−
−
4 x
6)(x
x
=
6
−
+
12
2)
or
x
=
0
=
0
=
−2
Solve
the
quadratic.
{
Continued
on
next
page
Chapter
When
ln x
When
ln
x
so
x
=
and
x
=
and
x
=
Check
6
ln(x
6
−
is
−
the
5)
are
only
solution.
equations
=
−
−
b
d
+
=
−
−
it
Example
log
(x
is
−
+ +
=
−
=
to
solve
a
+
log
equation
using
Since
=
exponents.
3
5
Answer
log
(2x
–
1)
=
3
5
3
b
5
=
2x
–
1
log
x
b
⇒
x
=
a
a
125
=
2x
=
x
Example
Solve
2x
–
1
126
=
63
log
x
+
log
2
(x
−
2)
=
3
2
Answer
x
log
+ log
2
(x
− 2)
=
3
2
[x (x
log
− 2 )] =
3
2
Using
the
first
law
on
page
2
(x
log
2x )
=
3
2
2
x
3
− 2x
=
2
b
Since
log
x
=
b
⇒
a
2
x
− 2x
=
8
2
x
(x
+ 2 )( x
x
x
=
− 2x
4
=
−2
is
Exponential
− 8
− 4)
or
the
x
=
0
=
=
only
0
4
solution
and logarithmic functions
x
must
be
positive.
x
=
a
=
easier
2)
Sometimes
Solve
x.
e
for
c
negative
QUESTION
these
a
solutions.
positive.
R
EXAM-STYLE
Solve
are
−2
ln(x
Exercise
1
5)
123.
+
Exercise
Solve
1
these
log
a
S
(x
equations
−
2)
=
2
for
x.
log
b
9
(2x
−
1)
=
3
c
−
=
3
Solve
2
these
a
equations
− +
log
c
(2x
−
3)
–
(4x
−
8)
–
log
log
(4x
−
5)
=
(x
−
5)
=
4
2
0
QUESTIONS
that
an
Hence
log
b
2
x
+
find
7
EXAM-STYLE
Given
=
x.
7
3
for
expression
or
x
+
=
for
otherwise
A
A
in
solve
terms
log
x
of
+
x.
log
2
(2x
+
7)
=
2
2
Y
ou
will
change
Solve
4
+
need
the
to
base
here
=
rst.
Solve
5
+
.
=
Applications
of
exponential
and
Extension
material
Worksheet
4
linear
logarithmic
Exponential
Models
of
exponential
and
growth
and
decay
functions.
areas
just
a
few
applications
of
exponential
of
appear
mathematics
to
decay
be
completely
growth
disconnected
and
to
form
use
that
are
CD:
decay
T
wo
Here
on
Reduction
functions
growth
exponential
-
might
be
exponentials
models.
and
probability.
Biology
But
●
Growth
●
Human
of
micro-organisms
in
a
consider
A
Spread
problem…
group
of
people
go
to
lunch
and
population
afterwards
●
this
culture
of
a
pick
up
their
hats
at
virus
random.
What
is
the
probability
that
Physics
no
●
Nuclear
●
Heat
chain
one
gets
their
own
hat?
reactions
It
can
be
shown
that
this
probability
transfer
1
is
.
Economics
Y
ou
●
Pyramid
Processing
power
e
to
one
of
you
the
basis
have
like
to
studied
explore
this
probability
once
fur ther
.)
of
of
Can
your
might
these
technolog y
as
●
wish
(Y
ou
pick
Computer
may
schemes
you
think
of
any
other
areas
Mathematical
computers
of
knowledge
that
are
surprisingly
Exploration.
●
Internet
trafc
growth
connected?
Chapter
Exponential
Example
The
growth
population,
A(t ),
in
thousands,
of
a
city
is
modeled
(0.02)t
by
of
the
function
years
after
A(t )
2010.
=
30e
Use
where
this
model
t
to
is
the
number
answer
these
questions:
a
What
b
By
was
what
each
the
population
percentage
is
the
of
the
city
in
population
2010?
of
the
city
increasing
year?
c
What
d
When
will
the
will
population
the
city’s
of
the
population
city
be
be
in
2020?
60 000?
Answers
t
0
a
A(0)
=
30e
=
30
is
so
The
population
in
2010
the
for
number
2010,
t
=
of
years
after
2010,
0
was
30 000.
(0.02)
b
A(1)
=
Write
30e
one
( 0.02 )
an
year
equation
after
for
the
population
2010.
30 e
( 0.02 )
=
e
Calculate
the
multiplying
factor.
30
= 1.0202...
The
at
population
2.02%
each
is
increasing
year.
In
2020,
t
=
10
( 0.02 ) ×10
A(10 )
c
=
=
In
30e
36.642...
2020
the
population
will
be
36 642
( 0.02 ) t
d
60
=
30e
When
population
is
60 000,
( 0.02 ) t
2
=
A(t)
e
Take
( 0.02 ) t
ln 2
=
ln e
ln 2
=
0 .02t
=
60
logarithmics
Bring
down
ln 2
t
=
Solve
0
t
The
after
=
34.657...
population
34.65
during
02
will
years,
be
that
60 000
is,
2044.
Exponential
and logarithmic functions
for
t.
the
of
each
exponent.
side.
Exponential
Example
A
decay
casserole
is
removed
from
the
oven
and
cools
according
to
the
model
−0.1t
with
the
equation
What
a
the
If
b
T (t)
temperature
the
85e
,
where
t
is
the
time
in
minutes
and
T
is
°C.
temperature
of
the
casserole
when
it
is
removed
from
oven?
the
the
is
=
in
temperature
casserole
to
of
the
reach
room
room
is
25 °C,
how
long
will
it
take
for
temperature?
Answers
0
T (0)
a
=
85e
=
The
85
the
temperature
casserole
0
is
of
oven,
casserole
t
=
is
removed
from
0
the
85 °C
T
= 25
25
0
the
1t
85e
b
When
=
25
if
the
temperature
room
is
Take
logarithms
of
the
25 °C.
5
1t
e
=
=
85
of
both
sides.
17
5
0
ln
1t
e
=
ln
17
5
0
1t
=
ln
17
1 .22377...
=
t
The
=
12.2
casserole
temperature
Exercise
1
The
(3
will
after
Solve
for
t.
sf)
reach
12.2
room
min.
T
sum
of
€450
is
invested
at
3.2%
interest,
compounded
annually
.
a
Write
after
b
2
In
How
i
b
how
formula
many
stages
people
many
after
How
a
for
the
value
of
the
investment
years.
early
infected
a
n
After
the
down
2
long
years
of
and
a
day
were
days
would
take
value
first
epidemic
the
exceed
there
number
rose
€600?
were
by
100
10%.
infected
ii
it
the
measles
each
people
will
for
after
250
a
week?
people
to
be
infected?
Chapter
3
Forest
fire
is
area
If
is
10
how
4
fires
left
spread
to
bur n
exponentially
.
unchecked
Ever y
15%
of
hour
the
that
the
remaining
bur nt.
hectares
long
Joseph
will
did
aircraft
are
a
his
it
bur nt
take
until
parachute
velocity
and
at
the
fire
10 000
jump
time
t
for
becomes
hectares
charity
.
seconds
out
are
his
control
bur ning?
After
after
of
jumping
out
parachute
of
the
opened
–1
was
v
where
m s
−0.063t
v
=
9
+
a
Sketch
b
What
29e
the
was
graph
of
Joseph’s
v
against
speed
at
t
the
instant
the
parachute
opened?
What
c
great
If
d
he
on
his
lowest
possible
speed
if
he
fell
from
a
ver y
height?
actually
landed
after
45
seconds
what
was
his
speed
landing?
How
e
was
long
when
the
did
it
take
parachute
him
to
reach
half
the
speed
he
had
opened?
b
5
Two
variables
When
of
The
a
n
=
and
x
and
=
32
n
are
and
connected
when
n
=
by
3,
x
the
=
formula
108.
Find
x
=
the
a
×
n
values
b
American
ear thquake
2,
x
geologist
to
Charles
Richter
dened
the
magnitude
of
an
be
I
M
=
log
S
where
M
is
the
ear thquake
taken
of
a
100
(measured
km
from
‘standard’
0.001
the
by
(as
the
a
decimal),
amplitude
epicenter
ear thquake.
The
of
the
I
of
is
a
the
intensity
seismograph
ear thquake)
intensity
of
a
and
standard
of
the
reading
S
is
the
the
Richter
Review
Evaluate
Scale
Solve
these
3
equations.
x−1
=
90
Exponential
is
b
5
Richter
Scale
287
2x+3
a
(
S)
fur ther
.
5
2
mm
Severity
exercise
log
in
intensity
ear thquake
millimetres.
Explore
1
magnitude
3x
=
3
and logarithmic functions
2x
c
2
×
3
x
=
5
Mild
0–4.3
Moderate
4.3–4.8
Intermediate
4.8–6.2
Severe
6.2–7.3
Catastrophic
7.3+
3
Solve
a
b
these
log
+
(x
equations.
+
6)
–
−
log
5
(x
=
+
2)
=
c
ln
d
Solve
(4x
–
7)
=
(
Solve
2
)
=
=
4
The
EXAM-STYLE
x
5
e
log
5
QUESTIONS
functions
f
and
g
are
defined
as
2x
f (x)
=
e
for
all
real
x
( )
=
for
x
>
0
a
State
the
b
Explain
ranges
why
of
both
f
(x)
and
g (x).
functions
have
inverses.
- 1
Find
expressions
c
Find
an
d
Solve
for
expression
the
for
inverse
( f
g)(x)
functions f
and
( g
°
the
equation
( f
f
–1
(x)
and
g
(x).
)(x)
°
g)(x)
=
( g
°
f
)(x)
°
0.08t
5
The
number,
where
t
is
a
Find
b
How
the
n,
the
the
of
insects
number
population
long
does
it
obser vations
Review
of
in
a
days
of
take
after
the
the
colony
,
is
given
by
obser vation
colony
after
population
50
to
n
=
4000e
commences.
days.
double
from
when
commenced?
exercise
✗
+
1
Solve
⎛
⎞
=
⎜
⎝
⎟
⎠
+
2
Find
the
exact
Give
your
value
x
satisfying
the
equation
+
=
answer
in
the
where
form
a,
b
∈
⎛ ⎞
3
Find
the
exact
value
of
+
−
⎜
⎝
EXAM-STYLE
⎟
⎠
QUESTION
4
Write
+
−
as
a
single
logarithm.
5
Solve
a
c
(
− ) =
=
b
+
d
(
− ) =
− +
−
=
EXAM-STYLE
6
If
m
=
log
QUESTION
4
and
n
=
log
x
a
log
4
8
8,
find
expressions
in
terms
of
m
and
n
for
x
b
log
x
2
c
log
x
16
d
log
32
8
Chapter
3(x−1)
7
The
function
Describe
a
f
is
defined
series
of
for
all
real
values
transformations
of
x
whereby
by
the
f
(x)
=
graph
e
+
2
≠
1.
of
x
y
=
f
(x)
can
EXAM-STYLE
be
obtained
from
the
graph
of
y
=
e
QUESTIONS
−1
8
Find
the
inverse
function
f
(x)
2x
a
f
(x)
=
if
3x
3e
f
b
(x)
=
10
f
c
(x)
=
log
(4x)
2
9
Solve
these
a
b
and
are
simultaneous
positive
real
equations
for a
and
b,
given
that
numbers.
1
log
64 +
log
a
b
=
8
log
a
=
ba
2
CHAPTER
4
SUMMARY
Exponents
Laws
of
●
●
●
exponents
+
×
=
÷
=
=
●
●
●
●
=
Exponential
●
functions
An exponential function
is
a
function
of
the
form
x
f
(x)
=
a
where
●
The
domain
●
The
range
is
●
The
graph
of
of
a
is
the
the
a
positive
real
exponential
set
of
all
number
function
positive
real
(that
is
is, a
the
>
set
0)
of
and
all
a
real
numbers.
numbers.
x
the
exponential
function f (x)
=
e
is
a
graph
−x
of
exponential
growth
of
exponential
decay
.
and
the
graph
y
of
f
(x)
=
e
is
a
graph
y
x
f(x)
=
e
–x
y
=
e
(0, 1)
1
(0, 1)
0
x
0
x
Continued
Exponential
and logarithmic functions
on
next
page
Logarithms
Properties
of
logarithms
●
If
then
●
=
●
=
●
b
●
●
is
undefined
is
for
any
base
a
if
b
is
negative
undefined
=
Logarithmic
●
To
find
and
an
then
functions
of
inverse
rearrange
a
to
function
make
y
the
●
Generally
if
switch
x
and
y
subject.
algebraically
,
then
x
y
=
log
y
=
ln x
x
is
the
inverse
of
y
=
a
a
x
●
is
the
inverse
of
the
exponential
function y
=
e
x
y
y
=
e
y
=
x
(0, 1)
y
=
In x
x
(1, 0)
x
●
log
(a
log
)
=
x
and
x
a
a
=
x
a
x
ln(e
lnx
)
=
x
and
e
=
x
log (10
Laws
x
log x
)
of
=
x
and
(10
)
=
x
logarithms
●
●
●
=
−
=
=
●
Change
of
base
formula
●
=
Chapter
Theory
of
The
“The
knowledge
beauty
greatest
music,
mathematics
standing
on
the
has
of
the
mathematics
simplicity
borderland
of
all
that
beautiful
Herbert
The
Beautiful
and
Have
solved
you
pleased
Was
it
with
just
solution
Look
ever
at
and
+
it
+
Turnbull
two
in
correct,
stylish,
solutions
simplify
=
x²
–
xy
=
x²
–
2yz
=
x²
–
(y²
=
x²
–
(y
–
xz
–
+
+
–
y
–
+
y²
mathematics
(x
+
even
to
y
or
+
the
z)(x
was
it
and
all
and
that
is
199
and
been
because
your
beautiful?
problem:
–
y
–
z)
z)
xy
–
2yz
science,
poetr y
(1885–1961)
Solution
z)(x
in
supreme
ar t.”
Mathematicians,
(x
wonderful
of
solutions
problem
was
efficient,
these
Solution
a
in
is
inevitableness
solution?
because
was
Expand
y
your
simple
Westren
Great
and
–
y²
–
yz
+
xz
–
yz
–
z²
(x
+
=
(x
=
x²
y
+
+
z)(x
(y
+
–
y
z))(x
–
z)
–
(y
+
z))
z²
+
(y
+
z)²
z²)
z)²
“Pure
mathematics
is,
in
its
way,
the
They
the
both
second
give
us
the
solution
same
seems
right
better.
answer
It’s
and
more
yet
somehow
elegant
poetr y
logical
than
the
first
ideas.”
and
Albert
insightful
of
Einstein
one.
(1879–1955)
Theory
of
knowledge:
The
beauty
of
mathematics
Simple,
beautiful
“The
essence
equations
of
mathematics
but
Stan
to
Gudder,
make
is
not
to
model
make
complicated
Professor
Here
that
of
are
simple
things
mathematics,
some
the
world
things
complicated,
simple.”
University
famous
of
Denver
equations
2
Einstein’s
Newton’s
equation:
second
E
law:
=
F
mc
=
ma
k
Boyle’s
law:
V
=
p
Schrödinger’s
equation:
Hψ
=
E ψ
m
m
1
Newton's
law
of
universal
gravitation:
F
=
2
G
2
r
Isn’t
it
using
star tling
that
These
equations
moon
and
inter net
human
the
mathematical
bring
and
universe
equations
have
him
helped
back,
understand
can
such
to
put
develop
the
be
as
described
these?
man
on
the
wireless
workings
of
the
body
.
These
are
just
five
equations
–
which
is
your
favorite?
Is
it
one
possible
day
that
discover
mathematics
the
ultimate
and
science
theor y
will
of
ever ything:
A
theor y
together
A
theor y
outcome
carried
Now
fully
known
that
of
has
any
explains
physical
predictive
experiment
and
links
phenomena?
power
that
for
could
the
be
out?
wouldn’t
”
that
all
Boyle's
that
Law
be
wonderful?
explains
why
bubbles
increase
in
size
as
they
rise
to
the
surface.
Chapter
Rational
CHAPTER
functions
OBJECTIVES:
1
The
2.5
reciprocal
x
function
x
≠
0,
its
graph
and
self-inverse
nature
x
The
rational
Ver tical
and
Applying
Before
Y
ou
1
e.g.
x
horizontal
rational
you
should
Expand
function
ax
+
b
cx
+
d
and
its
asymptotes
functions
to
real-life
situations
start
know
how
to:
Skills
polynomials.
Multiply
the
graph
1
polynomials
check
Expand
−4(2x
a
2
−2(3x
−
1)
and
−2(3x
−
1)
=
−6x
2
2
(x
+
−
polynomials.
5)
6(2x
b
2
+
1):
2
−
3)
2
c
−x (x
+
e
x (x
3)(x
−
7)
x
d
+
2
(x
+
3)
8)
3
+
3x (x
3x
the
Graph
1)
=
3x
+
3x
horizontal
2
Draw
these
lines
x
=
0,
y
=
0,
x
=
3,
x
=
−2,
y
=
−3,
on
one
graph.
y
x
and
ver tical
e.g.
Graph
lines.
the
=
x,
x
=
−1,
y
y
=
−2
=
−x,
=
=
–x
x
=
2
4
3
lines
2
y
y
y
=
y
=
x
2,
y
=
4
x
–2
y
=
3
and
y
on
the
x
same
=
=
–2
–1
–4
graph.
3
Recognize
and
describe
3
Describe
y
the
y
8
a
translation.
transformations
y
=
3
x
B
6
e.g.
Find
the
translations
that
map
4
2
that
map
y
=
x
3
onto
y
=
onto
x
functions
2
A
and
A
is
B
A
and
B
and
write
6
x
0
of
2
a
horizontal
units
to
the
shift
down
right.
of
B
=
x
A
and
equations
B
–4
2
y
the
2
2
Function A
B
is
a
units
is
y
ver tical
up.
is
y
=
Rational
(x
shift
Function
2
B
=
x
+
3.
functions
−
of
2)
3
A
–6
–2
A
x
0
–4
2
4
6
–8
If
you
have
sounds
and
quality
of
a
rough
8160
an
so
the
idea
MP3
on
you
recording
is
minutes
player,
can
that
of
a
you
on
setting
4GB
music.
do
fit
The
and
MP3
That’s
know
it?
the
player
how
many
answer
length
will
songs,
depends
of
hold
the
136
on
song.
hours
albums,
the
However,
or
approximately
2000
songs
of
4
minutes
or
1000
songs
of
8
minutes
or
4000
songs
of
2
minutes.
This
leads
us
to
the
function
=
where
s
is
the
number
of
songs
and
m
is
the
number
of
minutes
that
a
song
lasts.
This
function
is
an
example
of
the
reciprocal
function,
.
=
In
this
chapter,
reciprocal
you
will
functions
use
and
a
GDC
other
expressed
in
the
form
ver tical
domain
and
asymptotes
ranges
of
for
.
the
the
explore
the
functions
graphs
that
can
of
be
+
=
and
to
rational
Y
ou
will
examine
horizontal
+
graphs
of
these
functions
and
the
functions.
Chapter
.
Reciprocals
Investigation
Think
of
pairs
E.g.
24
×
and
add
1,
of
12
some
–
numbers
×
2,
8
more
×
24
12
8
3
y
1
2
3
8
and
your
0
Now
≤
y
tr y
pairs
≤
the
as
whose
3,
pairs
x
Show
graphing
3
of
×
product
product
8.
Copy
is
pairs
24.
the
table
numbers.
coordinates
on
a
graph
with
0
≤
x
≤
24
24.
same
idea
with
negatives,
e.g.
−12
×
−2
End
and
graph
these
the
Explain
what
behavior
you
notice
appearance
●
the
value
of
x
as
y
gets
bigger
●
the
value
of
y
as
x
gets
bigger
as
fur ther
either
the
end
behavior
of
of
a
about
graph
●
is
too.
your
it
is
and
followed
fur ther
in
direction.
graph.
Zero
does
not
have
1
➔
The
of
reciprocal
a
number
is
1
divided
by
that
number.
a
reciprocal
as
is
0
undened.
What
does
For
example,
the
reciprocal
of
2
is
your
GDC
show
for
Taking
the
reciprocal
of
a
fraction
tur ns
For
example,
the
reciprocal
of
is
1
reciprocal
A
number
down.
=
1
×
.
The
reciprocal
of
is
or
multiplied
by
its
reciprocal
0?
÷
=
of
➔
upside
÷
The
it
1
4.
equals
1.
For
example
3
×
=
1
Geometrical
Example
in
the
reciprocal
of
inverse
were
1
Find
quantities
propor tion
describedas
2
reciprocali
2
translation
in
a1570
of
Euclid’
s
Answer
Elements from
=
2
Write
as
an
improper
fraction.
2
5
5
2
Check:
Reciprocal
of
=
Tur n
it
upside
can
find
2
reciprocals
of
algebraic
terms
too.
The
The reciprocal
of
x
is
−1
or
x
is
−1
and
x
×
x
= 1
5
reciprocal
number
➔
2
×
down.
5
Y
ou
BCE.
5
1
2
300
also
or
a
of
a
variable
called
its
=1
multiplicative
Rational
functions
inverse.
Exercise
Find
1
A
the
reciprocals.
2
a
3
b
e
the
h
3
reciprocals.
6.5
a
−1
g
Find
d
f
2
−3
c
x
b
y
c
3x
d
4y
e
The
+
term
was
f
g
h
Multiply
3
i
each
quantity
by
its
6
a
b
is
the
reciprocal
of
the
reciprocal
of
is
the
reciprocal
of
the
reciprocal
of
function
y
b
What
c
Will
the
when
xy
x
=
is
happens
d
Find
e
What
f
Will
.
The
ever
x
The
reach
when
ever
48
the
480
ii
value
y
zero?
is
third
describe
4?
x?
of
y
4800
iii
edition
reach
(1797)
two
to
numbers
whose
product
This
the
is
1.
when
x
gets
48 000
iv
the
zero?
function
of
x
function
used
in
the
Investigation
480
ii
value
is
larger?
Explain.
48
i
to
reciprocal
reciprocal
use
back
24
i
to
happens
x
far
Encyclopaedia
you
y
as
Britannica
What
the
as
working.
What
Find
common
least
c
b
a
your
of
a
For
5
Show
at
4
reciprocal.
in
j
reciprocal
4800
iii
when
y
gets
48 000
iv
page
on
142.
larger?
Explain.
function
is
k
f (x)
=
x
where
k
Graphs
is
of
a
constant.
reciprocal
functions
Investigation
Use
your
GDC
to
–
draw
all
have
graphs
all
the
of
graphs
in
similar
reciprocal
this
Draw
a
graph
of
( x)
a
2
=
g ( x)
b
is
the
effect
of
Draw
a
graph
of
changing
( x)
a
the
value
of
is
the
effect
of
=
x
the
numerator?
2
=
g ( x)
b
3
=
h( x )
c
=
x
x
What
h( x )
c
x
1
2
3
=
x
What
functions
investigation.
1
1
shapes.
changing
the
sign
of
the
x
numerator?
4
3
Copy
a
and
complete
this
table
for
f ( x)
=
x
x
0.25
0.4
0.5
1
2
4
8
10
16
f (x)
b
What
c
Draw
do
the
you
notice
graph
of
about
the
the
values
function.
of
x
and
f (x)
d
Draw
f
What
in
the
the
table?
line
y
=
x
on
the
same
graph.
4
e
Reect
f ( x)
in
=
the
line
y
=
x
do
you
notice?
x
1
g
What
does
this
tell
you
about
the
inverse
function
f
?
Chapter
Asymptotes
The
on
graphs
page
closer
of
143
to
the
all
the
functions
consist
axes
but
of
f
(x),
two
never
g(x)
and
cur ves.
actually
h(x)
The
touch
in
the
cur ves
or
Investigation
get
cross
closer
and
them.
The
The
axes
are
asymptotes
to
the
graph.
is
word
derived
Greek
➔
If
a
cur ve
gets
continually
closer
to
a
straight
line
meets
it,
the
straight
line
is
called
=
b
is
an
asymptote
to
the
function
y
=
f
x
→ ∞,
f
(x )
=
symbol
‘not
together’.
f (x)
→ b
y
The
means
(x)
y
As
the
an asymptote
falling
y
from
asymptotos,
but
which
never
asymptote
→means
=
b
‘approaches’.
The
horizontal
line
k
➔
The
graph
of
any
reciprocal
function
of
the
form
y
has
=
y
a
x
ver tical
asymptote
x
=
0
and
a
horizontal
asymptote
y
=
b
is
=
The
graph
of
a
reciprocal
function
is
called
horizontal
of
the
0
graph
➔
a
asymptote
of
y
=
f(x).
a hyperbola
y
●
The
x-axis
is
the
horizontal
x
=
0, the
y-axis,
6
asymptote.
is
an
k
asymptote
y
=
x
●
The
y-axis
is
the
ver tical
4
y
=
–x
asymptote.
2
●
Both
are
the
all
domain
the
except
real
and
range
The
reciprocal
has
many
–4
4
6
=
0, the
x-axis,
in
The
applications
zero.
y
●
two
separate
par ts
–4
of
is
an
computer
graph
are
reflections
=
other
in
y
=
–6
related
y
=
−x
and
symmetr y
In
Chapter
1
you
y
=
for
saw
x
are
this
that
to
−x
number
●
par ticularly
x
of
those
each
science
asymptote.
algorithms,
y
the
function
x
numbers
lines
of
may
function.
to
draw
these
the
inverse
function
of
f
theor y.
wish
to
Y
ou
investigate
fur ther
.
(x),
you
reflect
its
graph
in
the
line
y
=
x.
If
you
reflect
f
(x)
=
in
the
line
y
=
x
you
get
the
same
graph
as
for
f
(x).
The
reciprocal
1
➔
The
reciprocal
function
is
a
self-inverse
function
function,
f(x)
=
,
is
x
one
The
equation
of
the
function
in
the
Investigation
on
page
142
of
the
simplest
is
examples
of
a
function
xy
=
24.
It
can
be
written
as
=
and
is
a
reciprocal
It
has
a
graph
Rational
simil ar
functions
to
the
one
shown
above.
function.
that
is
self-inverse.
The
design
of
the
Yas
Hotel
Asymptote
Architecture)
It
a
also
of
the
has
Formula
1
is
in
Abu
based
Dhabi
on
racetrack
(designed
mathematical
running
through
by
models.
the
center
hotel!
Example
✗
For
●
each
write
function:
down
the
equations
of
the
vertical
and
horizontal
asymptotes
●
sketch
●
state
the
the
graph
domain
and
range.
9
9
a
y
b
=
y
=
+ 2
x
x
Answers
a
Asymptotes
are
x
=
0
and
y
=
0
y
=
2
y
20
15
10
5
x
0
–6
–4
–2
2
4
6
–5
–10
–15
–20
Domain
range
b
y
x
∈
∈
,
,
Asymptotes
y
x
≠
are
≠
0,
0
x
=
0
and
The
y
is
6
f
graph
the
(x)
same
but
of
as
f
(x)
the
shifted
2
+
2
graph
units
of
in
4
the
y-direction.
2
x
–30
–20
–10
–2
–4
–6
Domain
range
y
x
∈
∈
,
,
y
x
≠
≠
0,
2
Chapter
Exercise
1
Draw
B
these
on
separate
a
graphs.
=
b
=
xy
c
=
8
Y
ou
need
to
On
the
same
graph
show
=
do
questions
4b
and
analytically
algebra
a
Sketch
the
graph
of
=
and
write
down
its
the
graph
of
3b
both
(using
and
sketching
Sketch
c
by
asymptotes.
b
able
=
3
be
and
2
to
and
using
transformations)
=
+
and
write
down
and
its
using
your
GDC.
asymptotes.
4
Identify
and
the
then
horizontal
state
their
and
ver tical
domain
and
asymptotes
of
these
functions
range.
It
=
b
=
+
5
The
Corr yvreckan,
world,
the
is
between
coast
west
The
of
and
heard
c
=
the
the
roar
16 km
of
third
largest
Flood
the
to
draw
graphs.
islands
Scotland.
the
speed
the
help
−
may
a
of
whirlpool
Jura
tides
resulting
and
and
in
Scarba
inflow
maelstrom
the
off
from
can
the
be
away
.
of
the
surrounding
water
increases
as
you
approach
the
center
and
is
modeled
by
where
=
s
is
−1
the
speed
the
center
a
Use
and
of
in
your
0
≤
s
the
water
in
m s
and
d
is
the
distance
from
metres.
GDC
≤
to
sketch
the
function
with
0
≤ d
≤
50
200.
−1
6
b
At
c
What
The
what
is
force
distance
the
(F )
is
speed
the
of
required
speed
the
to
10 m s
water
raise
an
?
100 m
object
from
of
the
mass
center?
1500 kg
is
[
modeled
by
where
=
Archimedes
believed
l
is
the
length
of
the
lever
in
is
to
have
a
place
said
metres
“Give
me
to
stand,
and
the
force
is
measured
in
Sketch
the
graph
with
≤
≤
≤
How
much
force
would
lever
enough
you
need
to
and
I
≤
shall
b
a
newtons.
long
a
and
apply
if
you
had
a
move
the
earth.
”
2 m
N
is
the
symbol
for
lever?
the
c
How
long
force
of
would
the
lever
need
to
be
if
you
could
manage
unit
a
newton.
Rational
functions
i
1000 N
ii
2000 N
iii
3000 N?
of
force,
the
.
Rational
Have
you
noticed
functions
the
way
the
sound
of
a
siren
changes
as
a
Sound
fire
engine
or
police
car
passes
you?
The
obser ved
frequency
frequency
measured
higher
than
the
emitted
frequency
during
the
approach,
it
at
the
instant
of
passing
by
,
and
it
is
lower
during
the
for
it
moves
the
obser ved
toward
you
This
is
frequency
called
of
the
sound
Doppler
when
the
effect.
The
source
is
of
per
second.
equation
traveling
is:
away
.
her tz
number
the
waves
time
in
is
(Hz),
identical
is
is
=
where
−1
●
330
is
●
f
the
is
the
speed
of
obser ved
sound
in
frequency
m s
in
Hz
1
●
f
is
the
emitted
●
v
is
the
velocity
f
is
a
rational
frequency
of
the
source
toward
you
function.
1
h (x)
➔
A
rational
function
is
a
function
of
the
form
since
where
g
and
h
are
this
of
the
course
form
g(x)
px
+
q
and
so
be
zero
a
value
divided
polynomials.
by
In
cannot
=
h(x)
we
will
can
be
restricted
investigate
to
linear
rational
zero
is
undened.
functions
functions f
(x)
where
+
=
Example
+
The
−1
A
vehicle
is
coming
towards
you
at
96 km h
(60
miles
per
hour)
units
must
sounds
its
hor n
with
a
frequency
of
8000 Hz.
What
is
the
frequency
sound
you
hear
if
the
speed
of
sound
is
330 m s
can
to
−1
−1
=
Conver t
96 000 m h
metres
96 000
−1
96 000 m h
be
the
equation.
same
Y
ou
per
hour
to
round
get
an
numbers
approximate
answer
.
second.
−1
=
=
26.7 m s
3600
Since
330
Observed
kilometres
per
all
the
?
Answer
96 km h
speed
of
in
−1
the
of
and
frequency
1
hour
=
3600
seconds
f
=
330
v
330 × 8000
=
330
=
26.7
8700 Hz (3 sf )
Chapter
Investigation
–
graphing
rational
functions
1
Use
a
your
GDC
to
show
sketches
of
y
,
=
y
=
y
x
x
1
1
1
=
2
x
+
3
2
and
y
=
x
Copy
b
and
+
3
complete
the
table.
Rational
Ver tical
Horizontal
Domain
function
asymptote
asymptote
Range
1
y
=
x
1
y
=
x
2
1
y
=
x
+
3
2
y
=
x
What
c
the
+
3
effect
ver tical
does
changing
the
denominator
d
What
do
you
notice
about
the
horizontal
e
What
do
you
notice
about
the
domain
the
range
the
ver tical
What
f
do
on
asymptotes?
and
the
value
of
asymptote?
you
horizontal
have
asymptote?
notice
about
and
the
value
of
the
asymptote?
k
Rational
functions
of
the
form
y
=
x
− b
1
is
A
rational
function
=
, where
will
have
that
is,
a
vertical
when
x
=
k
and
b
are
asymptote
consider
when
the
denominator
equals
zero,
b
detail
horizontal
Example
in
this
the
Knowledge
the
The
undened.
asymptote
will
be
We
will
0
constants,
end
of
in
more
Theor y
section
the
of
at
chapter
.
the x-axis.
1
a
Identify
b
State
c
Sketch
the
horizontal
and
ver tical
asymptotes
of
y
=
x
the
domain
the
and
function
3
range.
with
the
help
of
your
GDC.
Y
ou
may
explore
wish
the
Answers
of
a
The
x-axis
horizontal
x
=
3
is
( y
=
0)
is
the
asymptote.
the
ver tical
asymptote.
Since the numerator will never be 0,
the
graph
touches
The
of
the
x
=
functions
never
x-axis.
is
zero
3.
{
Rational
function
denominator
when
this
Continued
on
next
page
innity.
to
concept
b
Domain
Range
x
y
c
∈
∈
,
,
y
x
≠
≠
0
3
y
8
6
1
4
y
=
x
–
3
2
x
0
–2
–4
–2
–4
–6
–8
Exercise
1
Identify
C
the
horizontal
and
ver tical
and
range.
asymptotes
of
these
functions
Y
ou
and
state
their
domain
should
algebra
a
=
c
=
+
d
=
=
‘using
e
=
f
+
=
−
+
g
an
+
h
=
−
Sketch
the
each
domain
function
and
with
the
help
question
and
=
+
c
your
state
=
+
although
answers
to
check
with
a
e
−
=
1,
want
b
=
GDC
range.
d
your
may
GDC.
a
of
do
+
you
2
analytic
to
=
+
called
+
method’)
is
b
use
(this
Use
+
the
=
−
f
=
your
GDC
correct
with
viewing
+
window.
+
g
=
−
h
=
+
3
When
i
=
lightning
instantaneously
.
strikes,
the
But
sound
the
+
light
reaches
of
the
your
thunder
eyes
vir tually
travels
at
−1
approximately
the
.
331 m s
temperature
of
the
However,
sound
surrounding
air.
waves
The
are
time
affected
sound
by
takes
to
travel
one
kilometre
is
modeled
by
=
where
time
a
b
in
seconds
Sketch
If
you
the
are
and
c
graph
one
the
thunder,
On
the
is
of
the
t
is
temperature
for
kilometre
what
the
in
temperatures
away
and
it
temperature
is
of
t
is
the
+
degrees
from
3
−20 °C
seconds
the
Celsius.
to
40 °C.
before
you
surrounding
air?
hear
4
a
same
set
of
axes,
sketch y
=
x +
2
and
=
Compare
linear
the
two
function
graphs
and
its
and
make
reciprocal
connections
+
between
the
function.
b
Now
do
the
same
for
y
=
x
+
1
and
=
+
Chapter
Rational
functions
of
the
y
form
ax
+ b
cx
+ d
=
➔
Ever y
rational
called
a
function
of
the
form
has
graph
of
any
rational
function
has
Use
your
GDC
x
y
=
,
+
x
Copy
b
a
ver tical
and
a
+
asymptote.
Investigation
a
graph
+
=
horizontal
a
+
hyperbola.
The
+
=
y
graphing
show
x
+ 1
x
+
=
3
and
to
–
sketches
rational
y
and
x
complete
2x
=
the
+
of
2x
,
3
functions
y
1
=
3
x
+
3
table.
Rational
Ver tical
Horizontal
function
asymptote
asymptote
Domain
Range
x
y
=
x
y
+
3
x
+ 1
x
+
=
3
2x
y
=
+
x
2x
y
3
1
=
x
+
3
c
What
do
you
notice
about
the
horizontal
d
What
do
you
notice
about
the
domain
asymptotes?
and
the
value
of
the
ver tical
asymptote?
y
➔
The
ver tical
asymptote
occurs
at
the x-value
that
makes
the
4
denominator
zero.
3
➔
The
horizontal
asymptote
is
the
line
=
a
y
2
=
c
1
To
find
the
horizontal
asymptote
rearrange
the
equation
to
make
x
0
x
the
–6
subject.
–4
–2
–1
ax
y
+ b
d
=
x
–2
cx
y ( cx
cyx
+ d )
+ d
=
ax
− ax
=
b − dy
x
=
b
dy
cy
The
horizontal
that
is,
asymptote
when
a
cy
=
a
or
y
=
c
Rational
functions
–3
+ b
a
occurs
when
the
denominator
is
zero,
=
c
Example
x
For
the
function
y
+ 1
=
2x
a
sketch
b
find
c
state
the
the
ver tical
the
4
graph
and
domain
horizontal
and
asymptotes
range.
Answers
y
a
4
3
2
x
y
+
1
=
2x
–
4
1
x
0
–2
–4
–1
–2
–3
b
Ver tical
asymptote
x
=
2
When
2x
a
c
−
4
=
0,
1
Horizontal
asymptote
y
x
∈ ,
x
≠
=
2
a
=
=
1,
=
y =
2,
2
Domain
c
x
c
2
1
Range
y ∈ ,
y
≠
2
Exercise
1
Identify
and
the
then
a
D
state
the
Match
b
=
b
i
−
y
c
x
1
x
3
d
=
ii
8
6
4
2
2
x
–6
y
4
–4
+
=
6
–2
=
graph.
+
y
–2
functions
d
8
–4
these
+
=
−
=
−
the
of
range.
c
with
a
and
asymptotes
+
function
ver tical
=
the
and
domain
+
=
2
horizontal
0
–4
x
–2
–2
–4
–6
Chapter
y
iii
y
iv
8
8
6
6
4
4
2
2
x
0
x
0
–2
–4
–2
–2
–2
–4
–6
3
Sketch
and
each
using
your
GDC
and
state
the
domain
range.
a
function
+
=
b
c
+
=
+
−
+
=
+
+
Check
d
=
e
=
f
by
using
graph
g
=
4
Write
x
5
=
a
−4
Chris
and
a
design
and
in
has
T-shir ts
ver tical
at y
=
surfers
It
that
will
it
to
function.
asymptote
at
3
and
cost
will
the
GDC
=
a
for
garage.
estimate
−
asymptote
their
they
that
answer
your
set
up
$450
cost
a
to
$5.50
T-shir t
set
to
up
the
print
T-shir t.
Write
a
Write
linear
a
T-shir t
c
horizontal
Lee
T-shir ts.
b
function
business
equipment
a
i
−
rational
=
and
printing
each
h
your
=
What
rational
of
is
function
Remember
to
domain
x
giving
take
function
producing
the
C (x)
of
of
the
A (x)
the
total
cost
of
cost
into
account.
set-up
giving
the
average
producing x
cost
per
them.
A (x)
in
the
context
of
the
problem?
Explain.
d
Write
e
Find
down
the
ver tical
asymptote
of
A (x).
Sketch
this
the
value
Exam-Style
6
horizontal
have
rule
is
over
age
of
‘Take
plus
the
12.
the
context
for
of
A(x).
the
What
meaning
does
problem?
Question
Y
oung’s
the
in
asymptote
age
a
way
two,
of
the
Multiply
of
calculating
based
child
this
on
in
the
adult
years
number
by
doses
and
the
of
medicine
for
children
dose.
divide
adult
by
their
age
dose.’
This
is
modeled
by
the
function
where
=
dose,
a
is
the
adult
years.
Rational
functions
dose
in
mg
and
t
c
is
the
child’s
+
is
the
age
of
the
child
in
the
function.
a
Make
of
a
table
of
values
for
ages
2
to
draw
to
12
with
an
adult
dose
100 mg.
b
Use
your
c
Use
the
values
from
a
a
graph
of
the
function.
graph
to
estimate
the
dose
for
a
7
-year
old.
d
Write
down
e
What
does
Y
oung’s
7
The
a
new
cost
for
cost
a
a
Sketch
a
d
Since
e
Explain
the
A
it
of
the
the
horizontal
horizontal
asymptote.
asymptote
mean
for
electricity
costs
and
function
as
a
per
$550.
that
lasts
year
for
a
refrigerator
Determine
for
15
the
years.
total
is
$92.
annual
Assume
costs
electricity
.
that
gives
function
of
the
the
annual
number
cost
of
of
a
years
you
own
refrigerator.
graph
window?
f
of
refrigerator
refrigerator
c
of
purchase
Develop
the
value
refrigerator
include
b
the
equation
r ule?
average
A
the
of
Label
this
is
a
the
that
the
function.
What
is
an
appropriate
scale.
rational
meaning
function,
of
the
determine
horizontal
its
asymptotes.
asymptote
in
terms
of
refrigerator.
company
will
last
difference
Review
offers
at
in
least
a
refrigerator
twenty
years.
that
Is
costs
this
$1200,
but
refrigerator
says
wor th
that
the
cost?
exercise
✗
Extension
material
Worksheet
5
fractions
Exam-Style
1
Match
the
function
with
the
ii
CD:
aysmptotes
graph.
v
a
iii
=
iv
and
on
Continued
Question
i
-
+
=
vi
+
+
y
b
y
8
6
6
4
4
2
2
0
x
–2
–2
0
x
–2
–2
–4
–4
–6
–6
Chapter
Exam-Style
QuestionS
c
d
y
y
6
8
4
6
2
4
x
–2
x
0
–1
–3
3
–4
–2
–6
–4
e
y
f
y
6
6
4
4
2
2
x
0
–2
x
0
–2
–4
–4
–6
–6
2
Given
a
b
3
i
Sketch
ii
Determine
iii
Find
For
the
each
domain
the
of
and
=
=
c
+
=
+
function.
the
ver tical
domain
these
and
and
range
functions,
horizontal
of
write
the
asymptotes
of
the
function.
function.
down
the
asymptotes,
range.
a
b
y
y
6
8
5
4
f (x)
6
=
6
x
+
4
f (x)
=
–
3
4
x
2
x
0
–6
–2
–4
–2
x
0
–6
–4
–2
–4
–2
–6
–4
–8
–6
–8
y
c
y
d
6
8
4
6
2
4
–3
f (x)
2
f (x)
=
–
x
+
=
+
x
–
5
1
2
6
2
0
x
x
0
–8
Rational
functions
–6
–4
4
–4
–2
–6
–4
–8
–6
6
8
4
A
group
of
weekend
a
If
c
the
in
want
a
spa.
health
number
terms
b
Draw
c
Explain
The
students
represents
this
5
at
a
of
f
is
given
student
an
=
,
x
∈
,
x
Find
the
ver tical
iii
Write
b
Find
c
Hence
down
the
sketch
the
to
represents
show
the
cost
and
domain
of
−2
asymptote
asymptote
the
of
y
=
f
(x)
graph.
the
point P
at
which
the
of
the
intersection
graph
of
y
=
of
f
the
(x),
graph
with
showing
the
the
axes.
asymptotes
lines.
with
the
help
of
your
−
State
the
b
=
+
c
=
+
−
e
London
to
f
from
=
flies
of
GDC.
range.
=
distance
of
=
airline
graph
intersect.
function
and
the
Question
each
of
of
coordinates
An
and s
equation
range
2
a
exercise
Exam-Style
domain
the
points
dotted
Review
for
students.
on
≠
ii
d
voucher
+
horizontal
a
$300.
by
the
a
teacher
costs
Find
Sketch
each
write
of
their
voucher
function.
limitations
asymptotes
1
the
i
by
for
number
of
give
function.
a
cost
students,
the
any
function
(x)
of
graph
f
the
to
The
New
=
+
Y
ork,
which
is
−
a
5600 km.
a
Show
that
this
information
can
be
written
as
=
−1
where
and
b
is
Sketch
and
c
t
s
If
of
0
the
the
the
a
≤
the
is
t
average
time
graph
≤
in
of
speed
of
the
plane
in
km h
hours.
this
function
with
0
≤ s
≤
1200
20.
flight
takes
10
hours,
what
is
the
average
speed
plane?
Chapter
Exam-Style
3
People
with
amount
of
Questions
sensitive
time
skin
spent
in
must
direct
be
careful
sunlight.
about
The
the
relation
+
=
where
the
is
the
spend
a
Sketch
b
Find
s
i
in
=
relation
number
what
mayor
Bangkok.
of
s
this
is
The
the
=
and
that
≤
minutes
40
cost
in
population
s
the
sun
person
skin
with
=
value,
sensitive
gives
skin
damage.
≤
skin
scale
can
≤
be
≤
exposed
when
100
asymptote?
out
(c)
is
a
that
iii
represents
giving
s
without
when
of
horizontal
the
time
sunlight
is
Explain
percent
direct
of
ii
d
city
minutes
10
What
The
in
amount
this
the
c
in
time
maximum
can
4
m
for
face
Thai
is
a
person
masks
baht
given
with
during
for
giving
a
sensitive
flu
skin.
outbreak
masks
to
m
by
=
a
Choose
the
b
the
20%
of
the
The
suitable
scale
and
use
your
GDC
to
help
sketch
cost
of
supplying
50%
ii
iii
90%
population.
Would
this
5
a
function.
Find
i
c
it
be
model?
function
f
possible
Explain
(x)
is
to
supply
your
defined
all
of
the
population
using
answer.
as
f
(x)
=
,
+
a
Sketch
b
Using
your
the
equation
the
value
Rational
value
functions
of
sketch,
ii
the
≠
cur ve
i
iii
the
x
of
of
of
the
the
f
for
write
each
−3
≤
x
≤
5,
down
asymptote
x-intercept
y-intercept.
showing
the
asymptotes.
CHAPTER
5
SUMMARY
Reciprocals
●
The
●
A
of
reciprocal
number
a
number
multiplied
by
its
is
1
divided
reciprocal
by
that
equals
number.
1.
For
example
3
×
=
1
−1
●
The
of
reciprocal
x
is
or
−1
x
and
x
×
x
=1
The
●
If
reciprocal
a
cur ve
never
●
The
gets
meets
graph
continually
it,
of
function
the
any
closer
straight
line
reciprocal
is
to
a
straight
called
function
line
but
an asymptote
of
the
form
y
=
has
a
ver tical
asymptote
x
=
0
and
a
horizontal
asymptote
●
The
graph
y
=
of
0
a
reciprocal
function
is
called
y
a hyperbola
x
=
0, the
y-axis,
6
■
The
x-axis
is
the
horizontal
■
The
y-axis
is
the
ver tical
■
Both
the
domain
and
asymptote.
is
an
4
asymptote.
range
are
all
asymptote
y
the
real
=
–x
2
numbers
f
except
zero.
x
–2
–4
■
The
two
separate
par ts
of
the
graph
are
4
y
of
each
other
in
y
=
=
y
●
y
The
=
x
and
y
=
reciprocal
Rational
−x
are
lines
function
is
0, the
x
–4
−x
is
■
6
reflections
a
of
symmetr y
self-inverse
for
this
=
an
asymptote.
x
–6
function.
function
functions
●
A
rational
function
is
a
function
of
the
form
=
y
where
g
and
h
are
polynomials.
4
●
Ever y
rational
function
of
the
form
has
called
a
+
=
a
graph
3
+
hyperbola.
a
y
●
2
=
c
The
the
ver tical
asymptote
denominator
occurs
at
the x-value
that
makes
1
zero.
x
–6
●
The
horizontal
asymptote
is
the
line
–4
–2
=
–1
d
x
–2
=
c
–3
Chapter
Theory
of
knowledge
Number
Egyptian
systems
fractions
3
The
ancient
Egyptians
only
1
In
algebra:
=
4x
fractions
with
a
numerator
of
for
1
example:
,
3
etc.
meant
that
each
algebraic
Egyptian
expression
as
fraction.
4
instead
of
they
4
5
7
23
3x
4x
4x
24x
wrote
4
1
Write
an
3
This
4x
1
,
2
+
2x
1,
1
1
used
1
+
.
2
Their
fractions
were
all
in
the
4
Where
1
and
form
are
called
uni t
do
you
think
this
could
be
fractions .
n
useful?
2
Numbers
such
as
were
represented
as
7
What
2
sums
of
unit
fractions
(e.g.
1
=
7
the
twice
(so
same
fraction
+
4
could
not
1
of
these
).
be
it
possible
to
write
ever y
fraction
used
an
Egyptian
fraction?
1
=
7
limitations
28
as
2
the
fractions?
Is
Also,
are
1
+
7
was
not
allowed).
How
7
5
1
do
you
1
know?
For
example,
would
be
8
Write
these
+
2
as
unit
8
fractions.
5
5
2
6
6
8
5
7
In
an
Inca
quipu,
the
strings
represent
numbers
The
Rhind
1650
BCE
fractions
200
Theory
of
knowledge:
Number
systems
Mathematical
contains
copied
years
older!
a
from
Papyr us
table
of
another
dated
Eg yptian
papyrus
Is
there
a
dierence
between
25¢
zero
More
had
the
and
than
2000
systems
ninth
for
a
a
circle
this
became
Who
What
Make
Notice
Now
We
rst
some
How
par t
did
on
the
zero
before
of
in
keep
The
Hindu
a
the
name
place
rows’.
sifr
that
of
The
if,
tens,
Arabs
eventually
was
nothing?
the
subsets
and
the
1
be
is
{0}
were
What
not
something.
the
tentative
Mayan
and
equation
BCE.
and
{0,
1,
2,
another
9
3}.
is
{
x
=
3²
about
a
year
sure
+
what
Zeno’
s
use
Inca
of
of
to
}.
and
do
the
equation
3x
=
0.
zero?
with
paradoxes
(a
zero
good
and
they
topic
to
questioned
research)
how
depend
in
zero.
cultures
understand
In
and
remarked
the
cultures
number.
that?
subset
Greeks
could
appears
‘to
of
mathematician
al-Khwarizmi
(empty).
Solve
CE
ancient
nothing
1
Islamic
used
that
all
one
this.
have
The
of
be
and
absence
zero?
used
that
tr y
the
an
zero.
mean
list
Babylonian
number
sifr
used
was
a
no
word
this
CE,
should
circle
our
Does
ago,
Muhammad
calculation,
little
called
years
representing
centur y
philosopher
in
nothing?
zero?
ative?
What
happens
if
you
divide
zero
by
anything?
g
pens
The
Mayans
shell
symbol
represent
if
used
you
divide
zero
by
by
zero?
zero?
a
to
zero.
Chapter
Patterns,
sequences
and
series
CHAPTER
OBJECTIVES:
Arithmetic
1.1
geometric
series.
sequences
and
sequences
Sigma
and
series;
series;
sum
sum
of
of
nite
arithmetic
nite
and
series;
innite
geometric
notation.
Applications
n
The
1.3
binomial
theorem:
expansion
of
(
a +
b
)
,
n ∈ ;
⎛ n ⎞
Calculation
of
binomial
coefcients
using
Pascal’
striangle
and
⎜
⎝
Before
Y
ou
1
you
should
Solve
change
e.g.
the
Solve
and
how
to:
quadratic
subject
the
⎠
start
know
linear
⎟
r
of
a
equation
Skills
equations
and
1
formula.
n(n
–
4)
=
12
check
Solve
each
a
3x
b
p(2
c
2
2
–
5
–
equation.
=
p)
5x
=
+
7
–15
n
n
–
4n
=
12
4n
–
12
=
0
2)
=
0
+
9
=
41
2
–
n
2
(n
–
6)(n
n
e.g.
–2,
Make
ac
b
2
=
+
=
=
b
ac
n
b
–
=
the
for
a
6m
b
2pk
+
k
8k
=
30
6
subject
of
this
3
If
T
3
Substitute
known
e.g.
the
values
into
–
5
=
3
formula.
3
+
Solve
T
=
2x
(x
+
3y),
then
find
the
value
of
when
a
x
=
3
and
b
x
=
4.7
y
=
5
y
=
formulae.
and
–2
4
Using
formula
A
=
of
=
2
3p
–
10q,
x
find
the
value
A
if
p
4
A
=
3p
–
–
=
3(2)
A
=
3(16)
A
=
48
10(1.5)
–
q
=
1.5
4
Using
value
10q
4
A
and
15
the
of
formula
m
m
if
a
x
=
5
and
y
=
3
b
x
=
3
and
y
=
–2
c
x
=
–5
A
=
–
15
sequences
=
33
Patterns,
and
and
series
=
2
3
–
y
,
find
the
The
bacteria
in
this
petri
dish
are
growing
and
reproducing;
in
this
[
Bacteria
petri
case
the
total
measured
mass
The
at
as
mass
of
patter n
after
8
this
help
will
the
make
we
you
predict
●
work
●
predict
●
calculate
the
●
calculate
how
in
be
use
or
how
the
to
24
a
total
it
will
will
dish
the
so
At
10:00
8
a.m.
will
be
the
6
mass
grams,
a
is
the
on.
forms
the
a
numerical
mass
of
patter n.
bacteria
in
the
dish
near
patter ns.
and
distant
Patter ns
future.
can
For
to:
countr y
take
natural
it
and
at
mathematical
a
distance
long
hours.
mass
predict
about
of
two
in
hours.
patter ns
long
long
in
study
population
how
total
grams,
used
will
ever y
the
predictions
●
out
12
hours
can
the
so
bacteria
12
chapter
example,
be
could
hours,
us
doubles
3 grams,
12:00
This
In
mass
growing
dish
to
pay
resource
that
take
in
a
for
20
off
will
years
a
loan
last
bouncing
an
bank
ball
will
investment
to
travel
double
value.
Chapter
. Patterns
Investigation
Joel
He
decides
saves
week,
and
Copy
a
and
to
$20
so
and
how
star t
the
and
–
saving
saving
rst
sequences
money
money.
week,
$25
the
second
week,
$30
the
third
on.
complete
much
he
the
has
table
below
saved
in
to
total,
show
for
Week
Weekly
T
otal
number
savings
savings
1
20
20
2
25
45
3
30
75
how
the
much
rst
8
Joel
saves
each
week,
weeks.
4
5
6
7
8
b
How
much
will
c
How
much
money
d
How
long
T
ry
e
Let
let
Let
of
In
the
week
as
T
write
the
for
for
total
10th
save
him
week
his
the
to
the
amount
formula
represent
investigation
a
form
number
are
some
8,
11,
14,
400,
1,
4,
9,
in
week?
total
save
a
amount
of
in
the
total
of
money
In
of
the
rst
at
money
he
17th
year?
least
Joel
saves
week?
$1000?
saves
each
each
week,
week.
and
number
.
the
total
savings,
5,
10,
15,
amount
and
Patterns,
The
total
is
a
let
n
of
money
represent
Joel
the
has
saved.
number
25,
100,
25,
…
…
sequences
amounts
to
a
sequences.
…
and
series
of
money
of
Joel
money
saves
he
has
each
saved
sequence.
patter n
…
200,
amounts
according
number
20,
the
different
sequence
17,
16,
a
order
Here
800,
above,
sequence.
passes
par ticular
a
the
in
Joel
for
formula
represent
to
will
take
represent
form
A
a
it
save
weeks.
time
➔
write
M
n
T
ry
f
to
will
Joel
of
numbers
r ule.
arranged
in
a
➔
Each
individual
called
In
is
the
11,
Y
ou
a
can
or
element,
of
a
sequence
is
term
sequence
the
number,
third
also
8,
11,
term
use
is
the
14,
17,
…,
14,
and
notation
so
u
the
first
term
is
8,
the
second
term
on.
to
denote
the
nth
term
of
a
n
sequence,
So
u
for
=
where
8,
11,
8,
u
1
Y
ou
n
is
14,
=
a
positive
17,
11,
…
u
2
can
you
=
14,
integer.
could
and
say
so
on.
3
continue
the
patter n
if
you
notice
that
the
value
previous
term:
of
each
Sometimes,
term
is
three
greater
than
the
value
of
the
letters
8,
11,
14,
17,
20,
23,
represent
26
a
For
this
sequence,
you
could
write:
=
u
8
and
u
1
This
is
called
a
recursive
formula,
in
=
u
n+1
which
the
+
a
example,
a
,
t
n
on
the
value
of
the
previous
,
of
the
sequence
one-half
the
800,
value
400,
of
the
200,
100,
previous
terms
of
or
we
x
might
n
to
n
term.
represent
In
to
term
use
depends
the
u
sequence.
For
of
use
than
3
n
value
we
other
…,
the
value
of
each
term
a
the
nth
term
sequence.
is
term.
In
this
case,
=
u
800
and
=
+
1
Example
Write
a
recursive
a
9,
15,
b
2,
6,
21,
18,
27,
54,
formula
for
the
n th
term
of
each
sequence.
…
…
Answers
u
a
=
9
and
u
1
=
u
n+1
+
6
To
get
add
u
b
=
2
and
u
1
from
one
ter m
to
the
next,
you
from
one
ter m
to
the
next,
you
n
=
3u
n+1
To
6.
get
n
multiply
by
3.
Sometimes
Sometimes
term
of
of
a
In
the
a
term
it
is
more
sequence.
without
useful
With
having
a
to
write
general
to
know
a general formula for the
formula,
the
value
you
of
can
the
find
the
previous
nth
value
called
rule
the
for
1,
4,
9,
16,
25,
…
,
each
term
is
a
perfect
2
first
term
is
term’.
that
n,
square.
term
number
,
will
2
,
1
nth
term.
the
The
is
‘general
the
Remember
sequence
this
the
second
is
2
,
and
so
on.
A
general
formula
always
be
a
whole
2
for
the
nth
term
of
this
sequence
is
u
=
n
number
.
n
In
the
sequence
5,
10,
15,
20,
25,
…
,
each
term
is
a
multiple
of
We
could
not
3
5.
have
a
‘
th’
term,
or
a
4
The
first
term
is
5
×
1,
the
second
is
5
×
2,
and
so
on.
‘7.5th’
A
general
formula
for
the
nth
term
of
this
sequence
is
u
=
term.
5n.
n
Chapter
Example
Write
a
4,
a
general
8,
1
12,
1
,
,
6
for
the
n th
term
of
each
sequence.
…
1
,
3
formula
16,
1
,
b
9
…
12
Answers
a
u
=
4n
Each
ter m
is
a
multiple
of
4.
n
1
b
u
=
The
denominators
are
the
multiples
n
3n
of
Exercise
1
Write
down
3,
7,
11,
c
3,
4,
6,
u
a
15,
…
,
…
down
10
the
and
and
first
=
(
=
a
2,
c
64,
Write
1,
2,
4,
d
5,
–10,
8,
…
20,
–40,
…
6.0,
6.01,
6.012,
6.0123,
…
terms
in
)
each
u
b
sequence.
=
3
and
=
+
1
+
)
u
d
=
x
and
1
recursive
4,
b
(
4
four
+
a
sequence.
=
Write
each
f
+
1
3
in
u
terms
…
1
c
three
=
next
13,
Write
the
9,
2
A
a
e
3.
6,
32,
8,
for
each
…
16,
down
formula
sequence.
1,
b
8,
the
…
7,
d
first
four
terms
in
3,
9,
12,
each
27,
17,
…
22,
…
sequence.
T
o
nd
the
rst
term
n
a
u
c
u
=
3
=
2
u
b
n
6
a
a
2,
c
64,
The
n
=
1;
to
n
u
d
general
4,
6,
32,
2
,
2
3
=
n
nd
the
second
use
n
2
term
n
Write
1
+
substitute
1
n
e
−6n
n
n
5
=
8,
the
nth
8,
…
term
of
9,
each
b
1,
3,
27,
d
7,
12,
f
x,
2x,
is
known
17,
=
and
so
on.
sequence.
…
22,
…
4
,
,
3
for
…
16,
3
,
formula
4
…
3x,
4x,
…
5
sequence
1,
1,
2,
3,
5,
8,
13,
…
as
the
Fibonacci
sequence.
a
Find
b
Write
.
the
a
15th
term
recursive
Arithmetic
of
the
formula
Fibonacci
for
the
sequence.
Fibonacci
sequence.
sequences
[
Fibonacci,
as
In
the
sequence
8,
11,
14,
17,
…,
the
value
of
each
term
is
than
example
of
Patterns,
the
value
of
the
previous
an ari thmetic sequence ,
sequences
and
series
term.
or
This
sequence
arithmetic
of
known
Pisa
three
(Italian
greater
also
Leonardo
is
an
progression.
c.
1175 – c.
1250)
➔
In
an
arithmetic
sequence,
the
terms
increase
or
decrease
by
a
Examples
of
arithmetic
progressions
constant
value.
This
value
is
called
on
or
d.
The
common
difference
can
appear
the common dierence ,
be
a
positive
or
a
the
Ahmes
negative
Papyrus,
which
dates
value.
from
For
8,
about
1650
BCE.
example:
11,
14,
17,
…
In
this
sequence, u
=
8
and
d
=
3
=
35
and
d
=
–5
=
4
and
d
=
0.1
=
c
and
d
=
c
1
35,
30,
25,
20,
…
In
this
sequence, u
1
4,
4.1,
4.2,
4.3,
…
In
this
sequence, u
1
c,
2c,
3c,
4c,
…
In
this
sequence,
u
1
For
any
arithmetic
sequence,
u
=
u
n+1
We
can
find
difference,
In
an
d,
to
term
the
arithmetic
=
u
any
the
first
of
the
+
sequence
previous
d
n
by
adding
the
common
term.
sequence:
term
1
u
=
u
2
u
=
u
3
u
d
+
d
=
(u
2
=
+
d
=
=
d)
+
d
=
u
+
+
2d)
+
d
=
u
1
u
+
d
=
+
3d
+
4d
1
(u
4
2d
1
(u
3
5
+
1
u
4
u
+
1
+
3d)
+
d
=
u
1
1
…
…
=
u
u
n
➔
+
(n
–
1)d
1
Y
ou
can
find
formula:
Example
u
the
=
u
nth
+
n
1
12th
term
term
(n
–
of
an
arithmetic
sequence
using
the
1) d
a
Find
the
b
Find
an
of
expression
the
for
arithmetic
the
n th
sequence
13,
19,
25,
…
term.
Answers
a
u
=
13
and
d
=
6
Find
these
values
by
looking
at
the
1
u
=
13
+
(12
=
13
+
66
=
79
–
1)6
sequence.
12
u
For
n
=
the
12
12th
into
ter m,
the
substitute
for mula
12
u
=
u
n
b
u
=
13
+
(n
–
1)6
+
(n
–
1) d
1
For
the
nth
ter m,
substitute
the
n
=
13
+
6n
–
6
values
of
u
and
d
into
the
for mula
1
u
=
n
6n
+
7
u
n
=
u
+
(n
–
1) d
1
Chapter
Example
If
Find
the
number
of
terms
in
the
arithmetic
a
sequence
sequence
continues
84,
81,
78,
…,
and
there
term,
Answer
u
=
84
and
=
84
+
d
indenitely
12.
=
–3
Find
these
values
by
looking
at
it
is
is
no
an
nal
innite
sequence.
the
1
u
(n
–
1)(–3)
=
12
If
sequence.
a
sequence
ends,
or
n
Substitute the values of
u
has
and d into
a
‘last
term’
it
1
the formula u
= u
n
84
–
3n
There
+
3
=
87
3n
=
75
n
=
25
are
25
–
3n
terms
in
=
12
Solve
for
a
+ (n – 1)d
1
n.
the
sequence.
Exercise
For
1
each
sequence:
i
Find
the
ii
Find
an
a
3,
c
36,
e
5.6,
Find
2
B
5,
a
the
an
9,
the
nth
…
6.8,
…,
term.
25,
b
46,
6.2,
15,
for
…
41,
…
of
terms
in
d
100,
f
x,
each
255
x
55,
87,
+
a,
…
74,
…
x
2a,
+
…
sequence.
b
4.8,
d
250,
5m,
8m,
…,
80m
f
x,
5.0,
5.2,
…,
38.4
221,
192,
…,
–156
3x
+
3,
5x
+
6,
…,
19x
+
arithmetic
common
sequence,
u
=
48
and
u
=
75.
Find
the
first
term
and
12
difference.
Answer
u
+
3d
=
u
+
3d
=
75
the
3d
=
27
to
d
=
9
times.
9
48
To
get
from
the
9th
ter m,
12
u
,
to
9
12th
ter m,
u
,
you
would
need
12
u
=
9
u
+
(9
–
1)9
u
+
72
u
48
To
=
48
for mula.
=
–24
1
first
–24,
term
and
difference
the
=
1
is
add
find
Patterns,
of
the
is
the
sequence
common
9.
sequences
and
series
common
the
1
The
27
9
the
40,
Example
In
expression
2m,
e
term.
number
10,
c
6,
15th
first
dif ference
ter m,
use
the
3
nite
sequence.
is
Exercise
An
1
C
arithmetic
Find
the
common
EXAM-STYLE
In
2
an
sequence
has
first
term
19
arithmetic
sequence,
the
common
3
Find
the
value
of
x
4
Find
the
value
of
m
u
=
term.
or
sequence
This
In
a
2,
The
For
in
the
37
and
u
in
the
=
4.
21
and
the
arithmetic
first
term.
sequence
arithmetic
3,
sequence
x,
m,
8,
…
13,
3m
–
6,
…
sequences
18,
is
54,
an
…,
each
example
sequence ,
geometric
called
31.6.
term
of
is
three
times
the
previous
a geometric sequence ,
progression.
multiplying
is
6,
sequence
geometric
➔
difference
Geometric
the
term
QUESTION
Find
In
15th
difference.
10
.
and
the
the
common
previous
r,
term
ratio ,
common
ratio,
each
can
be
term
by
or
a
can
be
constant
obtained
value.
by
This
value
r
positive
or
negative.
example:
1,
5,
25,
125,
…
u
=
1
and
r
=
5
=
3
and
r
=
–2
=
81
1
3,
–6,
12,
–24,
…
u
1
81,
27,
9,
3,
…
u
and
=
1
2
k,
k
3
,
k
4
,
k
,
…
u
=
k
and
r
=
k
1
For
any
geometric
sequence, u
=
(u
n+1
sequence
For
any
u
by
multiplying
geometric
=
the
=
u
=
u
first
the
)r.
Y
ou
can
find
any
term
of
the
n
previous
term
by
the
common
ratio, r.
sequence:
term
1
u
2
×
r
×
r
1
2
u
3
=
(u
2
×
r)
×
r
=
u
1
×
r
1
2
u
=
u
4
×
r
=
(u
3
×
r
3
)
×
r
=
1
u
×
r
×
r
1
3
u
=
u
5
×
r
=
(u
4
×
r
4
)
×
r
=
1
u
1
…
…
n
=
u
u
n
➔
Y
ou
×
–
1
r
1
can
find
the
nth
n
formula:
u
n
=
u
(r
–
term
of
a
geometric
sequence
using
the
1
)
1
Chapter
Example
Find
the
9th
term
of
the
sequence
1,
4,
16,
64,
…
Answer
u
=
1
and
r
=
4
Find
these
values
by
looking
at
the
=
9
1
sequence.
9
u
–
1
8
=
1(4
)
=
=
1(65 536)
=
65 536
1(4
)
For
the
9th
term,
substitute
n
9
n
into
the
for mula
u
=
u
n
u
–
(r
1
)
1
9
Example
Find
the
12th
term
of
the
sequence
7,
–14,
28,
–56,
…
Answer
u
=
7
and
r
=
7((–2)
=
–2
Find
these
values
by
looking
at
the
1
sequence.
12
u
–
1
11
)
=
7((–2)
For
)
the
12th
ter m,
substitute
12
=
7(–2048)
n
=
–14 336
u
=
12
into
n
u
Exercise
For
each
16,
c
1,
10,
e
2,
6x,
8,
sequence,
4,
find
the
Example
a
…
100,
18x
for mula
)
b
…
d
– 4,
25,
12,
…
f
a
–36,
10,
7
,
4,
6
b,
a
ratio
7th
term.
…
5
,
the
…
2
b
and
a
3
b
,
…
geometric
sequence,
u
=
864
and
u
1
the
the
1
1
common
2
Find
–
(r
D
a
In
u
n
12
1
=
common
=
256
4
ratio.
Answer
4
u
=
u
4
–
1
(r
3
)
=
1
u
(r
)
Substitute
=
864(r
and
u
)
=
256
4
n
u
256
=
u
n
8
–
(r
1
=
=
864
27
8
r
=
3
Solve
27
2
r
=
3
Patterns,
sequences
and
series
for
1
)
3
r
=
4,
u
=
864,
1
3
256
n
1
r.
into
the
for mula
Example
For
the
that
geometric
the
nth
term
sequence
is
greater
5,
15,
than
45,
...
find
the
least
value
of
n
such
50 000.
Answer
u
=
5
and
r
=
3
1
Find
n
u
=
5
×
–
u
1
and
r
by
looking
at
the
1
3
n
sequence.
Substitute
u
=
5
and
r
=
3
into
the
1
n
for mula
u
=
u
n
You
can
of
n.
for
into
1
)
1
use
value
–
(r
the
GDC
First,
to
enter
help
the
find
the
for mula
GDC
u
a
function.
Let
help
x
represent
n,
as
shown.
Plus
and
GDCs
look
values
The
n
=
10,
since
u
>
of
9th
ter m
CD:
demonstrations
variable
Now
on
is
at
the TABLE
the
first
ter m
is
n
Alternative
the
n
to
see
Casio
are
on
for
the
TI-84
FX-9860GII
the
CD.
the
ter ms.
32 805,
and
the
10th
98 415.
50 000
10
Exercise
1
A
geometric
Find
2
A
For
that
4
the
the
each
the
a
16,
c
112,
A
sequence
first
geometric
Find
3
E
and
term
and
geometric
nth
24,
term
36,
–168,
that
find
and
sequence
first
geometric
Show
term
the
has
the
252,
greater
...
sequence
two
2nd
are
has
two
term
common
3rd
term
find
than
5th
term
3.2.
and
6th
term
144.
ratio.
least
value
of
n
such
1000.
1,
d
50,
2.4,
values
5.76,
55,
term
possible
possible
and
ratio.
the
b
first
50
–18
common
sequence,
…
there
the
is
has
9
60.5,
and
values
for
the
…
...
third
for
the
second
term
144.
common
ratio,
term.
Chapter
Find
5
the
value
EXAM-STYLE
Find
6
7x
.
–
the
2,
u
,
the
u
1
,
+
4,
+
,
u
1
u
(Σ)
+
sequence
18,
p,
40.5,
…
,
u
ways
of
…,
of
x
in
the
geometric
sequence
…
at
a
u
to
add
sequence
is
a
and
the
series
terms
gives
of
a
sequence.
a series
sequence.
n
+
u
3
Greek
geometric
notation
4
2
the
value
3x,
looks
3
in
QUESTION
terms
u
2
u
The
4x
section
Adding
p
positive
Sigma
This
of
…
+
+
u
4
letter
is
a
series.
n
Σ,
called
‘sigma’,
is
often
used
to
represent
sums
of
values.
When
➔
∑
means
the
sum
of
the
first
n
terms
of
a
sequence.
sum
of
Y
ou
read
this
‘the
sum
of
all
the
terms u
from
i
=
1
to
i
=
n’.
i
arithmetic
common
sequence
difference
sequence
is
=
u
6n
6.
+
8,
A
14,
20,
general
…
r ule
has
for
first
term
the nth
8
term
and
of
this
(
+
2
n
The
sum
of
the
first
five
terms
of
this
sequence
∑
is
This
To
to
means
‘the
calculate
5
into
sum
this
the
of
sum,
all
the
terms
substitute
expression
6n
+
2,
all
and
6n
the
+
2
from
integer
add
)
=
n
=
1
values
to
of
n
n
=
5’.
from
them:
(
∑
+
)
=
[6(1)
+
2]
+
[6(2)
+
2]
+
32
+
[6(3)
+
2]
+
[6(4)
+
2]
=
+
=
Example
8
[6(5)
+
14
+
+
2]
20
+
26
=
100
4
2
a
Write
the
expression
∑ (
x
b
Calculate
the
sum
of
x
3
)
as
a
sum
of
terms.
= 1
these
terms.
Answers
4
2
a
∑ (
x
x
3
)
= 1
2
=
(1
2
–
3)
+
(2
–
2
+
=
b
represent
values
in
(3
–2
–2
+
+
1
Patterns,
3)
Substitute
consecutive
integers
2
–
1
+
3)
+
6
+
6
+
(4
+
–
3)
beginning
13
13
=
sequences
with
18
and
series
x
=
with
4
x
=
1
and
ending
1
you
sigma
are
a
this
=
form,
The
you
using
notation
Example
8
a
Evaluate
the
2
∑ (
expression
a
=
)
3
‘Evaluate’
Answer
8
Substitute
a
3
2
∑ (
a
=
)
=
4
2
+
5
2
+
6
2
+
consecutive
nd
integers
the
tells
value
you
so
to
the
7
2
+
2
beginning
with
a
=
3
and
nal
ending
answer
will
be
a
8
3
+
2
with
=
8
+
+
=
the
128
+
+
32
+
a
=
8
64
256
504
Example
Write
16
number
.
series
3
+
15
+
75
+
375
+
1875
+
9375
using
sigma
notation.
Answer
n
u
=
–
The
1
3(5
ter ms
are
a
geometric
)
n
progression,
common
with
ratio
first
ter m
3
and
5.
6
This
n
∑ (
3
series
six
ter ms
of
the
Write
F
an
expression
a
1
+
2
b
9
+
16
c
27
d
240
e
5x
f
4
+
7
+
10
g
1
+
3
+
9
+
+
25
+
+
3
+
+
+
+
h
a
+
Write
2a
+
7x
+
+
+
+
23
120
6x
4
25
2
2
first
progression.
= 1
Exercise
1
the
))
geometric
n
is
1
(5
5
+
+
each
+
21
60
+
+
6
+
+
+
…
+
series
+
+
+
+
+
a
+
using
sigma
8
17
+
7.5
10x
55
59 049
5a
sum
of
terms.
8
7
5
11
r
a
a
n
3n
1
b
notation.
5
4a
as
series
15
4
+
7
19
9x
…
+
each
49
30
8x
13
27
3a
+
36
3
+
for
4
c
∑ (
5
(
2
n
)
d
x
Remember
,
a
1
r
=
n
3
nd
Evaluate.
7
5
9
n
1
8n
5
b
3
r
1
the
need
10
to
tells
you
value,
give
so
to
you
numerical
2
r
a
word
5
evaluate
3
the
1
c
m
m
1
d
x
7 x
4
answers.
4
Chapter
.
Arithmetic
series
Carl
Friedrich
Gauss
(1777–1885)
The
sum
of
the
terms
of
a
sequence
is
called
a
series.
The
sum
arithmetic
sequence
is
called
an ari thmetic series
said
the
terms
of
an
is
to
be
the
greatest
mathematician
For
so
example,
5
+
12
+
5,
19
12,
+
19,
26
+
26,
33
33,
+
40
40
is
is
an
an
arithmetic
arithmetic
sequence,
19th
how
series.
the
When
a
series
has
only
a
few
elements,
adding
the
individual
it
not
a
difficult
would
helpful
be
to
find
denotes
S
task.
ver y
a
the
However,
if
a
time-consuming
r ule,
sum
or
of
formula,
the
first
n
series
to
add
for
has
all
50
evaluating
terms
of
a
terms
these
or
terms.
100
It
sum
series.
For
a
of
of
the
F ind
out
worked
the
out
rst
terms
integers.
terms
will
arithmetic
centur y.
Gauss
100
is
often
of
be
series.
series
n
Remember
with
n
must
S
=
u
n
For
S
u
+
=
2
u
+
+
u
3
(u
1
we
u
arithmetic
n
If
+
1
an
+
4
d )
+
this
(u
1
reverse
…
+
+
n
order
of
would
be
the
same,
positive
2d )
+
be:
(u
1
the
a
n
would
+
be
integer
.
u
5
series
+
u
+
3d )
+
(u
1
the
terms
+
4d )
+
…
+
(u
1
in
the
+
(n
–
1)d)
1
equation,
the
value
of
the
Star t
sum
that
terms,
and
it
would
look
like
with
the
nal
this:
term
u
,
then
the
next-
n
S
=
u
n
+
(u
n
–
d )
+
(u
n
–
2d )
+
(u
n
–
3d )
+
(u
n
–
4d )
+
…
+
n
u
to-last
1
term
is
u
n
and
Adding
these
two
equations
for
S
ver tically
,
term
by
so
term,
n
2S
=
(u
n
This
is
+
(u
=
)
+
(u
u
)
+
u
1
added
)
+
(u
n
n
+
u
1
times,
)
+
(u
n
+
1
u
)
n
+
(u
1
+
u
)
n
so:
n
n(u
n
+
u
1
Dividing
+
n
1
2S
u
1
)
n
both
sides
by
2
gives:
=
(
+
+
)
Substitute
(
− )
for
u
, then
n
=
(
+
+
( − ) )
Y
ou
find
using
the
the
sum
of
the
first
=
+
)
or
Patterns,
sequences
=
(
terms
of
an
(
n
formula:
( − ) )
can
series
+
➔
(
=
and
series
+
( − ) )
arithmetic
+
…
+
(u
1
+
u
n
)
on.
–
d,
Example
Calculate
29
+
21
the
+
13
sum
+
of
the
first
15
terms
of
the
series
…
Answer
u
=
29
and
d
=
–8
1
15
S
=
( 2 ( 29 )
15
+
(15
− 1)
8)
(
For
)
the
sum
of
15
ter ms,
2
substitute
=
7.5(58
=
–405
–
112)
n
=
15
into
the
for mula
n
S
=
2u
(
n
+
(n
1
− 1
d
)
)
2
Example
a
b
Find
the
14
15.5
+
Find
the
number
+
17
sum
of
+
of
terms
18.5
the
+
in
…
the
+
series
50
terms.
Answers
a
u
=
14
and
d
=
1.5
Find
these
values
by
looking
1
at
u
= 50
the
To
sequence.
find
n,
substitute
the
n
u
=
14
+
(n
–
1)(1.5)
=
12.5 + 1.5n
values
you
know
into
the
n
12.5 + 1.5n = 50
1.5n
=
for mula
37.5
u
=
u
n
n
=
25
+
(n
–
1)d
1
Solve
for
n.
25
b
S
=
(14
25
+ 50
)
Substitute
2
the
=
12.5(64)
=
800
of
last
n
the
ter m
into
the
first
and
ter m,
the
value
for mula
n
S
=
(u
n
+
u
1
n
)
2
Exercise
1
Find
3
2
+
2.6
3
Find
–
3
+
94
the
3.4
sum
+
of
+
the
first
12
terms
of
the
arithmetic
series
the
first
18
terms
of
the
arithmetic
series
first
27
terms
of
the
arithmetic
series
first
16
terms
of
the
series
(3
...
of
88
sum
+
of
...
sum
+
the
5x)
sum
9
the
+
Find
(2
+
+
100
4
the
6
Find
G
+
of
–
the
...
the
4x)
+
(4
–
3x)
+
...
Chapter
EXAM-STYLE
5
6
Consider
QUESTION
the
a
Find
the
b
Find
the
Find
the
Write
120
number
sum
sum
Example
a
series
of
of
of
+
terms
the
the
116
+
in
112
the
...
+
+
28.
series
terms.
series
15
+
for
the
22
+
29
+
…
+
176
an
expression
S
,
sum
of
the
first
n
terms,
of
the
series
n
64
b
+
60
Hence,
+
56
find
+
…
the
value
of
n
for
which
S
=
0
n
Answers
a
u
=
64
and
d
=
–4
Substitute
the
values
for
1
u
n
S
=
d
into
the
for mula
1
( 2 ( 64 )
n
and
+
( n − 1) ( −4 ) )
n
S
2
=
(
n
2u
1
+ (n − 1)d
)
2
n
=
(128 − 4 n + 4 )
2
n
=
(132 − 4 n )
2
2
S
=
66 n − 2 n
n
Set
2
b
66n
−
2n
=
S
0
=
2n(33
–
n)
=
can
=
0
or
n
=
two
=
33
your
the
positive
1
An
series
has
u
=
4
and
S
1
the
value
EXAM-STYLE
2
a
n.
The
equation
the
GDC.)
the
When
equation
we
solve
usually
by
has
to
word
hence
question
use
your
answer
in
number
integer,
of
we
ter ms
must
disregard
n
be
=
a
0
Write
of
the
=
1425
30
common
difference.
QUESTION
an
expression
for
S
,
for
the
series
1
+
7
+
13
+
…
n
b
Hence,
find
the
value
of
n
for
which
S
=
833
n
3
a
Write
an
expression
for
S
,
for
an
arithmetic
series
n
with
u
=
–30
and
d
=
3.5
1
b
Hence,
find
the
value
of
n
for
which
S
=
105
n
4
In
Januar y
they
sell
2012,
600
a
How
b
Calculate
Patterns,
a
new
drinks,
many
then
drinks
the
total
sequences
coffee
and
will
700
shop
in
they
number
series
sells
March,
expect
of
and
to
drinks
500
sell
they
drinks.
so
in
on
In
in
Febr uar y
,
an
arithmetic
December
expect
to
sell
2012?
in
2012.
progression.
in
tells
you
previous
this
H
arithmetic
Find
for
this
solutions.
Since
Exercise
solve
solve
33
factoring,
n
and
also
0
using
n
0,
n
(You
par t.
5
In
an
and
the
6
In
arithmetic
the
of
common
an
ten
sum
the
first
the
ten
2nd
terms
term
is
is
–20.
four
Find
times
the
the
first
5th
term
term,
and
difference.
arithmetic
times
find
sequence,
the
the
series,
sum
common
of
the
the
sum
first
difference
3
of
the
terms.
and
the
first
If
12
the
value
terms
first
of
is
term
equal
is
to
5,
S
20
.
Just
Geometric
as
an
arithmetic
sequence,
a
a
the
following
series
geometric
geometric
Adding
series
is
the
is
series
sum
the
of
the
sum
of
terms
the
of
an
terms
arithmetic
of
sequence.
terms
of
a
geometric
sequence
gives
the
equation:
Multiply
this
2
=
S
u
n
+
u
1
r
+
u
1
+
=
u
n
r
+
u
1
u
r
+
+
+
u
u
1
4
r
+
u
1
–
2
n
r
+
u
1
3
r
n
…
1
2
rS
3
r
1
r
…
+
n
–
S
n
=
–
u
n
+
u
1
r
by
of
r
1
r
u
–1
n
r
+
u
1
Subtract
r
the
rst
1
equation
rS
sides
1
n
+
1
–
both
equation
from
the
both
sides
n
=
u
1
r
–
u
1
second.
1
n
S
(r
–
1)
=
u
n
(r
–
1)
1
Factorize
of
(
the
)
equation.
=
Y
ou
➔
Y
ou
can
find
the
sum
of
the
first
n
terms
of
a
geometric
may
nd
convenient
using
the
more
to
use
the
formula:
rst
it
series
formula
when
(
)
or
=
r
,
where
r
≠
>
1,
as
it
avoids
1
using
a
negative
denominator
Example
Calculate
the
sum
of
the
first
12
terms
of
the
series
1
+
3
+
9
+
...
Answer
u
=
1
and
r
=
3
Substitute
the
values
of
1
12
1
(3
S
1
)
u
,
r
and
n
into
the
for mula
1
=
12
n
3
1
u
1
S
(r
1
)
=
n
531 440
r
1
=
2
=
265 720
Chapter
Example
Geometric
a
Find
the
8192
b
+
number
6144
Calculate
+
the
of
terms
4608
sum
…
+
of
the
in
+
the
series
series
are
often
the
study
seen
in
1458.
of
fractals,
terms.
such
as
the
Koch
snowake.
Answers
6144
a
u
=
8192
r
and
3
=
=
Find
r
by
dividing
u
1
n
3
⎛
1
Substitute
the
values
⎠
n
the
for mula
u
=
u
n
n
729
3
⎛
=
1
⎟
4
⎝
⎠
6
3
⎛
=
3
6
3
6
729
=
729
and
4
=
4096
⎞
=
⎜
6
4096
4
⎝
⎟
4
You
could
=
also
solve
this
equation
⎠
using
logarithms
6
(see
=
Example
19).
7
Substitute
7
⎛
⎛
8192 ⎜ 1
3
⎜
⎜
⎝
values
of
u
,
r
1
⎟
⎟
4
the
⎞
⎞
and
n
into
the
for mula
⎟
⎠
⎝
S
into
)
1
6
n
know
1
⎞
⎜
4096
1
–
(r
=
8192
–
you
⎟
4
⎝
1458
1
⎞
⎜
b
u
4
1458 = 8192
n
by
2
8192
⎠
n
=
u
7
1
3
S
1
(r
r
4
)
1
[
You
⎛ 14 197 ⎞
8192
1
=
n
⎜
can
also
calculate
sums
Koch
snowake
using
⎟
16 384
⎝
the
seq
(and
sum)
functions
on
⎠
=
1
your
GDC.
4
=
28 394
Exercise
1
I
Calculate
the
value
of
S
for
each
geometric
series.
12
a
c
2
0.5
64
+
–
1.5
32
Calculate
+
+
4.5
16
the
–
+
8
value
…
0.3
b
…
+
d
of
S
for
(
each
+
0.6
+ ) +
+
1.2
(
+
+
)
…
+
(
+
)
+
series.
20
0.25
a
+
0.75
+
2.25
…
+
b
+
+
+
…
c
3
–
6
+
12
EXAM-STYLE
–
24
+
…
d
+
(
)
+
(
)
+
(
)
+
QUESTION
So
3
For
each
geometric
at
i
find
the
number
ii
calculate
far
we
of
arithmetic
1024
b
2.7
+
1536
10.8
+
2304
43.2
+
+
590.49
Patterns,
+
+
c
d
+
+
sequences
sum
and
a
looked
and
terms
geometric
the
have
series:
…
…
+
+
26 244
2764.8
there
of
other
Are
types
mathematical
+
sequences
196.83
and
and
+
sequences
+
series.
+
65.61
series
series?
+
…
+
0.01
used?
How
are
they
Example
GDC
For
the
geometric
series
3 + 3
2
+ 6 + 6
2
+
... ,
determine
the
Plus
value
of
n
for
which
>
S
help
on
CD:
demonstrations
least
and
Casio
Alternative
for
the
TI-84
FX-9860GII
500
GDCs
n
are
on
the
CD.
Answer
An
u
=
3
and
r
=
2
Substitute
the
known
old
Indian
fable
values
1
tells
into
n
the
S
us
that
a
prince
for mula.
n
3
2
(
1
)
Enter
S
=
>
the
S
500
was
so
new
game
taken
that
he
with
the
equation
n
n
2
1
into
the
of
chess
GDC.
asked
its
Remember:
inventor
On
the
the
number
GDC,
the
X
represents
ter ms,
and
reward.
The
man
f1(x)
said
represents
choose
‘n’,
his
of
to
he
would
like
S
n
one
Look
at
the TABLE
to
grain
rst
the
chess
sums
of
the
first
n
four
etc.,
number
This
to
of
the
first
of
the
first
13
12
456.29,
ter ms
the
ter ms
is
and
on
ask
the
doubling
each
seemed
that
the
time.
so
the
little
prince
is
agreed
approximately
on
ter ms.
third
sum
on
of
board,
grains
second,
The
rice
square
see
two
the
of
the
the
straight
away.
sum
Ser vants
star ted
bring
rice
to
approximately
the
–
and
648.29
to
the
prince’
s
surprise
soon
n
=
13,
since
S
>
the
great
grain
overowed
the
500
chess
13
board
to
ll
the
palace.
When
the
sum
of
a
geometric
series
includes
an
exponent n,
How
you
can
use
logarithms.
rice
many
did
have
Example
A
the
the
of
prince
give
the
man?
geometric
Find
to
grains
progression
value
of
n
such
has
that
first
S
term
=
of
0.4
and
common
ratio
2.
26 214
n
Answer
n
0
S
4
(2
1
)
=
= 26 214
n
2
1
n
0
4
(2
1
=
)
26 214
n
2
–
1
=
65 535
n
2
n
=
65 536
=
log
(65 536)
Express
this
using
logarithms.
2
log
n
65 536
log
n
Use
the
change-of-base
rule
and
your
=
=
2
GDC
to
find
this
value.
16
Chapter
Exercise
For
1
J
each
series,
determine
the
least
value
of
n
for
which
S
>
400
n
25.6
a
57.6
…
+
b
14
d
0.02
+
–
42
+
126
–
378
…
+
+
0.2
+
2
+
…
geometric
Find
+
+
A
38.4
+
c
2
+
the
series
common
has
third
ratio
term
and
the
1.2
and
value
of
eighth
term
291.6
S
10
In
3
a
geometric
series,
S
=
20
and
S
4
Find
the
common
EXAM-STYLE
=
546.5
7
ratio,
if
r
>
1
QUESTION
4
Find
a
the
common
ratio
for
the
geometric
+
series
Hence,
b
find
the
least
value
of
n
such
that
S
>
‘Hence’
+
+
tells
you
to
use
previous
800
your
answer
in
n
this
In
5
6
a
geometric
sum
of
In
geometric
a
the
.
the
sum
series,
first
of
6
terms
series,
the
first
Investigation
2
a
are
three
+
+
1
240
c
For
1
–
60
each
i
F ind
ii
Use
+
+
of
–
Find
the
the
If
r
three
sum
first
>
1,
and
terms
of
four
find
the
GDC
Do
2
you
Now
3
the
full
notice
use
ten
to
the
terms.
times
Extension
material
Worksheet
6
-
on
CD:
Finance
ratio.
infinity
series
series.
75
+
30
+
12
...
+
...
+
to
ratio,
r
calculate
values
any
your
is
and
seven
common
the
values
of
S
,
S
10
Write
304,
first
terms
the
sums
is
series:
common
your
first
converging
3.75
these
the
the
of
terms.
b
–
sum
series
…
15
of
1330.
the
geometric
0.5
sum
is
two
Convergent
Here
the
par t.
you
see
patterns?
GDC
to
on
Why
calculate
your
do
the
GDC
you
think
value
of
,
S
15
.
20
screen.
this
S
is
for
happening?
each
series.
50
Do
For
you
each
think
of
the
your
calculator
series
in
the
is
correct?
investigation
Explain
you
why
or
why
not.
should
Paradox
have
noticed
that
the
values
of
S
,
S
10
close.
This
series
has
is
because
when
a
and
S
15
are
ver y
20
Suppose
30-metre
a
common
ratio
of
|r|
<
1,
the
as
n
the
value
large.
is
term
increases.
actually
In
each
the
of
This
the
decreases
means
sum
approaching
We
call
series
2
+
approaching
Patterns,
some
geometric
4
1
+
as
0.5
n
sequences
+
as
ver y
you
such
0.25
ver y
series
+
closer
add
little.
constant
series
gets
and
(becomes
that,
changes
are
walking
down
a
…
,
large.
to
zero)
more
The
value
as
hallway.
sum
as n
terms,
is
gets
of
walk
the
half
might
to
reach
Will
you
ever
ver y
suspect
that
the
the
hallway.
you
these convergent series
you
Ever y
ten
seconds,
difference
you
between
you
geometric
sum
distance
How
the
get
long
end
of
there?
to
will
the
the
it
end
take
hallway?
If
you
tr y
to
find
S
on
the
GDC,
you
get:
50
50
=
4(1
–
0.5
)
=
4
Is
the
sum
decimals
the
actually
like
rounded
4?
NO!
The
3.999 999 999 99
value
Convergent
of
GDC
to
fit
rounds
on
its
the
last
screen,
so
digit
all
of
you
long
see
is
4.
series
The
sum
of
the
terms
of
a
geometric
series
is
IMPORTANT!
This
is
only
true
geometric
As
or
n
n
gets
→
ver y
large,
you
can
say
that
n
‘approaches
and
<
only
1,
then
as
n
→
∞,
r
→
0,
so
then
1
series,
when
(Remember
,
|r|
infinity’,
∞
n
If
for
<
r
if
<
| r |
| r |
<
<
1.
1,
1.)
We
can
write
this
as:
⎛
(
)
⎞
⎜
⎟
,
=
or
⎜
→∞
⎟
⎝
This
means
=
∞
⎠
that
as
n
gets
ver y
large
(it
approaches
infinity),
the
We
say
‘The
limit
of
value
of
the
series
is
n
.
approaching
The
series
is
converging
to
u
1
(1
r
)
as
the
1
value
.
We
write
this
as
S
,
and
call
it
‘the
sum
to
n
r
infinity’.
∞
approaches
innity
is
u
1
equal
to
’.
1
r
➔
For
a
geometric
series
with
|r|
<
1,
=
∞
Example
For
the
series
18
+
6
+
2
+
…,
find
S
,
S
10
and
S
15
∞
Answer
1
u
=
18
and
r
=
1
3
10
⎛
⎞
⎛ 1 ⎞
18 ⎜ 1
⎜
⎟
⎟
⎜
3
⎝
⎟
1
⎠
⎝
S
⎠
Substitute
=
u
=
18
and
r
=
1
10
1
1
n
u
3
(1
r
)
1
into
for mulae S
the
=
and
n
≈
26.999 542 75
1
r
u
15
⎛
⎛ 1 ⎞
18 ⎜ 1
⎜
⎜
⎝
⎟
3
S
=
∞
1
⎟
r
⎟
⎠
⎝
S
1
⎞
⎠
=
Write
down
all
the
digits
from
the
15
1
GDC
1
display.
3
≈
26.999 998 12
18
=
S
=
27
∞
1 ⎞
⎛
⎜
⎝
1
⎟
3
⎠
Chapter
Example
The
sum
sum
to
Find
of
the
infinity
the
first
first
is
three
terms
of
a
geometric
series
is
148,
and
the
256.
term
and
the
common
ratio
of
the
series.
Answer
3
u
1
S
(1
r
)
=
This
= 148
is
the
equation
for
S
3
3
1
r
u
1
S
=
Multiply
= 256
both
sides
of
this
equation
∞
1
r
3
by
(1
–
r
)
3
u
1
(1
r
)
3
= 256
1
(1
− r
)
The
left
side
of
this
equation
is
now
r
identical
to
the
left
side
of
the
S
3
equation.
3
256
(1
r
)
= 148
Set
148
the
right
equal
to
Solve
for
sides
each
of
these
equations
other.
37
3
1 − r
=
=
256
r.
64
37
27
3
r
= 1 −
=
64
64
3
r
=
3
4
Substitute
r
=
into
the
equation
4
u
u
1
= 256
1
S
=
=
256
∞
3
⎛
⎜
⎞
1
1
r
⎟
4
⎝
⎠
u
1
= 256
⎛
1 ⎞
⎜
⎝
⎟
4
⎠
4u
=
256
1
u
=
64
1
Exercise
1
Explain
a
2
K
how
convergent
Find
S
,
S
4
a
144
you
if
a
geometric
series
will
be
series.
and
S
7
+
know
for
each
of
these
series.
∞
48
+
16
+
...
b
500
+
400
+
320
+
...
What
real-life
c
80
+
8
+
0.8
+
...
d
situations
might
be
modeled
3
A
geometric
series
has
series?
and
=
S
∞
=
13.
Find
S
3
5
EXAM-STYLE
4
For
a
QUESTION
geometric
progression
with
u
3
Patterns,
sequences
and
series
=
24
and
u
6
=
3,
find
S
∞
by
covergent
In
5
a
geometric
progression,
u
=
12
and
S
2
EXAM-STYLE
A
6
The
7
sum
and
the
seven
.
of
such
If
a
a
250.
of
sum
to
has
Find
the
first
of
of
u
1
is
ratio
of
0.4
and
a
sum
to
term.
terms
of
4374.
the
$1000
annually
,
will
be
interest
a
geometric
Find
the
sum
series
of
is
the
3798,
first
in
is
the
patter ns
and
a
account
savings
makes
account
amount
at
the
star t
of
in
at
the
the
the
no
many
other
ten
after
situations,
pays
or
interest
at
deposits,
years?
each
(Multiply
account
which
withdrawals
annually
of
real-life
growth.
account
after
end
year.
in
population
compounded
the
amount
geometric
patterns
in
and
in
The
of
geometric
amount
1.04.)
first
five
interest
deposits
4%
much
When
common
the
infinity
compound
person
rate
a
arithmetic
examples
as
how
series
Applications
see
Find
terms.
and
We
64.
QUESTION
geometric
infinity
=
∞
(once
year
the
10
per
will
year),
be
104%
previous
years
the
of
amount
the
by
would
10
be
≈
1000(1.04)
Y
ou
can
each
think
year
as
of
a
$1480.24.
the
amount
geometric
of
money
sequence
in
with u
the
=
account
1000
and
at
r
=
the
end
of
1.04:
1
u
=
$1000
=
$1000(1.04)
=
$1040
=
$1040(1.04)
=
$1081.60
=
$1081.60(1.04)
1
u
2
u
3
u
≈
$1124.86
4
and
Now
so
consider
on.
what
happens
when
interest
is
compounded
In
more
than
once
each
year.
questions
the
is
term
like
‘per
sometimes
this,
annum’
used
Let
rather
A
=
the
amount
of
money
in
the
=
the
interest
n
=
the
number
of
times
t
=
the
number
of
years
p
=
the
principal
you
can
find
the
rate
(a
percentage,
(initial
amount
of
per
year
amount
money
‘annually’
account
or
r
than
written
that
of
in
as
a
interest
‘once
a
year’.
decimal)
is
compounded
money)
the
account
using
the
formula:
nt
r
⎛
A
=
p
⎞
+
⎜
⎝
⎟
n
⎠
Chapter
Example
APR
A
person
deposits
$1000
in
an
account
which
pays
interest
at
4%
means
percentage
compounded
quar terly
.
Assuming
the
person
makes
no
or
deposits,
how
much
will
be
in
the
rate’.
4%
additional
APR
withdrawals
‘annual
APR,
account
after
is
the
same
as
ten
4%
per
annum.
years?
Answer
4(10)
0.04
⎛
A
= 1000
⎞
1 +
⎜
Substitute
⎟
the
known
values
into
the
What
4
⎝
⎠
other
types
nt
r
⎛
for mula
A
=
p
1
⎞
of
+
⎜
mathematics
are
⎟
40
=
1000(1.01)
n
⎝
⎠
useful
≈
$1488.86
This
equation
annual
interest
up
4
into
and
so
1%.
If
times
the
interest
Example
The
is
year
10
rate
for
is
the
divided
each
quar ter,
interest
rate
compounded
(quar terly)
years,
will
because
(4%)
one
quar terly
ever y
of
rate
par ts,
interest
period
Population
works
this
be
is
four
for
a
quar terly
applied
40
times.
growth
population
of
a
small
town
population
at
the
star t
of
1980
population
at
the
star t
of
2020?
increases
was
by
12 500,
2%
what
per
is
year.
the
If
the
predicted
Answer
40
12 500(1.02)
≈
27 600.496
At
the
star t
population
previous
The
be
population
approximately
In
questions
of
years,
In
like
rather
Exercise
1
of
an
the
town
will
From
27 600
have
Example
than
as
a
23,
term
you
need
if
u
1
2
A
=
cup
five
cups
How
high
b
How
many
metre
Patterns,
star ting
2020,
40
the
of
the
population.
years
will
passed.
to
think
of
n
as
the
number
sequence,
u
=
15
cm
high.
3u
4
50
is
a
1
to
year,
102%
8
plastic
When
year’s
1980
be
L
arithmetic
u
each
number.
6
Find
of
will
12
cm
are
stacked
would
cups
high.
a
stack
would
high?
sequences
and
series
together,
of
it
20
take
the
cups
to
stack
is
stand?
make
a
stack
at
least
in
nance?
3
George
at
6%
or
4
deposits
APR.
deposits,
$2500
in
Assuming
how
an
he
much
account
makes
will
be
no
in
the
a
interest
is
compounded
annually
b
interest
is
compounded
quar terly
c
interest
is
compounded
monthly?
An
arithmetic
sequence
is
defined
which
pays
additional
account
by u
=
interest
withdrawals
after
12n
–
7
8
years
and
if
the
a
n
n
geometric
sequence
is
defined
by
v
=
–
1
0.3(1.2)
n
Find
the
least
number
of
terms
such
that v
>
u
n
5
In
a
geometric
sequence,
the
first
term
is
6
n
and
the
common
This
ratio
is
1.5.
In
an
arithmetic
sequence,
the
first
term
is
75
and
question
v
,
rather
than
n
common
difference
is
100.
After
how
many
terms
will
the
sum
the
terms
in
the
geometric
sequence
be
greater
than
the
sum
term
of
6
the
At
of
terms
the
the
beginning
fish
What
in
in
the
will
of
lake
be
the
arithmetic
2012,
is
a
lake
expected
number
sequence?
of
to
contains
in
the
,
to
of
the
the
nth
geometric
sequence.
increase
fish
u
n
represent
of
uses
the
200
at
a
lake
fish.
rate
at
The
of
the
number
5%
per
beginning
year.
of
2015?
7
The
population
increasing
continues
at
to
population
a
of
a
rate
city
of
increase
of
the
is
275 000
3.1%
at
city
per
this
to
year.
rate,
reach
people.
The
Assuming
how
long
500 000
will
population
the
it
is
population
take
for
the
people?
2
8
A
series
has
the
formula
S
=
3n
–
2n
n
Find
a
the
values
of
S
,
S
1
Find
b
the
values
of
u
,
u
1
Write
c
an
expression
and
S
2
3
and
u
2
for
3
u
n
n
9
A
series
has
the
formula
S
=
+
2
2
–
4
n
a
Find
the
values
of
S
b
Find
the
values
of
u
,
S
,
u
1
1
Write
c
an
expression
and
S
2
3
and
u
2
for
3
u
n
10
Two
species
population
1.25%
is
per
of
3%
species
at
population
species
B?
11.
annual
Mohira
interest,
which
of
rate
is
Mohira
a
remote
12 000
of
175
A
be
invests
in
an
of
spiders
$3000
Ryan
annual
interest,
neither
person
his
de posits,
account
than
how
B
is
an
the
at
a
rate
50 000
month.
than
of
and
When
will
population
account
which
but
has
much
Mohira
increasing
invests
3%
or
in
The
which
of
pays
yearly.
Assuming
withdrawals
is
each
account
yearly.
island.
species
g reater
compounded
$3000
and
population
species
interest,
pays
inhabit
A
The
compounded
also
in
a
invests
monthly.
have
spiders
month.
decreasing
the
11
of
is
in
3%
$3000
in
annual
an
account
compounded
made
more
has
pays
any
money
hers
after
additional
does
ten
Ryan
years?
Chapter
.
Pascal’s
triangle
and
the
binomial
expansion
Pascal’
s
named
Now
we
will
look
at
a
famous
mathematical
patter n
known
triangle.
Here
are
rows
1
to
7
of
Pascal’s
after
is
Blaise
as
Pascal
Pascal’s
triangle
(French,
triangle.
1623–62).
1
1
3
1
1
Any
number
numbers
in
immediately
3
4
7
4
35
triangle
above
1
6
21
Pascal’s
1
1
35
is
21
the
sum
7
of
1
the
Can
two
the
it.
8th
Y
ou
and
generate
adding
wanted
It
to
would
Here
are
the
numbers
pairs
find
be
ver y
the
of
the
in
the
numbers
numbers
to
in
triangle
get
the
time-consuming
numbers
in
the
4th
the
row
.
row?
make
row
star ting
next
15th
to
by
of
a
the
Or
at
But
the
the
what
27th
triangle
can
also
be
that
triangle,
1,
using combinations ,
found
row
will
in
what
the
be?
if
we
row?
big!
4,
6,
4,
1.
C
numbers
predict
top
n
These
you
numbers
or
is
commonly
r
the
⎛ n ⎞
C
n
function
on
the
GDC.
written
as
r
⎝
C
4
=
1
C
0
4
=
4
C
1
4
=
6
C
2
4
=
4
C
3
4
=
⎟
⎜
r
,
or
⎠
1
n
4
sometimes
as
C
r
⎛ ⎞
n
⎜
⎝
⎟
,
or
,
C
represents
the
number
of
ways
n
items
can
be
r
⎠
taken
r
labeled
at
a
A,
time.
B,
C,
For
D
example,
and
E.
If
suppose
you
reach
a
in
bag
contains
and
take
2
5
balls
balls
from
the
bag,
there
are
=
10
different
combinations
of
balls
you
could
select.
These
combinations
CE
DE.
are
AB,
AC,
AD,
AE,
BC,
BD,
BE,
CD,
Make
or
how
sure
to
use
you
the
know
nCr
⎛ ⎞
Y
ou
can
find
the
values
of
expressions
like
⎜
⎝
a
⎟
without
function
using
on
GDC.
⎠
calculator.
➔
The
is
number
found
of
combinations
of
n
items
taken
r
at
a
time
by:
!
is
the
sign.
⎛ ⎞
⎜
⎝
⎟
The
Patterns,
expression
,
=
⎠
factorial
(
where
n!
=
n
×
(n
–
1)
×
(n
–
2)
×
…
×
1
n!
is
called
)
factorial’.
your
sequences
and
series
‘n
Example
⎛ 7 ⎞
Find
the
value
of
⎜
⎝
⎟
5
using
the
formula,
and
check
with
your
GDC.
⎠
Answer
Substitute
7
5
n
=
7
and
r
=
5
7!
into
5!7
the
for mula.
5 !
Y
ou
7 × 6 × 5 × 4 × 3 × 2 ×1
Cancel
=
5 × 4 × 3 × 2 ×1
out
like
factors
may
used
2 ×1
the
numerator
and
the
see
dots
from
instead
multiplication
7 × 6
of
denominator.
signs.
42
=
=
2 ×1
=
For
2
example:
3
·
3
×
2
·
1
for
21
Remember,
we
can
also
2
×
1
⎛ 7 ⎞
⎜
⎝
=
⎟
5
find
21
⎠
the
value
using
Pascal’s
triangle.
Using
the
calculator:
On
the TI
Nspire,
Probability,
nCr
is
on
Combinations
the
menu.
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
Exercise
Find
each
value
using
the
⎛ ⎞
formula,
and
check
⎜
⎠
⎝
⎟
5
⎜
6
⎝
Investigation
Expand
as
a
7
each
of
the
–
(a
Do
GDC.
3
⎜
⎟
⎝
⎠
patterns
following
how
long
it
takes
you
in
expressions
to
+
⎠
polynomials
(write
+
2
you
Based
5
your
answers
notice
on
each
each
expression
expansion.
(a
+
3
b)
3
(a
+
b)
6
(a
+
b)
5
b)
at
do
2
b)
4
Look
your
⎛ ⎞
6
⎟
1
(a
4
with
polynomial).
Time
1
CD.
⎠
⎛ ⎞
9
the
C
3
⎟
C
4
on
TI-84
⎛ ⎞
2
⎝
are
the
FX-9860GII
M
1
⎜
Casio
Alternative
for
any
these
(a
+
and
note
similarities
patterns,
6
b)
any
to
patterns
Pascal’
s
predict
what
you
see.
triangle?
the
expansion
7
of
(a
+
b)
might
be.
Chapter
Binomial
We
will
expansion
look
at
what
happens
when
we
expand
an
expression
like
n
(a
+
In
b)
the
,
where
n
is
a
Investigation
positive
on
page
integer.
185,
you
expanded
these
expressions.
1
(a
+
b)
=
a
2
(a
+
b)
(a
+
b)
=
a
=
a
3
+
b)
(a
+
b)
you
=
a
=
a
2ab
+
3a
+
b
2
2
b
+
+
4a
+
5a
number
+
6a
b
+
10a
+
3
terms
3
b
each
b
2
b
at
of
+
2
4
closely
3
3ab
3
5
look
The
1
2
+
4
5
If
b
3
4
(a
+
2
4
4ab
+
2
2
b
+
10a
3
b
expansion,
is
one
b
4
+
5ab
you
greater
5
+
will
than
b
see
the
some
value
patter ns:
of
n
For
example,
when
n
The
2
powers
of
a
begin
0
until
you
reach
with
a
,
and
(a
a
=
1)
in
the
0
The
3
powers
of
b
the
powers
of
a
decrease
by
1
0
begin
with
b
last
n
=
4,
has
term.
5
the
expansion
terms.
0
(b
=
1),
and
the
powers
of
b
n
increase
by
1
until
you
reach
in
b
the
last
term.
4
(a
4
The
coefficients
are
The
coefficients
of
all
numbers
from
Pascal’s
triangle!
of
Pascal’s
the
formula
The
5
triangle.
for
exponents
+
are
can
numbers
find
combinations,
in
each
2
b
The
b)
Y
ou
b)
2
6a
n
(a
+
term
or
add
these
the
to
from
using
nCr
the
the
the
nth
power
of
on
the
a
3
+
4a
3
+
4ab
b
+
4
+
b
coefcients
1,
4,
row
triangle,
function
4
=
the
6,
4,
1
are
the
of
Pascal’
s
In
(a
4th
row
or
triangle.
GDC.
binomial.
5
3
For
example,
in
the
expansion
(a
+
b)
3
=
2
a
+
3a
2
b
+
3ab
+
b)
the
3
+
b
,
coefcients
the
exponents
in
each
term
add
to
3.
6
Y
ou
can
use
these
patter ns
to
expand
the
expression
(a
+
b)
10,
10,
5,
5th
row
of
1,
1
5,
are
the
Pascal’
s
.
triangle.
This
expansion
The
powers
The
coefficients
(1,
6,
15,
of
20,
will
a
have
will
will
15,
6,
These
(a
+
be
patter ns
the
6th
the
powers
row
of
of
b
will
Pascal’s
increase.
triangle
1).
6
=
b)
terms.
decrease,
6
Therefore,
7
and
5
a
+
6a
4
b
+
15a
obser vations
2
b
can
3
+
20a
help
3
b
you
2
+
to
15a
4
b
5
+
6ab
understand
6
+
b
the
+
general
➔
binomial
The
theorem
binomial
where
n
∈
for
theorem
expanding
states
that
powers
for
any
of
n
∈
a
positive
means
that
power
of
a
binomial,
,
⎛ ⎞
+
⎛ ⎞
=
)
⎜
⎝
⎟
⎛ ⎞
−
+
⎠
⎜
⎝
⎟
⎛ ⎞
⎠
−
+
⎜
⎝
⎟
+
⎠
+
⎜
⎝
⎟
used
in
many
mathematics
Y
ou
can
also
write
the
binomial
expansion
using
sigma
areas
of
beyond
⎠
binomial
including
Y
ou
are
the
➔
is
integer
.
Combinations
(
n
binomials.
can
theorem,
probability.
even
use
notation:
combinations
to
⎛ ⎛ ⎞
⎞
calculate
(
+
)
=
∑
Patterns,
=
sequences
⎜ ⎜
⎝
⎟
and
( )
⎠
series
( )
your
chance
⎟
of
winning
the
lotter y!
Example
5
Use
the
binomial
theorem
to
expand
(x
+
3)
.
Write
your
answer
in
its
simplest
form.
Answer
Substitute
⎛ 5 ⎞
5
(
x
⎛ 5 ⎞
5
+ 3)
=
⎜
⎝
⎟
0
x
⎟
1
⎝
)(1) + (5)(x
4
+
⎜
+
3
+
⎜
⎠
⎟
2
⎝
x
)(3) + (10)(x
⎛ 5 ⎞
2
2
3
+
⎠
⎜
⎝
3
3
15x
⎛ 5 ⎞
1
3
x
4
5
=
4
+
⎠
5
= (1)(x
⎛ 5 ⎞
0
3
x
⎟
3
1
+
⎝
2
)(9) + (10)(x
⎜
⎠
⎟
4
4
x
0
3
+
⎠
⎜
⎝
⎟
5
x
binomial
+
)(81) + (1)(1)(243)
You
should
find
these
with
270x
+
405x
+
theorem.
⎠
2
90x
the
5
3
1
)(27) + (5)(x
into
⎛ 5 ⎞
3
3
x
and
be
able
values
to
both
without
your
243
GDC.
Example
3
Use
the
binomial
theorem
to
expand
(2x
–
5y)
.
Write
your
answer
in
its
simplest
form.
Answer
⎛ 3 ⎞
3
(2x
5 y
=
)
⎜
⎟
0
⎝
3
⎛ 3 ⎞
0
(2x )
5 y
(
+
)
⎜
⎠
⎟
1
⎝
2
⎛ 3 ⎞
1
(2x )
5 y
(
+
)
⎠
⎜
⎟
2
⎝
1
(2x )
Be
careful
an
expression
5 y
)
+
⎜
⎝
⎟
3
0
(2x )
5 y
(
to
(1)(8x
)
2
)(1)
exponent
be
the
8x
+
(3)(4x
2
)(–5y)
+
(3)(2x)(25y
)
3
+
(1)(1)(−125y
applied
Exercise
Use
the
2
−
60x
.
needs
variable
y
2
+
150xy
)
3
−
to
both
and
the
coefficient!
3
3
(2x)
3
3
=
see
3
⎠
=
you
like
⎠
The
⎛ 3 ⎞
when
2
(
(2x)
125y
3
=
2
x
3
3
=
8x
N
binomial
theorem
to
expand
each
expression.
⎛
5
(y
1
+
4
3)
(2b
2
–
1)
(3a
3
+
2)
+
4
⎞
6
⎜
⎟
⎝
⎠
⎛
8
(x
5
+
(3a
6
–
2b)
7
8
⎝
Sometimes,
you
binomial.
Example
Y
ou
will
not
may
need
simply
the
be
entire
looking
the
⎟
⎜
⎠
⎝
expansion
for
one
of
a
4x
⎞
⎟
2 y
power
par ticular
+
⎠
of
term.
3
Find
2
+
⎜
a
⎛
⎞
4
y)
x
9
term
in
the
expansion
of
(4x
–
1)
3
Answer
In order to get x
the binomial, 4x, to the third power
. So
⎛ 9 ⎞
⎜
⎝
⎟
6
, raise the first term of
3
(
4 x
)
6
(
1)
the second term of
the binomial, –1,
⎠
will be raised to the sixth power
. You
3
=
( 84
)
(
64 x
)
could use
3
=
⎛ 9 ⎞
⎛ 9 ⎞
(1)
⎜
⎝
⎟
3
instead of
⎠
⎜
⎝
⎟
6
as these
⎠
5376x
values are equal.
Chapter
Example
n
In
the
expansion
of
(2x
+
1)
3
,
the
coefficient
of
the
x
term
is
80.
Find
the
value
Answer
n
⎛
You
⎛ n ⎞
⎜
⎝
⎟
3
3
(2x )
n
3
could
have
=
⎝
80 x
of
⎜
⎝
3
⎟
⎜
⎟
⎛
3)!
3
3
,
as
these
⎠
values
⎛
n ⎞
⎜
⎟
⎜
⎟
) (1)
= 80 x
Use
the
equal.
n!
3
(8x
are
⎠
for mula
⎠
⎝
r
=
r !( n
r
)!
⎠
⎞
n!
⎟
⎜
( 8 ) = 80
As
you
only
have
to
find
the
coefficient,
you
can
⎟
⎜
⎝
⎟
⎞
n!
⎜
⎝
⎟
n
⎛ n ⎞
instead
(3 )! ( n
n.
⎞
⎜
3
1
⎠
⎛
used
of
(3 )! ( n
3)!
⎠
3
leave
n ×
(n
− 1) ×
(
n
− 2) ×
(
n − 3) ×
(
n −
(8)
(3
× 2 × 1) × ⎡ ( n
n ×
− 3) ×
⎣
(n
(3
− 1) ×
(
n
(n
− 2) ×
× 2 × 1) × ⎡ ( n
⎣
−
the
x
=
80
=
10
Divide
both
sides
by
8
4 ) × ...⎤
⎦
n − 3) ×
(
of f
4 ) × ...
− 3) ×
(n
n ×
−
(
n −
4 ) × ...
Simplify
⎦
(n
by
canceling
out
like
factors
from
the
4 ) × ...⎤
− 1) ×
(
n
numerator
and
the
denominator.
− 2)
=
10
=
60
You
can
solve
polynomial
equations
such
as
this
6
using
n ×
(n
− 1) ×
3
n
(n
− 2)
3n
+
2n
–
60
=
0
=
5
O
5
1
Find
GDC.
2
–
n
Exercise
a
the
x
7
term
EXAM-STYLE
in
the
expansion
of
(x
–
of
(4y
4)
QUESTIONS
4
5
2
Find
the
y
term
3
Find
the
a
4
Find
the
constant
2
in
the
expansion
–
1)
4
b
6
term
in
the
expansion
of
(2a
–
3b)
9
term
in
the
expansion
of
(x
–
2)
The
is
6
5
In
the
expansion
of
(px
+
1)
‘constant
just
the
value
of
the
coefficient
of
the
x
term
is
160.
p
7
6
In
the
Find
expansion
the
value
EXAM-STYLE
of
of
(3x
+
q)
5
,
the
coefficient
of
the
Find
the
x
term
is
q
QUESTION
7
constant
term
in
the
expansion
of
8
Find
the
constant
term
in
the
expansion
of
In
the
QUESTION
n
9
EXAM-STYLE
expansion
of
(x
+
1)
3
,
the
coefficient
of
the
x
term
is
2
two
times
Patterns,
the
coefficient
sequences
and
of
series
numerical
3
,
term
Find
the
term’
the
x
term.
Find
the
value
of
n
81 648.
with
no
variables.
n
10
In
the
expansion
of
(x
+
2)
3
,
the
coefficient
of
the
x
term
is
two
4
times
the
coefficient
Review
of
the
term.
x
Find
the
value
of
n
exercise
✗
EXAM-STYLE
1
Consider
a
Write
b
Find
QUESTIONS
the
arithmetic
down
the
u
sequence
common
Find
c
3,
7,
11,
15,
...
difference.
the
value
of
n
such
that
u
71
2
The
first
three
a
Write
c
Find
=
99
n
terms
down
the
of
an
value
infinite
of
geometric
r.
Find
b
sequence
are
64,
16
and
4.
u
4
3
In
an
the
sum
to
arithmetic
infinity
sequence,
of
u
this
=
sequence.
25
and
u
6
4
a
Find
the
common
difference.
b
Find
the
first
of
Consider
a
Find
the
the
term
the
arithmetic
value
of
49
sequence.
sequence
x.
=
12
22, x,
Find
b
38,
...
u
31
5
Evaluate
the
expression
∑(
6
Consider
a
7
Find
Find
all
the
the
geometric
common
possible
geometric:
x,
EXAM-STYLE
12,
series
800
ratio.
values
9x,
)
=
of
+
such
+
Find
b
x
200
that
50
the
this
Find
9
A
the
b
x
top
row
store
in
the
How
are
In
an
first
a
...
the
EXAM-STYLE
2
Consider
a
Find
the
a
expansion
display
three
cans,
of
of
(2x
soup
and
each
+
3)
cans
row
stacked
has
two
in
a
pyramid.
more
cans
than
35
cans
in
the
bottom
row
,
how
many
rows
are
cans
63
is
series,
in
the
display
in
total?
the
first
term
is
4
and
the
sum
of
the
value
of
the
1000.
common
difference.
Find
b
the
17th
term.
QUESTION
the
u
are
exercise
terms
Find
is
it.
arithmetic
25
infinity
.
display?
many
Review
1
in
has
has
above
there
to
5
row
If
sum
QUESTION
term
grocer y
The
a
the
...
sequence
3
8
+
.
arithmetic
b
sequence
Find
the
3,
value
4.5,
of
n
6,
7.5,
such
...
that
S
=
840
n
Chapter
3
In
an
the
arithmetic
first
10
a
Find
the
first
b
Find
the
sum
EXAM-STYLE
4
series,
terms
In
a
is
the
tenth
term
is
25
and
the
sum
of
160.
term
of
and
the
the
first
common
24
difference.
terms.
QUESTIONS
geometric
sequence,
a
Find
the
common
b
Find
the
least
the
first
term
such
that
is
3
and
the
sixth
term
is
96.
ratio.
value
of
n
u
>
3000
n
5
In
an
arithmetic
difference
and
the
Find
In
a
50.
least
is
7
Find
is
first
EXAM-STYLE
the
first
geometric
ratio
greater
terms
the
a
value
geometric
first
In
common
the
sequence
6
is
sequence,
of
is
n
than
series,
term
is
sequence,
28
the
and
first
the
common
term
is
1
1.5
such
the
the
that
nth
3rd
the
term
term
is
nth
of
45
term
the
of
the
geometric
arithmetic
and
the
sum
sequence.
of
the
2735.
term
and
the
common
ratio, r,
if
r
∈
QUESTION
⎛
4
7
Find
the
term
in
x
in
the
expansion
⎞
of
⎜
⎝
⎟
⎠
8
8
In
the
expansion
of
(ax
+
2)
5
,
the
x
term
has
a
coefficient
of
.
Find
9
the
At
the
a
If
value
If
number
according
●
Each
6
In
This
a
positive
●
Y
ou
can
u
u
=
n
or
is
a
find
+
(n –
population
a
the
rate
of
beginning
continues
countr y
at
be
of
1.6%
this
of
a
countr y
was
annually
,
3.4
million.
estimate
the
2040.
rate,
expected
in
to
what
year
exceed
7
would
million?
sequences
is
a
patter n
of
numbers
arranged
in
a
par ticular
order
r ule.
number,
or
element,
of
a
sequence
is
called
a term
sequences
arithmetic
value
at
SUMMARY
individual
an
at
the
sequence
Arithmetic
●
of
and
to
the
grows
growth
population
Patterns
A
2010,
population
CHAPTER
●
of
population
population
the
a
beginning
the
countr y’s
b
of
sequence,
called
the
negative
the
nth
the
terms
common
increase
dierence ,
or
or
decrease
d.
The
by
a
constant
common
value.
difference
can
Patterns,
a
value.
term
of
an
arithmetic
sequence
using
the
formula:
1)d
1
Continued
be
sequences
and
series
on
next
page
Geometric
●
In
a
sequences
geometric
term
by
a
sequence ,
constant
value.
each
This
term
value
can
is
be
obtained
by
multiplying
the common ratio,
called
or
the
previous
r.
n
●
Y
ou
can
find
the
nth
term
of
a
geometric
sequence
using
the
formula: u
=
u
n
Sigma
(∑)
notation
and
–
(r
1
)
1
series
●
∑
means
the
sum
of
the
first
n
terms
of
a
sequence.
=
Y
ou
read
this
‘the
sum
of
all
the
terms u
from
i
=
1
to
i
=
n’.
i
Arithmetic
●
Y
ou
can
find
series
the
sum
of
the
=
(
+
or
)
=
can
+
of
an
arithmetic
series
using
the
formula:
( − ) )
series
find
the
sum
of
the
first
n
terms
of
a
geometric
series
using
the
formula:
(
terms
Geometric
Y
ou
n
(
●
first
)
=
or
(
)
=
,
where
r
≠
1.
Convergent
series
and
sums
to
infinity
●
For
a
geometric
series
with
< ,
=
∞
Pascal’s
●
The
triangle
number
of
⎛ ⎞
⎜
⎝
●
The
⎟
,
⎠
(
+
can
n
items
taken
r
expansion
at
a
time
is
found
by:
where
also
⎟
=
n
×
states
that
(n
for
⎛ ⎞
⎜
n!
–
1)
×
(n
–
2)
×
…
×
1
)
⎛ ⎞
=
)
theorem
⎝
Y
ou
of
binomial
●
the
combinations
=
binomial
(
and
⎠
write
−
+
⎜
⎝
the
⎟
power
of
a
⎛ ⎞
any
⎠
−
+
⎜
⎝
binomial
⎟
where n
∈,
⎛ ⎞
binomial,
+ +
⎠
expansion
⎜
⎝
using
⎟
⎠
sigma
notation:
⎛ ⎛ ⎞
(
+
)
=
∑⎜
=
⎜
⎝ ⎝
⎟
⎠
( )
( )
⎞
⎟
⎠
Chapter
Theory
of
knowledge
Whose
Pascal's
triangle
idea
is
named
after
was
it
the
This
Frenchman
Blaise
Pascal,
anyway?
who
is
not
the
long-standing
about
it
in
1654
in
his
Treatise
rst
case
of
a
relatively
wrote
on
mathematical
idea
being
the
attributed
to
a
par ticular
person.
This
Arithmetic Triangle
has
often
happened
mathematician
However,
the
proper ties
of
this
known
and
studied
other
par ts
before
of
in
the
Pascal's
India,
world
work,
China
for
China,
Throughout
have
time.
Pascal's
triangle
is
called
triangle'
after
a
been
13th
but
it
Do
you
was
known
In
the
to
the
given
or
centur y
,
the
credit
think
that
Omar
seen
have
been
's
in
Khayyám
Pascal's
triangle?
gl
used?
an
ngle
1
1
1
1
1
[
1
Theory
of
6
15
knowledge:
20
15
Whose
idea
for
mathematical
of
these
attributed
person?
triangle.
mathematicians
many
Persian
oet
a
public.
inventions.
this.
11th
the
years,
long
wrong
before
introducing
idea
centur y
ideas
mathematician,
an
'Yang
Hui's
well-known
and
centuries
discoveries
In
a
by
mathematical
mathematicians
when
published
patter n
impor tant
were
has
6
was
1
it
anyway?
Blaise
Pascal
(1623–62)
to
the
In
Fibonacci:
patterns
nature
The
Italian
Leonardo
sequence
in
If
mathematician
of
Pisa,
in
his
Liber
the
F ibonacci
Abaci,
published
1202.
set
begin
month
diagram
shows
how
the
this
problem:
sequence
was
with
each
which
becomes
month
on,
produced
F ibonacci
The
he
you
each
F ibonacci,
introduced
book
it
in
not
the
only
how
in
a
a
single
pair
pair
produces
productive
many
of
a
from
pairs
of
rabbits,
new
the
and
pair
second
rabbits
will
be
year?
mathematician
to
work
with
this
pattern.
Number
grows.
of
pairs
1st
month:
1
pair
of
original
two
rabbits
1
2nd
month:
still
1
pair
as
they
are
not
yet
productive
3rd
month:
2
pairs
new
4th
month:
3
pair
pairs
they
–
original
they
–
original
pair
produced
in
and
the
1
produce
produced
month,
pair
in
pair
,
pair
2
3rd
they
4th
3
month
The
the
number
of
F ibonacci
pairs
gives
sequence
5
1, 1, 2, 3, 5, 8, 13, 21, 34, 55,
,
,
where
, ...
each
term
i
een
sequence
Could
it
appears
be
that
mathematics
in
nature?
there
and
is
a
relationship
between
nature?
How
are
Pascal's
triangle
F ibonacci
sequence
the
of
and
related?
the
Hint:
look
at
{
sums
the
diagonals
in
the
triangle.
.1250)
Chapter
Limits
CHAPTER
derivatives
OBJECTIVES:
Informal
6.1
ideas
of
limit
and
convergence;
⎛
from
and
rst
principles
as
f ′( x )
= lim
f (x
+
⎜
h
⎝
normals,
interpreted
and
their
as
gradient
notation;
Denition
of
derivative
⎟
h→0
Derivative
limit
h) − f ( x ) ⎞
⎠
function
and
as
rate
of
change;
T
angents
and
equations
n
Derivative
6.2
of
x
(n
∈
);
Differentiation
of
a
sum
and
a
real
multiple
of
these
x
functions;
product
Derivatives
and
quotient
of
e
and
r ules;
ln x;
The
The
second
chain
rule
for
derivative;
composite
use
of
both
functions;
forms
of
the
notation,
2
d
y
and
f ″ (x)
2
dx
Local
6.3
maximum
gradients;
graphs
Before
1
f,
Factorize
and
an
behavior
f ″;
problems
+
of
involving
Points
functions,
Optimization
how
and
of
inexion
including
s,
zero
and
relationship
v
and
Expand
between
acceleration
1
the
a
Factorize:
2x
=
2x (x
4
+
2x
+
1)
a
9x
3
−
15x
2
+
3x
4x
b
2
2
non-zero
check
2
+
the
velocity
Skills
expression.
4x
with
applications
displacement
to:
2
2x
points;
start
know
3
e.g.
f ′
you
should
minimum
Graphical
Kinematic
6.6
Y
ou
of
and
binomials.
c
−
9
−
9x
2
x
−
5x
+
6
2x
d
−
5
4
e.g.
Expand
(2x
−
1)
4
(2 x
2
− 1)
Expand
each
binomial.
1
3
a
4
= 1( 2 x )
0
( −1)
3
+
4 (2 x )
1
1
1
2
+ 6( 2 x )
2
1
( −1)
+
4 (2 x )
+
4
2)
(3x
b
−
1)
3
(2x
c
1( 2 x )
2
1
3
3
3
1
Use
rational
exponents
to
rewrite
4
( −1)
1
4
= 16 x
3
− 32 x
4
6
4
1
expression
in
2
+ 24 x
− 8x
the
b
x
x
rational
exponents
to
rewrite
7
d
7
5
x
e
n
expressions
in
the
form
1
2
5
e.g.
=
2x
2
;
x
=
5
x
Limits
and
derivatives
x
cx
cx
c
3
6
Use
form
4
1
+ 1
a
3
3y)
3
n
+
+
( −1)
1
0
(x
1
( −1)
3
x
5
x
each
If
you
pluck
quieter
as
a
string
time
on
passes.
a
guitar
This
can
and
be
let
it
vibrate,
modeled
by
the
the
sound
gets
function
sin t
f
(t )
,
=
where
t
represents
time.
As
t
becomes
larger
and
larger,
t
sin t
becomes
closer
to
zero
–
this
is
the
limit
value
of
the
function.
t
sin t
We
write
this
as
=
lim
t →∞
calculus.
is
a
sine
Y
ou
wave,
Calculus
is
geometr y
problems.
variable
in
the
lear n
a
later
branch
together
that
evaluate
is
are
a
fundamental
concept
of
about
mathematics
the
limit
calculus
sine
and
that
process
uses
function,
whose
focus
to
In
on
algebra
looks
find
calculus
change.
then
takes
and
limits
Integral
repeated
limits
the
graph
chapter.
changing.
involve
basic
Limits
more
of
with
Differential
quantity
problems
to
will
0.
t
this
at
the
uses
and
two
rate
limits
chapter
differential
we
types
at
of
which
to
a
solve
will
lear n
calculus.
Chapter
.
In
Limits
this
section
convergence
basis
of
and
you
and
convergence
will
use
a
with
pair
the
notation.
concepts
The
of
concept
limits
of
and
limits
is
the
calculus.
Investigation
Work
investigate
limit
of
a
par tner
.
scissors
Round
–
Y
ou
and
Por tion
number
creating
a
need
copy
of
the
will
the
end
of
one
this
paper
of
sequence
rectangular
piece
of
paper
,
table.
you
the
Fraction
a
have
at
round
Decimal
(3 sf)
1
2
3
4
5
6
Round
1:
roughly
table.
as
Cut
the
equal
size.
Record
both
Round
a
2:
rectangular
the
fraction
Cut
T
ake
one
por tion
the
and
a
of
piece
piece
the
piece
paper
each,
original
decimal
‘spare’
of
(to
of
into
leaving
three
one
rectangle
three
paper
piece
you
signicant
into
three
pieces
now
on
When
cutting
of
paper
into
the
size
Each
of
you
add
one
piece
of
this
to
your
pieces
por tion
of
of
have,
in
Repeat
As
1
Record
the
the
you
the
same
same
total
way
as
of
the
original
rectangle
say
about
more
the
you
If
you
repeat
this
you
Limits
of
of
the
for
four
and
por tion
more
more
of
process
original
not
just
more
rounds
of
now
the
activity
say
the
rounds.
you
could
rounds
of
this
activity,
what
the
original
rectangle
you
forever
,
what
can
you
say
rectangle
you
number
of
rounds
approaches
innity.
Can
you
give
can
example
in
real
life
have?
about
grows
or
develops
the
like
por tion
complete
more
that
2
do
original
an
you
you
exact,
before.
process
complete
por tion
be
equal
the
and
rectangle.
to
equal
approximate.
gures).
As
size.
pieces
have
have
the
three
this?
have?
sequences
The
notation
lim u
=
L
n
n→∞
The
data
you
collected
in
the
investigation
forms
a sequence,
is
where
is
u
the
por tion
of
the
rectangle
you
have
after
round
1,
1
u
the
por tion
you
have
after
round
2,
and
so
on.
n
of
read
‘the
approaches
u
2
equals
limit
as
innity
L’.
n
Sequences
like
this
are
called
convergent
because
as
the
The
term
number
in
the
sequence
increases,
the
terms
in
ancient
the
approach
sequence.
a
We
fixed
can
value
write
known
this
as:
the limi t,
as
lim u
=
Greeks
the
used
sequence
of
as
L,
L
n
the
limits
to
using
the
idea
nd
of
areas,
‘method
of
n →∞
Sequences
that
are
not
convergent
are divergent
exhaustion’.
wish
What
is
Limits
the
limit
of
the
and
derivatives
sequence
formed
in
the
investigation?
to
Y
ou
research
may
this.
Example
Determine
If
a
sequence
0.3,
a
whether
1
c
0.33,
6
,
5
is
31
0.3333,
156
,
125
sequence
convergent,
0.333,
,
25
each
give
is
the
convergent
limit
…
of
the
or
divergent.
sequence.
b
2,
4,
8,
d
1,
−1,
16,
…
781
,
,
625
...
1,
−1,
. . .
3125
Answers
1
The
Convergent;
a
lim u
patter n
indicates
=
Other
n
n →∞
3
that
the
sequence
notations
for
is
recurring
decimals
approaching
0
include
0.333 3…,
or
3,
0
3
which
1
is
the
decimal
for m
of
3
Divergent
b
Each
ter m,
ter m
so
in
they
the
are
sequence
not
is
larger
approaching
than
a
the
previous
limit.
1
lim u
Convergent;
c
=
To
n
n →∞
compare
fractions
with
dif ferent
denominators,
4
use
a
GDC
0.248,
to
conver t
0.2496,
them
0.249 92,
to
decimals:
0.2,
0.24,
…
1
The
values
are
approaching
0.25
or
4
Divergent
d
The
two
Exercise
Determine
If
a
3,
5,
1
7,
3,
4,
3,
f
of
(x )
=
3,
is
the
1
limit
of
20
−
,
the
or
the
and
sequence
are
not
are
oscillating
approaching
a
between
fixed
value.
,
3.499,
182
,
27
162
divergent.
sequence.
3.499,
121
4
...
10 000
4,
convergent
3.49,
2
1000
4,
give
…
,
100
sequence
convergent,
,
Limits
lim
is
each
1
, −
10
5
whether
1
3
values
in
7A
sequence
1,
1
ter ms
1093
,
243
3.4999,
1640
,
1458
…
, ...
2187
…
functions
L
means
that
as
the
value
of
x
becomes
Y
ou
can
use
a
GDC
to
help
nd
the
x →c
sufficiently
f
(x),
close
becomes
to
close
c
(from
to
a
either
fixed
side),
value
L.
the
If
f
function,
(x)
does
not
limit
of
a
function
become
close
to
a
fixed
value
L,
we
say
that
the
not
and
Y
ou
can
examine
graph
the
the
values
of
limit
f (x)
does
function.
Graphically:
when
x
is
near
c
exist.
Numerically:
of
values
f (x)
when
and
x
is
Y
ou
can
examine
near
make
the
a
table
values
of
c
Chapter
Example
Use
a
Find
GDC
the
to
limit
examine
or
state
each
that
it
function
does
graphically
not
and
numerically
.
exist.
2
x
2
a
lim x
b
x →2
⎧1
1
lim
x →1
f
lim
c
(x )
;
where
f
(x )
=
for
x
≥
0
⎨
x →0
x
1
⎩
−1
for
x
<
Answers
0
y
2
2
a
lim x
Plot
the
graph
of
f (x)
=
x
a
7
f (x)
6
using
x →2
GDC,
and
look
at
the
values
of
2
f(x)
as x approaches 2
and
from
the
from
the
right
=
x
5
left.
4
3
2
1
Graphically,
f (x)
approaches
4
as
0
–4
x
approaches
Numerically,
2
from
either
–2
–3
x
–1
1
2
3
4
2:
as
x
side,
becomes
f (x)
close
becomes
to
close
to
4.
→
2
←
x
1.8
1.9
1.99
1.999
2.001
2.01
2.1
2.2
f 1(x)
3.24
3.61
3.960
3.996
4.004
4.040
4.41
4.84
4
←
2
To
build
variables
the
to
table
above
using
‘Ask’.
Enter
the
a
GDC,
values
for
enter
f1(x)
=
x
. Then
set
the
independent
x.
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
Casio
are
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
2
So,
lim x
=
4
x →2
The
graph
and
table
are
shown
on
the
same
screen.
2
2
For
f (x)
=
x
we
can
substitute
and
find
that
lim x
2
2
4
x 2
{
Limits
and
derivatives
Continued
on
next
page
y
2
x
b
1
lim
x →1
f (x)
x
approaches
2
as
x
approaches
1:
7
1
2
x
6
f(x)
–
1
=
x
–
1
5
4
3
2
1
0
–3
x
–1
2
1
x
Since
division
by
zero
is
not
defined,
f
(x )
is
=
when
x
−
1
=
0
or
x
=
1. Therefore
there
is
a
undefined
1
x
discontinui ty
in
the
graph
2
x
when
x
=
1.
Notice
that
f
(x )
(
1
=
x
+ 1
)(
x
− 1
)
=
x
=
1
x
x
+ 1,
when
x
≠
1
1
2
x
Even
though
f
(x )
1
is
=
x
becomes
close
to
1
from
undefined
when
x
=
1,
the
limit
exists
since
as
x
1
either
side,
f
(x)
becomes
→
close
to
2.
←
1
x
0.8
0.9
0.99
0.999
1.001
1.01
1.1
1.2
f (x)
1.8
1.9
1.99
1.999
2.001
2.01
2.1
2.2
→
2
←
2
2
x
x
Note
So,
lim
=
that
lim
=
x
+ 1
)(
x
− 1
)
x
lim
2
x →1
x →1
(
1
1
x
x →1
1
x
1
1
=
lim
(
x
+ 1
)
= 1 + 1
=
2
x →1
f (x)
c
lim
f
(x )
does
not
approach
the
same
value
as
x
y
where
x →0
approaches
⎧1
f
(x )
=
for
x
≥
0
from
the
left
and
right:
2
0
⎨
⎩
1
−1
for
x
<
0
0
–4
–3
–2
–1
x
1
2
3
4
–2
→
x
f (x)
So,
lim
f
(x )
does
Note
exist.
←
−0.2
−0.1
−0.01
−0.001
0.001
0.01
0.1
0.2
−1
−1
−1
−1
1
1
1
1
that
f (0)
=
1,
but
lim
f
(x )
does
not
exist.
x →0
x →0
not
0
This
is
close
to
because
−1
for
f
(x)
is
values
close
of
x
to
to
1
the
for
left
values
of
x
of
=
x
to
the
right
of
x
=
0
and
f (x)
is
0.
Chapter
Exercise
Use
and
a
7B
GDC
to
examine
numerically
.
Find
each
the
function
limit
or
graphically
state
that
it
does
Extension
material
Worksheet
7
look
not
An
on
CD:
algebraic
limits
exist.
3
2
x
2
lim ( x
1
at
-
+ 1)
− 4 x
+
x
lim
2
x →0
x →3
x
2
x
− 3x
1
+ 2
lim
3
x →4
x
2
⎧x
lim
5
lim
4
x →2
f
( x );
f
where
(x )
=
+ 3
x
for x
4
⎩
secant
a
circle
the
−x
+ 5
line
to
intersects
≥ 1
⎨
x →1
A
for x
circle
twice.
< 1
2
⎧x
lim
6
( x );
f
f
where
(x )
=
+ 3
for
x
≥
2
⎨
x →2
⎩
x
for x
<
2
A
tangent
to
a
circle
intersects
.
The
tangent
line
circle
and
line
the
once.
n
derivative
In
this
section
we
of
will
x
work
with
secant,
tangent
A
and
normal
lines.
We
will
define
the
derivative
tangent
cur ve
a
function
of
cer tain
and
lear n
some
r ules
for
finding
derivatives
the
functions.
–
secant
and
tangent
may
cur ve
than
Investigation
line
is
the
graph
of
f (x)
=
x
+
a
intersect
more
once.
lines
2
Here
to
of
y
1
6
Copy
1
the
graph
to
paper
and
draw
lines
AP,
BP,
CP,
DP,
5
EP
and
FP.
These
lines
are
called
secant
lines
to
the
4
2
graph
of
f (x)
=
x
+
1.
3
A
2
Copy
and
complete
the
F
2
table.
B
E
P
Gradient
x
0
–2
Point
Coordinates
Line
P
—
A
AP
B
BP
C
CP
or
1
–1
2
slope
—
Recall
of
a
that
line
points
the
(x
,
y
1
)
y
2
(x
DP
,
2
y
)
x
2
3
4
As
E
EP
F
FP
points
does
the
Draw
the
on
the
cur ve
gradient
line
at
of
get
the
point
P
closer
secant
that
and
lines
has
the
closer
seem
to
to
gradient
point
be
you
P,
what
value
approaching?
found
in
question
3.
2
This
Limits
line
and
is
called
derivatives
the
tangent
line
to
the
graph
of
f (x)
=
x
+
1
at
P
1
is
2
x
the
and
1
y
D
gradient
through
1
Lines
have
gradient
line
to
the
Newton
velocity
a
of
constant
a
cur ve
cur ve
at
worked
of
a
Gradient
a
with
a
given
that
moving
of
gradient,
at
point.
when
object
but
point
This
he
cur ves
is
gradient
is
the
the
wanted
whose
secant
other
to
do
concept
find
velocity
the
was
not.
of
The
the
that
tangent
Sir
Isaac
instantaneous
always
changing.
line
y
y
f(x
+
=
f(x)
h)
Q(x
+
h, f(x
+
h))
[
Sir
Isaac
Newton
1642–1727
,
English
mathematician,
is
one
f(x)
P(x, f(x))
of
the
mathematicians
credited
with
x
0
x
x
+
developing
h
calculus.
h
The
gradient
of
the
secant
line
PQ
is
written
as
The
f
(x
+ h) −
f
(x )
f
(x
+ h) −
f
(x )
f (x
+
expression
h)
−
f ( x)
is
=
(x
+ h) −
x
known
h
h
as
the
dierence
quotient
Example
2
Write
an
Simplify
expression
your
for
the
gradient
of
a
secant
line
for
f
(x)
=
x
+
1
expression.
Answer
2
f
(x
+
h) −
f
⎡( x
(x )
+
2
2
+ 1⎤
h)
⎣
−
⎦
(
x
Replace the x in x
)
+ 1
+ 1
=
with
h
x
+
h
to
write
an
h
expression
2
(
2
x
+ 2 xh
+
for
f
(x
+
h)
2
)
h
+ 1
−
(
x
)
+ 1
=
2
Expand
(x
+
h)
h
2
2 xh
+
h
Combine
=
like
ter ms.
h
h
(2x
+
h
Factorize.
)
=
h
=
Exercise
Write
an
function.
2x
Simplify.
+ h
7C
expression
Simplify
1
f (x)
=
3x
2
f (x)
=
2x
3
f (x)
=
x
+
for
your
the
gradient
of
a
secant
line
for
each
expression.
4
2
−
1
2
+
2x
+
3
Chapter
Gradient
Suppose
of
that
a
tangent
point
Q
slides
line
down
and
the
the
cur ve
derivative
and
approaches
y
point
y
P.
The
secant
lines
PQ
will
get
closer
to
the
tangent
line
at
Q(x
f(x
P.
h
As
Q
gets
closer
approaches
0
of
to
P,
the
h
gets
closer
gradient
of
to
the
0.
We
secant
can
line
take
to
get
=
f(x)
point
the
limit
the
gradient
+
+
h,
f(x
+
h))
h)
as
f(x)
of
the
tangent
line:
P(x, f(x))
f (x
+
h)
−
f ( x)
is
lim
f
(x
+ h) −
f
h→0
(x )
not
h
x
0
x
x
+
h
lim
a
h →0
constant.
It
is
a
h
h
function
that
gradient
of
f
gives
at
the
x
f ′(x)
f
➔
The
function
defined
(x
+ h) −
f
lim
by
is
h →0
is
read
as:
the
(x )
known
as
the
derivative
of
f
x.
f,
or :
h
dy
of
derivative
f.
The
derivative
is
defined
prime
of
is
read
by
dx
f
f
′( x )
=
(x
+ h) −
f
(x )
dy
lim
f
=
or
h →0
h
(x
+ h) −
f
(x )
as
the
h →0
dx
with
h
‘d
y
respect
d
change
that
in
to
x’
or
slope
change
in
This
is
x
2
the
derivative
of
f
(x)
=
x
+
1
and
hence
find
the
gradient
of
the
Δy
expressed
tangent
line
when
x
=
Δx
dy
y
2
⎡( x
+
2
+ 1⎤
h)
⎣
′( x )
=
−
⎦
(
x
Simplify
)
+ 1
lim
h →0
=
the
Example
3.
Evaluate
the
quotient
as
shown
in
h
lim
(2x
+ h
=
)
2x
limit
by
substituting
0
+ 0
h →0
for
f
′( x )
=
2x
′(3)
=
2(3)
f
=
h.
The
6
derivative,
f ′(x)
So
the
gradient
of
the
=
2x,
when
x
=
3
is
is
a
tangent
function
line
that
gives
6.
the
gradient
of
the
2
cur ve
at
Exercise
the
definition
find
the
gradient
1
f
(x)
=
2x
2
f
(x)
=
3x
3
f
(x)
=
x
−
3;
of
of
x
derivative
the
=
tangent
2
2
+
2x;
x
=
−3
2
Limits
f(x)
any
=
x
point
+
1
x
7D
Use
lim
x 0
dx
f
and
−
x
+
2;
x
derivatives
.
as
3
Answer
=
1
to
find
line
at
the
the
derivative
given
of
value
f
of
and
x
hence
is
y
.
Find
of
x’.
Recall
Example
‘derivative
lim
x
y
Some
rules
for
derivatives
n
Investigation
Use
1
the
denition
–
of
the
derivative
derivative
to
nd
the
of
f (x)
=
derivatives
x
of
Recall
2
f (x)
=
x
3
,
f (x)
=
the
denition
4
x
and
f (x)
=
x
of
derivative
is
n
Make
2
a
conjecture
about
the
derivative
of
f (x)
=
x
f ( x
f ′( x )
Express
your
conjecture
in
words
and
as
a
function.
+
h)
− f ( x )
= lim
h →0
h
5
3
We
is
Use
your
Use
the
conjecture
denition
investigated
tr ue
➔
for
any
Power
only
real
to
of
predict
derivative
positive
number
=
Example
the
derivative
see
if
your
values
of
f (x)
=
x
prediction
for n,
but
the
was
correct.
following
n.
rule
(x)
Use
f
to
integer
n
If
the
x
power
n−1
,
then
r ule
to
f
′(x)
find
=
the
nx
,
where
derivative
of
n
∈
each
function.
1
12
f
a
(x)
=
x
f
b
(x )
=
c
f
(x )
=
x
3
x
Answers
Use
12
f
a
(x)
=
12
f
′( x )
the
power
rule.
x
1
= 12 x
11
= 12 x
1
3
b
f
(x )
=
=
Rewrite
x
using
rational
exponents.
3
x
Use
the
power
rule.
3
−3 −1
′( x )
f
=
−3 x
−4
=
−3 x
Simplify.
=
−
4
x
1
2
c
f
(x )
=
x
=
x
Rewrite
1
−1
1
′( x )
=
rational
exponents.
−
1
2
f
using
1
x
2
=
x
Use
2
the
power
rule.
2
1
1
=
Simplify.
or
1
2
x
2
2x
Exercise
Find
the
7E
derivative
of
each
function.
1
5
1
f
(x)
=
8
x
2
f
(x)
=
x
3
f
(x )
=
4
x
1
5
3
4
f
(x )
=
x
5
f
(x )
=
6
f
(x )
=
3
x
x
Chapter
Using
the
power
derivative
of
a
➔
of
function
f
(x)
Constant
The
f(x)
➔
=
c
is
a
=
If
Sum
The
f
or
cf
is
any
or
(x)
of
the
The
any
a
=
the
is
two
process
real
zero.
which
r ules
of
number,
then
has
The
a
graph
gradient
of
of
the
c
is
any
real
number,
the
f
′(x)
=
0
constant
function
then
y ′
=
cf
′(x)
rule
constant
times
a
function
is
the
constant
times
the
u(x)
±
of
a
sum
rule
v (x)
f
′(x)
=
u ′(x)
±
v ′(x)
rule
function
or
then
that
difference
is
of
the
the
sum
or
difference
derivatives
of
the
of
two
or
more
terms.
each
function.
5
2
4x
+
2x
−
3
f
b
(x )
=
3
4 x
=
find
derivative
zero.
x
+ 8
3
f (x)
can
the
function.
3
c
we
finding
rule
where
dierence
=
line
dierence
Differentiate
f (x)
is
constant
multiple
Example
a
c
multiple
(x),
derivative
terms
following
dierentiation
where
horizontal
of
Sum
the
rule
of
derivative
derivative
called
c,
Constant
y
and
functions.
rule
Constant
➔
=
derivative
If
The
is
Constant
If
r ule
many
(x
−
2)
(x
+
4)
f
d
(x )
2
+ 2x
− 3
=
x
Answers
3
a
f
(x )
=
2
4 x
+ 2x
Find
− 3
3 −1
f
′( x )
=
4
(3x
of
2 −1
)
+ 2
(2x
)
the
the
derivative
constant
of
ter m
each
is
ter m.
Note
the
derivative
the
derivative
0.
0
2
= 12 x
+ 4 x
1
5
5
b
f
(x )
f
′( x )
=
3
x
+ 8
=
3x
Rewrite
+ 8
1
exponents.
−
3
5
3 ⋅
rational
4
−1
1
=
using
x
5
+
0
=
5
x
Find
the
derivative
of
each
ter m.
Note
5
of
3
the
constant
ter m
is
0.
3
=
or
4
5
5
5
4
Simplify.
x
5x
2
c
f
(x )
=
(x
− 2 )( x
+ 4)
=
x
+ 2x
− 8
First
expand
so
that
the
function
is
the
sum
or
n
2 −1
f
′( x )
=
2x
dif ference
1−1
+ 2 ⋅ 1x
− 0
=
2x
of
ter ms
in
the
for m
{
Limits
and
derivatives
ax
+ 2
Continued
on
next
page
3
2
4 x
f
d
(x )
3
+ 2x
− 3
=
2
4x
2x
=
3
+
x
Rewrite
−
x
x
2
′( x )
in
the
the
for m
function
is
the
sum
or
dif ference
of
ax
1
4 x
+ 2x
− 3x
2 −1
f
that
n
ter ms
=
so
x
=
4 ⋅ 2x
=
8x
1−1
−1−1
+ 2 ⋅ x
− 3 ⋅ ( −1) ⋅ x
3
2
+ 2 + 3x
=
8x
+ 2 +
2
x
3
2
8x
+ 2x
+ 3
or
2
x
Exercise
7F
Differentiate
each
function.
3
2
f
1
(x )
3
=
f
2
(x)
=
5
f
3
(x )
=
x
−
2
8
x
x
5
f
4
(x)
=
3
2
π x
f
5
(x)
=
(x
−
4)
f
6
(x )
=
x
− 4
x
3
3
4
f
7
(x )
=
f
8
(x )
=
(
3
10
f
(x )
=
x
(x )
=
x
normal
x
11
f
(x)
12
f
(x)
=
12
14
f
2
(x) = 3x
− 2x
−
x
+
3x
2
+ 5
=
2x
+
7
1
2
3
+ 2x
Equations
The
f
)
4
+
2
f
4 x
4
x
3
13
9
2
2
4 x
of
line
per pendicular
+ 1
tangent
at
to
a
point
the
(x)
2x (x
and
on
tangent
=
a
at
Normal
3x)
normal
cur ve
line
2
−
is
the
that
line
to
15
f
(x)
=
(x
+
3x)(x
−
1)
lines
line
point.
cur ve
[
y
=
f(x)
Sparks
created
grinding
wheel
tangent
the
Tangent
Example
Write
an
line
to
by
a
are
to
wheel.
cur ve
equation
for
each
line.
2
a
The
tangent
b
The
normal
c
The
tangent
line
line
to
to
the
cur ve
the
f (x)
f
cur ve
=
(x )
x
=
2
+
1
x
at
the
when
point
x
=
(1, 2)
9
27
and
normal
lines
to
the
cur ve
f
(x )
=
x
+
2
2x
when
x
=
3
3
d
The
tangent
to
f
(x)
=
x
2
−
3x
−
13x
+
15
that
is
parallel
to
the
[
tangent
at
(4,
Spokes
on
a
bic ycle
−21)
wheel
{
Continued
on
next
page
the
are
normal
to
rim.
Chapter
Answers
To
2
a
f (x)
=
x
+
find
line,
f
′(x)
=
the
gradient
of
the
tangent
The
symbol
∴
is
used
1
find
the
derivative
of
f
and
to
mean
‘therefore’.
2x
evaluate
m
=
when
x
=
1
′(1)
f
tangent
=
2(1)
=
2
Use
the
write
∴ y
− 2
=
2( x
− 1)
point
the
(1, 2)
equation
and
of
m
the
=
2
to
The
equation
line
through
,y
1
f
(x )
=
2
the
a
point
line.
(x
b
of
tangent
Rewrite
x
the
function
using
is
rational
with
y
=
gradient
m(x
x
1
(See
exponents.
1
)
m
1
y
).
1
Chapter
18,
2
=
2x
Section
3.11.)
If
has
1
1
2
f
′( x )
=
x
or
x
m
=
f
To
′(9 )
find
the
gradient
of
the
tangent
tangent
line,
find
the
derivative
of
f
and
1
=
evaluate
when
x
=
9.
a
line
gradient
9
m,
the
gradient
of
1
the
=
perpendicular
3
1
line
m
=
−3
Since
the
nor mal
line
is
.
m
is
normal
(See
per pendicular
f
(9 )
=
2
9
=
the
gradient
a
Use
− 9)
the
write
f
on
of
the
when
point
the
taking
the
the
gradient
of
(9, 6)
equation
nor mal
x
=
and
of
line
by
9
m
the
=
−3
to
tangent
line.
27
c
f
(x )
=
x
+
Rewrite
the
function
using
rational
2
2x
exponents.
27
2
=
x
+
x
2
27
f
′( x )
To
= 1 −
find
the
gradient
of
the
tangent
3
x
line,
m
=
f
find
evaluate
′(3)
the
derivative
when
x
=
of
f
and
3
tangent
27
= 1 −
3
3
Since
=
the
gradient
is
0,
the
tangent
line
0
is
horizontal,
so
the
normal
line
must
27
be
f
(3)
=
vertical.
3 +
2
2 (3
)
Find
9
a
point
on
the
lines
by
=
evaluating
f
when
x
=
3
2
∴
Normal
line
is
x
=
3
and
9
tangent
line
is
y
=
2
{
Limits
and
Chapter
derivatives
Continued
18,
line,
line.
point
evaluating
−3( x
by
reciprocal
tangent
Find
=
tangent
Section
the
opposite
− 6
the
6
find
∴ y
to
on
next
page
3.11.)
3
f
d
(x)
=
x
2
−
3x
−
13x
+
15
2
f
′( x )
=
3x
− 6x
f
′( 4 )
=
3( 4 )
− 13
2
Find
− 6( 4 ) − 13
the
when
= 11
x
gradient
=
of
the
tangent
line
4
2
3x
− 6x
− 13
− 6x
− 24
Set
= 11
to
2
3x
=
the
find
derivative
equal
x-coordinates
to
of
11
points
with
0
parallel
tangent
lines.
2
3( x
− 2x
3( x
− 8)
− 4 )( x
=
+ 2)
0
=
0
Recall
x
=
4, −2
Notice
x
=
4,
point
The
that
is
the
of
3
f
( −2 )
=
− 3( −2 )
the
values,
x-coordinate
x-coordinate
for
the
of
lines
the
given
that
have
parallel
the
same
gradient.
(4, −21).
of
the
point
parallel
line
of
is
−2
Evaluate
2
( −2 )
=
of
tangenc y
tangenc y
x
one
f
at
x
=
−2
to
find
the
− 13( −2 )
y-coordinate
of
the
point
of
+ 15
tangency.
=
21
Use
∴
y
− 21 = 11( x
the
write
Exercise
Find
point
(−2,
21)
and
m
=
11
to
+ 2)
the
equation
of
the
tangent
line.
7G
the
equations
of
the
tangent
and
normal
lines
to
the
graph
2
of
f
(x)
lines
2
=
by
Find
x
–
4x
at
the
point
(3,
–3).
Graph
the
function
and
the
hand.
the
equation
for
the
tangent
line
to
the
cur ve
at
the
given
point.
2
f
a
(x)
=
x
+
2x
+
1
at
(–3, 4)
b
f
(x )
=
d
f
(x )
=
2
x
+
4
at
x
=
1
2
x
8
+ 6
4
f
c
(x )
=
at
(3, 5)
x
+
at
x
3
Find
at
the
the
equation
given
x
=
1
x
for
the
normal
line
to
the
cur ve
point.
4
1
2
a
f
(x)
=
2x
–
x
–
3
at
(2, 3)
f
b
(x )
=
at
x
=
–1
2
x
x
4
3
2
c
f
(x)
=
(2x
+1)
at
(2, 25)
f
d
(x )
=
2
x
−
at
x
=
1
2
x
Exam-Style
4
Find
the
Questions
equations
for
all
the
ver tical
normal
lines
to
the
graph
of
3
f
5
(x)
The
=
x
–
3x
gradient
of
the
tangent
line
to
the
graph
of
2
f
(x)
=
2x
+
kx
–
3
at
x
=
–1
is
1.
Find
the
value
of
k.
Chapter
.
Y
ou
More
can
use
rules
a
GDC
for
to
derivatives
evaluate
a
derivative
of
a
function
1
3
at
a
given
value.
We
know
that
the
derivative
of
f
(x )
=
x
− 3x
4
3
3
2
2
is
f
′( x )
=
x
and
− 3
so
f
′( 4 )
=
(4 )
to
Click
Choose
and
the
display
the
enter
value
− 3
= 9
4
4
the
templates.
first-derivative
the
of
function,
template
variable
and
x
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
Since
line
the
to
calculator
is
approximate
using
the
a
value
Y
ou
can
it
will
graph
not
the
always
function
are
of
derivative
by
pressing
be
the
CD.
nd
the
derivative
exact.
and
menu
TI-84
the
at
a
of
x,
specic
use
the
value
context
find
menu
its
on
the
FX-9860GII
secant
T
o
derivative,
Casio
Alternative
for
:
show
of
its
the
point
to
coordinates,
dy
and
Analyze
Graph
|
5:
and
point
Y
ou
can
on
the
the
look
at
function
x-coordinate.
the
and
graphs
its
and
a
table
derivative.
To
of
values
graph f
and
f
time
the
first-derivative
template
to
write
the
and
derivatives
value
entering
the
Limits
a
of
will
x.
be
Y
ou
no
space
can
save
to
time
function.
by
there
′,
enter
use
the
graph.
This
for
edit
choosing
dx
the
then
f 1(x)
equation.
instead
of
re-typing
x
Investigation
–
the
derivatives
of
e
and
x
Use
1
a
GDC
Examine
to
the
functions
to
graph
f (x)
graphs
make
=
and
a
x
e
and
the
ln x
the
table
conjecture
of
derivative
values
about
the
of
for
f(x)
=
e
the
derivative
of
x
f(x)
=
Use
2
e
a
GDC
Examine
make
a
to
the
graph
f (x)
graphs
conjecture
=
and
ln x
the
about
and
table
the
the
of
derivative
values
derivative
of
for
f (x)
of
f (x)
the
=
=
ln x
functions
to
ln x
x
➔
Derivative
of
e
x
If
f
(x)
=
e
x
,
then
f
′(x)
=
x
e
Recall
y
=
that
ln x
are
y
=
e
and
inverses.
ln x
e
➔
Derivative
of
=
x
ln x
x
ln e
1
If
f
(x)
=
ln x,
then
f
′( x )
=
x
=
x
Example
Find
the
derivative
of
each
x
f (x)
a
=
The
function.
2
3e
f (x)
b
=
x
letter
e
ln x
c
f (x)
=
as
ln e
the
base
exponential
Answers
Use
x
f
(x)
=
the
constant
multiple
rule
and
3e
x
x
f ′(x)
=
fact
x
3 · e
=
that
the
derivative
of
e
(x)
=
x
+
is
e
Find
the
derivative
of
each
ter m.
2
1
f ′(x)
=
2x
2x
of
=
the
e
,
in
function
honor
Swiss
mathematician
3e
ln x
the
(x)
x
2
f
b
ofthe
x
f
a
isused
3x
+
Leonhard
Euler
(1707–83).
+ 1
Use
that
fact
that
the
derivative
of
or
+
1
x
x
ln
x
is
x
3x
c
f
(x)
f
′(x)
=
ln e
=
=
3x
Use
3
the
inverses
Then
Exercise
Find
the
fact
to
find
the
of
each
(x)
=
3
f
(x)
=
5
f
(x)
=
functions
are
first.
derivative.
function.
x
f
the
7H
derivative
1
that
simplify
4 ln x
2
f
(x) = e
4
f
(x)
=
6
f
(x)
=
x
+
4
2
3 x
ln e
ln 4 x
+
ln x
e
x
2e
+
3x
x
+
ln x
5e
+
1
x
+
4 ln e
Chapter
Write
an
equation
for
each
line
in
questions
7–10.
How
x
The
7
line
tangent
to
the
cur ve
f
(x)
=
are
exponential
functions
4e
–
7
at
x
=
used
in
ln 3
determining
the
2
x
8
The
normal
line
9
The
line
tangent
10
The
line
normal
to
the
f
cur ve
(x )
=
ln
(
e
)
at
the
point
concentration
(–3, 9)
drug
to
the
cur ve
f
(x)
=
ln x
at
x
2
Find
in
the
exact
questions
value
to
to
value
and
11
check
your
the
of
12
cur ve
the
and
f
(x)
=
derivative
then
use
a
e
in
a
a
patient’
s
body?
ln x
2x
at
=
of
+
the
e
–
given
GDC
to
3
at
x
value
find
an
=
2
of
x
approximate
work.
x
11
Find
f
′(3)
if
f
(x)
=
12
Find
f
′(8)
if
f
(x)
=
2e
−
5
3
Investigation
+
x
–
ln x
the
derivative
two
steps
1–4
let
u(x)
=
x
the
product
of
functions
4
For
of
7
,
v(x)
=
x
and
f(x)
=
u(x)·v(x)
n
1
The
function
2
F ind
f
can
written
as
f(x)
=
x
.
F ind
n
The
f ′(x).
3
F ind
u ′(x)
4
F ind
u ′(x) · v ′(x)
5
Is
f ′(x)
6
Using
ll
be
in
and
the
the
the
v ′(x).
of
same
three
as
If
u ′(x) · v ′(x)?
derivatives
blanks
derivative
functions
below
to
found
make
a
in
steps
true
2
and
3,
mathematical
the
f (x)
Is
a
is
two
=
of
the
the
sum
sum
of
of
the
two
derivatives
functions.
u(x)
+
similar
v(x)
rule
two
functions?
The
conjecture
then
true
f ′(x)
for
=
u′(x)
the
+
v ′(x).
product
of
statement.
4
f ′( x )
=
x
7
⋅ _______ +
x
⋅ _______
=
is
7
Complete
the
known
f(x)
=
as
the
the
investigation
product
rule.
Many
conjecture.
proofs
If
in
_______
u (x) · v (x)
then
f ′
(
x
)
=
____ ⋅ ____ +
are
straightforward,
but
the
____ ⋅ ____
proof
of
this
rule
uses
a
creative
step.
2
8
Use
help
the
function
conrm
f(x)
your
=
(3x
+
conjecture
1)(x
from
–
1)
step
to
7.
reject
or
Y
ou
an
can
research
example
complete
Limits
and
derivatives
of
the
a
the
proof
clever
proof.
and
step
nd
needed
to
4
For
functions
like
f
(x)
=
7
x
· x
2
and
f
(x)
=
(3x
+
1)(x
−
1)
you
can
Product
rewrite
the
function
But
other
and
use
the
power
as
(x)
r ule
to
take
the
The
for
functions
such
f
=
(3x
+
1)(ln x)
you
would
derivative
r ule
like
the
one
developed
in
the
conjecture
to
find
the
following
quotient
➔
of
The
r ules
two
are
used
to
find
the
derivative
of
the
product
or
the
the
functions.
product
f
(x)
=
then
f
′(x)
=
u(x) · v ′(x)
+
rst
factor
quotient
plus
factor
times
derivative
If
f
then
f
′( x )
the
factor
.
is
rule
derivative
quotient
v ( x ) ⋅ u ′( x ) − u ( x ) ⋅ v ′( x )
( x ) =
of
the
v(x) · u′(x)
rule
u( x )
times
the
factor
The
The
of
second
Quotient
➔
the
factors
second
the
rule
u(x) · v(x)
two
derivative
rst
If
of
derivative.
is
The
of
need
product
a
rule
derivative.
the
of
of
two
the
factors
denominator
=
2
v(x )
[v ( x
times
)]
of
the
the
minus
times
of
the
the
the
all
by
derivative
numerator
numerator
derivative
denominator
,
divided
denominator
Example
Find
the
of
each
(x)
=
c
f
(x)
=
squared.
function.
4
f
the
derivative
a
by
(3x
5x
+
1)(ln x)
b
f
(x)
=
d
f
(x)
=
(x
+ 3
x
2
3
+
3x
+
6)(2x
−
1)
+ 2
x
x
+ 1
2e
3
Answers
First
factor
Second
factor
f
a
f
(x )
=
(3 x
+ 1)
(x)
=
u(x) · v(x),
where
u(x)
=
3x
+
1
(ln x )
is
the
first
factor
and
v(x)
=
ln x
is
the
second
factor.
Derivative
of
First
second
Second
Derivative
Apply
the
product
rule.
factor
factor
⎛
f
′( x )
=
(3 x
1
+ 1) ⋅
+
⎜
⎝
3 +
(ln x ) ⋅
(3)
x
3x
u(x)
f
(x)
=
u(x)
u(x)
=
x
v ′(x)
+
v(x)
u ′(x)
+ 1 + 3x
ln
x
x
factor
4
b
=
or
=
′(x)
⎠
x
(x )
f
⎟
+ 3 ln x
First
f
first
1
=
of
⎞
Second
factor
3
(x
+ 3x
+ 6)
(2 x
− 1)
v(x),
4
First
where
3
+
3x
+
6
is
the
first
factor
Derivative
factor
of
4
f
′( x )
=
second
and
+ 3x
+ 6) ⋅
(2)
Apply
=
f
of
factor
2x
–
1
is
− 1) ⋅ ( 4 x
4
(2 x
the
second
factor.
4
′(x)
=
u(x)
v ′(x)
+
v(x)
u ′(x).
first
2
+ 9x
)
+ 12 ) +
3
− 4 x
Expand
3
+ 18 x
4
= 10 x
rule.
3
+ 6x
(8 x
product
3
+ (2 x
the
Derivative
Second
=
v(x)
3
(x
3
+ 20 x
the
brackets.
2
− 9x
)
2
− 9x
+ 12
Simplify.
{
Continued
on
next
Chapter
page
u( x )
5x
+ 3
f(x)
f
c
(x)
=
,
where
u(x)
=
5x
+
3
is
the
numerator
=
2
v(x )
x
+ 1
2
f
′(x)
and
=
v(x)
Derivative of
Apply
Denominator
numerator
=
x
+
1
is
the
denominator.
Derivative of
Numerator
the
quotient
rule.
denominator
(5 x
(2 x )
2
v ( x ) ⋅ u ′( x ) − u ( x ) ⋅ v ′( x )
(x
+ 1)
⋅
(5)
−
+ 3)
f
2
′(x)
=
2
2
(x
+ 1)
[v ( x )]
Denominator
2
squared
2
(5 x
+ 5 ) − (10 x
+ 6x )
Expand
the
numerator
so
that
you
can
combine
like
=
2
2
(x
+ 1)
ter ms.
Do
not
expand
the
denominator.
2
−5 x
− 6x
+ 5
Simplify.
=
2
2
+ 1)
(x
x
+ 2
u( x )
f
d
(x)
=
f
x
2e
(x)
=
,
where
u(x)
=
x
+
2
is
the
numerator
and
3
v(x )
f
x
′(x) =
v(x)
numerator
2e
–
3
is
the
denominator.
Derivative of
Derivative of
Denominator
=
Numerator
denominator
Apply
x
the
quotient
rule.
x
(2e
3)
⋅
(1)
−
(x
+ 2)
(2e
)
2
v ( x ) ⋅ u ′( x ) − u ( x ) ⋅ v ′( x )
x
( 2e
3
)
f
′(x)
=
2
[v ( x )]
Denomi nator
x
squared
x
( 2e
− 3) − ( 2 xe
x
+ 4e
)
Expand
the
numerator
so
that
you
can
combine
=
2
x
ter ms.
( 2e
3
x
Do
not
expand
)
x
−2 xe
− 2e
− 3
=
2
x
( 2e
3
)
Simplify.
Exercise
Find
the
7I
derivative
of
each
function
in
questions
1
to
8.
2
x
3
1
f
(x) =
x
2
f
(x)
= (2x
4
f
(x)
=
ln x
3
f
2
+
x
2
+
x)(x
+
4
x
(x) =
e
ln x
x
x
5
f
x
2
x
+ 4
e
(x) =
f
6
(x) =
x
e
+ 1
2
2
x
7
f
(x) =
x
3
e
(5x
+
4x)
f
8
(x)
=
3
x
Exam-Style
+ 1
Questions
x
9
The
function
Find
10
(x)
the
=
equations
x
+ 1
x
1
that
=
Limits
(x)
xe
has
a
horizontal
tangent
line
at
x
k
Write
f
f
and
are
derivatives
for
the
parallel
tangent
to
the
lines
line
x
to
+
the
2y
=
graph
10
of
=
k
1)
the
denominator.
like
The
product
quotients.
before
do
is
quotient
r ules
sometimes
are
more
not
needed
convenient
to
all
products
the
and
function
the
derivative.
If
it
is
more
convenient
to
rewrite
the
function
first,
so.
3x
2
a
for
rewrite
differentiating.
Example
Find
and
It
f
(x )
=
x (4 x
− 2x )
b
f
(x )
+
4
=
2
x
2
2
3x
9
c
f
(x )
=
d
f
(x )
+
2x
+ 1
=
2
3
4
x
x
Answers
2
f
a
(x )
=
x (4 x
− 2x )
1
2
2
=
x
(4 x
3
2
=
Rewrite
− 2x )
5
Use
the
4 ⋅
−1
3
2
x
− 2 ⋅
x
2
3
1
2
2
= 10 x
3x
f
(x )
− 3x
+
constant
multiple
and
2
power
b
exponents
3
−1
2
=
rational
− 2x
5
′( x )
expand.
2
4 x
5
f
using
and
rules
to
and
simplify.
Use
the
find
the
derivative
4
=
2
x
2
2
(x
f
′( x )
− 2 ) ⋅ (3 ) − (3 x
+
4 ) ⋅ (2 x )
=
quotient
rule.
2
2
(
x
2
2
)
2
(3 x
− 6 ) − (6 x
+ 8x )
=
2
2
(
x
2
)
2
−3 x
− 8x
− 6
=
2
2
(
x
2
)
4
9
3
c
f
(x )
=
=
3
9x
Rewrite
using
Rewrite
by
rational
exponents.
4
x
4
−
4
−1
3
f
′( x )
=
9 ⋅ −
x
3
7
12
3
=
−12 x
=
−
7
3
x
2
3x
f
d
(x )
+
2x
+ 1
=
2
x
2
3x
=
2x
+
1
2
2
x
2
x
′( x )
=
3 + 2x
+
0 − 2x
−2
=
then
use
rational
exponents.
x
−3
− 2x
2
−2 x
− 2
or
−
2
x
and
−2
−2
f
ter ms
x
−1
=
separating
+
3
x
3
x
Chapter
We
have
been
using
the
‘prime’
notation, f
dy
We
can
use
Leibniz
′(x),
to
denote
derivatives.
d
notation,
[ f
or
( x )]
and
we
can
also
use
dx
dx
dy
variables
other
than
x
and
y.
The
is
notation
read
as
‘the
dx
derivative
of
y
with
respect
to
x’,
or
‘d
y
by
d
x’,
or
simply
derivative
of
f
‘d
y
d
x’.
d
The
[ f
notation
( x )]
is
read
as
‘the
with
dx
respect
to
x’.
Example
[
Gottfried
Leibniz
d
[ (ln
Find
a
x )(7 x
2
2) ]
s (t )
If
b
=
(4t
− 1)
German
ds
2
,
Wilhelm
(1646–1716),
a
mathematician,
find
debated
dx
with
Isaac
dt
Newton
over
who
dA
2
If
c
A
πr
=
,
was
find
dr
the
develop
r =3
It
is
rst
to
calculus.
widely
believed
Answers
that
d
Use
[ (ln
x )(7 x
the
product
rule
to
find
the
Leibniz
Newton
2) ]
a
derivative
dx
of
(ln x)(7x
−
2)
developed
with
about
⎛
=
(ln x )(7 )
+
(7 x
1
⎝
x
ln
+ 7x
respect
⎞
to
x.
− 2)
⎜
7x
⎟
x
⎠
− 2
=
x
2
s (t )
b
=
2
(4t
Expand
− 1)
4
= 16t
2
− 8t
find
the
and
use
the
derivative
power
of
s
rule
with
to
respect
+ 1
to
t.
ds
3
=
64 t
− 16t
dt
2
A
c
2
r
Find
the
respect
dA
derivative
to
of
πr
with
r.
2 r
dr
The
dA
bar
tells
you
to
evaluate
the
2 (3)
derivative
dr
when
r
=
3
r 3
6
Exercise
7J
Differentiate
convenient
each
to
function
rewrite
the
in
questions
function
first,
1
to
do
12.
If
it
is
so.
3
2x
1
f
(x )
5x
2
=
2
f
(x )
=
4
f
(x )
=
(x
x
x
f
(x )
=
2e
2e
2
(x
)
2
x
2
(x )
= e
x
4
3
ln
5
x
+
6
f
(x )
=
x
5
4
x
Limits
and
derivatives
e
2
− 5 )( x
3
3
and
independently
+ 5)
more
the
calculus
same
at
time.
2
x
f
7
(x )
=
2
x
+ 1
8
f
(x )
= 3x
9
f
(x )
=
ln x
2
x
− 2x
+ 1
x
2
10
f
(x )
=
11
f
(x )
=
x (x
+ 1)
x
2
x
− 2x
+ 1
3
12
f
(x )
=
2
(x
− 3 x )(2 x
Exam-Style
13
Write
the
f
(x )
14
Write
the
+ 5)
Questions
equation
x
of
+ 3x
=
of
the
line
normal
to
the
graph
x
xe
− e
at
equation
x
of
=
the
1
tangent
line
to
the
graph
of
3
f
(x )
=
x
ln x
at
x
=
1
dc
2
15
If
c (n )
=
−4 .5n
+ 3 .5n
− 2,
find
dn
4
dA
3
16
If
A
=
πr
,
find
3
dr
dv
2
17
v (t )
If
=
2t
− t
+ 1,
find
dt
t =2
Exam-Style
Question
d
t
t
⎡ (e
18
)( t
+ 3) ⎤
⎣
can
be
written
as
e
(t
+ k ).
Find
k
⎦
dt
.
The
chain
rule
and
higher
order
The
symbol
is
used
°
to
show
a
composite
derivatives
3
function.
The
power
r ule
alone
will
not
give
the
correct
derivative
If
u( x )
=
x
for
and
v ( x)
=
2
−
x,
then
3
f
(x )
=
rather
(2 −
a
x )
.
power
This
of
is
because
another
the
function
function v ( x )
=
is
2 −
not
x.
a
The
power
of
function
x,
f
but
is
a
f ( x)
=
(u
=
u(v ( x ))
=
u
v )( x )
3
composite
v(x )
=
2 −
function,
(u v )( x )
or
u ( v ( x )),
where
u( x )
=
x
and
(2
−
x
)
3
x
=
(2
−
x
)
Chapter
Investigation
–
finding
the
derivative
composite
of
a
function
3
Let
1
f (x)
=
(2
−
x)
3
a
Expand
b
Y
ou
f (x)
derivative
=
of
(2
−
x)
Differentiate
each
term
to
nd
the
f
3
can
also
nd
the
derivative
of
f (x)
=
(2
−
x)
by
applying
3
the
power
Compare
rule
the
to
(2
−
x)
following
and
to
multiplying
your
answer
in
by
another
step
1
factor
.
and
nd
the
2
If
u(x)
=
x
and
2
missing
factor :
f ′(x)
=
3(2
−
x)
._____
v(x)
=
2x
+
1,
then
2
Repeat
2
the
process
for
f (x)
=
(2x
+
1)
f ( x)
a
Expand
b
Apply
f
and
nd
the
=
u(v ( x ))
=
u(2 x
=
(2x
derivative.
+ 1)
2
f ′(x)
the
=
power
2(2x
+
rule
to
(2x
+
1)
to
the
a
Expand
b
Apply
process
f
the
missing
factor :
2
and
for
nd
f (x)
the
=
(3x
2
+
1)
derivative.
2
If
2
the
power
rule
to
(3x
u(x)
=
2(3x
+
1)
to
nd
the
missing
factor :
Make
a
x
and
2
=
3x
+
1,
then
1)._____
f ( x)
4
=
2
+
v(x)
f ′(x)
+ 1)
1)._____
2
Repeat
3
nd
conjecture
about
nding
the
derivative
of
=
u(v ( x ))
=
u(3 x
a
2
composite
+ 1)
function.
2
4
Verify
5
To
find
the
➔
the
chain
The
If
that
f
your
conjecture
derivative
of
a
works
for
composite
f (x)
=
(x
function
2
+
x
we
=
3
2
(3 x
+ 1)
)
use
r ule.
chain
(x)
=
rule
Chain
u(v(x))
then
f
′(x)
=
The
u ′(v(x)) · v′(x)
rule
derivative
composite
is
➔
The
chain
r ule
can
also
be
written
the
dy
y
=
f
(u),
u
=
g (x)
and
y
=
f
(g(x)),
dy
=
then
derivative
outside
du
du
respect
function
function
the
function
is
in
the
form
f
(x)
=
u(x)
and
v(x),
then
find
the
derivative
of
derivative
respect
(x )
=
2
6
4 (5 x
+ 2)
f
b
(x )
=
4 x
x
+ 1
f
c
(x )
=
e
Answers
3
a
f
6
( x ) = 4(5 x
+ 2)
6
u( x ) = 4 x
u
is
the
outside
function.
3
v ( x ) = 5x
+ 2
v
3
f
′( x ) = 24(5 x
5
is
the
(
15 x
Apply
Derivative
with
chain
rule.
Derivative of
of
function
function.
)
outside
inside
2
+ 2)
inside
with
respect
function
respect
to
x
to inside
function
2
= 360 x
3
(5 x
5
+ 2)
Simplify.
{
Limits
and
derivatives
of
function
f.
2
3
f
remains
multiplied
the
by
the
u(v (x))
inside
a
the
(inside
same),
Identify
to
dx
inside
Each
of
function
⋅
with
dx
Example
a
as:
the
If
of
function
Continued
on
next
page
to
x
with
2
f
b
(x ) =
4x
Rewrite
+1
using
rational
exponents.
1
2
2
= (4 x
+ 1)
1
2
u(x ) =
x
u
is
the
outside
function.
2
v (x ) =
4x
v
+1
is
the
inside
function.
1
1
2
f
′( x ) =
2
(4 x
+ 1)
Apply
(8 x )
the
chain
rule.
2
Derivative of
function
with
Derivative
of
outside
respect
inside
function
to inside functi
ion
respect
4 x
with
to
x
4 x
=
or
Simplify.
1
2
2
(4 x
4 x
2
+ 1
+ 1)
2
x
f
c
(x )
= e
2
( x
)
= e
x
u( x )
= e
u
is
the
outside
function.
2
v(x )
=
v
x
is
the
inside
function.
2
(
f
′( x )
=
x
)
e
the
chain
rule.
Derivative
outside function
with
Apply
(2 x )
Derivative of
respect
of
to
inside
function
inside function
respect
with
to
x
2
x
=
2 xe
Exercise
Each
7K
function
Identify
Simplify.
u(x)
is
in
and
the
v(x),
4
f
1
(x )
=
form
then
f
(x )
find
=
the
u ( v ( x ))
derivative
5
(3 x
f
(x )
=
ln(3 x
2
f
(x )
=
4
f
(x )
=
f
(x )
6
f
(x )
=
8
f
(x )
=
10
f
(x )
= e
3
4 (2 x
+ 3x
+ 1)
3
)
2x
+ 3
4 x
5
f.
2
+ 2x )
5
3
of
3
= e
(ln x )
2
f
(x )
=
(9 x
2
4
3
7
+ 2)
2x
+ 3
3
3
9
f
Y
ou
(x )
= 5( x
can
find
rewriting
Maria
the
Agnesi
published
calculus
a
of
4
4 x
+ 3x )
the
derivative
function
into
(1718–99),
text
both
on
a
an
calculus
Isaac
of
some
form
Italian
that
Newton
functions
where
can
efficiently
apply
the
chain
by
rule.
mathematician,
included
and
you
more
the
Gottfried
methods
of
Leibniz.
3
a
Maria
also
studied
cur ves
of
the
form
y
whose
=
2
x
graphs
came
to
be
known
as
witches
of
2
+
a
Agnesi.
The
function
1
f ( x)
in
=
Example
13
is
an
example
of
such
a
graph.
2
x
+ 1
Chapter
Example
1
Use
the
chain
r ule
to
find
the
derivative
of
f
(x )
=
2
x
+ 1
Answer
1
f
(x )
=
Rewrite
2
x
2
′( x )
rational
exponents.
1
+ 1)
=
(x
=
−1( x
2
f
using
+ 1
2
+ 1)
⋅ 2x
Apply
the
chain
rule.
2x
=
−
2
2
(x
For
some
product
Simplify.
+ 1)
functions
or
the
quotient
Example
chain
r ule,
or
r ule
the
must
chain
be
r ule
combined
may
need
with
to
be
4
f
(x )
repeated.
2
a
the
=
x
1 −
2 (3 x
x
f
b
(x )
=
x
⎛
1)
e
f
c
(x )
=
⎞
ln
⎜
⎝
⎟
2
+ 1
x
⎠
Answers
1
2
a
f
(x ) =
x
1 −
2
x
=
x (1 −
2
x
Rewrite
)
using
rational
exponents.
1
1
2
f
′( x ) =
x
(1 −
x
2
)
( −2 x )
2
First factor
Derivative of
using
second
chain
Apply
chain
the
product
rule
to
find
rule,
the
using
the
derivative
of
factor
the
rule
second
factor.
1
2
(1
+
2
x
)
1
Derivative of
Second factor
first
factor
1
2
x
2
+ (1
=
x
Simplify.
2
)
1
2
(1
2
x
)
1
1
2
2
x
(1
2
+ (1
=
x
x
2
)
Find
2
)
2
(1
2
2
x
)
(1
2
x
a
common
denominator.
1
1
x
2
)
2
+ (1
x
)
=
1
2
(1
x
2
)
2
1
2
2x
1
=
2x
Simplify.
or
1
2
2
(1
1
2
x
x
)
4
2 (3 x
b
f
(x )
=
e
u( x )
=
e
1)
The
are
x
outside
shown.
and
inside
Note
the
functions
inside
function
4
v(x)
=
2(3x
−
1)
is
the
composition
4
v(x )
=
2 (3 x
− 1)
4
of
2x
and
3x
−
1.
4
2(3 x
f
′( x )
=
1)
3
e
8(3 x
Derivative of
1)
(3)
chain
rule
to
f
and
apply
the
outside function
inside function
respe
ect
the
the
Derivative of
with
Apply
to
the
res pect
to
it
again
when
finding
the
derivative
with
x
of
inside function
the
inside
function.
4
2(3 x
3
= 24(3 x
1)
1)
e
{
Limits
and
derivatives
Continued
on
next
page
x
⎛
f
c
(x )
=
⎞
ln
⎜
⎟
2
+ 1
x
⎝
⎠
2
1
(x
+ 1) ⋅ 1
x
⋅ (2 x )
Apply
f
′( x )
the
chain
rule
and
use
the
=
2
x
2
(
2
x
x
+ 1
)
quotient
rule
to
find
the
derivative
+ 1
Derivative of
Derivative of
the
of
the
the
inside
function.
inside function
outside function
with
with
respect
respect
x
to the
inside function
2
2
x
+ 1
2
x
+ 1
2x
=
Simplify.
2
2
x
(
x
+ 1
)
2
1
x
=
2
x (x
Exercise
Find
the
+ 1)
7L
derivative
of
2
1
f
(x )
=
f
(x )
=
x
each
function
in
questions
4
(2 x
2
− 3)
2
f
(x )
x
=
x
(x )
=
+ 3
2x
2 x
5
f
(x )
=
f
(x )
=
+ 1
2 x
e
3
+ e
f
6
(x )
=
ln(1 − 2 x
ln(ln x
)
f
8
(x )
=
x
x
e
1
9
f
(x )
)
2
2
7
10.
x
f
4
2
x
to
e
4
3
1
+ e
4
=
10
f
(x )
=
2
x
x
+ 3
2
x
− 3x
Exam-Style
− 2
Questions
2
x
11
12
For
the
the
a
Find
c
Hence
Find
the
f
f
cur ve
(x )
′(x).
Find
b
find
the
2 x
= e
equation
x-coordinate
of
the
of
f
′(2).
the
tangent
point(s)
on
line
the
to f
graph
when
x
=
2
of
3
f
(x )
=
x
ln x
where
the
tangent
line
is
horizontal.
1
13
Let
f
(x )
=
,
g(x )
= 1 − 2x
and
h( x )
=
(
f
g )( x )
3
x
Find
h (x)
and
show
that
the
gradient
of
h (x)
is
always
positive.
x
f (x)
g (x)
f ′(x)
g ′(x)
3
1
4
−3
2
4
2
−1
3
4
In
at
the
x
=
table
3
and
above,
x
=
4
the
are
values
of
f
and
g
and
their
derivatives
given.
a
Find
the
gradient
of
b
Find
the
gradient
of
(
f
g )( x )
when
x
=
3
1
when
x
=
4
2
[ g ( x )]
Chapter
Higher
order
derivatives
The
second
derivative
is
the
derivative
dy
The
derivative
f
′(x)
or
is
called
the
first
of
derivative
the
dx
d
as
of
y
with
respect
to
x.
We
are
sometimes
interested
rst
⎢
in
dx
derivative.
Writing
this
⎡ dy ⎤
helps
⎥
you
see
where
the
⎣ dx ⎦
2
the
gradient
of
the
first
derivative.
This
is
known
as
d
the
y
comes
notation
from.
2
dx
second
derivative
written
as
of
y
with
respect
to
x
and
can
be
2
d
f
″(x)
y
or
.
The
third
derivative
of
y
with
2
dx
The
3
d
respect
to
x
is
written
as
f
″′(x)
‘prime’
notation
is
not
ver y
useful
y
or
.
The
second
and
for
derivatives
of
order
higher
than
3
dx
three.
third
derivatives
are
examples
of
higher
order
derivatives
For
those
derivatives
(x).
For
example,
instead
(4)
f ″′′(x)
Find
b
If
we
4
a
the
first
three
derivatives
of
f
(x )
=
2
x
+ 3x
+
x
3
d
′( x )
=
x
2 x
2
f
+ 4 ,
x
find
f
′′( x ) .
If
c
y
=
4e
,
find
3
dx
2
d
2
d
s (t )
If
=
−16t
+ 16t
+ 32,
x =1
s
find
2
dt
Answers
4
a
f
2
(x )
=
x
+ 3x
′( x )
=
4 x
+
The
x
f
3
f
+ 6x
first
′ ( x ),
three
f
′′ ( x )
derivatives
and
f
are
′′ ( x )
+ 1
2
f
′′( x )
f
′′′( x )
f
′( x )
= 12 x
=
+ 6
24 x
2
b
=
x
Note
+ 4
2
=
(x
that
given,
1
so
the
you
first
derivative
only
was
dif ferentiate
once
2
+ 4)
to
get
the
second
derivative.
1
1
2
f
′′( x )
=
(x
2
+ 4)
(2 x )
2
x
=
2
x
+
4
2 x
=
y
c
4e
Find
the
first
three
derivatives
using
dy
2 x
=
4e
=
8e
2 x
⋅ 2
=
8e
the
chain
rule.
dx
2
y
d
2 x
2 x
⋅ 2
=
16e
2
dx
3
y
d
2 x
16e
=
2 x
⋅ 2
=
32e
=
32e
3
dx
3
d
Then
evaluate
when
x
the
third
derivative
y
2 (1)
=
32e
2
3
=
1
dx
x =1
2
d
s (t )
=
−16t
+ 16t
+ 32
=
−32t
+ 16
dt
2
d
s
=
−32
2
dt
Find
the
first
derivative
ds
Limits
and
derivatives
of
and
s
then
with
write
(n)
f
Example
we
the
respect
second
to
t.
write
f
(x).
of
writing
Exercise
7M
3
2
1
Find
2
If
3
If
the
second
derivative
5
f
(x )
f
of
(x )
=
4 x
4
= 3x
+
x
+ 2x
+ 1
,
find
f
′′′( x )
2
d
C
3n
C (n )
=
,
(3 + 2 n )e
find
2
dn
3
dy
d
4
=
If
4
,
y
find
3
dx
x
dx
6
4
d
d
y
y
3
=
If
5
ln( 4 x
) ,
find
6
4
dx
dx
dR
1
2
If
6
R (t )
=
t
),
ln( t
find
dt
2
t = −1
Exam-Style
Questions
3
What
7
for
n
is
≥
tr ue
about
the
nth
derivative
of
y
=
2
+ 3x
+ 2x
+ 4,
4?
x
Find
8
x
the
first
four
derivatives
y
of
= e
x
+ e
then
write
a
n
d
generalization
for
y
finding
of
this
function.
n
dx
1
Find
9
the
first
four
derivatives
y
of
=
then
write
a
generalization
x
n
d
for
y
finding
of
this
function.
n
dx
5
10
Find
.
The
the
Rates
of
section
we
of
diver
the
in
the
the
variable
will
motion
of
slope
change
gives
one
Example
A
of
derivative
change
and
gradient
slope
with
of
the
and
of
a
function
motion
function.
respect
to
It
f
in
also
another
(x )
a
2
= 3
x
line
gives
variable.
the
In
rate
of
this
study average and instantaneous rates of change
a
line
jumps
diver
from
above
a
platform
water
level
at
at
time
time
t
t
=
is
0
seconds.
given
The
distance
by
2
s (t )
a
=
−4.9 t
Find
the
+ 4 .9 t
+ 10 ,
average
where
s
veloci ty
is
of
in
metres.
the
diver
over
the
given
time
inter vals.
i
b
[1, 2]
Find
the
ii
[1.5, 1]
instantaneous
iii
[1.1, 1]
veloci ty
of
iv
the
[1.01, 1]
diver
{
at
t
=
1
second.
Continued
on
next
page
Chapter
The
Answers
of
a
Average
velocity
is
average
change
in
rate
s,
or
the
average
change
of
velocity,
is
the
(metres)
distance
−1
The
change
in
units
for
velocity
are
slope
m s
of
a
secant
line:
(seconds)
time
s(t
+
h)
−
s(t )
s(t
+
h)
−
s(t )
=
(t
s (2)
+
h)
− t
h
s (1)
1
i
=
2
−9
8 ms
The
Find
1
the
slopes
of
the
secant
instantaneous
lines
rate
s (t
)
s (t
2
s (1.5 )
on
1
=
ii
−7
35 m s
t
5
each
inter val.
Use
a
or
the
change
velocity,
slope
GDC
to
evaluate
the
of
a
v (t )
s (1)
1
=
1
1
−5
s (1.01)
s (1)
1
=
1
b
01
=
s ′( t )
949 m s
1
Instantaneous
v (t)
−4
velocity
s ′(t)
=
−9.8t
−9.8 + 4.9
=
Find
the
s
=
at
t
slope
of
the
tangent
line
to
1
Note that the slopes of
+ 4 .9
the secant lines
in part a approach the slope of
1
s ′(1)
the
−4.9 m s
=
tangent line in part b
Example
During
one
month,
the
temperature
of
the
water
in
a
pond
is
modeled
t
3
by
the
function
measured
a
Find
15
b
in
the
days
Find
degrees
average
of
the
C (t )
the
rate
=
20 + 9 te
,
where
t
is
measured
in
days
and
C
is
Celsius.
rate
of
change
in
temperature
in
the
first
month.
of
change
in
temperature
on
day
15.
Find
the
slopes
Answers
a
Average
rate
C (15 )
of
change
C (0 )
=
line
≈
15
0
on
the
of
the
inter val
secant
[0, 15]. The
0606 °C/day
change in temperature
0
units
for
are
change in time
°C
b
Instantaneous
rate
of
change:
t
⎛
=
9t
⎜
1 ⎞
−
e
⎟
+ e
⋅ 9
3
⎠
t
t
−
−
3
−3te
3
+ 9e
−5
C ′(15) = − 3 ⋅ 15e
−5
+
9e
5
= − 36e
≈ − 0
On
is
day
degrees
15
Limits
and
243 °C/day
the
dropping
at
Find
C
−
3
⋅ −
⎝
=
/ day
the
slope
t
3
C ′( t )
temperature
a
Celsius
rate
per
derivatives
of
0.243
day
.
=
at
t
=
15.
of
the
tangent
line
to
+
h)
lim
h→0
39 m s
1
iv
tangent
line:
slopes.
s(t
s (1.1)
s,
the
1
1
iii
of
is
t
2
1
of
)
1
s (1)
−
s(t )
=
h
s
(t )
Exercise
Use
a
7N
GDC
to
Exam-Style
1
A
ball
is
ground
help
function
values.
Question
thrown
t
evaluate
ver tically
seconds
after
it
upwards.
is
thrown
Its
is
height
in
modeled
metres
by
the
above
the
function
2
h (t )
=
−4 .9 t
Find
the
height
b
Find
the
average
t
seconds
c
=
0
Find
the
the
ball
The
+ 1 .4
of
to
the
rate
t
=
ball
of
2
t
what
amount
of
=
1
when
change
rate
second,
these
water
t
=
of
0
seconds
the
height
and
of
when
the
ball
t
=
2
seconds.
from
seconds.
instantaneous
when
Explain
2
+ 19 .6 t
a
values
in
a
of
t
change
=
tell
tank
2
of
the
seconds
you
about
after t
height
and
the
minutes
t
=
3
of
seconds.
motion
is
of
the
modeled
ball.
by
2
t
⎛
the
function V ( t )
=
4000
⎞
,
1−
⎜
Answer
the
following
Find
the
amount
a
when
b
Find
tank
t
=
the
20
from
c
Find
tank
d
The
that
t
the
rate
t
your
=
0
number
t
is
measured
in
litres.
in
whole
the
tank
number.
when t
=
0
minutes
and
=
of
0
=
the
change
minutes
of
to
t
the
=
amount
20
of
minutes.
water
in
Explain
the
the
answer.
20
rate
minutes.
amount
minutes
of
nearest
water
instantaneous
when
Show
from
3
the
when
of
of
V
⎠
minutes.
average
meaning
to
where
⎟
60
⎝
to
bacteria
of
change
Explain
water
t=40
in
of
a
in
of
the
the
the
amount
meaning
tank
is
of
never
of
water
your
in
the
answer.
increasing
minutes.
science
experiment
on
day t
is
0.25t
modeled
a
Find
over
b
Find
by
the
the
Find
The
rate
0
to
of
10
instantaneous
at
the
cost
100e
average
the
bacteria
4
=
inter val
bacteria
c
P (t)
any
time
(in
day
10.
dollars)
days
of
the
the
number
of
bacteria
experiment.
rate
of
change
of
the
number
of
rate
of
change
of
the
number
of
Explain
of
of
t.
instantaneous
on
change
the
producing
meaning
n
units
of
of
a
your
answer.
product
is
modeled
2
by
the
function
a
Find
the
when
b
average
the
=
0.05n
rate
production
of
=
105
units
and
when
n
=
100
units
to
=
Find
the
any
Find
when
n
101
instantaneous
number
the
n
of
=
100
units.
+
of
changes
the
5000
C
with
from n
production
respect
=
level
100
to
n
units
changes
to
from
units.
rate
units
instantaneous
+10n
change
level
n
for
c
C (n)
of
change
of
C
with
respect
to
n
of
change
of
C
with
respect
to
n
n.
rate
Explain
the
meaning
of
your
answer.
Chapter
Motion
If
an
in
object
a
line
moves
along
a
straight
line,
its
position
Y
ou
from
an
origin
at
any
time
t
can
be
modeled
by
line
function ,
displacement
s(t).
The
can
use
a
horizontal
or
ver tical
a
to
model
motion
in
a
line.
function
For
s(t)
>
0,
the
object
is
to
the
right
2
s(t)
=
+
−4.9t
4.9t
+
10
from
Example
16
is
an
example
of
of
a
displacement
diver
The
is
position
origin
diving
➔
the
is
at
when
water
platform
The
function.
10
t
The
=
level,
metres
instantaneous
0,
so
ini tial
or
the
above
rate
of
s(0)
posi tion
=
diver
water
10
is
of
the
metres.
initially
the
s(t)
<
origin
on
a
origin
0,
or
position
or
the
below
is
above
object
the
origin.
s (t
=
change
+ h) −
s
of
displacement
is
For
the
v(t)
=
For
s ′( t )
is
is
moves
after
by
in
a
s (t)
straight
leaving
=
2t
a
to
object
the
right
v(t)
<
0,
the
down
object
or
to
left.
v(t)
at
=
0,
the
object
rest.
ini tial
veloci ty
v(0).
line
fixed
with
point.
a
displacement
The
of
displacement
s
metres
function
2
−
21t
+
60t
+
3,
for
t
≥
0.
This
a
Find
the
velocity
b
Find
the
initial
of
the
par ticle
at
any
time
is
an
area
position
and
initial
velocity
of
the
Find
when
the
par ticle
is
at
d
Find
when
the
par ticle
is
moving
kinematics,
rest.
which
left
and
when
the
par ticle
is
a
about
the
is
motion
moving
known
par ticle.
as
c
of
t
mathematics
Draw
the
3
e
0,
moving
The
given
initial
up.
For
seconds
>
moving
the
is
the
h
is
t
The
For
of
(t )
lim
h →0
particle
left
level.
or
A
origin.
the
s(0).
is
Example
the
to
function ,
veloci ty
(t )
is
of
objects.
right.
motion
diagram
for
the
par ticle.
Answers
a
v (t )
=
s ′( t )
v (t )
=
6t
s (0 )
=
2( 0 )
Velocity
is
the
derivative
of
displacement.
2
− 42 t
+ 60,
3
b
t
≥
0
2
− 21( 0 )
The
+ 60 ( 0 ) + 3
=
3 m
when
2
v (0)
=
6(0)
6t
− 42t
t
=
position
is
the
displacement
0.
1
−
42(0)
+
60
=
60 ms
2
c
initial
+ 60
=
The
initial
velocity
The
par ticle
is
at
is
rest
the
velocity
when
when
velocity
is
t
=
0.
0.
0
Set
the
velocity
function
equal
to
0
and
solve
for
t.
2
6( t
− 7t
6( t
+ 10 )
− 2 )( t
− 5)
t
The
par ticle
is
=
0
=
0
=
2,
at
5
rest
at
2
seconds
and
5
seconds.
{
Limits
and
derivatives
Continued
on
next
page
signs
d
of
+
v
Make
+
t
0
The
2
par ticle
(5, ∞)
moving
right
because
for
(2, 5)
3
e
s (2)
=
v (t)
for
>
(0, 2)
− 21( 2 )
3
t
=
=
seconds
0.
The
par ticle
because
=
0
t
=
+ 60 (5 ) + 3
v (t)
<
the
(0 , 2)
t
=
1
v(1) =
6(1 − 2)(1 − 5) =
(+)( − )( − ) = +
(2 , 5)
t
=
3
v(3) =
6(3
(+)(+)( − ) =
55
28
the
find
diagram.
− 2)(3
the
sign
− 5) =
of
Choose
a
v(t).
−
t
=
v(6)
6
=
6(6
−
2)(6
− 5)
=
(+)(+)(+)
=
+
the
displacement
or
position
of
the
par ticle
m
the
par ticle
changes
direction.
=
28
m
=
Use
these
plot
the
positions
motion.
and
the
Although
initial
the
position
motion
to
is
2
actually
on
the
line,
we
draw
it
above
the
line.
on
a
line
with
displacement
function
2
−
Find
a
and
55
moves
3
t
inter val
7O
par ticle
=
each
is
s
s (t)
on
values
0.
5
t
A
rest
the
in
Find
+ 60 ( 2 ) + 3
− 21(5 )
0 3
1
at
Put
2
2 (5 )
Exercise
is
velocity.
value
when
s (5)
par ticle
for
and
2
2( 2 )
diagram
when
(5, ∞)
left
sign
5
is
seconds
moving
a
6t
the
+
9t
centimetres
initial
position
for
and
t
≥
the
0
seconds.
initial
velocity
for
the
par ticle.
b
Find
when
c
Draw
a
A
ball
is
par ticle
motion
Exam-Style
2
the
is
diagram
at
rest.
for
the
par ticle.
Question
thrown
ver tically
upwards.
The
height
of
the
ball
in
feet,
2
t
seconds
t
≥
0
a
Find
b
Show
is
c
after
it
is
released,
is
given
by s (t)
=
−16t
+
40t
+
4
for
seconds.
20
the
that
is
a
of
the
Solve
ii
height
of
of
the
the
ball.
ball
after
2
seconds
second
time
when
the
height
of
the
ball
is
feet.
Write
i
the
height
feet.
There
20
initial
down
ball
the
an
is
equation
20
that t
must
satisfy
when
the
height
feet.
equation
algebraically
.
ds
d
i
Find
ii
Find
the
iii
Find
when
iv
Find
the
dt
3
A
par ticle
initial
velocity
the
velocity
maximum
moves
along
a
of
of
height
line
the
the
of
with
ball.
ball
the
is
0.
ball.
displacement
function
t
s (t )
=
,
where
s
is
in
metres
1
t
and
t
is
in
seconds.
t
e
a
Show
that
(t )
=
t
e
b
Hence
find
when
the
par ticle
is
at
rest.
Chapter
➔
The
instantaneous
rate
of
change
of
v (t
function ,
acceleration
a (t )
=
the
velocity
+ h) − v
= v ′( t )
a(t)
>
0
the
velocity
of
the
object
is
increasing.
For
a(t)
<
0
the
velocity
of
the
object
is
decreasing.
For
a(t)
=
0
the
velocity
is
the
=
s ′′( t )
h
For
For
the
(t )
lim
h →0
Example
is
constant.
displacement
3
function
from
Example
18,
2
s(t)
=
we
found
2t
−
21t
+
60t
+
3,
with
s
in
metres
and
t
≥
0
seconds,
2
a
b
that
v (t)
Find
the
t
seconds
=
1
to
Find
the
seconds.
3
6t
average
t
=
=
t
−
42t
+
60.
acceleration
=
4
of
the
par ticle
from
seconds.
instantaneous
Explain
acceleration
the
meaning
of
of
the
your
par ticle
at
answer.
Answers
a
Average
acceleration
is
2
1
change
in
change
v (4 )
(m s
velocity
in
The units for acceleration are m s
)
(seconds)
time
v (1)
Use
2
=
4
b
Instantaneous
a(t)
a
GDC
to
evaluate.
−12 m s
1
=
a (t )
=
acceleration
v′(t)
v ′( t )
= 12t
− 42
2
a (3 )
=
This
−6 m s
means
decreasing
each
3
the
6
second
velocity
metres
at
per
is
Note
second
does
time
is
seconds.
Speed
is
the
that
not
value
of
velocity
.
negative
mean
slowing
velocity
absolute
a
is
an
down.
acceleration
object
It
in
means
motion
that
the
decreasing.
Velocity
tells
us
how
fast
an
For
object
is
moving
and
the
direction
in
which
it
is
moving.
Speed
more
value’
us
only
how
speeding
and
up
fast
or
it
is
moving.
slowing
acceleration.
Limits
and
derivatives
down
To
determine
you
can
if
an
compare
object
the
in
signs
motion
of
on
‘absolute
tells
is
velocity
see
Section
Chapter
2.7.
18,
Investigation –
velocity,
and
1
Copy
and
change
complete
of
velocity.
Velocity
a
and
the
acceleration
speed
tables.
Speed
is
Recall
the
acceleration
are
that
absolute
both
acceleration
value
of
Velocity
b
positive.
is
the
velocity.
is
positive
and
acceleration
−2
Let
acceleration
Time
be
Velocity
2
2 m s
Let
acceleration
Time
Speed
be
Velocity
−2 m s
Speed
−1
−1
(sec)
(m s
0
−1
)
(m s
10
1
(sec)
)
(m s
0
10
−1
)
(m s
10
1
12
)
10
8
2
2
3
3
4
4
Velocity
c
and
acceleration
are
both
Velocity
d
is
negative.
is
negative
and
acceleration
positive.
2
−2
Let
acceleration
Time
be
−2 m s
Velocity
(sec)
(m s
−10
1
−12
(m s
Time
be
Velocity
2 m s
Speed
−1
)
(sec)
10
(m s
0
−10
1
−8
−1
)
(m s
)
10
2
2
3
3
4
4
State
acceleration
−1
)
0
Let
Speed
−1
2
is
negative.
whether
the
object
is
speeding
up
or
slowing
down.
If
a
Velocity
and
b
Velocity
is
acceleration
are
both
the
speed
object
positive
and
acceleration
is
Velocity
and
d
Velocity
is
acceleration
are
both
an
is
increasing ,
negative.
the
c
of
positive.
object
is
speeding
negative.
up.
negative
and
acceleration
is
positive.
If
3
Complete
the
the
speed
object
If
a
velocity
and
acceleration
have
the
same
sign
then
is
an
is
decreasing ,
the
the
object
of
statements:
object
is
slowing
___________________.
down.
If
b
velocity
object
When
velocity
motion
When
is
is
is
acceleration
have
opposite
signs
then
the
___________________.
and
speeding
velocity
motion
and
and
slowing
acceleration
have
the
same
have
different
sign,
the
object
in
up.
acceleration
signs,
the
object
in
down.
Chapter
Example
For
the
displacement
3
function
from
Example
18,
2
s(t)
=
we
found
2t
−
a
Find
21t
+
60t
+
3,
with
s
in
metres
and
t
≥
0
seconds,
2
that
the
whether
t
b
=
3
v(t)
=
speed
the
6t
of
−
the
par ticle
42t
+
60
par ticle
is
and
at
speeding
t
a(t)
=
up
3
=
12t
−
seconds
or
42
and
slowing
determine
down
when
seconds.
During
0
speeding
≤
t
up
≤
10
and
seconds,
when
it
is
find
the
slowing
inter vals
when
the
par ticle
is
down.
Answers
2
a
v(3)
=
6(3)
−
42(3)
+
To
60
−1
=
given
−12 m s
−1
speed
=
|−12|
find
=
the
12 m s
the
speed
time,
find
absolute
of
the
the
par ticle
velocity
and
at
a
take
value.
−2
a(3)
=
The
t
=
b
<
42
is
seconds
=
−6 m s
speeding
since
v(t)
up
<
0
at
and
The
t
=
par ticle
3
have
0.
Compare
and
−
par ticle
3
a(t)
12(3)
the
signs
of
velocity
acceleration.
Use
since
the
the
from
12t
of
v
0
of
it,
a
42
2
5
≤
t
in
3.5
par ticle
in
v (t)
the
<
speeding
0
because
>
a (t)
is
inter val
because
and
in
18.
a
sign
diagram
a(t)
=
0
=
⇒
t
0
=
3.5
value
on
the
inter val
10.
a
v (t)
the
>
up
a(1)
a (t)
<
0,
(5, 10)
v (t)
>
slowing
(0, 2)
0
because
and
a (t)
>
0
Limits
and
derivatives
down
a (t)
<
(3.5, 5)
v (t)
0.
=
t
=
12(1)
<
0
(3.5, 10)
a(4)
seconds
and
inter val
seconds
value
in
each
inter val:
1
−
42
=
−30
=
6
(−)
seconds
0.
particle
the
and
inter val
and
in
velocity
10
(2, 3.5)
seconds
The
is
inter val
because
and
for
––––––––––++++++++++++++
0
the
diagram
align
this
≤
(0, 3.5)
The
acceleration
sign.
10
Check
t
and
at
+++++––––––––+++++++++++
0
signs
sign
when
−
Place
t
velocity
same
up
a(t).
Find
signs
speeding
Example
Below
for
is
0,
=
t
=
12(4)
4
−
42
(+)
Exercise
Use
1
a
A
7P
GDC
to
par ticle
help
moves
4
s(t)
=
2t
−
6t
Find
at
the
Find
A
in
of
the
par ticle
a
=
the
displacement
function
for
the
for
t
≥
0
velocity
seconds.
and
acceleration
of
the
at
time
t
=
2
seconds
and
explain
the
answer.
velocity
par ticle
moves
+
Write
Find
with
and
is
acceleration
speeding
along
a
line
up
with
and
the
equal
zero.
slowing
Then
find
down.
displacement
function
2
−t
12t
an
par ticle
b
values.
t
your
3
s(t)
line
centimetres,
time
when
when
2
a
acceleration
meaning
c
,
expressions
par ticle
b
along
function
2
Write
a
evaluate
36t
+
expression
at
the
−
time
20,
for
in
metres,
the
for
velocity
0
≤
and
t
≤
8
seconds.
acceleration
of
the
t
initial
position,
velocity
and
acceleration
for
the
par ticle.
c
Find
when
Then
d
Find
find
the
inter vals
when
inter vals
par ticle
on
changes
which
acceleration
on
which
the
is
0
direction
the
for
par ticle
par ticle
0
is
for
≤
t
≤
8
0
≤ t
travels
up
8
right
seconds.
speeding
≤
and
seconds.
and
Then
left.
find
slowing
down.
Exam-Style
3
A
diver
QuestionS
jumps
from
a
platform
at
time t
=
0
seconds.
Look
The
distance
of
the
diver
above
water
level
at
time t
is
again
at
the
diver
given
in
Example
16.
2
by
s (t)
=
a
Write
diver
an
at
4.9t
+
expression
time
Find
when
the
c
Find
when
velocity
height
A
Show
of
the
that
par ticle
10,
for
where
the
s
is
in
velocity
metres.
and
acceleration
of
the
t
b
d
4
+
−4.9t
diver
hits
the
equals
water.
zero.
Hence
find
the
maximum
diver.
the
moves
diver
is
along
a
slowing
line
down
with
at t
=
0.3
displacement
seconds.
function
1
2
s (t )
=
t
− ln( t
+
1),
t
≥
0
where,
s
is
in
metres
and
t
is
in
seconds.
4
a
b
i
Write
ii
Hence
i
Write
time
ii
an
expression
find
an
when
for
the
expression
the
velocity
par ticle
for
the
is
at
of
the
par ticle
at
time t
rest.
acceleration
of
the
par ticle
at
t
Hence
show
that
velocity
is
never
decreasing.
Chapter
.
The
derivative
and
graphing
Although
named
One
of
the
most
powerful
uses
of
the
derivative
is
the
after
Car tesian
René
the
graphs
of
functions.
In
this
section
you
how
to
connect
f
′
and
f
′′
to
the
graph
of
function
in
x
on
results
an
is
increasing
in
interval
an
if
on
increase
an
an
in
increase
y.
inter val
A
in x
if
an
function
results
in
is
a
positive
increase
1642–1727)
decreasing
decrease
in
rst
his
y
use
book
Third
x-axis
used
(English
and
an
x-axis.
is
attributed
negative
linearum
Enumeration
Degree,
a
Newton
y-axis
with
the
coordinates.
Enumeratio
and
mathematician,
of
ter tii
Cur ves
used
and
In
both
positive
of
an
and
coordinates.
down
increasing
of
or
negative
Write
numbers
Newton
ordinis,
Example
he
f
Isaac
A
(French
1596–1650),
will
only
see
Descar tes
was
to
mathematician,
analyze
plane
the
or
inter vals
on
which
the
function
is
decreasing.
y
a
y
b
5
4
3
y
c
5
3
4
2
3
1
2
2
x
1
1
1
–2
x
0
–5
–4
–3
–2
1
–1
2
3
4
5
x
0
–5
–4
–3
–2
1
–1
2
3
–3
–4
Answers
y
a
Decreasing
for
x
<
0
increase
5
Increasing
for
x
>
0
in
x
4
decrease
increase
3
in
y
in
y
2
increase
1
in
x
x
0
–5
b
Increasing
for
all
real
–4
–3
–2
1
–1
numbers
2
4
3
5
y
5
4
3
increase
in
y
2
1
increase
in
x
0
–5
c
Increasing
for
Decreasing
x
for
<
0
0
<
and
x
<
2
x
>
2
–4
–3
–2
1
–1
x
2
3
y
3
2
1
x
1
–2
–3
–4
Limits
and
derivatives
2
2
➔
When
a
function
is
decreasing,
the
tangent
lines
to
the
y
cur ve
have
negative
slope.
When
a
function
is
increasing,
5
the
tangent
lines
to
the
cur ve
have
positive
slope.
It
follows
4
3
that:
2
If
f
′(x)
>
0
for
all
x
in
(a, b)
then
f
is
increasing
on
1
(a, b).
x
0
If
f
′(x)
<
0
for
all
x
in
(a, b)
then
f
is
decreasing
–5
on
–4
–3
–2
1
–1
2
4
3
5
(a, b).
Example
Use
the
derivative
of
f
to
find
the
inter vals
on
which
f
is
increasing
or
A
is
decreasing.
stationary
a
point
point
where
2
x
3
a
f (x)
=
4
2
2x
−
3
3x
−
12x
b
f
(x ) =
f (x)
c
=
f ′(x)
x
=
0
2
x
1
A
of
cri tical
f
is
a
number
point
where
Answers
3
f
a
(x )
=
f ′(x)
2
2x
− 3x
=
0
or
f ′(x)
is
− 12 x
undened.
2
f
′( x )
=
6x
− 6x
− 12
Find
the
derivative
Find
the
critical
of
f.
2
6x
− 6x
− 12
=
0
2
6( x
setting
−
x
− 2)
=
f
− 2 )( x
+ 1)
=
x
signs
of
=
2,
Make
is
increasing
since
f
a
sign
on
′(x)
(−∞, −1)
>
f
decreasing
′(x)
<
diagram
for
f
′(x).
and
can
use
on
inter val
notation
to
0
describe
is
and
2
We
f
0
− 1
–1
(2, ∞)
to
x.
+
x
f
for
0
+
f'
equal
by
0
solving
6 (x
′(x)
numbers
(−1, 2)
the
inter vals.
since
0
2
x
f
b
(x )
4
=
2
x
1
2
(x
f
′( x )
2
− 1)( 2 x ) − ( x
− 4 )( 2 x )
Find
the
derivative
Find
the
critical
of
f.
=
2
2
(x
1)
6x
=
2
2
(x
f
′(x)
=
0:
6x
=
0
x
=
0
1)
f
′(x)
undefined
2
(x
when:
setting
f
′
equal
numbers
to
0
and
by
solving
2
−
1)
=
0
for
=
0
undefined.
x,
and
by
finding
where
f
′
is
2
x
−
1
x
=
±1
Make
signs
of
f'
–
–
+
a
Notice
x
–1
0
sign
diagram
for
f
′.
+
that
f
and
f
′
are
not
1
defined
circles
at
on
x
=
the
remember
±1.
sign
Use
open
diagram
to
this.
{
Continued
on
next
page
Chapter
f
is
increasing
(−1, 0)
f
is
since
on
f
′(x)
decreasing
(1, ∞)
since
f
(−∞, −1)
>
on
′(x)
on
0.
(0, 1)
<
We
and
cannot
(−∞, 0)
since
and
x
0.
say
=
f
is
or
not
that
f
is
increasing
decreasing
defined
at
on
x
=
(0, ∞)
−1
or
1.
3
f
c
(x )
=
′( x )
=
x
2
f
3x
2
3x
=
Find
the
Find
the
=
of
+
f'
x
is
f
′
equal
0
and
by
solving
x.
a
sign
diagram
for
f
′
0
increasing
on
(−∞, 0)
Even
and
x
(0, ∞).
=
the
is
though
0,
we
zero
at
Exercise
f
is
defined
cannot
inter val
x
increasing
Write
numbers
to
+
Make
f
f.
0
for
signs
critical
of
0
setting
x
derivative
include
because
=
at
0,
x
so
=
at
the
f(x)
0
in
gradient
is
not
0.
7Q
down
the
inter vals
on
which
f
is
increasing
y
1
or
decreasing.
y
2
4
y
3
1
2
3
0
–2
2
x
–1
–1
1
1
–2
–3
0
–4
–3
–2
x
–1
1
2
3
x
4
–1
–1
–4
–2
–5
–3
–1
–6
–4
–2
In
questions
which
f
4–9,
use
is
increasing
=
x
the
or
derivative
f
(x)
4
5
f
f
to
find
all
inter vals
on
decreasing.
4
4
of
(x)
=
x
x
+ 2
x
3
2
−
2x
6
f
(x )
Use
=
a
a
GDC
graph
of
to
the
look
at
function
3
to
x
3
7
f
(x )
=
8
f
(x)
=
x
x
e
9
f
(x )
=
2
x
x
1
y
Exam-Style
Question
4
10
The
graph
of
the
derivative
of
f
is
y
shown.
=
f'(x)
3
Write
down
the
inter vals
on
which
f
is
2
decreasing
and
increasing.
1
0
–3
–1
–1
–2
–4
Limits
and
derivatives
x
4
verify
your
results.
A
function
when
the
has
a
relative
function
maximum
changes
from
(or
point
increasing
local
to
maximum)
decreasing.
Note
A
function
has
a
relative
minimum
point
(or
local
not
when
the
function
changes
from
decreasing
to
that
change
relative
minimum
and
of
extrema
a
maximum
points
are
called
the
then
function.
is
number
the
point
neither
a
minimum
➔
The
first
derivative
test
is
used
to
locate
relative
extrema
of
f
is
defined
at
a
critical
number
c
does
at
a
x
=
c,
(c, f (c))
relative
nor
a
f.
relative
If
f ′(x)
sign
increasing.
critical
Relative
if
minimum)
maximum.
then:
relative
maximum
If
1
f
′(x)
changes
from
positive
to
negative
at x
=
c,
then
f
neither
has
a
relative
maximum
point
at
(c, f
a
relative
(c)).
minimum
relative
If
2
f
′(x)
changes
from
negative
to
positive
at x
=
c,
then
f
nor
maximum
relative
minimum
has
a
Example
Use
the
relative
minimum
point
at
(c, f
(c)).
first
derivative
test
to
find
the
relative
extrema
for
the
functions
in
Example
22.
2
x
3
a
f
(x)
=
4
2
2x
−
3
3x
−
12x
f
b
(x )
=
f
c
(x)
=
x
2
x
1
Answers
3
a
f
(x )
=
2x
′( x )
=
6x
=
6( x
2
− 3x
− 12 x
2
f
signs
of
− 6x
− 12
− 2 )( x
+ 1)
+
f'
+
Use
–1
x
Since
x
=
f
′(x)
f
′(x)
−1
2
changes
there
is
changes
relative
a
from
relative
from
minimum
x
3
f
( −1)
=
2 ( −1)
positive
=
to
maximum
negative
at
to
negative
at
x
positive
=
at
at
−1.
x
=
Locate
Since
2
for
there
is
(2)
=
=
So
for
f
′
from
relative
changes
extrema
for
f
by
looking
′
a
Evaluate
f
at
x
=
−1
and
x
=
2
to
find
− 12( −1)
maximum
and
minimum
values.
7
3
f
diagram
22.
the
sign
the
=
sign
2.
2
− 3( −1)
the
Example
2( 2 )
2
− 3( 2 )
− 12( 2 )
−20
the
relative
minimum
maximum
point
is
point
is
(−1, 7)
and
the
relative
(2, −20).
2
x
f
(x )
4
=
b
2
x
1
6x
f
′( x )
=
2
2
(x
signs
of
1)
–
f'
–
x
–1
+
0
+
1
There
Since
f
′(x)
changes
from
negative
to
positive
at
x
=
would
not
be
relative
extrema
at
0
x
=
−1
f
′(x)
and
x
=
1
even
if
the
sign
of
2
0
there
is
a
relative
minimum
at
x
=
0.
f
(0 )
4
=
=
4
had
changed,
since
f
is
undefined
2
0
1
at
So
the
relative
minimum
point
is
x
=
−1
and
x
=
1.
(0, 4).
{
Continued
on
next
Chapter
page
3
f
c
(x )
=
′( x )
=
x
2
f
signs
of
3x
+
f'
+
x
Note
0
that
condition
f
has
no
relative
the
derivative
does
=
0.
It
sign
Exercise
In
x
=
0.
7R
questions
extrema
at
for
1
to
each
8,
use
the
first
derivative
test
f
(x)
=
to
find
the
relative
function.
2
1
3
2x
−
4x
−
3
2
f
(x)
=
x
4
f
(x)
=
x
6
f
(x)
=
x
8
f
(x )
−
12x
−
2x
−
5
5
4
3
3
f
(x )
=
x
3
5
f
(x)
=
x (x
+
2
2
3)
x
e
2
x
1
f
7
(x )
=
− 2x
+
=
2
(x
➔
x
+ 1)
If
f
′′(x)
>
0
for
all
x
in
If
f
′′(x)
<
0
for
all
x
in
The
points
on
a
graph
inflexion
points .
A
point
′′(x)
and
if
f
=
0
(a, b)
then
f
is
concave
up
(a, b)
then
f
is
concave
down
where
point
The
f
+ 1
on
′′(x)
graph
the
the
concavity
graph
changes
is
concave
of
on
on
changes
f
is
an
(a, b).
(a, b).
are
called
inflexion
sign.
down
for
(−∞, 0).
The
tangent
y
lines
4
y
=
shown
in
red
have
decreasing
gradients.
This
f(x)
means
that
f ′
is
decreasing ,
so
its
derivative
f ′′
is
negative.
0
x
The
–2
graph
is
concave
up
for
(0, ∞).
The
tangent
–2
lines
shown
in
blue
have
increasing
gradients.
This
–4
means
that
f ′
is
increasing ,
so
its
derivative
f ′′
is
positive.
The
point
concavity
Limits
and
to
=
have
0
a
is
derivatives
must
also
not
a
relative
be
not
changes
change
′(x)
sufficient
extrema
at
extrema
x
since
f
(0, 0)
at
x
=
is
an
0.
inexion
point
since
f
changes
sign
at
x
=
0.
true
that
f
′(x)
Example
For
the
functions
function
is
from
concave
up
Example
and
22,
concave
use
the
down.
second
Find
derivative
the
inflexion
to
find
the
inter vals
where
the
points.
2
x
3
a
f
(x)
=
4
2
2x
–
3
3x
−
12x
f
b
(x )
=
c
f
(x)
=
x
2
x
1
Answers
3
a
f
(x )
2
2x
=
− 3x
− 12 x
2
f
f
′( x )
′′( x )
12 x
6x
=
− 6x
= 12 x
− 6
=
− 12
Find
the
Find
where
second
derivative
of
f.
− 6
f
″(x)
=
0.
0
1
x
=
2
signs
of
–
f ''
+
Make
a
sign
diagram
for
f
″
1
x
2
1
⎛
f
is
concave
down
⎜
⎛
is
concave
up
on
1
f
″(x)
<
0
and
⎠
⎞
,∞
⎜
⎝
since
⎟
2
⎝
f
⎞
−∞,
on
since
⎟
2
f
″(x)
>
0.
⎠
1
Since
f
″(x)
changes
sign
at
x
,
=
there
is
an
inflexion
2
3
1
⎛
point
⎞
there.
⎜
⎝
2
1
⎛
=
f
2
⎞
⎛
2
1
⎞
⎛
3
⎟
⎜
⎠
⎝
2
1
13
⎞
=
12
⎟
⎜
⎠
⎝
2
⎟
⎜
⎠
⎝
−
2
1
Evaluate
⎟
f
at
So
the
inflexion
point
1
is
find
the
2
y-coordinate
⎛
to
x
2
⎠
of
the
inflexion
point.
13 ⎞
,
⎜
⎝
⎟
2
2
⎠
2
x
f
b
(x )
4
=
2
x
1
6x
f
′( x )
=
2
2
(x
1)
2
(x
f
″(x)
2
− 1)
(x
′′( x )
−6 (3 x
− 1)( 2 x )]
+ 1)
Find
the
second
derivative
of
f.
=
2
f
2
2
( 6 ) − ( 6 x )[2 ( x
=
=
0
f
′′( x )
2
+ 1)
(x
=
2
3
(x
1)
2
−6 (3 x
2
4
is
undefined
1)
when
To
− 1)
=
0
− 1 =
0
find
a
where
sign
f
diagram
″(x)
=
0
for
and
f
″
you
where
f
must
″(x)
is
0
3
undefined.
2
(x
make
3
1)
x
2
−6(3 x
+ 1)
=
0
x
=
±1
1
2
x
=
−
3
No
real
signs
of
f
solutions
–
''
Even
+
x
x
–1
is
concave
=
down
on
(−∞, −1)
and
(1, ∞)
since
f
″(x)
<
f
is
concave
there
f
″(x)
are
changes
no
sign
inflexion
at
points. This
because
up
on
(−1, 1)
since
f
″(x)
>
f
(x)
is
undefined
at
x
=
±1.
0,
In
and
±1
1
is
f
though
–
this
case
the
concavity
is
changing
on
0.
either
side
of
a
vertical
{
asymptote.
Continued
on
next
Chapter
page
3
c
f
(x )
=
x
′( x )
=
3x
′′( x )
=
6x
6x
=
0
x
=
0
2
f
f
signs
of
f
–
''
Find
the
second
Find
where
f
Make
0
f
is
concave
down
on
f
is
concave
up
(0, ∞)
(−∞, 0)
since
f
″(x)
<
0,
a
sign
on
since
f
″(x)
>
0.
Since
f
f
signs
at
x
=
0,
there
is
an
inflexion
So
(0)
the
point
there.
Find
the
the
point
is
(0, 0).
7S
questions
where
0.
inflexion
Exercise
In
=
1
to
6,
function
inflexion
use
is
the
second
concave
up
derivative
and
to
concave
(x)
=
2x
−
3
f
(x)
=
x
5
f
(x)
=
2xe
4x
3
−
3
2
f
(x)
=
−x
4
f
(x)
=
x
2
−
6x
the
inter vals
down.
4
f
find
points.
2
1
3
+
4x
4
+
12x
1
x
f
6
(x )
=
2
x
Exam-Style
+ 1
Questions
24
7
Let
f
(x )
=
2
x
+ 12
48 x
Use
a
that
fact
that
f
′( x )
to
=
2
show
that
the
second
2
(x
+ 12 )
2
144 ( x
derivative
is
f
′′( x )
4)
=
2
(x
b
8
i
Find
the
relative
ii
Find
the
inflexion
The
graph
of
the
3
+ 12 )
extrema
second
points
of
of
derivative
the
the
graph
graph
of
of
f
f
of
y
f
is
shown.
Write
down
the
inter vals
4
on
which
f
is
concave
up
and
concave
3
down.
Give
inflexion
the
x-coordinates
of
any
0
–1
–2
–4
–5
and
derivatives
=
f ''(x)
1
points.
–1
Limits
y
2
–3
diagram
at
3
=
0.
for
f
″
x
=
0
to
find
the
″(x)
y-coordinate
(0)
=
f.
and
Evaluate
f
″(x)
of
+
x
changes
derivative
x
of
the
inflexion
point.
The
the
first
and
graph
of
asymptotes
Example
Sketch
second
the
to
derivatives
function.
help
We
complete
of
can
the
a
function
also
use
tell
us
much
intercepts
about
and
graph.
the
intercepts
graph
and
of
each
function.
asymptotes
to
help
Use
the
information
draw
the
graph.
you
found
in
Examples
22–24
and
2
x
3
a
f
(x)
=
2x
4
2
−
3
3x
−
12x
f
b
(x )
=
f
c
(x)
=
x
the
x-intercepts,
2
x
1
Answers
3
a
f
(x)
=
2x
2
−
increasing:
3x
−
12x
(−∞, −1)
decreasing:
and
(−1, 2)
relative
maximum:
relative
minimum:
(−1, 7)
(2, −20)
1
⎛
concave
⎜
⎟
2
⎝
⎛
⎞
−∞,
down:
concave
(2, ∞)
1
⎠
⎞
, ∞
up:
⎜
⎝
To
⎟
2
find
and
⎛
inflexion
set
the
function
equal
to
0
equal
to
0
⎠
1
13 ⎞
2 x
,
point:
⎜
⎝
solve:
3
2
− 3 x
− 12 x
= 0
⎟
2
2
⎠
2
x (2 x
x-intercepts:
y-intercept:
(0, 0),
(−1.8, 0),
− 3 x
− 12)
= 0
(3.31, 0)
(0, 0)
3
x
0
or
x
x
0
or
x
9
4(2)( 12)
y
2(2)
relative
maximum
x
0
–3
1.81, 3.31
–1
To
find
the
y-intercept
evaluate
To
find
the
x-intercepts,
f
(0).
–5
Inexion
point
–10
–15
–20
relative
increasing
concave
decreasing
down
minimum
increasing
concave
up
2
x
b
f
(x )
4
=
2
x
1
increasing:
(−∞, −1)
decreasing:
relative
(0, 1)
and
minimum:
concave
down:
concave
up:
and
(−1, 0)
(1, ∞)
(0, 4)
(−∞, −1)
and
(1, ∞)
and
(−1, 1)
set
the
function
solve:
2
x
4
2
inflexion
points:
none
0
x
4
0
x
2
2
x
x-intercepts:
y-intercept:
(2, 0),
(0, 4)
1
(−2, 0)
To
find
the
y-intercept
evaluate
{
f
(0).
Continued
on
next
Chapter
page
ver tical
asymptotes:
x
=
To find the vertical asymptotes, find where the
±1
denominator equals 0 (check to see that the numerator is
not 0 for that same value):
2
x
horizontal
asymptote:
y
=
− 1
= 0
⇒
x
=
±1
We learned that the horizontal asymptote of
1
ax
of
y
the form
a function
+ b
=
is found by using the leading
cx
+
d
a
coefficients,
y
. This
=
method
works
for
any
rational
c
function
the
where
degree
of
the
the
degree
of
the
numerator
is
equal
denominator.
1
y
=
⇒
y
= 1
1
y
Limit
notation
can
be
used
to
describe
the
8
asymptotes.
showing
relative
us
The
that
horizontal
for
large
asymptote
positive
y
=
values
1
is
of x,
y
minimum
gets
2
of
0
x,
close
y
to
gets
1,
and
close
to
for
1.
small
Using
negative
limit
values
notation
to
x
–2
–3
–2
say
this
we
write:
lim f ( x )
=
1
and
x →∞
lim
f (x)
=
1
x → −∞
–4
For
the
ver tical
asymptote
x
=
1,
as
x
gets
–6
close
to
1
from
the
left
side
of
1, y
grows
–8
large
decreasing
concave
and
positive
without
bound,
down
up
down
x
gets
y
grows
Using
close
to
large
limits
1
from
and
to
the
right
negative
say
this
we
f
(x)
=
side
without
write:
3
c
and
as
increasing
of
1,
bound.
lim
f (x)
=
∞
x →1
x
and
lim
f (x)
=
∞
+
x →1
increasing:
(−∞, ∞)
Similarly,
no
relative
lim
f (x)
+
down:
concave
up:
(−∞, 0)
x → −1
(0, ∞)
point:
(0, 0)
x-intercept:
(0, 0)
y-intercept:
(0, 0)
y
8
6
4
2
Inexion
point
0
–5
–4
–3
–2
–1
x
1
2
3
4
–2
–4
–6
–8
increasing
concave
Limits
and
x
=
−1
down
concave
derivatives
we
write:
lim
x → −1
and
concave
inflexion
for
extrema
up
5
=
∞
f (x)
=
−∞
to
Exercise
In
7T
questions
and
second
intercepts
1
to
4,
sketch
derivative
and
to
the
graph
analyze
key
of
the
function.
features
f
(x)
=
3x
the
3
+
x
+ 2
x
4
the
graph.
first
Find
any
asymptotes.
2
1
of
Use
10x
−
8
2
f
(x)
=
x
4
f
(x)
=
(3
6
f
(x )
2
+
x
−
5x
−
5
4
f
3
(x )
=
x
x
e
f
5
(x )
−
x)
2
e
x
=
1
=
2
2
Given
it
is
the
graph
possible
Example
a
Given
to
Given
of
f
that
graphs
of
any
sketch
a
one
of
graph
the
of
three
the
+ 1
functions f,
other
two
f
′
or
f
″,
functions.
that
graphs
b
x
of
the
′
the
f
graph
and
f
graph
and
f
shown
is
a
graph
of
f,
shown
is
a
graph
of f
sketch
the
y
″
′,
sketch
the
″
0
x
–2
–3
5
6
Answers
a
The
y
y
y
=
=
increasing
at
x
=
at
x
=
changes
and
has
a
from
decreasing
2. This
2
and
means
relative
changes
that
f
minimum
from
′(x)
equals
negative
to
zero
–2
x
–1
5
6
The
graph
that
f
′′(x)
of
is
f
′′(x)
is
the
f
′′(x)
must
f
is
always
always
concave
positive.
positive.
derivative
of
f
up. This
means
Since
′(x),
a
linear
function,
f '(x)
=
y
b
Since
y
=
to
f
f ''(x)
–3
y
of
f(x)
0
y
graph
=
from
f ''(x)
f
′(x)
be
a
positive
equals
positive
to
zero
constant.
when
negative,
the
x
=
−1
graph
and
of
f
changes
has
a
f '(x)
relative
Since
from
f
maximum
′(x)
equals
negative
to
point
zero
when
when
positive,
the
x
x
=
=
−1.
5
and
graph
of
changes
f
has
a
x
–6
y
=
f(x)
–2
relative
Since
f
minimum
′(x)
has
x
=
2,
the
x
=
2.
Since
negative
x
>
2,
f
a
graph
for
f
x
′′(x)
is
<
is
point
when
relative
of
f
′′(x)
concave
2.
Since
positive
x
=
5.
minimum
equals
down
f
for
is
x
when
zero
for
x
concave
>
when
<
2,
up
f
′′(x)
is
for
2.
Chapter
Exercise
7U
Exam-Style
The
1
graph
y
Questions
of
y
=
f
(x)
is
given.
y
Sketch
a
graph
of
y
=
f
′(x)
and
y
=
f
=
f(x)
′′(x).
0
–3
The
2
graph
Sketch
a
of
the
graph
derivative
of
y
=
f
(x)
of
f,
and
y
y
=
=
f
f
′(x),
is
x
–2
given.
y
′′(x).
y
=
f '(x)
x
–3
The
3
is
graph
given.
of
the
Sketch
second
a
graph
derivative
of
y
=
f
(x)
of
f,
and
y
y
=
=
f
f
–1
′′(x),
y
′(x).
0
–2
–4
x
–1
y
.
More
on
extrema
and
=
f ''(x)
optimization
problems
Y
ou
have
seen
how
to
use
the
second
derivative
to
determine
See
concavity
and
inflexion
points
for
a
graph
of
a
function.
The
Chapter
second
Section
derivative
This
is
of
a
called
function
the
can
second
also
be
used
derivative
to
find
relative
extrema.
test
If
Second
derivative
f
″(c)
then
If
f
′(c)
near
c,
=
0
and
the
>
0
near
c,
test
second
derivative
of
f
up
exists
So
then
f
is
near
f
concave
c
has
a
relative
minimum.
1
If
f
″(c)
>
0,
then
f
has
a
relative
minimum
2
If
f
″(c)
<
0,
then
f
has
a
relative
maximum
3
If
f
″(c)
=
0,
the
at
x
=
c
If
at
x
=
f
″(c)
then
second
derivative
test
fails
first
derivative
test
must
be
used
to
locate
f
is
Limits
f
has
a
extrema.
and
derivatives
c,
concave
near
maximum.
near
c
the
So
relative
0
and
down
the
<
c
relative
2.6.
17,
Example
Find
the
relative
extreme
points
of
each
function.
Use
the
second
derivative
test
whenever
possible.
3
a
f
(x)
=
2
x
−
5
3x
−
2
b
f
(x)
=
3x
3
−
5x
Answers
3
f
a
(x )
=
2
x
− 3x
− 2
2
f
f
′( x )
′′( x )
= 3x
=
− 6x
6x
Find
the
first
and
Find
the
values
second
derivative
of
f.
− 6
2
3x
− 6x
=
0
of
x
where
the
first
derivative
equals
zero.
3x ( x
− 2)
x
=
0
=
0, 2
f
′′( 0 )
=
−6
f
′′( 2 )
=
6
f
(0 )
=
f
(2)
=
>
−2
relative
<
0
0
⇒
⇒
⇒
relative
relative
( 0, −2 )
is
maximum
first
minimum
rel
lative
⇒
is
second
f
′( x )
of
the
implies
a
relative
maximum
and
f
″
>
0
implies
a
relative
minimum.
find
a
the
the
function
relative
where
minimum
the
and
extrema
occur
maximum
to
values.
minimum
5
=
zero
derivative.
0
Plus
(x )
each
<
help
on
CD:
demonstrations
f
at
″
GDC
b
derivative
f
Evaluate
( 0, −6 )
the
a
maximum
−6
Evaluate
3
3x
and
GDCs
Casio
are
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
− 5x
4
2
= 15 x
2
− 15 x
= 15 x
(x
+ 1)( x
− 1)
Find
the
first
and
Find
the
values
second
derivative
of
f.
3
f
′′( x )
=
60 x
− 30 x
4
2
15 x
− 15 x
=
0
=
0
=
0,
2
15 x
(x
+ 1)( x
− 1)
x
f
′′( 0 )
=
f
′′( −1)
f
′′(1)
signs
of
=
f
0
=
⇒
second
−30
30
>
<
0
0
⇒
'
x
⇒
x
where
the
first
derivative
zero.
± 1
derivative
relative
e
relative
–
–1
equals
of
test
fails
maximum
Evaluate
first
the
second
derivative
at
each
zero
of
the
derivative.
f
″
=
0
implies
the
f
″
<
0
implies
a
relative
second
maximum,
derivative
f
″
>
0
implies
a
relative
minimum.
test
fails,
minimum
and
0
0
1
{
Continued
on
next
Chapter
page
Since
is
no
at
there
is
relative
that
no
sign
change
minimum
or
in
f
′
at
x
=
0,
Since
there
use
maximum
f
point.
′
the
the
first
( −1)
f
(1)
=
2
⇒
( −1
, 2)
is
a
relative
maximum
is
a
r elative
minimum
find
=
−2
⇒
(1
,
− 2)
the
GDC
at
the
on
and
CD:
GDCs
Exercise
Find
Use
the
the
Casio
are
on
=
test
test
to
failed
see
if
at
the
x
=
sign
0,
of
0.
function
demonstrations
Plus
x
relative
help
derivative
derivative
changes
Evaluate
f
second
where
minimum
the
and
extrema
occur
maximum
to
values.
Alternative
for
the
TI-84
FX-9860GII
the
CD.
7V
relative
second
extreme
derivative
points
test
of
each
whenever
function.
possible.
2
1
f
(x)
=
3x
3
f
(x)
=
x
5
f
(x)
=
(x
2
−
18x
4
−
48
2
f
(x)
=
(x
4
f
(x)
=
xe
6
f
(x )
3
−
2
−
1)
x
4x
1
4
−
1)
=
2
x
We
have
been
finding
relative
or
local
+ 1
extrema
of
functions.
We
can
The
also
find
the
absolute
or
global
extrema
of
a
function.
of
the
of
extrema
are
the
greatest
and
least
values
function
over
domain.
Absolute
extrema
occur
at
either
the
a
relative
function
minimum
endpoints
of
a
an
the
inter val
at
an
derivatives
the
Relative
never
endpoint
function.
on
near
point.
extrema
and
values
function
critical
Limits
and
function.
of
are
extrema
maximum
or
extrema
its
the
entire
relative
Absolute
occur
of
a
Example
D
a
Identify
each
minimum,
b
Find
the
a
labeled
relative
absolute
point
as
an
maximum
maximum
absolute
or
and
maximum
minimum,
or
or
B
neither.
minimum
2
for
f
(x)
=
x
−
2x
on
−1
≤
x
≤
2.
A
C
Answers
a
A
is
neither.
The
points
values
and
the
those
value
absolute
A
on
the
greater
below
of
graph
than
the
the
the
endpoint
be
of
a
black
function
maximum
cannot
above
value
nor
relative
the
line
at
an
the
of
black
the
have
A.
So
values
A
absolute
extrema
line
function
is
have
at
less
neither
A
than
an
minimum.
since
A
is
at
an
function.
D
B
A
C
B
is
a
relative
B
maximum.
cannot
values
of
function
C
is
an
absolute
minimum
and
relative
C
is
its
is
an
absolute
at
of
of
entire
The
maximum.
absolute
function
maximum
greater
than
since
the
there
value
are
of
the
B.
C
is
minimum
the
least
since
value
the
of
value
the
of
the
function
over
domain.
value
the
an
the
absolute
function
minimum.
D
an
be
of
the
function
function
over
its
at
D
entire
is
the
greatest
value
domain.
2
b
f
(x)
f
=
x
′( x )
2x
− 2
x
−
2x
=
2x
=
0
on
−1
≤
x
≤
2
− 2
Find
the
critical
numbers
where
f
′(x)
=
0.
= 1
2
f
( −1)
=
( −1)
− 2 ( −1)
=
Evaluate
3
2
f
(1)
=
number
(1)
− 2 (1)
=
the
in
function
the
at
each
inter val. The
endpoint
largest
and
value
is
critical
the
−1
maximum
and
the
smallest
is
the
minimum.
2
f
(2)
The
=
(2)
− 2( 2 )
absolute
=
0
maximum
of
2
f
(x)
=
x
absolute
−
2x
on
−1
minimum
≤
is
x
≤
2
is
3
and
the
−1.
Chapter
Exercise
Identify
7W
each
maximum
labeled
or
point
minimum,
a
in
questions
relative
1
and
maximum
2
as
or
an
absolute
minimum,
or
neither.
C
1
2
C
A
A
B
B
D
Find
on
the
the
absolute
given
maximum
and
minimum
of
the
function
inter val.
3
3
f
(x)
=
(x
−
4
f
(x)
=
8x
5
f
(x )
=
2)
for
0
≤
x
≤
4
2
−
x
for
−1
≤
x
≤
7
3
3
2
x
−
x
for
−1
≤
x
≤
2
2
Many
practical
minimum
an
area
or
For
1
For
minimize
o p t i m i za t i o n
➔
problems
values.
optimization
Assign
2
Write
3
Find
an
a
≤
x
≤
since
occur
Limits
that
or
and
in
maximum
want
proble m s
to
are
the
to
are
the
equals
you
or
maximize
c al le d
derivatives
be
optimized
of
two
quantities
to
be
sketch.
(minimized
or
variables.
or
derivative
have
a
feasible
of
the
minimum
derivative
′ (x)
a
and
for
the
equation
problem
to
be
zero.
maximum
f
draw
sensible
remember
when
quantities
possible
terms
that
second
b,
given
When
where
optimized
Verify
to
equation
values
situation
first
Such
may
problems:
variables
maximized)
c o st.
finding
we
problems
determined.
4
involve
example,
=
that
or
0
test.
the
If
at
maximum
domain
endpoints
minimum
or
or
the
an
on
a
is
must
closed
endpoint.
using
such
be
the
that
tested
inter val
can
Example
The
product
number
of
plus
two
three
positive
times
the
numbers
second
is
48.
Find
number
is
a
the
two
numbers
so
that
the
sum
of
the
first
minimum.
Answer
x
=
the
first
y
=
the
second
S
=
x
positive
number
positive
Assign
=
+ 3 y
Write
48 ⇒
y
to
the
quantities
to
be
deter mined.
number
an
equation
for
the
sum,
the
quantity
to
be
minimized.
48
xy
variables
=
x
S
=
x
⎛
48 ⎞
⎜
⎟
+ 3
144
=
x
⎝
x
Use
the
other
given
infor mation
to
rewrite
the
+
equation
x
⎠
for
the
sum
using
only
two
variables.
144
S ′( x )
Find
= 1
the
derivative
of
the
equation
to
be
minimized
2
x
and
then
find
the
critical
numbers,
where
the
144
derivative
1 −
=
equals
0.
0
2
x
2
x
= 144
x
Since
only
=
±12
the
x
=
numbers
are
positive
consider
12.
288
S ′′( x )
=
Use
3
the
second
derivative
test
to
verify
that
the
x
critical
number
12
gives
a
minimum.
Note
that
the
288
S ′′
(12 )
=
>
0
⇒
relative
minimum
first
derivative
test
could
also
be
used.
3
12
48
y
=
48
⇒
y
=
=
4
Find
The
numbers
Example
A
second
number.
four th
enclose
are
12
and
4.
rectangular
The
the
12
x
plot
side
the
of
of
farmland
the
maximum
plot
is
area.
is
enclosed
bordered
Find
the
by
by
a
180 m
stone
maximum
of
wall.
fencing
Find
material
the
on
three
dimensions
of
sides.
the
plot
that
area.
Answer
Make
to
w
be
a
sketch
and
assign
variables
to
the
quantities
deter mined.
w
l
Write
A
=
an
equation
for
the
area,
the
quantity
to
be
lw
maximized.
2w
+
l
=
180
⇒
l
=
180
–
2w
2
A
=
(180
–
2w)w
=
180w
–
Use
the
other
equation
A′(w )
= 180 − 4 w
180 − 4 w
=
0
Find
and
the
then
for
=
the
infor mation
area
derivative
find
derivative
w
given
to
rewrite
the
2w
the
equals
using
of
the
critical
only
two
equation
numbers,
variables.
to
be
maximized
where
the
0.
45
{
Continued
on
next
Chapter
page
Use
A′′(w )
=
the
critical
A′′( 45 )
=
−4
<
l
=
180
–
2w
l
=
180
–
2(45)
A
=
A
45 m
second
derivative
test
to
verify
that
the
−4
90(45)
by
0
⇒
relative
=
45
gives
a
maximum.
maximum
Find
⇒
=
number
the
length
and
the
area.
90
4050
90 m
plot
will
have
the
maximum
2
area
of
4050 m
Exercise
1
The
so
sum
that
The
of
the
second
2
7X
two
sum
of
number
sum
number
of
is
positive
is
one
200.
the
a
numbers
first
is
number
20.
and
Find
the
the
two
square
numbers
root
of
the
maximum.
positive
Find
the
number
two
and
twice
numbers
so
a
second
that
their
positive
product
is
a
maximum.
3
A
rectangular
and
the
built
pen
from
figure.
400
What
is
par titioned
feet
of
into
fencing
dimensions
as
should
two
sections
shown
be
in
used
y
so
that
the
area
will
be
a
maximum?
x
Example
Find
area
the
of
dimensions
192
square
of
an
open
centimetres
box
that
with
has
a
a
square
base
maximum
and
surface
volume.
Answer
Make
to
a
the
sketch
and
quantities
assign
to
be
variables
deter mined.
h
x
Write
an
equation
for
the
volume,
the
x
quantity
to
be
maximized.
2
V
=
x
h
Since
2
x
is
+ 4 xh
the
the
box
sum
is
of
open,
the
the
area
of
surface
the
area
bottom
= 192
2
of
the
box,
x
,
and
the
area
of
the
2
192
h
x
=
four
lateral
faces,
4xh.
4 x
2
⎛ 192
x
⎞
2
V (x )
=
x
⎜
⎟
4 x
⎝
Use
this
to
rewrite
the
equation
for
⎠
the
area
using
only
two
variables.
1
3
=
48 x
x
−
4
{
Limits
and
derivatives
Continued
on
next
page
x
Find
the
derivative
of
the
equation
3
2
V
′( x )
=
48 −
x
to
be
maximized
and
then
find
the
4
critical
numbers,
where
the
derivative
3
2
48 −
x
=
0
equals
0.
4
3
2
x
=
48
4
2
=
x
x
64
±8
=
The
feasible
is
x
=
V
′
′( x )
critical
number
8.
Use
3
=
that
2
′
′(8)
second
derivative
test
to
verify
=
the
critical
number
8
gives
a
maximum.
3
V
the
x
−
(8 )
=
−12
<
0
2
⇒
relative
maximum
2
192 −
h
2
x
192
=
⇒
h
=
4 x
The
Find
4
the
height
of
the
box.
4 (8)
dimensions
maximum
by
− 8
=
area
of
the
are
box
8 cm
with
by
8 cm
4 cm.
Example
10 000
The
cost
C
of
ordering
and
storing
x
units
of
a
product
is
C (x )
=
x
.
+
A
deliver y
tr uck
can
x
deliver
at
most
200
units
per
order.
Find
the
order
size
that
C
the
will
minimize
the
cost.
Answer
10 000
C (x )
=
x
+
where
x
is
the
number
of
units.
is
function
to
be
minimized.
of
the
x
10 000
C ′( x )
= 1 −
2
x
Find
the
derivative
equation
to
be
minimized
10 000
and
1 −
=
then
find
the
critical
numbers,
where
the
0
2
x
derivative
equals
0.
10 000
= 1
2
x
2
x
= 10 000
x
The
the
no
=
±100
feasible
order
more
absolute
critical
must
than
number
include
200
at
units,
minimum
for
1
is
least
we
≤
x
x
=
need
≤
100.
one
unit
to
Since
and
find
the
Since
the
endpoints
must
be
function
and
is
zeros
considered
defined
of
for
the
the
on
a
closed
derivative
minimum
in
inter val,
the
the
inter val
value.
200.
{
Continued
on
next
Chapter
page
10 000
C (1)
= 1 +
= 10 001
1
10 000
C (100 )
= 100 +
=
200
⇐
minimum
cost
100
10 000
C ( 200 )
=
200 +
=
250
200
The
100
minimum
cost
occurs
when
there
are
units.
Exercise
7Y
3
1
An
open
Find
2
the
box
with
that
the
3
=
how
3
A
square
dimensions
Suppose
C(x)
a
of
base
the
average
box
cost
and
that
of
open
top
minimize
producing x
x
–
3x
many
–
9x
items
+
30.
should
moves
on
a
If
at
be
most
10
produced
the
at
4
A
time
t
is
maximum
Exam-Style
✗
the
volume
surface
units
horizontal
items
to
cylinder
given
by
line
distance
s(t)
=
t
so
of
an
of
32 000 cm
.
area.
item
is
given
by
be
that
its
produced
the
cost
position
in
for
a
the
day
,
day?
from
2
–
between
can
minimize
3
origin
Find
a
2
par ticle
the
has
12t
the
+36t
par ticle
–10
for
and
0
the
≤
t
≤
7.
origin.
Questions
is
inscribed
in
a
cone
with
radius
6
centimetres
10
and
a
height
10
Find
an
of
the
h,
–
h
centimetres.
expression
height
of
for
the
r,
the
radius
of
the
cylinder
in
terms
r
10 cm
cylinder.
h
b
Find
c
Find
an
expression
of
the
volume,
V
,
of
the
cylinder
in
terms
of
h
2
d
dV
V
and
2
d
5
Hence
Let
x
be
6 cm
dh
dh
find
the
the
radius
number
of
and
height
thousands
of
of
cylinder
units
of
with
an
maximum
item
volume.
produced.
The
revenue
for
2
selling
a
x
The
units
profit
is
r (x )
p(x)
=
4
r(x)
–
x
and
c(x)
the
Write
cost
an
of
producing
expression
for
x
units
is
c (x)
=
2x
p(x).
2
d
dp
b
Find
p
and
2
dx
dx
c
Hence
the
find
the
number
of
units
that
should
profit.
Review
exercise
✗
1
Differentiate
with
respect
to
x
a
4x
3
4
3
3
2
+3x
–
2x
+
6
x
b
c
4
x
2
d
(x
Limits
3
–
1)(2x
and
x
4
x
+ 7
2
–
x
derivatives
4x
+
x)
e
f
e
be
produced
in
order
to
maximize
ln x
3
g
4
(x
+
1)
ln(2x
h
+3)
i
2
x
x
2
4 x
2x
2
j
2e
x
(3x
k
+
1)(e
)
l
x
e
6
3
1
⎛
2
m
3
2x
5
e
ln
o
⎜
⎟
x
⎝
Exam-Style
⎞
2x
x
n
⎠
Questions
3
2
Let
f
(x)
=
2x
–
6x
3
a
Expand
b
Use
(x
+
h)
f
the
formula
f
′( x )
=
(x
+ h) −
derivative
c
The
graph
d
Write
e
Find
of
down
the
of
f
f
is
f
(x)
is
6x
inter val
show
that
h
–
decreasing
″ (x)
(x )
to
h →0
the
f
lim
6.
for
p
<
x
<
q.
Find
the
values
of
p
and
q.
.
on
which
f
is
concave
up.
2
x
3
Find
at
the
the
equation
point
of
the
normal
line
to
the
f
cur ve
(x )
=
1
4 xe
(1, 4).
3
4
5
Find
the
coordinates
at
which
A
graph
a
the
of
Write
tangent
y
=
f
down
(x)
f
on
the
line
is
(2),
f
is
graph
of
parallel
f
(x)
to
the
=
2x
3x
line y
=
+
5x
1
–
2
y
given.
′ (2)
and
f
″ (2)
in
order
4
from
y
=
f(x)
3
the
greatest
value
to
the
least
value.
2
b
Justify
your
answer
from
par t
a
1
3
6
A
cur ve
has
the
equation
y
=
x
(x
–
4)
0
x
1
2
3
4
5
–1
2
dy
a
d
y
–2
Find
i
ii
2
dx
b
c
For
this
A
the
x-intercepts
iii
the
coordinates
Use
your
the
answers
indicating
par ticle
from
moves
origin
a
Find
the
b
Find
when
c
Show
Use
a
Find
is
the
the
of
to
the
par t
the
coordinates
points
b
to
features
a
you
by
s (t)
function
par ticle
is
of
for
=
relative
minimum
to
limit
examine
or
state
each
that
a
graph
found
line
20t
–
in
such
100
of
the
cur ve,
par t b
that
ln
t,
t
its
≥
displacement
1.
to
the
par ticle
is
left.
always
increasing.
it
function
does
not
graphically
and
numerically
.
exist.
2
2
1
a
1
lim
x →2
b
x
2
point
s
moving
the
the
inflexion.
sketch
horizontal
given
velocity
of
of
exercise
GDC
the
O
the
ii
along
velocity
that
Review
1
find:
i
clearly
7
cur ve
dx
x
lim
x →3
c
x
2
x →4
x
16
d
lim
x →1
x
4
+ 3
lim
x
1
Chapter
Exam-Style
2
A
10
foot
Question
post
and
per pendicular
the
on
a
top
the
of
to
each
ground
a
25
the
foot
post
ground.
pole
and
between
are
the
two
an
expression
for
y
ii
Write
an
expression
for
z
write
an
in
in
expression
30
a
as
apar t
single
of
terms
of
L(x),
and
and
shown
terms
for
feet
lengths y
by
poles,
Write
Hence
of
attached
i
iii
stand
Wires
z
r un
stake
in
the
at
from
a
point
figure.
x
x
the
total
length
of
z
25 ft
y
wire
used
for
both
poles.
10 ft
dL
b
Find
i
30
x
–
x
dx
Hence
ii
10
find
foot
CHAPTER
the
pole
7
in
distance
order
to
x
the
stake
minimize
should
the
be
placed
amount
of
from
wire
the
used.
SUMMARY
n
The
tangent
line
and
derivative
f
The
function
defined
(x
+ h) −
f
is
h →0
derivative
is
defined
by
f
′( x )
=
Power
rule
If
=
n
●
f
(x)
=
y
=
Sum
If
f
cf
or
(x)
More
derivative
of
f.
(x
+ h) −
f
(x )
dy
h
f
=
or
dx
(x
+ h) −
f
(x )
lim
h →0
h
then
f
′(x)
=
nx
,
where
n
∈
rule
where
c
multiple
(x),
u(x)
rules
±
is
c
real
number,
then
f
′(x)
=
0
is
any
real
number,
then
y ′
=
cf
′(x)
rule
v (x)
for
any
rule
where
dierence
=
the
n−1
,
c,
Constant
If
x
Constant
If
f (x)
as
lim
h →0
known
h
f
The
x
(x )
lim
by
of
then
f
′(x)
=
u ′(x)
±
v ′(x)
derivatives
x
Derivative
e
of
x
If
f
(x)
=
e
Derivative
x
,
then
of
ln
f
′(x)
=
e
′( x )
=
x
1
If
f
(x)
=
ln x,
then
f
x
The
If
f
product
(x)
The
=
rule
u(x) · v (x)
quotient
then
f
′(x)
f
u(x) · v ′(x)
+
v(x) · u′(x)
rule
v ( x ) ⋅ u ′( x ) − u ( x ) ⋅ v ′( x )
u( x )
If
=
then
( x ) =
f
′( x )
=
2
v(x )
[v ( x
)]
Continued
Limits
and
derivatives
on
next
page
The
The
If
chain
f
chain
(x)
The
=
rule
and
higher
order
derivatives
rule
u(v(x))
chain
r ule
then
can
f
′(x)
also
=
be
u ′(v(x)) · v′(x)
written
as:
dy
If
y
=
f
(u),
u
=
g (x)
and
y
=
f
(g(x)),
dy
⋅
dx
Rates
The
of
change
instantaneous
and
rate
of
s (t
function,
(t )
=
The
change
+ h) −
s
of
=
rate
of
function ,
derivative
change
a (t )
=
of
the
function
is
decreasing,
When
a
function
is
increasing,
= v ′( t )
the
tangent
tangent
lines
lines
′(x)
>
0
for
all
x
in
(a, b)
then
f
is
increasing
f
′(x)
<
0
for
all
x
in
(a, b)
then
f
is
decreasing
is
used
to
first
f
If
f
f
′(x)
at
if
f
′′(x)
More
For
f
=
on
for
is
0
Write
all
negative
in
f
the
curve
curve
have
have
negative
positive
slope.
slope.
It
locate
on
(a, b).
on
relative
(a, b).
extrema
of
f.
If
f
is
defined
at
a
to
negative
at x
=
c,
then
f
has
a
relative
maximum
to
positive
at x
=
c,
then
f
has
a
relative
minimum
then
′′(x)
on
down
f
is
(a, b).
points .
changes
and
concave
A
The
up
on
points
point
on
(a, b).
on
the
a
If
f
graph
graph
of
′′(x)
<
where
f
is
an
0
for
the
all
x
in
concavity
inflexion
point
sign.
optimization
problems
problems:
to
given
draw
a
to
quantities
and
quantities
to
be
determined.
sketch.
be
optimized
(minimized
or
maximized)
in
terms
variables.
the
0
that
of
the
you
are
or
the
at
sensible
equation
have
domain
since
=
(a, b)
inflexion
equation
that
If
′ (x)
from
variables
values
tested
positive
concave
and
an
Verify
test.
from
x
possible
two
Find
test
extrema
Assign
f
to
the
(c)).
called
derivative
to
(c)).
optimization
of
(c, f
are
When
s ′′( t )
then:
changes
0
then
changes
(c, f
at
>
c
changes
′(x)
′′(x)
(a, b)
derivative
number
point
If
=
that:
point
the
f
If
the
h
If
is
(t )
If
The
the veloci ty
graphing
a
critical
and
velocity
+ h) − v
lim
When
follows
is
s ′( t )
h →0
line
displacement
v (t
The
a
dx
h
instantaneous
acceleration
in
du
(t )
lim
h →0
motion
du
=
then
a
is
feasible
be
such
that
or
for
optimized
minimum
maximum
an
or
to
a
or
≤
the
maximum
x
≤
b,
minimum
problem
equals
using
remember
on
a
situation
where
the
zero.
closed
the
that
first
the
inter val
or
second
derivative
endpoints
must
can
when
occur
be
endpoint.
Chapter
Theory
of
knowledge
Truth
Inductive
Inductive
in
reasoning
.
looks
reasoning
generalization.
this
mathematics
Use
at
inductive
par ticular
reasoning
cases
to
to
make
a
make
a
conjecture
about
problem.
Copy
the
the
circles
points
on
and
each
table.
circle.
Draw
Count
all
possible
the
number
chords
of
connecting
non-overlapping
The
regions
in
the
interior
of
each
the
results
in
the
comple
table.
If
you
based
of
points
on
the
Number
of
on
the
for
the
regions
number
circle
your
obvious
pattern
Number
you.
for
ted
conjecture
most
y
alread
are
s
circle
Record
two
first
circle
of
regions
formed
formed,
2
find
2
not
3
you
that
true
will
it
was
for
n
=
6.
4
4
5
.
Describe,
regions
in
words,
any
patterns
you
obser ve
for
the
number
of
formed.
How
does
a
have
to
before
it
.
Make
that
a
conjecture
are
formed
mathematical
many
by
about
the
number
connecting
n
of
points
non-overlapping
on
a
circle.
Write
is
Use
your
to
it
as
a
Can
we
know
predict
the
number
of
regions
repeat
we
it
ever
is
true
all
possible
chords
connecting
six
points
on
a
by
formed
inspecting
when
know
regions
expression.
conjecture
pattern
true?
always
.
times
circle
the
are
pattern?
drawn.
.
Draw
a
circle
with
six
points.
Draw
all
the
possible
Does
we
connecting
the
points
to
check
your
prediction
this
mean
chords
from
question
should
never
4.
use
inductive
reasoning?
Theory
of
knowledge:
Truth
in
mathematics
Deductive
reasoning
n
In
Section
7.1
we
conjectured
that
the
derivative
of
f (x)
=
x
n−1
is
f '(x)
=
nx
.
5
We
confirmed
reasoning
In
to
deductive
base
that
the
provide
conjecture
validity
reasoning
deductive
we
reasoning
to
was
our
reason
on
tr ue
from
basic
for f (x)
=
.
x
We
can
use
deductive
conjecture.
the
axioms,
general
to
the
definitions
specific.
and
In
mathematics
we
theorems.
n
Use
the
definition
of
derivative
n−1
f '(x)
=
nx
f '(x)
=
lim
and
the
binomial
theorem
to
show
that
if
f (x)
=
x
then
+
for
n
∊
n
f
(x
+
h)
–
f
Apply
(x)
the
denition
of
derivative
to
f(x)
=
x
and
n
h→0
then
h
n
(x
=
+
use
the
binomial
theorem
to
expand
(x
+
h)
n
h)
–
x
lim
h
h→0
n
n
(
)
[
=
0
n
x
0
h
+
n
n
)
(
1
n−1
x
)
(
1
h
+
2
n−2
x
n
(
2
h
+...+
)
1
x
n –1
n−1
h
+
)
(
n
0
x
n
h
] –
n
x
lim
h
h→0
n
n
[x
=
n
(
n−1
+
nx
h +
)
2
n−2
x
(
2
h
+...+
)
n−1
n
xh
n –1
+
h
] –
Simplify
n
where
possible
x
lim
h
h→0
n
n
=
Collect
(
n−1
nx
h +
)
2
n−2
x
(
2
h
+...+
)
n−1
xh
n –1
like
terms
n
+
h
lim
h
h→0
n
n
[nx
h
=
(
n−1
+
)
2
(
n−2
x
h +...+
Factorize
)
n –1
n−2
xh
n−1
+
]
h
lim
h
h→0
(
n−1
[nx
lim
+
Simplify
n
n
=
)
2
(
n−2
x
h +...+
)
n−2
xh
n –1
n−1
+
]
h
h→0
n
n
(
n−1
=
nx
+
Evaluate
)
2
(
n−2
(x
)(0) +...+
)
n –1
n−2
(x )(0)
the
limit
n−1
+
(0)
n−1
f '(x)
=
nx
Can
we
now
true
for
all
say
for
cer tain
that
the
conjecture
will
be
+
The
A
classic
'math
physicist
a
astronomer
,
a
mathematician
physicist
were
∊
?
disagreed:
Why
or
“No!
why
not?
What
kind
of
reasoning
was
joke'
Some
An
n
Welsh
sheep
are
black!”
the
mathematician
using?
and
traveling
re
through
Wales
by
train,
when
is
they
saw
middle
of
a
a
black
sheep
in
at
least
eld.
astronomer
said:
sheep
are
at
least
one
“All
side
Welsh
eld,
the
sheep,
The
one
of
which
is
black!”
black!”
Descriptive
CHAPTER
5.1
OBJECTIVES:
Population,
sample,
presentation
histograms
outliers;
width,
5.2
of
data:
with
class
data:
boundaries
and
measures.
Cumulative
Before
use
Y
ou
1
should
Draw
e.g.
a
Draw
distributions
of
modal
mid-inter val
bar
continuous
(tables);
and
frequency
whisker
values
for
plots;
calculations,
tendency:
range,
mean,
interquar tile
cumulative
median,
range,
frequency
mode;
variance,
in
the
inter val
quar tiles
standard
and
deviation.
graphs.
how
Skills
to:
1
char t
for
the
number
check
Draw
a
bar
char t
families
of
30
for
this
frequency
of
f
Favorite
children
data;
class.
char t.
a
box
and
start
know
bar
discrete
inter vals;
Central
frequency;
you
sample,
frequency
equal
grouped
Statistical
random
percentiles. Dispersion:
5.3
statistics
students
color
in
the
frequency
table
below
.
Red
6
Blue
8
y
f
Children
12
1
Pink
8
10
10
12
3
5
4
3
5
2
ycneuqerF
2
e.g.
c
Find
the
mean,
a
the
median
x
2
mode
mean
of
2,
and
b
3,
the
3,
5,
4
3
of
Mean
=
Mode
=
c
Median
3
=
5
Descriptive statistics
6
median.
mode
6,
7,
2
and
9
35
=
7
b
5
children
2 + 3 + 3 + 5 + 6 + 7 + 9
a
4
2
Number
the
Black
4
1
Find
9
6
0
2
Purple
8
=
7
a
Find
the
mean
of
4,
7,
7,
8,
6
b
Find
the
mode
of
5,
6,
8,
8,
9
c
Find
the
median
i
6,
4,
8,
7,
ii
5,
7,
9,
11,
11,
iii
6,
8,
11,
of
2,
4
5
13,
11,
15
14,
17
table.
Statistics
are
visible
all
around
us.
Averages
(such
as
the
y
mean,
10
mode
and
median)
and
char ts
(such
as
bar
graphs,
line
graphs
9
and
pie
char ts)
it.
Y
ou
to
media.
have
ever yday
used
We
probably
ever ywhere
use
statistics
made
conversation
or
some
–
from
ever y
business
day
statistical
thinking.
to
without
such
to
realizing
statements
Statements
spor ts
ycneuqerF
fashion
are
as
in
‘I
your
8
7
sleep
6
for
about
eight
hours
per
night
on
average’
and
‘Y
ou
are
earlier’
are
more
5
likely
to
pass
the
exam
if
you
star t
preparing
1875
1900
1925
1950
1975
2000
2025
x
Year
actually
statistical
in
nature.
Statistics
Statistics
is
concer ned
a
●
designing
●
representing
experiments
and
and
analyzing
other
data
collection
information
to
drawing
●
estimating
the
from
present
or
this
the
future.
know
help
This
chapter
explains
these
techniques
and
how
them
to
real-life
science
situations.
used
to
of
data.
organize
It
is
and
data.
chapter
how
your
you
on
to
can
your
do
do
GDC,
them
by
most
but
if
hand
understanding.
The
of
the
you
too,
it
will
emphasis
to
is
apply
the
tools
calculations
data
predicting
of
analyze
In
conclusions
set
aid
understanding
●
is
with:
on
the
understanding
results
Statistical
you
tables
examinations
your
and
obtain,
–
are
you
interpreting
in
context.
not
will
allowed
need
to
in
use
GDC.
Chapter
Investigation
–
what
test
32
students
a
test
1,
1,
2,
2,
2,
3,
3,
4,
7,
7,
7,
8,
8,
8,
8,
9,
10
What
should
How
could
How
should
Should
What
Can
.
you
you
of
of
use
display
an
you
draw
any
much
Data
is
either
the
called
and
to
that
give
from
How
your
do
you
involves
is
in
a
your
this
single
class.
scores
data.
you
that
or
can
Comparing
and
data
variable,
Y
ou
will
you
6,
seen
as
to
for
their
were:
7,
7,
of
see
this
the
scores?
grades?
gather
example,
charts,
in
and
or
data
pen
school?
computer
do
you
Chapter
is
weights
10.
classified
as
data
data
that
describes
can
be
counted
that
give
quantitative
include:
How
many
How
long
to
How
own?
and
height
averages
measured.
Questions
color?
the
find
data.
information
qualitative
to
letter
heights
Quantitative
data.
give
travel
7,
picture
draw
quanti tative
sometimes
favorite
of
results
6,
scores?
get
brand
6,
better
data
is
What
Their
6,
a
the
the
include:
What
5,
Quantitative
is
our
data?
conver ting
analysis
categorical
Questions
data
scores?
10.
5,
this
the
information
data
categories
of
5,
data
Qualitative
with
analysis
quali tative
Qualitative
out
5,
with
conclusions
with
bivariate
5,
the
about
students
more
do
do
average?
do
analysis
the
called
➔
organize
you
4,
we
scores?
scored
4,
teacher
Univariate
all
and
the
you
should
Univariate
is
took
0,
should
pens
does
do
it
you
take
own?
you
to
school?
many
computers
have
you
owned?
[
Is
the
data
from
our
test
scores
qualitative
or
Discrete.
quantitative?
How
➔
Quantitative
and
➔
A
Here
we
A
can
be
split
up
into
two
categories: discrete
are
of
discrete
working
CDs
that
quantitative
accuracy
variable
has
exact
numerical
with values of , , , ,…, for
you
have
or
continuous
depends
instr ument
do
pairs
you
of
own?
on
the
the
number
variable
accuracy
of
can
of
the
example,
children
be
values.
in
measured
the
your
and
family
.
its
measuring
used.
[
Continuous
variables,
such
as
length,
weight
and
time,
may
have
Continuous.
What
the
fractions
many
shoes
continuous
quantitative
number
➔
data
or
decimals.
Descriptive statistics
is
the
train?
speed
of
What
is
and
sample?
a
When
we
the
dierence
think
of
the
term
between
population ,
a
we
population
usually
think
Population
of
people
in
our
town,
region,
state
or
countr y
.
Sample
➔
In
statistics,
defined
➔
A
part
the
Ever y
2
The
the
sample
the
we
are
population
a
has
have
has
an
called
of
two
for
all
data
a sample.
individuals
members
driven
It
is
from
a
a
decisions.
subset
the
of
of
population.
characteristics:
equal
essentially
includes
studying
is
selection
must
oppor tunity
the
same
of
selection.
characteristics
A
each
of
a
The
number
b
The
length
c
The
time
d
The
number
the
test
continuous
.
population
population.
Classify
Are
ter m
that
individual
Exercise
2
of
samples
1
1
g roup
population,
Random
as
the
the
of
of
following
fish
the
taken
caught
catch
friends
scores
either
by
an
discrete
or
continuous
data.
angler.
fish.
to
of
as
at
the
a
fish.
that
star t
the
of
angler
the
took
chapter
with
him.
discrete
or
data?
Presenting
A
data
bar
char t
column
A
frequency
table
is
an
easy
way
to
view
is
sometimes
called
a
graph.
your
y
data
Y
ou
quickly
can
and
also
look
show
for
8
patter ns.
discrete
data
in
a bar chart
ycneuqerF
6
4
2
0
x
20
21
22
23
Cars
Example
A
student
inter vals
counted
for
30
how
22,
22,
24,
22,
23,
22,
27,
26,
25,
28,
a
this
bar
many
minutes.
22,
Draw
25
26
27
28
minute
23,
Display
24
per
data
char t
in
a
for
His
cars
passed
results
21,
23,
23,
27,
26,
22,
20,
21,
20.
this
house
in
one-minute
were:
21,
frequency
his
21,
21,
22,
23,
25,
27,
26,
23,
table.
data.
{
Continued
on
next
page
Chapter
Answer
Tally
Number
of
cars
Tally
each
cor rect
per
data
item
in
the
Frequency
row.
Write
the
totals
in
minute
the
20
||
frequenc y
column.
2
21
5
The
22
number
times
23
21
appears
5
7
||
in
the
data.
6
|
24
|
1
25
||
2
26
|||
3
27
|||
3
28
|
1
A
y
bar
char t
discrete
is
data
suitable
and
may
for
have
8
gaps
between
the
bars.
ycneuqerF
6
Use
the
ver tical
frequenc y
and
scale
the
for
the
horizontal
4
scale
2
0
21
22
23
Cars
When
in
number
of
cars
per
x
20
➔
for
minute
a
you
24
per
25
27
28
minute
have
grouped
26
a
lot
of
data,
frequency
you
can
organize
it
into
groups
table
Why
For
continuous
data,
you
can
draw
a histogram.
It
is
in
to
a
bar
Example
char t
but
it
doesn’t
have
gaps
between
the
are
ages
no
gaps
continuous
data?
bars.
Only
The
there
similar
of
200
members
of
a
tennis
club
frequency
are:
histograms
20,
22,
23,
24,
25,
25,
25,
26,
26,
26,
26,
28,
28,
29,
29,
29,
30,
30,
30,
30,
30,
30,
30,
32,
32,
33,
33,
33,
34,
34,
34,
34,
34,
34,
34,
34,
35,
35,
class
with
inter vals
equal
will
be
examined.
35,
35,
36,
36,
36,
36,
36,
37,
37,
37,
38,
38,
38,
39,
39,
39,
40,
40,
40,
41,
41,
41,
42,
42,
42,
42,
42,
42,
42,
42,
43,
43,
43,
43,
43,
43,
44,
44,
44,
44,
44,
44,
45,
45,
45,
45,
45,
45,
45,
45,
46,
46,
46,
46,
46,
46,
46,
46,
47,
47,
47,
47,
47,
47,
47,
47,
47,
48,
48,
48,
48,
48,
48,
48,
48,
48,
49,
49,
49,
49,
49,
49,
49,
49,
50,
50,
50,
50,
50,
50,
51,
51,
51,
51,
51,
51,
51,
52,
52,
52,
52,
52,
53,
53,
53,
53,
53,
53,
53,
53,
53,
54,
54,
54,
54,
55,
55,
55,
55,
55,
56,
56,
56,
57,
57,
57,
57,
57,
57,
57,
57,
57,
57,
Having
each
us
58,
58,
58,
59,
59,
59,
60,
60,
60,
60,
63,
63,
64,
64,
64,
64,
65,
65,
68,
69.
60,
61,
61,
61,
62,
62,
62,
63,
a
age
63,
a
grouped
frequency
Descriptive statistics
table
and
histogram
for
the
data.
{
Continued
on
next
page
line
would
table
deep!
Draw
one
50
for
give
lines
Answer
Age
20
≤
age
<
25
25
≤
age
<
30
30
≤
age
<
35
35
≤
age
<
40
Tally
Frequency
||||
4
Equal
12
||
25
is
class
in
the
inter vals
class
25
of
≤
5
age
years
<
30
20
18
|||
40
≤
age
<
45
26
|
45
≤
age
<
50
42
||
50
≤
age
<
55
31
|
55
≤
age
<
60
24
Numbers
of
||||
the
like
60
≤
age
<
65
on
bars
the
or
edges
an
x-axis
scale.
19
||||
No
gaps
between
the
bars
Y
ou
65
≤
age
<
70
||||
can
use
draw
GDC
histograms.
Chapter
to
17,
See
Section
ycneuqerF
45
5.4.
30
15
GDC
help
on
CD:
demonstrations
20
25
30
35
40
45
50
55
60
65
70
Plus
Exercise
day
the
they
T ime
Casio
the
TI-84
are
on
FX-9860GII
the
CD.
B
EXAM-STYLE
of
and
GDCs
Age
All
Alternative
for
x
0
1
a
4
QUESTION
IB
students
studied
in
a
school
mathematics.
were
The
asked
results
how
are
many
given
in
minutes
the
a
table.
spent
studying
0
≤
t
<
15
15
≤
t
<
30
30
≤
t
<
45
45
≤
t
<
60
60
≤
t
<
75
75
≤
t
<
90
mathematics
(min)
Number
of
21
32
35
41
27
11
students
a
Is
b
Use
this
data
your
represent
continuous
GDC
this
to
help
or
discrete?
you
draw
a
fully
labeled
histogram
to
data.
Chapter
EXAM-STYLE
QUESTION
Number
The
2
following
table
shows
the
age
distribution
of
of
Age
mathematics
teachers
teachers
the
work
a
Is
b
How
many
High
School?
Use
c
to
The
3
who
data
your
at
Caring
discrete
or
represent
following
to
this
School.
20
≤
x
<
30
5
30
≤
x
<
40
4
40
≤
x
<
50
3
50
≤
x
<
60
2
60
≤
x
<
70
3
continuous?
mathematics
GDC
High
help
teachers
you
draw
work
a
at
fully
Caring
labeled
histogram
data.
histogram
shows
data
about
frozen
y
chickens
60
in
≤
supermarket.
w
<
2,
the
b
Draw
c
How
w
mass
the
<
of
3
and
frozen
grouped
many
so
in
are
grouped
such
that
on.
chickens
frequency
frozen
kg
chickens
discrete
table
are
for
there
or
continuous
this
in
data?
histogram.
the
rebmuN
Is
≤
masses
fo
a
2
The
snekcihc
1
a
supermarket?
50
40
30
20
10
0
x
1
2
3
Mass
The
4
histogram
on
the
right
shows
how
many
minutes
4
5
6
(kg)
y
it
5
takes
for
a
Is
b
Represent
c
What
get
data
is
to
retur n
discrete
the
the
data
or
in
shor test
home
after
school.
ycneuqerF
the
students
continuous?
a
grouped
time
that
a
frequency
student
table.
could
take
4
3
2
1
to
home?
0
x
5
10
15
20
T ime
.
A
a
Measures
measure
set
of
data
tendency
The
➔
of
central
lies.
are
of
the
The
central
tendency
three
mode,
the
tells
most
30
35
40
45
(min)
tendency
us
where
common
mean
25
and
the
the
middle
measures
of
of
central
Another
median.
word
is
‘average’.
mode
The
mode
is
the
value
that
occurs
most
frequently
in
a
set
of
data.
The
mode
often.
that
in
a
list
Remember
mo st
of
numbers
that
does.
mo de
is
the
star ts
number
with
the
that
same
occurs
first
most
two
letters
There
than
number
Descriptive statistics
more
occurs
once
then
there
that
be
mode.
than
for
can
one
in
is
set
no
more
the
no
of
If
set,
mode
numbers.
Example
Find
the
mode
of:
9,
3,
9,
41,
17,
17,
44,
15,
15,
15,
27,
40,
13
Answer
The
mode
When
the
is
(15
presented
biggest
the
occurs
with
a
the
most
frequency
at
3
times).
table,
the
mode
is
the
group
with
frequency
.
Example
Find
15
mode
or
modal
class
for
a
these
frequency
tables.
b
Goals
Frequency
0
4
1
7
5
2
3
10
≤
t
<
15
6
3
3
15
≤
t
<
20
7
4
1
20
≤
t
<
25
6
T imes
0
≤
≤
Frequency
t
t
<
<
5
1
10
5
Answers
a
The
mode
is
1
goal.
Common
1
2
The
No.
The
The
mode
No.
The
is
b
The
15
modal
≤
t
<
Exercise
1
Find
a
class
The
is
is
7.
biggest
is
frequenc y
is
7.
3.
most
common
frequenc y
3.
mode
table
20.
errors:
mode
is
from
called
a
the
grouped
modal
frequenc y
class.
C
the
modes
7,13,18,24,
of
these
sets
of
data.
A
9,3,18
set
of
bimodal
b
8,11,9,14,9,15,18,6,9,10
c
24,15,18,20,18,
d
−3,
two
4,
0,
−2,
12,
2,
7,
4,
2,
1,
9,
0,
0,
3.5,
Find
the
mode
of
each
a
3,
0,
it
is
has
modes.
,
4
5
1
3
,
2
2
if
22,24,26,18,26,24
1
e
data
,
11
2
frequency
table.
b
Goals
Frequency
Height
Frequency
0
4
140
≤
h
<
150
6
1
7
150
≤
h
<
160
6
2
3
160
≤
h
<
170
5
3
3
170
≤
h
<
180
10
4
1
180
≤
h
<
190
8
Chapter
The
The
the
mean
arithmetic
most
➔
The
of
is
mean
common
mean
is
numbers
usually
measure
the
in
a
sum
set
Sum of
Mean
of
of
of
the
mean
data
It
representative
score
for
teacher
the
us
is
a
the
single
usually
value.
year
always
the
mean
or
average
and
is
tendency
.
numbers
divided
by
the
number
data values
=
gives
set.
the
data.
Number of
The
called
central
number
not
For
may
gives
data values
a
member
example,
be
85.73%
scores
that
that
of
your
even
are
indicates
the
data
average
though
whole
a
center
set
but
lower
case
Gr eek
l etter
μ
is
the
pronounced
Σ
(which
the
tells
sum
pronounced
a
and
mathematics
N
is
us
‘mu’,
to
here)
Is
‘sigma’
‘nu’.
your
numbers.
symbol
for
is
often
a
misunderstanding
t he
between
population
is
nd
of
There
The
μ
the
population
mean
and
mean.
sample
mean.
The
population
mean
x
Population
μ
mean
=
uses
Greek
letters
whereas
the
N
sample
where
Σx
is
the
sum
of
the
data
values
and
N
is
of
data
Example
Find
the
values
in
the
uses
and
x
n.
uses
only
the
population
mean
of
89,73,84,91,87,77,94
b
2,
GDC
help
on
CD:
demonstrations
3,
mean.
population.
a
3,
Our
the
course
number
mean
4,
6,
Alternative
for
the
TI-84
7
Plus
and
GDCs
Casio
are
on
FX-9860GII
the
CD.
Answers
x
a
89
73 84
91 87 77
94
Y
ou
N
can
calculate
the
7
mean
from
a
list
on
595
your
85
GDC.
In
one-
7
variable
x
b
2 3 3
4
6 7
N
6
the
25
statistics
GDC,
the
can
also
calculate
is
6
Descriptive statistics
the
mean
mean
4.16
x .
The
GDC
calculates
Y
ou
on
from
a
frequency
table.
Σ x
also
and
n.
Example
Find
the
mean
of
each
set
of
data
a
displayed
below
.
b
(g)
Grade
0
11
1
10
2
19
3
10
(a)
Age
Frequency
Frequency
10
≤
t
<
12
4
12
≤
t
<
14
8
14
≤
t
<
16
5
16
≤
t
<
18
3
Answers
a
(g)
Frequency
fg
0
11
0
1
10
10
2
19
38
Grade
Add
f
×
The
a
3rd
column;
fg
means
g.
total
of
the
fg
column
is
the
sum
This
of
all
of
the
it
3
10
is
the
as
appears
in
the
IB
30
The
total
of
the
f
column
is
the
Formula
T
otal
formula
grades.
50
78
number
of
booklet:
grades.
n
∑
fg
∑
Mean
f
x
i
i
78
i
=
=
=
μ
1.56
= 1
=
n
∑
50
f
∑
i
b
When
(a)
Age
f
data
are
grouped,
we
i
can
fm
Mid
calculate
point
(m)
all
10
≤
t
<12
4
11
44
12
≤
t
<
8
13
104
14
the
f
= 1
of
the
the
spread
mean
data
by
values
around
the
assuming
are
that
equally
midpoint.
This
method
small
and
that
is
≤
t
<
16
5
15
75
16
≤
t
<
18
3
17
51
often
T
otal
20
say,
Mean
=
mean’.
not
mean
=
=
‘guess’
work
out,
–
it
as
in
13.7
this
example
or
with
GDC.
Lana’s
mathematics
she
on
get
does
20
your
Example
It
274
274
f
∑
questions
‘estimate
the
means
fm
to
why
examination
14
leads
inaccuracies
the
fifth
test
test
grades
in
order
are
to
87,
get
a
93,89and
mean
of
85.
90
What
for
the
score
must
term?
Answer
μ
x
=
Select
the
mean
for mula.
N
87
90
+
93
+
89
+
85
+
x
Substitute
the
infor mation
into
the
=
for mula.
450
=
354
x
=
96
Lana
must
+
x
score
Solve
96
on
her
fifth
for
Answer
x.
the
question.
test.
Chapter
Exercise
1
Find
D
the
mean
driving
speed
for
6
different
−1
road
if
their
speeds
are
−1
66 km h
cars
on
the
−1
,
57 km h
same
−1
,
71 km h
−1
,
69 km h
,
−1
58 km h
and
54 km h
Ronald
.
F isher
(1890–1962)
2
The
price
$1.61,
of
buying
$1.96
and
music
$2.08
from
per
different
track.
What
sites
was
was
the
seen
mean
as
$1.79,
price?
in
the
UK
Australia
called
3
A
computer
repair
ser vice
received
the
following
number
of
and
the
day
over
a
period
of
30
5
6
9
7
4
2
4
7
8
3
4
9
8
2
3
5
9
7
8
9
7
5
6
7
7
4
6
2
4
practical
the
b
Constr uct
calls
data
per
discrete
a
or
biology
The
table
table
and
find
the
mean
number
could
of
as
day
.
sunshine
below
in
the
≤
m
first
<
the
100
(m)
Minutes
0
shows
be
the
30
m
<
60
16
60
≤
m
<
90
20
120
36
120
m
≤
<
m
a
Is
the
b
What
<
150
data
days
of
of
the
minutes
year
in
of
statistics.?
of
Newtown.
or
continuous?
modal
Find
the
mean
number
of
minutes
of
Kelly’s
get
on
five
6
the
mean
new
Find
fifth
The
player
the
must
8
in
82,
order
76
to
and
88.
achieve
What
an
score
average
of
must
84
she
on
be
of
their
the
players
team
new
family
and
in
a
the
spor ts
mean
team
goes
is
up
80.3 kg.
to
81.2 kg.
player.
must
vacation
300 km,
they
eleven
the
travel
on
210 km,
on
mean
total
of
the
Tigger’s
number
Descriptive statistics
of
mean
drive
an
time.
On
275 km
sixth
day
average
the
and
in
last
shots
8
rounds
of
first
to
250 km
five
240 km.
order
that
he
of
took
formula
days,
How
finish
per
day
they
many
their
golf
in
is
the
71
8
shots.
rounds?
to
travel
km
vacation
What
is
technology
seen
papers.
all
time?
The
the
95,
QUESTIONS
Onceonly
220 km,
of
joins
mass
complete
on
are
test
mass
EXAM-STYLE
7
scores
tests?
The
A
test
both
sunshine.
and
5
of
class?
using
c
or
16
discrete
the
else
considered
father
,
Calculation
is
social
Who
12
≤
≤
number
in
astronomy,
f
30
90
used
analyze
problems
and
science.
inventor
4
He
to
agriculture,
continuous?
frequency
of
days.
6
Is
often
calls
statistics
a
is
‘Father
statistics’.
per
lived
and
on
may
exam
After
9
8
points.
10
matches,
After
points
did
Billy’s
mean
13
with
sales
Jean
The
➔
at
a
she
sales
end
combine
basketball
in
the
price
of
of
the
her
last
for
$320.
their
player
matches
score
mean
the
a
more
12
3
a
mean
was
score
29.
of
How
27
many
games?
computers
Their
week.
had
average
boss
What
tells
is
is
$310
them
the
to
mean
and
Jean
sold
combine
after
Billy
their
and
sales?
median
The
a
set
median
of
the
two
Example
Find
the
13,
7,
is
data
numbers
2,
3
in
the
are
a
arranged
data
middle
number
set
is
in
in
the
middle
order
even,
then
of
when
size.
the
If
the
the
median
numbers
number
is
the
in
of
mean
of
numbers.
median
of
5,
19,
23,
39,
23,
42,
23,
7,
12,
13,
14,
19,
23,
23,
23,
29,
39,
42,
55
14,
12,
55,
23,
29.
Answer
2,
5,
23,
The
of
median
numbers
value
is
of
this
Write
the
numbers
in
order.
There are 15 numbers. Our middle
set
number will be the 8th number:
23.
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
Y
ou
Casio
are
on
can
median
nd
Common
➔
If
there
are
a
lot
of
numbers
and
it
is
difficult
to
find
the
TI-84
FX-9860GII
the
on
Alternative
for
CD.
the
your
GDC.
error.
This
the
formula
does
not
give
⎛ n +1 ⎞
middle
member
we
can
use
the
formula
Median
=
th
⎜
⎝
member,
where
n
is
the
number
of
members
in
the
⎟
2
the
median.
It
the
position
of
set.
median
Exercise
Find
a
2,
the
3,
median
4,
5,
6,
of
7,
2,
the
3,
c
9,
3,
4,
6,
7,
2,
3,
d
8,
1,
2,
4,
5,
9,
12,
e
12,
in
the
the
data
set.
E
The
1
gives
⎠
4,
9,
1,
20,
7,
following.
4
b
2,
19th-centur y
Gustav
5,
5,
2,
7,
3,
8
median
data,
0
0,
2,
4,
1.5,
8.4
Fechner
into
the
although
formal
French
and
astronomer
had
used
it
German
psychologist
popularized
the
analysis
of
mathematician
Pierre-Simon
Laplace
previously.
5
Chapter
2
Su
has
been
collection.
3
counting
Find
the
the
number
median
of
number
tracks
of
on
tracks
the
on
CDs
Su’s
Number
of
tracks
7
8
9
10
11
12
13
Number
of
CDs
3
2
2
1
3
5
3
Find
of
the
the
mode,
mean
and
median
of
our
test
scores
in
her
CDs.
at
the
star t
chapter.
Summary
of
measures
of
central
tendency
Advantages
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The
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for
or
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values
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asked
the
Does
of
●
qualitative
choose
popular
can
Extreme
Disadvantages
the
Not
use
data
all
be
members
set.
necessarily
may
most
not
more
unique–
than
one
answer
.
●
item.
When
the
no
data
values
set,
repeat
there
is
in
no
mode.
●
When
one
there
mode,
interpret
●
Mean
The
mean
describes
Most
popular
elds
such
measure
as
in
●
Affected
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set
engineering
data.
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Uses
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side
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is
the
it.
the
only
middle
data
●
Not
●
Less
as
mean?
is
popular
used
fur ther
answer
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the
the
comparing
mean.
sets
the
data.
strongly
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when
of
affect
data.
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sets
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affect
●
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set.
value
one
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set
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extreme
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a
difcult
science.
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of
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computer
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and/or
salaries
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it
more
business,
In
the
is
as
mean.
in
calculations.
Investigation
What
will
amount
Copy
to
happen
to
and
all
to
data
the
the
measures
measures
values,
complete
calculate
–
this
mean,
or
Y
ou
and
central
central
multiply
table.
mode
of
of
each
should
median
tendency
data
use
value
your
each
6,
7,
8,
4
data
to
in
data
Now
in
by
each
the
Multiply
set
piece
each
piece
original
same
amount?
Mode
Median
14,
15,
16,
20
of
of
data
2.
and
complete
happens
to
4
the
a
If
you
add
b
If
you
multiply
.
the
same
GDC
Mean
12,
add
the
set.
the
copy
what
we
set
14,
Add
10,
if
by
time.
Data
Data
tendency
to
Measures
of
mean,
each
each
Measures
central
the
mode
data
data
of
following
sentences
and
median
to
of
explain
the
original
data
set.
value………………………………………
value
by
2………………………………….
dispersion
tendency
(mean,
median,
mode)
explore
the
When
middle
of
a
data
set.
Measures
of
dispersion
describe
the
spread
data
the
data
around
a
central
you
value.
give
set
at
you
least
measure
➔
The
is
range
the
difference
between
the
largest
and
describe
a
of
of
should
one
central
smallest
tendency
and
one
of
values.
dispersion.
The
be
range
is
affected
data
For
is
the
by
measure
values.
It
of
dispersion
doesn’t
tell
you
to
calculate
how
the
but
it
can
remaining
distributed.
example,
lowest
score
is
0
10
simplest
extreme
–
=
for
is
0
the
test
and
the
scores
at
highest
the
star t
score
is
of
10.
the
chapter,
Therefore
the
the
range
10.
Quartiles
The
less
of
median
than
data
(25%)
the
into
of
of
a
of
median,
four
the
set
data
half
equal
separates
greater.
par ts.
the
data
Quarti les
Each
of
these
into
two
separate
par ts
halves
the
–
half
original
contains
set
one-quar ter
data.
Chapter
➔
F irst
quar tile
The
of
first
the
One
also
has
into
quar ter
quar tile
of
and
called
the
is
quarti le
way
the
the
data.
the
data
value
lies
three-four ths
the
25th
symbol
one-quar ter
below
lies
the
above.
percentile
and
rst
It
is
often
Q
1
Second
quar tile
The
the
second
median
also
Third
quar tile
called
The
third
way
in.
the
It
has
the
the
50th
quarti le
quar tile
also
the
called
set
of
name
data
for
and
is
percentile.
is
three-quar ters
of
and
the
symbol
another
entire
Three-four ths
third
is
is
quarti le
of
the
data
one-four th
75th
of
lies
lies
percentile
the
below
above.
and
Q
3
1
Q
3
=
(n
+
1)th
value
and
Q
1
=
(n
4
the
Y
ou
get
statistical
a
minimum,
maximum,
median
the
first
the
third
The
(or
where
n
is
of
data
sense
of
values
a
data
in
the
set’s
data
set.
distribution
by
examining
a five
second
quar tile),
quar tile,
quar tile.
shows
near
value
summary ,
or
1)th
4
number
can
This
+
3
the
GDC
the
extent
to
which
the
data
is
located
near
the
median
extremes.
calculates
these
five
values
in
One-Variable
GDC
help
on
CD:
demonstrations
Alternative
for
the
TI-84
Statistics.
Plus
and
GDCs
Here
is
a
five
Minimum
statistical
F irst
quar tile
65
Y
ou
you
do
tells
70
not
80.
you
and
know
half
of
First
that
test
scores
quar tile
the
a
set
Third
80
ever y
the
for
Median
70
that
above
summar y
=
middle
70
but
below
and
50%
of
test
quar tile
third
the
and
=
80
CD.
are
=
tells
are
90
between
90.
Y
ou
can
median
on
the
Maximum
half
quar tile
scores
on
100
median
80
are
FX-9860GII
scores.
90
score,
are
of
Casio
Descriptive statistics
a
nd
and
GDC.
the
quar tiles
See
Chapter
17,
5.7
5.8.
and
Sections
➔
The
the
difference
between
interquarti le
the
third
(IQR)
range
=
and
Q
first
−
Q
3
The
IQR
is
sometimes
interquar tile
range
is
called
‘the
quar tiles
is
called
.
1
middle
half ’.
Here
the
20.
Y
ou
can
GDC
to
use
the
calculate
interquar tile
See
Chapter
Section
➔
A
five
statistical
summar y
can
be
represented
graphically
as
a
box
and
whisker
17,
5.9.
Sometimes
whisker
the
range.
a
plot
box
is
and
just
plot.
called
a
box
plot.
Range
Whisker
Interquartile
Range
Whisker
The
diagram
drawn
Min
X
Q
m
Q
1
Max
X
to
example
3
should
scale,
on
be
for
graph
(Median)
paper
.
The
is
first
and
shown
by
minimum
plot
third
a
are
shows
quar tiles
ver tical
at
the
the
data
line
ends
from
are
in
of
at
the
the
page
the
box,
ends
and
whiskers.
of
the
the
box,
the
maximum
This
box
and
median
and
whisker
268.
Y
ou
can
whisker
draw
box
plots
on
and
a
x
60
70
80
90
100
110
GDC.
See
Sections
GDC
help
on
screenshots
Plus
and
GDCs
Extreme
➔
An
or
distant
outlier
is
data
any
values
value
at
are
Chapter
5.5
CD:
for
Casio
are
on
and
17,
5.6.
Alternative
the
TI-84
FX-9860GII
the
CD.
called outliers.
least
1.5
IQR
above Q
3
or
below
Q
1
Chapter
Example
a
Find
and
the
the
range,
the
median,
interquar tile
range
the
of
lower
this
set
quar tile,
of
the
upper
quar tile
scores.
Y
ou
18,
27,
34,
52,
54,
59,
61,
68,
78,
82,
85,
87,
91,
93,
may
explore
b
Show
the
data
in
a
box
and
whisker
wish
to
some
of
100
misuses
Check
c
if
18
is
=
100
an
outlier.
Answers
a
Range
–
18
=
82
Range
=
largest
value
smallest
18,
27,
34,
52,
54,
87,
91,
93,
100
59,
61,
68,
78,
82,
85,
Write
the
data
There
are
15
in
data
in
–
value
order.
Median
n 1
=
2
=
15
th
8th
1
th
value
=
numbers
value
2
the
∴
68
n
=
set.
15.
1
Q
=
(n
+
1)th
value
1
4
1
=
(15
+
1)th
=
4th
value
=
52
4
3
Q
=
(n
+
1)th
value
3
4
3
=
(15
+
1)th
=
12th
value
=
87
4
IQR
=
Q
–
Q
3
=
87
–
52
=
35
1
Lower
Upper
quartile
quartile
b
Median
0
c
10
20
Q
–
30
40
50
1.5(IQR)
60
70
=
52
–
=
–0.5
80
90
1.5(35)
=
100
52
–
110
52.5
Outliers
are
more
than
1
∴
18
is
not
an
1.5
outlier.
×
below
IQR
Q
or
above Q
1
Exercise
1
The
on
30, 75,
Find a
in
31
of
the
55,
60,
range,
and
a
snow
Januar y
.
125,
quar tile
data
QUESTION
depths
years
3.
F
EXAM-STYLE
e
box
b
the
and
Descriptive statistics
at
a
All
75,
the
ski
resor t
data
65,
is
65,
45,
median,
interquar tile
whisker
in
plot.
c
the
plot.
are
collected
ever y
year
for
12
centimetres.
120,
the
range
70,
110.
lower
of
the
quar tile,
data
set
d
the
and
upper
show
the
of
statistics.
EXAM-STYLE
Here
2
76
are
QUESTIONS
Albie’s
79
76
test
scores
74
75
for
the
71
year.
85
82
82
79
81
Y
ou
Find a
the
range,
the
b
median,
c
the
lower
can
draw
the
d
upper
quar tile
and
e
the
interquar tile
range
of
the
data
scores
Here
3
are
ever y
and
the
hour
10, 11,
Find a
12,
the
quar tile
Use
4
c
the
the
of
6
Match
e
data
data
in
in
in
°C
21,
the
25,
median,
box
and
below
quar tile,
box
at
27,
interquar tile
a
a
a
and
hill
d
to
the
c
28,
whisker
resor t
in
Montana
the
whisker
find
a
upper
the
lower
of
quar tile,
the
data
d
the
8
4
plots.
taken
upper
plot.
range,
quar tile
b
and
the
e
median,
the
interquar tile
data.
each
3
whisker
set.
9
box
10
11
plot
with
the
correct
histogram.
c
x
2
and
plot.
b
1
to
and
29.
range
a
0
GDC
hours.
22,
b
the
plot
the
7
eleven
18,
range,
lower
range
5
for
box
the
temperatures
14,
and
Show the
show
a
histograms
set
box
of
use
quar tile,
5
6
7
8
9
x
10
0
1
2
3
4
5
6
7
8
9
x
10
0
1
2
3
4
5
6
7
8
9
10
y
y
i
ii
iii
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
7
6
5
4
3
2
1
0
x
1–2
3–4
5–6
7–8
9–10
0
x
1–2
.
➔
Cumulative
To
calculate
the
3–4
5–6
7–8
0
9–10
x
1–2
the
data
cumulative
values
5–6
as
you
go
frequency
add
up
the
frequencies
is
cumulative
cumulative
frequency
diagram
or ogive
is
most
useful
when
often
diagram
called
calculate
the
median,
quar tiles
and
percentiles
of
a
large
set
a
tr ying
cumulative
to
9–10
along.
frequency
A
7–8
frequency
A
of
3–4
frequency
of
graph.
grouped
or
continuous
data.
Chapter
Example
50
batteries
were
tested
to
see
how
long
they
lasted.
The
T ime
results
(in
hours)
are
shown
in
the
table.
Draw
a
0
frequency
diagram
and
find
the
median
and
(h )
f
<
5
3
10
5
cumulative
≤
h
interquar tile
5
≤
h
<
range.
10
≤
h
<
15
8
15
≤
h
<
20
10
20
≤
h
<
25
12
25
≤
h
<
30
7
30
≤
h
<
35
5
Answer
Add
(h)
T ime
f
frequency
0
≤
h
<
a
‘cumulative
5
3
3
10
5
8
8
16
cumulative
frequenc y
f
5
≤
h
<
≤
h
<
15
by
column
adding
Cumulative
3
10
frequenc y’
to
the
table.
Work
out
the
Cumulative
up
as
you
go.
frequency
3
3
batteries
than
15
≤
h
<
20
10
26
20
≤
h
<
25
12
38
5
3
+
5
=
8
8
≤
h
<
30
7
45
30
≤
h
<
35
5
50
y
8
3
+
5
+
8
=
16
10
3
+
5
+
8
+
10
12
3
+
5
+
8
+
=
10
lasted
less
hours
batteries
than
25
5
10
lasted
less
hours
26
+
12
=
38
38
batteries
lasted
less
50
than
ycneuqerf
40
25
hours
Q
3
7
3
+
5
+
8
+
10
+
12
+
7
=
45
37.5
5
3
+
5
+
8
+
10
+
12
+
7
+
5
=
50
30
evitalumuC
M
20
Q
1
12.5
10
Plot
the
upper
limit
frequencies. The
0
of
first
the
two
time
points
inter vals
are
(5,
against
3)
and
the
(10,
cumulative
8)
x
10
20
30
40
n
T ime
=
50
(h)
50
Median
=
=
25th
data
value
2
Draw
a
line
For
across
from
25
large
data
sets,
on
n
the
the
ver tical
axis
to
the
graph
median
is
the
then
th
2
value.
Median
=
19
hours
down
to
Read
of f
the
time
Q
and
axis.
Q
3
from
the
1
Values
graph
in
the
same
and
Q
=
25,
Q
3
IQR
=
(25
–
13)
hours
=
12
hours
IQR
=
of
quar tiles
from
a
13
=
Q
−
GDC
may
be
from
values
from
a
different
Q
1
read
cumulative
frequency
Descriptive statistics
median
1
3
the
way.
graph.
Exercise
1
The
of
G
cumulative
100
frequency
plot
shows
the
reach
in
cm
boxers.
a
Estimate
the
b
What
is
c
What
does
the
median
reach
interquar tile
the
of
a
boxer.
range?
interquar tile
range
tell
you?
y
100
ycneuqerf
75
evitalumuC
50
25
0
x
60
65
70
75
Reach
2
The
table
computer
Show
this
Length
below
80
85
(cm)
shows
the
length
of
40
flash
drives
in
a
store.
data
f
on
a
cumulative
Upper
frequency
class
Length
diagram.
Cumulative
Sometimes
(mm)
(l
boundar y
mm)
frequency
continuous
6–10
0
10.5
l
≤
10.5
0
11–15
2
15.5
l
≤
15.5
2
16–20
4
20.5
l
≤
20.5
6
given
this.
at
in
groups
Plot
the
the
upper
boundar y,
21–25
8
25.5
l
≤
25.5
14
26–30
14
30.5
l
≤
30.5
28
data
midpoint
is
like
points
class
usually
the
between
classes.
31–35
6
35.5
l
≤
35.5
34
36–40
4
40.5
l
≤
40.5
38
41–45
2
45.5
l
≤
45.5
40
Chapter
3
The
a
table
below
distribution
T ime
(min)
Number
and
under
6
6
and
under
18
8
and
under
24
10
and
under
40
12
and
under
60
14
and
under
78
16
and
under
92
18
and
under
100
a
scale
for
2
your
the
The
of
minutes
data
in
Find
the
2
of
in
<
30
the
IB
on
the
axis,
interquar tile
to
eat
lunch.
ver tical
plot
axis
and
draw
the
table
and
a
represented
of
p
and
8
≤
t
8
<
in
range.
the
form
of
q
12
12
≤
36
t
<
16
16
≤
p
students
has
table.
Frequency
m
<
30
2
30
≤
m
<
40
3
40
≤
m
<
50
5
50
≤
m
<
60
7
60
≤
m
<
70
6
70
≤
m
<
80
4
80
≤
m
<
90
2
100
1
a
Constr uct
b
Draw
c
Use
a
students
horizontal
mathematics
≤
<
students
t
<
20
q
QUESTION
20
m
frequency
diagram.
the
be
t
the
100
estimate
values
≤
10
24
Marks
≤
on
ii
Frequency
shown
to
can
a
T ime
EXAM-STYLE
for
frequency
graph
median
below
.
class
1 cm
by
students
4
i
of
0
Use
90
cumulative
taken
under
cumulative
A
the
times
and
1 cm
4
shows
the
2
Using
b
for
a
cumulative
cumulative
your
graph
to
the
median
ii
the
upper
iii
the
interquar tile
Descriptive statistics
frequency
lower
quar tiles
range.
table.
diagram.
estimate
i
and
frequency
the
semester
averages
EXAM-STYLE
5
For ty
The
QUESTIONS
students
results
Distance
throw
are
shown
(m)
0
≤
6
Constr uct
b
Draw
a
c
If
top
<
a
20
cumulative
cumulative
20
≤
your
20%
of
graph
the
to
interquar tile
e
Find
the
median
spor ts
day
.
<
40
40
≤
d
<
60
60
15
≤
d
<
10
80
80
≤
d
<
100
2
table.
the
are
considered
qualifying
for
the
final,
distance.
range.
distance
the
school
diagram.
students
estimate
the
shows
d
frequency
frequency
Find
graph
the
9
d
The
at
4
a
use
javelin
below
.
d
Frequency
the
the
time
thrown.
that
students
listen
to
music
during
school.
y
200
ycneuqerf
150
evitalumuC
100
50
0
5
10
15
20
T ime
a
30
35
40
x
45
Estimate
i
the
median
ii
the
interquar tile
iii
the
time
top
b
25
(minutes)
The
a
time
that
students
listen
to
music
range
student
must
listen
to
music
to
be
in
the
10%.
minimum
listening
time
represent
this
listening
is
45
time
minutes.
is
zero
Draw
a
and
box
the
maximum
and
whisker
plot
to
information.
Chapter
EXAM-STYLE
The
7
of
QUESTION
cumulative
220
frequency
diagram
below
shows
the
heights
sunflowers.
y
220
200
180
ycneuqerf
160
140
120
evitalumuC
100
80
60
40
20
x
0
150
140
160
170
Height
a
Find
b
The
go
to
The
c
the
median
smallest
those
tallest
190
(cm)
height
25%
are
garden
10%
180
of
sent
shops?
go
to
a
sunflower.
to
home
garden
Between
hotel
what
displays.
shops.
heights
How
many
How
are
go
many
they?
to
the
Y
ou
hotels?
What
is
the
smallest
sunflower
that
goes
to
a
may
wish
to
hotel
explore
different
visual
display?
representations
The
d
middle
half
of
the
sunflowers
are
sold
immediately
.
of
How
statistics.
many
The
e
of
is
this?
height
the
of
the
shor test
is
tallest
sunflower
136 cm.
Draw
a
is
195 cm
box
and
and
the
whisker
height
plot
to
Extension
material
Worksheet
8
central
represent
.
The
of
range
and
but
The
a
the
1
least
three
Squaring
mean
2
the
do
Squaring
many
3
The
the
of
sunflowers.
standard
is
calculated
combines
spread.
between
range
It
all
is
each
the
the
are
from
good
only
values
in
arithmetic
value
between
deviation
and
each
the
value
measures
two
a
data
data
set
mean
mean
and
of
values.
to
produce
the
squared
value.
the
mean
value
advantages:
makes
not
each
cancel
adds
more
this
from
the
in
term
values
may
be
relatively
subsequent
Descriptive statistics
below
to
weighting
mean
is
positive
weighting
extra
mathematics
measure
one
difference
cases
fur ther
of
interquar tile
variance
measure
Squaring
and
each
differences
at
heights
Variance
spread
➔
the
so
that
the
mean.
the
larger
is
above
differences.
appropriate
more
since
the
In
points
significant.
manageable
statistical
values
when
calculations.
using
this
has
-
on
CD:
Measures
tendency
and
of
spread
Because
the
differences
are
squared,
the
units
of
variance
are
Y
ou
not
the
same
as
the
units
of
the
should
GDC
to
The
standard
deviation
is
the
square
root
of
the
variance
the
same
units
as
the
the
standard
and
deviation
has
a
calculate
population
➔
use
data.
and
data.
variance.
➔
The
formulae
for
the
variance
and
standard
deviation
are:
n
2
2
σ
=
Population
variance
x
=
i 1
n
n
2
σ
=
Population
standard
deviation
x
=
i 1
n
Example
Thir ty
Their
4,
farmers
were
responses
5,
6,
Calculate
5,
3,
the
asked
how
many
farm
workers
they
hire
during
a
typical
har vest
season.
were:
2,
8,
mean
0,
4,
and
6,
7,
8,
4,
standard
5,
7,
9,
8,
deviation
6,
for
7,
5,
this
5,
4,
2,
1,
9,
3,
3,
4,
6,
4
data.
Answer
Solution
–
‘by
hand’
The
2
Workers
Frequency
(x)
(f )
(fx)
(x
μ)
(x − μ)
2
IB
syllabus
covers
deviation/variance
‘Calculation
using
only
of
standard
technology’. This
is
f (x − μ)
how
0
1
0
−5
25
25
1
1
1
−4
16
16
1
2
2
4
−3
9
18
3
3
9
−2
4
12
4
6
24
−1
1
6
5
5
25
0
0
0
you
discrete
would
Calculate
2
Subtract
3
Square
4
Add
5
calculate
variable
the
the
each
these
Divide
‘by
standard
deviation
for
a
mean.
mean
of
from
the
squared
this
the
hand’.
total
each
results
results
by
the
obser vation.
from
step
2.
together.
number
of
obser vations.
2
This
6
4
24
1
1
4
7
3
21
2
4
12
6
Take
gives
the
3
24
3
9
27
9
2
18
4
16
32
30
To
calculate
the
150
root
to
get
the
standard
σ
mean:
5
fx
n
30
To
square
σ
152
150
variance,
positive
deviation,
8
the
calculate
the
standard
deviation:
2
152
30
2.25
f
(x
)
n
{
Continued
on
next
Chapter
page
Solution
using
Enter
data
Add
the
a
new
Press
Press
lists
called
page
|
workers
to
1:Stat
your
and
freq.
document.
Calculations
.
Leave
the
a
dialogue
number
opens
of
another
box.
lists
as
dialogue
1
and
box.
press
.
Choose
number
from
the
drop
down
box
for
X1
List
from
the
drop
down
box
for
the
Frequency
Press
The
screen.
Y
ou
can
standard
‘σ x:
and
freq
list.
.
information
The
|
Statistics…
opens
This
in
calculator
6:Statistics
1:One-Var
This
GDC
σ
x’
(the
shown
scroll
will
up
deviation
not
and
is
population
fit
onto
down
the
to
value
standard
a
see
single
it
shown
all.
as
deviation).
n
σ
=
2.25
Y
ou
the
(3
sf)
should
value
always
σx
in
use
this
GDC
course,
never
use
Plus
value
help
and
standard
mean
CD:
Casio
Alternative
for
the
TI-84
FX-9860GII
sx.
GDCs
The
on
demonstrations
the
and
deviation
gives
an
idea
shows
of
the
how
much
shape
of
variation
the
are
on
there
the
is
CD.
from
the
y
distribution.
Low
●
Low
ver y
standard
close
to
deviation
the
shows
that
the
data
points
tend
to
sd
be
mean.
High
●
High
standard
deviation
indicates
that
the
data
is
spread
out
sd
over
x
a
large
range
Properties
●
Standard
of
of
values.
standard
deviation
is
only
deviation
used
to
measure
spread
or
dispersion
The
around
the
mean
of
a
data
standard
is
●
Standard
deviation
is
never
●
Standard
deviation
is
sensitive
used
widely
to
outliers.
A
single
outlier
data
standard
deviation
and
in
tur n,
distor t
●
For
data
spread,
●
If
all
zero
with
the
of
spread.
approximately
greater
values
of
because
a
each
Descriptive statistics
the
data
the
standard
set
value
are
is
same
mean,
the
greater
the
deviation.
the
equal
same,
to
the
the
standard
mean.
science,
the
spor ts
representation
in
can
business,
the
to
negative.
describe
increase
deviation
set.
deviation
is
and
medicine.
Exercise
Use
1
your
Find
sets
2
Find
sets
3
GDC
the
of
7,
a
H
mean,
12,
the
standard
32,
49
c
35,
65,
84,
27,
20,
b
and
standard
30,
40,
deviation
table
the
illustrates
the
shoe
deviation
sizes
of
4
5
6
7
8
f
9
14
22
11
17
50,
for
28,
30,
of
their
73
the
following
60
the
following
shoe
of
children
below
.
Find
in
the
the
families
mean
and
in
3
4
5
6
7
f
5
12
8
3
0
0
1
below
shows
studying
Spanish
standard
deviation.
f
9
could
the
number
remember
a
class
standard
2
5–9
a
ballet
class.
sizes.
1
Words
in
QUESTIONS
number
table
44
students
Children
The
for
66
standard
shown
19,
b
Size
The
deviation
numbers.
44,
is
and
37
variance
EXAM-STYLE
5
25,
27,
The
exercise.
variance
a
Find
4
this
numbers.
9,
of
for
of
in
words
a
year
of
29
children
deviation.
that
the
group.
students
Find
the
Pafnuty
Lvovich
(1821–94)
Russian
10–14
11
15–19
10
was
20–24
20
25–29
10
how
standard
30–34
12
35–39
6
40–44
3
1
50–54
1
theorem
the
value
deviation
applied
to
Several
statistical
advances
Russia
19th
45–49
a
mathematician.
Chebyshev’
s
shows
Chebyshev
wish
any
were
and
can
Y
ou
research
a
be
set.
made
France
centur y.
to
data
of
in
in
the
may
some
more.
55–59
2
60–64
3
65–69
0
70–74
1
75–79
1
Chapter
6
A
sur vey
208
was
randomly
chosen
of
bedrooms
Number
of
houses
whether
the
41
60
52
32
15
8
mean
c
Write
down
the
standard
EXAM-STYLE
A
random
used
to
using
per
(t
is
discrete
number
of
or
table.
continuous.
bedrooms
deviation
the
of
the
per
house.
number
of
house.
many
houses
standard
have
deviation
a
number
above
the
of
bedrooms
greater
mean.
QUESTIONS
sample
collect
their
T ime
in
6
the
how
shown
5
down
one
are
in
4
Write
than
results
bedrooms
3
b
Find
The
of
2
State
d
data
number
1
a
per
the
of
houses.
Number
bedrooms
7
conducted
of
data
167
on
phones.
people
the
The
who
amount
results
are
own
of
time
mobile
they
displayed
in
phones
spent
the
per
was
day
table.
spent
day
0
≤
t
<
15
15
≤
t
<
30
30
≤
t
<
45
45
≤
t
<
60
60
≤
t
<
75
75
≤
t
<
90
minutes)
Number
of
21
32
35
41
27
11
people
Use
your
values
day
8
in
of
on
The
graphic
the
these
figure
the
net
display
mean
below
of
and
mobile
a
standard
to
calculate
deviation
of
approximate
the
time
spent
per
phones.
shows
small
calculator
the
lengths
in
centimetres
of
fish
found
trawler.
12
10
hs
8
fo
rebmuN
6
4
2
0
20
40
60
80
100
120
Length (cm)
a
Find
the
b
Write
c
i
total
down
Write
number
an
of
estimate
down
an
fish
of
estimate
in
the
for
the
net.
mean
the
length.
standard
deviation
of
the
lengths.
ii
How
many
fish
(if
any)
have
length greater than
three
Extension
material
Worksheet
8
central
standard
deviations
Descriptive statistics
above
the
mean?
-
on
CD:
Measures
tendency
and
of
spread
Investigation
–
the
eect
of
multiplying
adding
the
or
data
set
on
a
standard
deviation
Here
is
a
set
a
Calculate
b
Now
add
100,
Calculate
d
Explain
e
Now
the
to
What
4,
0,
all
g
What
the
will
Eect
to
the
10,
the
to
constant
in
k
standard
but
If
the
you
a
list,
5,
5,
these
in
the
101,
104,
deviation
and
of
why
values
10,
1,
4,
6
numbers.
series:
to
get
104,
102,
106.
2,
in
8,
this
this
the
new
set.
happens.
original
list
by
2
to
12.
mean?
deviation.
the
variance?
changes
a
add/subtract
numbers
3,
mean?
standard
happen
of
you
of
9,
of
notice
6,
0,
numbers
105,
the
standard
18,
2,
deviation
the
105,
to
happens
Calculate
all
you
multiply
8,
4,
standard
103,
what
f
If
the
happens
c
➔
numbers:
100
109,
What
get
of
the
to
Why?
the
constant
value
arithmetic
deviation
multiply/divide
all
original
mean
remains
the
k
to/from
all
may
to
use
in
your
be
expected
these
rules
the
increases/decreases
the
numbers
Y
ou
data:
exam.
See
by
question
3
non-GDC
review
in
the
same
in
the
list
by
a
exercise.
constant
value
deviation
Review
are
k,
both
the
arithmetic
multiplied/divided
by
mean
and
the
standard
k
exercise
✗
1
2
Find
a
of
7,
A
1,
the
class
home
8,
mode,
2,
6,
collected
as
shown
b
the
5,
the
in
median,
10,
data
the
c
the
mean
on
table
the
number
3
4
5
6
7
8
9
10
f
3
9
10
2
3
1
1
0
1
a
Calculate
the
mean
b
Calculate
the
median.
c
Write
The
17.5
mean
years
again
and
at
the
the
d
range
of
pets
in
their
below
.
2
down
and
3
Pets
EXAM-STYLE
3
3,
number
of
pets.
mode.
QUESTION
age
and
the
of
a
the
group
standard
school
standard
of
reunion
deviation
of
friends
at
deviation
after
their
10
the
is
end
0.4
years.
of
school
years.
What
is
They
the
is
all
new
meet
mean
ages?
Chapter
EXAM-STYLE
4
A
farmer
results
QUESTIONS
grows
are
two
shown
different
45
Type
50
Mass
Find
5
The
the
a
of
is
six
92
sweetcor n
and
the
season’s
B
60
grams
median,
mean
another
55
in
of
below
.
Type A
40
types
b
the
range,
numbers
and
the
is
other
c
71.
four
the
interquar tile
One
number
numbers
are
is
range
for
each
Y
ou
could
type.
46,
all
be
asked
to
calculate
the
the
mean
using
the
formula
or
a
GDC.
Y
ou
same.
will
Find
a
the
total
of
all
six
numbers.
b
a
If
c
each
find
6
the
Draw
a
of
a
the
six
mean
of
numbers
the
cumulative
new
is
set
decreased
of
frequency
only
be
standard
by
expected
deviation
or
to
calculate
variance
GDC.
9
numbers.
graph
for
the
data
in
the
table.
‘Estimate
Height
150
≤
h
<
155
155
≤
h
<
160
160
≤
h
<
165
165
≤
h
<
170
170
≤
h
<
175
graph’
you
(cm)
f
4
22
56
32
5
Estimate
the
median
from
c
Estimate
the
interquar tile
your
A
dice
and
is
six
rolled
written
The following
Number
Frequency
a
Calculate
b
Find
100
on
times.
from
dice
your
has
a
number
shows
the
frequencies
for
1
2
3
4
5
6
26
10
20
k
29
11
the
i
graph.
between
it.
table
value
of
each
number.
k
the
median
the
midday
ii
the
interquar tile
range.
o
8
The
table
gives
mountains
in
November.
Temperature
12.5
≤
t
<
27.5
6
27.5
≤
t
<
42.5
3
42.5
≤
t
<
57.5
5
57.5
≤
t
<
72.5
8
72.5
≤
t
<
87.5
6
102.5
2
87.5
f
≤
t
<
Descriptive statistics
temperature
Find
the
as
(
F)
median
in
and
the
Omani
IQR.
one
your
that
show
and
graph.
the
ver tical
working
graph.
range
Each
means
horizontal
your
7
from
should
lines
b
the
using
on
Review
1
Calculate
9,
11,
the
12,
13,
EXAM-STYLE
2
June
for
3
r uns
last
exercise
median
13,
17,
and
19,
interquar tile
21,
21,
25,
27,
a
cats’
year
33,
35
home.
The
numbers
of
kittens
per
litter
were
4
5
6
7
8
9
f
3
7
11
12
6
3
a
Find
the
mean
b
Find
the
standard
numbers
season
30,
of
QUESTION
Kittens
The
range
of
number
of
kittens
per
litter.
deviation.
tennis
racquets
broken
by
410
players
in
a
were.
Broken
2
3
4
5
6
7
8
9
10
3
11
43
90
172
13
64
10
4
racquets
f
Find
EXAM-STYLE
4
The
the
a
in
the
b
median
the
of
hours
students
study
mathematics
0
1
2
3
4
5
6
f
2
5
4
3
4
2
1
Find
the
mean.
the
each
night
is
table.
Hours
a
c
QUESTION
number
given
mode
mean,
median,
mode,
standard
deviation
and
variance.
Find
b
5
The
the
range,
histogram
school
in
lower
below
quar tile
shows
the
and
the
heights
interquar tile
of
the
range.
students
in
a
high
Per u.
y
90
80
70
ycneuqerF
60
50
40
30
20
10
0
140
150
160
Height
a
Write
b
Constr uct
for
down
the
a
the
170
modal
grouped
mean
180
190
x
(cm)
height
class
height.
frequency
of
the
table
Per uvian
and
calculate
an
estimate
students.
Chapter
EXAM-STYLE
6
A
school
words
the
QUESTION
with
they
table
150
can
remember
in
of
Cumulative
of
11
16
21
32
17
33
p
18
q
19
38
137
20
13
150
Write
down
the
Find
the
median
Find
the
mean
CHAPTER
8
Univariate
●
Univariate
●
Data
is
either
value
the
of
number
number
information
quali tative
data
data
can
that
or
be
a
Find
ii
single
you
split
the
value
of
q
variable.
gather
up
into
and
is
classified
as
data.
two
categories: discrete
continuous.
discrete
continuous
depends
on
the
Continuous
have
statistics,
par t
of
or
the
group
the
you
groups
in
term
to
a
the
as
exact
can
numerical
be
measured
measuring
length,
values.
and
instr ument
weight
and
time,
its
accuracy
used.
may
population
we
are
includes
studying
is
selection
called
of
for
a
all
data
sample.
individuals
members
driven
It
is
from
a
of
a
decisions.
subset
the
of
population.
data
a
lot
grouped
continuous
similar
a
have
a
of
such
population
Presenting
has
variable
decimals.
that
population,
When
variable
accuracy
variables,
fractions
defined
the
number
quanti tative
quantitative
For
in
words.
involves
quantitative
●
given
analysis
A
●
are
SUMMARY
A
the
results
French
words.
●
A
p
of
of
●
●
The
many
99
analysis
Quantitative
and
how
students
11
c
In
see
15
b
●
to
minute.
students
i
●
tested
one
Number
words
a
●
is
below
.
Number
of
students
bar
of
data,
you
frequency
data,
char t
you
but
can
it
can
organize
it
into
table.
draw
doesn’t
a histogram.
have
gaps
It
is
between
bars.
Continued
Descriptive statistics
on
next
page
Measures
of
central
●
The
mode
is
the
value
●
The
mean
is
the
sum
of
numbers
in
a
set
of
of
Sum of
Mean
that
The
number
the
●
If
in
of
mean
there
the
most
numbers
frequently
divided
by
in
the
a
set
of
data.
number
data.
=
is
median
numbers
occurs
the data values
Number of
●
tendency
a
the
set
number
of
data
numbers
of
are
the
a
in
two
lot
data values
of
present
are
a
in
the
arranged
data
middle
set
is
in
middle
order
even,
when
of
then
size.
the
the
If
the
median
is
numbers.
numbers
and
it
is
difficult
to
find
the
middle
⎛ n +1 ⎞
member
we
can
use
the
formula
Median
=
th
⎜
where
n
is
the
number
of
members
in
the
member,
⎟
2
⎝
⎠
set.
Advantages
●
Mode
The
be
mode
used
can
when
choose
do
not
affect
the
●
asked
Does
of
data
the
popular
values
mode.
for
qualitative
or
Extreme
Disadvantages
●
Not
–
to
not
the
use
data
all
necessarily
may
be
members
set.
more
unique
than
one
answer
.
●
most
item.
When
the
no
data
values
set,
repeat
there
is
in
no
mode.
●
When
one
there
mode,
interpret
●
Mean
The
mean
describes
middle
of
of
Most
such
the
a
set
data.
popular
as
computer
●
Uses
●
It
is
measure
business,
all
in
●
elds
engineering
and
Affected
is
it
more
is
and/or
by
than
difcult
to
compare.
extreme
values.
science.
members
unique
–
of
there
is
the
data
only
set.
one
answer
.
●
Useful
when
comparing
sets
of
data.
●
Median
The
median
describes
middle
of
of
median
the
a
Extreme
set
values
as
do
strongly
not
as
affect
they
the
do
the
mean.
●
data.
Useful
●
Not
●
Less
as
popular
used
in
as
mean.
fur ther
calculations.
when
comparing
sets
of
data.
●
It
is
unique
–
there
is
only
one
answer
.
●
As
the
50%
of
median
the
is
data
the
is
middle
either
value,
side
of
it.
Continued
on
next
Chapter
page
Measures
●
The
is
range
F irst
of
dispersion
the
difference
quar tile
The
of
between
first
the
data
lies
into
the
below
lies
percentile
the
data.
the
above.
and
largest
is
quartile
way
four ths
the
rst
It
often
is
and
value
One
has
and
called
the
values.
one-quar ter
quar ter
quar tile
also
smallest
of
three-
the
symbol
the
25th
Q
1
Second
quar tile
The
the
second
median
also
Third
quar tile
The
way
the
It
called
third
in.
is
of
the
the
also
the
called
set
of
name
data
for
and
three-quar ters
of
and
the
symbol
another
is
percentile.
is
quarti le
quar tile
is
entire
50th
Three-four ths
third
has
quarti le
the
data
lies
one-four th
75th
lies
percentile
of
the
below
above.
and
Q
3
3
1
Q
=
(n
+
1)th
value
and
Q
1
=
(n
number
The
the
●
A
1)th
value
where
n
is
the
4
4
●
+
3
of
data
difference
values
between
interquarti le
five
box
statistical
and
in
the
the
data
third
set.
and
first
quar tiles
is
called
(IQR).
range
summary
can
be
represented
graphically
as
a
plot.
whisker
Range
Whisker
Min
Interquartile
Q
X
Range
Whisker
m
Max
Q
1
X
3
(Median)
●
An
outlier
is
any
value
at
least
1.5
IQR
above Q
or
below
3
Cumulative
●
To
of
calculate
the
data
Variance
●
The
a
the
differences
cumulative
values
of
as
you
go
frequency
spread.
between
It
all
is
each
add
up
the
frequencies
along.
standard
combines
variance
1
frequency
and
measure
Q
the
the
deviation
values
in
arithmetic
value
and
the
a
data
mean
mean
set
of
to
the
produce
squared
value.
Continued
Descriptive statistics
on
next
page
●
The
has
●
standard
the
The
same
deviation
units
formulae
for
as
the
is
the
the
square
root
of
the
variance
and
data.
variance
and
standard
deviation
are:
n
2
2
σ
=
Population
variance
x
=
i 1
n
n
2
σ
=
Population
standard
deviation
x
=
i 1
n
Eect
of
If
you
add/subtract
in
a
list,
the
standard
If
you
value
are
constant
changes
a
arithmetic
deviation
both
the
mean
all
by
original
value
k
data:
to/from
all
increases/decreases
the
the
arithmetic
multiplied/divided
the
constant
remains
multiply/divide
k,
to
the
by k
numbers
but
the
same .
numbers
mean
and
in
the
the
list
by
standard
a
constant
deviation
k
Theory
of
knowledge
Facts
in
and
statistics
Statistics
as
F ind
out
what
misconceptions
it
What
Is
What
main
how
led
did
it
its
a
relatively
advances
Florence
moder n
have
Nightingale
branch
been
used
made
of
in
statistics
mathematics
the
past
and
to.
Francis
easy
is
is
to
Galton
mix
the
up
μ
invent?
and
difference
x ?
between
a
sample
and
a
population?
Do
different
median,
Were
mode)
measures
discovered?
measures
Could
of
express
of
central
different
central
Where
do
mathematics
(mean,
proper ties
tendency
they
make
tendency
come
of
invented
the
or
from?
alter native,
equally
tr ue
formulae?
What
Darrell
1954)
does
Huff's
this
book
attempted
spin-doctors
“Statistical
efficient
for
What
Do
of
How
to
to
you
you
agree
knowledge:
will
as
G.
mathematical
with
the
Wells
with
and
one
the
think
Facts
Lie
expose
thinking
do
about
Statistics
tricks
‘self-defense’
citizenship
Theory
us
the
H.
tell
of
day
ability
of
the
‘honest
be
to
as
(Nor ton,
statistical
men’.
necessar y
read
and
(1866–1946)
H.
G.
Wells
meant?
him?
misconceptions
in
tr uths?
statistics
for
write.”
data?
400
years.
How
easy
Criticize
these
is
it
to
lie
with
statistics?
graphs:
4.8%
3.3%
We
so
are
much
now
3.1%
b etter
than
were
3.1%
doing
in
we
“There
kinds
1990s
1970s
1980s
1990s
ease
incr
lies:
40
author
Mark
in
lies,
statistics.”
US
huge
of
damned
and
a
three
Current
lies,
t
Wha
are
the
Twain
attributed
this
to
30
the
b er
num
of
the
19th
British
s
frog
Minister,
century
Prime
Benjamin
Disraeli
0
May
Take
their
a
sur vey
favorite
Use
of
your
style
situation
friends
about
Here
subject.
Microsoft
different
September
(or
you
Excel
char ts
draw
to
to
show
the
by
hand).
Tr y
the
changing
value
Show
the
3D
the
y-axis
y-axis
scale
star ts
some
use,
of
the
and
‘tricks’
how
that
they
mislead:
produce
graphs
are
could
Showing
or
hides
or
of
data.
highlights
the
large
or
This
change
repor ted.
Using
a
nonlinear
expecting
char ts.
unsuitably
amount
being
at.
an
small
a
linear
scale.
scale
Anyone
would
be
misled.
See
what
happens
to
a
subject
that
may
have
zero
votes
on
a
pie
Not
Keep
char t.
also
can
can
be
be
ver y
used
■
How
can
■
How
can
to
distor t
statistics
we
helpful
be
decide
in
our
used
providing
an
them
Making
the
scale
at
all.
uninformed.
bars
of
a
histogram
three-dimensional.
It
difference
data
look
Statistics
showing
between
makes
the
values
larger.
influential
inter pretation
of
reality
but
perceptions.
or
whether
misused
to
accept
to
the
assist
and
statistical
mislead
evidence
us?
that
is
presented
to
us?
Chapter
Integration
CHAPTER
OBJECTIVES:
1
n
Indenite
6.4
integration
as
antidifferentiation.
Indenite
integral
of
x
(n
∈
),
x
x
and
e
.
The
Integration
composites
by
Denite
the
revolution
1
you
should
Write
a
a.
the
any
or
of
these
with
substitution
boundar y
analytically
the
of
the
the
condition
and
x-axis).
to
using
Areas
linear
form
function
technolog y.
the
of
given
terms.
Areas
cur ves.
under
Volumes
T
otal
displacement
distance
s,
velocity
v
of
and
how
in
Skills
to:
sigma
notation
as
a
1
check
Write
as
a
sum
of
terms.
6
e.g.
2
(2i
)
b
i 1
1)
cur ves
traveled.
a
(2i
term.
x-axis.
involving
4
b
constant
5
sum
+
f (g(x))g′(x) dx
determine
between
ax
start
know
series
a
and
problems
acceleration
Before
both
cur ve
about
Kinematic
6.6
with
integrals,
(between
Y
ou
inspection,
Antidifferentiation
6.5
of
(3k
2)
k 2
[2(2) 1] [2(3) 1] [2( 4 ) 1]
5
i 2
3
2
c
5 7 9
[(i )
g(x
)]
i
d
[ f
(x
)( x
j
)]
j
i 1
j 1
4
e.g.
f
(x
)
=
f
(x
j
) +
f
(x
1
) +
(x
f
2
) +
f
(x
3
)
4
2
Find
the
area.
j =1
a
2
Use
geometric
formulae
to
find
area.
b
5 mm
4 mm
e.g.
Area
of
trapezium:
7 cm
9 mm
1
=
(b
+ b
1
8 cm
)h
2
2
3
10 cm
Find
the
volume.
1
8 cm
=
(10 + 8)( 6 )
a
b
4 m
2
2
= 54 cm
6 cm
14 ft
10 cm
3
Use
geometric
formulae
e.g.
to
Volume
find
of
volume.
sphere:
2 m
4
r
3
Integration
32
4
3
V
3
(2)
3
3
m
3
6 ft
We
know
we
derivative
on
the
Can
a
find
the
Suppose
find
the
find
the
of
a
moving
function
function).
Now
object
(carr ying
consider
function
for
out
the
a
by
taking
the
differentiation
reverse
moving
process.
object,
if
you
function?
velocity
function,
velocity
displacement
velocity
the
the
displacement
displacement
you
know
of
can
s (t)
function
such
that
is
given
s ′(t)
=
by v (t)
2t
+1.
=
2t
+
1.
Working
We
need
backwards,
differentiation
to
we
2
t
+
t
2t
+
1
2
find
one
possible
displacement
function
is s (t)
=
t
+
t
+
t,
since
integration
d
2
(t
2
t )
2t
1.
Why
do
we
say
that
s (t)
=
t
is
one
possible
dt
displacement
function?
2
The
function
The
process
chapter
you
integration
line,
.
area
s (t)
of
t
+t
finding
will
can
and
=
lear n
be
is
an
called
to
the
solve
the
of
problems
v (t)
=
2t
called integration.
is
process
of
integration
involving
and
motion
+
In
1.
this
how
on
a
volume.
Antiderivatives
Suppose
antiderivative
antiderivative
about
used
an
derivative
of
a
and
the
function f
is
indefinite
given
by
2x
+3.
integral
Working
2
backwards,
we
find
that
f
may
be
the
function
f
(x)
=
+
x
3x,
since
d
2
(x
3x )
2x
3.
But
there
are
other
functions
that
have
the
same
dx
2
2
derivative,
such
as
f
(x)
=
x
x
2
+
3x
+
1
or
f
(x)
=
x
+
3x
–
6
+
3x
+
+
3x
–
+
3x
1
since
2
x
d
d
6
2x
+
3
2
2
(x
3x
1)
2x
3
(x
and
3x
6)
2x
3
2
dx
dx
x
Chapter
2
The
are
functions
all
called
f
(x)
=
2
x
+
3x,
f
antiderivatives
(x)
of
=
2x
2
x
+
+
3x
+
1
and
f
(x)
=
x
+
3x
–
6
3.
A
2
Any
function
of
the
form
f
(x)
=
x
+
3x
+
C,
where
C
is
an
an
arbitrar y
f
constant,
is
an
antiderivative
of
2x
+
function
F
is
called
antiderivative
if
F ′ (x)
=
of
f (x).
3.
n
Investigation
Copy
and
–
complete
f (x)
antiderivatives of
the
table
Antiderivative
of
below.
The
rst
x
entr y
is
completed
for
you.
f
1
2
x
x
C
2
2
x
3
x
4
x
n
Write
a
Show
Are
general
expression
or
r ule
for
the
antiderivatives
of
x
1
2
–3
whether
there
any
your
r ule
values
of
gives
n
the
where
correct
your
antiderivatives
rule
does
not
for
x
and
x
apply?
1
n +1
n
The
antiderivatives
of
x
are
given
by
x
Just
C,
+
as
the
process
of
n + 1
nding
where
C
is
an
arbitrar y
constant
and
n
≠
–1.
is
a
derivative
called
dierentiation,
Example
the
an
Find
the
antiderivative
of
each
process
of
nding
antiderivative
is
called
function.
antidierentiation.
1
4
10
a
x
b
c
3
x
5
x
Answers
1
1
10 +1
x
a
1
11
+C
=
10 + 1
n +1
x
+C
Apply
the
rule
11
n
where
n
=
+ C ,
x
+ 1
10.
1
5
=
b
Rewrite
using
rational
exponents.
x
5
x
1
n +1
Apply
1
the
rule
+ C ,
x
1
−5 +1
x
−4
+C
=
x
−5 + 1
n
+C
4
where
1
n
=
+ 1
–5.
Simplify.
=
+C
4
4 x
3
4
c
3
x
4
=
Rewrite
x
using
rational
exponents.
1
n +1
3
1
7
1
1
=
x
x
+ C ,
Remember
1
n
+ 1
2
7
x
x
x
x
x
x
3
1
4
rule
+C
where
the
4
+C
3
Apply
4
x
4
n
=
1
4
3
3
7
4
1
4
=
x
7
Integration
+C
Simplify.
4
4
etc.
Exercise
Find
the
9A
antiderivative
of
each
function.
1
7
4
x
1
–2
x
2
1
2
x
3
x
4
2
1
1
3
5
x
5
x
6
7
8
12
4
x
x
1
7
3
x
9
1
3
x
10
11
12
5
3
x
Antidifferentiation
is
also
known
2
x
as indefini te integration
and
is
If
denoted
with
an
integral
symbol,
dx.
For
F ′(x)
=
f (x),
we
write
example,
f (x) dx
=
F (x)
+
C
1
3
x
4
dx
=
x
+
C
4
The
expression
1
3
means
that
the
indefinite
integral
(or
antiderivative)
of
4
x
is
x
+
C.
f (x) dx
is
called
an
4
indefini te
These
r ules
will
help
you
find
indefinite
f (x) dx
➔
Power
integral.
integrals.
is
read
as
rule
‘the
n
antiderivative
of
n+1
x
dx
=
x
+
C,
n
≠
1
f
with
respect
to
x’
or
n 1
‘the
➔
Constant
integral
➔
=
kx
Constant
+
C
(x) dx
multiple
=
k
f
to
Sum
or
x’.
rule
=
F (x)
+
C
(x) dx
Integrand
➔
with
Variable
f (x) dx
kf
f
rule
respect
k dx
of
dierence
Constant
rule
of
( f
(x)
±
g (x)) dx
=
f
(x) dx
±
g (x) dx
integration
Example
Find
the
indefinite
integral.
6
a
x
5
dx
4
d
(3u
4 dt
b
c
3x
e
(x
dx
2
+
3
6u
+
2) du
+
x
) dx
Answers
1
6
a
x
6+1
dx
=
x
+
C
Apply
the
power
rule
Apply
the
constant
with
n
=
6.
6 + 1
1
7
=
x
+
C
7
b
4 dt
=
4t
+
C
variable
of
rule. The
integration
is
dt
tells
you
that
the
t.
{
Continued
on
next
Chapter
page
5
5
3x
c
dx
=
3
x
⎛
dx
Apply
the
Apply
the
constant
multiple
rule.
⎞
1
5 +1
=
3
x
⎜
+C
5 +1
⎝
power
rule
with
n
=
5.
⎟
1
⎠
1
6
=
x
+
3C
3C
1
is
some
arbitrar y
constant
C.
We
usually
just
1
2
show
the
final
arbitrar y
constant.
1
6
=
x
+
C
2
4
2
(3u
d
+
6u
+
4
=
3u
3
=
3
u
⎛
+
du
+
6u
du
+
2 du
sum
rule.
du
+
2 du
Apply
the
constant
Apply
the
power
2
6
u
⎛
⎞
1
u
⎜
rule
and
rule.
constant
rule,
with
2 +1
+
⎟
4 + 1
multiple
⎞
1
4 +1
⎝
the
2
du
4
=
Apply
2) du
6
u
⎜
⎟
2 + 1
⎝
⎠
+
2u
+
variable
C
of
integration
u.
⎠
3
5
=
3
u
+
2u
+
2u
+
We
C
actually
get
a
constant
of
integration
for
each
5
ter m,
but
C
+
C
1
+
C
2
is
some
arbitrar y
constant
C.
3
1
3
(x
e
1
+
x
) dx
=
(x
+
3
) dx
x
Rewrite
using
rational
exponents.
1
1
+1
1
1+1
=
x
3
+
x
+
C
Apply
the
power
rule
to
each
ter m.
1
1 + 1
+ 1
3
4
1
3
2
=
x
3
+
2
Exercise
Find
the
x
+C
4
9B
indefinite
integral
in
questions
1
to
10.
Y
ou
can
check
the
1
3
1
x
dx
answers
dt
2
to
your
2
t
indenite
4
5
x
3
dx
4
answer
to
2
(3x
+
2x
+
1) dx
by
your
2 du
4
5
integrals
differentiating
see
and
that
checking
it
equals
dx
6
3
x
(t
3
4
2
7
the
+
t
) dt
8
(
integrand.
2
x
+
1) dx
0
dt
4
9
(5x
3
+
12x
+
6x
–
2) dx
10
4
3
11
Let
f
(x)
=
x
+
.
2
x
Find
f
a
′(x)
b
f
(x) dx
5
12
Let
g(x)
Find
a
Integration
=
30
x
g ′(x)
.
b
g (x) dx
dt
=
1
×
dt
=
t
dt
Y
ou
saw
moving
at
the
object
beginning
is
given
of
by
this
v (t)
=
section
2t
+
1,
that
then
if
the
the
velocity
of
a
displacement
of
the
2
par ticle
is
s(t)
=
+
t
t
+
C,
for
some
arbitrar y
constant
2
now
the
write
this
general
Suppose
time
t
=
(2t
solution
you
1
as
is
are
6.
1) dt
for
also
Y
ou
+
(2t
given
can
=
+
t
Y
ou
can
2
+
t
+
C,
where
t
+
t
+
C
is
called
1) dt.
that,
then
C.
for
find
this
par ticle,
the
position
at
at
=
C
2
s (t)
=
t
+
t
+
C
2
+
s (1)
=
1
6
=
2
C
=
4
+
1
+
C
C
2
Therefore,
s (t)
=
+
t
t
+
4.
The
fact
that
the
position
time t
1
is
2
6
is
called
a
boundary
of
solution
(2t
Example
+
1)
condi tion
dt,
given
and
the
t
+
t
+
4
boundar y
is
a
particular
condition.
2
a
If
f
′(x)
b
The
=
3x
+
2x
and
f
(2)
=
–3,
find
f
Sometimes
(x).
boundar y
cur ve
y
=
f
(x)
passes
through
the
point
(32, 30).
The
the
cur ve
is
given
by
f
′(x)
is
=
condition
gradient
1
of
a
given
as
an
ini tial
.
5
3
condi tion.
This
represents
a
x
Find
the
equation
of
the
cur ve.
condition
dP
The
c
rate
of
growth
of
a
population
of
fish
is
given
by
=
150
t
when
t
is
zero.
For
dt
example,
for
0
≤
t
≤
5
years.
The
initial
population
was
200
fish.
Find
of
fish
at
t
=
4
you
are
the
told
number
if
that
the
initial
years.
displacement
4,
this
means
is
that
Answers
displacement
is
4
2
f
a
′(x)
=
3x
+
2x
when
t
=
0.
2
f
(x)
=
(3x
+
2x)
dx
Apply
the
power
rule
to
find
the
2
3
f
(x)
=
x
x
+
2
3
f
(2)
3
C
=
=
2
8
f
+
=
(x)
=
+
C
+
C
(3x
+
2x) dx.
2
4
+
Use
the
fact
that
f
(2)
=
–3
to
find
C.
C
15
3
∴
general solution for
2
+
x
2
+
x
–
15
1
b
f
′(x)
=
5
3
x
1
f
(x)
=
dx
5
Rewrite
with
rational
exponents
3
x
and
3
5
=
x
apply
the
power
rule
to
find
the
1
dx
general
solution
dx
for
5
2
3
x
5
5
f
(x)
x
=
+
C
2
{
Continued
on
next
page
Chapter
2
5
5
f
(32)
30
(32 )
=
=
C
10
=
+
+
C
Use
the
fact
that
through
the
Rewrite
with
the
point
cur ve
(32,
30)
passes
to
find
C.
C
20
2
5
5
∴
f
(x)
=
+
x
20
dP
=
c
150
t
dt
P(t)
=
t dt
150
and
find
the
rational
general
exponents
solution
for
1
t
150
t
150
2
=
dt
dt
3
2
P(t)
=
100t
+
C
3
2
P(0)
=
100 ( 0 )
200
=
0
+
C
=
200
+
The
C
initial
means
C
population
P(0)
=
200.
was
Use
200
this
to
fish
find
C.
3
2
P(t)
100t
=
+
200
3
2
P(4)
=
100 ( 4 )
=
There
t
=
4
+
Find
200
P
when
t
is
4.
1000
are
1000
fish
when
years.
Exercise
9C
Exam-Style
Questions
5
1
The
derivative
The
graph
Find
an
of
f
of
the
function f
passes
expression
is
through
for
f
given
the
by
point
f
′(x)
=
4x
+
8x.
(0, 8).
(x).
dy
4
4
2
It
is
given
that
=
x
+
of
a
x
and
that
y
=
10
when
x
=
1.
dx
Find
y
in
terms
of
x.
–1
3
The
velocity
v
m s
,
moving
object
at
time
t
seconds
is
given
2
by
v (t)
When
Find
=
t
=
an
–
3t
3,
2t.
the
displacement,
expression
for
s
in
s,
terms
of
of
the
object
is
12
metres.
t
3
4
The
rate
at
which
the
volume
of
a
sphere
is
increasing
in
cm
dV
2
is
given
by
=
2π (4t
+
4t
+
1),
for
0
≤
t
dt
3
volume
Find
was
the
Integration
π
cm
volume
of
the
sphere
when t
=
3.
≤
12.
The
initial
–1
s
Exam-Style
Question
–1
The
5
velocity
by
v (t)
=
a
Find
b
The
20
v
m s
–
of
a
moving
object
at
time
t
seconds
is
given
5t
–2
its
initial
Find
.
The
acceleration
an
More
power
m s
displacement
expression
on
r ule
in
for
s
s
is
in
integration
metres.
terms
indefini te
for
5
of
t.
integrals
tells
us
that
Why
do
we
say
that
1
n
x
n+1
dx
=
x
+
C,
n
≠
−1.
The
r ule
does
not
work
1
when
is
n + 1
undened?
0
–1
n
=
–1
because
it
would
result
in
dividing
by
0.
So
what
is
x
dx?
1
0
Is
the
same
as
1
–1
Y
ou
have
seen
?
0
0
d
(ln x)
that
dx
=
=
x
for
x
>
0,
Why
so
or
why
not?
x
1
➔
dx
=
ln x
+
C,
x
>
0
x
d
x
Also
(e
x
)
=
e
,
so
dx
x
➔
e
x
dx
=
Example
Find
the
e
+
C
indefinite
integral.
t
4
e
dx
a
dt
b
x
2
Answers
⌠
a
4
⌡
1
⌠
dx
⎮
=
4
⌡
x
Apply
dx
⎮
the
constant
multiple
rule.
x
1
=
4ln x
+
C,
x
>
0
Use
the
fact
that
dx
=
ln
x
+
C,
x
x
>
0.
Integration
rules
t
e
1
t
b
dt
e
=
Apply
dt
the
constant
multiple
rule.
1
dx
2
=
ln
x
+
C,
x
>
0
2
x
x
Use
1
the
fact
that
e
x
dx
=
e
+
C.
t
=
e
+
C
x
e
x
dx
=
e
+
C
2
2
3x
2
For
some
integrals
such
as
(x
+ 2x
+ 1
2
+
1)
dx
dx,
x
2t–1
and
ln (e
expanding
you
can
) dt
the
you
may
bracket,
integrate.
The
have
to
rewrite
separating
next
the
example
the
terms
shows
integrand
or
by
simplifying
before
how
.
Chapter
Example
Find
the
indefinite
integral.
2
3x
2
(x
a
+
2x
+ 1
2
+
1)
2t–1
dx
dx
b
ln(e
c
) dt
x
Answers
2
2
(x
a
+
1)
4
dx =
2
(x
+
2x
Expand
+ 1) dx
and
then
integrate
each
ter m.
1
2
5
=
x
3
+
x
5
+
x
+
C
3
2
3x
+
2x
+ 1
b
dx
x
2
⎛ 3x
2x
=
+
x
⎝
⎮ ⎜
x
1
3x
⎟
x
⌠ ⎛
=
⎞
1
+
⎜
dx
Separate
the
ter ms.
Simplify
and
⎠
⎞
+ 2 +
dx
then
integrate
each
⎟
x
⌡ ⎝
⎠
ter m.
3
2
x
=
+
2x
+
ln x
+
C,
x
>
0
2
2t–1
ln(e
c
x
) dx
=
(2t
–
1) dx
Simplify
2
=
Exercise
Find
the
t
–
t
+
ln x
C
are
using
the
fact
that
e
and
inverses.
9D
indefinite
integral.
2
x
1
dx
2
3e
dt
4
e
dx
x
1
ln x
3
dx
4t
3
2x
2
6x
5
2
(2x
5
+
3)
dx
6
dx
x
2
u
ln
7
3
e
du
8
dx
10
(x
x
e
–
1)
dx
2
1
9
x
+
x
+ 1
dx
2
x
Y
ou
can
rule
Now
we
look
at
indefinite
integrals
of
functions
that
with
the
linear
function ax
+
verify
right-hand
the
equation
showing
1
+
b)
that
and
you
get
1
n 1
n
(ax
side
b
of
➔
each
differentiating
are
the
compositions
by
dx
=
( ax
a
+
b)
C
the
integrand.
n 1
1
ax + b
e
➔
ax + b
dx
e
=
+
C
Note
that
ln(ax
+
b)
a
is
1
⌠
➔
1
dx
⎮
+ b) + C ,
x
>
ax
+ b
a
when
b
−
ax
⌡
dened
b
ln( ax
=
+
b
>
0
or
x
>
–
a
a
Integration
Example
Integration
Find
the
indefinite
rules
integral.
n
(ax
a
(3x
+
1)
2x+5
dx
e
b
dx
+
b)
dx
=
1
3
4
dx
c
dx
d
4
4 x
(6 x
2
+ 3)
1
1
n1
( ax
a
Answers
1
⎛
+
b
C
+
1)
( ax
Find
dx
=
e
+
1
5
=
(3 x
⎟
1
+ 1
n
⎝
⎠
+
⎜
3
⎝
ax
⎞
+ 1)
for
C
a
=
3,
b
=
1
and
n
=
=
b
4.
⎟
5
1
⎠
ln(ax
1
Check
5
=
C
C
dx
1 ⎛
+
a
+ b )
⎜
a
ax + b
dx
⎞
n +1
4
(3x
+
b)
1
ax
e
1
a
n 1
(3 x
+
+ 1)
by
x
d
1
⎡
5
1
⎤
dx
⎣
+
C
(5(3x
+ 1)
>
–
a
4
=
(3 x + 1)
⎢
b)
b
C
15
+
a
dif ferentiating.
(3))
⎥
15
15
⎦
4
=
1
1
2 x +5
b
e
2 x +5
dx
=
e
+
1)
ax + b
Find
e
+ C
(3x
+
C
for
a
=
2
and
b
=
5.
a
2
Check
d
by
⎡ 1
dif ferentiating.
2x + 5
dx
dx
4 x
=
2
⎣
[e
2x + 5
=
(2) ]
e
2
⎦
Apply
dx
3
4 x
2
the
Find
− 2)
⎢
+ C , x
rule.
Check
3
d
1
4 x
by
dif ferentiating.
2
⎦
(
ln(ax + b) for a = 4 and b = –2.
>
⎥
4
ln
multiple
a
1
⎤
ln( 4 x
⎣
constant
2
1
⎡ 1
3
=
2x +5
=
⎥
1
3
c
=
1
⎤
e
⎢
− 2) + C ,
x
⎡ 3
2)
⎢
4
dx
2
⎣
3
⎤
ln(4x
>
4
1
⎛
⎞
=
(4)
⎜
⎥
4
⎦
⎝
⎟
4x
2
⎠
3
=
4x
2
1
–4
d
4
(6 x
dx
=
(6x + 3)
dx
Rewrite
using
rational
exponents.
+ 3)
1
1
⎛
n +1
Find
1 ⎛
1
3
=
(6 x
+ 3)
⎜
6
⎝
a
+ C
⎝
⎞
+ b )
+
C
⎟
n
+ 1
⎠
⎟
3
⎠
for
a
=
Check
1
=
( ax
⎜
⎞
−
+
6,
by
b
=
3
and
n
=
– 4.
dif ferentiating.
C
3
18( 6 x
+ 3)
d
1
⎡
=
3
−
(6x
d
⎡
dx
⎣
⎤
1
⎢
3
18( 6 x
+ 3)
dx
⎣
⎤
+ 3)
⎢
⎥
18
⎦
⎥
1
1
⎦
4
=
−
( −3(6x
+ 3)
(6))
=
4
18
(6x
+ 3)
Chapter
Exercise
Find
the
9E
indefinite
integral
in
questions
1–10.
1
x
2
(2x
1
+
3
5)
dx
(–3x
2
1
+
5)
3
2
dx
dx
3
e
6
4e
3
2x+1
dx
4
5x
dx
5
+ 4
7
dx
2x
1
⎛
–
3)
dx
(
8
7x
+ 2
)
dx
9
⎜
4
4 x
2
7
6(4x
7
e
⎞
+
⎟ dx
3x
⎝
5 ⎠
2
10
3
3( 4 x
dx
5)
Exam-Style
Questions
3
Given
11
f
a
that
f
(x)
=
′(x);
(4x
f
b
+
5)
find
(x) dx.
–3t
12
The
velocity
The
displacement
metres
The
We
f
v
when
of
t
a
=
use
of
0
substitution
par ticle
the
at
par ticle
seconds,
g ′(x) dx.
Example
Find
the
a
(3x
The
t
at
express
is
given
time t
s
in
is
by
s.
terms
v (t)
=
Given
of
e
+
that
s
6t.
=
4
t
method
the substi tution method
(g (x))
time
next
to
example
evaluate
shows
integrals
you
of
the
form
how
.
indefinite
integral.
3
2
4
+
5x)
(6x
+
5) dx
b
2
x
3x
3
12 x
(2x
–
3) dx
2
3x
2
4 x
c
+1
x e
dx
d
4
3x
3
dx
x
Answers
2
a
(3x
4
+
5x)
(6x
+
This
5) dx
f
integral
(g(x))
is
of
the
for m
g′(x) dx,
2
where
g(x)
=
3x
+
5x
and
g′(x)
=
6x
+
5.
du
4
4
u
=
dx
=
u
du
du
2
Let
u
=
3x
+
5x,
then
=
6x
+
5
and
substitute.
dx
dx
1
Simplify
and
integrate.
5
=
u
+
C
5
2
1
Substitute
2
=
(3x
3x
+
5x
for
u.
5
+
5x)
+
C
5
{
Integration
Continued
on
next
page
Check
by
dif ferentiating.
⎡ 1
d
2
5
(3 x
dx
⎣
⎤
+ 5x )
⎢
⎥
5
⎦
1
2
=
4
(5(3x
+
5x)
(6x
+
5))
5
2
=
3
(3x
+
5x)
(6x
+
5)
2
x
b
4
3x
(2x
–
3) dx
This
f
integral
(g(x))
is
of
the
for m
g′(x) dx,
2
where
g(x)
=
x
–
3x
and
g′(x)
=
2x
–
3.
1
du
du
2
3
=
u
dx
Let
u
=
x
–
3x,
then
=
2x
–
3
and
substitute.
dx
dx
1
=
3
u
du
Simplify
and
integrate.
4
3
3
=
u
+ C
4
4
3
2
=
(x
2
3
− 3x )
+ C
Substitute
x
–
3x
for
u.
4
Check
by
dif ferentiating.
1
4
⎡
d
⎤
3
2
⎢
dx
3
⎛
⎞
4
2
3
(x
3 x )
=
⎥
4
⎜
4
⎣
(x
⎦
3
− 3 x )
(2 x
− 3)
⎟
3
⎝
⎠
1
⌠ ⎛ 1
2
4 x
c
+1
x e
dx
=
⌡ ⎝
⎞
×
⎮ ⎜
8x
3
(x
–
3x)
(2x
–
3)
x
=
3 x
(2x
–
3)
2
4 x
+1
e
2
d
⎟
8
2
3
2
=
If
g(x)
=
+
4x
1
then
g ′(x)
=
8x.
Rewrite
the
⎠
2
1
4 x
integrand
+1
e
=
( 8x )
so
that
it
is
in
the
for m
c
f
(g(x))g ′(x) dx.
dx
8
du
u
1
du
e
dx
2
=
Let
u
=
4x
+
1,
then
=
8x
and
substitute.
dx
dx
8
1
u
e
=
du
Simplify
and
integrate.
8
1
u
e
=
+ C
8
2
1
4 x
+1
e
=
2
+ C
Substitute
4x
+
1
for
u.
8
3
2
12 x
3x
dx
d
4
3
3x
This
integral
is
of
the
for m
x
3
du
2
12 x
f
(g(x))g′(x) dx,
where
3x
dx
4
=
dx
dx
3
3x
4
x
g(x) = 3x
3
– x
3
and
g(x) = 12x
2
– 3x
u
1
=
du
du
4
Let
u
u
=
3x
3
−
x
,
3
then
=
12x
2
−3x
dx
=
lnu
=
ln(3x
+
C,
u
>
0
and
4
4
3x
x
)
x
+
C,
Simplify
3
–
substitute.
3
–
>
and
integrate.
0
4
Substitute
3x
3
–
x
for
u.
Chapter
With
f
practice
(g(x))g ′(x) dx
about
of
f
you
u
what
is
the
with
the
factor
to
be
able
inspection.
would
other
Exercise
Find
by
you
respect
may
the
indefinite
That
choose
of
find
for
is,
u,
you
may
check
integrand
integrals
to
and
just
see
be
that
then
of
the
able
the
form
to
think
derivative
mentally
integrate
u
9F
indefinite
integral
in
questions
1–10.
2
3x
2
+ 2
2
(2x
1
+
5)
(4x) dx
dx
2
3
x
+ 2x
4
3
2
(6x
3
+
5)
3x
5x
4x
4
dx
x
e
dx
x
2x
3
e
dx
5
2
(x
3x
2
x
7
dx
6
2
1)
3
2
2x
4
(2x
+
x
5)
dx
1
8
dx
4
2
x
x
2
4
3
4
(8x
9
–
2
4 x)(x
–
x
3x
3
)
dx
dx
10
3
x
Exam-Style
4 x
Questions
8x
Let
11
f
′(x)
=
.
2
4 x
Given
that
f
′(0)
=
4,
find
f
(x).
+ 1
3
2
The
12
gradient
passes
.
This
of
through
Area
section
and
is
a
cur ve
the
is
point
given
(1, 5e).
defini te
about
the
by
f
′(x)
Find
=
an
3x
x
e
.
The
expression
cur ve
for f
(x).
Indenite
integrals
functions
that
are
a
family
a
constant.
of
integrals
definite
integral,
Denite
differ
integrals
by
are
real
numbers.
b
In
which
is
written
as
f
(x) dx,
and
its
relationship
the
next
section
we
will
learn
to
about
the
relationship
between
a
the
area
under
a
denite
cur ve.
how
to
– area and
the
definite
indenite
evaluate
without
Investigation
and
a
a
integrals
denite
and
integral
GDC.
integral
y
5
2
1
Consider
the
area
bounded
by
the
function
f (x)
=
x
+
1,
x
=
2
0,
f(x)=
x
+
1
4
x
=
2
and
the
x-axis,
which
is
shaded
in
green
in
the
graph.
3
a
i
Write
down
the
width
of
each
of
the
four
rectangles
shown
2
in
the
R
graph.
4
R
3
ii
Calculate
the
height
of
each
of
the
four
rectangles.
R
R
2
1
iii
F ind
the
sum
of
the
areas
of
the
four
rectangles
to
nd
–0.5
a
lower
bound
of
the
area
of
the
shaded
0
{
Integration
0.5
1
1.5
2
x
region.
Continued
on
next
page
Write
this
down
the
width
of
the
four
rectangles
shown
in
graph.
5
2
f(x)=
ii
Calculate
iii
F ind
the
the
sum
height
of
the
of
each
areas
of
of
the
the
four
four
rectangles.
rectangles
x
+
1
4
to
nd
3
R
4
an
upper
bound
on
the
area
of
the
region.
2
R
3
Use
c
a
GDC
to
evaluate
the
defini te
integral
R
2
R
2
1
2
+ 1) d x .
x
Compare
your
result
with
your
answers
x
–0.5
0
0.5
1
1.5
2
0
in
par ts
and
a
b
The
What
do
you
think
the
denite
integral
might
GDC
an
approximation
method
GDC
help
on
CD:
demonstrations
Plus
and
Casio
to
are
on
the
the
TI-84
values
the
could
Now
2
we
not
1;
will
F ind
use
we
a
geometric
could
only
consider
the
area
of
use
some
the
formula
to
nd
geometric
regions
shaded
whose
region
the
area
formulae
area
under
to
can
the
denite
of
the
region
approximate
be
line
so
the
CD.
are
question
of
FX-9860GII
values
We
determine
Alternative
for
integrals,
GDCs
uses
represent?
found
f(x)
=
from
not
the
always
GDC
exact.
in
the
area.
geometrically.
2x
+
2
y
between
x
=
–1
and
x
=
2
using
a
geometric
formula.
Then
6
write
down
a
denite
integral
you
think
may
represent
the
area.
4
Evaluate
the
integral
on
a
GDC
and
compare
answers.
x
3
We
refer
to
the
area
between
a
function
f
and
the
x-axis
y
the
area
under
the
curve .
If
f(x)
is
a
=
2
2
as
=
2x +
2
non-negative
0
–3
function
that
4
for
gives
Verify
and
it
then
on
the
that
following
a
a
≤
≤
area
your
by
x
b
then
under
answer
nding
writing
the
down
write
the
down
cur ve
from
question
area
using
a
denite
a
from
3
denite
x
=
a
works
integral
and
to
for
geometric
f (x)
=
–
x
=
+
3
from
=
1
to
b
=
1
x
the
evaluating
=
=
5
formula
In
mathematics,
is
a
coordinate
x
4
–4
graph
a
on
plane,
a
so
4
4
x
3
–2
y
x
x
–1
cur ve
2
x
–2
integral
GDC.
1
a
f
the
cur ves
include
lines.
4
2
1
y
x +
1
3
2
x
–1
0
2
b
f (x)
=
16
x
y
from
x
=
–4
5
to
x
=
4
2
y
=
√16
–
x
3
2
1
x
–4
–3
–2
–1
0
1
2
3
4
Chapter
In
the
investigation
you
approximated
the
area
under
a
Approximations
for
area
under
2
cur ve
areas
f
(x)
of
=
x
four
+
1
from
rectangles.
x
=
0
to
Using
x
=
2
sigma
by
summing
notation,
we
2
the
f(x)
can
=
x
+
different
1
from
x
numbers
=
0
of
to
x
=
2
for
rectangle.
4
Upper
#
express
this
∑
as
f
(x )Δ x ,
i
where
f
(x )
i
represents
Rectangles
Lower
sum
the
sum
i
i =1
4
height
of
each
rectangle
and
represents
Δ x
the
width
To
5.75
of
i
each
3.75
10
4.28
5.08
50
4.5872
4.7472
100
4.6268
4.7068
4.658 67
4.674 67
rectangle.
get
better
approximations
of
the
area
we
can
use
500
more
rectangles.
Using
an
infinite
number
of
rectangles,
Exact
n
area
=
2
lim
∑
n →∞
14
f
(x )Δ x ,
i
leads
to
the
exact
2
area.
(x
+
1) dx
=
≈
4.66667
i
i =1
3
0
n
If
a
function
f
is
defined
for
a
≤
x
≤
b
and
lim
∑
f
(x )Δx
i
Notice
that
both
the
lower
and
upper
i
n →∞
i =1
sums
exists,
We
we
call
say
this
that
limit
f
is
the
on
integrable
defini te
a
≤
integral
x
≤
and
appear
to
approach
4.66667.
b.
denote
it
b
b
n
The
as
lim
f
∑
(x )Δ x
i
n →∞
=
f
(x) dx
or
y dx.
The
number
a
symbol
is
an
is
i
elongated
S
and
is
i =1
a
a
also
called
the
lower
called
the
upper
limi t
of
integration
and
the
number
b
When
f
is
a
non-negative
sum.
The
a
≤
x
≤
b,
German
f
denite
notation
introduced
y
function
b
for
indicate
integration.
integral
➔
to
is
a
of
limi t
used
(x) dx
gives
y
the
=
by
was
the
mathematician
f(x)
Gottfried
Wilhelm
a
Leibniz
towards
the
b
area
under
the
cur ve
from
∫
f(x)dx
a
end
x
=
a
to
x
=
of
the
b
17th
b
a
0
x
b
centur y
.
f(x) dx
is
a
Example
Write
GDC.
as
from
a
a
definite
Whenever
a
integral
possible
find
that
the
gives
area
the
area
using
a
the
shaded
geometric
region
formula
to
and
evaluate
verify
your
b
to
it
integral
of
f (x)
with
x’.
using
a
answer.
y
b
y
of
‘the
to
respect
down
read
3
2
f(x)
=
2
1
+
x
2
1
f(x)
=
2
–
|x|
1
x
–3
–2
–1
1
0
2
Answers
–2
–1
0
1
2
x
3
The
function
intersects
the
2
x-axis
(2
a
–|x|)
dx
=
at
–2
and
2
and
for ms
4
a
triangle.
So
the
limits
of
–2
1
Area
=
integration
(4
2
×
2)
=
are
–2
and
2. The
4
area
for mula
for
a
1
triangle
is
A =
(b × h )
2
{
Integration
Continued
on
next
page
The
1
region
is
bounded
by
the
2
dx
b
≈
3.14
2
2
1 +
function
x
f
(x ) =
,
the
2
–1
1
x-axis
x
GDC
help
on
CD:
demonstrations
Plus
and
Casio
=
are
on
the
and
x
x
ver tical
=
1.
So
lines
the
Alternative
for
the
limits
TI-84
of
integration
are
–1
FX-9860GII
and
GDCs
and
–1
+
the
1. The
area
cannot
be
CD.
deter mined
from
a
geometric
for mula.
Exercise
Write
down
region
area
9G
and
using
a
definite
evaluate
a
integral
it
using
geometric
that
your
formula
gives
GDC.
to
the
area
Where
verify
your
of
the
shaded
possible
find
the
answer.
y
4
y
1
f(x)
x +
=
3
1
2
3
3
f(x)
=
x
–
4x
2
1
1
–3
–1
–1
0
1
2
3
4
5
x
0
3
4
–1
x
–2
6
–2
–3
y
y
4
4
f(x)
=
3
2
f(x)
3
=
√9
–
x
2
2
1
1
x
–2
–1
1
0
2
3
4
0
5
–4
–3
–2
x
1
–1
2
3
4
y
y
1
4
3
f(x
x +
2
3
3
2
1
f(x)
=
x
1
1
x
0
–1
1
2
3
–1
x
–1
0
1
2
3
4
5
6
7
b
When
f
is
a
non-negative
function
for
a
≤
x
≤
b,
f
(x)dx
gives
y
the
y
=
2x
+
2
a
area
under
the
cur ve
from
x
=
a
to
x
=
6
b
4
Consider
what
happens
when
f
is
not
non-negative.
–1
2
(2x
i
+
2) dx
–3
–4
The
area
of
the
shaded
triangle
is
4,
–3
–2
–1
0
x
1
2
3
–2
but
–1
–4
(2x
+
2) dx
=
–4
since
f
(x)
<
0
when
–3
<
x
<
–1.
–3
Chapter
2
y
(2x
ii
+
2) dx
6
–1
2
4
(2x
+
2) dx
=
9
is
the
area
of
the
shaded
triangle
2
–1
y
since
f
is
a
non-negative
function
for
–1
≤
x
≤
=
2x
+
2
2.
–4
–3
–2
x
0
–1
1
2
3
–2
2
(2x
iii
+
–4
2) dx
–3
2
(2x
+
2) dx
=
5
because
it
is
equal
y
y
to
=
2x
+
2
–3
6
–1
2
4
(2x
+
2) dx
+
(2x
–3
+
2) dx
=
–
4
+
9
=
5.
This
is
the
–1
2
A
2
negative
of
the
area
of
the
region
labeled A
plus
the
1
area
of
the
region
labeled
–4
A
–3
2
–2
–1
x
0
1
2
3
–2
A
1
This
illustrates
one
b
➔
the
proper ties
c
f
(x) dx
=
(x) dx
+
a
Example
graph
of
definite
–4
integrals.
b
f
a
The
of
f
(x) dx
c
of
f
consists
of
line
segments
as
shown
in
the
y
figure.
8
Evaluate
(8, 4)
4
f
(x) dx
using
geometric
formulae.
3
0
(2, 2)
(3, 2)
2
1
0
x
1
2
–1
–2
–3
–4
(6, –4)
Answer
8
f
(x) dx
=
A
–
A
1
+
2
A
Find
the
area
of
the
trapezium
A
3
,
minus
the
area
1
0
the
triangle
A
,
plus
the
area
of
the
triangle,
2
1
=
1
(4
+ 1)( 2 ) −
2
2
3
1
(3)( 4 ) +
y
(1)( 4 )
(8, 4)
2
4
=
5
=
1
–
6
+
2
3
(2, 2)
(3, 2)
2
1
A
A
1
3
0
x
1
2
–1
A
2
–2
–3
–4
(6, –4)
Integration
A
of
➔
Some
properties
of
b
kf
defini te
integrals
b
(x) dx
=
k
f
a
(x) dx
a
b
b
(f
(x)
±
g (x)) dx
=
b
f
a
(x) dx
±
a
g (x) dx
a
a
f
(x) dx
=
0
a
b
a
f
(x) dx
=
–
f
(x) dx
Y
ou
a
do
learn
b
that
(x) dx
=
f
(x) dx
+
f
need
to
a
Example
go
the
with
these,
proper ties.
c
2
that
5
f
(x) dx
=
4,
=
2
f
0
4
g(x) dx
numbers
(x) dx
just
a
the
b
c
f
Given
not
b
(x) dx
=
12,
2
6,
evaluate
these
g(x) dx
=
–3
and
0
definite
integrals
without
using
your
GDC.
0
2
2
2
(3f
a
(x)
–
g (x)) dx
g (x) dx
b
0
+
2
5
f
(x) dx
5
4
−1
1
f
c
(x) dx
g (x) dx
d
f
e
(x
+
3) dx
2
2
0
−3
Answers
2
(3f
a
(x)
–g(x)) dx
0
2
=
2
3f
(x) dx
–
g(x) dx
0
Apply
proper ty
2.
Apply
proper ty
1.
0
2
2
=
3
f
(x) dx
–
g(x) dx
0
0
=
3(4)
=
15
–
(–3)
and
evaluate.
2
2
g(x) dx
b
Substitute
+
f
2
(x ) dx
Apply
5
proper ty
3
to
the
first
ter m
and
5
=
0
–
f
proper ty
(x) dx
4
to
the
second
ter m.
2
Substitute
=
0
–
=
–12
and
evaluate.
12
5
f
c
(x) dx
0
5
2
=
f
(x) dx
4
+
=
16
f
(x) dx
Apply
proper ty
5.
2
0
=
+
12
Substitute
and
{
evaluate.
Continued
on
next
page
Chapter
4
2
Apply
g(x) dx
d
+
proper ty
5.
g(x) dx
2
0
4
=
g(x) dx
0
4
So
g(x) dx
2
4
2
=
g(x) dx
–
g(x) dx
0
=
6
=
9
Rear range
ter ms.
0
–
(–3)
Substitute
and
evaluate.
–1
1
f
e
(x
+
3) dx
Apply
proper ty
1.
2
3
The
graph
of
f
(x
+
3)
is
a
result
of
–1
1
=
f
(x
+
3) dx
translating
the
graph
of
f
(x)
to
the
2
3
left
3
units. The
limits
of
integration,
2
1
=
f
x
=
0
x
=
–3
x
=
2
are
translated
to
(x) dx
2
0
1
these
=
and
and
x
=
integrals
–1.
are
So
the
values
of
equal.
(4)
2
=
2
Exercise
✗
9H
y
The
graph
of
f
consists
of
line
segments
as
shown.
(6, 4)
(8, 4)
4
Evaluate
the
definite
integrals
in
questions
1
and
2
using
3
geometric
formulae.
2
8
1
f
1
(x) dx
4
0
x
1
f
2
(x) dx
–2
(3, –2)
0
6
Given
that
f
(x) dx
=
–3,
f
–3
6
10
1
(x) dx
=
8,
g (x) dx
=
4,
and
1
1
10
g (x) dx
=
8
evaluate
the
definite
integrals
in
questions
6
6
6
1
2
3
f
(x )
g(x )
1
dx
2
g (x) dx
4
10
10
10
g (x) dx
5
f
6
(x) dx
10
1
10
10
f
7
(x) dx
f
8
(x
–
4 ) dx
5
6
4
10
( g(x)
9
+
3) dx
3g(x
10
–1
6
2
–1
8
Integration
+
2) dx
3–10.
Exam-Style
Questions
2
Given
11
that
5
h(x) dx
=
–2
and
h(x) dx
0
=
6,
deduce
the
value
of
2
5
5
h(x) dx
a
(h(x)
b
+
2) dx
2
0
4
Let
12
f
be
a
function
such
that
f
(x) dx
=
16.
0
4
1
Deduce
a
the
value
of
f
(x) dx
4
0
b
b
If
i
f
(x
(
f
–
3) dx
=
16,
write
down
the
value
of
a
and
of
b
a
4
If
ii
(x)
+
k) dx
=
28,
write
down
the
value
of
k.
0
.
Fundamental
Theorem
of
Calculus
y
=
f(x)
y
The
quotient
,
the
slope
of
a
secant
line,
gives
us
an
Secant
x
approximation
for
the
slope
of
a
tangent
line
line.
Tangent
The
product
give
us
(Δy)(Δx),
the
area
of
a
rectangle,
line
helps
Δy
an
approximation
for
the
area
under
a
Δy
cur ve.
Slope
of
tangent
line
≈
Δx
In
much
are
the
inverse
same
sense
operations,
as
division
Isaac
and
Newton
multiplication
and
x
0
Gottfried
Δx
Leibniz
and
independently
definite
This
➔
fact
is
integrals
established
Fundamental
If
f
is
a
came
are
in
the
Theorem
continuous
to
realize
inverse
that
following
of
differentiation
operations.
theorem.
Calculus
function
on
the
inter val a
≤
x
≤
b
and
F
is
b
The
an
antiderivative
of
f
on
a
≤
x
≤
b,
notation
[F ( x )]
a
then
b
means
F(b)
F(a).
b
f
( x ) dx
= [ F ( x )]
=
F (b ) −
F ( a ).
a
a
2
2
Consider
the
definite
integral
(x
+
1) dx
that
you
When
applying
the
Fundamental
0
Theorem
evaluated
using
a
GDC
in
the
investigation
in
Calculus,
although
F
the
can
last
of
be
any
member
of
the
family
of
section.
functions
of
the
antiderivatives
of
f,
2
This
gave
the
area
under
the
cur ve
f
(x)
=
x
+
1
we
between
x
=
0
and
x
=
choose
to
use
the
‘simplest’
one,
2.
that
is,
one
where
the
constant
of
2
2
Y
ou
found
(x
+
1) dx
≈
4.67.
integration
is
C
=
0.
We
can
do
this
0
because,
for
any
C,
b
f ( x ) dx
=
[F
( x)
+
C
]
a
=
[F(b)
=
F(b)
+
–
C]
–
[F(a)
+
C]
F(a)
Chapter
Using
the
Fundamental
Theorem
of
Calculus
we
get:
2
2
1
2
(x
1) dx
3
x
x
0
3
0
1
3
x
1
3
(2
) 2
1
3
(0
+
x
is
the
‘simplest’
antiderivative
3
1
) 0
2
of
3
3
x
+
1.
Evaluate
x
+
x
at
and
x
=
0,
then
nd
the
3
Example
Evaluate
4.67
the
definite
integral
1
without
using
3
a
GDC.
3
1
2
(u
a
–
1) du
dt
b
4x
c
(x
–
1) dx
t
–2
2
1
Answers
1
1
Find
⎡ 1
(u
a
–
1) du
=
⎣
‘simplest’
antiderivative
of
u
–
1.
u
⎢
–2
the
⎤
2
u
⎥
2
⎦
-2
1
2
Evaluate
⎛
1
⎞
2
(1
=
) − 1
⎜
⎝
⎛
2
1
1
⎞
2
−
( −2 )
⎟
⎜
⎠
⎝
−
⎜
–
u
at
u
=
1
and
u
2
=
–2,
then
find
the
dif ference.
⎠
9
–
⎟
2
u
2
( −2 )
⎟
⎞
1
=
⎝
⎛
(2
+
2)
=
–
2
⎠
3
a
1
3
dt
b
Recall
that
ln a
–
ln b
=
ln
= [ln t ]
2
b
t
2
3
=
ln 3
–
ln 2
=
ln
2
3
3
2
4x
c
3
(x
–
1) dx =
4
(x
2
–
x
Rewrite
) dx
1
1
3
⎡ 1
=
1
4
3
x
4
⎥
4
⎣
⎡⎛
=
4
⎢
⎣
3
1
⎝
⎣
✗
Evaluate
⎞
−9
4
Exercise
⎞
3
(3
)
3
⎜
⎝
1
1
)
4
4
⎢
⎦
4
(3
⎜
⎡ ⎛ 81
=
⎤
x
⎢
⎟
⎠
⎛
−
⎠
⎛
−
⎟
1
⎝
⎝
1
4
⎞⎤
1
4
(1
3
)
−
4
(
1
3
⎞⎤
)
⎟
⎠
⎥
⎦
136
=
−
⎜
1
⎜
⎟
3
⎠
⎥
3
⎦
9I
the
definite
integrals
1
in
questions
1–8.
1
2
2x
1
dx
(u
2
0
–
2)
du
–1
8
1
2
3
3
2
x
1
dx
1
Integration
⎞
3
4
⎜
2
⎛
0
⎝
x
=
2
difference.
14
≈
x
3
3
x
⎟
⎠
dx
the
integrand
in
order
to
integrate.
2
3
e
1
x
4e
5
dx
dx
6
The
force
between
x
0
e
electric
9
1
2
(t
7
+
3)(t
+
1)
dt
charges
depends
x
on
the
3
dx
8
amount
of
and
distance
the
charge
x
4
0
the
between
Exam-Style
the
charges.
Questions
How
are
denite
2
integrals
It
9
is
given
that
f
(x)
dx
=
used
to
8
calculate
the
work
0
2
Write
a
down
the
value
of
done
3f
(x)
when
charges
dx.
are
separated?
0
2
2
Find
b
the
value
of
(f
(x)
+
x
)
dx
0
k
1
10
Given
dx
=
ln 6,
find
the
value
of
k
x
2
Now
we
linear
look
at
function
Example
Evaluate
definite
ax
+
b
or
integrals
the
that
involve
substitution
compositions
with
the
method.
the
definite
integral
without
using
a
GDC.
1
5
1
⎛
a
⎜
⎞
2 x
e
3
dx
+
2
x
⎝
(2x
b
–3)
dx
⎟
–1
⎠
1
1
3
2
3x
c
+ 16
dx
(2x
d
3
+
1)
(4x) dx
0
0
Answers
5
1
⎛
1
⎞
2 x
ax + b
+
e
a
⎜
that
e
ax+b
dx
=
e
+
C.
a
x
⎝
Recall
dx
⎟
2
⎠
1
5
2x
=
–2
(e
+
x
)
dx
1
5
⎡ 1
1 ⎤
2 x
e
=
⎢
⎣
⎛
⎥
x
2
1
−
2
5
1
⎜
⎠
⎝
e
1 ⎞
2 (1 )
e
⎟
1
−
⎟
2
1
⎠
4
10
=
⎛ 1
−
⎜
⎝
1
1 ⎞
2(5)
e
=
⎦
2
−
2
e
+
2
5
10
5e
2
− 5e
+ 8
or
10
{
Continued
on
next
Chapter
page
1
n
Recall
3
(2x
b
–
3)
that
(ax
+
dx
b)
=
dx
–1
1
⎡ 1 ⎛
4
(2 x
3)
⎜
⎢
2
⎣
1 ⎛
⎞⎤
1
=
a
4
⎝
⎠⎦
⎛ 1
4
8
1
625
8
8
=
=
⎝
⎞
⎟
+
C.
⎠
⎞
⎛ 1
⎟
⎜
⎠
⎝
4
⎞
( 2( −1) − 3)
⎜
⎝
n+1
( ax + b )
n + 1
1
( 2(1) − 3)
=
1
⎜
⎟⎥
⎟
8
⎠
–78
3
c
3x
n
dx
+ 16
Recall
that
(ax
+
b)
dx
=
3
0
1
=
(3x + 16)
2
dx
1 ⎛
1
n+1
a
3
⎝
⎞
+
( ax + b )
⎜
0
C
⎟
n + 1
⎠
3
⎡
1 ⎛ 2
⎢
⎜
⎞⎤
2
=
3
⎢
⎣
(3 x
+ 16 )
⎟⎥
3
⎠⎥
⎦
⎝
2
⎛
=
0
3
3
2
(3(3) + 16)
⎜
2
–
(3(0) + 16)
9
⎞
⎟
⎠
⎝
3
3
3
⎛
2
⎞
⎜
=
25
–
16
⎟
3
2
122
2
2
Recall
25
25 )
=
125
and
=
3
⎝
9
⎠
9
2
3
16
16
)
=
64.
1
and
1
2
(2x
d
3
+
1)
(4x) dx
0
x = 1
du
du
3
=
2
u
dx
Let
u
=
2x
+
dx
Change
u = 3
3
⎡ 1
3
=
u
du
=
4
4x
and
substitute.
⎣
the
limits
of
integration
and
then
you
can
⎤
u
⎢
u = 1
evaluate
the
integral
in
ter ms
of
u.
When
⎥
4
⎦
1
2
x
=
1
0,
u
=
2(0
)
+
1
=
1,
and
when
x
=
1,
2
4
=
=
dx
x = 0
u
4
[(3)
–
(1)
]
=
=
2(1
)
+
1
=
3
20
4
Exercise
✗
Evaluate
9J
the
definite
integrals
in
questions
1–8.
What
are
some
4
1
1
x
dt
t
+
1
e
applications
dx
center
3
of
1
3
(–2x
+
1)
x
dx
mass
How
2
of
+ 2
–1
(e
(centroid)?
can
denite
–x
+
e
) dx
integrals
be
used
to
–1
–1
nd
2
2
centroid
6x
dx
4
(x
3
+
x)
(2x
1
0
4
1
8t
6
2
x
dt
2
2t
3t
3
the
2
Integration
4x e
2
0
3
dx
+
1) dx
cur ved
area?
of
a
Exam-Style
Questions
y
2
The
9
diagram
Write
a
down
shaded
Find
b
shows
par t
an
of
the
integral
graph
which
of
f
(x)
=
represents
–2x
the
(x
–
area
2)
of
the
region.
the
area
of
the
shaded
region.
x
0
y
1
10
The
diagram
The
area
shows
par t
of
the
graph
of
y
=
x
Find
.5
We
to
the
the
area
sums
He
such
the
the
of
two
are
of
ln
4
1
units.
0
k.
two
of
of
area
under
used
Riemann
Georg
the
x
k
a
cur ve
to
sums
Riemann.
limits
of
sums.
{
Investigation:
Consider
the
area
Georg
Area
between
Riemann
(1826–66)
between
the
two
two
curves
cur ves
y
2
f(x)
2
curves
rectangles
named
existence
is
cur ves.
mathematician
the
region
concept
areas
area
German
proved
value
between
of
shaded
between
extend
approximate
after
the
exact
Area
now
The
of
=
x
+
3x
22
and
20
g(x)
=
x
–
2
from
x
=
–1.5
to
x
=
3.5.
18
16
14
12
10
2
f(x)
=
x
+
3x
8
6
g(x)
=
x
–
2
4
2
x
–4
–6
Continued
on
next
Chapter
page
Copy
1
of
and
the
complete
ve
the
rectangles
Inter val
table
shown
to
in
give
the
the
dimensions
and
area
of
each
graph.
Width
Height
Area
Notice
–1.5
≤
x
<
–0.5
–0.5
≤
x
<
0.5
1
f(–1)
–
g(–1)
=
–2
–
(–3)
=
1
1(1)
=
1
that,
regardless
whether
f
positive,
0.5
≤
x
<
or
zero,
of
the
≤
x
<
≤
x
g
are
the
height
rectangle
given
by
is
f(x),
2.5
the
2.5
and
negative
1.5
always
1.5
of
<
top
g(x),
3.5
cur ve,
the
minus
bottom
cur ve.
Approximate
2
areas
Write
3
of
the
down
the
area
between
the
cur ves
by
nding
the
sum
of
the
rectangles.
a
denite
integral
you
think
can
be
used
to
nd
the
exact
area
2
between
x
=
the
the
Compare
the
If
y
y
and
1
a
cur ves
f(x)
=
x
+
3x
and
g(x)
=
x
–
2
from
x
=
–1.5
to
3.5.
Evaluate
➔
two
≤
integral
on
answer
are
your
to
GDC.
your
approximation
continuous
on
a
≤
x
≤
b
from
and
y
2
x
≤
b,
question
≥
y
1
then
the
area
between
and
y
y
1
for
2.
all
x
in
2
from
x
=
a
to
x
=
y
b
2
b
is
given
by
(y
–
y
1
) dx
2
a
y
–
1
y
2
y
1
Height
of
each
rectangle
=
‘top
cur ve’
–
‘bottom
cur ve’
= y
–
y
1
Width
Area
of
of
each
each
rectangle
rectangle
=
=
( y
dx
–
y
1
Sum
of
the
areas
of
an
2
dx
x
0
) dx
2
infinite
number
of
rectangles
from x
=
a
b
y
2
to
x
=
b
and
exact
area
between
the
two
curves
=
(y
–
1
y
) dx
2
a
Example
2
a
Graph
Write
Solve
the
region
down
this
an
bounded
expression
problem
by
the
that
without
cur ves
gives
using
a
the
y
=
x
area
–
of
2
and
the
y
=
region
–x.
and
then
find
the
area.
GDC.
x
–
2
b
Sketch
Write
Find
the
graph
down
the
an
area,
of
the
region
expression
using
a
that
bounded
gives
the
by
the
area
curves
of
the
f
(x)
=
2e
2
and
g(x)
=
x
Integration
4x.
region.
GDC.
{
–
Continued
on
next
page
Answers
Find
2
a
x
–
2
=
by
2
x
+
x
the
–
2
=
+
setting
2)(x
–
1)
=
=
–2,
Points
the
two
cur ves
the
for
equations
x.
Substitute
equal
the
and
x-values
into
0
either
x
of
0
solving
(x
intersection
–x
equation
to
get
the
y-coordinates.
1
of
intersection:
(–2, 2)
and
(1, –1)
2
y
The
graph
of
y
=
x
–
2
is
the
graph
2
4
of
y
=
x
translated
down
2
units. The
2
y
=
–
x
2
3
graph
2
(0,
0)
of
y
and
=
–x
is
a
gradient
line
with
–1. The
y-intercept
graphs
(–2, 2)
intersect
1
0
–3
at
(–2,
2)
and
(1,
–1).
x
–2
3
–1
(1, –1)
y
=
–x
–3
1
1
2
Area
=
((–x)
–
(x
2
–
2)) dx
=
(–x
–
x
+
y
2) dx
=
–x
is
greater
than
or
equal
to
2
–2
–2
y
=
x
–
2
for
–2
≤
x
≤
1,
so
the
‘height
1
1
⎡
=
1
3
−
x
⎤
2
−
x
by
(–x)
rectangle’
is
represented
⎥
3
⎣
2
1
⎛
⎦
1
3
(1)
−
2
2
⎞
2
−
(1)
+ 2(1)
⎜
3
⎝
⎛
=
each
+ 2x
⎢
=
of
2
1
1
3
2
⎝
⎞
⎛ 8
⎟
⎜
⎠
⎝
+ 2
−
⎜
1
⎛
−
⎜
⎠
⎝
⎟
( −2 )
2
GDC
to
2).
+ 2( −2 )
⎠
9
=
⎠
2
y
b
–
⎞
2
−
(x
⎟
3
⎞
− 2 − 4
3
( −2 )
−
⎟
1
3
–
4
Use
a
and
to
points
3
find
of
help
the
sketch
the
x-coordinates
intersection.
Write
graphs
of
the
down
at
x
least
2
f(x)
=
4
significant
digits
since
these
2e
values
will
be
used
to
compute
the
area.
1
0
–2
x
1
–1
2
–1
–2
2
g(x)
–3
=
x
–
4x
–4
x
2
2e
2
=
x
–
4x
x
f(x)
x
≈
–0.5843,
=
2e
2
is
greater
than
or
equal
to
4.064
2
g(x)
4.064
Area
=
((2e
–0.5843
=
x
–
4x
for
–0.5843
≤
x
≤
4.064,
x
2
2
)
–
(x
–
4x)) dx
≈
14.7
so
the
‘height
represented
of
each
rectangle’
is
by
x
(
2
GDC
)
2
–
(x
–
help
on
CD:
2e
demonstrations
Plus
and
GDCs
Casio
are
on
4x).
Alternative
for
the
TI-84
FX-9860GII
the
CD.
Chapter
Exercise
In
9K
questions
Write
Find
1–4,
down
the
an
area
graph
the
expression
using
a
region
that
bounded
gives
the
by
area
the
of
given
the
cur ves.
region.
GDC.
1
1
2
1
y
=
–
2
x
+
2
and
y
=
x
–
2
2
2
2
2
f
(x)
3
y
4
g(x)
=
x
and
g(x)
=
x
3
=
2x
–
4,
y
=
x
between
x
=
–2
and
x
=
2
2
=
x
+
1
Exam-Style
✗
and
h(x)
=
3
+
2x
–
x
Question
4
5
Consider
the
a
Find
the
b
i
Find
ii
Hence
function
f
(x)
=
2
x
–
x
.
x-intercepts.
f
′(x).
find
the
coordinates
of
the
minimum
and
maximum
points.
c
i
Use
ii
Sketch
your
answers
to
par ts
a
and
b
to
sketch
a
graph
of
f
2
Write
d
questions
cur ves.
Find
6
y
7
f
lnx
f
(x)
=
x
8
f
(x)
=
e
9
y
y
and
sketch
an
using
and
of
x
find
a
=
1
the
graph
–
–
that
x
on
gives
area
of
expression
your
=
g(x)
expression
g
down
area
=
an
and
6–9,
Write
the
graph
down
between
In
a
the
that
of
the
same
the
the
region
gives
area
axes.
of
the
region
region.
bounded
the
area
of
by
the
the
given
region.
GDC.
2
2
–
3x
+
1
and
g (x)
=
x
x
h (x)
y
=
–
x
x
+
–
x
6
2
Question
the
Sketch
b
i
functions
the
Write
graphs
down
between
Find
The
2
–
1
a
c
3
1
and
Consider
ii
=
2
Exam-Style
10
+
2
and
=
x
–x
f
this
line
x
of
an
and
f
(x)
f
=
x
and
and
g
on
expression
g (x)
the
for
=
2
same
the
area
x
axes.
of
the
region
g
area.
=
k
divides
the
area
of
the
region
from
par t b
in
half.
i
ii
Write
down
from
par t
Find
the
Integration
an
expression
b
value
of
k
for
half
the
area
of
the
region
Now
we
a
≤
look
at
cases
where
y
and
y
1
≤
x
b,
but
y
is
not
greater
are
continuous
or
Use
equal
to
y
1
a
≤
x
≤
b.
In
on
2
than
for
all
x
this
case
you
must
find
all
the
nd
points
of
determine
determined
which
by
the
cur ve
points
is
of
above
the
other
in
the
inter vals
the
the
above
in
formed
(x)
down
=
of
and
which
the
cur ve
other
an
expression
2
f
help
x-coordinates
points
determine
cur ve
Write
to
intersection
intersection.
is
Example
GDC
intersection
of
and
a
in
2
10x
+
that
gives
3
x
–
the
area
of
the
region
and
g(x)
=
x
of
–
2x.
Find
the
by
inter vals
the
points
between
2
3x
the
intersection.
area.
Answer
2
10x
+
x
3
–
Find
2
3x
=
x
–
of
x
=
–2,
0,
the
points
of
intersection
2x
f
and
g.
2
2
g(x) = x
– 2x is greater than or
0
2
2
2
((x
– 2x) – (10x + x
– 3x
3
equal to f(x) = 10x + x
3
– 3x
)) dx
for
2
–2
≤
inter val
x
≤
the
0,
so
on
‘height
this
of
each
GDC
help
on
CD:
Alternative
2
2
+
((10x
+
3
x
–
3x
demonstrations
2
)
–
(x
–
2x)) dx
rectangle’
is
represented
2
(x
0
–
2x)
–
(10x
+
x
=
24
f(x)
=
10x
+
and
Casio
TI-84
FX-9860GII
3
–
2
the
by
Plus
2
for
3x
).
GDCs
are
on
the
CD.
3
x
–
3x
is
greater
2
than
–2x
or
equal
for
0
inter val
≤
x
the
rectangle’
to
≤
Exercise
In
1–4,
bounded
3
=
x
write
by
the
down
two
2
–
x
is
so
x
on
of
3x
)
this
each
represented
3
–
=
by
2
–
(x
–
2x).
9L
questions
region
+
2,
‘height
2
(10x
g(x)
1
y
2x
2
f
(x)
=
(x
3
f
(x)
=
xe
4
g(x)
an
expression
cur ves
and
then
to
find
find
the
the
area
of
the
area.
2
and
y
=
2x
–
3x
3
–
1)
and
g(x)
=
x
–
1
2
x
3
and
4
=
–
x
g(x)
=
x
–
x
2
+
Exam-Style
10x
4
–
9
and
h(x)
=
x
2
–
9x
y
Question
1
2
5
The
cur ves
shown
in
the
figure
are
graphs
of
f
(x)
=
x
6
,
4
2
g(x)
a
=
–
x
and
i
Find
ii
Show
h (x)
the
=
2x
–
coordinates
that
the
line
4
4.
Q
of
point
passing
2
Q
through
points P
and
Q
is
0
x
3
4
5
–2
1
2
tangent
to
f
(x)
=
x
at
point
Q
–4
4
b
i
Find
the
coordinates
ii
Hence
write
region
and
down
then
of
an
find
point P
correct
expression
the
for
to
4
the
significant
area
of
the
digits.
shaded
area.
Chapter
.
Volume
of
revolution
Solids
of
revolution
manufacturing
A
solid
of
is
revolution
formed
by
rotating
a
plane
First
an
axis
of
consider
Imagine
a
used
items
in
such
figure
as
about
are
many
pistons
and
crankshafts.
revolution
rectangle
rotating
the
per pendicular
rectangle
360°
to
about
the x-axis.
the
x-axis.
y
y
[
0
0
Pistons
x
x
The
solid
that
cylindrical
in
is
formed
is
called
a
disk.
The
disk
is
shape.
y
dx
2
V
=
πr
=
πy
h
cylinder
2
y
dx
[
0
Crankshaft
x
Investigation –
volume
of
revolution
y
Consider
the
triangle
formed
by
the
line
4
f(x)
f (x)
=
0.5x
Copy
and
and
the
x-axis
complete
between
the
table
x
to
=
0
give
and
the
x
=
6
=
0.5x
3
2
dimensions
1
and
volume
rectangles
of
each
shown
of
in
the
the
disks
gure
formed
are
when
rotated
the
360°
x
1
2
3
4
5
6
7
8
9
–1
about
the
x-axis.
The
last
row
in
the
table
below
–2
is
completed
for
you.
–3
y
Inter val
Radius
Height
Volume
height
4
0
≤
x
<
1
1
≤
x
<
2
2
≤
x
<
3
=
dx
3
2
radius
=
y
1
3
≤
x
<
4
4
≤
x
<
5
x
–2
4
8
–1
2
5
≤
x
<
f(6)
6
=
3
6
–
5
=
1
π (3
)(1)
≈
28.27
–2
–3
2
F ind
the
sum
of
the
volumes
of
the
six
disks
in
y
question
1.
Is
this
sum
greater
or
less
than
the
exact
volume
3
of
the
solid
formed
by
rotating
the
triangle
about
the
x-axis?
2
Write
the
down
exact
a
denite
volume
of
integral
the
solid
you
of
think
can
revolution
be
used
formed
to
nd
when
1
the
0
x
3
triangle
is
rotated
about
the
x-axis.
Evaluate
the
integral
on
–1
–2
a
GDC
and
compare
it
to
your
estimate
in
question
2.
–3
When
is
a
and
the
cone.
triangle
Use
compare
Integration
it
a
is
rotated
geometric
to
the
value
about
the
formula
of
your
to
x-axis
nd
the
the
denite
solid
volume
integral
in
formed
of
the
cone
question
3.
4
5
8
y
➔
If
y
=
f
(x)
is
continuous
on
a
≤
x
≤
b
and
the
region
bounded
dx
by
y
=
f
(x)
and
the
x-axis
between
x
=
a
and
x
=
b
is
rotated
y
360°
about
the
x-axis
then
the
volume
of
the
solid
formed
=
f(x)
is
y
given
by
0
2
V
π (f
=
x
b
b
2
(x))
dx
π y
or
a
dx.
a
y
Radius
Height
of
of
disk
disk
(height
(width
of
of
‘representative
‘representative
2
Volume
of
Sum
the
disk
πr
=
rectangle’)
rectangle’)
= y
=
d
x
2
h=
πy
dx
0
of
volumes
of
an
innite
number
of
disks
from x
=
a
to
x
=
a
b
b
x
b
2
and
exact
volume
of
the
πy
solid
dx.
a
Example
Use
a
definite
integral
to
find
the
volume
of
the
solid
formed
when
the
region
bounded
2
by
f
(x)
using
=
a
9
and
x
geometric
the
x-axis
is
rotated
360°
about
the
x-axis.
Verify
your
answer
formula.
Answer
Sketching
y
4
a
graph
and
‘representative
rectangle’
is
helpful.
Radius
of
disk
is
the
height
dx
2
2
y
√9
=
–
2
of
x
representative
rectangle,
9
x
.
1
Height
of
disk
is
the
width
of
representative
rectangle,
x
–4
–3
–2
0
–1
1
2
3
4
5
dx.
b
The
limits
Use
a
of
integration
are
the
x-intercepts,
–3
and
3.
2
V
π y
=
dx
a
3
2
2
π
=
(
9
)
x
dx
GDC
to
evaluate
the
integral.
−3
≈
To
113
verify:
4
4
3
V
πr
=
3
π(3
=
When
)
about
=
the
region
is
rotated
y
x-axis,
36 π
a
≈
the
3
3
sphere
is
2
for med.
113
4
3
Volume
of
sphere
πr
=
3
x
0
–4
1
4
–2
Exercise
Use
the
a
definite
region
x-axis.
1
f
9M
(x)
Verify
=
integral
bounded
4
your
and
by
to
find
the
answers
the
x-axis
the
given
using
volume
cur ves
is
of
geometric
between
x
=
0
the
solid
rotated
formed
360°
about
when
the
formulae.
and
x
=
5
Chapter
f
2
(x)
=
6
–
2x
and
the
x-axis
between
x
=
0
and
x
=
3
Ibn
al-Haytham
(965–1040),
mathematician
who
lived
a
mainly
in
2
f
3
(x)
4
=
x
and
the
x-axis
Egypt,
the
is
credited
integral
of
a
with
calculating
function
in
order
to
2
f
4
(x)
16
=
x
and
the
x-axis
between
x
=
0
and
x
=
4
nd
the
f
5
(x)
=
x
and
the
x-axis
between
x
=
2
and
x
=
the
3-D
volume
shape
Use
a
a
paraboloid
created
by
–
rotating
a
4
parabola
Example
of
about
its
axis
of
symmetr y.
definite
integral
to
find
the
volume
of
the
solid
formed
when
the
region
under
the
cur ve
2
y
=
x
between
x
=
0
and
x
=
2
is
rotated
about
the
x-axis.
Give
your
answer
in
terms
of
π.
Answer
y
b
5
2
V
πy
=
dx
4
a
2
3
2
π(x
=
2
)
2
dx
y
2
=
x
0
1
2
4
πx
=
dx
dx
–4
–3
–2
0
–1
1
x
2
4
3
0
2
⎡ 1
=
5
Sketching
⎤
π
⎢
⎣
⎥
5
⎦
graph
and
‘representative
rectangle’
is
helpful.
0
Radius
⎛ 1
=
a
x
π
1
5
(2
(0
rectangle,
⎟
5
disk
is
the
height
of
representative
2
)
⎜
⎝
of
⎞
5
)
5
x
.
⎠
Height
of
disk
is
the
width
of
representative
32
rectangle,
dx.
=
5
The
Exercise
limits
of
integration
are
0
and
2.
9N
✗
In
questions
solid
1–4,
formed
about
the
by
use
a
definite
rotating
the
integral
region
to
find
bounded
the
by
volume
the
given
of
the
cur ves
x-axis.
3
1
f
(x)
2
y
3
f
=
x
and
the
x-axis
between
x
=
1
and
x
=
2
2
=
x
+
1
and
the
x-axis
between
x
=
0
and
x
=
1
2
(x)
=
3x
–
x
and
the
x-axis
1
4
y
=
and
the
x-axis
between
x
=
1
and
x
=
4
y
Exam-Style
Question
1
x
5
The
diagram
shows
par t
of
the
graph
of
y
=
4
.
e
The
shaded
1
(4
1
x
region
to
a
x
=
Write
the
b
between
ln
4
is
down
solid
This
graph
a
360°
definite
of
y
=
about
and
e
the
integral
is
equal
to
kπ.
=
x
)
e
the
x-axis
that
Find
from
x
=
0
x-axis.
represents
the
formed.
volume
Integration
the
rotated
y
4
the
value
of
k.
volume
of
0
ln 4
x
Exam-Style
Question
y
1
The
6
shaded
region
in
the
diagram
is
bounded
by y
=
,
x
x
=
1,
x
=
a
and
the
x-axis.
The
shaded
region
is
rotated
360°
1
about
the
x-axis.
y
=
√x
Write
a
of
.
the
The
b
down
solid
a
definite
integral
that
represents
the
volume
formed.
volume
of
Definite
the
solid
formed
integrals
3π.
is
with
Find
linear
the
value
of
0
a
motion
1
a
Extension
material
on
Worksheet
9
volumes
of
and
other
Another
application
function
over
solids
the
of
-
More
CD:
revolution
problems
of
definite
integrals
is
finding
the
change
in
a
Recall
time.
that
if
displacement
then
Suppose
x
displacement
function
for
a
par ticle
moving
along
velocity
=
=
s(t),
v(t)
=
a
s′(t)
and
acceleration
2
horizontal
line
is
given
by
s(t)
=
t
–
4t
+
3
for
t
≥
0,
where
t
is
=
measured
in
seconds
and
s
is
measured
in
metres.
The
a(t)
=
v ′(t)
=
s″(t).
initial
2
displacement
time
0
of
seconds
the
the
par ticle,
par ticle
s(0)
is
3
=
0
–
metres
4(0)
to
+
the
3
=
right
3,
tells
of
the
us
that
at
origin.
The
2
2
–
metre
to
s(2)
=
4(2)
+
3
=
–1
tells
us
that
at
time
2
seconds
the
par ticle
displacement
is
function
1
the
left
of
the
tells
us
the
origin.
distance
and
direction
2
a
v(t)
Consider
dt.
Since
the
antiderivative
of
velocity
is
par ticle
is
from
an
displacement
origin
0
at
any
time
t
2
2
we
have
v(t)
dt
[ s ( t )]
=
=
0
s(2)
–
s(0)
=
–4.
0
Note
t
=
that
t
=
=
2t
–
4,
0
and
s(t)
–1
0
1
2
3
4
5
6
7
4
v(t)
metres
=
2.
gives
from
us
the
change
in
displacement
from
time
0
to
2
us
that
at
2
seconds
negative
the
par ticle
is
4
metres
to
the
at
left
it
was
at
0
t
=
changes
to
2,
so
par ticle
changes
of
direction
where
when
seconds.
the
tells
0
Velocity
positive
This
=
8
t
It
v(t)
2
when
t
=
2.
seconds.
t
➔
v(t)
dt
=
s(t
)
–
s(t
2
)
is
the
change
in
displacement
from t
1
to
t
1
2
t
5
Now
consider
v(t) dt
=
s(5 )
–
s(0 )
=
8
–
3
=
5.
This
tells
us
that
0
at
0
5
seconds
the
par ticle
is
5
metres
to
the
right
of
where
it
was
at
seconds.
t
t
=
=
5
0
s(t)
–1
0
1
2
3
4
5
5
6
7
8
metres
Chapter
Note
that
distance
the
change
traveled
traveled
is
traveled
to
the
total
the
of
right
9
in
displacement
between
4
or
0
and
metres
13
5
of
traveled
metres
5
metres
seconds.
as
to
The
the
shown
is
not
total
left
the
total
distance
plus
9
metres
below
.
metres
t
t
=
=
5
0
s(t)
–1
0
1
4
2
3
4
5
6
7
8
metres
y
We
will
consider
this
in
terms
of
the
area
under
the
cur ve
of
6
v (t)
=
2t
–
v(t )
=2t
–
4
4.
5
4
Let the area of
and the area of
the triangle below the x-axis be A
the
1
3
5
A
triangle
above
the x-axis
be
A
.
v (t)dt
is
the
negative
of
2
2
2
plus
A
0
A
1
1
2
0
x
1
5
2
3
4
5
6
–1
1
1
A
1
|v (t)|dt
=
–A
+
A
1
=
–
(2)(4)
+
(3)(6)
=
–4
+
9
=
5.
–2
2
2
2
0
–3
–4
This
To
gives
find
we
us
the
need
the
total
the
displacement
distance
sum
of
the
from
traveled
areas
A
time
from
and
to
time
A
1
0
.
5
0
We
seconds.
to
5
can
|v(t)|
seconds
find
this
by
means
the
absolute
value
modulus
of
or
2
5
evaluating
v(t).
|v (t)|dt
0
y
5
1
v (t) dt
=
A
+
A
1
1
=
(2)(4)
6
+
(3)(6)
=
4
+
9
=
v(t )
13
=|2t
–
4|
2
2
0
2
4
This
gives
us
a
total
of
13
metres
traveled
from
time
0
to
5
seconds.
3
A
2
2
A
1
➔
If
v
the
is
the
total
velocity
function
distance
for
traveled
a
particle
from t
to
1
t
moving
is
given
along
a
line,
1
by:
0
2
distance
=
x
1
t
2
3
4
5
6
|v (t)| dt.
t
Example
The
displacement
function
for
a
par ticle
moving
along
a
horizontal
line
is
given
by
2
s (t)
=
8
+
2t
–
a
Find
the
b
Find
when
c
Draw
d
Write
total
and
a
t
for
velocity
the
distance
use
≥
of
0,
where
the
particle
motion
definite
then
t
diagram
traveled
the
is
par ticle
is
integrals
t
at
moving
for
to
on
motion
measured
the
find
the
time
right
in
seconds
and
s
is
measured
in
t.
and
when
it
is
moving
left.
par ticle.
the
par ticle’s
inter val
diagram
to
0
≤
t
verify
change
≤
4.
the
Use
in
a
displacement
GDC
to
and
evaluate
the
the
Integration
integrals
results.
{
metres.
Continued
on
next
page
Answers
a
v(t)
=
b
2
2t
–
2
–
=
2t
v(t)
=
Find
t
c
=
s′(t)
0
when
velocity
equals
zero.
1 s
Moves
right
Moves
left
s(0)
=
8
for
0
when
and
<
t
s(1)
t
>
=
<
1
The
1
v(t)
9
par ticle
<
v(t )
t
t
=
=
moves
right
when
v(t)
>
0
and
left
when
0.
+
1
0
t
=
0
1
s(t)
0
1
2
3
4
5
6
7
8
9
Find
d
Change
in
dispacement
at
t
=
0
and
t
=
1.
displacement
t
4
Change
=
(2
–
2t)dt
=
in
displacement
=
v(t)dt
– 8 m
t
0
4
Total
distance
=
t
|2
–
2t|dt
Total
distance
=
|v(t)|dt
t
0
=
9
t
=
10
m
metres
t
4
t
=
=
1
0
s(t)
0
1
2
3
4
8
5
6
7
8
metres
1
9
metre
Show
s(4)
par ticle
at
0
of
8
0
GDC
to
traveled
metres
metres
help
on
to
and
CD:
GDCs
Casio
are
on
the
from
demonstrations
Example
the
metres
par ticle
9
10
Plus
on
diagram.
the
left
At
of
4
seconds
where
it
the
was
seconds.
The
and
is
=
1
left
time
metre
for
0
a
to
to
the
total
4
right
distance
seconds.
Alternative
for
the
TI-84
FX-9860GII
the
CD.
–1
The
velocity
function
v,
in
m s
,
of
a
par ticle
moving
(8, 4)
4
along
a
line
in
traveled
is
shown
in
displacement
on
the
inter val
the
figure.
and
0
≤
t
the
≤
Find
total
16.
the
par ticle’s
distance
)dnoces
change
3
2
1
rep
t (seconds)
0
sertem(
2
4
–1
–2
v
(14, –2)
(15, –2)
–3
–4
{
Continued
on
next
Chapter
page
Answer
4
Let
A
,
A
and
A
2
triangles
be
the
areas
of
the
two
3
and
Change
the
in
)dnoces
1
trapezium.
displacement
rep
16
sertem(
=
v(t)dt
0
–
A
+
A
1
–
)t(v
=
A
2
3
3
2
A
2
1
t (seconds)
0
2
4
–1
A
A
3
1
–2
–3
1
1
=
–
(4)(4)
1
+
(8)(4)
–
2
2
Total
(4
+
1)(2)
=
3 m
–4
2
distance
)dnoces
16
=
|v(t)|dt
A
+
A
1
+
3
(4)(4)
=
(8)(4)
29
of
(4
+
1
A
A
1
A
2
3
t (seconds)
0
2
4
6
8
10
12
14
16
18
1)(2)
2
m
9O
questions
inter val,
+
2
Exercise
Each
1
+
2
2
)t(v
1
1
=
A
2
3
sertem(
=
rep
0
4
where
the
t
is
1–3
gives
a
measured
a
Find
b
Draw
a
velocity
c
Write
definite
motion
displacement
of
displacement
in
the
par ticle
diagram
integrals
and
the
seconds
for
to
the
find
total
function
and
at
s
is
and
time
measured
in
metres.
time t.
par ticle.
the
par ticle’s
distance
change
traveled
on
the
in
given
time
inter val.
Use
a
GDC
diagram
to
to
evaluate
verify
the
the
integrals
and
then
use
the
motion
results.
2
1
s(t)
=
2
s(t)
=
t
–
6t
+
8;
0
≤
t
≤
4
1
3
t
2
–
3t
+
8t;
0
≤
t
≤
6
3
3
3
s(t)
=
4
The
(t
–
2)
;
0
≤
t
≤
4
(5, 6)
–1
line
is
function
shown
displacement
and
the
the
inter vals.
≤
t
≤
12
b
0
≤
t
≤
5
c
0
≤
t
≤
12
figure.
total
,
of
Find
a
the
distance
par ticle
moving
par ticle’s
traveled
for
along
change
each
in
of
the
6
5
4
3
2
1
t (seconds)
0
v
2
m s
sertem(
a
in
rep
following
in
v,
)dnoces
a
velocity
2
4
–1
–2
(9, –2)
Integration
(11, –2)
Exam-Style
Questions
–1
The
5
velocity
,
v,
in
m s
of
a
par ticle
moving
in
a
straight
2
line
is
given
a
Find
b
The
c
the
by
an
Find
the
=
t
–
acceleration
initial
Find
v (t)
9,
where
of
the
displacement
expression
distance
for
of
s,
is
the
par ticle
the
the
traveled
t
time
at t
par ticle
=
is
times
seconds.
1.
12
displacement,
between
in
metres.
in
2
terms
seconds
of
t
and
(4, 4)
8
4
seconds.
line
is
function,
shown
Find
b
Write
the
the
the
m s
of
a
par ticle
moving
along
figure.
acceleration
down
in
time
when
t
=
sertem(
a
in
v,
3.
inter val(s)
on
which
the
par ticle
is
)t(v
traveling
to
the
Find
total
3
2
rep
a
✗
velocity
)dnoces
–1
The
6
1
t (seconds)
0
4
–1
–2
(13, –2)
(15, –2)
–3
right.
–4
c
the
distance
traveled
for
0
≤ t
≤
16.
The
Definite
integrals
can
be
used
in
situations
other
than
linear
of
We
can
use
definite
integrals
to
find
the
cumulative
effect
integral
of
a
rate
motion.
of
change
is
the
total
any
change
from
t
to
1
var ying
rate
of
t
:
2
change.
t
2
F′ (t)dt
=
F(t
)
–
F(t
2
Example
A
culture
bacteria.
of
bacteria
The
).
1
t
rate
at
is
star ted
which
with
the
an
number
inital
of
population
bacteria
of
100
changes
over
a
one-
0.273 t
month
r
is
period
measured
Find
the
can
in
be
modeled
bacteria
population
of
per
by
the
function
r(t)
=
e
,
where
day
.
bacteria
20
days
after
the
culture
was
star ted.
Answer
0.273t
r(t)
is
=
the
R(t),
e
is
a
rate
derivative
that
bacteria
gives
at
of
a
the
time
t.
of
change.
function,
number
It
say
of
Therefore
20
r(t)dt
=
R(20)
–
R(0)
Notice
0
is
the
change
bacteria
Since
from
the
in
the
day
initial
0
number
to
day
of
show
20.
population
was
that
that
results
100
in
the
the
a
units
integral
number
of
bacteria.
20
bacteria,
the
population
after
20
0.273 t
20
0.273t
days
is
100
+
e
dt
or
dt
bacteria
( days )
(per
0
about
e
day
≈
857
bacteria
)
0
957
bacteria.
You
could
using
the
method
get
the
same
following
(see
{
next
result
longer
page).
Continued
on
next
page
Chapter
0.273t
R(t)
=
e
dt
Find
the
R′(t)
=
function
r(t).
R(t),
such
that
Recall that
1
1
0.273t
=
e
0
dt
+
C
ax + b
e
273
ax + b
dx =
e
+ C.
a
1000
0.273t
=
e
+
C
273
000
0.273(0)
100
=
e
+
C
Use
273
R(0)
1000
C
=
100
the
=
initial
100
condition
to
find
that
C.
Notice
how
much
–
273
more
convenient
it
26300
to
=
obtain
the
same
273
result
1000
using
20
26300
0.273t
R(t)
=
e
+
0.273t
273
100
273
+
e
0
26300
000
0.273(20)
R(20)
=
e
+
≈
Exercise
Write
an
answer
1
The
1,
≈
R(20).
9P
expression
these.
rate
2000
Find
957
273
273
Use
of
to
a
involving
GDC
to
1,
definite
evaluate
consumption
Januar y
a
of
2010
oil
(in
in
integral
the
a
that
can
be
used
to
expression.
cer tain
billions
of
countr y
barrels
from
per
Januar y
year)
is
t
modeled
years
Find
2
The
by
since
the
the
function
Januar y
total
number
1,
C ′(t)
18.4e
20
,
where
t
is
the
number
spectators
of
oil
who
over
enter
the
a
10-year
stadium
period.
per
hour
for
2
football
0
≤
t
1.5
There
at
0
t
=
From
are
hours.
many
There
is
is
modeled
hours.
hour.
How
3
≤
game
The
no
cubic
midnight
to
the
function
game
spectators
36.5
by
spectators
The
of
a.m.
in
snow
is
the
at
the
snow
function r(t)
r(t)
in
begins
are
cm
8
time t
=
stadium
on
a
=
1375t
measured
stadium
1.5
driveway
accumulates
in
modeled
by
the
function
s(t)
=
the
the
at
for
per
gates
game
the
open
begins?
driveway
3
+
t
midnight.
4
rate
–
hours.
on
0.01t
a
3
people
when
when
(–
a
of
2000.
consumption
of
=
0.13t
at
2
–
0.38t
–
0.3t
+
0.9)
,
5te
3
where
How
4
is
measured
many
Water
The
t
cubic
begins
rate
at
in
cm
leaking
which
it
hours
of
snow
from
is
and
a
are
tank
leaking,
s
in
on
.
cm
the
driveway
holding
measured
4000
in
modeled
by
the
function
r (t)
=
–133
8
gallons
gallons
per
a.m.?
of
.
1
How
much
Integration
water
is
in
the
tank
at
the
end
60
of
20
water.
minute,
t
be
at
minutes?
can
957.
dt
is
Review
exercise
✗
1
Find
the
indefinite
integral.
3
3
4
3
(4x
a
–
8x
+
6) dx
x
b
dx
dx
c
4
x
4
5x
3x
4x
dx
d
2
e
e
dx
3
f
x
i
(3x
l
2x
4
(x
+
1)
dx
2
6x
ln x
1
2
dx
g
2x
dx
h
+
1)(6x) dx
x
+ 3
x
2
2 x
2e
dx
j
3
k
2x
dx
5
e
dx
x
e
2
Find
the
definite
integral.
16
2
2
e
4
4
2
(3x
a
–
6) dx
dt
b
dx
c
t
x
4
0
1
1
2
1
2
3 x
3
6x e
d
1
3
dx
(3x
e
–
1)
dx
dx
f
0
Exam-Style
2x
0
–1
+ 1
Questions
2
3
The
diagram
shows
par t
of
the
graph
of
f
(x)
=
x
–
1.
Regions
A
2
y
y
and
B
are
=
x
–
1
shaded.
3
a
Write
down
b
Calculate
c
Write
an
expression
for
the
area
of
region B
2
the
area
of
region
B
1
B
down
an
expression
for
the
total
area
of
shaded
regions
0
A
and
B.
Region
d
(Y
ou
B
is
need
rotated
not
evaluate
about
the
the
x-axis.
Write
down
x
1
–2
expression.)
A
2
an
–2
expression
evaluate
4
A
cur ve
the
with
gradient
for
the
volume
of
the
solid
formed.
(Y
ou
need
not
expression.)
equation
function
is
y
=
f ' (x)
f
=
(x)
3x
passes
–
2.
through
Find
the
the
point
equation
of
(2,
the
6).
Its
cur ve.
5
5
Given
that
f
5
(x) dx
=
20,
1
deduce
the
value
of
5
1
f
a
(x) dx;
[f
b
(x)
+
2] dx
4
1
1
−1
6
A
par ticle
moves
along
a
straight
line
so
that
its
velocity
, v
m s
2t
at
time
t
seconds
displacement,
for
s
in
terms
s,
of
is
of
given
the
by
v (t)
par ticle
=
is
4e
8
+
2.
metres.
When
Find
t
an
=
0,
the
expression
t
k
1
7
Given
dx
2x
=
ln 5,
find
the
value
of
k
1
1
Chapter
Review
exercise
Exam-Style
1
Find
the
Questions
volume
of
the
solid
formed
the
x-axis
when
the
region
bounded
2
by
f
(x)
=
4
–
x
and
is
rotated
360°
about
the
x-axis.
–1
2
A
par ticle
moves
along
a
horizontal
line
with
velocity v
m s
2
given
by
v (t)
=
2t
–
11t
+12
where
t
≥
0.
2
a
b
c
Write
down
terms
of
an
expression
par ticle
is
of
and
value
Find
to
the
the
time
the
acceleration, a
m s
,
in
t.
The
a
for
total
5
moving
of
to
the
left
for a
<
t
<
b.
Find
the
value
b
distance
the
par ticle
travels
from
time
2
seconds
seconds.
3
3
a
Find
the
equation
of
the
tangent
line
to f
(x)
=
x
–
2
at
x
=
–1.
3
b
The
tangent
Find
the
c
Graph
d
Write
and
f
and
an
the
line
intersects
coordinates
the
tangent
9
line
f
(x)
this
tangent
expression
CHAPTER
of
for
=
x
–
2
at
a
second
point.
point.
line.
the
and
area
then
enclosed
find
the
by
the
graphs
of
f
area.
SUMMARY
Antiderivatives
and
the
indefini te
integral
1
●
n
Power
rule:
x
n 1
dx
x
C ,
n
1
n 1
●
Constant
rule:
●
Constant
multiple
●
Sum
or
More
k
dierence
on
dx
=
kx
rule:
rule:
indefinite
+
kf
(
f
C
(x) dx
(x)
±
=
k
f
g(x)) dx
(x) dx
=
f
(x) dx
±
g (x) dx
integrals
1
●
dx
ln x
C ,
x
0
x
x
●
e
x
dx
e
1
n
●
( ax
b)
dx
1
n 1
( ax
b)
C
a
n 1
1
ax b
●
e
ax b
dx
e
C
a
1
●
1
dx
ax
b
b
ln( ax
a
b) C ,
x
a
Continued
Integration
on
next
page
Area
and
definite
integrals
b
y
●
When
f
is
a
non-negative
function
for
a
≤
x
≤
b,
f
(x)dx
y
a
gives
the
area
under
the
cur ve
from x
=
a
to
x
=
=
f(x)
b.
b
∫
f(x)dx
a
●
Some
properties
of
defini te
b
integrals
b
0
kf
(x) dx
=
k
f
a
x
b
(x) dx
a
a
b
b
b
(f
(x)
±
g (x)) dx
=
f
a
(x)dx
±
g (x) dx
a
a
a
f
(x) dx
=
0
f
(x) dx
=
–
a
b
a
f
a
b
c
f
(x) dx
f
is
is
an
(x) dx
+
f
a
Fundamental
f
b
=
a
If
(x) dx
b
a
(x) dx
c
Theorem
continuous
antiderivative
function
of
f
on
a
of
on
≤
Calculus
the
x
≤
inter val a
b,
≤
x
≤
b
and
F
then
b
b
f
( x ) dx
=
∫
[F
( x )]
=
F (b ) −
F (a )
a
a
Area
●
If
between
y
and
y
1
a
≤
are
two
curves
continuous
on
a
≤
x
≤
b
and
y
2
x
≤
b,
≥
y
1
then
the
area
between
y
and
y
1
for
all
x
in
2
from
x
=
a
to
x
=
b
2
b
is
given
by
(y
–
y
1
) dx
2
a
Volume
●
If
y
by
=
y
f
=
of
(x)
f
(x)
revolution
is
continuous
and
the
on
x-axis
a
≤
x
≤
between
b
x
and
=
a
the
and
region
x
=
b
is
bounded
rotated
y
dx
360°
y
about
the
x-axis
then
b
(x))
of
the
solid
formed
is
=
f(x)
given
y
2
πy
or
a
dx
0
x
a
Definite
and
volume
b
2
π ( f
by
the
integrals
other
wi th
linear
motion
problems
t
●
v(t)dt
=
s(t
)
–
s(t
2
)
is
the
change
in
displacement
from t
1
to
1
t
2
t
●
If
v
is
the
velocity
function
for
a
par ticle
moving
along
a
line,
t
the
total
distance
traveled
from
t
1
to
t
is
given
by:
distance
=
|v(t)|dt
2
t
Chapter
Theory
of
knowledge
Know
The
your
method
The
ancient
was
formalized.
one,
the
Greeks
ancient
circumscribed
Let
be
a
the
To
of
used
find
Greeks
polygons
areas
of
limits!
exhaustion
concepts
an
of
estimate
for
constr ucted
with
the
calculus
the
regular
increasing
regular
long
area
of
a
with n
of
calculus
circle
inscribed
numbers
polygons
before
of
radius
and
sides.
sides
inscribed
in
a
n
circle
of
radius
one
and
be
A
the
areas
of
the
circumscribed
n
polygons.
The
ancient
Greeks
found
that
both
and
lim A
lim
a
n
n
n→∞
were
equal
What
Can
π.
to
conclusion
you
n→∞
think
of
were
they
other
able
to
draw
applications
of
from
limits
these
in
real
Although
Newton
vs.
development
culmination
of
of
calculus
centuries
mathematicians
all
over
of
was
work
the
tr uly
world.
mathematicians
and
a
resolved,
today
generally
believed
Gottfried
Isaac
Wilhelm
are
recognized
for
the
of
calculus.
One
and
develop
conflicts
in
Leibniz
over
one
of
Moder n-day
of
the
centur y
,
calculus
first
and
is
due
them
was
to
calculus
the
emerged
effor ts
in
the
of
invented
whether
such
as
Augustin-
the
Cauchy
(French),
Ber nhard
or
(German),
Karl
any
Weierstrass
plagiarism
one
most
histor y
Riemann
discovered
of
actual
mathematical
which
independently
Leibniz
Louis
argument
that
calculus
mathematicians
famous
is
Newton
19th
development
it
The
another.
(German)
debate
fully
by
did
(English)
the
never
Newton
17th-centur y
life?
Leibniz
was
The
facts?
(German),
IS
AA
C
and
involved.
others.
What
are
some
possible
consequences
N
when
people
seek
personal
acclaim
for
their
Did
the
work
of
Suppose
that
Newton
and
Leibniz
calculus
independently
of
Would
this
offer
suppor t
need
to
the
to
calculus
was
discovered
or
solve
cer tain
that
it
problems
or
idea
purely
that
from
one
real-world
another.
arise
did
the
develop
T
work?
mathematicians
EW
these
from
V
S
intellectual
was
curiosity?
invented?
GO
TT
FR
I
LE
Theory
of
knowledge:
Know
your
limits!
ED
IB
W
ILH
N
I
Gabriel’s
Consider
the
horn
solid
formed
when
the
region
bounded
by
1
f
(x)
,
=
x
=
1
and
x
=
a,
a
>
1
is
rotated
about
the
x-axis.
x
If
a
→
The
∞,
the
volume
solid
of
the
is
known
as
Gabriel’s
horn .
solid
y
of
revolution
about
the
3
x-axis
is
given
by
2
b
π∫
y²dx.
It
can
be
a
1
shown
that
the
surface
x
1
area
of
the
solid
is
b
by
2
3
4
a
given
-1
2π∫
1
+
(y')²dx
a
-2
Use
a
GDC
to
find,
-3
to
four
places,
above
table.
decimal
the
for
volume
the
Then
surface
given
make
area
and
surface
values
a
a.
conjecture
approach
as
=
a
1
a
a
Volume
of
π∫
area
of
Write
about
the
solid
them
what
approaches
in
a
the
described
copy
volume
dx
and
Surface
area
=
2π∫
1
1
1
a
²
the
infinity
.
x
1
of
[
1
+
4
]
dx
x
x
10
100
1000
10 000
100 000
1 000 000
a
→
∞
Volume
Based
take
on
to
How
the
fill
up
much
→
results
Surface
in
your
Gabriel’s
paint
table,
how
area
much
→
paint
would
it
hor n?
would
it
take
to
cover
its
surface?
N
Paradoxes
A
result
that
example
of
a
dees
logic
paradox.
is
called
Research
a
paradox.
some
other
Gabriel’
s
horn
examples
is
one
of
paradoxes.
LM
Chapter
Bivariate
CHAPTER
Linear
5.4
analysis
OBJECTIVES:
correlation
coefcient
r;
of
bivariate
scatter
data;
diagrams,
Pearson’
s
lines
of
best
product–moment
t;
mathematical
correlation
and
contextual
interpretation.
The
5.4
equation
of
the
regression
line
of
y
on
x;
use
of
the
equation
for
prediction
purposes.
Before
Y
ou
1
you
should
Calculate
start
know
simple
how
positive
to:
Skills
exponents
1
Evaluate:
4
e.g.
Evaluate
3
5
.
a
2
b
3
c
7
4
3
check
3
=
3
×
3
×
3
×
3
=
81
3
3
2 ⎞
⎛
e.g.
Evaluate
.
⎜
7
⎟
5
⎝
1 ⎞
⎛
⎠
d
⎜
⎛
⎝
2
2
×
2
×
2
5
×
5
×
5
4
=
⎟
5
⎠
3
2 ⎞
⎜
⎟
2
⎝
3
=
3
⎠
5
3
⎛
e
⎜
⎞
⎟
4
⎝
⎠
8
=
3
f
0.001
125
2
Conver t
numbers
to
exponential
form
2
State
the
value
n
e.g.
Find
2
2
n
given
2
=
8.
equations:
n
a
×
×
2
=
2
=
16
=
243
=
343
=
625
8
n
b
3
3
2
=
n
8
c
7
d
5
n
n
=
3
n
e
(–4)
=
–64
n
⎛
f
⎝
Bivariate analysis
1 ⎞
⎜
⎟
2
⎠
1
=
8
of
n
in
the
following
In
1956,
an
convincing
cancer.
He
Australian
case
of
souther n
was
was
of
skin
In
Chapter
5
population :
A
sample
Suppose
height
The
the
x
is
This
the
cancer
a
that
and
the
rate
Oliver
well
of
Lancaster,
exposure
of
skin
correlated
nor ther n
result
dealt
with
por tion
are
weight
bivariate
the
was
consists
we
sampling
that
between
with
states
before
careful
data
in
latitude,
in
and
the
collection
the
and
Australia
higher
hole
made
sunlight
cancer
had
the
to
hence
rates
and
among
with
than
ozone
first
skin
the
layer!
His
comparison
rates.
we
it
statistician,
link
strongly
sunlight:
ones.
discover y
a
obser ved
Caucasians
amount
for
units
data
of
of
univariate
all
of
the
of
are
adult
adult
contains
We
measurements
defined
of
a
interest.
population.
interested
y
the
analysis.
in
studying
males.
males
all
the
of
Sampling
uni t
Variable(s)
Population
adult
males
height
univariate
adult
males
weight
univariate
adult
males
height,
bivariate
and
the
pairs
(x, y)
weight
of
height
➔
and
Bivariate
pairs
of
weight
of
analysis
variables
is
the
males
in
concer ned
(x, y)
in
a
our
with
data
sample.
the
relationships
between
set.
Chapter
In
this
data
chapter
using
using
a
we
will
graphs,
scale
to
describe
Investigation
The
bell
and
soon
its
tower
show
was
The
the
beyond
of
began
name.
the
–
Pisa
in
associations
to
So
of
in
a
the
built
side
–
in
as
an
two
sets
equation
of
and
relationship.
of
Pisa
1178
hence
below
millimetre
1975
metres
of
tower
was
one
between
relationship
strength
cathedral
tenths
2.9642
a
leaning
leaning
metres.
leaning
for
measurements
lean
2.9
look
representing
the
from
tower
the
ver tical.
Y
ear
1975
1976
1977
1978
1979
1980
1981
1982
1983
1984
1985
1986
1987
Lean
642
644
656
667
673
688
696
698
713
717
725
742
757
Does
If
it
so,
look
how
like
fast
the
is
the
Is
there
evidence
Is
there
an
Can
you
.
lean
lean
that
the
Scatter
the
tower
increasing
the
approximate
predict
of
lean
in
increasing
with
changes
formula
lean
is
the
for
with
time?
time?
signicantly
calculating
the
with
time?
lean?
future?
diagrams
Correlation
One
way
to
view
data
is
by
showing
it
on
way
➔
Scatter
investigate
that
(also
diagrams
both
the
possible
relate
to
the
called
scatter
relationship
same
plots)
are
between
used
two
to
diagrams
horizontal
a
ver y
and
specific
are
similar
ver tical
axes
pur pose.
A
to
to
associated
to
or
affects
variables
The
purpose
The
To
line
plot
scatter
graphs
data
in
that
points.
diagram
they
use
However,
shows
how
allow
they
much
have
us
from
a
the
graph.
the
a
dots
to
prediction
variable
one
is
doing
to
make
about
based
we
know
a
one
on
about
another.
relationship
draw
of
‘event’.
between
two
variables
is
called
scatter
data
The
can
diagram
table
as
patter n
give
us
dots
plot
the
y)
values
of
y
on
formed
(x,
variable.
their correlation
For
➔
are.
variables
another
➔
how
related
two
what
variable
a
measure
correlations
Scatter
is
a scatter diagram
the
Pisa
think
leaning
that
increases
by
time.
some
tower
example,
the
we
lean
with
Time
is
the
independent
indication
of
the
correlation.
Dependent
variable.
The
depends
on
lean
variable
The
be
independent
on
the
variable
horizontal
axis
should
with
variable
on
the
is
0
ver tical
axis.
Bivariate analysis
time,
the
so
dependent
the
Independent
variable
the
the
amount
of
lean
dependent
x
variable.
➔
A
general
upward
trend
in
the
patter n
of
dots
shows posi tive
correlation.
y
The
value
of
the
dependent
variable
increases
as
the
value
of
the
7
6
independent
variable
increases.
5
4
3
2
1
0
➔
A
general
negative
The
downward
in
the
patter n
of
dots
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
1
2
3
4
5
6
7
8
shows
correlation.
dependent
variable
trend
x
1
variable
decreases
as
the
y
independent
7
increases.
6
5
4
3
2
1
0
➔
Scattered
close
to
points
with
no
trend
may
indicate
x
correlation
zero
y
7
6
5
4
3
Scatter
Here
diagrams
are
allow
differing
us
to
amounts
assess
of
the
positive
strength
of
a
correlation.
2
1
correlation:
0
y
y
y
10
10
10
9
9
9
8
8
8
7
7
7
6
6
6
5
5
5
4
4
4
3
3
3
2
2
2
1
1
1
0
x
1
Strong
y
2
3
4
5
positive
increases
as
6
7
8
9
10
correlation:
x
x
0
x
1
2
3
Moderate
4
5
6
positive
7
8
9
10
correlation
0
x
1
Weak
2
3
4
positive
5
6
7
8
9
10
correlation
increases
Chapter
Here
are
differing
amounts
of
negative
y
correlation:
y
y
10
10
10
9
9
9
8
8
8
7
7
7
6
6
6
5
5
5
4
4
4
3
3
3
2
2
2
1
1
1
0
x
1
2
Strong
3
4
5
6
negative
7
8
9
0
x
10
correlation:
1
y
decreases
2
3
Moderate
4
5
6
negative
7
8
9
0
10
x
1
x
Not
3
4
5
6
7
8
9
10
correlation
Weak
as
2
negative
correlation
increases
all
relationships
are
linear.
y
10
9
8
The
points
on
this
graph
are
7
6
approximately
linear.
5
4
3
2
1
0
x
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
6
7
8
9
10
y
10
9
8
7
The
points
on
this
graph
would
6
be
represented
by
a
cur ve.
5
4
There
is
a
non-linear
relationship
3
between
the
2
variables.
1
0
x
Causation
➔
A
correlation
that
Here
is
one
an
students’
words,
has.
to
causes
the
example:
the
vocabular y
the
Now
do
between
larger
it
with
is
the
easy
each
shoe
have
other,
data
sets
does
not
necessarily
mean
other.
shoe
to
two
see
a
sizes
strong,
size,
that
but
of
the
positive
larger
shoe
they
grade
are
size
students
correlation.
the
and
highly
school
In
vocabular y
vocabular y
correlated.
and
other
the
student
have
The
the
nothing
reason
is
Y
ou
may
‘causation
there
is
a
confounding
factor ,
age.
The
older
grade
for
school
exploration.
students
will
have
Bivariate analysis
larger
shoe
sizes
and
often
a
larger
vocabulary
.
to
use
versus
correlation’
stimulus
that
wish
as
an
the
Example
a
Represent
this
data
on
a
scatter
diagram.
x
1
2
3
4
4
6
6
6
7
8
y
1
3
3
5
6
7
5
6
8
9
b
Is
c
Describe
the
relationship
the
type
linear
and
or
non-linear?
strength
of
the
relationship.
Answers
a
y
10
9
8
7
6
5
4
3
2
1
0
x
2
b
This
is
4
a
6
linear
8
relationship.
Compare
the
There
c
is
a
strong,
the
examples
scatter
diagram
with
earlier.
posi tive
correlation.
Exercise
1
0A
Describe
a
correlation
shown
y
0
d
the
b
e
x
each
of
these
scatter
diagrams.
y
0
x
y
0
by
c
x
y
0
x
y
0
x
Chapter
2
For
the
following
data
i
is
the
correlation
ii
is
the
relationship
iii
is
the
association
a
sets:
positive,
linear
strong,
negative
or
or
is
moderate,
weak
b
y
6
5
5
4
4
3
3
2
2
1
1
0
0
x
1
2
3
4
5
6
7
8
9
10
y
d
10
6
7
8
9
10
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
6
7
8
9
10
y
5
4
4
3
3
2
2
1
1
0
x
1
2
3
4
5
6
7
8
9
0
10
f
y
10
9
9
8
8
7
7
6
6
5
5
4
4
3
3
2
2
1
1
x
1
and
2
3
4
5
complete
6
7
dependent
then
as
variable
independent
correlation
dependent
Bivariate analysis
then
9
these
independent
correlation
8
as
variable
x
y
10
0
5
6
5
the
4
7
6
If
3
8
7
b
2
9
8
the
x
1
10
9
If
y
7
6
a
zero?
8
7
Copy
or
9
8
3
association
10
9
e
no
non-linear
10
c
there
0
10
x
sentences.
and
the
dependent
variables
independent
variable
show
a
positive
increases
the
…………………
and
the
dependent
variables
independent
variable
…………………
show
a
negative
increases
the
This
4
table
shows
Year
cm
in
Tennessee
from
2000
to
2003
2004
2005
2006
2007
2008
42
51
39
44
31
33
30
28
21
this
b
Describe
c
In
data
the
general,
table
on
a
scatter
diagram.
correlation.
what
shows
Friend
a
has
happened
group
of
to
friends
the
rainfall
with
their
since
the
year
mathematics
Ted
Tom
T
od
May
Ray
Kay
Jay
Mathematics
85
75
66
80
70
95
90
60
Science
75
65
40
72
55
88
80
40
1
Draw
a
scatter
2
Describe
the
diagram
correlation
Investigation
Construct
a
–
scatter
tower
to
of
represent
in
terms
leaning
diagram
Pisa
for
of
this
the
investigation
data
at
of
direction
Pisa
from
the
science
and
form.
(continued)
the
star t
of
Extrapolation
this
scores.
data.
strength,
tower
2000?
and
Tim
leaning
2008.
2002
Show
a
in
2001
a
This
rainfall
2000
Rainfall
5
the
means
estimating
chapter
.
a
b
Describe
c
What
the
value
a
point
that
is
larger
than
correlation.
(or
is
at
happening
to
the
lean
as
the
smaller
than)
the
data
you
have.
years
Extrapolating
here
means
assuming
increase?
that
d
Research
the
latest
developments
on
the
effor ts
to
save
Comment
.
The
on
the
the
line
leaning
dangers
of
tower
of
best
of
the
trend
of
the
lean
will
remain
the
same.
Pisa.
extrapolation.
fit
y
➔
A
line
find
to
of
the
best
or
fit
direction
show
the
trend
of
trend.
an
This
line is
drawn
on
a
association
between
line
fit
of
best
can
scatter
two
then
diagram
variables
be
used
to
and
to
(x, y)
make
➔
To
predictions.
draw
will
a
line
balance
number
of
of
the
points
reference
point:
the
point
best
fit
number
below
one
by
of
eye
the
point
draw
points
line.
for
the
a
line
above
An
that
the
line
with
improvement
line
to
pass
is
the
to
through.
0
have
This
a
is
The
mean
and
is
calculated
by
finding
the
mean
of
x
mean
written
x-values
and
the
mean
of
the
point
is
the
as
(
x
y
)
y-values.
Chapter
Example
Is
there
fast
a
relationship
between
the
grams
of
fat
and
the
total
calories
in
food?
Meal
T
otal
fat
Hamburger
(g)
T
otal
calories
9
260
Cheeseburger
13
320
Quar ter
Pounder
21
420
Quar ter
Pounder
30
530
31
560
Big
with
Cheese
Burger
T
oasted
Sandwich
31
550
Chicken
Wings
34
590
Chicken
25
500
28
560
20
440
5
300
Crispy
F ish
F illet
Grilled
Chicken
Grilled
Chicken
Light
a
Find
the
mean
number
of
grams
b
Find
the
mean
number
of
calories.
c
Constr uct
d
Plot
of
the
best
a
scatter
mean
diagram
point
on
your
for
of
fat.
this
scatter
data.
diagram
and
use
it
to
draw
a
line
fit.
Answers
247
a
Mean
grams of
fat
Mean
=
grams
of
fat
11
Total
=
grams
of
fat
Hence
=
22.45
Number
of
meals
(x,
y )
=
(22.45,
457.27)
5030
b
Mean
no. of
calories
=
Mean no. of
calories
11
Total
=
457.27
no. of
Number
c
and
calories
=
of
meals
d
A
‘line
of
best
t’
Calories
is
The
line
of
best
fit
does
not
also
called
a
have
600
regression
to
pass
through
(0, 0).
It
must
through
Mean
point
400
the
mean
point
and
The
British
and
statistician
have
the
same
number
of
data
Francis
points
either
side
of
Galton
coined
regression
centur y.
100
0
10
20
Grams
Bivariate analysis
30
of
fat
40
(1822–
it.
1911)
200
scientist
( x, y )
roughly
300
line
pass
500
in
the
the
term
19th
Exercise
1
The
0B
table
width
of
a
Length
Width
118
125
136
145
25
30
38
50
36
42
52
48
58
62
mean
a
scatter
(cm)
table
and
draw
a
line
of
best
fit
through
gives
the
heights
and
weights
of
ten
sixteen-
Abe
Bill
Chavo
Dee
Eddie
Fah
Grace
Hanna
Ivy
Justin
182
173
162
178
190
161
180
172
167
185
73
68
60
66
75
50
80
60
56
72
(kg)
b
Constr uct
i
the
a
mean
table
mean
in
diagram
shows
the
the
0
a
line
of
scatter
in
hours
best
spent
fit
through
studying
and
mathematics.
5
6
7
8
1
3
7
9
9
8
10
14
point.
a
grades
of
4
mean
mean
draw
3
the
Constr uct
weight.
2
1
b
and
mean
1
grade
Find
the
number
students’
in
a
ii
point.
studying
Increase
height
scatter
below
increase
your
diagram
students.
Find:
Hours
point.
point.
following
a
the
and
mm.
105
mean
Weight
The
in
length
95
Name
3
measured
the
80
Constr uct
your
leaf
between
78
b
Height
relationship
50
the
year-old
tree
the
35
Find
The
shows
mango
a
your
2
below
What
diagram
and
draw
a
line
of
best
fit
through
the
wish
to
the
What
can
Y
ou
in
correlation.
nancial
d
of
explore
extrapolation
Describe
c
risks
extrapolation?
may
point.
are
you
say
about
the
effect
of
the
number
of
or
climate
hours
models.
spent
The
equation
mean
of
mathematics
the
line
and
of
the
best
increase
fit
in
grade?
through
the
point
Raw
data
with
rough
diagram
➔
studying
rarely
fit
appears
The
line,
a
straight
predictions.
to
equation
can
be
‘fit’
of
used
line
exactly
.
Typically
,
a
the
for
straight
line
of
you
line,
best
prediction
Usually
,
have
the
fit,
a
set
line
also
of
you
of
must
data
best
called
be
satisfied
whose
scatter
fit.
the regression
pur poses.
Chapter
Example
Miss
and
Lincy’s
final
10
exam
students’
are
scores,
shown
out
of
100,
for
their
classwork
below
.
Student
Ed
Craig
Uma
Phil
Jenny
James
Ron
Bill
Caroline
Steve
Classwork
95
66
88
75
90
82
50
45
80
84
F inal
95
59
85
77
92
70
40
50
Abs
80
Caroline
was
absent
for
the
final.
a
Find
the
mean
classwork
b
Find
the
mean
final
c
Constr uct
d
Find
e
Use
the
the
a
scatter
equation
equation
exam
of
the
the
not
include
her
grades
in
finding
the
mean
point.
score.
diagram
of
Do
score.
and
draw
regression
regression
a
line
of
best
fit
through
your
mean
point.
line.
line
to
estimate
Caroline’s
score
for
the
final
exam.
Answers
Classwork
a
Mean
classwork
score
total
=
Number
of
students
675
Mean
classwork
score =
= 75
9
Final exam
b
Mean
final exam
total
score =
Number
of
students
648
Mean
final exam
score =
= 72
9
c
100
erocs
80
Mean
point
maxe
60
lani F
40
20
0
20
40
60
Classwork
80
100
score
y
y
2
d
Using
the
mean
point
and
Uma’s
Use
m
where
x
x
2
results,
we
have
(x
, y
1
)
=
1
, y
1
, y
2
)
=
) is the mean point
1
(88, 85)
2
and (x
, y
2
m
1
(75, 72)
(x
(x
1
=
85
72
88
75
=
=
) is any point on
2
the line. Use y
1
y
= m(x
1
for the equation of
The
equation
of
the
line
x
)
1
the line.
is:
Using
y
–
72
=
1(x
–
best
y
=
x
–
y
=
80
t
–
3
Caroline’s
=
77
exam
Caroline’s classwork
score
is
estimated
to
be
77.
score was 80.
Let
x
=
80.
data
the
of
predict
value
range
given
than
Bivariate analysis
of
data
is
within
the
called
interpolation.
generally
line
to
3
a
e
the
75)
more
It
is
reliable
extrapolation.
Exercise
0C
EXAM-STYLE
1
QUESTIONS
T
omato
plants
scientist
wishes
the
disease.
percentage
Percentage
Draw
the
Find
c
Use
2
how
designs
(x
to
a
disease
the
an
called
temperature
experiment
leaves
(y)
leaves
diagram
with
a
of
the
which
at
An
agricultural
greenhouse
she
different
monitors
affects
the
temperatures.
70
72
74
76
78
80
12.3
9.5
7.7
6.1
4.3
2.3
°F)
diseased
in
occurring
blight.
regression
line
passing
through
point.
equation
your
of
equation
the
to
regression
estimate
the
line.
number
of
diseased
leaves
75 °F
.
Market
sales
see
scatter
the
risk
diseased
of
mean
b
at
a
at
to
She
of
Temperature
a
are
research
figures
Price
for
of
new
(thousands
Sales
of
new
real
estate
homes
of
homes
this
the
mean
house
b
Find
the
mean
number
c
Draw
the
mean
d
Find
e
Use
the
different
year
Find
scatter
of
£)
a
a
investments
reveals
prices
the
over
following
the
past
year.
160
180
200
220
240
260
280
126
103
82
75
82
40
20
price.
of
diagram
sales.
with
a
regression
line
passing
through
point.
equation
your
of
equation
the
to
regression
estimate
the
line.
number
of
new
homes
Extension
material
Worksheet
10
bivariate
priced
at
£230 000
Understanding
Example
A
study
years
can
18
of
r un
was
the
were
on
CD:
More
analysis
sold.
regression
line
was
a
that
-
done
young
one
to
investigate
person
kilometre.
collected.
The
and
Data
the
the
time
from
equation
relationship
of
y
in
minutes
children
the
between
in
between
regression
the
which
the
line
the
ages
was
age
of
x
in
child
7
found
The
and
to
the
be
y-intercept
height
when
x
=
of
0,
is
the
and
line
might
1
y
x.
= 20
Inter pret
the
slope
and
not
y-intercept.
always
have
a
2
meaning.
with
your
Be
careful
interpretation
Answer
of
1
In
the
context
of
the
question,
The
slope
is
.
What
this
means
can
say
that,
on
average,
as
Sometimes
ages
one
year
their
time
the
value
a
that
child
intercept.
is
2
we
the
for
ever y
increase
of
1
in
x
there
x
=
0
is
impossible
to
1
or
is
r un
a
kilometre
goes
down
by
a
decrease
of
in
represents
y.
30
2
a
seconds
(half
a
dangerous
minute).
extrapolation
For
is
this
not
question,
relevant,
children
the
since
cannot
run
y-intercept
0-year-old
one
The
y-intercept
that
when
x
is
is
0,
20,
y
is
which
means
the
range
of
outside
the
data.
20.
kilometre.
Chapter
Example
A
biologist
trees
She
x
per
wants
calculates
State
the
to
hectare
the
study
and
the
the
equation
gradient
and
relationship
number
the
of
the
of
between
birds
y
per
regression
y-intercept
and
the
number
of
hectare.
line
to
inter pret
be
y
=
8
+
5.4x.
them.
Note
that
all
these
interpretations
follow
Answer
a
The
slope
expect
The
an
is
y-intercept
birds
per
and
1
each
x
She
2
A
them
a
of
police
par ticular
homework
up
with
chief
=
equation
A
of
per
is
wants
+
y
group
The
additional
per
an
tree,
you
can
hectare.
area
with
no
trees
averages
8
not
the
to
x
a
relevant,
that
same
student
of
has
the
be
the
with
the
been
y
=
is
the
and
does
why
.
per
of
=
40
between
a
days
week.
line y
of
of
number
relationship
convicted
are
number
the
regression
crime
–
0.3x
the
and
the
knows.
0.5
+
relationship
person
on
spor ts
person
to
data
explain
plays
investigate
person
and y-intercept
collected
equation
person
up
If
has
found
x
slope
smokes
6x
between
per
day
the
and
number
the
of
number
of
sick.
the
equation
of
the
reg ression
line
2.4x
y
of
exam
of
came
the
wants
to
his
to
shop
regression
mathematics
investigate
and
each
line
is y
science
year
=
–5
the
number
of
x.
+
teachers
100x
wished
to
compare
scores.
science
regression
salesman
that
equation
their
a
the
comes
the
that
researching
year
customers
A
was
skateboard
The
y
cigarettes
doctor
7
ever y
birds
that
student
the
criminals
doctor
what
teacher
of
y
for
means
relevant.
number
days
if
times
The
5
state
of
A
that
additional
which
number
packs
4
y
came
The
3
means
5.4
score
line
Bivariate analysis
The
gradient
line
is
which
x
science
year
hours
8,
scenario,
social
per
is
of
0D
inter pret
A
This
hectare.
Exercise
For
5.4.
average
y
y
=
and
–10
the
+
mathematics
0.8x
score
x
gave
the
pattern:
the
y
of
the
amount
increases
increases
by
1
by
when
unit.
.
The
Least
term
contexts.
between
regression
The
the
but
shorter
than
height
the
him;
of
sons
is
Let
the
us
and
visit
is
a
the
illustrate
of
best
fit
is
then
for
to
a
to
1.0.
A
tends
We
can
mean
the
one
‘by
father
have
of
problem
constr uct
mean
draw
to
taller
have
than
mean.
again.
sons
him.
The
We
term
scatter
a
line
know
number
line
of
of
that
years
diagram
Inaccuracies
the
of
fitting.
the
draw
point.
to
a
other
related,
tends
the
to
relationship
are
sons
curve
and
the
two
between
point
point
differently
towards)
sorts
Pisa
The
tall
to
(moves
of
quite
examine
sons.
correlation
tower.
drawn
and
many
tower
positive
have
used
than
regresses
find
statistics
father
through
only
in
first
less
used
data,
line)
we
is
short
the
regression
fathers
leaning
of
the
(regression
because
now
strong,
lean
a
used
was
of
slope
‘regression’
there
is
method
heights
course,
The
squares
best
to
fit
occur
through
and
the
line
eye’.
y
There
is
another
involving
way
to
improve
our
line,
actual
residuals
data
point
(x
y )
i
Residual
=
y
–
i
predicted
data
i
y
p
point
(x
p
0
➔
A
the
The
graph
residual
above
is
residual
the
the
of
is
a
ver tical
distance
regression
positive
if
the
between
a
data
point
and
equation.
data
point
y
is
graph.
residual
is
negative
if
the
data
point
is
residual
below
Negative
the
)
p
x
Positive
The
y
residual
graph.
Zero
The
residual
through
the
is
0
only
data
when
the
graph
point.
0
The
residual
passes
equation
of
the
regression
line
x
of yonx
y
The
least
squares
regression
line
uses
our
previous
formula
(3, 5)
5
y
–
y
=
m(x
–
x
1
),
but
now
uses
the
method
of
least
squares
to
1
r
4
find
a
suitable
value
for
the
slope,
m
(1, 3)
3
➔
The
least
squares
regression
line
is
the
one
that
has
p
the
2
smallest
possible
value
for
the
sum
of
the
squares
of
q
the
1
residuals.
(2, 1)
2
In
the
diagram
we
aim
to
make
p
2
+
q
r
x
0
2
+
as
close
to
zero
as
1
2
3
4
5
possible.
Chapter
A
rather
complicated
formula
emerges:
The
earliest
the
The
formula
for
of
regression
finding
the
gradient,
or
slope
method
line
of
regression
least
squares
was
that
was
(m)
published
a
form
of
by
Legendre
in
1805, and
is:
by
Gauss
four
years
later
.
Legendre
S
xy
m
➔
=
,
and
where
Gauss
both
applied
the
method
2
(S
)
x
to
the
problem
astronomical
(∑
S
=
xy
x
y
)(∑
determining ,
obser vations,
from
the
orbits
)
and
xy −
∑
of
of
bodies
about
the
Sun.
n
2
2
(S
(∑
2
)
=
x
∑
x
x
)
∑
is
n
‘S’
an
Example
the
and
data.
least
regression
diagram
squares
line
on
Greek
regression
through
page
the
formula
points
(1, 3),
to
find
(2, 1),
the
and
equation
(3, 5)
of
from
is
letter
used
instruction
the
Use
the
−
∑ xy
sum
as
to
sum
means
of
all
the
xy
of
regression
values.
the
the
345.
Answer
(∑
S
=
xy
x
)(∑
y
)
2
x
xy
∑
y
xy
The
x
ter ms
in
The
line
of
on
n
6
=
×
1
3
3
1
2
1
2
4
the
for mula
y
x,
which
can
9
20
be
used
to
estimate
y
3
=
3
5
15
9
6
9
20
14
given
2
2
(∑
2
x
The
)
=
x
∑
of
)
each
2
(S
sum
x
x
column
n
2
6
=
14
3
=
The
line
2
equation
of
the
regression
is
S
xy
y
–
y
=
2
(S
x
(
x
x
)
)
2
y
–
3
=
(x
–
2)
The
mean
point
(
x ,
y
)
is
(2, 3).
2
y
Now
=
x
you
+
1
have
regression
line
seen
how
works,
the
from
formula
now
on
for
you
the
can
equation
use
your
of
the
GDC
to
find
it.
See
GDC
Sections
➔
Y
ou
should
use
your
GDC
to
find
the
equation
of
the
5.16.
regression
line
Bivariate analysis
in
examinations.
Chapter
5.15
17,
and
Example
The
table
in
US
to
twelve
of
shows
dollars
Use
a
this
Write
c
Use
distance
Changi
in
km
air por t,
and
airfares
Distance
Fare
Singapore,
576
178
370
138
612
94
1216
278
409
158
1502
258
946
198
998
188
189
98
787
179
210
138
737
98
destinations.
your
b
the
from
GDC
data
down
your
1000 km
to
with
the
sketch
the
a
scatter
line
of
best
equation
of
your
equation
to
estimate
diagram
fit.
line
the
of
cost
best
of
fit.
a
flight.
Answers
a
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
y
b
=
0.117x
+
83.3
You
will
answers
cost
c
=
(0.117
=
×
1000)
+
Cost
83.3
=
three
$(0.117
Dollars
$200.30
usually
to
and
have
to
round
significant
× distance
cents
two
Casio
are
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
your
figures.
+ 83.3)
decimal
places.
Exercise
Y
ou
0E
should
use
your
GDC
for
all
of
medicine
by
a
this
exercise.
It
1
A
patient
is
given
concentration
intervals.
will
exist
The
in
his
blood
doctors
between
the
is
believe
drip
feed
measured
that
a
at
linear
and
its
the
hourly
this
the
relationship
variables.
x
(hours)
0
1
2
3
4
5
a
Show
the
b
Write
down
c
Find
the
y
2.4
data
on
the
4.3
a
5.0
scatter
equation
concentration
of
6.9
9.1
diagram
of
the
sensible
–
relationship
value
The
after
we
from
8
don’t
will
process
to
predict
hours
know
continue
of
outside
tr ying
the
from
whether
to
to
range
be
predict
of
your
6
11.4
with
be
equation
data
Concentration
not
concentration
linear
.
a
T ime
would
a
is
called
extrapolation
13.5
line
of
best
fit.
line.
medicine
in
the
blood
after
3.5
hours.
Chapter
2
The
of
table
below
Malaysian
shows
the
Ringgits
for
value
the
of
first
Jai’s
seven
car
in
years
thousands
after
it
was
purchased.
Age
(yrs)
Cost
a
(MYR
Show
best
1000)
the
price
0
1
2
3
4
5
6
7
30
25
21
19
18
15
12
10
of
the
car
on
a
scatter
diagram
with
a
line
of
fit.
b
Write
down
c
Estimate
the
equation
of
the
regression
line.
1
the
cost
of
his
car
after
4
years.
2
d
Suppose
Jai
equation
50
3
The
and
takes
cannot
good
be
care
used
to
of
the
car.
estimate
Explain
the
cost
why
of
the
the
car
after
years.
table
the
below
number
shows
of
hours
Person
Months
ten
people
that
who
they
bought
exercised
gym
in
the
membership
past
week.
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Nick
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Nancy
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7
8
9
1
5
12
2
10
4
6
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3
5
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3
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8
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of
membership
Hours
of
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Show
b
Find
c
If
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4
Sarah’s
Their
Age
he
use
diagram
height,
HEIGHT
=
height
of
a
the
diagram
regression
member
last
for
2
years
concer ned
has
3
the
to
that
she
following
60
86
90
91
94
95
showed
hormone),
a
least
+
and
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strong
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positive
squares
0.3833
is
50
best
fit.
how
uses
the
The
old
if
shor t
of
many
for
her
Sarah’s
line
doctor
there
and
hours
why
.
association
regression
prediction,
many
Explain
regression
AGE.
years
doctor’s
how
seems
record
57
she
of
estimate
membership?
51
71.95
line
line.
estimate
48
when
a
months,
36
a
with
week.
equation
after
and
Check
scatter
your
are
(cm)
and
a
exercised
you
prediction.
was
between
found
wants
is
no
line
to
to
predict
inter vention
for
then
age.
height:
this
comment
on
procedure.
Revisit
the
a
Find
b
Draw
the
been
(months)
(growth
5
has
parents
Sarah’s
this
on
equation
exercised
A scatter
be
data
paediatrician
Height
age
the
the
hours
d
exercise
data
the
a
scatter
mean
c
Find
d
Use
from
mean
the
leaning
diagram
tower
of
Pisa.
with
a
regression
line
passing
point.
equation
your
the
point.
of
equation
Bivariate analysis
to
the
regression
estimate
the
line.
lean
in
1990.
through
.
Up
to
Measuring
this
point
relationship
determined
We
have
then
line
for
Now
it
as
positive
called
the
will
the
used
seek
a
scatter
between
or
two
negative;
correlation
equation
prediction
we
have
(correlation)
also
found
we
correlation
of
the
diagram
to
variables.
zero
weak,
regression
if
see
We
there
is
moderate
line
of
y
if
there
is
a
have
no
or
on
correlation.
strong.
x
and
We
used
the
pur poses.
to
classify
the
strength
of
the
correlation
Karl
numerically
.
There
are
several
scales
that
are
in
use;
we
will
study
Pearson
1936)
correlation
coefficient
developed
by
Karl
founded
rst
statistics
The
Pearson
product
moment
correlation
byr)
variables
X
is
a
measure
of
the
It
is
of
widely
linear
Y,
used
giving
in
the
dependence
a
value
correlation
sciences
between
between
as
a
two
between
+1
measure
and
of
−1
the
in
coefficient
is
used
to
strength
variables.
determine
how
nearly
the
relationship
fall
on
a
straight
line,
or
how
nearly
linear
they
are.
the
A
not
linear
,
will
have
a
regression
coefficient
of
r
=
1.000.
the
0.01
physical
or
gives
better.
us
margin
the
of
sciences
That
we
means
confidence
to
would
that
say
a
like
to
have
coefficient
that
a
of
a
‘confidence
r
=
relationship
is
0.990
or
linear
level’
of
higher
within
a
not
coefcient
adequately
represent
of
the
the
y
between
the
r-value
Pearson’
s
the
from
correlation
coefcient
indicates
strength
relationship
two
x
0
positive
Here
r
=
r
are
variables.
y
The
correlation
strength
relationship
0.01%.
y
Perfect
this
Normally
does
in
then
perfect
correlation
correlation
variables
the
is
points
1911.
inclusive.
between
regression
College
two
In
The
depar tment
University
London
and
university
coecient
at
(denoted
the
Pearson.
world’
s
➔
(1857–
a
=
linear
x
0
No
correlation
r
=
0
some
negative
correlation
more
data
sets
and
their r
the
sets.
x
0
Perfect
1
data
of
between
r
=
linear
−1
values:
0.7
r
=
0.3
Chapter
For
negative
correlation,
the
value
of
r
is
also
negative
r
r
➔
The
formula
=
=
–0.3
–0.7
for
finding
Pearson’s
correlation
coefficient
is
S
xy
r
=
S
S
x
y
where
2
(∑
S
=
xy
x
)(∑
y
2
xy −
∑
(∑
)
,
S
=
∑
x
x
x
)
−
Y
ou
these
n
n
the
2
(∑
2
S
=
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∑
y
y
)
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n
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quick
way
to
inter pret
r-value
0
<
0.25
0.5
|r|
≤
<
|r|
<
0.75
Example
Sue
wants
number
orchids
formula
Plant
of
<
|r|
is:
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0.5
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0.75
≤
weak
Strong
1
to
determine
spoons
grown
to
≤
≤
r-value
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0.25
|r|
the
of
from
inter pret
Spoons
of
food
a
the
plant
plant.
the
Use
she
of
the
uses
correlation
and
Pearson’s
the
extra
correlation
Increase
of
in
the
orchids
A
1
2
B
2
3
C
3
8
D
4
7
Bivariate analysis
number
the
of
coefficient
number
y
{
between
relationship.
plant
x
strength
food
should
recognise
and
Continued
on
next
page
formulae
previous
from
section.
Y
ou
should
use
your
Answer
GDC
2
(∑
S
=
∑
xy
x
)(∑
y
calculater
2
Plant
x
y
xy
x
y
A
1
2
2
1
4
B
2
3
6
4
9
in
)
the
examinations.
xy
n
10 × 20
60 −
=
to
We
have
formula
shown
and
the
table
= 10
4
C
3
8
24
9
64
D
4
7
28
16
49
here
to
help
understand
you
where
the
2
value
2
S
x
x
x
Total
10
20
60
30
comes
from.
126
See
GDC
Chapter
17,
n
Section
5.16.
2
10
=
30
5
4
2
2
S
y
y
y
n
2
20
126
26
4
Regression
S
r
=
=
S
≈
S
x
5
y
correlation
0.877
positive
correlation
means
the
number
of
spoons
of
help
on
CD:
Alternative
different
sets
to
there
see
for
the
the
number
of
Plus
and
GDCs
extra
orchids
Casio
r-value
if
are
on
the
CD.
a
you
0.877
to
explore
other.
are
to
For
predict
are
example,
positively
expectancy
correlated,
we
know
correlated.
who
is
likely
So
to
we
that
when
succeed
can
IB
predict
scores
one
and
based
college
college
admission
at
universities,
their
on
students
with
high
IB
formulae
appear
officials
they
want
will
scores.
statistical
methods
complicated
at
first
sight,
making
useful
and
evaluating
the
r-value
are
quite
straightforward.
would
for
the
analyzing
table
a
achievement
be
the
and
GDP.
the
Which
While
between
correlation.
variables
choose
the
indicates
countr y’
s
two
For
might
increases.
of
life
If
data,
may
connection.
example,
relationship
strong
of
FX-9860GII
wish
The
two
TI-84
be
increases,
compare
plant
demonstrations
food
to
that
GDC
as
allow
26
you
A
and
10
xy
From
business
this
performance?
point
on
we
will
be
using
technology
to
find
the r-value.
Chapter
Exercise
1
Nine
their
0F
students
results.
between
the
a
Find
two
French
the
sets
and
r-value
of
a
Spanish
and
test.
describe
the
The
C
D
E
F
G
H
I
French
56
56
65
65
50
25
87
44
35
Spanish
87
91
85
91
75
28
92
66
58
thinks
that
social
income
psychologist
and
gives
scores.
B
A
table
correlation
A
Subject
2
sat
education.
She
there
found
is
that
a
correlation
people
with
Y
ou
between
can
that
higher
people
years
income
have
more
years
of
education.
The
results
of
her
shown
Years
3
(thousand
of
Find
b
What
can
c
What
does
a
shows
$)
A
B
C
D
E
F
G
H
I
J
125
100
40
35
41
29
35
24
50
60
19
20
16
16
18
12
14
12
16
17
education
a
Does
the
car
the
with
it
more
education
higher
income.
below
.
Person
Income
of
phrase
sur vey
have
are
also
r-value.
you
say
the
stop
age
about
sign
more
(in
of
the
the
slowly
months)
strength
r-value
as
of
it
a
the
correlation?
indicate?
gets
car
of
older?
and
its
The
table
stopping
below
distance
(in
1
metres)
Age
from
a
speed
(months)
of
40 km h
9
15
24
30
38
46
53
60
64
76
28.4
29.3
37.6
36.2
36.5
35.3
36.2
44.1
44.8
47.2
Stopping
distance
4
(m)
a
Find
the
b
What
c
Describe
happens
Kelly
has
focus
on
on
the
her
to
the
stopping
strength
always
school
Kelly’s
r-value.
been
studies.
grades
of
told
Kelly
first,
the
and
to
distance
as
the
car
gets
older?
correlation.
stop
wants
chatting
to
decides
see
to
if
on
the
there
sur vey
10
computer
will
be
any
friends.
and
effect
Here
are
results.
An
GPA
3.1
2.4
2.0
3.8
2.2
3.4
2.9
3.2
3.7
3.5
14
16
20
7
25
9
15
13
4
14
A
grade
points,
Chat
a
B
is
wor th
is
3
4
points,
time
a
C
is
2
points,
a
D
is
0
1
(h/week)
point,
and
points.
a
Find
b
Describe
c
the
Based
F
average
r-value.
on
the
the
grade
is
grade
point
sur vey
,
decreased?
Bivariate analysis
called
the
correlation.
would
Kelly’s
grade
increase
if
the
chat
GPA.
time
an
The
average,
5
Mo
get
10
had
on
always
with
friends
been
some
to
see
told
work,
the
to
and
effect
stop
so
on
playing
he
computer
decided
GPA.
The
to
games
conduct
results
are
a
in
and
sur vey
the
of
table
below
.
GPA
Game
2.7
3.8
1.5
3.6
2.2
3.8
2.0
1.9
2.5
3.0
10
24
25
17
5
26
14
30
22
7
time
(h/week)
a
Find
b
Describe
c
Based
time
the
r-value.
the
on
correlation.
the
sur vey
,
would
Mo’s
grade
increase
if
his
game
Extension
material
Worksheet
10
bivariate
6
Find
on
CD:
decreased?
and
inter pret
Review
the
r-value
for
the
leaning
tower
of
Pisa
-
More
analysis
data.
exercise
✗
1
Phrases
i,
between
i
High
ii
Low
iii
No
iv
Low
v
High
Which
ii,
iii,
two
iv,
and
v
represent
positive
positive
of
the
correlation
linear
linear
correlation
correlation
correlation
negative
linear
negative
phrase
variables
best
shown
correlation
linear
in
correlation
represents
each
of
the
the
relationship
scatter
between
diagrams
the
two
below?
y
y
a
b
10
10
8
8
6
6
4
4
2
2
0
0
x
2
c
descriptions
variables:
4
6
8
10
x
2
4
6
8
10
2
4
6
8
10
y
y
d
10
10
8
8
6
6
4
4
2
2
0
x
2
4
6
8
10
0
x
Chapter
EXAM-STYLE
y
QUESTIONS
60
2
The
and
following
the
table
number
of
gives
the
amount
kilometres
of
traveled
fuel
after
in
a
car’s
filling
the
fuel
tank,
tank.
)sertil(
Distance
Amount
220
tank
500
680
850
(km)
of
20
fuel
55
in
276
leuF
0
traveled
40
43
30
24
10
6
(litres)
0
x
200
a
Copy
the
scatter
diagram
and
plot
the
remaining
400
Distance
The
mean
amount
on
the
of
Sketch
c
A
the
This
table
took
them
Age
T ime
a
Plot
b
Find
c
Draw
d
How
to
to
2
28
of
best
fit
left
the
r un
in
through
Use
the
ages
of
your
ten
the
line
mean
of
point.
best
policemen
fit
to
estimate
the
and
the
time
that
it
100 m.
45
45
50
10.9
11.1
10.8
12.0
11.2
12.1
12.6
13
12.7
13.6
data
a
on
mean
line
the
of
a
scatter
age
and
best
would
fit
you
diagram.
the
mean
through
expect
a
point.
30-year-old
policeman
to
take
exercise
QUESTIONS
6
the
number
of
push-ups
3
4
5
6
Push-ups
7
8
5
3
2
2
a
scatter
a
Show
the
b
What
happens
points
c
Find
the
equation
d
Find
the
r-value
heights
(m)
that
David
can
do
each
minutes.
2
and
h
(kg)
on
to
the
use
of
regression
it
a
to
along
push-ups
a
the
line
time
of
best
fit.
increases?
line.
describe
sample
with
as
of
the
11
relationship.
students
are:
1.47
1.54
1.56
1.59
1.63
1.66
1.67
1.69
1.74
1.81
52
50
67
62
69
74
59
87
77
73
67
regression
regression
Bivariate analysis
the
of
1.36
the
height
of
and
diagram
number
weights
w
down
whose
mean
100 m?
shows
the
time.
the
1
Use
plotted
39
for
b
is
35
table
Write
point
32
long
a
This
mean
25
the
Weight
the
tank.
Minutes
The
litres.
and
24
the
Height
is
y
421 km,
23
r un
minute
tank
is
x
22
EXAM-STYLE
The
the
350 km.
fuel
shows
Review
1
line
of
in
traveled
(km)
diagram.
traveled
amount
3
fuel
scatter
b
car
distance
600
points.
is
line
1.6 m.
to
line
of
w
estimate
on
the
h.
weight
of
someone
EXAM-STYLE
3
A
psychologist
IQ
of
a
child
the
IQ
of
a
sample
a
Write
b
Find
c
Use
a
down
the
the
The
and
of
8
your
children
mother.
and
She
mothers:
103
108
111
123
94
96
89
102
98
94
116
117
correlation
of
to
took
line
line
IQ
predict
between x
the
IQ
of
the
a,
explain
how
accurate
mathematics
are
of
Test
given
tests.
2
We
from
Test
72
32
68
55
80
45
77
Test
2
31
38
16
34
27
41
22
37
Describe
c
Copy
on
your
Find
e
If
the
tend
that
it
a
Height
was
y
(cm)
Plot
these
represent
2 cm
on
b
Write
c
Plot
d
i
e
Find
the
f
Draw
g
Using
1.
‘Students
score
of
a
diagram.
best
score
will
on
get
with
Test
a
high
score
2’.
fit.
of
in
measured
40
in
Test
for
Test
1,
what
can
2?
the
first
8
weeks
3
4
5
6
7
8
23.5
25
26.5
27
28.5
31.5
34.5
36
37.5
of
values
on
the
line
the
the
on
of
on
a
scatter
horizontal
diagram
axis
and
taking
1cm
to
1cm
to
represent
axis.
value
point
of
on
the
the
mean
scatter
correlation
this
equation
your
was
know
2
down
the
line
student
to
1
Comment
ii
…….
the
like
0
mean
Write
your
sentence
achieved
ver tical
down
the
think
diagram.
from
a
plant
week
the
you
bought:
pairs
1
scatter
the
of
this
pot
x
Week
a
have
student
of
when
to
equation
predict
height
from
a
1
on
correlation
complete
another
we
The
and
Test
d
results
the
of
below:
54
Plot
y
mother
would
1
b
and
x.
estimate
result
results
on
be.
two
the
to
coefficient
y
100.
par t
to
of
Test
a
measures
98
likely
percentage
their
between
94
answer
could
of
relationship
91
an
is
IQ
the
87
the
with
investigate
the
regression
students
we
to
regression
estimate
Eight
if
y
IQ
child
Using
this
x
IQ
Mother’s
5
wants
the
Child’s
4
QUESTIONS
of
Name
coefficient, r,
for
it L.
these
readings.
result.
the
regression
regression
equation
point.
diagram.
on
estimate
the
the
line
of
scatter
height
y
on
x.
diagram.
of
the
plant
after
1
4
weeks.
2
h
DJ
tall
uses
after
the
30
equation
weeks.
to
claim
Comment
that
on
a
his
plant
would
be
62.8 cm
claim.
Chapter
EXAM-STYLE
6
QUESTIONS
Personality
10
teenagers.
called
is
to
around.
bossy
average
They
The
of
also
repor ted
those
items
a
is
is
a
stealing
is
a
of
a
group
personality
measure
ask
the
of
behaviors
problems
behavior
of
about
person
how
how
is.
variable
nice
that
cheerful,
Each
of
person
stubbor n,
teenager’s
repor ted.
measure
various
swearing,
behavior
This
the
assessed
questions
cooperative
created
on
cheating,
The
and
studied
researchers
‘agreeableness’.
be
polite,
of
researchers
behavior
in
and
the
last
problems.
six
fighting.
repor ted.
The
months,
Each
youths
including
teenager’s
sum
Agreeableness
Behavior
factor
problems
Par ticipant
George
4.3
5
Bill
3.0
22
Ronald
3.4
10
Jimmy
3.3
12
Gerald
2.9
23
Laura
4.0
21
Hilar y
4.7
2
Nancy
2.4
35
Eleanor
2.9
12
Elizabeth
4.7
4
a
Constr uct
b
What
c
Find
d
Describe
e
Copy
the
the
and
Write
g
Michelle
behavior
produces
nine
a
show
agreeableness
the
factor
regression
line.
increases?
on
the
to
sentence
have
_________
equation
absent
her
for
‘Teenagers
of
the
the
behavior
regression
behavior
agreeableness.
who
more
problems’.
line.
problems
Estimate
were
her
questions
score
for
but
the
problems.
clothing
are
the
factor y
total
shown
recorded
production
in
the
the
cost y
following
number x
dollars.
of
The
coats
it
results
26
44
65
43
50
31
68
46
57
y
400
582
784
625
699
448
870
537
724
Write
Use
down
your
the
equation
regression
line
as
b
Inter pret
the
meaning
c
Estimate
the
cost
d
The
factor y
number
order
to
of
sells
coats
make
Bivariate analysis
a
of
the
of
a
the
of
i
boxes
the
profit.
for
to
the
producing
that
regression
model
answer
the
ii
of
y
on
x.
following:
the
y-intercept.
coats.
$19.99
factor y
line
gradient
70
for
table.
x
a
the
was
and
days
and
correlation.
tended
4.5
day
,
the
complete
down
scored
as
diagram
r-value.
f
Each
scatter
happens
agreeable
7
a
each.
should
Find
produce
the
in
smallest
one
day
in
CHAPTER
●
Bivariate
pairs
of
Scatter
analysis
The
●
To
draw
concer ned
(x, y)
in
(also
diagrams
between
relationship
graph.
is
a
with
data
the
relationships
between
set.
diagrams
relationship
●
SUMMARY
variables
Scatter
●
10
a
two
between
scatter
The
called
variables
two
diagram
patter n
scatter
that
variables
plot
formed
plots)
by
the
the
both
is
(x,
to
to
investigate
the
same
the
possible
‘event’.
their correlation
values
can
used
relate
called
y)
dots
are
give
from
us
the
some
data
table
as
indication
of
dots
on
a
the
correlation.
The
independent
variable
on
the
variable
ver tical
should
be
on
the
horizontal
axis
with
the dependent
axis.
y
elbairav
tnednepeD
0
independent
x
variable
●
A
general
upward
●
A
general
downward
●
Scattered
●
A
correlation
the
The
●
A
points
line
the
trend
with
no
between
of
of
best
best
association
then
be
used
fit
or
in
patter n
the
trend
two
dots
patter n
may
data
of
of
indicate
sets
does
shows posi tive
dots
shows negative
correlation
not
correlation.
close
necessarily
correlation.
to zero
mean
that
one
causes
to
the
line
rises
■
If
the
line
falls
■
Strong
the
Weak
of
positive
clustered
near
two
from
and
best
drawn
variables
on
and
a
to
scatter
show
diagram
the
to
trend.
find
This
the
line
direction
of
best
fit
of
can
left
left
to
to
right
right
negative
then
then
there
there
correlations
is
is
a posi tive
a negative
have
data
correlation.
correlation.
points
ver y
close
fit.
and
or
line is
predictions.
from
positive
line
trend
make
If
to
fit
between
■
■
in
other.
line
an
trend
negative
on
the
correlations
line
of
best
have
data
points
that
are
not
fit.
Continued
on
next
Chapter
page
●
To
draw
the
a
line
number
below
the
one
point
and
is
mean
of
of
line.
for
best
points
An
the
the
by
eye
the
draw
line
improvement
line
calculated
of
fit
above
by
to
pass
finding
line
that
the
have
a
through.
This
is
mean
of
will
balance
number
to
the
is
a
with
of
reference
points
point:
the mean point
the x-values
and
the
y-values.
y
(x, y)
0
●
x
The
can
equation
be
Least
●
A
used
of
for
the
is
of
prediction
squares
residual
line
the
best
fit,
also
called
the regression line ,
pur poses.
regression
ver tical
distance
between
a
data
point
and
the
graph
of
a
regression
equation.
●
The
the
●
least
sum
The
squares
of
the
formula
regression
squares
for
of
finding
line
the
the
is
the
one
that
has
the
smallest
possible
value
for
residuals.
gradient,
or
slope
(m)
of
a
regression
line
is
S
xy
m
=
,
where
2
(S
)
x
2
(∑
S
=
xy
∑
x
)(∑
y
)
2
and
xy −
(S
)
(∑
2
=
x
∑
x
Y
ou
should
use
your
)
−
n
n
●
x
GDC
to
find
the
equation
of
the
regression
line
in
examinations.
Continued
Bivariate analysis
on
next
page
Measuring
●
The
of
Pearson
the
correlation
product
correlation
inclusive.
It
is
dependence
moment
between
widely
between
used
two
two
in
correlation
variables X
the
sciences
coecient
and
as
a
Y,
(denoted
giving
measure
a
of
value
the
byr)
is
a
between
strength
of
measure
+1
and
−1
linear
variables.
S
xy
●
The
formula
for
finding
Pearson’s
correlation
coefficient
is
r
=
S
S
x
y
where
2
2
(∑
S
=
∑
xy
x
)(∑
y
xy −
)
(∑
2
,
S
=
x
∑
x
−
A
quick
way
to
r-value
0
<
0.25
0.5
|r|
≤
<
|r|
<
0.75
≤
≤
<
|r|
the
r-value
S
=
∑
y
y
)
−
n
is:
Correlation
0.25
|r|
inter pret
2
and
n
(∑
)
y
n
●
x
0.5
0.75
≤
1
Ver y
weak
Weak
Moderate
Strong
Chapter
Theory
of
knowledge
Correlation
shows
Correlation
For
example,
is
Causation
For
■
If
we
nd
a
and
high
that
strong
pregnant
weight,
how
the
when
example,
bir th
as
the
because
two
time
correlation
of
causation?
two
one
variables
you
go
to
between
at
should
heavier
closely
value
achievement
women
or
age
tr y
babies
to
directly
bed
a
24,
variables
increases,
baby’
s
boost
achieve
their
more
the
weight
we
with
does
affect
affects
should
var y
so
the
each
each
other.
other.
other.
number
of
hours
sleep.
at
suggest
baby’
s
you
EFFECT
bir th
highly?
EFFECT
Sometimes
always.
It
correlated
between
a
is
cause
easy
are
two
and
to
also
effect
assume
that
connected
events
does
are
not
closely
events
causally
.
mean
related,
that
But
that
are
but
not
closely
correlation
one
has
caused
the
CAUSE
other.
For
example,
if
your
cat
stays
out
all
night
and
then
gets
A
gets
sick
night
be
a
because
may
vir us
not
or
it
stays
cause
the
out
all
night.
sickness.
The
But
being
cause
is
outside
more
necess
What
h
■
knowledge:
Correlati
or
ation?
asks
questions:
relationship
p
exists
of
to
of
■
w
What
etween
v
ri
l
connects
or
separates
them
from
other?
each
correlation
variab
bacteria.
these
Theory
all
likely
Correlation
EFFECT
cau
Which
is
●
Bullying
●
Stress
causal
harms
and
mental
which
is
correlated?
●
health.
Watching
too
television
of
watching
major
spor ting
be
hazardous
The
temperature
to
the
and
real
the
number
of
vendors
out
on
that
TV
raises
blood
being
violent
Surgeons
pressure
with
video
game
skills
perform
in
simulated
surger y.
day.
●
●
people
life.
ice
better
cream
to
on
hear t.
●
●
leads
violence
events
in
can
much
in
Swedish
speakers
are
obese
healthier
than
Dutch
adults.
speakers.
●
Deep-voiced
men
have
Anscombe's
Anscombe’s
caution
without
Quartet
against
first
is
children.
Quartet
a
applying
generating
more
group
of
four
individual
more
data
sets
statistical
that
provide
methods
to
a
useful
ncis
Fra
data
be
com
Ans
evidence.
(191
F ind
the
mean
variance
Set
y
and
x,
the
the
mean
r-value
1
of
for
Set
y,
the
each
variance
data
of
x
and
01)
tish
Bri
the
an.
istici
stat
set.
2
Set
3
Set
4
x
y
x
y
x
y
x
y
4
4.26
4
3.1
4
5.39
8
6.58
5
5.68
5
4.74
5
5.73
8
5.76
6
7.24
6
6.13
6
6.08
8
7.71
7
4.82
7
7.26
7
6.42
8
8.84
8
6.95
8
8.14
8
6.77
8
8.47
9
8.81
9
8.77
9
7.11
8
7.04
10
8.04
10
9.14
10
7.46
8
5.25
11
8.33
11
9.26
11
7.81
8
5.56
12
10.84
12
9.13
12
8.15
8
7.91
13
7.58
13
8.74
13
12.74
8
6.89
14
9.96
14
8.1
14
8.84
19
12.5
Write
and
of
of
8-20
down
their
Using
each
you
regression
your
set
what
GDC,
of
think
lines
sketch
points
on
a
the
will
the
graphs
look
graph
different
Draw
Explain
the
regression
line
on
each
graph.
like.
what
you
notice.
of
graph.
Chapter
Trigonometry
CHAPTER
OBJECTIVES:
3.1
The
circle:
3.2
Denition
radian
measures
of
angles;
length
of
an
arc;
area
of
a
sector
sin
cos θ
of
and
sin θ
in
terms
of
the
unit
circle;
denition
of
tan θ
as
;
cos
exact
values
of
trigonometric
ratios
of
0,
,
6
2
3.3
The
Pythagorean
3.6
Solution
of
identity
triangles;
the
,
and
,
4
3
their
multiples.
2
2
θ
cos
+
cosine
sin
r ule;
θ
=
the
1
sine
rule,
including
the
ambiguous
1
case;
area
of
a
triangle
ab sin C;
applications
2
Before
Y
ou
1
you
should
Use
start
know
proper ties
Pythagoras’
of
how
to:
triangles,
Skills
including
1
check
Find
the
value
the
x
in
each
diagram.
theorem.
a
e.g. Find
of
value
of
b
x
x°
(2x)°
41°
96°
in
each
diagram.
x°
(x
x°
–
49°
a
x°
+
96°
+
x°
=
180°
38°
=
180°
38°
x
=
–
96°
–
38°
46°
c
ABC
b
is
isosceles,
so
∠A
=
d
x°
∠C
(4x)°
B
∠A
+
∠B
+
∠C
=
180°
=
180°
(x
x°
+
53°
+
x°
+
20)°
53°
56°
2x°
=
180°
–
53°
=
127°
x
=
63.5°
x°
A
C
e
c
Using
f
Pythagoras,
x
2
x
2
=
6
2.4
2
+
9
x
6
24
5.6
19
9
x
Trigonometry
20)°
Sometimes
tree
or
a
directly
.
and
For
the
a
method
a
tree
or
also
rock
find
the
of
these
distance
reference
on
formation.
angle
dimensions
width
a
(such
canyon)
dimensions
as
the
that
we
using
height
can’t
of
a
measure
trigonometr y
triangulation.
accurately
the
know
the
can
find
of
to
or
of
to
point
measures
and
need
Sur veyors
example,
needs
as
we
mountain
the
the
far
Then,
distance
formed
across
by
a
side
canyon,
of
standing
between
these
points
the
on
a
sur veyor
canyon,
the
near
two
known
and
the
such
side,
he
points,
point
on
the
Some
far
side.
Using
trigonometr y
,
this
is
enough
information
to
use
the
distance
the
far
to
the
other
side,
without
ever
having
to
cross
mathematicians
find
the
phrase
to
‘measure
side
of
the
canyon.
instead
of
of
angle’
‘size
of
angle’.
.
This
sizes
then
Right-angled
chapter
of
star ts
angles
extends
and
to
trigonometr y
.
by
looking
side
areas
triangle
of
at
lengths
the
of
triangles
trigonometry
relationships
right-angled
and
real-life
between
triangles,
the
and
applications
of
Some
‘right
of
people
say
triangle’
instead
right-angled
triangle.
Chapter
Star t
by
looking
at
the
right-angled
triangle
with
ver tices
at
the
Angles
points
A,
B
and
C.
The
angles
at
these
ver tices
are
can
described
Â,
ˆ
B
ˆ
C,
and
be
called
in
various
respectively
.
A
The
side
AB,
the
side
opposite
the
right-
ways.
This
could
be
ABC
angle,
is
called
the
of
hypotenuse
triangle
called
the
angle
at
the
ˆ
A
c
right-angled
could
ˆ
C A B;
∠CAB.
also
can
labeled
with
B
a
this
A;
∠BAC;
Angles
be
Greek
In
called
ˆ
B A C;
C
be
triangle.
b
triangle,
notice
that
the
side
labeled a
(side
BC )
is
opposite
letters
like
θ
Â,
(theta).
the
side
AB )
is
their
labeled
opposite
opposite
b
(side
ˆ
C.
It
at
these
is
is
opposite
convenient
to
ˆ
B,
and
identify
the
side
sides
in
labeled
relation
c
(side
to
angles.
Trigonometric
Look
AC )
two
ratios
right-angled
triangles.
D
A
59°
59°
31°
31°
B
C
ABC
and
DEF
is
DEF
E
each
have
F
angles
measuring
59°,
31°
and
90°.
Some
larger
than
ABC.
Triangles
with
the
same
three
the
are
called
triangles ,
simi lar
and
their
corresponding
sides
are
in
two
of
a
right-angled
propor tions.
triangle
of
For
ABC
and
the
the
legs
triangle.
AC
BC
DF
,
=
AB
any
their
AB
similar
sides
will
corresponding
The
fact
define
These
that
the
ratios
●
any
the
be
in
the
the
sides
of
var y
DF
regardless
same
be
right-angled
how
ratios.
in
In
large
other
propor tion
similar
according
longest
side
the
triangles
ratios
to
to
the
of
small
words,
each
form
– sine,
sizes
or
are,
their
other.
equal
cosine
the
they
ratios
and
angles
helps
us
tangent.
in
the
triangles.
opposite
side
side
●
the
triangle
(often
the
opposite
(sometimes
side
next
(sometimes
a
triangle.
abbreviated
right
to h
or
H)
is
the
longest
side
angle
(hypotenuse)
h
o
●
of
=
trigonometric
hypotenuse
is
the
will
is
EF
and
AC
triangles,
right-angled
and
BC
,
DE
sides
three
right-angled
In
EF
=
and
DE
The
DEF:
hypotenuse
For
call
shor test
the
sides
same
textbooks
angles
Trigonometry
to
the
angle
marked θ
abbreviated
the
angle θ
abbreviated
to o
is
to a
or
called
or
A)
is
called
the
opposite
O)
the
i
adjacent
side
a
(adjacent)
(opposite)
➔
For
any
right-angled
triangle
opposite
θ
sine
with
A
angle θ :
an
mnemonic
made-up
O
is
word
a
or
=
hypotenuse
phrase
H
that
remember
adjacent
helps
a
list
you
or
a
A
H
cosine
θ
=
O
formula.
hypotenuse
H
Remember
opposite
with
tangent
θ
these
i
O
the
mnemonic
=
A
adjacent
Look
at
this
right-angled
A
SOH-CAH-TOA
triangle,
with
Â
highlighted.
We
BC
A
sin
A
AB
AC
c
cos
A
A
tan
b
AB
c
BC
a
=
AC
astronomer
Ar yabhata,
about
use
trigonometric
ratios
to
find
unknown
side
lengths
in
right-angled
circled
between
sine,
cosine
and
in
ABC :
invent
a
sin
the
and
the
different
tangent
orbits.
triangle
India
triangles.
Ear th
In
that
planets
stars
Relation
in
CE,
and
Sun,
angles
born
476
believed
can
these
ratios.
b
in
Y
ou
for
=
B
tan
and
trigonometric
The
a
C
sin,
c
cos
b
the
abbreviations
=
=
use
a
in
He
began
to
trigonometr y
order
to
calculate
=
c
the
distances
from
b
planets
cos
to
the
Ear th.
=
A
c
a
sin
i
a
c
c
so
=
=
b
cos
b
b
c
a
but
tan
C
=
a
B
b
sin
= tan
so
cos
sin
➔
tan
cos
Although
Example
mathematicians
studied
For
this
triangle,
find
the
length
of
side
have
triangles
a.
for
thousands
years,
34°
the
of
term
‘trigonometr y’
was
6
rst
by
used
Pitiscus
a
{
Continued
in
1595
Bar tholomaeus
(German,
1561–1613).
on
next
page
Chapter
Be
sure
you
are
in
Answer
opposite
a
degree
tan
34°
=
=
Use
adjacent
the
=
6
tangent
ratio.
6
The
a
mode
side
opposite
the
angle
of
34°
tan 34°
T
o
is
the
opposite
side,
and
the
change
to
degree
side
and
a
=
6
tan 34°
≈
4.05
adjacent
to
34°
has
length
6.
choose
You
can
find
the
value
of
tan
&
using
your
5:
Settings
34°
Status
2:Settings
GDC.
1:General
Use
the
move
To
enter
tan
μ
press
and
then
select
to
select
tan.
key
Angle
and
Degree.
and
then
to
Press
select
4:Current
If
you
know
the
and
you
want
you
will
need
lengths
to
find
of
the
the
sizes
sides
of
of
the
a
right-angled
angles
of
the
triangle
triangle,
–1
–1
use
the
inverse
this
triangle.
trigonometric
functions
sin
,
–1
and
cos
to
tan
Example
GDC
Find
the
size
ˆ
B
of
in
help
on
CD:
demonstrations
Plus
and
GDCs
Casio
are
on
Alternative
for
the
the
CD.
9 cm
5 cm
−1
sin
is
'arc
called
sine,'
−1
B
cos
is
'arc
cos'
is
'arc
tan'
−1
tan
Answer
opposite
sin B
5
=
=
hypotenuse
The
side
opposite
ˆ
B
has
length
of
9
5 cm
of
and
9 cm.
the
Use
hypotenuse
the
sine
has
a
length
ratio.
⎛ 5 ⎞
ˆ
B
–1
–1
=
sin
⎜
⎝
⎟
9
≈
To
33.7°
enter
sin
press
μ
and
then
select
⎠
–1
sin
In
this
exercise,
missing
angles
calculator
is
and
set
Trigonometry
you
in
will
side
be
solving
lengths).
DEGREE
right-angled
Always
mode.
make
triangles
sure
that
(finding
your
TI-84
FX-9860GII
Exercise
For
each
A
question,
use
the
diagram
and
the
information
given
to
b
A
find
all
your
the
unknown
answers
correct
angles
to
3
and
sides.
significant
All
lengths
figures
are
where
in
cm.
C
Give
necessar y
.
a
c
1
a
=
3
c
5
a
7
If
12,
=
c
20
ˆ
B =
4.5,
=
=
11,
Â,
55°
=
35°
2
b
=
37,
Â
4
b
=
48,
c
6
a
=
8.5,
=
=
b
40°
60
=
B
9.7
x
2
a
of
=
x,
2x,
and
Special
Look
b
at
=
5x
the
–
1
angles
and
Â
c
isosceles
1
+
1
(x
∈
)
find
the
∈
means
that
x
is
value
an
integer
.
triangles
right-angled
To
C
x
ˆ
B
and
right-angled
this
=
solve
triangle.
the
triangle,
you
need
to
find
the
A
length
AB,
Using
2
angles
Pythagoras’
2
+
1
and
1
2
=
c
Â
ˆ
B
and
theorem
2
,
so
c
=
2,
and
c
=
AB
=
1
Using
c
the
tangent
BC
tan
A
ratio
=
=
AC
=
1
–1
Â
B
This
and
Now
look
at
the
values
of
=
is
ˆ
B
the
tan
an
=
(1)
=
45°
isosceles
triangle,
so
Â
=
ˆ
B,
45°.
trigonometric
ratios
of
this
triangle.
1
2
45°
➔
sin
45°
=
=
2
2
1
cos
45°
=
=
2
45°
1
tan
45°
=
=
1
1
Now
half
To
look
of
2
1
an
solve
Using
a
this
right-angled
equilateral
the
triangle,
Pythagoras’
2
+
at
2
=
2
triangle,
which
so
a
you
need
theorem
=
1
A
triangle.
to
find BC,
3,
and
a
Â
and
ˆ
B.
gives
2
,
C
is
=
BC
=
3
a
2
B
Chapter
Using
the
cosine
AC
cos
A
ratio,
=
=
AB
⎛
⎞
–1
Â
=
cos
⎜
⎝
ˆ
B
=
180°
Here
are
=
⎟
–
90°
the
60°
⎠
–
60°
values
=
of
30°
all
the
trigonometric
ratios
for
this
30°–60°–90°
triangle.
1
➔
sin 30°
=
sin
60°
=
cos 30°
60°
=
cos
60°
=
√3
2
3
3
tan 30°
=
=
tan
Example
Find
the
60°
=
30°
=
3
3
1
value
exact
of
x
in
this
triangle.
When
an
answer
you
60°
exact
is
asked
should
for
leave
the
5 cm
square
in
your
not
root
or
answer
change
it
radical
and
to
a
x
rounded
Square
Answer
x
tan 60°
=
5
Exercise
1
Use
the
perfect
3 cm
called
diagram
to
a
a
=
12,
b
b
=
9,
c
c
d
b
=
6,
c
e
a
=
,
solve
are
in
each
right-angled
triangle.
Give
are
exact
b
c
Â
4.5,
=
=
ˆ
B
=
C
24
the
c
all
means
unknown
and
angles.
The
diagram
sides
a
=
=
exact
c
60°
will
not
B
10
values
of
x,
y
and
z
P
z
x
30°
Q
Trigonometry
here
nd
45°
8
squares
cm.
always
Find
not
surds.
'Solve'
Lengths
A
2
are
B
answers.
=
of
that
3
5
x
roots
numbers
=
=
decimal.
8
R
y
S
be
to
scale
ABC
3
has
Â
=
ˆ
C
60°,
=
90°,
BC
=
x
+
2,
and
Sketch
the
triangle
2
AB
=
–
x
4.
rst.
a
Find
the
exact
value
b
Find
the
exact
length
Triangle
4
ABC
has
ˆ
B
=
of
x.
of
side
ˆ
C
45°,
=
AC
90°,
AC
=
4x
–
1
and
65°
y
2
BC
=
+
x
2.
z
a
Find
the
value
of
x.
b
Find
the
exact
length
of
side
AB
x
In
5
to
the
diagram
1dp.
find
Lengths
the
are
in
values
of
w,
x,
y
and
45°
z,
cm.
w
4
9
.
Applications
triangle
In
the
last
triangles
how
to
section,
using
apply
right-angled
trigonometry
you
sine,
these
of
found
cosine
lengths
and
and
tangent.
trigonometric
ratios
angles
In
to
this
in
right-angled
section,
solve
you
problems
will
in
see
real-life
situations.
Let’s
➔
begin
The
with
angle
some
of
terminology
.
elevation
is
The angle of depression
the
is
angle
the
‘up’
angle
from
‘down’
horizontal.
from
horizontal.
C
Angle
of
elevation
Angle
of
depression
A
B
Horizontal
D
Example
An
obser ver
elevation
the
of
nearest
stands
the
top
100
of
m
the
from
the
building
base
is
of
65°.
a
building.
How
tall
is
The
the
angle
of
building,
to
metre?
{
Continued
on
next
page
Chapter
Answer
T
Star t
Let
by
O
sketching
represent
obser ver
on
the
of
base
represent
Mark
the
the
the
the
diagram.
position
ground
B
building
top
65°
a
the
of
the
angle
of
of
the
represent
and T
building.
elevation.
65°
B
100
O
BT
tan
65°
=
,
so
You
are
finding
the
height
of
the
100
building,
BT
=
The
the
Y
ou
➔
100 tan
building
nearest
also
65°
is
≈
214 metres
need
four
east
(E)
to
solve
and
clockwise
using
40°
40°
E,
west
of
to
compass
from
using
compass
points
are
points
nor th
(N),
and
bearings.
south(S),
(W).
bearings
give
directions
as
angles
measured
nor th.
compass
which
east
tall,
problems
cardinal
Three-figure
N
BT.
metre.
The
When
length
214.45...
points
means
nor th.
for
W
direction,
20°S,
south
of
which
you
will
means
see
20°
west.
notation
NW,
such
which
nor th
and
as:
means
45°
between
west.
N
N
N
N 40°E
NW
40°
45°
45°
20°
W 20°S
S
S
S
When
035°
using
which
clockwise
bearings
means
from
for
35°
nor th.
direction,
110°,
110°
you
which
will
see
notation
means
clockwise
from
such
as:
270°,
nor th.
from
270°
N
which
nor th.
is
the
means
Notice
same
as
270°
that
‘due
clockwise
a
N
N
035°
110°
270°
270°
110°
S
Trigonometry
S
S
bearing
west’.
of
Example
Two
ships
Ship
A
Ship
B
Find
leave
sails
sails
the
nearest
dock
due
on
at
nor th
a
the
for
bearing
distance
same
30 km
of
between
050°
the
time.
before
for
ships
dropping
65 km
when
anchor.
before
they
dropping
are
anchor.
stationar y
,
to
the
kilometre.
Answer
Sketch
B
the
dock
Ship
A
at
30
a
A
diagram.
from
stops
Point
which
at
A
the
and
D
represents
ships
ship
B
set
sail.
stops
B.
65
You
need
to
find
the
length
AB,
the
50°
distance
D
are
C
B
between
the
ships
when
they
stationar y.
There
are
no
right-angled
in
diagram,
triangles
The
the
so
draw
them
angle
DBE
is
in.
found
50°
using
the
A
The
hypotenuse
of
each
right-angled
alternate
triangle
30
is
the
path
of
one
of
angle
the
65
proper ty.
ships.
50°
40°
angle
D
Add
any
angles
you
know
from
proper ties.
E
BE
sin
40°
=
Find
BE.
Find
DE.
65
so BE
=
65 sin 40°
≈
41.781...
DE
cos 40°
=
65
so
DE
=
65 cos 40°
BC
=
DE
AC
=
BE
C
=
–
=
49.7928
Store
these
values
in
your
GDC.
49.7928...
30
=
11.7812...
49.7929
Add
B
the
new
infor mation
to
the
diagram.
11.7812
50°
A
30
65
50°
40°
D
E
2
AB
2
=
(49.7929...)
2
+
(11.7812...)
Use
Pythagoras’
Use
the
theorem
in
ABC.
Use
so
AB
=
.1677...
values
you
in
The
ships
are
exact
the
intermediate
approximately
steps,
52 km
apar t,
values
stored.
to
the
nearest
km.
for
the
and
round
nal
only
answer
.
Chapter
Exercise
1
B
C
Isosceles
triangle
AB
=
=
CB
ABC
15 cm,
as
a
Find
the
height
b
Find
the
sizes
has
side
AC
=
10 cm
and
shown.
of
the
triangle.
15
of
BÂC
15
ˆ
AB C
and
A
2
ABE
as
a
b
fits
shown.
Find
BC
the
Find
Give
exactly
28 cm
lengths
the
your
=
inside
sizes
and
of
square
DE
=
segments
ˆ
AE D,
of
answers
the
ˆ
EB A
correct
to
D
ABCD,
C
10
8
E
C
8 cm.
AE
and
BE
ˆ
AE B
and
28
3 sf.
A
3
An
obser ver
sea
level
standing
on
the
top
of
a
ver tical
cliff
120
B
m
above
If
sees
a
ship
in
the
water
at
an
angle
of
depression
of
a
not
How
far
is
the
ship
from
the
base
of
the
diagram
cliff ?
given
4
A
5
the
Anya
in
From
of
angles
a
A
then
Buildings
X
From
elevation
12 m
the
the
and
width
diagonals
nor th,
then
Find
above
top
70 m
por t
and
depression
her
18 mm.
of
tur ns
the
and
distance
rectangle.
walks
and
another
bearing
3
km
from
her
of
the
ground
Building
apar t,
what
B
is
in
Building
across
the
the
A,
street
height
of
the
is
angle
40°.
Building
If
B?
to
to
sails
must
and
a
and
Y
the
15 km
the
are
point
the
sails
top
of
on
ship
the
of
a
on
a
bearing
sail
across
on
base
35 km
to
the
roof
Building
of
Y
of
retur n
street
Building
bearing
105°.
from
is
is
55°
35°.
to
each
X,
and
How
047°.
How
directly
Building
Y
of
far,
the
tall
ship
and
on
por t?
other,
the
The
95 m
angle
of
angle
are
It
of
the
is
to
two
a
sure
side
Jacob
is
walking
north
along
a
straight
road
when
he
spots
a
good
check
answers
buildings?
9
is
a
field
to
his
right
on
a
bearing
of
018°.
After
walking
he
notices
the
tower
is
now
on
a
bearing
of
walking
Trigonometry
north,
how
close
will
he
pass
to
the
make
the
shor test
opposite
angle
the
and
the
another
066°.
If
tower?
side
is
opposite
he
the
continues
to
nal
tower
longest
240 metres
idea
your
that
smallest
in
by
own.
QUESTION
bearing,
apar t.
due
N35°W
.
are
leaves
tur ns
what
8
of
buildings
ship
25 mm
between
window
EXAM-STYLE
7
star t
your
point.
elevation
the
length
2 km
direction
star ting
6
has
walks
the
the
QUESTION
rectangle
Find
with
question,
drawing
EXAM-STYLE
is
9°.
largest
angle.
10
From
her
position
at
ground
level,
Hayley
notices
that
the
angle
Unless
of
elevation
of
the
top
of
a
building
is
40°.
When
she
tells
20 metres
closer
to
the
building,
the
new
angle
of
the
elevation
is
you
11
the
is
A
car
A
passenger
ahead
the
the
of
traveling
at
angle
more
12
height
an
of
time
in
the
at
the
angle
a
car
of
sees
a
elapse
of
the
before
the
ground
is
level.
speed
bridge
elevation
elevation
will
building.
constant
otherwise,
55°.
assume
Find
question
moves
of
the
a
straight
spanning
5°.
bridge
on
Ten
is
car
the
seconds
17°.
How
passes
highway
.
highway
later,
much
directly
under
bridge?
The
diagram
ABCDEFGH.
Find
these
shows
AD
=
a
right
24 cm,
rectangular
DH
=
F
prism,
9 cm,
and
HG
=
G
18 cm.
angles.
18
a
HÂD
b
AB
E
c
HÂG
ˆ
E
H
9
AGD
d
A
.
Using
in
The
angle
the
coordinate
axes
trigonometry
θ
in
a
Car tesian
coordinate
system
has
its
ver tex
at
the
In
origin,
as
shown
in
the
diagram.
A
positive
some
measured
anticlockwise
from
people
use
other
anticlockwise
x
O
the
An
are
positive
three
angles
α,
β
and
the
the
that
positive
side.
side
is
its
the
The
called
terminal
like
side.
this,
ver tex
at
the
δ
origin
y
angle
called
angle
with
Here
is
ini tial
i
of
the
along
x-axis
‘counterclockwise’
instead
of
the x-axis.
lies
Some
textbooks
angle
side
is
D
24
y
and
its
initial
y
side
along
x-axis
is
standard
a
the
said
positive
to
be
in
posi tion
b
d
O
x
O
x
O
x
The
of
rst
the
are
four
Greek
alpha
gamma
γ
α,
letters
alphabet
beta
and
β,
delta
Chapter
δ.
2
This
diagram
shows
a
circle
with
equation x
2
+
y
=
1
y
B
The
one
center
unit.
of
the
This
is
circle
called
is
a
at
the
uni t
origin
and
its
radius
is
y
circle.
i
A
In
the
Now
diagram,
take
a
the
closer
angle
look
θ
at
is
positive.
acute
x
0
angles
in
B
the
first
quadrant
OA
and
OB
OA
=
of
the
unit
circle.
1
are
radii
of
the
unit
circle
so
A
i
OB
Next,
=
use
1
0
the
right-angled
acute
angle
triangle
θ
to
form
1
x
a
BOC.
y
Using
the
trigonometric
ratios
in
BOC,
,
so
x
=
cos θ,
B(cos i, sin i)
y
and
sin
,
so
y
=
sin θ
1
1
So
point
B
has
y
coordinates
(cos θ,
sin θ).
i
A
0
Example
Find
then
the
exact
give
x
C
x
these
coordinates
values
to
of
point
D,
y
three
D
significant
figures.
1
59°
A
0
x
1
Answer
The
exact
(cos 59°,
To
3
sf
the
Example
In
the
coordinates
of
point
D
are
AÔD
is
a
positive
angle.
sin 59°)
coordinates
of
D
are
(0.515,
0.857)
Use your GDC to find the values of
cos 59° and sin 59°.
diagram,
find
the
exact
coordinates
of
point
P
y
P
1
30°
0
A
x
1
Answer
⎛
The
exact
coordinates
of
P
are
3
1
,
⎜
AÔP
is
in
the
page
368
for
exact
2
values
of
sine
⎠
quadrant. Therefore,
30°
the
coordinates
point
Trigonometry
P
are
the
first
⎟
2
⎝
See
⎞
⎟
⎜
of
(cos 30°,
sin 30°).
and
cos
30°.
Exercise
1
Use
the
y
D
diagram
to
find
the
coordinates
of
point P
for
each
P
given
value
a
θ
=
20°
b
θ
=
17°
of
θ.
Give
your
answers
to
3
significant
figures.
1
(1, 0)
i
A
2
c
θ
=
60°
d
θ
=
74°
e
θ
=
90°
Use
the
0
diagram
from
question
1
to
find
the
value
of
x
θ
The
for
the
given
coordinates
of
point
P
.
Give
your
answers
diagram
always
the
nearest
These
0.913)
have
b
P (0.155,
0.988)
c
P (0.707,
0.707)
d
P (0.970,
0.242)
3
3
Use
the
value
diagram
of
θ.
a
θ
=
70°
b
θ
=
38°
not
be
to
scale.
degree.
P (0.408,
a
will
to
Give
to
find
your
the
area
answers
of
to
3
coordinates
been
rounded
signicant
AOP
for
significant
to
gures.
the
y
given
figures.
P
The
dashed
line
is
the
1
θ
c
=
height
24°
of
the
triangle.
(1, 0)
i
A
θ
d
=
30°
0
Now
look
at
are
obtuse
the
second
angles
angles
in
the
second
(between
quadrant
in
the
90°
quadrant.
and
unit
180°).
These
Here
is
x
y
angles
an
obtuse
angle
in
circle.
B
When
you
are
working
with
obtuse
they
relate
angles
it
is
sometimes
i
1
helpful
to
think
of
how
to
angles
in
the
first
A
0
quadrant
(acute
angles).
Investigation
This
diagram
angle
of
θ
shows
from
x
1
–
obtuse
point
B
at
a
angles
positive
angle
of
30°
from
OA,
and
point
C
at
a
positive
OA.
y
F ind
the
What
Use
C
value
are
the
the
of
θ
coordinates
symmetr y
of
the
of
point
unit
B?
circle
B
to
write
down
the
coordinates
of
i
point
D
30°
30°
0
C
A
x
{
Continued
on
next
Chapter
page
Now
look
at
the
triangles
formed
by
the
sides
OB
and
OC
and
the
x-axis.
y
(–x
y)
C
B
(x
y)
150°
D
F
0
EOC
is
triangles
of
A
30°
E
point
congruent
with
B
to
FOB.
hypotenuse
are
(x,
y),
coordinates
of
x
then
Both
length
the
are
1.
Y
ou
30°–
can
coordinates
of
60°–
also
see
point
C
The
B
are
(cos 30°,
sin 30°),
90°
that
are
3
1
coordinates
of
point
C
are
2
(cos 150°,
sin 150°),
which
are
the
same
as
the
coordinates
coordinates
2
the
the
y).
or
So
if
(–x,
(–cos 30°,
sin 30°),
or
3
1
2
2
Draw
diagrams
showing
40°
and
140°
25°
and
155°
68°
and
112°
Label
meet
From
the
the
the
coordinates
unit
circle.
each
of
the
What
investigation
of
these
points
do
you
pairs
where
of
the
angles
in
.
the
non-horizontal
unit
circle.
sides
notice?
you
should
supplementar y
angles.
now
understand
an
impor tant
Supplementar y
proper ty
of
add
➔
For
supplementar y
angles
α
and
β,
sin α
=
sin β,
to
will
see
illustrated
graphically
For
any
angle
θ,
sin θ
=
sin (180°–
θ),
and
cos θ
=
– cos (180°–
study
θ)
and
This
proper ty
Exercise
these
–cos β
=
proper ties
➔
angles
180°.
and
Y
ou
cos α
up
will
be
useful
later
on.
in
when
graphs
cosine
Chapter
of
you
sine
functions
13.
E
y
1
Use
C
to
the
for
3
diagram
the
given
significant
to
find
values
of
the
θ.
coordinates
Give
your
of
points B
and
answers
figures.
C
a
θ
=
30°
b
θ
=
57°
c
θ
=
45°
B
180°–
i
1
1
i
D
d
θ
=
13°
e
θ
=
85°
Trigonometry
0
A
x
2
Use
the
the
diagram
given
tenth
of
in
positions
a
question
of
point
1
C.
to
find
Give
the
your
value
of
answers
θ
for
to
each
the
of
nearest
degree.
a
C (–0.332,
0.943)
b
C (–0.955,
0.297)
These
have
c
C (–0.903,
0.429)
d
C (–0.769,
0.639)
3
3
Find
the
angle
4
that
a
15°
b
36°
c
81°
d
64°
Find
sine
one
of
has
the
acute
sin A
=
0.871
b
sin A
=
0.436
c
sin A
=
0.504
d
sin A
=
0.5
look
at
the
acute
same
and
a
Next,
each
line
(to
4
sf),
and
state
the
rounded
signicant
to
gures.
obtuse
sine.
one
with
angle
coordinates
been
obtuse
value
equation y
=
for Â
mx:
y
y
=
Any
mx
line
with
equation
y
=
mx
This
has
gradient
m,
and
passes
form
the
is
a
special
through
origin.
of
the
equation
y
=
ax
+
of
b
standard
a
or
line
y
=
mx
+
c
x
Now
point
look
B
in
at
what
the
first
happens
when
the
line
intersects
the
unit
circle
at
quadrant.
y
y
=
mx
B
x
Chapter
In
A
the
first
quadrant
right-angled
the
triangle
line
is
forms
formed
angle θ
an
with
with
the
segment OB
y
x-axis.
(par t
of
y
the
line
This
y
=
mx)
illustrates
as
its
some
=
mx
hypotenuse.
impor tant
proper ties
involving
the
right
B(cos i, sin i)
triangle
First,
and
using
the
line
y
2
theorem
1
gives
2
(cos θ)
+
mx
Pythagoras’
2
(sin θ)
=
=
1
2
.
The
usual
way
of
writing
(sin θ)
i
sin i
2
and
2
(cos θ)
is
2
θ
sin
and
θ,
cos
which
gives
x
cos i
2
2
θ + cos
sin
Suppose
you
θ = 1
want
to
find
the
gradient
of
the
line y
=
mx.
y
(x
This
line
passes
through
the
points
gradient
of
a
line
0),
and
B(sin θ,
(x
,
y
1
you
can
find
the
)
2
=
Here
, y
2
cos θ).
The
O(0,
gradient,
m,
using
the
)
1
coordinates
0
of
points
O
and
sin
m
cos
The
sin
0
=
x
B
=
tan θ
cos
0
gradient
of
a
line
rise
is
run
➔
These
three
proper ties
2
sin
1
are
tr ue
for
any
angle θ
2
cos
= 1
Proper ty
sin
tan
2
is
For
number
known
as
1
the
=
Pythagorean
cos
3
also
any
x-axis,
line
the
y
=
value
mx
of
which
m
(the
forms
an
gradient
angle
of
the
of
θ
line)
with
is
Identity.
the
tan θ
i
Example
Proper ty
Find
the
the
gradient
of
the
line
which
forms
a
positive
angle
of
130°
with
Answer
y
=
mx
Gradient
=
tan θ
130°
(1, 0)
0
The
gradient
tan 130°
≈
often
useful
calculations.
x-axis.
y
is
number
x
of
–1.19
Trigonometry
the
line
is
You
can
GDC.
find
this
value
using
your
in
2
Example
Find
in
the
the
gradient
of
the
line
shown
y
diagram.
y
=
mx
60°
0
x
Answer
y
Find
y
=
the
for med
mx
‘standard
by
equivalent
of
120°
this
to
position’
line. The
a
positive
angle
angle
obtuse
60°
is
angle
120°.
60°
This
0
line
standard
The
gradient
for ms
an
angle
of
120°
in
x
of
the
line
position.
is
3
sin120°
sin60°
2
=
cos 120°
=
cos 60°
1
2
3
=
Exercise
1
Find
the
answers
=
–1.73
F
gradient
to
three
of
the
line
significant
y
=
mx
in
each
diagram,
giving
your
figures.
a
b
y
y
117.5°
56.3°
x
0
x
y
y
=
=
mx
mx
c
d
y
y
135°
42.3°
0
x
y
=
mx
x
y
=
mx
Chapter
2
Find
the
point
P.
equation
Find
the
of
the
value
line
of
θ
passing
to
a
the
through
nearest
the
origin
and
degree.
b
y
y
P(0.471, 0.882)
P(0.674, 0.738)
i
i
x
x
y
y
=
=
mx
mx
c
d
y
y
=
mx
P(–0.336, 0.942)
i
i
x
0
x
0
y
=
mx
P
y
e
y
f
P(1.59, 3.76)
i
P(–0.8, 0.6)
x
0
i
Extension
material
Worksheet
11
on
CD:
x
y
=
mx
and
.
Y
ou
The
can
Look
at
use
sine
ABC.
The
to
Angle
sums
differences
rule
trigonometry
per pendicular
-
to
solve
alti tude
triangles
(height),
h,
that
of
are
the
not
right-angled.
triangle
is
AD,
A
BC
c
b
h
In
the
right-angled
triangle
ABD
sin B
=
B
This
In
gives
the
h
=
c sin B
right-angled
triangle
ACD
sin C
=
This
gives
Equate
the
c sin B
h
=
b sinC
values
=
of
b sin C.
Trigonometry
h
to
give
D
C
in B
Rearranging
this
equation
sin C
gives
=
b
The
ratios
side
are
Now
of
sine
draw
the
from
sin C
sin B
=
=
a
equal,
➔
each
altitude
A
ratios
are
of
c
angle
to
the
length
of
the
opposite
equal.
sin
the
the
The
as
B
to
side
again.
c
AC,
Y
ou
and
from
should
C
find
to
AB
that
and
these
find
ratios
b
before.
sine
rule
This
in
For
any
ABC,
where
a
is
the
length
of
the
side
opposite
the
is
the
length
of
the
side
ˆ
opposite B
,
and
c
is
the
length
of
is
given
Formula
booklet
Â,
that
b
fomula
you
use
in
the
the
examination.
side
opposite
sin
A
ˆ
C,
sin B
=
a
Y
ou
at
can
least
(the
one
length
of
the
a
r ule
and
side
sin
c
sine
angle
Example
Find
the
its
or
b
to
solve
opposite
the
size
c
=
or
b
use
a
sin C
=
of
A
=
sin B
triangles
side,
an
sin C
if
and
you
one
know
other
the
size
of
measurement
angle).
missing
angles
and
sides
in
this
triangle,
giving
your
answers
Be
to
3
sure
you
are
in
sf.
degree
mode
A
T
o
9.4 cm
98°
change
mode
to
degree
press
and
c
choose
Status
B
12 cm
C
|1:
5:
Settings
|2:
Settings
&
General
Use
the
key
to
Answer
move
Using
the
sin 98°
sine
sin
r ule
B
ˆ
You
and
need
the
to
find
length
the
angles
B
ˆ
and
C,
to
select
Angle
Degree.
and
c.
and
then
Press
select
=
12
4:
9.4
9.4 sin
sin
so
ˆ
B
Current
98
12
ˆ
B
50.9 (3 sf )
–
Â
ˆ
ˆ
C
=
180
–
B,
so
ˆ
C
=
31.1305533...
The
is
sum
of
the
angles
in
any
triangle
180°.
ˆ
C
=
31.1°
sin 98°
(3
sf)
sin 31.13055...
Use
=
12
12 sin 31.13055 …
c
the
sine
rule
once
more
to
find
c.
c
Don't
round
the
intermediate
=
GDC
sin 98°
steps,
just
the
nal
help
on
CD:
demonstrations
ˆ
for
c
=
6.26 cm
(3
B ,
Alternative
values
for
the
TI-84
ˆ
C
and
c
Plus
and
Casio
FX-9860GII
sf)
GDCs
are
on
the
CD.
Chapter
In
Example
10,
the
triangle
with
A
all
measures
Always
sure
check
the
labeled
your
shor test
would
final
side
is
look
answers
opposite
to
like
this:
be
the
9.4
98°
6.26
smallest
angle
opposite
50.9°
the
largest
longest
side
is
angle.
31.1°
B
12
Example
Find
the
and
the
triangle,
C
missing
angles
rounding
your
and
sides
answers
in
to
2
A
this
40.5 cm
decimal
places.
39°
C
c
77°
a
B
Answer
You
Â
=
sin
180°
–
77°
77°
sin
–
39°
=
,
a
so
a
=
angle
Â,
and
the
lengths
a
and
c
the
sine
rule
to
find
a
and
c.
sin 77°
a
37.36 cm
sin 77°
find
=
Use
40.5
to
40.5 sin 64°
64 °
=
need
64°
(2
dp)
sin 39 °
=
40.5
c
Check:
40.5 sin 39°
c
Shor test
side
(26.16)
is
opposite
the
smallest
=
sin 77°
so
c
=
26.16 cm
Example
A
of
ship
is
032°.
How
far
(2
dp)
angle
(39°).
angle
(77°).
Longest
side
(40.5)
is
opposite
the
largest
sailing
Later,
did
due
the
the
nor th.
captain
ship
travel
The
captain
obser ves
between
that
sees
the
these
a
lighthouse
bearing
two
of
10 km
the
away
lighthouse
on
is
a
bearing
132°.
obser vations?
Answer
N
Draw
A
is
a
the
diagram
position
lighthouse,
and
to
model
from
B
is
the
which
his
situation.
the
second
captain
position.
first
L
is
spots
the
the
position
132°
B
of
the
You
lighthouse.
have
point
A
to
to
find
point
d,
the
distance
the
ship
travels
from
B.
L
d
10
32°
A
Angle
ABL
=
180°
–
132°
=
48°
{
Trigonometry
Continued
on
next
page
ˆ
L=
180
–
Â
sin 100°
ˆ
B
–
=
100°
Ptolemy
sin 48°
Use
=
d
10
d
the
sine
rule
to
find
sin
CE),
in
work
100°
his
sine
48°
from
=
Almagest,
values
0°
to
for
wrote
angles
90°.
He
13.251....
also
The
ship
travels
Give
approximately
your
answer
to
a
included
13.25 km
between
points
A
and
degree
B
of
Solve
similar
to
the
accurac y.
sine
Exercise
a
sensible
theorem
1
90–168
13-volume
=
sin
d
(c
d.
10
rule.
G
each
significant
triangle
ABC.
Give
your
answers
correct
to
3
'Solve'
a
means
nd
triangle
figures.
A
unknown
all
sides
and
angles.
c
b
B
a
C
a
b
=
24 cm,
c
a
=
4.5 cm,
e
c
=
5.8 cm,
EXAM-STYLE
2
An
Julia
then
her
second
Adam
of
a
the
a
walks
from
4
Use
sees
Â
=
=
the
tree
in
2 km
the
is
27°,
83°
Â
ˆ
B =
a
has
r ule
field
due
=
55°
b
ˆ
c = 2.5 cm, Â = 40°, C = 72°
d
b
=
60,
ˆ
B =
15°,
ˆ
C
=
X
125°
43°
on
find
S40°E
and
cm,
the
and
length
from
is
the
of
where
notices
far
base
that
tree
angles
of
sides XY
she
the
is
and
is
both
now
her
as
XZ
standing.
tree
from
68.2°,
68.2°
Z
20
Y
She
S75°E
first
and
road?
Adam’s
How
20
standing
flagpole
50°.
to
How
the
are
From
base
south
position.
Kevin
flagpole.
of
3.6 cm,
sine
positions
elevation
ˆ
B =
47°,
triangle
new
and
top
b
=
QUESTION
isosceles
shown.
3
Â
is
position,
36°.
high
35 metres
is
From
the
apar t,
the
angle
Kevin’s
on
of
opposite
elevation
position,
the
sides
of
angle
of
flagpole?
Chapter
A
T
riangles
are
often
used
in
triangle
you
Left:
The
made
up
Hearst
Tower
in
New
Y
ork
City
cannot
isosceles
A
builder
Crossbars
give
can
structure.
strengthen
a
rectangular
frame
by
making
diagonal
form
Investigation
T
ry
to
nd
draw
that
triangle
there
are
–
corners
ABC,
with
actually
Â
two
to
triangles.
ambiguous
=
32°,
a
possible
triangles
=
3
cm,
and
triangles
c
that
=
t
5
cm.
this
Y
ou
should
description:
B
B
5
3
5
3
32°
32°
A
b
C
A
The
given
F ind
measurements
the
size
of
the
do
b
not
angle
C
C
describe
in
each
a
of
unique
the
triangle.
triangles
(call
them
C
1
and
C
).
What
is
the
relationship
between
these
two
angles?
2
Using
This
is
when
these
known
you
are
Example
In
all
the
given
for
C,
nd
ambiguous
two
sides
and
ABC,
Â
possible
=
40°,
cases.
a
=
14
Give
a
cm,
your
Answer
sin 40°
angle
case ,
B
and
and
it
the
length AC
can
non-included
for
sometimes
angle
of
a
each
triangle.
happen
triangle.
triangle
giving
angles
as
and
c
=
answers
20
cm.
correct
Solve
to
Use
1
triangle,
dp.
your
degree
sin C
this
GDC
in
mode.
=
14
20
Round
20 sin 40°
sin C
=
to
1
dp.
=
Supplementary
14
ˆ
C
have
66.7°
the
same
angles
sine
1
value. The
ˆ
C
=
180°
–
66.7°,
2
ˆ
so C
=
two
possible
113.3°
2
values
for
C
give
two
ˆ
B
=
180°
–
40°
–
66.7°
=
73.3°
1
ˆ
possible
values
for
B
ˆ
B
=
180°
–
40°
–
113.3°
=
26.7°
2
{
Trigonometry
–
its
and
triangles.
struts
Right:
rigid
change
is
shape.
of
is
architecture.
Continued
on
next
page
rigidity
to
a
B
1
And
sin 40
finally,
find
two
sin 73.3
values
14
for
b
cor rect
to
b
1
1
dp.
73.3°
o
14
b
20
sin 73.3
14
=
1
o
sin 140
b
=
20.9 cm
1
66.7°
40°
o
sin 40
sin 26.7
20.9
A
C
1
=
14
b
2
B
o
14
b
2
sin 26.7
=
2
o
sin 40
b
=
9.8 cm
26.7°
2
20
14
The
ambiguous
case
does
not
occur
ever y
time
you
solve
a
triangle.
113.3°
40°
A
➔
There
can
be
an
ambiguous
case
when
you
use
the
sine
r ule
[
●
you
are
given
two
sides
and
a
non-included
acute
C
9.8
if:
If
you
2
draw
triangles
●
the
side
two
Exercise
opposite
given
the
given
acute
angle
is
the
shor ter
of
the
you
Use
the
is
what
see.
H
given
information
to
find
the
missing
sides
and
angles
of
triangle
ABC.
Give
all
possible
solutions
with
answers
to
1
these
do
not
in
involve
a
this
sides.
Some
1
these
angle
the
ambiguous
dp.
case.
All
a
lengths
Â
=
are
30°,
a
in
=
cm.
4,
and
c
=
7
b
ˆ
B =
50°,
b
=
17,
and
c
=
21
A
c
Ĉ
=
20°,
b
=
6.8,
and
c
=
2.5
d
Â
=
42°,
a
=
33,
and
c
f
Â
=
70°,
a
=
25,
and
b
=
25
10 m
e
Â
=
70°,
a
=
25,
and
b
=
28
g
Â
=
45°,
a
=
22,
and
b
=
14
ˆ
B =
h
56°,
b
=
45,
and
c
=
=
26
50
6 m
E
B
10
2
Look
at
this
diagram.
a
Find
BE,
CE
b
Find
the
c
Explain
C
and
DE
17 m
case
of
how
the
EXAM-STYLE
3
A
at
ship
a
is
sizes
distance
Draw
b
How
far
c
How
far
ship
is
the
relates
ˆ
BCD,
to
the
ˆ
BDC,
ˆ
AB D
and
ˆ
CB D
ambiguous
r ule.
of
due
D
must
must
is
the
west
20 km
diagram
lighthouse
What
diagram
ˆ
BCE,
QUESTION
a
d
angles EÂB,
this
sine
sailing
a
of
to
the
the
on
a
sail
ship
at
bearing
a
of
time
the
bearing
model
ship
again
second
when
this
the
of
lighthouse
this
are
point
16 km
lighthouse
two
a
lighthouse
230°.
the
beyond
distance
the
of
sees
situation.
before
sail
captain
from
from
16 km
is
16 km
before
the
away?
the
ship?
the
apar t?
Chapter
.
Y
ou
The
cannot
cosine
use
the
rule
sine
r ule
to
solve
triangles
like
these:
X
D
6.56
8.9
3.63
80°
13.2
Z
8.28
E
F
a
Y
A
Consider
In
the
triangle
triangle
ACD,
ABC,
with
Pythagoras’
altitude
theorem
h
from
A
to
side
BC
c
gives
b
h
2
2
b
=
2
h
+
(a
–
2
x)
=
2
h
+
a
2
–
2ax
+
x
B
In
triangle
2
+
=
–
x
C
c
2
h
a
2
x
2
so
D
ABD,
2
h
x
=
2
c
–
x
2
Substitute
2
b
for
2
=
in
2
c
–
x
+
a
2
=
h
the
first
equation
2
+
to
get
2
a
–
2ax
+
x
2
c
–
2ax
x
In
triangle
ABD,
cos
B
,
=
so
x
=
c cos B
c
By
substituting
2
b
This
➔
2
=
for
+
c
–
equation
2ac
is
The
cosine
For
ABC,
length
side
you
get
2
a
the
x,
cosB
one
form
of
the
cosine
rule
rule
where
of
the
opposite
a
is
side
the
length
opposite
ˆ
C:
ˆ
B
,
of
the
and
c
side
is
the
opposite
length
Â,
of
b
the
is
Y
ou
2bc·cos
dot
2
a
2
=
b
b
c
2
=
2
c
a
a
Trigonometry
also
see
written
A,
means
where
as
the
multiply.
–
2bc cos A
or
–
2ac cos B
or
cosine
rule
is
in
2
+
2
=
A
2
+
The
2
might
2bc cos
c
2
+
b
–
2ab cos C
the
Formula
booklet.
Example
Find
a
and
the
missing
angles
in
this
triangle.
A
8.9 cm
80°
13.2 cm
C
a
B
Answer
2
a
2
=
2
13.2
+
8.9
2
a
=
a =
–
2(13.2)(8.9)
cos 80°
Use the cosine rule.
2
13.2
+ 8.9
− 2
(13.2 ) ( 8.9 )
cos 80°
14.6 cm
sin 80°
sin B
Use
=
the
sine
rule.
8.9
a
8.9 sin 80°
sin B
=
14.6
so
ˆ
B
=
36.9°
(1dp)
ˆ
C =
180°
When
to
–
you
80°
use
rearrange
➔
–
Cosine
36.9°
the
the
63.1°
cosine
formula
r ule
like
to
find
angles,
it
is
sometimes
helpful
this:
rule
b
A
=
+ c
− a
=
B
bc
+
c
a
−
=
A
+
C
b
−
=
Example
Find
angles
A,
B
and
C
A
6.56 mm
3.63 mm
B
8.28 mm
C
{
Continued
on
next
page
Chapter
Answer
2
2
( 3.63 )
cos
A
+
(
2
6.56 )
(
Use
8.28 )
the
cosine
=
2
b
( 3.63 ) (
2
6.56 )
cos
A
rule.
2
2
+ c
a
=
2bc
2
⎛
–1
 =
2
( 3.63 )
+
(
6.56 )
2
(
−
⎞
8.28)
⎜
cos
⎟
⎜
⎟
2
⎝
Â
=
105°
( 3.63 ) (
2
B
2
+
( 8.28 )
2
−
( 6.56 )
=
2
rule
(You
use
sine
rule
2
3.63
8.28
could
here
instead.)
2
the
6.56
–1
=
cos
2
so
ˆ
B
ˆ
C =
=
Now
be
Cosine
( 3.63 ) ( 8.28 )
2
ˆ
B
⎠
(3 sf)
( 3.63 )
cos
6.56 )
=
49.9°
180°
–
25.1°
look
solved
105°
–
49.9°
(3 sf)
again
more
Example
Two
3.63 8.28
at
Example
quickly
5
using
from
the
cosine
11.2.
This
problem
can
r ule.
ships
leave
dock
at
the
same
time.
30 km
before
dropping
anchor.
Ship
65 km
before
dropping
anchor.
Find
when
Section
they
are
stationar y
,
to
the
B
Ship
sails
the
nearest
A
on
sails
a
due
bearing
distance
nor th
of
between
for
050°
the
for
ships
kilometre.
Pythagoras’
theorem
Answer
is
a
special
case
of
B
Draw
a
diagram.
the
cosine
rule.
See
A
what
happens
equation
to
when
the
you
65
use
30
the
cosine
rule
50°
with
P
2
AB
2
=
2
30
+65
–
2(30)(65)
×
cos50°
Use
the
2
2
AB
=
=
The
the
30
a
2
+ 65
2
( 30 ) ( 65 )
cos50 °
51.17
ships
are
nearest
51
km).
Trigonometry
kilometres
apar t
(to
cosine
2
=
b
rule:
2
+
c
–
2bc cos 50°
an
angle
of
90°.
Exercise
1
Use
I
the
given
information
to
find
all
sides
and
angles
in
each
There
triangle.
Â
a
2
c
a
e
ˆ
C
A
=
=
a
64°,
3.6,
=
for
a
far
B
C
he
the
Ship
the
A
what
c
and
b
and
He
of
a
the
=
=
=
1
dp.
72
86
walks
to
lengths
=
20,
b
d
ˆ
B =
31°,
f
a
45,
=
on
a
walking
again
get
a
5 km
continues
stops
All
b
2.4
before
back
to
port
bearing
the
away
N27°E
leaves
parallelogram
diagonals
of
distance
same
and
walk
of
15 km
is
c
to
=
a
b
are
33,
=
=
10,
c
=
c
many
8 km
straight
real-life
applications
triangle
trigonometr y.
of
41
=
=
058°.
another
heading
c
and
and
of
are
metres.
and
50,
bearing
for
in
14
58
He
on
back.
camp?
QUESTIONS
lengths
is
and
then
103°.
lengths
Town
75,
break,
The
Town
43,
camp
must
the
answers
4.9,
=
diagonals
Find
5
=
The
Find
4
=
a
of
EXAM-STYLE
3
your
leaves
bearing
How
b
b
70°,
hiker
stops
Give
of
sides
from
town
and
and
sails
was
ship
of
and
due
49 km.
B
cm
the
towns
sails
6
Town
A,
between
port
are
form
an
and
A,
in
and
east
The
9
angle
of
62°.
cm.
parallelogram.
towns
B
acute
the
direction
A
C
and
N36°W
.
are
20 km
apart.
C.
for
28 km.
ships
are
Ship
then
B
leaves
36 km
from
apart.
On
sailing?
E
6
The
Its
pyramid
other
ABCDE
faces
are
has
a
square
congruent
base
isosceles
with
sides
triangles
15
with
cm.
equal
sides
24
of
24
Find
cm.
these
angles.
B
ˆ
A BD
a
C
ˆ
b
E DC
c
EÂC
A
15
.
Look
Y
ou
D
Area
at
of
triangle
can
find
a
triangle
ABC
the
with
area
of
base
the
b
and
triangle
height
using
h.
the
B
formula:
c
a
h
1
area
bh
=
2
A
In
ADB,
sin A
=
D
C
b
,
so
h
=
c
sin A
1
Substituting
for
h
in
the
area
formula
gives
area
bc sin A
=
2
Notice
height
that
of
in
the
this
formula
you
do
not
need
to
know
the
triangle.
Chapter
➔
The
area
of
any
triangle
ABC
1
area
or
area
=
2
Example
a
Find
given
by
the
formula:
1
bc sin A
=
is
1
ac sin B
or
area
=
2
ab sin C
2
the
area
of
triangle
ABC
C
7.8 cm
82.7°
8.4 cm
A
B
E
2
b
The
area
of
this
triangle
is
50
cm
.
8.2 cm
Find
θ
angle
i
D
13.7 cm
F
Answers
1
a
Area
1
( 8.4 ) ( 7.8 )
=
sin 82.7°
Area
=
2
ab
sin
C
2
2
=
32.5 cm
(3
sf)
1
(8.2) (13.7 )
b
sin
50
2
A
50
sin
b
=
8.2 cm
1
(8.2 )
(13.7 )
2
c
i
100
C
=
In
0.8901...
(8.2 )
the
rst
centur y
CE,
(13.7 )
Hero
(or
Heron)
of
13.7 cm
1
= sin
0.8901
a
Alexandria
developed
a
method
B
= 62.9 (3 sf )
different
nding
Exercise
1
Find
the
J
area
a
of
each
triangle.
All
lengths
in
using
lengths
of
c
56.5°
13.4
115°
25.1
9
6.8
8
32°
d
e
f
7.88
86°
46
8.74
30
41
58°
10.98
Trigonometry
46°
area
triangle
cm.
10
b
9.4
are
the
the
of
only
for
a
the
sides.
2
2
The
triangle
Find
the
shown
value
has
an
area
of
100 m
.
θ
of
15 m
i
18 m
3
The
triangle
shown
has
an
area
2
of
324
cm
.
x
Find
the
value
of
x
57.4°
33.9 cm
EXAM-STYLE
4
a
Find
b
Hence,
QUESTIONS
the
largest
angle
in
this
triangle.
The
command
term
10.2 cm
find
the
area
of
the
triangle.
‘hence’
tells
you
to
17.2 cm
use
your
par t
16.4 cm
a
to
answer
answer
help
par t
from
you
b
2
5
The
triangle
Find
the
shown
value
of
has
an
area
of
30
cm
.
2x
+
3
x.
30°
4x
+
5
2
6
The
area
Two
Find
of
sides
two
a
of
triangle
the
Radians,
Angles
can
be
20 mm
triangle
possible
.
is
are
lengths
arcs
measured
in
for
.
8 mm
the
and
and
third
11 mm.
side.
sectors
radians
instead
of
degrees.
The
Why
One
use
radians?
complete
arbitrar y
tur n
measure.
is
Babylonians
believed
360°,
but
Radians,
the
number
however,
are
360
is
a
directly
were
somewhat
related
year
to
360
and
360°
measurements
within
a
circle.
In
this
section,
you
will
see
that
to
there
days
in
hence
a
used
represent
one
how
revolution.
radians
One
of
by
as
are
radian
the
an
is
dened
central
arc
the
connected
angle
which
radius
is
of
the
arc
length
size
subtended
the
the
as
to
same
length
and
Two
sector
radians
area
is
the
angle
subtended
equal
to
twice
size
by
the
an
of
the
arc
radius
central
with
of
the
a
length
circle.
A
circle.
central
angle
B
B
subtended
2r
is
r
an
by
angle
an
with
arc
its
r
r
i
ver tex
at
the
center
i
A
A
O
O
of
the
circle
and
its
r
r
sides
the
passing
endpoints
through
of
the
arc.
θ
θ
=
1
=
2
radians
radian
Chapter
One
complete
length
to
the
tur n
around
The
Therefore,
is
2π
the
circumference
the
angle
of
circle
the
is
subtended
circumference
which
by
an
arc
equal
in
circle.
subtends
the
=
2πr
circumference
of
the
circle
radians.
Arc
length
=
circumference
=
2πr
r
i
Any
=
2r
radians
central
length
of
angle
the
arc
in
a
the
angle
➔
Arc
length
=
angle
r
is
the
a
fraction
subtends
as
a
of
2π,
so
fraction
you
of
can
the
calculate
the
circumference.
rθ
=
radius
measured
is
where
circle
in
and
θ
is
the
central
O
radians.
r
i
r
2
Similarly
,
The
area
area
of
the
of
the
formula
a
sector
for
the
with
a
area
central
of
a
circle
angle θ
is:
will
area
be
a
= πr
fraction
of
the
circle.
➔
Area
of
sector
=
where
r
is
the
radius
of
the
circle
and
θ
is
the
central
angle,
in
radians.
Example
a
Find
a
the
length
central
angle
of
of
the
2.6
arc
which
radians
subtends
(see
diagram)
i
in
b
a
circle
Find
the
of
radius
area
of
7
the
Arc
length
=
7(2.6)
=
18.2 cm
Arc
length
=
rθ
2
2.6
(7
2
)
r
2
b
Sector
area
=
=
2
Trigonometry
2.6
radians
7 cm
sector.
Answers
a
=
cm.
63.7 cm
Sector
area
=
2
The
abbreviation
for
radians
is
rad.
In
the
example
above,
Another
2.6
radians
could
be
written
as
2.6
rad.
If
you
see
an
angle
angles
no
units,
such
as
‘sin
2.6’,
you
can
assume
that
the
angle
way
of
writing
with
is
2.6
radians.
in
radians
c
is
2.6
stands
where
for
the
c
circular
measure.
Example
A
circle
has
subtended
radius
by
an
2.5 mm.
arc
of
Find
length
the
size
of
the
central
angle
9 mm.
Answer
9
=
2.5θ
Arc
length
=
rθ
9
2.5
3.6 rad
Example
In
this
circle,
arc
AB
=
7.86
cm
and
the
area
of
Some
sector
2
AOB
=
23.58 cm
.
Find
the
central
angle
θ
and
the
radius
their
r
A
farmers
crops
patterns.
real-life
are
in
plant
circular
What
other
applications
there
for
circles,
i
B
arcs
r
O
and
sectors?
Answer
2
2
r
r
2
23.58
=
,
so
47.16
=
θr
Sector
area
=
2
2
7.86
7.86
rθ,
=
Arc
=
so
length
=
rθ
r
7.86
2
47.16
=
(r
)
=
7.86r,
so
Substitute
for
θ
in
the
previous
r
equation.
47
r
16
=
7
=
86
6 cm
7.86
7.86
,
=
so
θ
=
1.31
rad
Use
the
=
result
r
6
Exercise
1
Find
of
2
1.7
Find
of
the
K
length
radians
the
3.25
of
in
length
radians
the
a
of
at
arc
circle
the
the
which
with
arc
radius
which
center
subtends
of
a
a
central
angle
5.6 cm.
subtends
circle
an
with
angle
diameter
24
cm.
Chapter
An
3
of
An
4
is
circle.
arc
a
circle
of
sector
arc
circle
In
at
the
by
value
an
radius
an
θ
of
angle
arc
if
of
50 cm.
of
the
2.4
Find
length
circle
radians
the
12.5 mm
has
area
at
at
radius
the
and
the
center
2.5 mm.
center O
perimeter
AOB
WX
with
the
the
subtends
with
subtends
radius
EXAM-STYLE
6
subtended
Find
AB
of
An
5
θ
angle
a
3
an
cm.
angle
Find
of
5.1
the
radians
area
and
at
the
center P
perimeter
of
of
a
sector WPX
QUESTION
circle
with
center.
If
center
the
P
the
length
of
arc
arc
QR
QR
subtends
is
27.2
an
cm
angle
and
the
of
θ
area
of
2
7
sector
PQR
Circle
O
The
has
centers
intersect
the
is
at
217.6
radius
of
A
the
and
cm
,
find
4 cm,
and
circles
B,
find
are
the
θ
and
circle
8
cm
blue
the
P
radius
has
apar t.
shaded
of
radius
If
the
area
the
6
circle.
A
cm.
circles
O
in
b
P
a
diagram.
B
Degrees
Y
ou
have
and
radians
already
seen
that
one
full
rotation
around
a
circle
gives
a
Any
central
to
angle
360°.
Y
ou
2π,
of
can
and
use
you
this
know
fact
to
that
one
convert
full
rotation
between
is
radians
also
and
angle
which
equal
degrees.
is
a
multiple
is
assumed
measured
360°
=
2π,
=
1
so
180°
=
of
to
in
π
be
radians
π.
so
you
don’t
need
to
180
and
radian
write
‘rad’.
1
radians
=
180
➔
To
conver t
degrees
to
radians
multiply
by
➔
To
conver t
radians
to
degrees
multiply
by
Example
Exact
a
Conver t
these
angles
to
radians:
30°,
45°,
are
Give
exact
Conver t
these
angles
to
degrees:
rad,
5
Give
exact
Trigonometry
written
multiples
rad
9
answers.
{
values
as
answers.
2
b
radian
60°
Continued
on
next
page
of
π
Answers
30°
a
180
180
180
2
180
180
180
5
9
these
values
Conver t
Give
20°
Conver t
Give
Multiply by
=
b
72°
180
=
Example
3
180
=
9
a
=
=
=
5
4
60
60
2
=
b
180
=
by
6
45
=
Multiply
180
60°
=
45
=
=
45°
30
= 30
angles
to
these
values
3
to
significant
angles
to
radians:
one
to
43°,
136°
figures.
degrees:
decimal
70°,
1
rad,
2.3
rad
place.
Answers
43°
a
=
=
=
=
1
rad
rad
(3
sf)
=
2.37
rad
(3
sf)
180
180
= 1
180
=
1.22
=
b
sf)
136
136
(3
180
=
=
180
rad
70
0.750
180
70
136°
=
180
70°
43
43
57.3°
(1
dp)
180
2.3
rad
=
2.3
=
Exercise
1
Conver t
Give
a
2
a
3
these
75°
Give
b
these
56°
Give
b
these
exact
angles
240°
angles
to
3
radians.
c
to
angles
c
to
80°
d
330°
d
230°
radians.
significant
107°
figures.
324°
degrees.
b
to
values.
a
dp)
values.
answers
Conver t
(1
L
exact
Conver t
131.8°
c
d
Chapter
Conver t
4
Give
1.5
a
In
Section
often
radians.
11.1,
you
in
to
degrees.
significant
0.36
45°,
used
is
3
b
30°,
It
angles
to
rad
triangles:
are
these
answers
rad
looked
60°
and
to
2.38
c
at
some
90°.
trigonometr y
helpful
figures.
remember
angles
angles,
they
these
3.59
d
‘special’
These
and
rad
can
and
also
angles,
right-angled
their
be
so
in
rad
multiples,
expressed
you
do
not
T
o
in
have
to
change
mode
choose
do
the
conversion
ever y
time.
The
table
shows
some
special
degrees
and
their
equivalents
in
radians
as
multiples
of
radian
and
5:Settings
&
angles
Status
in
to
press
|
2:Settings
|
π
1:General
Angle
in
Use
30°
45°
60°
90°
120°
135°
150°
180°
210°
the
degrees
move
to
select
Angle
in
3
2
to
Angle
and
Radian.
5
7
5
key
225°
Press
and
then
π
radians
4
6
3
2
3
4
6
6
4
select
to
Angles
which
are
multiples
of
30°,
45°,
60°
and
90°
are
4:Current
return
to
the
usually
document
written
When
in
you
whether
To
exact
find
solve
the
set
Example
are
cosine
your
form
using
trigonometr y
angles
sine,
radians,
radian
given
and
GDC
problems
in
degrees
tangent
to
π
radian
values
you
or
must
be
careful
to
note
radians.
for
angles
measured
in
mode
GDC
help
on
CD:
demonstrations
Plus
The
diagram
shows
a
circle
O
and
radius
5
GDCs
cm.
5 cm
Find
to
3
the
area
of
significant
the
shaded
region,
O
figures.
1.46
rad
D
Answer
2
(1.46 )
Area
of
sector
OCD
(5
)
=
Area
of
the
shaded
region
=
area
2
2
=
18.25
sector
OCD
–
area
cm
1
Area
of
OCD
Area
=
ab
2
1
(5) (5)
=
sin
(1.46 )
2
≈
12.42335...
Shaded
area
1
=
18.25
(5) (5)
–
2
2
≈
5.83 cm
Trigonometry
(3
sf)
sin
Casio
the
(1.46 )
sin C
of
OCD
of
are
on
TI-84
FX-9860GII
with
C
center
and
Alternative
for
the
CD.
Exercise
1
Find
M
the
exact
value
of
Find
c
tan
d
the
value
of
each
ratio.
cos
b
2
trigonometric
sin
a
each
sin
trigonometric
ratio,
to
3
significant
figures.
cos
a
0.47
EXAM-STYLE
3
The
=
1.3
A
4
The
c
tan
2.3
d
cos
0.84
shows
the
circle,
center A,
radius
4.5
cm,
and
radians.
C
1.3
1.25
QUESTIONS
diagram
BÂC
sin
b
a
Find
the
area
of
b
Find
the
length
c
Find
the
area
ABC
BC
rad
4.5
of
the
shaded
region.
B
diagram
shows
the
circle,
center O,
A
with
11 m
3 m
radius
3 m,
AB
=
11
and
AÔB
=
0.94
radians.
0.94
Find
the
shaded
B
area.
O
5
The
6
diagram
cm,
QR
=
shows
11.2
the
cm
P
circle,
and
PÔQ
center O,
=
1.25
with
radius
radians.
a
Find
the
area
of
POQ
b
Find
the
area
of
QOR
c
Find
θ
d
Find
the
6 cm
Q
1.25
i
(PÔR).
M
O
11.2 cm
length
of
arc
PMR
R
Review
exercise
✗
1
In
triangle
Find
2
In
a
the
length
triangle
Find
ABC,
of
XYZ,
ˆ
X ZY.
Â
=
ˆ
B =
45°.
The
length
of
AC
is
7
cm.
AB
XY
=
b
8
cm,
Find
XZ
=
16
cm,
and
ˆ
XY Z
=
90°.
YZ
Chapter
EXAM-STYLE
3
A
straight
point
with
with
the
Find
QUESTIONS
line
passes
through
coordinates
(5,
2).
the
origin
The
line
(0,
0)
forms
and
an
through
acute
the
angle
of θ
x-axis.
the
value
tan θ
of
Z
4 cm
4
The
diagram
shows
triangle
XYZ,
with
XZ
=
4
cm,
XY
=
10
cm,
30°
ˆ
X
and
=
30°
X
Find
the
area
of
triangle
10 cm
Y
XYZ
B
A
5
The
diagram
AÔC
=
2.5
Find
a
shows
a
circle,
center O
and
radius
10
cm.
2.5
radians.
the
length
of
arc
C
ABC
10 cm
O
Find
b
the
Review
1
An
the
How
top
tall
of
the
shaded
sector.
exercise
obser ver
sees
area
is
standing
of
the
the
100 m
building
from
at
an
the
base
angle
of
of
a
building
elevation
of
36°.
building?
y
2
The
diagram
shows
par t
of
a
unit
circle
(radius
1
unit)
C
D
with
center
a
Angle
b
Point
Find
O
AOB
C
=
has
angle
32°.
Write
coordinates
down
the
(0.294,
coordinates
of
B
B
0.956).
E
AOC
A
x
0
c
Angle
COD
EXAM-STYLE
=
54°.
Find
the
coordinates
of
D
QUESTIONS
ˆ
3
The
diagram
shows
triangle
XYZ,
with
X=
Y
42.4°,
13.2 cm
ˆ
Z
=
82.9°
and
XY
=
13.2
cm.
X
42.4°
a
Find
ˆ
Y
b
Find
XZ
82.9°
Z
4
The
diagram
shows
triangle
PQR,
with
ˆ
Q
=
118°,
PQ
=
9.5 m
P
9.5 m
and
QR
=
11.5 m.
Q
a
Find
PR
b
Find
ˆ
P
118°
11.5 m
R
Trigonometry
2
5
This
diagram
a
Find
ˆ
ACB,
b
Find
AB
shows
the
given
triangle ABC,
that
it
is
an
which
obtuse
has
an
area
of
10
B
cm
angle.
5.83 cm
A
6
Two
ships
Ship
A
sail
from
the
same
por t
P
at
the
same
C
4
N
time.
A
sails
on
a
bearing
of
050°
for
a
distance
of
24 km
before
24 km
droping
anchor.
50°
Ship
B
sails
on
a
bearing
of
170°
for
a
distance
of
38 km
before
P
droping
Find
170°
anchor.
the
distance
between
the
two
ships
when
they
are
stationar y
.
38 km
B
7
The
diagram
shows
quadrilateral
ABCD,
with
AB
=
7
B
cm,
9 cm
BC
=
9
cm,
CD
=
8
cm,
and
AD
=
15
cm.
Angle
ACD
=
82°,
7 cm
C
y°
angle
CAD
=
x°,
and
angle
ABC
=
y°
A
a
Find
the
value
of
x.
b
Find
AC
c
Find
the
value
of
y.
d
Find
the
82°
x°
8 cm
area
of
triangle
ABC
15 cm
D
8
The
diagram
Angle
DAC
shows
=
0.93
a
circle
radians,
a
Find
BC
b
Find
DB
c
Find
the
length
d
Find
the
perimeter
with
and
center A
angle
and
BCA
=
radius
1.75
12
cm.
B
radians.
D
E
of
arc
DEC
1.75
of
the
region
BDEC
0.93
12 cm
A
C
Chapter
CHAPTER
11
SUMMARY
Right-angled
For
any
triangle
right-angled
triangle
opposite
●
sine θ
an
angle θ:
adjacent
O
=
=
;
hypotenuse
cosine θ
A
=
=
;
hypotenuse
H
opposite
tangent θ
with
trigonometry
H
O
=
=
adjacent
A
H
O
●
i
A
●
The
trigonometric
ratios
of
‘special
angles’
are:
angle
30°
sine
cosine
tangent
2
√3
measure
1
3
1
3
60°
=
°
2
2
3
3
1
1
2
1
1
2
1
=
°
2
2
1
2
2
1
= 1
=
45°
3
3
1
=
°
2
2
Applications
of
45°
3
1
right-angled
triangle
trigonometry
angle
of
elevation
horizontal
●
The
angle
of
elevation
is
The
angle
of
depression
The
four
east
(E),
the
angle
‘up’
from
horizontal.
angle
●
cardinal
and
compass
west
Three-figure
is
the
angle
points
‘down’
are
nor th
from
(N),
south
depression
(S),
(W).
bearings
of
horizontal.
N
give
directions
as
clockwise
from
nor th.
40°
Using
the
coordinate
axes
in
trigonometry
O40°
●
For
●
For
●
For
●
These
supplementar y
any
any
three
sin
sin θ
θ°,
angle
=
sin θ
proper ties
2
1
θ,
angle
angles
α
and
sin (180°–
=
β,
θ),
sin (180°–
are
tr ue
sin α
for
and
θ),
=
cos θ
and
any
sin β,
=
and
cos α
=
– cos (180°–
cos θ
=
=
N40°E
–cos β
θ)
–cos (180°–
θ)
angle θ:
2
θ
+
cos
θ
=
1.
sin
2
tan
=
cos
y
3
For
any
line
y
=
mx
which
forms
an
angle
of
θ
=
mx
with
i
the
x-axis,
the
value
of
m
(the
gradient
of
the
line)
is
tan
θ
x
Continued
Trigonometry
on
next
page
The
●
sine
For
any
rule
ABC,
B
a
is
the
length
of
the
b
is
the
length
of
the
c
the
c
side
opposite
Â,
side
opposite
ˆ
B
,
opposite
ˆ
C,
side
and
is
length
A
sin B
There
the
b
can
sine
=
A
sin B
sin C
c
be
r ule
c
=
or
=
sin
●
b
C
b
sin C
=
a
the
A
a
sin
of
a
an
ambiguous
case
when
you
use
if:
B
○
you
are
given
two
sides
and
a
non-included
B
5
3
acute
○
the
angle
side
5
32°
opposite
the
given
acute
angle
is
A
the
3
32°
b
C
A
shor ter
The
●
of
cosine
The
cosine
2
a
=
b
=
a
+
c
+
c
+
b
2
A
C
states
that:
–
2bc cos A
B
–
2ac cos B
–
2ab cos C
or
c
a
or
2
a
2
b
cos
b
sides.
2
2
=
given
2
2
2
c
two
rule
r ule
2
2
b
●
the
A
C
b
2
+ c
− a
=
2bc
2
2
a
cos
B
2
+ c
− b
=
2ac
2
2
a
cos
C
2
+ b
− c
=
2ab
Area
●
The
of
a
area
triangle
of
any
triangle
ABC
1
area
sin
A
or
area
=
2
Radians,
For
●
a
sector
Arc
by
the
formula:
1
ac
sin B
or
area
with
of
and
central
sector
2
sectors
angle
=
ab sin C
=
2
arcs
length
given
1
bc
=
is
θ
radians
in
a
circle
of
radius
r :
rθ
●
Area
of
sector
=
●
To
conver t
degrees
to
radians,
multiply
by
●
To
conver t
radians
to
degrees,
multiply
by
Chapter
Theory
of
knowledge
Units
of
measurement
Mathematics
However,
is
this
often
considered
language
actually
a
‘universal
has
many
language’.
forms.
There
Angles
can
be
measured
in
different
units:
degrees
are
number
radians.
Why
do
we
need
more
than
one
unit
other
or
systems
as
of
well
as
the
decimal
measurement?
(base-10)
Actually
,
different
we
don’t
forms
of
need
different
measurement
units
have
of
measurement.
developed
in
But
which
we
different
Another
par ts
of
the
world
and
at
different
times.
system
which
The
idea
of
thousands
of
sexagesimal
to
the
to
360
The
fact
360°
years
a
that
the
full
to
(base-60)
the
circle
ancient
number
orbit
is
of
thought
date
Babylonians,
system.
the
to
Ear th
It
may
about
the
is
is
around
be
Sun
used
a
binar y
related
is
Where
the
The
322
tablet
1800
script
numbers
numbers
arranged
and
in
show
triples
dates
BCE.
from
Scholars
–
into
are
moder n
Old
have
written
in
digits,
base
are
columns
Pythagorean
the
Babylonians
these
more
years
before
were
than
the
using
1000
time
of
Pythagoras.
What
is
a
Pythagorean
Why
is
called
Theory
of
2.
is
the
system
commonly
close
Babylonian
translated
the
What
this
triple?
tablet
Plimpton
knowledge:
Units
322?
of
measurement
and
60.
discovered
that
do
measure
base
cuneiform
all
binar y,
base
back
who
also
impor tant
used?
days.
Plimpton
times,
in
system
use.
we
in
60?
The
radian
more
sensible
related
type
unit
of
to
the
angle
of
unit
in
1870s.
How
are
angles,
measurement
the
Today
,
geometr y
,
for
measurements
mathematicians,
the
measurement
term
radian
in
a
had
circle.
been
related
to
be
are
a
much
closely
Although
used
wasn’t
measurement
and
to
radians
‘radian’
trigonometr y
radians
as
seems
previously
much
is
this
by
used
until
commonly
used
calculus.
the
measurements
in
a
circle?
Who
measures
angles
in
gradians?
[
Angles
are
not
the
only
area
in
which
different
units
The
was
measurement
A
look
that
at
the
the
are
units
‘universal
universal
as
we
common.
of
might
like
to
of
distance
and
mass
mathematics’
may
radian
used
papers
currency,
language
term
of
will
not
by
show
be
in
written
James
Thomson
in
the
early
in
Belfast.
1870s
as
think.
D
EE
SP
IT
M
LI
driver
that
the
speed
limit
is
30,
but
give
no
units.
30
Which
speed
is
actually
faster?
30
Would
rather
you
be
a
millionaire
in
the
UK
US,
or
China?
6000
Which
kg
7
elephant
is
(US)
11
000
pounds
heaviest?
Do
to
tons
you
be
think
truly
What
sor t
been
sent
it
is
possible
for
any
language
‘universal’?
of
mathematical
into
communicate
deep
with
information
space,
other
to
has
perhaps
intelligent
life-
forms?
Chapter
Vectors
CHAPTER
Vectors
4.1
of
a
zero
OBJECTIVES:
as
vector ;
vector
,
vectors;
The
4.2
scalar
Vector
column
the
base
between
4.3
displacements
the
plane
representation
vector
vectors
product
in
–v;
i,
of
j
the
and
sum
multiplication
and
two
k;
by
position
vectors;
in
three
and
a
dimensions;
difference
scalar ;
of
components
two
magnitude
vectors;
of
a
the
vector ;
unit
vectors.
perpendicular
vectors;
parallel
vectors.
Angle
vectors.
equation
of
a
line
in
two
and
three
dimensions.
The
angle
between
two
lines.
Coincident
4.4
whether
Before
Y
ou
1
two
parallel
lines
you
should
Use
and
lines.
Point
of
intersection
know
in
OABCDEFG
is
A
lies
on
D
the
y-axis
and
lies
down
the
A
coordinates
lines.
Determining
start
coordinates
units.
two
intersect.
how
three
to:
Skills
dimensions.
1
e.g.
of
a
cube
x-axis,
on
the
coordinates
with
C
lies
z-axis.
of
A,
sides
B
on
2
has
the
2
Write
and
The
check
cuboid,
length
units.
y-axis
A
3
OABCDEFG
units,
lies
and
D
on
on
F
OC
the
the
4
is
such
units
x-axis,
C
that
and
on
OA
OD
the
z-axis.
Give
the
coordinates
of
G
has
(2, 0, 0).
G
B
has
coordinates
C
(2, 2, 0).
D
D
F
has
coordinates
F
A
b
B
c
E
d
F
e
H,
F
4
the
midpoint
2
C
(2, 2, 2).
a
GF
B
O
E
3
E
O
B
A
A
2
Use
e.g.
of
a
Pythagoras’
Find
the
triangle
theorem.
length
with
of
the
other
2
Find
the
length
of
the
hypotenuse, x,
sides
4 cm,
7 cm.
x
3
2
x
x
2
=
=
7
2
+
65
Vectors
4
=
=
65
8.06 cm
6
hypotenuse,
x
of
3
Use
e.g.
QR
the
In
=
cosine
triangle
11 cm
Calculate
2
PR
=
PQ
=
PQ
=
a
triangle
BC
6 cm,
=
correct
PR
ABC,
15 cm
Calculate
95°.
of
In
to
AB
and
the
the
=
angle
length
9 cm,
ABC
of
nearest
=
110°.
AC
cm.
2
+
QR
–
2PQ
×
QR
×
cos
95°
b
In
triangle
ABC,
AB
=
8.6 cm,
AC
=
9.7cm.
2
=
6
=
145.49.
=
ˆ
Q
length
2
2
PR
PQR,
and
the
3
r ule.
+
11
12.1 cm
–
.
2
×
.
(3
6
×
11
×
cos
95°
BC
=
3.1cm
Diagram
sf)
and
B
NOT
accurately
3.1 cm
drawn
8.6 cm
C
9.7 cm
A
Calculate
angle
ABC
to
the
nearest
degree.
Chapter
Some
quantities
require
one
can
piece
temperature
is
be
of
described
information.
37 °C,
the
length
by
a
For
of
single
number
example,
the
–
normal
Amazon
river
is
they
only
body
6400 km,
the
–3
density
of
water
magnitude
However,
direction
vectors.
you
I
They
some
to
you
are
are
alone
and
quantities
wish
what
used
used
is
are
fly
represent
not
them.
340 km,
direction
a
only
Such
from
this
you
in
quantities
are
determined
by
called scalars
yourself
extensively
to
These
require
define
to
distance
you
.
1000 kg m
completely
the
tell
Vectors
(size)
If
that
until
is
of
to
branch
quantities
quantities
London
piece
need
magnitude
as
also
called
Paris
and
is
I
tell
useless
in!
Physics
such
are
information
travel
of
to
but
called
Mechanics.
displacement,
force,
Y
ou
weight,
velocity
interested
in
velocities.
The
questions
and
vectors
final
where
the
then
leads
Vectors
basic
on
as
and
will
the
Mathematics
be
of
this
able
to
of
see
vocabular y
basic
operations
has
these
and
a
are
and
of
applications
in
notation
This
of
geometr y
both
chapter
vectors
of
may
wish
to
primarily
number
problems.
and
we
displacements
chapter
three-dimensional
concepts,
to
In
representations
exercise
you
two-dimensional
with
momentum.
deals
and
vectors.
explore
the
vectors
in
role
of
Mechanics.
.
If
you
have
A
Vectors:
travel
you
basic
ki lometres
concepts
north
and
ki lometres
east,
how
far
traveled?
simple
question
perhaps
–
but
a
question
with
two
sensible
answers:
●
One
answer
to
this
question
is
to
say
that
you
have
3 km
traveled
F inish
7
kilometres.
through
(4
This
+
3
=
is
7
the
total
distance
that
you
have
moved
kilometres).
4 km
Start
●
A
second
answer
to
this
question
is
to
say
that
you
have
3 km
traveled
F inish
5
kilometres.
This
2
theorem
(
value
has
been
found
using
Pythagoras’
2
4
=
+ 3
5
kilometres).
This
value
is
called
the
4 km
displacement .
position
and
It
measures
your
final
the
difference
between
your
5 km
initial
position.
Start
Vectors
➔
A
and
is
vector
direction.
➔
A
of
As
can
have
scalars
be
velocity
and
something
the
For
at
rate
is
traveling
if
This
of
hour
and
is
whereas
think
per
size
(magnitude)
are
displacement
size
a
and
no direction.
velocity
.
Examples
displacement
also
to
tr ue
how
velocity
changes
car
then
but
and
speed.
refers
something
we
has
and
distance
Speed
has
vectors
that
distance
meanings.
which
that
of
quantity
above,
kilometres
is
fast
refers
its
that
this
of
is
to
position.
traveling
the
speed.
that
car
star ting
place,
the
was
line
then
star ting
If
a
speed.
instance,
90
car’s
If
at
quantity
are
seen
different
a
Examples
is
scalar
scalars
same
direction,
and
its
line
the
would
car
per
be
was
one
hour
around
finish
velocity
after
kilometres
traveling
line
when
it
a
track
were
in
retur ns
where
the
to
the
same
the
zero.
traveling
hour
we
down
would
a
straight
say
that
road
its
in
a
velocity
westerly
is
90
west.
Chapter
Representation
Vectors
of
the
on
are
line
the
Consider
represented
represents
line
of
(indicted
the
points
vectors
using
the
by
size
an
directed
of
the
arrow)
A(2, 3)
and
line
segments
vector
shows
B(5, 7)
quantity
the
on
where
and
direction
the
the
the
of
Car tesian
length
direction
the
vector.
plane:
y
8
B
6
4
A
2
0
x
2
To
in
3
describe
the
is
(or
positive
the
y)
the
the
called
x
direction
the
describe
from
and
horizontal
component .
movement
This
movement
are
The
4
4
A
B
units
(or
we
in
could
the
of
impor tant.
this
We
say
the
movement
can
‘move
positive y
x) component,
direction
both
to
6
4
units
direction’.
is
and
therefore
3
the
vertical
the
use
a
The
length
vector
of
to
this.
vector
can
be
represented
in
a
variety
of
ways:
In
In
the
diagram
the
line
AB
represents
the
vector
AB
where
a
column
⎛ x ⎞
the
vector
arrow
over
the
letters
indicates
the
direction
of
the
movement
to
B).
The
components
of
the
vector
are
here
represented
,
⎟
y
⎝
x
A
⎜
(from
represents
vector
a
using
movement
column
the
⎠
in
the
form.
positive
x
and
y
direction
3
AB
=
4
in
Vectors
For
can
the
a
movement
also
example
we
be
represented
could
use
a
to
using
a
lower
represent
the
case bold
vector
AB
letter.
the
positive
y
direction.
.
3
a
=
AB
=
4
B
Bold
letters
difcult
hand
to
so
are
write
by
instead
when
a
4
writing
you
underline
to
show
should
the
that
letter
it
is
vector
.
A
3
So
a
hand
Vectors
is
written
as
a
by
a
Finally
the
vector
can
be
represented
in uni t vector form .
We
can
3
write
as
4
3i
+
4j
where
i
and
j
are
vectors
of
length
1
unit
in
the
j
directions
of
x
and
y
respectively
.
i
and
j
are
called
base
vectors.
i
The
vector
positive
As
x
well
3i
+
4j
therefore
direction
as
objects
and
that
4
means
in
the
move
a
movement
positive
along
a
flat
y
of
3
units
in
the
direction.
surface
in
two
dimensions,
k
also
think
space.
way
about
We
as
can
above
objects
that
represent
but
we
a
move
vector
introduce
around
in
the
three
in
dimensions
letter k
j
three-dimensional
for
the
in
vector
a
similar
of
length
i
one
So
unit
now
in
the
we
z-direction.
have
three
components.
3
2
=
3i
−
2j
+
k
would
therefore
represent
a
movement
of
3
units
1
in
the
unit
➔
positive
in
the
The
x-direction,
positive
unit
2
units
in
the
negative
y-direction
and
1
z-direction.
vector
in
the
direction
of
the x-axis
is
i.
⎛ 1 ⎞
⎛ 1 ⎞
In
two
dimensions
i
=
⎜
and
⎜
⎝
in
three
dimensions
⎟
0
i
⎟
0
=
⎜
⎠
⎟
⎜
⎟
0
⎝
➔
The
unit
vector
in
the
direction
of
the y-axis
is
j.
In
⎠
two
⎛ 0 ⎞
⎛ 0 ⎞
dimensions
j =
⎜
and
⎜
⎝
in
three
dimensions
j =
⎟
1
⎠
In
three
z-axis
dimensions
⎟
⎜
⎝
➔
⎟
1
⎜
the
unit
vector
in
the
⎟
0
⎠
direction
of
the
is
⎛ 0 ⎞
⎜
k
⎟
0
=
⎜
⎟
⎜
⎟
1
⎝
The
⎠
vectors
Example
Write
a
=
k
are
called
base
vectors
6 ⎞
⎜
⎟
⎜
⎝
b
j,
⎛
a
i,
Write
i
+
in
unit
vector
form.
⎟
7
5k
⎠
in
column
vector
form.
Answers
a
a
=
6i
⎛
7j
1⎞
⎜
b
b
=
⎟
Here
0
⎜
⎟
⎜
⎟
⎝
5
the
coefficient
component
is
of
the
j
zero.
⎠
Chapter
The
The
magnitude
magni tude
of
AB
of
a
is
vector
the
length
of
the
B
vector
Other
and
is
denoted
by
|AB
names
magnitude
Magnitude
is
found
for
|.
by
using
Pythagoras’
are
theorem.
modulus,
a
length,
norm
4
⎛ 3 ⎞
and
2
AB
If
=
⎜
⎝
then
⎟
4
|AB
|
=
size.
2
+ 4
3
=
=
25
5
⎠
A
3
a
2
➔
AB
If
=
In
three
=
b
ai
+
bj
then
|AB
|=
a
2
+ b
dimensions
this
becomes
a
b
➔
AB
If
=
2
=
ai
+
bj
+
c k
then
|AB
|
a
=
2
+ b
2
+ c
c
Example
When
Find
the
magnitude
of
these
5 ⎞
⎛
a
OP
=
⎜
problems
‘uniform
deal
of
acceleration’
⎟
⎜
2
b
⎟
with
vectors
3 ⎞
⎛
physicists
and
‘free
fall
under
⎟
⎜
12
⎟
⎜
1
⎝
gravity’
to
consider
magnitude
2
( −5)
|OP |=
2
+ 12
need
⎠
Answers
a
they
direction
=
169
=
the
and
of
the
13
acceleration
⎛
3
⎜
⎟
b
(
−2 )
14
=
=
3.74
⎠
fur ther
.
A
unit
vector
form.
⎛
⎛
x
=
⎜
b
⎝
y
=
⎟
3
1⎞
⎛ 0 ⎞
2 ⎞
⎜
⎜
⎝
⎠
c
⎟
7
z
⎟
1
=
⎜
⎟
⎠
⎜
⎟
1
⎝
2
Write
a
AB
Vectors
these
=
2i
in
+
column
3j
wish
to
(3 sf)
⎟
1
Write these in
a
may
2
+ 1
explore
Exercise
1
+
⎟
⎜
⎝
3
Y
ou
2
2
=
2
⎜
vector
.
⎞
b
vector
CD
=
⎠
form.
−i
+
6j
− k
c
EF
= k
this
concept
Write
3
the
vector
vectors
form
and
a,
b,
c,
column
d
and
e
vector
in
the
diagram
in
both
unit
form.
a
c
d
e
b
Find
4
the
magnitude
⎛
1⎞
⎜
⎟
⎝
3
a
b
4
Find
5
the
3
Equal,
➔
vector.
2.8
2i
+
5j
d
⎠
of
each
2i
c
+
2j
+
k
2
d
e
3
Two
vectors
are
j −
k
6
negative
− 5j
3
1
2i
b
e
4.5
vector.
4
5
each
c
magnitude
2
a
3
of
and
parallel
if
equal
they
vectors
have
the
same
direction
and
the
It
does
where
same
magnitude;
their
i,
j,
k
components
are
equal
too,
and
column
vectors
are
the
Car tesian
these
vectors
equal.
are
Consider
matter
the
so
plane
their
not
in
following:
–
they
are
still
equal.
B
Vectors
AB
and
PQ
are
pointing
in
the
same
Q
direction
(are
parallel)
and
have
the
same
If
magnitude.
Therefore
AB
=
two
equal
PQ
vectors
in
are
length
then
A
their
be
P
components
the
will
same.
⎛
Here
The
two
vectors
AB
and
MN
have
the
same
=
AB
PQ
=
⎝
magnitude
So
AB
≠
but
MN
different
directions.
2
⎜
B
⎞
⎟
5
⎠
M
.
A
N
⎛ 2 ⎞
Here
AB
=
⎜
⎝
⎟
5
⎛
and
MN
=
2 ⎞
⎜
⎝
⎠
⎟
5
and
so
AB
=
–MN
.
The
direction
vector
MN
➔
is
called
Y
ou
can
the
negative
write
AB
as
vector
–
of
a
⎠
not
is
just
impor tant,
its
length.
BA
Chapter
D
Vectors
AB ,
CD
EF
and
are
all
parallel
but
B
⎛
AB
have
different
=
2
⎜
magnitudes.
⎝
⎞
=
⎟
5
2i
+
5j
⎠
1
AB
Here
CD
=
AB
and
=
2EF
4
⎛
CD
2
=
⎜
F
⎝
⎞
=
⎟
10
4i
+
10j
⎠
A
⎛
EF
=
⎝
⎞
⎟
2.5
=
1i
+
2.5j
⎠
E
1
AB
Here
1
⎜
CD
=
and
AB
=
2EF
C
2
➔
Two
So,
vectors
AB
and
quantity
.
AB
Vectors
are
RS
This
and
parallel
are
can
if
one
parallel
also
be
if
is
a
AB
written
scalar
=
k
RS
as a
=
multiple
where
k
of
is
the
a
other.
scalar
kb
GH
B
⎛
of
29
but
different
directions.
So
AB
≠
GH
AB
=
⎝
⎛
H
GH
=
We
⎞
⎟
5
5
⎜
⎝
A
2
⎜
2i
+
5j
⎞
⎟
2
=
⎠
=
–5i
+
2j
⎠
cannot
multiply
AB
G
by
Example
The
diagram
shows
several
vectors.
a
c
e
d
b
Write
each
of
the
other
vectors
in
terms
of
the
vector
a
Answer
From
the
diagram
we
can
see
the
following:
a
=
⎛
3 ⎞
⎜
⎟
5
⎝
⎛ 1.5 ⎞
,
b
=
⎝
⎠
⎛ 3 ⎞
d
=
⎜
⎝
⎟
5
⎠
⎛
,
⎜
e
=
⎟
2.5
6
⎜
⎝
c
=
6
⎜
⎝
⎠
⎞
⎟
10
,
⎠
⎞
⎟
10
⎛
,
,
⎠
{
Vectors
Continued
on
next
page
a
scalar
to
get
GH
therefore
1
b
=
–
b
a
is
parallel
a,
to
in
the
opposite
2
direction
c
=
c
–2a
is
in
twice
d
=
d
–a
is
=
e
2a
is
the
in
with
e
For
m
what
=
3i
+
half
opposite
the
magnitude
direction
to
a,
with
magnitude
the
the
in
twice
Example
with
the
opposite
same
the
the
direction
to
a,
magnitude
same
direction
as
a,
with
magnitude
values
t j
–
of
6k
t
and
and
n
s
=
are
9i
–
these
12j
+
two
vectors
parallel?
s k
Answer
For
parallel
vectors
3i
+
t j
–
6k
=
k
3i
+
t j
–
6k
=
9ki
(9i
3
=
m
–
–
=
kn
12j
+
sk)
Multiply
12k j
+
skk
From
i
components
From
j
components
From
k
out
and
equate
coefficients.
9k
1
k
=
3
1
So t
=
–12
=
×
–4
3
1
–6
=
s
×
⇒
s
=
–18
components
3
Exercise
1
The
B
diagram
shows
several
vectors.
c
a
f
e
b
d
a
Write
the
b
2
vector
How
Which
⎛ 0
a
=
0
=
g
=
–i
a
the
or
and
vectors
7
c,
d,
e
and
f
in
terms
of
b
b
related?
vectors
b
are
parallel
⎛
1⎞
= ⎜
⎟
1⎞
⎝
⎠
7
to i
+
7j?
⎛
c
⎠
0
05 ⎞
0
03
= ⎜
⎝
⎟
⎠
10 ⎞
⎜
⎝
a
of
these
⎟
⎛
d
are
of
⎜
⎝
each
⎟
70
+
e
=
60i
+
420 j
f
= 6i
–
42j
⎠
7j
Chapter
For
3
what
a
r
=
b
a
4i
value
+
t
⎛
=
For
v
5
=
8
t i
In
the
of
each
and
are
s
=
b
=
5j
+
of
t
8k
–
two
vectors
parallel?
12j
7 ⎞
⎟
10
⎝
values
cube
14i
⎜
⎠
–
these
⎛
and
⎟
what
t
⎞
⎜
⎝
4
t j
of
and
and
⎠
s
w
OABCDEFG
are
=
the
these
5i
+
j
two
+
vectors
parallel?
s k
length
G
edge
is
one
unit.
D
Express
these
vectors
in
terms
of
i,
j
and
k
OG
a
E
O
B
6
b
BD
c
AD
d
OM
A
where
M
Repeat
question
OA
units,
=
5
Position
is
5
OC
the
midpoint
given
=
4
that
units
of
GF
OABCDEFG
and
OD
=
3
is
a
cuboid
where
units.
vectors
y
Posi tion
are
vectors
vectors
giving
the
position
15
of
a
point,
relative
to
a
fixed
origin, O
P(–5, 12)
The
point
P
with
coordinates
(–5, 12)
has
10
5 ⎞
⎛
position
OP
vector
=
⎜
⎝
=
⎟
12
−5i
+
12j
⎠
5
x
–6
➔
The
OP
point
=
⎛ x
⎞
⎜
⎟
⎝
y
P
=
The
x i
+
coordinates
(x, y)
has
position
–2
0
vector
y j
⎠
Resultant
Consider
with
–4
vectors
the
points
A(2, 3)
diagram
shows
the
and
B(6, 6).
position
vectors
of
A
and
B.
⎛ 4 ⎞
We
can
see
that
the
vector
AB
8
=
⎜
⎝
⎟
3
B
⎠
6
Remember
We
can
also
see
that
to
move
from
A
to
B
that
AB
we
4
should
A
could
describe
this
movement
as
either
from
A
to
B
or
first
going
from
A
to
a
vector
,
not
then
Vectors
from
O
to
B
as
O
coordinate
and
written
as
going
2
directly
be
O
2
4
6
8
pair
.
a
Thus
The
we
vector
Recall
and
could
AB
is
called
AO
=
–
OA,
AB
=
–
OA
that
AO
=
the
+
OB.
resultant
of
the
vectors AO
and
OB.
hence
=
➔
AB
write
To
find
we
can
vector
Find
A
the
OB
OA
vector
position
AB
between
vector
of
A
two
from
points
the
A
and
B
position
B
and
the
–
resultant
subtract
of
Example
Points
the
OB
+
B
have
vector
coordinates
(–3,
2,
0)
and
(–4,
7,
5).
AB
Answer
A(–3, 2, 0)
B(–4, 7, 5)
O
First
we
write
⎛
⎜
OA
down
the
position
vectors
OA
and
OB.
3 ⎞
⎟
2
=
⎜
⎟
⎜
⎟
0
⎝
⎠
⎛
4 ⎞
⎜
OB
=
⎟
7
⎜
⎟
⎜
⎟
5
⎝
AB
=
⎠
OB
–
OA
=
⎛
4 ⎞
⎛
3 ⎞
⎛
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎠
⎝
7
⎜
⎜
2
5
⎝
Similarly
vector
are
if
PR
given
we
know
then
each
relative
to
a
the
point
⎟
5
⎟
⎜
⎟
⎜
⎠
⎝
0
vector
of
=
1⎞
⎟
⎟
5
PQ
points
⎠
and
Q
the
and
Q
R
P.
R
Now
P
QR
=
QP
+
=
PR
–
PR
PQ
Chapter
Example
2 ⎞
⎛
0 ⎞
⎛
⎜
Given
that
=
XY
⎟
⎜
1
⎜
⎟
⎜
⎟
XZ
and
⎟
=
10
⎜
⎟
⎜
3
⎝
⎟
⎠
1
⎝
Find
the
vectors
YZ
a
⎠
ZY
b
Answers
a
YZ
XZ
=
XY
−
⎛
0 ⎞
⎛
2 ⎞
⎛
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
=
10
1
⎜
⎜
1
⎝
b
ZY
⎛
2 ⎞
⎛
⎜
⎟
⎜
=
−11
⎜
⎜
Exercise
1
P
has
Find
⎟
=
11
⎟
⎜
⎟
⎜
⎟
⎟
3
⎠
⎝
2
⎠
⎝
⎠
2 ⎞
⎟
=
11
⎟
⎜
⎟
⎜
⎠
⎝
⎟
⎟
2
⎝
2 ⎞
2
⎠
C
coordinates
the
vectors
(7,
PQ
4),
Q
has
coordinates
(2,
3).
QP
.
and
5
2
Point
A
has
position
vector
C
has
position
1
3
4
Write
these
vectors
.
bj
+
c k
c
the
vector
from
(2, –3, 5)
to
(1, 2, –1)
d
the
vector
from
(1, 2, –1)
to
(2, –3, 5)
⎛
1⎞
⎜
⎟
LN
⎟
AB
TS
⎛
1⎞
⎜
⎟
y
,
=
.
⎟
⎜
⎟
BC
3i
origin
Find
LM
⎠
+
4j
−
k
and
TU
=
2x
3
=
and
AC
i
−
1
4j
4
=
.
⎜
⎟
⎜
⎟
⎠
z
2
the
Vectors
the
3
that
⎝
Find
to
⎟
⎜
⎝
⎠
=
CB
4 ⎞
2
NM
0
Given
(1, –5, 6)
⎜
⎟
vectors:
(2, –3, 5)
⎛
and
2
⎠
form.
joining
⎝
+
column
d
vector
is
as
AC
c
ai
these
the
⎜
6
Write
b
P
⎟
3
⎝
OP
where
vector
⎜
a
⎜
5
in
position
⎠
BA
b
has
⎟
4
⎝
AB
B
vector
⎜
a
1⎞
⎛
,
2 ⎞
⎛
and
values
of
the
x
constants
x,
y
and
y
z
+
2k,
find
US.
The
following
example
demonstrates
how
to
show
that
three
points
Collinear
are
points
all
lie
collinear
in
Example
Show
2i
+
a
straight
line.
that
the
3j
k
points
and
A,
4i
B
7j
+
and
7k
C
with
position
respectively
are
vectors
i
2j
+
3k,
collinear.
Answer
Star t
AB
OB
=
+
=
OA
−
(3
−3i
(
+
2))j
5j
+
(
1
any
3)k
=
(4
=
3i
AB
OC
=
1)i
5j
2i
+
lie
The
that
3j
+
points
points,
for
example,
using
any
two
other
(7
points,
3)k
for
AC
example
A,
on
=
scalar
a
points
the
BC
parallel
that
A,
same
B
AC,
line.
we
6i
could
10j
multiple
showing
both
parallel
+
of
8k
found
which
both
that
to
have
AB
is
AB
and
a
and
AC
are
BC.
points
and
4i
A,
7j
B
+
and
7k
C
with
position
respectively
are
vectors
i
2j
+
3k,
collinear.
QUESTION
A,
B
and
C
have
coordinates
(2, 3, –3),
(5, 1, 5)
and
respectively
.
AB
a
Find
b
Show
Show
are
contain
the
k
(8, –1, 13)
3
the
D
EXAM-STYLE
2
2))j
AC
point
must
Show
of
repeat
Note
they
C
joining
4k
since
Exercise
1
(
and
common
and
7
AB
Hence
vector
AB
4k
AC
=
and,
(
+
the
OA
−
+
finding
two
Now
AC
by
(−2−1)i
that
that
A,
the
B
and
points
C
P
are
collinear.
(1, 2, 4),
P
1
(–2, 1, 4)
and
P
2
(–5, 0, 4)
are
3
collinear.
Given
that
is
P
also
collinear
with
4
x-coordinate
P
,
P
1
of
is
P
2,
find
the
y
and
and
2
P
and
the
3
z-coordinates.
4
4
The
position
vectors
of
respectively
.
Find
the
and
ratio
AB
find
the
A,
B
value
:
and
of
x
C
so
are
given
that
A,
B
by
3i
and
C
+
4j,
are
xi,
i
−
2j
collinear
BC
Chapter
Distance
➔
If
A
=
between
(x
,
y
1
,
z
1
)
two
then
a
points
OA
=
=
x
1
in
i
+
space
y
1
j
+
z
1
k
1
A
and
if
B
=
(x
,
y
2
,
z
2
)
b = OB
then
= x
2
i
+
y
2
j
+
z
2
k
2
b
AB
AO
=
–
a
OB
+
a
B
= OB
=
b
=
(x
− OA
−
b
a
O
−
x
2
)i
+
(y
1
−
y
2
)j
+
(z
1
−
z
2
)k
1
2
Distance
Example
Find
the
distance
AB
(x
=
x
2
2
)
( y
1
y
2
)
2
(z
1
z
2
)
1
vector
from
between
A(1, 3, 4)
the
two
to
B(4, 2, 7)
and
hence
determine
the
points.
Answer
OA
=
AB
=
i
+
3j
OB
−
+
4k
First
OB
and
=
4i
+
2j
+
write
position
=
(4i
=
3i
+
j
=
2j
+
+
|AB
Exercise
1
Find
the
A
|
7k)
(3 )
vector
the
EXAM-STYLE
Point
vectors.
(i
+
3j
+
4k)
2
+ ( −1)
19
=
=
2
+ (3)
4.36
(3 sf)
E
determine
2
AB
from
distance
A(–1, 5, 1)
between
QUESTION
has
position
the
⎛
5 ⎞
⎜
⎟
2
vector
⎜
⎜
to
B(4, 5, –1)
two
and
hence
points.
6
,
B
has
position
⎟
⎜
position
vector
and
6
⎠
8 ⎞
⎛
has
0
vector
⎟
4
⎝
C
as
3k
9 + 1 + 9
=
points
OA
2
Distance
the
7k
⎟
10
⎜
⎟
⎜
⎟
1
⎝
3
Show
that
If
position
the
find
two
Given
Vectors
that
ABC
vector
possible
EXAM-STYLE
4
triangle
⎠
a
is
of
values
isosceles,
a
of
point
and
calculate
(2, –3, t)
is
such
the
that
angle
CAB
|a|=
7,
t
QUESTION
a
=
xi
+
6j
−
2k
and
|a|
=
3x,
find
two
possible
values
of
x
a
a
u =
5
,
2
4
v =
2a
a
6
and
Find
Given
b
are
the
two
value
vectors
of
|a
+
b
=
2a
c
b
is
per pendicular
b
b
is
vector
find
the
|u|
=
|v|,
find
the
possible
values
of
a
and
b|
=
to
|a|
=
5.
when
−3a
a
and
|b|
=
12
vectors
uni t
To
that
a
Unit
A
.
2
a
unit
length
of
a
vector
vector
the
in
of
the
vector
a,
length
same
1
in
a
given
direction
namely
|a|,
as
and
direction.
a
vector a
then
first
multiply
find
the
1
vector
since
a
it
.
by
is
a
This
scalar
vector
will
multiple
of
be
a
in
and
the
same
will
be
direction
one
unit
long
since
it
1
is
×
➔
A
the
length
vector
of
of
the
length
original
1
in
the
vector.
direction
of
a
is
found
by
using
the
formula
Using
length
and
➔
this
k,
in
then
A
method
the
formula
Example
a
Find
can
direction
multiple
vector
we
of
this
of
by
length
also
k
a.
find
We
a
vector
would
of
first
any
find
length,
the
unit
say
vector
k.
in
the
direction
of
a
is
found
by
using
the
k
the
unit
vector
in
the
same
direction
as
the
vector
3i
+
4j
3
b
Find
a
vector
of
length
10
in
the
same
direction
as
)
(
1
Answers
2
a
The
vector
3i
+
4j
has
length
3
2
+ 4
=
25
=
5
1
Therefore
a
vector
of
length
1
will
be
3
(3i + 4 j)
=
5
⎛
b
The
vector
3
⎜
⎝
1
has
length
10
.
The
vector
⎠
the
vector
of
length
10
is
This
10
can
3
10
1
be
simplified
10
10
=
3
10
10
if
3 ⎞
⎝
⎟
1
has
length
1.
⎠
3 ⎞
⎟
1
required:
1
⎛
⎜
10
⎛
j
5
⎜
10
10
Therefore
5
⎞
⎟
1
4
i +
=
10
3
1
Chapter
Exercise
F
3
1
Show
that
4
i
5
1
2
Show
that
2
i
3
3
Find
a
unit
is
j
a
unit
vector.
5
2
j
3
k
is
a
unit
vector.
3
vector
parallel
to
4i
–
3j
⎛
1⎞
⎜
4
Find
a
unit
vector
parallel
to
the
vector
⎟
5
⎜
⎟
⎜
⎟
4
⎝
5
Find
a
unit
vector
in
the
direction
of
the
⎠
vector
between
the
Show
points
P
(1, 0, 1)
and P
1
that
magnitude
6
ai + 2aj is
7
Find
a
1
unit
vector
EXAM-STYLE
of
long.
Given
magnitude
that
5
a
that
>
is
0,
find
the
parallel
to
value
2i
⎛
QUESTION
–
Find
a
vector
of
magnitude
7
in
the
direction
of
of
a
j
−1 ⎞
⎜
8
⎟
−3
⎜
⎟
⎜
⎟
2
⎝
9
Find
a
unit
vector
in
the
b
2 sin
Addition
Addition
of
direction
as
1
and
tan
subtraction
of
we
vectors
vectors
⎛ 3 ⎞
⎛ 5 ⎞
Suppose
have
two
vectors
u =
⎜
⎝
⎟
0
and
v =
⎜
⎝
⎠
⎟
4
⎠
u
v
Now
u + v
along
is
vector
Vectors
inter preted
u’
followed
⎠
.
same
2 cos
a
the
(3, 2, 0).
2
geometrically
by
‘move
as
along
first
‘move
vector
v’.
is
1.
➔
The
resultant
formed
head
when
to
u
vector,
u
and
u +
v
v,
are
is
the
placed
third
next
side
to
of
each
the
triangle
other
tail.
+
v
v
u
Notice
also
that
vector
addition
is
commutative
The
since
u +
v =
v
+
u.
This
gives
rise
to
the
word
‘commute’
exchange
of
vector
means
to
parallelogram
or
switch.
addition.
In
mathematics,
the
commutative
u
proper ty
u
+
order
means
without
you
can
affecting
switch
the
the
outcome.
v
By
⎛ 8 ⎞
The
considering
calculations,
u
resultant
vector
u +
v
in
this
case
is
⎜
⎝
⎟
4
subtraction,
the
following
which
of
addition,
multiplication
and
division
.
would
you
say
are
commutative
⎠
operations?
Notice
that
adding
the
this
can
easily
be
found
arithmetically
by
10
corresponding
⎛ 5 ⎞
u
+
v =
⎜
⎝
⎛ 3 ⎞
+
⎟
0
⎜
⎠
⎝
⎟
4
Subtraction
components
⎛ 5 + 3 ⎞
=
⎜
⎠
of
⎝
⎟
0 + 4
⎛ 8 ⎞
=
⎜
⎠
⎝
consider
the
⎠
–
by
v
is
inter preted
‘move
along
and
5
+
10
10
÷
10
–
10
×
5
5
and
and
5
5
÷
–
10
10
5
and
5
×
10
vectors
two
vectors
u =
⎜
⎝
u
5
⎟
4
⎛ 5 ⎞
Again
+
together.
geometrically
negative
v’
or
u
as
+
⎛ 3 ⎞
and
⎟
0
v =
⎜
⎠
⎝
‘move
along
⎟
4
⎠
vector u’
followed
(–v).
u
u
u
–
v
–
resultant
vector
here
is
u
–
v
and
in
this
specific
case
subtracting
that
the
u
we
can
easily
corresponding
find
this
arithmetically
by
Subtraction
⎛ 5 − 3 ⎞
⎛ 3 ⎞
⎛
v
= ⎜
⎝
➔
⎟
0
−
⎠
Vectors
⎜
⎝
⎟
4
are
⎠
=
⎜
⎝
⎟
0 − 4
⎠
subtracted
=
–
u
⎝
by
not
2⎞
=
≠
⎜
2 ⎞
⎜
is
commutative.
components.
⎝
−
v)
⎠
⎛
u
(–
⎟
4
v
⎛ 5 ⎞
+
.
⎝
again
=
2 ⎞
is
⎜
Notice
v
v
⎛
The
–
u
–
v
⎟
4
⎠
⎟
−4
⎠
adding
a
negative
vector.
Chapter
Q
The
zero
Consider
PQ
a
+
the
QR
return
vector
+
to
triangle
RP
the
must
PQR
be
starting
equal
point.
to
zero
This
is
as
the
written
overall
as PQ
journey
+
QR
+
results
RP
=
in
0
P
R
The
zero
vector
is
in
bold
type
to
indicate
that
it
is
a
Equi li brium
vector
.
name
⎜
⎜
⎟
=
in
⎝
0
the
two
dimensions
and
⎜
⎠
where
⎟
0
in
three
a
are
in
–
their
resultant
Y
ou
explore
may
the
a
+
a
=
b
2i
–
b
b
3j
–
+
3k
and
a
b
=
2b
c
4i
–
–
2j
k,
find
the
vectors:
3a
Answers
a
b
a
b
+
–
2b
c
b
a
–
=
(2
+
=
6i
–
=
(4
–
=
2i
+
3a
=
1
Given
these
2
4)i
5j
j
2i
–
+
+
+
2)i
(2(4)
=
Exercise
(–3
+
(–2))j
+
(3
+
(–1))k
2k
+
(–2
–
(–3))j
+
(–1
–
3)k
4k
–
5j
3(2))i
–
+
(2(–2)
–
3(–3))j
+
(2(–1)
–
3(3))k
11k
G
that
a = 2i
–
j, b = 3i
+
2j, c = –i
+
j
and d = 3i
a
+
b
d
a
+
b
Given
+
a =
d
⎛
2 ⎞
⎜
⎟
⎝
−3
b
b
+
e
a
–
c
b
c
c
+
a
f
d
–
⎛ −4 ⎞
,
b =
⎜
⎝
⎠
⎟
5
b
+
a
⎛ −5 ⎞
and
c =
⎠
⎜
⎝
⎟
−3
,
find
these
⎠
+
b
b
b
–
c
(a
c
2
d
a
+
Vectors
find
d
1
a
+3j,
vectors.
a
is
wish
3b
–
c
e
3c – 2b
+
5a
+
c)
vectors.
to
concept
equilibrium
that
balance
⎟
zero.
Example
of
dimensions.
0
a
state
number
⎟
forces
⎜
Given
the
⎛ 0 ⎞
⎛ 0 ⎞
0
for
is
fur ther
.
of
Given
3
that
a = 3i
–
j – 2k and b = 5i
– k,
The
find
these
method
combined
a
a
+
c
2a
b
–
b
b
b
–
2a
d
4(a
–
forces
b)
+
2(b
+
c)
by
the
vectors
p =
3i –
5j and
q
=
adding
–i
+
since
the
vectors
x,
y
–
3p
=
q
4p
b
–
3y
=
7q
2p
c
+
z
=
x
⎛
The
vectors
b
a and
6
⎛
and
y
b
are
such
that
a
=
⎜
⎝
⎞
⎞
⎜
+
y
in
− 3
⎠
that
a
=
in
b,
find
the
values
of
x
and
his
⎜
vectors
BCE).
the
polymath
Greek
Aristotle
Dutch
Mathematician
(1548–1620)
treatise
the
Principles
which
led
development
was
not
Caspar
The
of
the
been
of
used
the
it
to
a
Ar t
of
of
breakthrough
a and
b
are
such
⎛ t
and
⎜
⎟
⎜
⎟
u
⎝
s ⎞
⎜
⎟
t
that
until
around
Mechanics.
1800
that
y
⎛ 3 ⎞
6
time
called
has
⎠
It
Given
and
Stevin
Weighing
⎟
−2 x
is
and
more
⎟
x
=
⎝
the
or
0
Simon
5
them
law
the
two
z where
and
(384–322
2x
a
of
4j,
philosopher
find
calculating
action
parallelogram
known
Given
4
of
vectors.
(Danish-Norwegian,
and
Jean-Rober t
Argand
⎟
3s
(Swiss,
⎜
⎟
⎜
⎟
⎝
⎠
Wessel
1745–1818)
t
+
s
1768–1822)
formalize
the
general
star ted
to
concept
of
a
⎠
‘vector’.
Given
that
3a
=
Geometrical
When
you
addition,
are
2b,
Example
the
values
of
s,
t
and
u
proofs
not
given
subtraction
geometrical
find
specific
and
scalar
vectors
you
multiples
to
can
still
deduce
use
vector
some
results.
In
triangle
OXY,
of
OX,
and
OY
A,
XY
B
and
C
are
respectively
,
the
midpoints
OX
=
x
and
OY
=
y
X
OA
a
,
OB
,
XY ,
OC
and
x
CO
in
terms
of
x
and
y
A
C
b
Find
x
an
and
expression
y.
between
What
the
is
line
for
the
XY
AB
in
terms
of
relationship
and
the
line
O
AB?
Y
y
B
2
c
P
is
the
point
such
that
OP
=
OX
XP
+
.
Find
OP
3
d
What
can
you
conclude
about
the
position
of
P?
Answers
1
1
a
OA
OX
=
2
1
OB
2
XY
=
Use
XO
infor mation
from
the
diagram.
1
OY
=
x
=
2
y
=
2
+
OY
=
−x
+
y
=
y
−
x
Use
{
vector
addition.
Continued
on
next
page
Chapter
1
OC
=
OX
XC
+
=
x
Use
XY
+
vector
addition.
2
From
the
diagram,
1
=
x
+
( y
−
1
x)
2
XC
XY.
=
2
1
1
=
x
+
y
x
2
2
1
1
=
x
+
1
y
=
(x
2
2
+
y)
2
1
CO
=
−OC
=
(x
+
y)
2
1
1
AB
b
=
AO
OB
+
=
x
+
AO
y
=
OA
2
2
1
=
( y
−
x)
2
1
Since
XY
=
y
−
x
AB
and
=
( y
−
x)
2
then
XY
the
and
The
line
in
lines
AB
the
are
is
half
same
the
length
direction
therefore
as
of
XY.
parallel.
2
OP
c
=
OX
XB
+
3
2
=
x
+
(XO
+
OB
Use
)
vector
addition.
3
XO
2
=
x
+
+
OX
y)
3
2
1
=
=
1
(−x
1
x
+
y
3
3
1
=
(x
+
y)
OP : OC
=
2 : 3
3
So
2
P
d
lies
of
the
way
along
the
line
OC
3
Exercise
1
In
this
of
PQ
H
triangle
and
a
=
OA
OA,
=
b
AP,
=
BQ
=
3OB,
N
is
the
P
midpoint
OB.
A
N
Show
that
a
a
AP
=
a
c
PQ
=
4b
e
ON
=
a
b
AB
=
b
d
PN
=
2b
f
AN
=
2b
−
a
O
Vectors
−
+
2a
2b
−
a
b
B
Q
A
2
In
this
Show
triangle
a
=
OA
,
b
OB
=
and
AC : CB
=
3 :1.
that
a
3
AB
a
=
b
−
a
AC
b
=
(b
−
a)
C
4
1
1
CB
c
=
(b
−
a)
OC
d
=
OABC
is
a
+
4
4
3
4
a
OA
trapezium.
=
a,
OC
=
O
b
B
b
4
c,
and
a
A
O
CB
=
3a.
D
is
the
midpoint
of
AB.
c
D
Show
that
OB
a
=
c
+
3a
AB
b
=
c
+
2a
B
C
3a
1
OD
c
=
2a
+
1
c
OC
d
=
2a
−
c
2
2
A
4
ABCDEF
is
a
regular
hexagon
with
center O.
FA
=
a
and
FB
=
b
a
Express
a
each
of
these
in
terms
of
a
and/or
b
F
i
AB
ii
FO
iv
BC
v
FD
iii
B
FC
b
O
What
b
geometrical
facts
can
you
deduce
about
the
lines
AB
E
and
Using
c
C
FC?
vectors,
determine
whether
FD
and
AC
are
parallel.
D
5
In
the
diagram
OA
=
a
and
OB
=
b.
M
is
the
A
midpoint
of
OA
and
P
lies
on
AB
such
that
a
2
AP
AB
=
M
3
Show
P
that
2
a
AB
=
b
−
a
and
AP
=
X
O
(b
−
B
b
a)
3
MA
=
a
and
MP
=
If
d
Prove
X
is
a
b
−
a
6
3
2
c
1
2
1
b
point
that
such
MPX
is
that
a
OB
=
straight
BX,
show
that
MX
=
2b
−
a
line.
Chapter
.
We
Scalar
often
solving
need
product
to
find
the
θ
between
two
vectors
when
problems.
Investigation
Consider
and
angle
two
=
OB
b
–
vectors
=
5i
+
going
to
use
cosine
=
OA
a
=
3i
rule
+
4j
12j
B
b
A
a
O
Y
ou
are
F ind
the
vector
F ind
the
lengths
Recall
the
the
cosine
to
nd
the
θ
angle
between
the
two
vectors.
AB
of
cosine
OA,
r ule
2
OB
and
2
| OA |
rule
+ | OB
and
AB
apply
(|OA |,
it
to
|OB |
this
and
|AB |).
situation
2
|
|
AB
|
=
cos θ
2 | OA | × | OB
|
2
2
⎛ | OA |
2
+| OB |
⎞
| AB |
−1
θ
F ind
by
nding
cos
⎜
⎟
2 | OA |× | OB |
⎝
Y
ou
should
Now
nd
repeat
that
this
θ
=
⎠
14.3º.
process
using
=
OA
a
=
a
i
+
a
1
OB
=
b
=
b
i
+
b
1
At
step
3
it
and
j
2
is
a
possible
b
1
cos θ
j
2
+
a
1
to
simplify
the
expression
you
obtain
to
b
2
2
=
| a | | b |
Or
a
alternatively
b
1
a
b
1
+
1
a
b
2
,
is
+
a
1
called
the
=| a || b | cos
b
2
2
scalar
product
of
the
two
vectors
a
=
a
2
i
+
a
1
j
2
The
and
b
=
b
+
i
b
1
It
can
be
2
found
coefficients
of
k
➔
of
together
Scalar
If
a
=
j
also
by
multiplying
together
and
then
and
the
(if
adding
coefficients
in
three
them
all
of
i
together,
dimensions)
the
the
i
product
+
a
1
j
known
and
b
=
b
i
+
1
b
j
if
a
=
a
i
+
a
1
j
2
+
then
a · b
=
a
2
a
k
3
and
b
1
+
1
a
scalar
b
2
=
a
b
1
Vectors
+
1
a
b
2
+
2
a
b
3
.
3
dot
that
the
product
b
=
b
i
1
+
b
j
2
+
b
is
2
k
then
3
a · b
a · b
the
coefficients
Notice
2
as
product.
commutative,
Similarly
is
up.
product
a
scalar
j
=
b · a
that
is
➔
The
scalar
between
Example
If
a
=
i
+
product
the
a
b
=
|a||b|cos θ
where
θ
is
the
angle
vectors.
4j
–
2k
and
b
=
2i
+
4j
+
6k,
find
a
b
Answer
a
b
=
=
(1
2
×
+
2)
+
16
(4
–
×
12
4)
=
+
(–2
×
The
6)
not
6
result
a
is
a
scalar
number,
vector.
Y
ou
can
GDC
to
product
Finding
If
you
you
do
can
the
not
angle
know
between
the
angle
θ
between
|a||b|cosθ
=
or
cosθ
θ
find
rather
Example
Find
b
=
the
5i
of
the
two
your
scalar
vectors.
vectors
two
vectors
a
and
b
then
+
than
resor ting
b
=
a
to
nd
use
use
a
a · b
two
also
to
b
the
full
cosine
r ule
each
time.
angle
between
a
and
b
given
that
a
=
3i
+
4j
and
12j.
Answer
Using
a · b
a · b
=
3
×
|a|
=
5,
|a|
|b|
=
5
|a||b|cosθ,
+
|b|
cosθ
⇒
63
4
=
×
12
=
63
13
=
5
×
13
×
=
65cosθ
=
65cosθ
cosθ
63
cosθ
=
65
63
1
θ
=
cos
65
o
=
14.25
Chapter
Special
properties
Perpendicular
An
impor tant
of
the
scalar
product
vectors
fact
is
that
two
vectors
are
per pendicular
if
and
only
Perpendicular
if
their
scalar
product
is
are
This
is
because
θ
if
=
vectors
zero.
90º
also
called
then
or thogonal.
o
a · b
=
|a||b|
cos 90
=
|a||b|
×
0
Notice
=
i,
0
j
that
and
k
since
are
all
perpendicular
➔
For
vectors
perpendicular
a · b
=
0
i · j
=
j · k
Parallel
If
two
j · i
=
=
k · j
i · k
=
=
k · i
=
0
vectors
vectors
a
and
b
are
parallel
then
o
a · b
➔
=
|a||b|
=
|a||b|
For
parallel
Coincident
cos 0
vectors
a · b
=
|a|
|b|
vectors
Since
Given
a
vector
i
and
j
and
all
one
unit
in
i · i
=
k
are
a
length
o
a · a
=
|a||a|
cos 0
j · j
=
k · k
=
1
2
=
a
2
➔
For
coincident
vectors
a · a
=
a
In
1686
Newton
published
Philosophiae
Exercise
I
Naturalis
1
Given
that
a
a · b
b
b · c
a
=
2i
+
4j,
b
=
i
–
5j
and
c
=
–5i
–
2j,
find
Mathematica,
he
detailed
laws
of
a · a
d
c · (a
applying
+
+
to
two
Given
that
u
⎛
1⎞
⎛
4 ⎞
⎛
⎜
⎟
⎜
⎟
⎜
0
=
⎜
,
v
3
=
⎟
⎜
⎜
⎟
⎜
⎠
⎝
5
⎝
know
we
how
to
and
w
⎟
⎜
⎟
⎜
⎠
⎝
,
b
u · (v
d
2u · w
e
(u
–
resultant
to
of
nd
forces
⎟
⎠
w)
c
u · v
–
are
v) · (u
+
perpendicular
.
may
wish
to
u · w
explore
–
and
find
⎟
Y
ou
u · v
into
6
1
a
force
directions
⎟
3
=
a
perpendicular
that
w)
fur ther
.
Vectors
these
In
and
1⎞
the
motion.
a) · b
resolve
2
which
b)
need
(c
in
three
understanding
c
e
Principia
these
laws
3
Determine
parallel
or
whether
these
pairs
of
vectors
are
per pendicular,
neither.
⎛ 1 ⎞
⎛ 2 ⎞
a
a
=
2i
+
4j
and
b
=
4i
–
2j
c
b
=
⎜
⎝
c
u
⎛
8 ⎞
⎛
⎜
⎟
⎜
2
=
⎜
⎟
⎜
and
v
⎜
⎟
⎟
⎜
⎟
⎠
⎝
0
⎜
OZ
=
⎟
⎜
⎝
4
Find
5
Given
find
6
⎜
+
⎟
2
=
Find
2
1
c
2i
one
and
EXAM-STYLE
8
9
Consider
a
AB
b
AB
c
the
Find
the
1⎞
⎛
⎟
⎜
c
2i
and
b
=
3i
–
2j
–
k
=
2i
–
8j
and
m
=
–i
+
4j
1⎞
⎜
if
5k,
⎟
1
such
=
a
b
5
i
2i
+
+
a · d
the
j
+
7j
=
2k
and
–9,
vectors
a
c
=
b · d
=
and
b
b
i
=
+
j
11
if
3i
+
+
k,
and
|a|
2j
c · d
=
–
k
=
6.
3,
these
vectors,
giving
your
answer
in
place.
⎛ 4 ⎞
⎜
b
⎠
⎝
–
and
6
between
2i
⎠
=
=
that
⎟
0
3 ⎞
⎛
and
⎠
⎜
⎝
⎟
1
⎠
5j
points
A(2,4),
of
2
the
angle
between
B(1,9)
and
⎜
⎟
⎜
between
these
⎛ 2 ⎞
⎟
⎜
⎜
C(3,2).
⎟
⎜
6
⎝
⎠
–7j
+
k
Find
⎝
i
+
j
–
AC
vectors.
4
⎞
⎟
2
⎜
⎟
⎜
⎟
2
1
⎠
and
⎟
and
⎜
and
⎟
of
⎛
⎟
3
b
⎟
AB
pairs
⎞
3
and
⎟
2
⎝
–
b)
⎟
angles
⎜
⎜
k
AC
·
cosine
2
+
AC
and
the
⎜
2j
QUESTIONS
⎛
a
–
decimal
⎜
⎝
5j
n
–
⎠
⎛ 2 ⎞
and
⎠
+
=
between
angles
to
⎟
⎝
d
a · b
⎞
⎜
3i
angle
the
2
⎛
a
=
and
degrees
f
3i
⎟
⎝
vector
the
CD
⎠
a
⎟
⎛
and
3b) · (2a
that
the
Find
|b|
7
(a
=
⎟
⎝
⎠
⎝
a
1
⎛ 2 ⎞
=
d
0
⎜
⎜
⎟
0
AB
⎠
⎛ 0 ⎞
and
⎜
g
⎟
2
⎠
⎟
=
⎜
⎝
⎠
1
⎜
OX
=
⎟
1
=
⎛ 1 ⎞
e
1
d
4 ⎞
2
⎝
and
⎟
⎠
⎝
⎠
k
Chapter
EXAM-STYLE
QUESTION
1
⎛
⎟
⎜
10
Points
A,
⎛ 2
⎟
⎜
⎜
2 ⎞
⎜
⎟
⎟
⎜
⎟
⎜
⎠
⎝
C
form
a
triangle.
Their
position
vectors
1
are
⎟
⎜
⎟
4
respectively
.
⎠
Find
⎟
⎟
1
⎠
a
the
lengths
b
the
exact
c
the
area
of
the
value
of
the
sides
of
the
AB
and
cosine
AC
of
the
angle BAC
triangle.
⎛ 1⎞
⎜
11
Find
the
angle
between
⎟
and
1
⎜
the
x-axis.
⎟
⎜
⎟
1
⎝
EXAM-STYLE
12
The
position
13
a
Show
b
Find
λ
Find
vectors
relative
that
the
if
⎠
QUESTION
respectively
,
OA
A
and
to
and
length
the
OB
of
vectors
an
B
are
origin
are
4i
+
4j
–
4k
and
i
+
2j
+
3k
O
per pendicular.
AB.
2i
+
λ j
+
k
and
i
–
2j
+
3k
are
per pendicular.
EXAM-STYLE
14
Let
a
=
5i
QUESTIONS
– 3 j
per pendicular
15
Let
a
+
7k,
to
b
a
p ⎞
⎛
⎟
⎜
2
and
b
⎟
⎜
⎜
⎟
⎜
value
that
+
λk.
Find
λ
such
that
a
+
b
is
2 ⎞
⎟
⎟
.
⎟
⎝
⎠
of
Vector
Suppose
j
3
p
.
+
p
=
⎜
the
i
b
⎜
⎝
Find
=
⎛
=
a
p
⎠
such
that
a
equation
straight
line
+
b
of
passes
and
a
a
b
are
per pendicular.
line
through
a
b
A
point
that
A
where
the
line
is
A
has
a
parallel
position
to
a
vector
vector
a
and
b
a
Now
if
we
let
R
be
any
point
on
the
line,
0
then
AR
is
Any
point
parallel
R
on
the
to
b
line
L
can
be
found
by
b
A
star ting
vector
a
at
to
the
origin
reach
the
then
line.
moving
Now
through
there
must
the
R
be
a
some
number
t
such
that
AR
=
t b
r
0
Hence
r
=
Vectors
OR
,
⎜
⎝
1
4
⎝
and
⎛
and
3
⎜
B
=
OA
+
AR
=
a
+
tb
➔
The
is
vector
the
general
given
Example
a
Find
to
Find
c
Find
and
vector
a
vector
of
a
parallel
vector
line
is
of
point
to
the
given
a
by
point
on
the
line.
t
r
on
line
is
=
a
+
the
t b
where
line, a
and b
called
is
the
is
r
a
a
parameter.
the
the
b
position
position
direction
of
equation
vector
vector
the
–i
equation
+
vector
3j
–
of
the
line
through
(1, –1, 3)
of
the
line
through
the
and
parallel
k
equation
points
A(1, 0, –4)
B(–2, 1, 1).
the
acute
angle
between
these
two
lines.
Answers
a
a
=
i
The
r
b
=
–
j
+
3k
vector
(i
OA
j
+
b
=
equation
3k)
+
–i
+
+
3j
–
1
and
OB
=
1
OB
Write
OA
–
r
⎛
1⎞
⎛
⎜
⎟
⎜
0
=
equation
⎜
⎜
+
AB
of
of
the
line
the
⎟
⎜
⎠
⎝
⎠
1
vectors
3
are
3
1
and
To
5
find
a
b
=
+
3
between
×
35
×
the
these
1
=
cos θ
=
35
11
two
angle
1
+
–1
×
lines
between
cos θ
direction
vectors.
5
r
the
=
a
equation
+
t b,
direction
⇒
angle
|a||b|cos
In
11
the
their
=
line.
⎟
find
–3
direction
⎟
1
×
the
5
–1
in
1
⎜
direction
Using
is
⎟
t
⎟
The
B.
is
c
and
3 ⎞
4
⎝
A
5
the
position
3
Hence
of
1
=
the
vectors
1
=
k
k)
AB
–
2
4
3j
is
t (–i
0
=
and
b
is
the
vector.
cos θ
1
11
35
1
–1
θ
=
cos
=
11
35
87.1°
Chapter
Exercise
1
Find
J
the
through
equation
point
B
a
=
⎜
⎝
b
a
=
2
c
a
=
⎜
5 ⎞
⎜
⎟
2
=
⎛
3 ⎞
⎛
⎜
⎟
⎜
b
⎟
⎜
d
2
a
Find
the
3
=
a
⎟
⎜
⎟
⎟
2
⎠
–
j
⎝
+
(4, 5)
c
(3, 5, 2)
an
passing
k
b
a
=
and
b
a
=
and
of
2
point
=
5 ⎞
⎜
⎟
2
3
=
⎜
⎠
⎜
b
⎜
⎟
⎜
⎠
⎝
–
line
B
+
r
(5, –2)
r
(–3, 5, 1)
=
r
(2, 1, 1)
p
=
⎜
=
⎟
1
⎜
given
Vectors
b
as
and
given.
t
⎠
=
⎟
1
⎜
⎝
r
=
t
–2i +
q
so
⎛
⎜
5
3j +
that
given
line.
⎟
3
⎝
⎟
of
the
4 ⎞
1⎞
−
on
+
⎠
⎜
⎜
2i
lies
⎟
2
⎛
+
⎠
⎜
point
⎛ 1 ⎞
+
⎛
equation
and
vector
vector a
5k
t
⎠
1⎞
⎟
0
⎟
⎜
⎟
⎜
⎠
⎝
3
direction
position
to
⎠
the
⎝
Find
per pendicular
(1, –1, 0)
⎠
⎜
the
and
⎟
whether
(4, 5)
Find
(0, 0, 1)
⎟
b
⎝
5
d
(5, –2)
⎠
⎛ 5 ⎞
d
and
⎟
4k
⎝
c
(4, –2)
with
⎛ 2 ⎞
b
b
1
3j
Determine
a
through
2
=
⎟
1
⎝
4
passes
⎛ 4 ⎞
0
i
which
⎟
0
⎝
⎟
=
=
line
1⎞
⎛
b
⎞
⎜
a
the
⎟
6
⎝
⎛
⎜
d
k
1⎞
⎜
⎠
⎜
the
⎛
b
⎟
⎛
a
of
–
(2, –4, 5)
through
⎝
c
2j
(3, –2)
equation
⎜
⎝
=
equation
⎛ 3 ⎞
a
⎠
points.
a
Find
⎠
⎜
vector
two
given.
1
=
⎟
3i
as
passing
3 ⎞
8
⎝
b
and
⎟
0
⎝
2
vector
vector a
⎠
⎜
⎠
⎜
position
to
1⎞
⎛
b
parallel
⎟
2
⎝
⎠
⎛
⎝
=
line
1⎞
⎛
b
⎟
the
with
⎛ 3 ⎞
a
of
⎟
⎟
2
j
−
the
3k
⎠
+
line
t (−2j
−3k)
through
the
point
(2,
4,
5)
8k
the
point
(p, 10, q)
lies
on
this
line.
in
the
6
Find
the
point
7
Are
vector
the
parallel
lines
r
or
=
per pendicular,
⎜
+
⎟
1
4
⎝
s
⎠
r
=
⎜
+
⎟
1
5
⎝
s
⎠
r
=
⎜
2 ⎞
⎜
⎟
1
4 ⎞
⎟
s
1
⎠
⎜
=
⎜
s
1
⎠
⎜
⎜
⎝
Find
+
⎟
7
the
s
⎠
9 ⎞
⎜
⎟
10
⎜
⎝
angle
=
⎟
1
⎜
⎝
equations
coincident,
6 ⎞
⎜
⎟
3
⎝
⎠
⎛ 1 ⎞
+
t
⎠
⎟
t
⎠
⎜
⎝
⎟
2
⎠
8 ⎞
⎛
+
t
3
⎠
⎜
⎟
=
⎜
⎟
6
⎝
⎛ 2 ⎞
r
⎟
3
⎠
⎛ 1⎞
+
t
=
⎜
⎠
⎝
⎟
1
⎜
⎟
6
and
r
⎛
4 ⎞
⎛
⎜
⎟
⎜
7
⎜
⎟
⎜
⎝
⎝
⎠
EXAM-STYLE
points
1⎞
⎟
+ t
3
⎟
⎜
⎟
⎜
⎟
⎟
2
2
⎟
=
1
r
lines.
1
4
⎟
⎠
of
2 ⎞
⎜
⎟
3
2
10
=
and
⎠
⎜
⎝
⎜
0
t
⎜
1
⎛ 4 ⎞
+
pairs
⎛
t
⎝
⎠
these
0
⎠
2
4
2
⎛ 5 ⎞
r
between
1
⎝
⎠
4 ⎞
1
3
1
⎠
⎝
⎠
QUESTIONS
A
and
B
have
coordinates
(–2, –3, –4)
⎛
1⎞
⎜
respectively
.
The
line
l
has
equation
r
⎜
⎜
and
⎜
+
t
⎝
that
point
A
lies
on
⎟
2
⎟
⎜
⎟
⎜
2
Show
(–6, –7, –2)
⎛ 1 ⎞
⎟
1
=
1
a
the
these?
⎛
+
⎛ 5 ⎞
r
⎠
⎜
⎝
The
⎛
⎛ 2 ⎞
=
⎠
⎛
=
=
2
2
9
of
2
⎟
⎝
⎛ 5 ⎞
=
r
through
2
1
b
none
⎛ 1 ⎞
+
⎟
⎝
r
or
=
r
⎟
3
⎝
⎛ 2 ⎞
a
vector
⎝
1
8
passing
2
r
r
line
these
4 ⎞
⎛
+
⎟
⎝
e
ver tical
by
⎠
⎜
2
r
1
d
a
2
⎛
⎝
⎛ 5 ⎞
c
⎛
⎝
⎛ 2 ⎞
b
of
represented
⎛ 3 ⎞
a
equation
(–6, 5).
⎟
⎟
6
⎠
⎝
⎠
l
1
b
Show
that
AB
is
per pendicular
to
l
1
10
The
OC
figure
=
5 m
shows
and
a
OD
cuboid
=
in
which OA
=
G
2 m,
F
3 m.
5 m
Take
in
O
the
as
the
origin
direction
OA,
and
OC
unit
and
vectors
OD
i,
j
and
k
respectively
.
D
C
B
E
a
Express
these
vectors
in
terms
of
the
unit
3 m
vectors.
O
2 m
i
b
iii
|OF
Find
Hence
the
|
ii
the
find
A
AG
ii
Calculate
i
c
OF
value
|AG
scalar
the
of
|
product
angle
of
between
OF
the
and
AG
diagonals OF and
AG
Chapter
Relative
11
vectors
to
i
a
+
fixed
5j
–
Find
the
vector
b
Find
the
cosine
c
Show
that,
A
(1
and
Find
e
Hence
to
the
work
are
out
of
for
all
7)i
+
value
find
3j
points
+
angle
A
and
B
have
position
respectively
.
OAB
values
(5
–
of
the
μ
of
8)j
for
point
point
+
,
the
(–2
which
on
of
given
the
vector
where
the
lines
Example
the
point P
+
8)k
OP
foot
two
with
lies
on
position
the
line
through
is
of
per pendicular
the
to
AB
per pendicular
from
vectors
equations
lines
of
two
different
lines,
you
can
cross.
3
0
Two
6k
AB
Intersection
you
the
–
B
d
O
+
O,
8i
AB
a
vector
If
point
2k and
have
equations
r
=
1
+
s
6
1
and
1
r
1
1
2
=
2
0
+
t
0
4
.
8
In
three
two
Show
that
the
lines
intersect
and
find
the
coordinates
of
the
point
dimensions,
lines
will
either
of
1
intersection.
intersect
the
value
–
of
if
the
Answer
variables
Two
vectors
their
corresponding
are
equal
if
and
r
r
1
components
are
equal.
is
intersect
if
there
consistent
2
a
value
of
t
and
x
r
3
y
=
1
=
0
+
s
3
+
value
s
of
s
such
⇒
y
=
z
=
s
x
6
1
=
1
1
+
be
parallel
2
will
s
have
vectors
1
2
+
multiples
+
y
=
2
+
z
=
8t
4t
8
be
skew
lines
are
=
–
s
=
6
(1)
2
+
4t
(2)
+
s
=
Equate
and
8t
(3)
(1)
gives
s
=
the
components
solve
the
are
not
resulting
simultaneous
so
equations.
the
Equation
the
not
and
consistent
–1
if
values
s
of
other
parallel
3
are
6
3
0
=
each
⇒
z
direction
that
x
4
t
they
0
=
y
=
2
–
r
scalar
r
all
equations
that
2
r
1
z
=
1
in
a
three
x
is
lines
do
not
3
intersect.
Substituting
3
=
2
+
4t
s
=
3
so
into
t
equation
(2):
A
=
Q
4
B
Substituting
–
1
+
3
=
s
=
3
8t
into
so
equation
t
(3):
=
4
Since
the
value
of
s
and
the
O
value
of
t
AB
consistent
for
all
three
equations
the
must
intersect.
{
Vectors
and
PQ
are
skew
two
–
lines
P
are
Continued
on
next
page
they
never
meet.
Substituting
s
=
3
into
r
To
:
find
the
point
of
1
r
⎛ x
⎞
⎛
3 ⎞
⎜
⎟
⎜
⎟
y
=
⎟
⎜
⎜
0
=
⎜
1
⎟
⎜
⎠
⎝
3
+
⎜
⎟
value
1
⎟
⎜
⎠
⎝
y
z
=
=
=
of
+
3
=
6
0
+
3
=
3
+
Therefore
of
=
the
⎟
⎛ x
⎞
⎛ 6 ⎞
⎜
⎟
⎜
y
2
2
=
are
(6,
⎜
⎟
⎜
⎟
⎜
⎠
⎝
give
of
the
of
in
intersection.
3,
of
the
point
2).
⎜
⎟
4
⎜
⎟
⎟
⎜
⎟
⎠
⎝
Alter natively
we
could
4
8
0
z
⎝
substitute
⎠
into
x
=
=
2
the
+
4
⎜
⎛
0
+
8
=
3
This
gives
the
1
and
is
a
⎞
=
⎟
4
2
useful
way
of
checking
⎠
the
Exercise
same
⎠
⎜
⎝
Find
t
⎞
⎟
4
coordinates
=
of
2
1
⎝
z
value
r
6
⎛
y
1
to
⎛ 0 ⎞
1
+
⎟
r
vector
2
⎟
⎜
into
position
⎠
coordinates
intersection
=
r
3
s
1
the
1
3
–1
the
⎟
point
x
substitute
⎟
⎜
1
z
⎝
intersection
⎛ 1⎞
answer.
K
the
coordinates
of
the
point
where r
=
4i
+
2j
+
λ(2i
–
4j)
1
intercepts
=
r
11i
+
16j +
μ(i +
2j).
2
2
The
vector
equations
of
two
lines
are
given
by r
=
⎛
4 ⎞
⎜
⎟
1
⎝
⎛
and
r
=
⎜
2
⎝
position
6 ⎞
3
The
line
+
l
t
⎠
⎟ .
⎜
⎝
vector
EXAM-STYLE
⎠
s
⎜
⎝
⎟
2
⎠
⎛ 9 ⎞
⎟
3
2
⎛ 8 ⎞
+
of
6
The
lines
intersect
at
the
point
P.
Find
the
⎠
P
QUESTIONS
has
equation
1
r
=
⎛
5 ⎞
⎛
⎜
⎟
⎜
1
⎜
⎜
+
2 ⎞
⎟
t
1
⎟
⎜
⎟
⎜
⎠
⎝
⎟
⎟
2
⎝
The
1
line
l
⎠
has
equation
2
r
⎛
3 ⎞
⎜
⎟
2
=
⎜
⎜
⎛ 2 ⎞
⎜
+
s
⎜
⎟
⎜
4
⎝
Show
⎟
1
⎟
⎟
⎟
2
⎠
⎝
that
the
⎠
lines
l
and
1
the
point
of
l
intersect,
and
find
the
coordinates
of
2
intersection.
Chapter
4
Find
where
the
lines
with
equations r
=
i
+
j
+
t (3i
–
j)
and
1
=
r
–i
+
s j
intersect.
2
5
Show
that
the
two
straight
lines
r
⎛ 3 ⎞
⎛
⎜
⎜
⎟
0
=
⎜
1
⎜
+
1⎞
⎟
t
1
⎟
⎜
⎟
⎜
⎝
⎛ 1 ⎞
⎜
r
⎟
⎟
⎜
+
1
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎠
⎝
0
6
The
L:
l
M:
a
m
b
7
3i
=
are
skew
.
2j
–
+
vector
s (–i
6k
lines
of
the
has
have
+
the
point
that
L
5k
20j
that
their
line
+
M
+
3j
equations
–
5k)
+
t (3i
–
4j
L
and
M
–
3k)
meet
and
lines
L
equation
and
r
M
are
⎛
⎜
⎜
=
⎟
9
⎝
A
The
point
B
The
points
Find
Point
8
P
Find
c
Hence
The
b
a
=
A
–
coordinates
and
lies
B
on
lie
of
L
on
a
+
A
–
the
and
such
2
⎜
⎟
⎜
⎟
⎟
k
Determine
have
vector
⎠
⎝
line
⎠
where
a
is
where
b
a
constant.
is
a
constant.
L.
b.
OP
is
per pendicular
to
L
P
distance
OP
.
position
respectively
,
a
2
(b, 13, –1),
that
of
exact
B
vector
⎟
t
⎟
(5, 7, a),
the
and
coordinates
find
2j
coordinates
values
the
points
3i
has
the
b
has
position
1⎞
3
point
the
per pendicular.
⎛ 6 ⎞
⎜
The
find
intersection.
⎜
a
⎠
⎠
and
14i
Show
The
L
–
Show
of
⎝
1
lines
=
⎠
⎟
s
⎜
⎝
2
⎛ 1⎞
4
=
2
and
⎟
5
vectors
relative
equation
of
to
a
the
a
=
fixed
2i
–
j
+
2k
and
origin O
line L
,
passing
through
1
A
and
The
B.
line
L
has
vector
equation
r
=
7i
+
3k
+
s (2i
+
j
+
2k).
2
b
Show
that
the
lines
L
and
L
1
vector
c
Find
d
Find,
of
the
the
point
length
of
of
the
intersect
and
find
the
position
2
intersection
line
C
AC
Extension
to
the
nearest
degree,
the
acute
angle
between
the
lines
a
L
1
Vectors
and
L
2.
material
Workesheet
line
in
12
three
-
on
CD:
Equation
dimensions
of
.
Applications
Vectors
are
quantities
Example
The
applicable
such
as
to
of
vectors
real-life
situations
displacements
and
that
include
vector
velocities.
position
vector
of
a
boat,
A,
t
hours
after
it
leaves
a
harbour
is
30
given
by
r
=
t
.
second
boat,
B,
is
passing
near
the
harbour.
Its
15
position
A
1
vector
50
at
time
t
is
given
by
r
=
10
+
t
2
5
a
How
far
leaves
b
How
c
Are
apar t
the
fast
the
change
are
the
two
boats
at
the
time
the
10
first
boat
harbour?
is
each
boats
boat
in
traveling?
danger
of
colliding
if
one
of
the
boats
does
not
course?
Answers
0
a
At
t
=
0
boat
A
is
at
the
‘origin’
with
position
vector
0
50
and
boat
B
has
position
vector
2
them
b
The
is
50
speed
their
therefore
5
the
distance
between
5
+ 5
of
=
the
direction
=
2525
boats
vectors
is
–
50.2
found
this
is
km.
by
calculating
each
boat’s
the
magnitude
velocity
of
vector.
30
For
boat
A
the
vector
that
it
will
pass
through
in
one
hour
is
2
which
has
length
15
2
30
+ 15
=
1125
=
33.5
km.
–1
Therefore
boat
A
has
a
speed
of
33.5 km h
10
For
boat
B
the
vector
that
it
will
pass
through
in
one
hour
is
2
which
has
length
10
10
2
+ 10
=
=
200
14.1
km.
–1
Therefore
For
c
the
that
boats
the
B
to
has
a
speed
collide
position
there
vectors
x
components:
30t
=
50
y
components:
15t
=
5
Therefore
Exercise
1
boat
the
boats
of
+
+
will
the
10t
not
14.1 km h
would
10t
two
⇒
t
t
=
need
boats
=
.
to
be
are
a
value
the
of
t
such
same.
2.5 h
1 h
collide.
L
The
position
vector
of
ship
The
position
vector
of
buoy
What
of
S
is
B
30
is
km
20
nor th
km
and
nor th
60
and
km
45
east.
km
east.
is
a
the
position
b
the
exact
of
the
distance
ship
from
relative
the
to
ship
the
to
buoy
the
buoy?
Chapter
2
A
par ticle
with
P
is
constant
vector
⎛ x
⎞
⎜
⎟
y
⎝
at
the
origin
velocity
and
O
at
time t
arrives
at
=
the
0.
The
point
par ticle
Q
with
moves
position
⎛ 20 ⎞
=
⎜
⎠
⎝
⎟
−8
a
the
velocity
b
the
position
the
same
m
4
seconds
later.
Find
⎠
of
P
of
P
if
velocity
it
continues
for
6
more
moving
past
this
point
with
seconds.
–1
3
Another
par ticle
It
through
passes
Find
the
speed
Find
the
distance
e
Will
In
this
the
vector
3 p.m.
on
velocity
of
traveling
the
b
4i
time
the
by
r
point
⎛
⎜
⎜
⎟
3
+
⎜
⎟
Find
b
Show
c
Find
+
are
the
the
by
the
3i
in
t
=
3
is
(4i
5j) m s
–
j) m
at
3i
+
3j
3j.
ships
of
Ship
+
of
2j
=
is
time
in
hours.
1 km.
a
A
’s
and
and
it
cliff
looking
position
is
given
relative
traveling
by
4i
+
out
3j
with
and
to
sea
to
a
a
it
is
Find
will
collide
if
one
does
not
will
collide.
helicopters
X
and
Y
at
time t
seconds
are
given
0,
and
y
⎟
2
⎟
7
t
the
in
of
distance
9
metres.
the
two
respectively
.
2
given
1
a
vector
–1)
are
two
helicopters.
helicopters
between
do
the
not
meet.
helicopters
when t
=
10.
method
that
the
points A(1,
2,
3),
B(–2,
3,
5)
collinear.
QUESTION
points
and
1
exercise
using
the
t
s.
kilometres
top
position
ships
⎞
speed
that
the
C(7,
Vectors
given
on
two
⎟
⎜
EXAM-STYLE
2i
when
displacement
of
4
a
Show
1
⎜
Review
2
two
−1
t
⎟
Distances
and
which
of
3
Prove
are
B’s
the
vector
–
formulae
⎜
1
which
O
traveling.
velocity
position
(12i
collide?
given
Ship
velocity
QUESTION
⎛ 11 ⎞
=
a
at
position
the
x
at
3j.
whose
from
a
ships
is
constant
par ticle.
T
standing
two
+
A
with
course
EXAM-STYLE
The
is
shore
with
change
the
of
distances
man
the
of
represents
obser ving
point
a
a
point
par ticles
question
unit
and
two
moving
the
d
At
✗
is
c
A
4
T
–3i
–
A,
5j
B
+
and
8k
C
with
position
respectively
vectors
form
a
5i
j
+
right-angled
6k,
triangle.
0.
EXAM-STYLE
3
Given
QUESTION
that
a
⎛
5 ⎞
⎛
⎜
⎟
⎜
=
and
1
b
1⎞
⎟
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎠
⎝
4
+
b
Two
and
a
lines
b
with
are
point
EXAM-STYLE
5
A
triangle
P.
the
vectors
per pendicular.
equations
r
⎛
0 ⎞
⎜
⎟
=
Find
the
+
6
⎛ 7 ⎞
⎛ 3 ⎞
⎛
⎜
⎜
⎜
s
⎟
⎜
⎜
⎟
⎜
⎠
⎝
1
⎟
and
3
⎜
⎝
the
that
⎠
1
at
show
5
3
⎝
a
,
3
=
⎜
r
=
2
⎟
of
+
1
⎜
⎟
⎜
⎠
⎝
1
coordinates
⎟
2 ⎞
⎟
t
intersect
4
⎟
⎜
⎟
⎜
⎠
⎝
⎟
⎟
2
1
⎠
P.
QUESTIONS
has
its
a
Find
AB
and
b
Find
AB
·
c
Show
ver tices
at
A(–2,
4),
B(1,
7)
and
C(–3,
2).
AC
AC
3
that
cos BÂC
=
2
5
6
6
Two
lines
L
and
L
1
are
given
by
1
2
2
2
s
2
3
a
P
b
Show
is
the
point
on
L
when
s
=
4.
2
the
12
1
Find
and
0
t
1
11
7
position
vector
of
3
P
1
that
P
is
also
on
L
2
2
⎛
7
The
line
L
has
vector
equation
r
⎟
⎜
−3
=
1
⎛ 1 ⎞
⎞
⎜
+ t
is
parallel
to
L
2
a
and
passes
.
⎟
⎜
⎟
⎜
⎟
⎜
⎟
3
L
⎟
3
⎜
through
2
the
point
B(2,
2,
4).
1
Write
down
a
vector
equation
for
L
in
the
form
r
=
a
+
s b
2
⎛
3
⎜
A
third
line
L
is
per pendicular
to
3
L
and
is
represented
by
1
r
⎟
11
=
⎛ 7 ⎞
⎞
⎜
⎜
⎜
+ q
⎜
⎟
⎜
7
⎝
b
Show
c
Find
that
the
x
=
of
the
point
C,
the
intersection
of
L
BC
e
Find
|BC
|
in
⎠
and
1
Find
⎟
⎟
1
⎝
⎠
–3.
coordinates
d
⎟
x
⎟
the
form
a
b
where
a
and
b
are
integers
to
L
3
be
found.
Chapter
EXAM-STYLE
8
(In
this
question
hours.)
Ship
QUESTION
At
A
’s
distances
noon
a
position
are
lighthouse
at
time
t
is
measured
keeper
given
in
km
obser ves
by
1
two
4
3
and
B’s
position
at
time
t
is
given
by
r
2
9
Show
a
will
In
that
A
occur
order
to
and
and
B
the
prevent
will
collide,
position
collision,
and
vector
at
find
of
12:15
the
ship
A
and
B.
12
5
time
point
in
17
the
A
4
4
Ship
ships
time
of
when
this
collision.
changes
its
16
direction
.
to
17
Find
b
the
Review
distance
between
A
and
B
at
12:30.
exercise
2
3
1
Find
the
size
of
the
angle
between
the
two
vectors
and
Give
your
answer
EXAM-STYLE
2
The
3
OP
of
the
OQ
,
3
A
tent
defined
1
=
are
OR
and
1
and
OABCDE
section
that
is
ˆ
PQR
c
triangular
prism
b
is
an
a
the
position
vectors
.
Find
5
0
QP
by
2
=
1
3
PQR
QR
4
degree.
a
nearest
triangle
2
=
the
5
QUESTIONS
vertices
to
.
equilateral
the
triangle
area
with
with
a
of
triangle
constant
sides
of
PQR
cross-
2 m.
The
E
tent
2 m
4 m
is
4 m
long.
The
base
OADC
is
horizontal.
Suppor t
poles
are
to
be
C
D
laid
along
Take
OA
O
and
the
as
the
OC
diagonals
origin
BC
and
respectively
,
and
unit
k
is
BD
B
vectors
a
unit
i
and
vector
j
in
the
directions
ver tically
of
2 m
4 m
upwards.
O
A
2 m
i,
a
OC
i
Hence
c
Calculate
d
find
i
|BC
iii
the
Hence
OB
ii
b
vectors
the
|
values
scalar
find
BD
|
product
the
k
of
|BD
ii
and
and
OD
iii
BC
j
angle
BC
of
BD
and
between
the
.
suppor t
poles.
2
4
Given
is
a
that
scalar
a
=
+
variable,
a
the
values
b
the
angle
Vectors
xi
of
x
(x
−
2)j
+
k
and
b
=
x
i −
2x j
−
12x k
find
for
between
which
a
and
a
b
and
b
when
are
x
=
per pendicular
−1.
where
x
EXAM-STYLE
5
The
QUESTIONS
points
P
and
Q
have
position
⎛
1⎞
⎜
⎟
⎛ 1 ⎞
⎜
1
vectors
⎜
⎜
⎜
⎟
⎜
⎠
⎝
3
⎝
⎟
5
and
⎟
⎟
⎟
5
⎠
O
OP
a
Show
that
b
Write
down
is
the
per pendicular
vector
to
equation
PQ
of
the
line L
,
which
passes
1
through
The
line
P
and
has
L
Q
equation
r
2
2
1
1
3
2
Show
c
that
the
lines
L
and
L
1
vector
of
their
Calculate,
d
lines
to
and
L
the
All
intersect
and
find
the
position
2
of
intersection.
nearest
degree,
the
acute
angle
between
the
L
1
6
point
2
2
distances
in
this
question
are
in
metres
and
time
is
in
seconds.
An
at
is
insect
point
at
a
flying
with
point
Find
The
is
A
at
a
constant
coordinates
(0,
height.
0,
6).
At
Two
time t
=
seconds
0,
the
later
insect
the
is
insect
B
vector
insect
AB
continues
to
fly
in
the
same
direction
at
the
same
speed.
b
Show
that
x
0
y
vector
t
of
=
3
vector
of
the
insect
is
given
by
0
0,
the
a
bird
bird
at
takes
time
off
t
is
from
given
the
by
ground.
⎛ x
⎜
⎜
down
the
coordinates
of
the
position
⎞
⎛
⎟
⎜
⎜
=
⎟
+ t
18
⎟
⎜
⎟
⎜
⎠
⎝
z
⎝
The
⎛ 36 ⎞
y
Write
time t
⎜
c
at
1
6
time
position
t
0
z
At
the
⎟
.
4
⎟
⎜
⎟
⎜
⎠
⎝
0
star ting
3 ⎞
⎟
⎟
1
position
⎠
of
the
bird.
d
Find
The
the
bird
speed
reaches
of
the
bird.
the
insect
the
bird
e
Find
the
time
f
Find
the
coordinates
at
point C
takes
of
to
reach
the
insect.
C
Chapter
CHAPTER
Vector:
●
A
A
is
vector
scalar
The
unit
a
of
is
Examples
●
SUMMARY
basic
Examples
●
12
concepts
quantity
vectors
a
are
quantity
of
scalars
vector
in
that
size
(magnitude)
displacement
that
are
the
has
has
size
distance
direction
and
of
and
but
and
direction.
velocity
.
no direction.
speed.
the x-axis
is
i.
⎛ 1 ⎞
⎛ 1 ⎞
In
two
dimensions
i
=
⎜
and
⎜
⎝
in
three
dimensions
i
⎟
0
⎟
0
=
⎜
⎠
⎟
⎜
⎟
0
⎝
●
The
unit
vector
in
the
direction
of
the y-axis
is
⎠
j.
⎛ 0 ⎞
⎛ 0 ⎞
In
two
dimensions
⎜
and
j =
⎜
⎝
in
three
dimensions
j =
⎟
1
⎠
In
three
the
dimensions
z-axis
is
k,
the
⎟
⎜
⎝
●
⎟
1
⎜
unit
vector
in
the
direction
⎟
0
⎠
of
where
⎛ 0 ⎞
⎜
k
⎟
0
=
⎜
⎟
⎜
⎟
1
⎝
●
The
●
If
⎠
vectors
i,
j,
k
are
called
base
vectors
⎛ a ⎞
2
AB
=
⎜
⎟
then
=
ai + b j
=
ai + b j + c k
|AB
|
2
a
=
b
b
⎛ a ⎞
⎜
⎟
2
If
AB
=
b
⎜
then
|AB
|
=
a
2
b
2
c
⎟
⎜
⎟
c
●
Two
vectors
magnitude;
vectors
●
Y
ou
●
Two
So,
●
can
The
their
write
and
can
i,
AB
are
RS
also
point
equal
if
they
have
the
j,
k
components
as
–
BA
same
are
direction
equal
too,
and
and
so
the
their
parallel
are
be
with
if
one
parallel
written
as
if
a
coordinates
is
a
AB
=
(x,
scalar
=
k
RS
multiple
where
k
of
is
the
To
find
vector
the
of
A
column
a
other.
scalar
quantity
.
kb
y)
has
posi tion
vector
OP
=
⎛ x
⎞
⎜
⎟
⎝
●
same
equal.
vectors
AB
This
are
are
resultant
from
the
vector
AB
position
between
vector
of
two
points
A
and
y
B,
=
x i +
y j
⎠
subtract
the
position
B.
Continued
Vectors
on
next
page
●
If
A
=
(x
,
y
1
,
z
1
)
then
OA
=
a
=
x
1
i
+
y
1
j
+
z
1
k
1
A
and
if
B
=
(x
,
y
2
,
z
2
)
then
b
= OB
= x
2
i
+
y
2
j
+
z
2
k
2
b
AB
=
AO
–
a
OB
+
a
B
= OB
− OA
b
=
b
−
a
O
=
−
(x
x
2
)i
+
(y
1
−
y
2
)j
+
(z
1
−
z
2
Distance
AB
(x
=
A
vector
of
length
1
●
A
vector
of
length
k
●
The
and
resultant
placed
u
next
+
)
in
( y
y
2
the
in
each
+
is
the
head
of
of
of
third
to
z
2
direction
v,
other
(z
direction
the
u
2
)
1
subtraction
vector,
to
2
1
●
Addition
x
2
)k
1
2
)
1
a
is
found
by
using
the
formula
a
is
found
by
using
the
formula
k
vectors
side
of
the
triangle
formed
when u
and
v
are
tail.
v
v
u
●
Vectors
Scalar
●
Scalar
If
a
=
are
subtracted
a
negative
vector.
product
a
i
+
a
if
a
1
j
and
b
=
a
b
2
b
1
+
a
i
a
1
b
2
+
+
a
+
b
a
k
●
For
perpendicular
●
For
parallel
●
For
coincident
a · b
=
a
b
1
b
=
b
3
b
=
vectors
vectors
then
and
i
+
1
b
+
a
1
j
b
2
+
b
2
k
.
2
then
3
3
a
product
j
2
+
b
3
The
j
2
a
2
●
scalar
i
1
=
1
=
adding
product
Similarly
a · b
by
a
b
=
|a||b|
a
b
=
cos θ
where
θ
is
the
angle
between
the
vectors.
0.
|a||b|.
2
Vector
●
The
a
equation
vector
point
vectors
on
direction
equation
the
line,
vector
a
a
a
of
of
is
a
a
parallel
=
a
a
line
line
is
given
to
.
the
r
=
a
+
position
line.
t
is
t b
where
vector
called
of
the
r
a
is
the
point
general
on
the
position
line
and b
vector
is
of
a
parameter.
Chapter
Theory
of
knowledge
Separate
Mathematics
■
List
the
is
or
often
different
connected?
separated
fields
of
into
different
mathematics
topics,
you
can
or
fields
think
of
knowledge.
of.
■
Algebra
and
Geometry
lly
o
and
used
al
proper ties.
Making
Making
connections
connections
between
seemingly
different
e
mathematician
René
Descar tes
(1596
French
one
of
t
e
key
etr y.
arated,
ces
each
mutual
forces,
Joseph
Proving
Louis
and
have
Lagrange,
Pythagoras’s
marched
have
together
1736–1813,
their
French
progress
been
have
united,
towards
they
is
a
right-angled
prove
Pythagoras’
show
that
a²
+
b²
can
theorem
we
=
c²
knowledge:
links
Separate
or
bra
alge
etry
geom
when
a
b
of
the
need
the
Theory
see
triangle.
are
lent
theorem
c
to
have
mathematician
ween
b et
T
o
slow
perfection.”
You
Here
been
connected?
used
e
sam
to
and
they
le
tack
.
lem
prob
G
EOMETRIC
PROOF
c
a
Draw
and
cut
out
four
triangles
identical
to
this
one.
b
a
Arrange
them
to
make
a
square
with
side
lengths
a
+
b
like
b
this:
a
c
What
is
the
area
of
the
white
square
in
the
center?
b
c
c
b
c
a
b
b
Rearrange
square
with
What
the
triangles
the
are
same
the
to
make
side
areas
of
length,
the
a
a
another
two
like
white
c
a
this:
a
squares?
c
The
area
of
the
central
square
in
the
rst
b
b
diagram
areas
must
of
the
be
two
equal
to
squares
the
in
sum
the
of
the
second
b
diagram.
A
That
is
c²
LGEBRAIC
=
a²
+
a
b²
PROOF
a
Use
the
same
diagram,
but
look
at
the
triangles
b
instead
a
of
the
squares.
c
b
c
Use
these
with
side
two
methods
lengths
a
+
for
nding
the
area
of
the
large
square
b.
c
b
c
Method
.
Square
Method
.
Calculate
and
Methods
large
1
and
2
the
add
both
side
the
this
give
lengths:
area
to
of
c²,
(a
the
the
+
four
area
expressions
for
a
b)²
congr uent
of
the
the
triangles
b
a
square.
area
of
the
square.
Equating
these
gives
b²
+
2ab
+
a²
=
2ab
+
c²
⇒
a²
+
b²
=
c²
c
V
ECTOR
PROOF
a
Represent
a,
b
and
the
sides
of
the
right-angled
triangle
by
vectors
c
b
Because
they
form
a
triangle,
a
So
+
(a
b
+
=
b)
c
(a
+
b)
b
+
b
=
c
c
Expanding
this
gives
a
a
+
a
a
+
b
b
=
c
Which
proof
a
b
=
b
a
=
0,
because
a
and
b
are
a
or
a²
did
you
of
prefer?
perpendicular
So
method
c
a
+
b
b
=
c
Which
was
the
c
easiest?
+
b²
=
c²
Which
was
the
most
beautiful?
Chapter
Circular
CHAPTER
functions
OBJECTIVES
sin
3.2
Denition
3.2
Exact
cosθ
of
sinθ
and
in
terms
of
the
unit
circle;
tanθ
denition
of
as
and
multiples
cos
values
of
trigonometric
ratios
of
0,
6
2
3.3
The
Pythagorean
3.3
Double-angle
3.3
Relationship
3.4
The
3.4
Composite
3.4
T
ransformations
3.4
Applications
3.5
Solving
circular
both
Y
ou
1
the
the
sin 30°
e.g.
Find
of
for
of
the
4
3
their
2
and
cos x
form
= 1
cosine
ratios
and
f (x)
tan x
=
a sin (b(x
+
c))
+
d
functions
equations
in
a
nite
inter val,
analytically
how
values
of
to:
Skills
cer tain
1
exact
value
of
sin 30°.
value
of
tan
check
Find
the
exact
value
of
a
sin 45°
b
tan 60°
c
cos 150°
d
sin 225°
0.5
exact
2
tan
sin
trigonometric
trigonometric
and
sine
sin x,
,
2
ratios.
=
the
cos
,
start
know
exact
Find
functions
trigonometric
trigonometric
e.g.
between
functions
you
should
Find
identities
graphically
Before
identity
,
Find
exact
value
sin
−1
of
a
=
the
tan
b
c
2
Work
with
the
graphing
functions
cosπ
sin
d
of
your
GDC.
3
e.g.
Use
the
graphing
functions
of
Use
to
GDC
to
find
the
x-intercepts
the
of
3
of
f
(x)
=
x
functions
of
your
GDC
find
the
x-intercepts
of
the
graph
of
the
each
graph
graphing
your
function.
2
−
3x
+
2.
3
a
x
≈
−0.732,
1,
4
e.g.
Use
the
graphing
functions
GDC
to
(x)
=
2x
2
−
x
+
5
b
f
Use
solve
the
the
graphing
functions
solve
each
Circular
7
=
functions
ln(x
2ln x.
x
≈
of
your
−
3)
GDC
equation.
−
=
equation
2
4x
(x)
of
to
your
f
2.73
0.0303,
1.38
a
4
−
=
+
b
x
2
=
3
−
x
The
London
opened
carr y
to
up
average
The
the
to
3.5
Eye
Eye,
25
on
public
the
in
people.
million
makes
south
the
It
is
bank
year
a
visitors
2000.
major
each
approximately
of
the
River
Each
tourist
of
Thames,
the
32
attraction,
was
capsules
and
has
can
an
year.
one
revolution
ever y
30
minutes.
It
is
Circular
1.1
Functions
y
0.30103
about
135
capsules
height
metres
travels
above
tall
in
the
a
at
the
circle
boarding
highest
in
a
point.
complete
platform
can
A
passenger
revolution.
be
modeled
in
The
by
one
of
150
the
passenger’s
the
function
x
0
1
2
30
0
h t
67.5 cos
where
h
t
is
the
15
30
height
67.5
,
in
[
metres,
and
t
is
the
time
in
minutes
after
a
This
of
is
the
which
passenger
boards
the
capsule.
This
is
an
example
of
a
the
graph
function
models
passenger ’s
function,
which
you
will
study
in
this
chapter.
the
circular
above
the
height
boarding
platform.
Chapter
.
In
this
➔
Using
section,
The
unit
the
we
uni t
will
circle
circle
continue
has
its
to
center
work
at
the
with
the
unit
circle.
Remember
y
origin
that
B(cos i, sin i)
(0, 0)
and
a
radius
length
of
1
the
unit.
unit
circle
2
equation
The
terminal
side
of
any
angle
θ
position
will
meet
the
at
a
point
with
y
=
unit
x
In
coordinates
this
(θ)
is
diagram,
in
AÔB
standard
(cosθ, sinθ).
position.
The
Look
at
some
angles
in
standard
position
in
the
unit
point
A
the
angle
positive
in
θ
opens
x-axis),
degrees
or
in
in
then
θ
a
counterclockwise
is
positive.
These
direction
angles
can
(from
be
r
3
3
B
r
A(1, 0)
A(1, 0)
3
x
x
0
y
7r
6
A(1, 0)
335°
A(1, 0)
x
0
0
7r
B(cos
If
the
x
7r
, sin
6
B(cos 335°, sin 335°)
)
6
angle
positive
θ
opens
x-axis),
in
then
θ
a
is
clockwise
direction
(from
the
negative.
4r
(cos (–
4r
)
B
3
, sin
(–
))
3
A(1, 0)
0
A
0
–80°
x
x
4r
3
B(cos –80°, sin –80°)
Circular
functions
the
point
B
has
coordinates
measured
y
0
and
(cosθ, sinθ).
r
(1, 0),
the
radians.
B(cos 45°, sin 45°)
has
circle.
coordinates
If
1.
A(1, 0)
0
circle
2
+
in
i
standard
x
has
If
we
know
numerical
the
unit
the
sine
and
coordinates
cosine
to
the
values
point
for
where
an
angle,
the
we
angle
can
give
meets
circle.
y
1
B
(
,
2
2
)
√2
√2
2
2
B
135°
A(1, 0)
0
A(1, 0)
x
30°
Investigation
–
sine,
unit
Y
ou
can
cosine
also
use
values
Sketch
each
sketch
(not
tangent
Angles
of
in
of
the
angle
each
unit
angles
your
in
x
0
standard
GDC)
to
help
and
to
help
you
terminal
position
you
understand
sides
on
on
unit
the
on
the
the
the
x-
circle.
sine,
sine
and
Use
cosine
and
y-axes.
your
and
angle.
degrees:
90°
2
180°
3
270°
4
360°
5
−90°
6
−180°
9
π
12
4π
in
lie
the
determine
1
Angles
tangent
circle
circle
whose
cosine
radians:
0
7
8
2
3
3
10
11
2
In
2
Chapter
the
exact
Y
ou
and
will
11,
you
values
now
of
used
sine,
extend
right
triangles
cosine
this
to
and
to
help
tangent
include
other
you
for
find
30°,
special
45°
angles
and
in
60°.
degrees
radians.
Angle
measure
degrees,
0°,
0
Sine
Cosine
Tangent
0
1
0
radians
radians
1
1
3
3
It
=
is
impor tant
for
you
30°,
2
6
3
3
2
to
remember
values,
1
2
1
=
45°
4
= 1
2
2
required
1
=
2
2
will
know
be
them
using
your
GDC.
3
60°,
3
you
to
1
without
3
as
1
=
2
2
2
these
3
1
1
90°,
0
undened
2
Chapter
In
Chapter
same
sine
cosine
For
11,
you
sin 30°
will
trigonometric
Consider
x-axis.
also
that
found
supplementar y
that
these
angles
the
values,
sine
and
angles
positive,
use
you
in
the
the
and
=
the
sin150°,
unit
in
each
and
circle
to
quadrant
coordinates
cosine
quadrant,
cos150°
find
have
the
opposite
other
see
that
the
that
unit
angles
form
circle
in
cosine
in
the
the
values
angles
the
different
with
same
‘related’
angle
the
quadrants
have
For
is
(–x
cosine
(x
y)
y)
the
and
related
angles
quadrant,
is
cosine
i
in
the
the
values
first
sine
are
and
both
positive.
x
i
(–x, –y)
third
sine
are
with
cosine
second
i
angles
−cos 30°.
represent
i
quadrant,
=
values.
sine
the
can
on
negative.
For
angles
have
values.
angles
As
sine
For
discovered
Y
ou
values.
example,
Now
,
you
value.
(x
y)
For
and
angles
quadrant,
both
positive,
negative.
in
the
the
and
fourth
cosine
the
is
sine
is
negative.
➔
For
any
θ,
angle
,
where
cosθ
≠
0.
It
follows
will
the
be
that,
positive,
tangent
will
Example
Find
a
for
and
be
for
in
the
angles
first
in
and
the
third
second
quadrants,
and
four th
the
tangent
quadrants,
negative.
three
sine
angles
other
angles
with
the
same
value
as:
35°
b
cosine
c
tangent
35°
35°.
Answers
a
To
find
angles
with
the
same
sine:
Angles
with
the
same
sine
values
meet
y
the
145°
unit
circle
at
points
with
the
same
35°
y-coordinates.
–325°
–215°
T
o
find
angles
with
the
same
sine
values,
x
draw
a
horizontal
line
across
the
unit
circle.
These
a
sin 35°
=
sin 145°
=
sin (−215°)
=
35°
sin (−325°)
x-axis.
{
Circular
functions
Continued
on
next
page
angles
angle
all
with
form
the
To
b
find
angles
with
the
same
cosine:
Angles
meet
–325°
35°
same
T
o
x
with
the
find
unit
325°
same
circle
cosine
at
points
values
with
the
x-coordinates.
angles
values,
–35°
the
unit
draw
with
a
the
vertical
same
line
cosine
across
the
circle.
cos 35° = cos 325° = cos (−35°) = cos (−325°)
These
a
To
c
find
angles
with
the
same
35°
angles
angle
all
with
form
the
tangent:
x-axis.
y
Tangent
values
are
positive
in
the
first
–325°
and
third
quadrants.
35°
T
o
find
angles
with
the
same
tangent
x
values,
draw
a
line
through
the
origin
of
215°
the
unit
circle.
–145°
These
a
35°
angles
angle
all
with
form
the
tan 35° = tan 215° = tan (−145°) = tan (−325°)
x-axis.
This
➔
last
For
example
any
sin θ
=
2
angle
sin(180
−
=
cos(−θ
tan θ
=
tan(180
Sketch
illustrate
some
useful
proper ties.
θ:
cos θ
Exercise
1
helps
θ)
)
+
θ)
A
each
angle
in
standard
position
on
the
unit
a
75°
b
110°
c
250°
d
330°
e
−100°
f
−270°
g
−180°
h
40°
Sketch
each
angle
in
standard
position
on
the
unit
circle.
circle.
These
a
b
c
a
4
the
a
three
given
60°
Find
g
35°
other
radians.
−2π
h
3
angles
(in
degrees)
with
the
same
sine
as
angle.
b
three
given
in
the
measured
d
f
Find
are
e
3
angles
other
200°
angles
c
(in
−75°
degrees)
with
d
the
115°
same
cosine
as
angle.
b
130°
c
295°
d
−240°
Chapter
Find
5
the
a
three
given
50°
Find
6
other
the
angles
100°
b
three
given
degrees)
other
c
angles
(in
220°
radians)
the
same
c
three
given
−25°
d
with
the
same
4.1
rad
−3
d
other
angles
(in
radians)
with
the
as
rad
same
cosine
as
1
b
rad
c
2.5
rad
d
Find
three
given
other
angles
(in
radians)
with
the
same
tangent
as
angle.
a
1.3
b
rad
c
Example
Given
−5
d
rad
a
sine
angle.
a
the
as
Find
8
tangent
b
the
with
angle.
a
7
(in
angle.
that
sin 50°
cos 50°
=
0.766
(to
cos 130°
b
3
significant
c
figures),
sin 230°
find
the
values
of
cos (−50°)
d
Answers
2
sin
a
2
50°
+
2
cos
50°
=
1
Use
2
θ
sin
+
cos
‘Pythagorean
Section
2
(0.766)
cos
=
1
50°
=
1,
the
found
in
11.3.
1
2
50°
=
2
+
2
cos
θ
identity’
−
Substitute
(0.766)
solve
for
sin 50°
=
0.766,
then
cos θ
2
cos 50°
=
cos 50°
=
1
0.766
±0.643
(3 sf)
b
(–0.643, 0.766)
(0.643, 0.766)
130°
It
is
a
sketch
good
of
strategy
the
angles
to
make
on
a
a
unit
50°
circle. This
makes
the
relationships
x
between
cos
130°
=
−0.643
the
angle
Circular
functions
to
see.
(3 sf)
{
easier
Continued
on
next
page
c
y
Use
similar
sketches
to
help
answer
(0.643, 0.766)
par ts
c
and
d
230°
50°
x
(–0.643, –0.766)
These
sin 230°
=
related
angles
all
−0.766
make
with
d
an
the
angle
of
50°
x-axis.
(0.643, 0.766)
50°
x
–50°
(0.643, –0.766)
cos
(−50°)
Exercise
1
Given
each
a
=
0.643
B
that
sin 70°
=
cos
b
cos 70°
=
0.342
that
(−70°)
cos
c
sin
=
and
=
,
sf),
find
sin
b
cos
Given
sin
d
find
sinA
=
0.8
cosA
=
0.6,
find
each
−
A)
b
cos (−A)
c
cos (360°
d
sin (180°
+
A)
e
tan A
f
tan (−A)
g
sin (360°
−
A)
h
tan (180°
terms
that
of
a
tan θ
e
sin (π
a
+
sinθ
and
θ)
=
a
and
cosθ
=
b,
value.
sin (180°
+
cos
a
Given
290°
value.
d
and
each
sin
c
that
250°
4
3
cos
3
(to
Questions
Given
a
and
value.
sin 110°
Exam-Style
2
0.940
−
A)
A)
find
each
value
in
b
b
sin (π
f
cos (−θ)
−
θ)
c
cos (π
g
sin (2π
+
θ)
−
θ)
d
tan (π
+
θ)
h
cos (θ
−
π)
Chapter
.
Solving
equations
using
the
uni t
circle
Suppose
we
want
to
solve
an
equation
such
as
sinx
=
.
We
know
that
sin 30°
,
=
but
we
also
know
that
sin 150°
,
=
sin
=
,
and
=
sin
.
So
what
is
the
value
of
x
in
the
equation
sinx
=
?
In
x,
fact,
so
there
we
looking
need
for.
●
Is
●
What
x
are
an
more
We
the
number
information
need
measured
is
infinite
in
to
know
degrees
about
two
or
of
values
the
we
values
could
of
x
substitute
that
we
are
things:
radians?
domain?
Now
suppose
we
want
to
solve
the
equation
sinx
=
,
for
−360°
≤
x
≤
for
which
sinx
360°.
There
are
two
positions
on
the
unit
circle
=
,
so
we
will
find
the
angles
at
those
positions
which
are
within
our
150°
x
=
−360°
≤ x
≤
360°.
30°
–210°
The
domain
–330°
equation
−330°,
has
four
−210°,
Example
solutions
30°,
within
the
given
domain.
150°
2
Solve
the
equation
cos x
=
,
–2π
≤
x
≤
2π
2
Answer
3r
You
–5r
know
3
=
4
Draw
other
–3r
5r
4
4
this
a
position
you
the
within
5
=
,
4
3
Circular
3
,
4
functions
5
,
4
4
ver tical
same
Once
on
x
.
2
4
4
2
cos
on
cosine
have
unit
the
ter minal
line
the
unit
found
find
circle
both
find
domain
at
help
the
with
value.
circle,
sides
to
all
that
these
positions
the
have
angles
their
positions.
for
Example
Solve
the
equation
tan x
=
3,
0
≤
x
≤
720°
Answer
60°
tan
60°
3
=
420°
Draw
find
a
line
the
circle
You
through
other
with
can
this
find
the
position
same
the
origin
on
the
tangent
angles
to
unit
value.
420°
and
240°
600°
600°
by
around
x
=
60°,
240°,
Exercise
✗
Solve
1
420°,
unit
another
600°
equation
for
−360°
≤
x
≤
360°.
=
cos x
b
=
c
tan x
f
tan
=
1
2
sin x
d
=
0
cos
e
2
x
=
x
=
Solve
2
each
equation
for
−2π
θ
≤
2π
≤
sinθ
a
rotation
circle.
C
each
sin x
a
making
the
=
b
tanθ
e
2tan
=
0
c
cosθ
f
sinθ
=
2
sinθ
d
=
−1
θ
=
6
cosθ
=
Although
Solve
3
each
equation
for
−180°
≤
θ
≤
720°.
number
2
cosθ
a
=
1
b
sinθ
been
the
itself
had
studied
for
=
2
centuries,
the
use
2
sinθ
c
=
−cosθ
d
3tan
x
−
1
=
8
of
the
π
symbol
introduced
Solve
4
each
equation
for
−π
≤
x
≤
sin x
=
1
b
2sin x
2
10
c
sin
Example
+
3
=
=
5
d
4cos
Jones
by
(Welsh,
2
1675–1749).
2
x
1706
π
William
a
in
was
x
+
2
=
5
2
Solve
the
equation
sin(2x)
=
,
0°
≤
x
≤
360°.
2
Answer
If
0°
≤
x
≤
360°,
then
0°
≤
2x
≤
720°
We
know
that
2
135°
sin45°
45°
=
sin135°
=
2
495°
405°
To
find
another
unit
2x
x
=
=
45°,
135°,
22.5°,
405°,
67.5°,
495°
202.5°,
247.5°
the
other
rotation
angles,
around
make
the
circle.
These
angles
of
not
2x,
represent
the
value
of
the
value
x.
Chapter
Example
2
Solve
the
equation
2sin
x
+
5sinx
−
3
=
0,
0
≤
is
a
x
2π
≤
Answers
2
2sin
x
+
(2sin x
5sin x
−
−
1)(sin x
3
=
+
0
3)
This
=
Solve
0
by
‘quadratic-type’
equation.
factorizing.
1
sin x
=
or
sin x
=
The
−3
2
−1,
can
cannot
disregard
be
less
sinx
=
than
–3.
6
Exercise
1
we
sine
,
=
6
✗
so
of
5
x
value
Solve
D
each
equation
for
−180°
≤
x
≤
180°.
a
cos (2x)
=
b
6sin (2x)
d
sin
−
2
=
1
c
sin
−
Solve
each
2
=
2
cos
0
equation
for
−π
≤
θ
≤
2
=
3cos
π
a
sin (2θ)
=
b
tan (3θ)
d
sin
=
1
2
c
2
=
Solve
each
=
2cos
QUESTION
equation
for
0
≤
θ
≤
2π
2
x
−
5cos x
−
3
=
0
2sin
b
2
c
tan
.
In
this
x
+
3sin x
+
1
=
0
2
x
+
2tan x
+
1
=
0
section,
Y
ou
we
will
are
look
already
equation
identity
,
is
an
sin
x
=
6sin x
−
5
identi ties
at
special
familiar
2
trigonometric
sin
d
Trigonometric
identi ties.
This
2
a
1
EXAM-STYLE
3
cos
kinds
with
of
one
equations
called
impor tant
2
x
identity
,
+
cos
x
=
because
1.
it
is
tr ue
for
ALL
values
of
x.
Another
identity
with
which
you
are
familiar
is
tanx
=
,
definition
Circular
of
tangent,
functions
which
is
also
tr ue
for
ever y
value
of
x
the
Double-angle
The
diagram
identity
shows
the
for
cosine
θ
−θ
angles
and
drawn
in
standard
position
B(cos i, sin i)
1
in
the
unit
circle.
i
i
The
length
of
segment
CD
is
equal
to
the
length
of
segment
BD,
1
and
we
have
BC
We
can
found
BD
=
see
BD
that
using
=
the
AB
=
1
we
put
[1]
2θ.
r ule
The
in
[1]
length
of
segment
BC
can
be
ΔABC:
AC
2(AB)(AC)cos(2θ)
−
1
2(1)(1)cos(2θ)
−
=
2
−
cos(2θ)
2
two
and
2sinθ
2sinθ.
BC =
2
+
have
C(cos (–i), sin (–i))
2
+
2
BC =
so
=
cosine
=
2
If
CD,
2
BC
we
+
sinθ
=
∠BAC
2
BC
Now
CD
=
[2]
expressions
for
[2]
equal,
we
.
BC
find
2
Squaring
both
sides
gives
us
θ
4sin
=
2
2cos(2θ).
−
2
Rearranging
this
equation
gives
us
2cos(2θ)
=
2
−
4sin
θ.
2
Finally
,
➔
we
The
for
divide
by
We
will
We
know
values
use
this
cos(2θ)
get
=
1
−
identity
+
1
2sin2θ
to
help
us
2
θ
=
−
is
2sin
an
θ
identi ty,
as
it
is
tr ue
θ
of
2
sin
to
cos(2θ)
equation
all
2
other
2
θ
cos
find
=
1,
so
2
θ
=
1
=
1
−
sin
identities.
−
cos
θ.
2
Using
substitution,
Rearranging
this
we
cos(2θ)
get
equation
gives
2(1
−
cos
θ).
us
2
cos(2θ)
=
θ
2cos
−
1.
2
We
can
substitute
sin
2
θ
+
2
2cos(2θ)
=
2
cos
θ
−
(sin
The
➔
three
The
=
=
cos
equations
1
into
this
equation
to
get
2
θ
+
2
cos(2θ)
θ
cos
2
cos
θ),
which
gives
us
2
θ
−
we
double-angle
sin
have
θ
just
identi ties
found
for
are:
cosine:
2
cos(2θ )
=
1
−
2sin
θ
2
=
2cos
θ
−
2
=
cos
θ
1
2
−
sin
θ
Chapter
Double-angle
Now
we
will
identity
find
a
double-angle
2
We
know
that
+
From
the
identity
(2θ)
cos
2
for
sine.
=
1,
so
2
(2θ)
cos
sine
2
(2θ)
sin
for
=
1
−
(2θ).
sin
double-angle
[1]
identity
for
cosine,
2
cos(2θ)
=
1
−
θ
2sin
2
2
cos
(2θ)
1
sin
=
(1
−
2
θ)
2sin
[2]
2
−
2
(2θ)
=
(1
−
2sin
2
θ)
Equate
2
1
−
2
(2θ)
sin
=
2
−
4sin
−
2
4sin
The
sin
cosθ
=
θ
sin
(2θ)
=
sin
2
1
(2θ)
sin
−
sin
θ
2
=
cos
θ
(2θ)
sin(2θ)
=
double-angle
Example
4sin
2
θ
cos
+
2
θ)
2
θ
2sinθ
➔
=
2
θ (1
[2]
4
θ
4sin
and
2
θ
2
4sin
−
4
θ
4sin
1
[1]
identity
for
sine
sin(2θ)
is
=
2sinθ
T
ake
square
both
sides
roots
of
cosθ
3
Given
that
sinx
=
,
and
0°
<
x
<
90°,
find
the
exact
values
of
4
a
cos x
b
sin(2x)
c
cos(2x)
d
tan(2x).
Answers
2
a
sin
2
x
+
cos
x
=
1
Pythagorean
identity.
2
3
2
+
cos
x
=
Substitute
1
the
value
of
sin x.
4
9
7
2
cos
x
=
1
−
=
16
Remember
,
16
7
cos x
an
7
=
Take
the
square
root
acute
cosine
16
must
positive.
sin(2x)
=
sin(2x)
=
2sin x
⎛
3
⎜
⎝
⎟
4
3
sin(2x)
⎞
⎠
cos x
⎛
7
⎜
identity.
⎞
⎟
⎜
Double-angle
Substitute
the
value
of
sin x
and
⎟
4
⎝
⎠
cos x.
7
=
8
{
Circular
functions
x
is
of
4
b
if
angle,
Continued
on
next
page
be
the
Y
ou
2
c
cos(2x)
=
1
−
2sin
Use
x
a
double-angle
could
use
any
of
identity.
the
three
identities
for
2
⎛
cos(2x)
=
1
3
9
⎞
2
−
⎜
⎝
=
⎟
4
1
Substitute
−
the
value
of
cos(2x).
sin x.
8
⎠
1
cos(2x)
=
8
2x
sin
d
tan(2x)
Definition
=
cos
⎛
of
tangent.
2x
3
7
⎜
⎞
⎟
⎜
⎟
8
⎝
tan(2x)
⎠
Substitute
=
the
values
of
sin (2x)
and
1 ⎞
⎛
cos (2x).
⎜
⎟
8
⎝
⎛
=
3
⎠
⎞
7
⎜
⎟
⎜
⎟
8
⎝
tan(2x)
Example
⎠
=
3
8 ⎞
⎛
⎜
⎟
1
⎝
⎠
7
4
Given
cosθ
that
=
3
,
and
5
a
sinθ
<
θ
<
2π,
find
the
exact
values
of
2
cos(2θ).
b
Answers
2
a
sin
2
θ
+
θ
cos
=
1
Pythagorean
identity.
2
⎛
4
⎞
2
sin
θ
+
⎜
⎝
=
⎟
5
1
Substitute
the
value
of
cosθ
⎠
Remember
,
16
9
3
2
sin
θ
=
1
−
if
<
=
θ
<
2π,
the
2
25
25
angle
3
sinθ
9
=
Take
the
square
root
will
four th
be
in
the
quadrant.
The
of
5
25
cosine
the
is
sine
positive
is
but
negative.
2
b
cos(2θ)
=
cos(2θ)
=
2cos
θ
−
1
Use
a
double-angle
identity.
2
⎛
4
32
⎞
1
⎜
⎝
1
=
Substitute
the
value
of
cos θ
⎟
5
25
⎠
7
cos(2θ)
=
25
Notice
that,
cos (2θ)
in
Example
without
ever
8,
we
finding
could
the
find
measure
the
of
values
the
of
sin θ
and
angle θ
Chapter
Exercise
E
Exam-Style
Questions
Given
1
that
sinθ
=
,
and
0°
<
θ
<
90°,
find
the
exact
Y
ou
value
of
answer
each.
sin(2θ)
a
should
cos(2θ)
b
of
questions
tan(2θ)
c
all
nding
be
able
these
without
the
size
of
the
Given
2
that
cosx
=
,
and
90°
<
x
<
180°,
find
each
value.
angle.
sin(2x)
a
cos(2x)
b
tan(2x)
c
Given
3
that
cosθ
=
,
and
0
θ
<
π,
<
find
each
value.
tan θ
a
sin(2θ)
b
cos(2θ)
c
d
tan(2θ)
each
value.
Given
4
that
sinx
=
,
and
180°
<
x
<
270°,
find
sin(2x)
a
cos(2x)
b
Exam-Style
tan(2x)
c
d
sin(4x)
Question
Given
5
that
tanθ
=
,
and
0
<
θ
<
π,
find
each
value.
sin θ
a
cos θ
b
Given
6
that
sin(2x)
=
,
and
cos(2x)
x
<
tan(2x)
b
d
cos(2θ)
<
a
sin(2θ)
c
,
find
each
value.
c
sin(4x)
d
cos(4x)
Given
7
that
tanx
=
,
and
0°
<
x
<
90°,
find
each
value
in
terms
of
a
b
sin x
a
Y
ou
and
can
cos x
b
also
Example
use
identities
c
when
sin(2x)
working
with
d
cos(2x)
equations.
There
Solve
the
equation
sin (2x)
=
sin x
for
0°
≤
x
≤
are
more
360°.
trigonometric
Do
not
use
your
GDC.
identities.
are
Answer
they?
identities
sin(2x)
=
2(sinx)(cosx)
What
What
are
used
sinx
in
=
sinx
Use
double-angle
=
0
Rear range.
=
0
Factorize.
other
areas
identity.
mathematics?
2(sinx)(cosx)
−
(sinx)(2cosx
sinx
=
0
If
sinx
=
If
2cosx
or
sinx
−
2cosx
0,
then
x
1)
−
1
=
0°,
=
0
180°,
360°
1
so
x
x
=
=
−
60°,
0°,
1
0,
then
cosx
=
300°
60°,
Circular
=
180°,
functions
300°,
360°
,
to
of
Example
2
Prove
that
(1
+
tan
2
x)
cos
(2x)
=
1
−
tan
x
Answer
2
(1
+
2
x)
tan
×
cos(2x)
=
1
−
tan
x
2
⎛
2
sin
⎞
x
sin
x
2
⎜
1+
⎟
2
cos
⎝
x
(
2cos
x
− 1
)
= 1 −
Rewrite
⎠
− 1 + 2sin
x
sin
=
x
1 −
Multiply
x
cos
of
the
the
left
equation.
2
x
+ 2sin
2
=
x
2
Simplify.
2
x
+
cos
x
Example
for
on
x
side
2
through
2
cos
In
cos x.
2
x
−
2
sin
and
2
x
cos
sin x
x
2
sin
2
cos
using
2
cos
all
of
x
1
10,
values
values
=
we
of
and
Divide
x.
is
ended
up
with
Therefore,
also
an
the
a
known
original
identity
,
by
2.
identity
,
equation
though
it
is
not
which
is
tr ue
one
is
tr ue
for
you
all
must
lear n.
When
you
‘proving
show
Solve
a
to
be
tr ue
in
this
way
,
it
is
called
identities’.
Exercise
1
equations
F
each
sin (2x)
equation
=
for
0°
≤
x
≤
180°.
cos x
sin (2x)
b
=
cos (2x)
2
c
(sin x
+
2
cos x)
=
0
cos
d
x
=
2
Solve
each
equation
for
−180°
≤
θ
≤
180°
2
a
2sinx cosx
=
b
sinx(1
−
d
cos(2x)
b
2cos
d
sin(4x)
b
sinx
sinx)
=
cos
x
2
c
cos
2
x
=
+
sin
x
=
sinx
3
Solve
each
equation
for
0
≤
x
≤
π
2
a
tanx
=
sinx
c
cos(2x)
x
−
1
=
4
Solve
=
each
cosx
equation
for
0
≤
θ
≤
=
sin(2x)
π
2
a
(sin(2x)
+
cos(2x))
2
=
2
2
c
5
cos
x
Prove
−
1
=
=
1
cos
x
2
=
cos(2x)
each
2sin
d
x
identity
.
2
a
(sinx
+
cosx)
=
1
+
sin(2x)
b
( )
=
c
4
e
cos
d
4
x
−
sin
x
=
cos(2x)
Chapter
Exam-Style
The
6
Questions
expression
Find
the
value
2sin 3x
of
cos 3x
can
be
written
in
the
form
sin kx.
k
2
The
7
expression
Find
.
In
the
value
sections,
relationships
tangent
y
=
to
help
Y
ou
in
y
with
and
=
we
have
section,
tanx.
know
written
in
the
form
1 −
bsin
2
x
cos
x.
Y
ou
GDC
to
functions
used
different
understand
the
be
circular
this
cosine
already
the
you
and
functions
In
can
b
between
values.
cosx,
Sine
of
Graphing
previous
used
cos 4x
the
angles
you
the
unit
and
will
circle
their
see
will
also
help
you
practice
solve
find
sine,
how
trigonometric
to
cosine
these
and
values
functions y
graphing
=
can
be
sinx,
these
equations.
functions
the
exact
sine
values
for
many
angles,
as
seen
table.
Angle
measure
degrees,
(x)
radians
Sine
value
(sinx)
Angle
measure
degrees,
(x)
Sine
radians
(sinx)
1
7
0°,
0
radians
210°,
0
2
6
1
1
2
3
3
2
3
3
60°,
−1
270°,
3
−
2
4
2
2
=
2
240°,
=
4
2
−
4
2
45°,
1
5
225°
30°,
6
value
2
2
3
5
90°,
1
2
300°,
2
3
3
2
7π
315°,
120°,
3
3
1
4
2
5
Circular
1
6
2
1
π
functions
2
0
2π
−
2
2
11
360°,
6
2
=
330°,
2
2
150°,
180°,
4
=
135°,
1
−
2
0
If
we
let
y
=
sin x,
we
can
plot
these
values
as
coordinates
on
a
graph.
y
1.0
0.5
0
x°
90
180
270
360
450
540
–0.5
–1.0
“
Graphing
same
set
the
of
equation
axes,
we
y
see
=
sinx
on
this
“
this:
If
we
measure
graph
has
y
the
x
in
radians,
same
the
shape.
y
1.0
1.0
.5
.5
x°
–90
We
can
same
see
sine
Similarly
,
along
“
y
that
the
values
if
with
=
x
90
cosx,
we
the
graph
that
let
y
=
graph
with
x
we
of
cosx,
of
the
measured
the
found
in
we
function y
using
can
the
plot
function y
=
degrees:
=
unit
the
r
r
3r
2
2
2
sinx
is
generating
5r
3r
2
the
circle.
cosine
values
we
know
,
cosx
“
y
=
y
cosx,
with
x
measured
in
radians:
y
1.0
1.0
.5
.5
0
r
x
x
0
r
r
2
2
–.5
–.5
–1.0
–1.0
r
2
2r
5
3r
2
Chapter
➔
If
we
compare
the
sine
are
the
and
cosine
functions,
we
see
many
similarities.
y
●
The
cur ves
horizontal
sa m e
positions
on
size
the
and
a xes
shape,
d iffer.
only
T he
the ir
s ine
y
=
cos
x
c ur ve
x
0
passes
through
the
o rig in
passes
through
the
po int
( 0,
( 0,
0) ,
and
the
c o sine
cu r ve
1).
y
●
The
the
functions
same
are
cycle
of
periodic,
values
which
over
means
and
over.
that
they
repeat
The period,
y
=
sin
x
or
x
length
look
of
at
apart,
one
two
the
cycle,
points
is
360°
or
2π.
This
means
whose x-coordinates
y-coordinates
of
those
two
are
that
360°
points
if
2π)
(or
would
you
be
the
same.
●
Both
functions
minimum
an
this
y
is
value
ampli tude
between
=
the
case)
−1,
have
of
a
−1.
1.
Each
The
case).
the
or
We
ver tical
of
value
these
amplitude
axis
maximum
this
one-half
maximum
horizontal
to
in
of
a
of
is
the
also
distance
the
wave
say
from
1
and
functions
minimum
can
of
has
difference
(y
=
value
that
a
a
0,
(y
the
in
=
1
or
amplitude
maximum
to
a
minimum.
We
can
use
equations,
earlier
in
the
graphs
much
as
we
this
chapter.
the
equation
of
y
=
used
sin x
the
and
unit
y
=
cos x
circle
to
to
help
help
us
us
solve
to
Consider
sinx
,
=
−360°
≤
x
≤
360°.
By
graphing
a
horizontal
line
y
=
on
the
same
set
of
axes
as
sin x
=
y
=
sin x,
we
can
see
that
there
y
y
=
sin x
1
y
1
=
2
x
These
x
=
points
−330°,
Circular
correspond
−210°,
functions
30°,
to
the
150°.
values
are
four
points
solve
equations
where
Example
The
Solve
the
equation
cosθ
=
0.4,
−360°
≤
θ
≤
GDC
helpful
Give
your
answers
to
the
nearest
can
be
ver y
360°.
in
solving
tenth.
equations
and
Answer
T
o
with
cosine
change
sine
functions.
to
degree
mode
and
choose
Status
5:Settings
|
2:Settings
2:Graphs
Use
&
the
to
Angle
and
and
GDC
then
select
select
to
help
and
GDCs
Enter
y
=
set
y
0.4
an
view
There
is
four
−293.6°,
−66.4°,
66.4°,
GDC,
on
CD:
four
within
|
Casio
are
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
and
window
Be
sure
DEGREE
to
your
mode!
intersection
this
solutions.
these
=
graph.
in
to
and
the
equation
Graph
θ
into
are
points
the
cos x
appropriate
the
GDC
so
=
return
document.
demonstrations
Plus
to
Graphing
4:Current
the
|
Geometr y
key
move
&
domain,
will
Use
have
6:Analyze
4:Intersection
intersection
to
find
points.
293.6°
Chapter
Example
Angle
Solve
the
equation
sinx
=
0.25x
−
0.3,
−2π
≤
x
measures
are
in
2π.
≤
radians
Give
your
answers
to
three
significant
figures.
Answer
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
T
o
Casio
are
on
mode
|
=
sinx
the
=
0.25x
−
and
to
sure
0.3
into
the
the
three
=
−2.15,
Circular
−0.416,
functions
2.75
window
is
mode!
intersection
points
domain,
equation
will
have
solutions.
Use
6:Analyze
Graph
these
GDC
three
this
|
4:
to
select
Press
then
select
Intersection
intersection
to
points.
to
return
GDC,
graph.
your
are
within
x
the
|
and
appropriate
RADIAN
There
so
an
view
Be
in
set
&
Geometr y
key
Angle
the
and
and
Graphing
4:Current
y
&
to
and
y
radian
2:Settings
move
Radian.
Enter
to
5:Settings
2:Graphs
Use
TI-84
CD.
press
choose
Status
the
FX-9860GII
the
change
Alternative
for
find
document.
to
Exercise
In
G
questions
Give
your
1
to
4,
answers
use
to
your
the
GDC
nearest
to
solve
each
equation.
each
equation.
degree.
sin x
1
=
,
−360°
≤
x
≤
360°
2
cos θ
3
sin θ
4
sin x
=
=
,
≤
θ
θ
≤
360°
−0.9,
=
cos(x
In questions
Give
−180°
your
5
0°
−
to
≤
20),
8,
answers
0°
use
to
≤
≤
360°
x
≤
your
three
540°
GDC
to
solve
significant
figures.
sin θ
5
=
−2π
,
≤
x
≤
2π
1
cos θ
6
=
,
−π
≤
x
≤
2π
2
e
7
cos x
8
sin x
=
−π
−x,
≤
x
2π
≤
2
=
x
Tangent
−
1
,
−2π
≤
the
cosx
Now
1
sine
that
tr y
List
0°,
a
3
On
a
2π
cosine
already
similar
±45°,
315°,
graphing
functions,
values
±60°,
330°,
of
we
grid
the
function
for
the
angles:
120°,
135°,
plot
angle
(measured
value
of
values
Connect
y
there
=
for
sinx
and
=
tanx
180°,
210°,
225°,
240°,
values
sometimes
not
points
in
values
as
degrees),
points.
and
let
Let
the
the
x-axis
y-axis
tanx.
tangent
you
do
values
for
the
angles
have
on
paper
to
the
±90°
graph
or
of
a
270°?
function
exist?
on
your
grid
sketch
the
graph
of
tanx.
Graph
your
Are
that
the
no
do
y
150°,
these
the
feature
with
360°.
paper
,
the
are
began
to
represent
Why
tan x
knew.
approach
tangent
piece
–
represent
for
5
we
the
What
4
and
±30°,
300°,
2
≤
function
Investigation
For
x
the
function
y
=
tanx
on
your
GDC,
and
compare
it
to
sketch.
your
graphs
similar?
Chapter
If
you
had
tangent
been
using
function
radians,
would
look
rather
like
than
degrees,
the
graph
of
the
this:
y
3
2
1
x
3r
r
2
r
r
3r
2
2
2
r
–2
–3
➔
Like
the
sine
periodic.
the
The
the
sine
an
Example
Solve
the
of
and
the
functions,
ver tical
not
pair
of
exist.
cosine
The
has
same
no
tangent
at
function
values
cycle
of
of
x
values
is
where
repeats
asymptotes.
function
functions,
It
the
asymptotes
ver tical
tangent
amplitude.
the
is
180°
tangent
maximum
(or π
radians).
function
or
does
minimum
Unlike
not
values.
equation
your
cosine
are
does
each
period
have
Give
There
function
between
and
answers
tanθ
to
=
1
three
−
x
,
−2π
significant
≤
θ
≤
2 π.
figures.
Answer
Be
sure
your
RADIAN
There
are
within
this
equation
GDC
five
have
on
CD:
demonstrations
Plus
and
GDCs
θ
=
−4.88,
Circular
−1.90,
functions
0.480,
2.25,
4.96
Casio
are
on
is
intersection
domain,
will
help
GDC
in
mode!
so
five
the
solutions.
Alternative
for
the
TI-84
FX-9860GII
the
points
CD.
Exercise
In
questions
Give
your
1
tan x
2
tan θ
=
3
tan θ
=
4
tan x
In
H
=
=
1
your
4,
answers
2,
,
≤
cos x,
0°
0°
to
≤
8,
answers
x
360°
≤
θ
≤
x
≤
≤
use
to
GDC
nearest
θ
≤
your
the
−180°
−1.5,
5
use
to
−360°
questions
Give
to
≤
to
solve
each
equation.
each
equation.
degree.
360°
360°
720°
your
three
GDC
to
solve
significant
figures.
tan θ
5
=
−2π
,
≤
x
2π
≤
6
tan θ
7
tan x
=
2x
8
tan x
=
4
=
π,
−π
−
≤
3,
θ
0
π
≤
≤
x
2π
≤
2
.
−
x
,
−2π
≤
x
≤
Translations
2π
and
trigonometric
Investigation
–
stretches
of
functions
transformations
of
sinx
and
cosx
Using
your
GDC
and
y
=
cos
do
you
What
do
they
Describe
Now,
mode,
the
same
notice
have
the
this
and
about
in
graphs
are
=
sin x
for
sin x
2
y
=
cos x
and
y
=
2 cos x
3
y
=
cos x
and
y
=
cos (2x)
4
y
=
sin x
+
=
cos x
axes.
and
y
=
sin
=
cos
each
these
two
functions?
these
tr y
to
pairs
explain
of
why
this
is
happening.
functions.
x
of
and
3
y
of
different,
x
and
graphs
3
sin x
y
common?
process
y
the
=
=
functions
y
y
the
1
5
graph
2
how
repeat
radian
on
x
What
in
2
Chapter
In
the
last
section,
we
looked
at
the
basic
trigonometric
functions
Y
ou
y
=
sin x,
y
=
cos x
and
y
=
tan x.
Now
we
will
study
need
familiar
of
these
functions.
begin
by
looking
at
the
graphs
of
the
sine
and
be
with
features
sine
Let’s
to
ver y
transformations
cosine
and
of
the
the
basic
cosine
functions,
cur ves.
and
reviewing
some
vocabular y
relating
to
these
functions.
y
y
1
y
=
y
sin x
r
2
2
These
the
functions
functions
These
graphs
that
you
of
have
have
these
a
Chapter
rather
an
2π
of
period
than
ampli tude
functions
transformed
see
–2
r
3
the
can
graphs
(or
r
2
360°,
if
we
–1
r
r
2
were
2r
2
graphing
radians).
of
be
of
x
r
2
degrees
functions
The
book;
in
3r
cos x
0
x
3r
=
1.
transformed
other
in
functions
the
same
earlier
in
way
this
1.
Translations
➔
The
function
standard
y
sine
The
cur ve
The
function
=
sin(x)
+
d
is
a
vertical
translation
of
the
cur ve.
shifts
up
if
d
is
positive,
down
if
d
is
negative.
A
standard
positive,
It
is
“
or
This
left
the
graph
=
cosine
impor tant
period
y
to
if
c
sin(x
cur ve.
is
note
a
c)
is
The
a
horizontal
cur ve
shifts
to
translation
the
right
if
of
c
the
that
of
vertical
known
shift’.
a
translation
a
trigonometric
translation.
horizontal
translation
is
negative.
amplitude
shows
−
does
“
not
change
the
function.
This
graph
shows
a
horizontal
translation.
The
2
sine
units.
cur ve
The
has
green
been
shifted
arrow
shows
up
The
of
the
cur ve
has
been
shifted
units
2
the
the
direction
sine
right.
The
green
arrow
shows
translation.
direction
of
the
translation.
y
r
y
3
=
sin x
y
y
=
sin
(
x
–
)
2
y
=
sin x
+
2
1
2
1
x
r
y
=
x
r
r
2
Circular
functions
r
3r
r
2
sin x
r
r
3
r
the
to
as
is
a
also
‘phase
➔
The
function
standard
cur ve
The
function
positive,
As
or
with
the
“
the
The
y
sine
if
graph
shows
cur ve
3
units.
is
d
is
is
a
vertical
−
positive,
c)
is
The
a
a
down
if
cur ve
shifts
to
d
is
direction
of
the
cosine
been
green
the
negative.
of
translation
the
right
does
not
if
c
the
is
the
period
function.
translation.
“
The
shifted
arrow
change
graph
below
translation.
The
shows
cosine
a
horizontal
cur ve
has
been
3
shows
shifted
the
of
translation
horizontal
translation
vertical
has
The
+
negative.
the
a
d
cur ve.
c
of
if
cos(x
cur ve,
cosine
down
up
=
cosine
left
cos(x)
cur ve.
shifts
amplitude
This
=
cosine
The
standard
y
units
translation.
to
the
left.
4
3r
y
y
y
y
=
r
2
r
r
2
–1
=
cos
x
(
+
)
4
0
x
r
y
cos x
cos x
0
r
=
r
2r
r
2
2
y
=
cos x
–
x
r
r
r
r
2
–1
3
–3
–4
Now
consider
the
graph
of
the
tangent
function.
y
3
2
1
x
3r
r
r
r
2
2
2
3r
r
2r
2
–2
–3
Remember
amplitude,
that
this
because
function
there
are
has
no
a
period
of
maximum
There
are
ver tical
asymptotes
at
x
=
(or
at
x
=
±90°,
x
=
±270°,
or
(or
180°).
minimum
It
has
no
points.
π
,
etc.
etc).
Chapter
As
with
the
sine
translations
We
can
=
tan(x
in
−
Sketch
the
the
c)
Example
On
and
not
combine
equations
y
do
+
cosine
change
horizontal
form
y
=
functions,
the
period
and
sin(x
y
=
the
ver tical
−
c)
+
d,
and
horizontal
tangent
function.
translations
y
=
cos(x
−
by
c)
looking
+
d,
at
and
d
the
graph
same
set
of
of
y
=
sinx.
axes,
sketch
the
sinx
+
1
y
b
=
sin
graph
of:
2π ⎞
⎛
a
ver tical
of
c
3
⎝
2π ⎞
⎛
x
⎜
y
=
sin
+
x
⎟
⎜
⎠
⎝
1
⎟
3
⎠
Answers
a
y
=
sinx
+
1
y
The
2
basic
function
1
unit
sine
is
cur ve
shown
in
is
shown
in
blue. This
red,
is
a
the
translated
ver tical
shift
of
upward.
1
x
3r
r
2
2
y
=
sin
3r
r
2π ⎞
⎛
b
r
x
⎜
⎟
3
⎝
⎠
y
Again,
the
basic
sine
cur ve
is
shown
in
red,
the
new
2
function
is
shown
in
blue. This
is
a
horizontal
shift
2
1
of
units
to
the
right.
3
x
4r
3
r
r
3
r
3
3
3
r
r
3
r
3
5
3
–2
2π ⎞
⎛
c
y
=
x
sin
+
⎜
1
⎟
3
⎝
⎠
This
is
a
combination
of
the
translations
from
par ts
y
a
and
b,
with
the
new
function
shown
in
blue.
2
The
1
basic
units
3
x
5r
4r
3
3
r
r
r
2r
3
3
r
r
5
–1
3
3
–2
Circular
functions
sine
cur ve
(shown
in
red)
has
2
3
3
r
to
the
right,
and
1
unit
up.
been
shifted
Example
Write
a
an
equation
Write
a
sine
for
each
function,
as
directed.
equation.
y
0
–2r
3r
r
x
r
r
r
3r
–1
2
2r
2
–2
b
Write
a
cosine
equation.
y
0
–2
c
x
r
r
2
Write
–1
2
one
sine
3r
2
and
one
2r
2
cosine
equation.
y
1
0.5
r
–2r
5r
4r
3
3
r
0
2r
x
r
2r
3
3
r
4r
5r
3
3
2r
3
3
–0.5
–1
Answers
a
y
=
sinx
−
2
You
a
can
see
minimum
shifted
b
y
=
cos
x
is
value
value
down
2
of
a
sine
of
cur ve
−1
–3.
It
and
has
with
a
been
units.
+
4
this
maximum
You
can
see
this
is
a
cosine
cur ve
which
has
been
shifted
to
the
left
by
4
c
y
=
cos
x
+ 0.5
3
You
can
see
this
as
a
cosine
cur ve
Because
which
has
been
shifted
to
the
right
of
or
,
and
up
=
sin
sine
there
might
also
view
this
as
a
are
and
may
so
cosine
be
similar
,
many
sine
+ 0.5
the
cur ves
You
x
shapes
0.5.
3
y
the
by
6
correct
cur ve
which
has
been
shifted
to
equations
for
the
the
left
by
,
and
up
graph
of
a
sine
or
0.5.
6
cosine
function.
Chapter
Exercise
For
for
I
questions
–2π
≤
1
to
x
≤
2π.
−
5
1
y
=
sinx
3
y
=
tan
8,
sketch
the
graph
of
the
function
2
y
=
cosx
4
y
=
sin
2
+
5
y
=
cos
y
6
=
sin
=
cos
−
1.5
y
8
=
tan
For questions
9
to
12,
write
an
equation
y
1
0
x
r
r
2r
r
2r
y
3
2
1
x
–2r
r
0
y
6
4
2
0
3r
r
2
2
x
r
3r
2
2
2r
–2
y
0
–2
3r
r
–1
–2
–3
Circular
functions
x
r
2
r
3r
2
2r
for
the
2
+
–2
−
y
7
4
function
shown.
Vertical
➔
The
stretches
functions
vertical
When
the
=
asin x
of
stretches
the
stretch,
y
g raph
ever y
value
of
of
and
the
a
y-value
y
sine
the
acos x
and
func tion
in
=
are
cosine
func tio ns.
und ergoe s
original
a
function
ver tic al
is
multiplied
by
y
=
asin
a.
x
0
If |a|
>
1,
the
function
will
appear
to
stretch
away
from
the
x-axis.
If
0
the
y
<|a|
x
y
<1,
the
function
will
appear
to
compress
closer
=
sin
x |a|>1
to
x-axis.
y
If
a
is
negative,
the
function
will
also
be
reflected
over
the
y
=
sin
x
x-axis.
With a
In
by
stretch,
function
will
change
function
will
not
the
a
ver tical
graph
factor
minimum
function
below
,
of
3.
is
are
from
of
ampli tude
1
to
|a|.
The
the
sine
period
or
of
cosine
0
x
y
the
=
asin
x
|a|<1
change.
the
The
values
the
sine
cur ve
maximum
at
y
=
−3.
has
been
values
The
are
stretched
at y
amplitude
=
of
3,
ver tically
and
the
the
new
3.
y
3
2
y
y
sin x
=
=
3
sin x
1
x
–2
3r
r
2
2
3r
r
–3
The
the
next
been
The
are
graph
x-axis.
shows
of
multiplied
maximum
at
The
All
y
=
the
by
a
ver tical
y-values
in
stretch
the
that
is
standard
also
a
cosine
reflection
cur ve
about
have
−0.5.
values
are
at
y
=
0.5,
and
the
minimum
values
−0.5
amplitude
of
the
new
function
is
0.5.
y
y
=
cos
x
y
=
–0.5cos x
0
3r
2
r
x
r
r
2
2
r
r
2
–1
Chapter
Example
Sketch
On
a
the
the
y
graph
same
=
set
of
of
y
=
cosx.
axes,
0.25cosx
sketch
y
b
=
the
graph
of:
−2cosx
Answers
a
y
=
0.25cosx
y
The
basic
cosine
cur ve
is
shown
in
blue,
the
new
1.0
function
stretch
is
shown
factor
in
red. This
is
a
ver tical
stretch
of
0.25.
x
3
r
r
2
r
2
–1.0
b
y
=
−2cosx
Again,
y
new
the
basic
function
is
cosine
shown
cur ve
in
is
shown
in
blue,
the
red.
2
Ever y
y-value
multiplied
r
r
2
2
Horizontal
➔
The
the
to
blue
give
function
the
red
has
been
function.
2r
2
2
stretches
functions
horizontal
When
in
−2
x
0
–2r
by
the
y
=
sin(bx),
stretches
graph
of
of
a
y
the
=
cos(bx)
sine,
function
and
cosine
y
=
and
undergoes
a
tan(bx)
tangent
represent
functions.
horizontal
stretch,
ever y
x-value
in
the
original
function
is
multiplied
.
by
We
could
divided
also
by
that
ever y
x-value
in
the
original
function
is
b
Multiplying
changes
say
(or
the
dividing)
period
of
a
the
x-values
by
trigonometric
a
number
in
this
way
function.
y
y
●
If
|b|
>
1,
the
period
will
be
shor ter,
and
the
function
=
sin b x
y
=
sin
x
will
x
appear
●
If
0
<
will
●
If
b
to
compress
|b|
<1,
appear
is
to
the
toward
period
stretch
negative,
the
the
will
away
y-axis.
be
from
function
will
longer,
and
the
function
|b|>1
the y-axis.
also
be
reflected
over
the
y-axis.
y
When
a
sine
or
cosine
function
undergoes
a
horizontal
y
=
sinb
y
=
sin
x
stretch,
the
period
of
the
function
from
360°
to
Circular
functions
will
change
from
2π
to
,
x
or
x
|b|<1
“
In
the
blue
graph
has
a
below,
period
the
sine
cur ve
in
“
π
of
In
the
in
blue
graph
has
function
y
y
=
sin
below,
a
period
has
also
the
of
been
sine
4π.
cur ve
The
reected
(2x)
about
1.0
the
y-axis.
y
0.5
y
=
sin (–0.5x)
y
=
sin x
1
x
3
–2r
r
2
3
2
2
r
2
x
–3
y
➔
For
a
function
in
the
=
r
r
3
sin x
form
y
=
tan (bx),
the
period
will
y
change
from
π
to
,
or
from
180°
to
4
The
graph
to
The
period
the
of
right
this
shows
function
is
the
function y
=
2
tan(0.5x).
2π
x
–3r
Example
r
r
r
r
3r
–4
Sketch the graph of:
y = sin (0.5x)
a
b
y = tan (2x)
c
y = 2 cos (3x)
Answers
2
y
a
=
The
sin(0.5x)
period
of
this
function
is
,
or
0.5
y
4π
1
x
–3r
–2
r
r
2
3r
4r
y
b
=
The
tan (2x)
period
of
this
function
is
2
y
4
2
x
r
r
r
2
3r
r
2
2r
2
–2
–4
2
c
y
=
The
period
of
this
The
amplitude
function
is
.
2 cos (3x)
3
is
2.
y
2
r
0
r
2r
x
Chapter
Exercise
For
J
questions
1
to
8,
sketch
the
graph
of
the
function
from
–2π
≤
x
≤
2π.
y
1
=
0.5 sin x
y
2
=
−4 cos x
y
3
=
tan
y
4
=
sin
(–2x)
y
5
=
2 cos
=
−2.5 sin (0.5x)
y
8
=
questions
9
to
12,
write
an
equation
for
the
function
shown.
y
y
3 sin (3x)
For
=
−cos
y
6
y
7
8
7
6
5
0
–6
4
–4r
r
x
2
4r
6r
–1
3
2
1
0
r
x
2
–1
3r
–2
–3
–4
–5
–8
y
y
6
3
2
4
1
x
0
–2r
2
2r
–1
3r
x
–6r
–4r
–2r
2r
4r
6r
–2
–4
–6
.
Combined
and
In
y
this
=
For
cosine
section,
asin(b(x
−
functions
transformations
we
c))
of
will
+
d
this
wi th
sine
functions
be
and
looking
y
type,
=
at
functions
acos(b(x
there
are
−
four
c))
+
of
the
form
d
possible
transformations
happening:
●
a
represents
cosine
●
b
a
represents
function.
a
to
Circular
functions
will
stretch.
be
period
The
equal
horizontal
The
equal
ver tical
function
of
to
stretch,
the
amplitude
of
the
sine
or
|a|.
which
sine
or
affects
cosine
the
period
function
will
of
be
the
●
c
represents
shift
●
d
to
a
the
represents
up
if
d
is
horizontal
right
a
if
c
is
ver tical
positive
translation,
positive
translation,
or
down
The
function
y
=
2 sin
same
axes
as
the
basic
sine
if
d
or
to
or
is
shift.
the
left
shift.
The
if
The
c
function
is
will
negative.
functions
shifts
negative.
or
is
shown
in
blue
on
the
cur ve
(shown
in
red).
3
2
1
x
3
This
function
function
y
function
in
two
●
=
sinx
blue.
changes
There
has
to
has
●
been
There
amplitude
has
the
of
have
a
1.
been
2
and
four
been
a
period
4π.
of
transformations
two
changes
to
The
to
become
the y-values
the
and
x-values.
ver tical
All
the
multiplied
has
of
undergone
There
been
translation
have
an
a
stretch
of
y-values
in
by
2,
then
horizontal
scale
the
factor
standard
increased
stretch
of
2
by
scale
and
sine
a
ver tical
function
1.
factor
2
and
a
horizontal
translation
of
.
All
the
x-values
in
the
original
sine
function
have
been
multiplied
by
2
(divided
by
),
then
decreased
by
The
function
y
=
3 cos
is
shown
in
blue
on
the
y
3
same
axes
as
the
basic
cosine
cur ve
(shown
in
red).
2
This
function
has
an
amplitude
of
3
and
a
1
period
of
π.
undergone
The
four
function
y
=
cos x
transformations
to
has
become
x
2r
the
●
function
There
All
has
the
have
in
blue.
been
y-values
been
a
–1
ver tical
in
the
multiplied
stretch
standard
by
of
scale
cosine
factor
3.
function
3.
●
There
has
been
a
horizontal
stretch
of
scale
factor
,
a
reflection
about
the
y-axis,
and
a
horizontal
translation
of
.
All
the
x-values
in
the
original
cosine
function
have
been
divided
by
−2,
then
increased
by
When
sketching
step-by-step
functions
like
these
by
hand,
it
is
best
to
take
a
approach.
Chapter
Example
2
Sketch
the
graph
of
the
function
y
=
5 cos
3
x
+
2
Answers
This
a
function
ver tical
The
will
shift
of
maximum
function
will
have
amplitude
of
5
and
−2.
and
be
an
3
The
y
minimum
and
−7,
values
of
=
horizontal
−2,
which
axis
is
of
the
the
wave
ver tical
will
be
translation.
the
respectively
.
y
4
2
Mark
these
maximum
and
minimum
values
1
0
–2r
r
(shown
in
red)
and
(shown
in
green).
the
axis
of
the
wave
x
r
–1
2r
3r
–3
These
will
be
helpful
guidelines
when
–4
–5
graphing
the
function.
–6
–8
2
Period
=
2
=
b
=
2
This
function
horizontal
3
= 3
2
3
will
shift
2
of
have
a
period
of
3π
and
a
−π.
y
4
2
1
The
0
–2r
r
x
r
–1
2r
standard
where
3r
x
=
0,
maximum
cosine
so
this
where
x
cur ve
has
function
a
maximum
will
have
a
−π
=
–3
–4
As
the
3π
units
period
is
3 π,
there
will
be
another
maximum
–5
–6
to
maximum
the
right,
points
where
on
the
x
red
=
2π.
Mark
these
line.
–8
Use
your
knowledge
of
the
features
of
the
cosine
y
cur ve
4
and
to
the
mark
points
other
on
points,
the
axis
such
of
as
the
the
minimum
wave.
2
1
0
–2r
r
–1
x
r
2r
3r
Midway
–3
there
is
a
between
two
minimum
maximum
values,
value.
–4
–5
Midway
between
the
maximum
and
minimum
–6
values,
–8
there
(horizontal)
will
be
points
on
Circular
functions
green
axis.
{
the
Continued
on
next
page
y
Connect
these
points
and
sketch
the
function.
4
2
You
0
–2
may
want
to
erase
the
guidelines
when
your
x
r
r
2r
3r
sketch
is
complete.
–4
–6
–8
Example
Find
one
the
amplitude
cosine
and
equation
for
period,
the
then
function
write
one
shown
in
sine
the
equation
and
diagram.
y
2
1
0
–2
x
r
r
2
3r
4r
–1
Answer
3 −
The
amplitude
(
−1)
is
=
2
The
amplitude
is
one-half
the
dif ference
between
the
2
maximum
and
the
minimum
value.
3 + ( −1)
The
ver tical
shift
is
= 1
2
The
period
is
4π
The
period
function
to
find
is
a
this
5
1
= 2 sin
x
2
horizontal
one
looking
point
point
to
to
at
a
a
distance
c ycle. The
the
it
takes
easiest
horizontal
maximum
minimum
for
way
distance
point
or
the
to
from
from
a
point.
+ 1
+
the
by
maximum
minimum
y
is
complete
For
a
sine
function,
the
horizontal
translation
the
x-coordinate
is
4
found
the
by
looking
horizontal
gradient. This
standard
sine
at
axis
of
the
wave,
cor responds
to
the
this
a
point
a
point
on
positive
(0,
0)
on
the
cur ve.
In
of
with
function,
one
such
point
is
5
,1
,
so
the
4
5
horizontal
translation
is
4
{
Continued
on
next
Chapter
page
The
y
= 2 cos
1
x
4
and
amplitude,
+ 1
+
2
sine
cosine
function
have
the
same
For
a
period,
cosine
translation
maximum
to
the
standard
looking
point
(0,
on
1),
cosine
point
function,
by
point
you
at
the
can
the
translation.
find
cur ve. This
is
In
the
horizontal
x-coordinate
a
this
of
a
cor responds
maximum
function,
on
one
the
such
4
ver tical
which
cur ve.
,1
is
and
,
so
the
horizontal
translation
is
4
Remember,
that
➔
there
can
be
For
sine
written
and
translations
Exercise
For
for
for
be
a
cosine
will
more
given
than
sine
functions
differ
by
write
one
one
or
of
possible
cosine
the
equation
function.
same
one-fourth
correct
the
cur ve,
period
the
of
horizontal
the
function.
K
questions
the
may
1
to
4,
sine
and
one
cosine
equation
function.
y
y
1
2
3
2
0
–3r
–2r
x
r
r
2r
1
0
–2r
x
r
r
2r
–2
–3
–4
–5
y
y
3
4
5
4
3
3
2
2
1
1
0
–2r
x
r
–1
–2
x
–2
3
r
r
2
For
at
2
questions
least
one
5
full
to10,
Circular
2r
–3
2
a
neat
sketch
of
the
function
make
7
2
3
over
cycle.
5
r
functions
6
8
2
3r
3r
.
Many
Modeling
real-life
functions.
average
of
situations
Examples
➔
be
used
To
to
create
to
model
a
the
amplitude
●
the
ver tical
●
the
horizontal
●
the
period.
and
sine
to
Example
Create
a
modeled
at
using
heights,
section,
how
we
sine
cosine
sine
sunrise
will
and
functions
and
times,
use
our
cosine
cosine
and
knowledge
functions
data.
function
of
the
model
for
data
you
need
to
know:
function
translation
its
left
has
the
same
horizontal
of
the
amplitude,
translation
cosine
ver tical
is
translation
one-four th
of
the
cur ve.
model
measured
be
and
translation
but
the
sine
tide
this
look
function
period
period
In
cosine
●
The
can
include
temperatures.
transformations
can
wi th
off
a
for
this
buoy
data,
in
the
which
ocean
shows
over
an
the
depth
18-hour
of
the
period,
water
star ting
at
midnight.
T ime
0:00
2:00
4:00
6:00
8:00
10:00
12:00
14:00
16:00
18:00
6.7
8.3
9.1
8.1
6.4
5.6
6.7
8.4
9.2
8.2
Water
depth
(m)
Answer
Enter
and
data
depth),
GDC.
will
The
be
depth
the
into
on
will
lists
then
(label
plot
the
independent
the
be
x-axis,
the
these
data
time
on
the
variable,
and
the
dependent
time,
Be
water
variable
sure
your
RADIANS
GDC
is
in
mode.
on
y-axis.
From
the
graph,
minimum
which
occurs
maximum
Use
the
value
these
at
5.6
metres,
10:00. The
value
values
the
is
is
to
9.2
metres.
estimate
amplitude.
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
{
Continued
on
next
Casio
are
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
page
Chapter
The
an
data
is
clearly
apparent
Now
To
tr y
to
develop
horizontal
The
find
the
a
with
trigonometric
model,
estimate
translations
amplitude
and
periodic,
the
water
height
rising
and
falling
in
patter n.
minimum
is
of
the
one-half
function
the
model
amplitude,
this
period,
data.
and
ver tical
and
function.
the
ver tical
distance
between
the
maximum
values.
9.2
Estimated
to
amplitude
5.6
=
= 1.8
metres
2
The
vertical
translation
is
the
value
half-way
between
the
maximum
and
Y
ou
minimum
can
also
nd
the
values.
ver tical
translation
9.2 + 5.6
Ver tical
translation
=
by
= 7.4
subtracting
the
2
amplitude
The
period
is
the
horizontal
distance
the
function
takes
to
from
maximum
one
cycle.
The
maximum
values
are
at
4:00
and
16:00,
so
estimate
to
be
12
the
estimate
data,
points
either
the
seem
of
these
Substitute
the
easiest
to
have
horizontal
way
is
to
for
estimates
the
by
a
the
To
create
maximum
where
x
=
4
a
cosine
point.
and
The
where
model
a
the
amplitude
minimum
value.
for
plotted
x
=
16.
Use
translation.
equation
y
=
a cos(b(x
–
c))
+
d
⎞
(
⎜
for
point
horizontal
into
2π
= 1.8cos
translation.
look
maximum
x-values
these
⎛
y
or
hours.
to
Finally
,
value,
the
adding
period
a
complete
x
− 4
)
+ 7.4 .
⎟
12
Enter
axes
this
as
equation
the
data
into
the
GDC
and
graph
the
function
on
the
same
points.
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
Y
ou
are
on
–
create
you
TI-84
CD.
function
it
the
FX-9860GII
the
could
sine
T
ry
Casio
Alternative
for
a
instead.
should
get
2
y
The
function
appears
to
be
a
ver y
good
model
for
the
=
1.8sin
data.
12
Y
ou
could
Circular
tr y
to
functions
make
minor
adjustments
to
get
a
‘better
fit’.
+
7.4
x
1
Example
The
y
a
=
following
a cos (b(x
set
−
of
c))
+
data
can
be
modeled
by
the
function
d
x
1
2
3
4
5
6
7
8
9
10
11
y
4
7.6
9.4
7.6
4
2.2
4
7.6
9.4
7.6
4
Use
the
data
horizontal
b
Write
c
Graph
d
Use
the
the
estimate
function
function
regression
data,
the
period,
amplitude,
and
ver tical
and
translations.
cosine
the
the
to
and
on
which
the
same
function
graph
this
on
models
axes
your
function
on
as
the
the
GDC
the
data.
data
to
get
same
points.
a
sine
axes
as
model
the
for
data
points.
Answers
9.4
a
Amplitude
2.2
=
= 3.6
2
9.4 + 2.2
Ver tical
translation
= 5.8
=
2
Horizontal
Period
=
9
translation
−
⎛
b
3
6
+
5.8
⎞
(
⎜
3
6
2
= 3.6cos
⎝
=
=
x
3)
⎟
Be
sure
your
GDC
is
in
⎠
RADIAN
mode!
c
Use
the
Sine
Regression
under
the
menu.
the
Be
GDC
contain
function
STAT
sure
which
the
CALC
to
tell
lists
data
(x,
y).
d
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
Casio
are
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
Chapter
Exercise
L
What
For
each
set
of
real
situations
data,
modeled
Use
a
the
data
to
life
estimate
the
period,
amplitude,
and
can
by
ver tical
functions?
and
horizontal
what
translations.
adjustments
Write
b
a
cosine
function
in
the
form
y
=
acos (b(x
−
c))
+
model
the
Graph
d
Use
the
to
be
made
data.
to
c
function
on
the
same
axes
as
the
data
account
for
points.
uctuations
the
might
d
need
to
be
periodic
regression
function
on
your
GDC
to
get
a
in
the
sine
data?
model
as
the
for
the
data
data,
and
graph
this
function
on
the
same
axes
points.
1
x
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
y
11.8
8.5
2.2
5.5
11.8
8.5
2.2
5.5
11.8
8.5
2.2
x
5
10
15
20
25
30
35
40
45
50
55
y
12.5
9.3
12.5
18.9
21.9
18.9
12.5
9.3
12.5
18.9
21.9
x
2
4
6
8
10
12
14
16
18
20
22
y
1.8
2.1
1.8
1.3
0.7
0.5
0.7
1.3
1.8
2.1
1.8
When
to
you
make
have
function
(t )
be
you
can
use
that
function
Use
this
8
ii
19
Use
to
function
to
height
estimate
of
the
a
passenger
height
of
a
above
the
passenger
boarding
above
the
platform
on
the
London
Eye.
platform
boarding
after
function
+ 67.5
⎟
⎠
the
after
minutes
this
15 )
30
model
minutes
⎞
(t
⎜
used
i
2π
= 67.5cos
⎝
b
data,
⎛
a
model
function
h
can
to
predictions.
Example
The
a
to
boarding.
estimate
how
long
it
takes
for
a
passenger
to
first
reach
a
height
of
100 m.
Answers
a
i
8
minutes
after
boarding:
⎛
h
(8)
passenger
is
30
t
=
8
in
the
function.
⎞
(8
⎜
⎝
The
2π
= 67.5cos
Substitute
− 15 )
+ 67.5
≈ 74.6
⎟
⎠
approximately
74.6
metres
above
the
platform.
{
Circular
functions
Continued
on
next
page
19
ii
minutes
after
boarding:
2 p
⎛
h
(19 )
= 67.5cos
passenger
is
≈ 112.7
+ 67.5
⎟
30
⎝
The
⎞
− 15 )
(19
⎜
⎠
approximately
112.7
metres
above
the
platform.
⎛
b
h
(t )
≈
9.90
Exercise
The
depth
+ 67.5 = 100
Set
the
the
height.
function
equal
to
100,
for
⎟
30
⎠
M
depth
by the
15 )
minutes
EXAM-STYLE
1
⎞
(t
⎜
⎝
t
2π
= 67.5cos
QUESTIONS
of
the
function
of
the
water
d (t )
water
at
the
= 5.6sin
in
end
(
metres,
of
a
pier
0.5236 ( t
and
t
is
can
− 2.5)
the
be
estimated
+ 14.9
, where
)
number
of
d
hours
is
the
after
midnight.
2
a
What
b
Estimate
the
depth
of
the
water
at
midnight.
c
Estimate
the
depth
of
the
water
at
14:00.
d
At
The
is
the
what
time
average
is
the
a
the
What
is
day
What
does
c
(1
How
is
it
this
the
function?
water
first
in
a
reach
city
its
can
highest
be
depth?
modeled
( ) = ( ( − ) ) + ,
Jan
=
the
of
of
temperature
temperature
year
first
b
will
high
the function
T
period
1,
in
14
degrees
Jan
expected
=
Celsius,
14,
high
and d
is
the
by
where
day
of
etc,...)
temperature
in
this
city
on
the
Febr uar y?
the
highest
expected
temperature,
and
on
what
day
occur?
many
days
each
year
are
expected
to
remain
below
freezing?
3
A
Ferris
height
wheel
of
46
20
minutes
a
If
a
will
b
he
Write
t
How
d
For
40
is
a
wheel
the
to
the
child
of
make
for
to
boarding
length
park
minimum
Ferris
riding
function
after
high
what
amusement
and
on
after
sine
minutes
c
the
gets
be
a
an
metres
for
child
at
if
time
one
the
full
of
a
1
maximum
metre.
It
takes
rotation.
when t
=
0,
how
high
minutes?
model
he
height
wheel
10
reaches
the
Ferris
has
will
height
the
child
wheel.
been
the
of
riding
child
be
for
3
higher
minutes?
than
metres?
Chapter
EXAM-STYLE
4
The
owner
discovers
day
of
a
QUESTION
of
of
he
sells
January
,
Assuming
Let
x
How
of
c
ice
a
cream
shop
minimum
and
a
of
tracks
5
maximum
his
gallons
of
37
annual
of
ice
gallons
sales
cream
on
the
and
on
first
the
first
day
July
.
function,
b
an
the
annual
create
represent
many
an
the
gallons
sales
can
equation
to
be
modeled
model
this
by
a
cosine
situation.
month.
would
he
expect
to
sell
on
the
first
day
April?
During
what
month
would
he
expect
to
sell
30
gallons
of
Extension
material
Worksheet
13
-
temperatures
ice
cream
Review
in
one
on
CD:
Modeling
project
day?
exercise
✗
1
Given
find
2
cos 70°
values
a
cos 110°
b
cos 250°
c
cos (−290°)
Given
find
3
that
the
that
the
sin 40°
values
a
sin 140°
b
sin 320°
c
sin (−140°)
Solve
each
=
0.342
(to
three
significant
figures),
of
=
0.643
(to
three
significant
figures),
of
equation
for
−360°
≤
x
≤
360°.
a
b
c
EXAM-STYLE
4
Solve
the
QUESTIONS
equation
sin
2x
+sin
x
=
0,
for
0
≤
x
≤
π
The
5
The
graph
of
f
,
for
0
≤
x
≤
9,
is
maximum
shown.
and
a
Given
that
the
function
can
be
written
in
the
a
has
at
a
(6,
minimum
11)
at
form
(2,
graph
1)
( ) = sin ( ( − ) ) + ,
y
i
find
the
ii
explain
values
of
a,
c
and
d.
12
why
10
8
b
Write
down
the
inter val
for
which
( ) >
6
4
2
0
Circular
functions
1
2
3
4
5
6
7
8
9
x
2
6
Given
that
cos x
=
,
and
x
is
an
acute
angle,
find
5
a
sin
x
tan
b
x
sin
c
2x
2
7
Sketch
the
graph
of
the
function
f
x
x
−3
≤
x
≤
Review
1
Solve
a
2
Solve
a
=
0.75
The
equation
cosx
of
f
,
=
−2π
for
cosx
b
3
−180°
≤
x
≤
360°.
−0.63
≤
θ
c
tanx
c
=
=
3x
−
1
≤
x
≤
7,
is
0
for
−2.8
2π
≤
cos
graph
for
b
3
5.
equation
each
2sin
2,
exercise
each
sinx
1
5
for
3cos
shown.
A(2, 7)
7
a
Given
that
the
function
can
be
written
in
the
form
6
f
( x ) = acos bx +
c,
5
find
b
the
Write
values
down
of
the
a,
b
and
solutions
c.
4
to
the
equation
3
2
1
4
The
depth
of
the
water
at
the
end
of
a
fishing
pier
is
given
by
0
the function
of
the
water
in
metres,
x
1
,
where
D
is
the
depth
2
–1
B(4, –1)
and
t
is
the
number
when
the
depth
of
hours
after
midnight.
Low
high
tide
tide
Find
b
Sketch
c
At
d
the
a
what
Fishing
The
at
values
graph
time
8
is
4:00,
10:00,
of
P
of
the
does
when
and
of
the
of
depth
the
of
water
the
is
water
6 m,
is
and
14 m.
Q
function
the
prohibited
metres.
longest
day
15
hours
21
December,
of
h ( x ) =
x
How
of
with
of
water
a
Find
b
How
1
on
the
1
D,
first
for
0
≤
reach
a
t
≤
24.
height
of
many
year
9.35
of
the
water
hours
in
a
city
shor test
hours
of
daylight
depth
each
is
day
day
is
is
21
June,
of
the
less
fishing
prohibited?
with
year
is
daylight.
can
( x − ) + B, where
be
x
modeled
is
the
day
by
of
the
the
function
year
Jan).
value
many
when
The
hours
A
=
the
daylight.
The number
(i.e.
at
metres?
than
5
occurs
a
8
occurs
of
hours
A
and
of
of
B
daylight
would
you
expect
on
1
Feb?
Chapter
CHAPTER
Using
●
the
The
unit
of
unit.
1
will
13
unit
circle
meet
SUMMARY
The
the
circle
has
its
center
terminal
unit
side
circle
at
at
the
of
a
origin
point
(0, 0)
angle θ
any
with
in
and
a
radius
standard
coordinates
length
position
(cosθ, sinθ).
B(cos i, sin i)
i
A(1, 0)
x
0
●
For
any
θ,
angle
,
where
cosθ
≠
0.
●
For
any
θ:
angle
■
sinθ
■
cosθ
=
cos(−θ)
■
tanθ
=
tan(180
=
sin(180
θ)
−
Trigonometric
θ)
+
identi ties
2
●
The
●
The
for
cos(2θ
equation
all
values
)
θ
= 1 − 2sin
is
an
identi ty,
as
it
is
tr ue
θ
of
double-angle
identity
for
cosine
for
sine
are:
2
cos(2θ)
= 1
−
2sin
θ
2
θ − 1
= 2cos
2
=
●
The
cos
double-angle
Graphing
●
The
sine
Both
Like
the
There
does
●
of
period
of
and
amplitude.
function
sine
cur ve.
the
have
axes.
graphs
The
maximum
=
2 sin θ cos θ
of
the
functions
value
of
1
same
size
are periodic
and
a
and
shape
with
minimum
but
period
value
of
different
360°
–1,
and
It
y
same
at
cycle
the
values
of
tangent
of
values
x
function
Circular
functions
2π.
an
where
repeats
the
is periodic.
function
between
each
tangent
functions,
has
=
function
no
sin (x)
cur ve
the
+
d
is
a
up
180°
tangent
maximum
shifts
is
or
d
is
radians).
function
minimum
vertical
if
(or π
does
not
have
values.
translation
positive,
Unlike
down
of
if
the
d
is
standard
negative.
Continued
or
asymptotes.
the
The
functions,
asymptotes
cosine
The
on
cosine
The
ver tical
sine
sin (2θ )
functions
functions
a
is
1.
and
exist.
pair
an
●
not
of
have
ver tical
The
the
positions
sine
are
θ
identity
cosine
functions
ampli tude
sin
circular
and
horizontal
●
2
θ −
on
next
page
●
The
function
standard
left
●
if
The
c
The
is
cur ve.
standard
●
The
of
c)
is
=
cos (x)
The
y
=
+
cur ve
cos (x
c)
y
=
asin x
and
a
is
will
and
to
of
translation
the
right
if
c
is
by
from
d
is
=
acos x
value
to
to
the
|a|.
right
vertical
the
y-value
of
down
the
positive,
graph
in
the
if
the
if
translation
are
When
c
d
of
standard
is
negative.
the
is
stretches
of
a
function
original
a
ampli tude
1
positive,
shifts
ever y
the
the
if
of
translation
horizontal
functions.
stretch,
change
vertical
up
a
y
stretch,
multiplied
of
The
the
sine
period
or
of
cosine
the
function
change.
functions
horizontal
When
cosine
ver tical
ver tical
not
shifts
cur ve
functions
function
horizontal
a
is
The
negative.
sine
is
shifts
−
cur ve.
d
is
a
a
cur ve
c
With
The
−
The
if
the
left
function
●
sin (x
cur ve.
cosine
undergoes
will
y
function
positive,
=
negative.
function
cosine
●
y
cosine
the
y
=
sin (bx),
stretches
graph
of
a
of
y
=
the
cos (bx)
sine,
function
and
cosine
y
and
undergoes
a
=
tan (bx)
tangent
represent
functions.
horizontal
stretch,
ever y
x-value
in
the
original
function
is
multiplied
by
.
●
When
a
stretch,
sine
the
or
cosine
period
of
function
the
function
2π
to
undergoes
will
a
horizontal
change
from
,
or
from
360°
to
●
For
a
function
in
the
form
y
=
from
π
tan (bx),
,
or
from
180°
sine
and
translations
Modeling
●
●
To
create
a
cosine
will
with
cosine
amplitude
■
the
ver tical
■
the
horizontal
■
the
period.
sine
of
the
by
and
function
the
of
the
one-fourth
sine
of
sine
same
the
cur ve,
period
cosine
model
for
data
the
of
horizontal
the
function.
functions
you
need
to
know:
function
translation
function
and
period
functions
differ
the
translation
wi th
functions
■
The
change
to
transformations
cosine
For
will
Combined
●
period
to
and
the
to
translation
has
the
period
the
same
but
left
of
its
amplitude,
horizontal
the
cosine
ver tical
translation
is
one-four th
cur ve.
Chapter
Theory
of
knowledge
‘pure’
and
‘applied’.
areas?
A
a
is
t
weigh
as
g,
sprin
the
t
weigh
d
pulle
we
sed,
relea
e
oscillat
up
it
h
at
t
heigh
by
given
is
s
second
t
t,
weigh
the
of
t
heigh
ion
oscillat
the
down
time
and
let
g
restin
n
Whe
n.
show
is
If
on
ded
suspen
will
h(t)
.
down
and
t
weigh
The
5cm
s.
second
b
and
e
Ignor
c))
d
pulle
full
two
s
value
the
Find
down
one
tes
ever y
of
c
the
ion
frict
is
comple
and
ion
oscillat
a,
–
(t
(b
sin
a
=
ts
eec
air
and
of
e.
tanc
resis
s
far
ws
as
the
of
atics
This
question
ignore
the
should
will
is
an
effects
oscillate
reduce
example
pure
friction
and
air
indefinitely
.
But
in
until
of
of
the
weight
comes
mathematics.
resistance,
real
to
life,
the
the
If
refer
we
,
weight
they
n;
oscillations
are
and
rest.
do
What
is
the
studying
point
pure
problems
like
of
Should
mathematics
applied
this,
which
when
we
only
study
mathematics,
could
have
reality.”
some
Albert
the
in
Theory
of
results
real
are
unrealistic
practical
life?
knowledge:
use?
Einstein,
Sidelights
on
Relativity
Pure
vs.
applied
mathematics
There are 10
Pure
mat
without
types of
ha
mathematic
and
solve
people in
this world: those who
pr
understand binary
physics,
econo
and those who don't.
Application
Practical
applications
George
Boole,
an
English
mathematician,
pure
mathematics
after
the
rst
−
ideas
s
developed
1850s.
wer
his
This
Boolean
system
Logic
was
system
later
used
in
in
the
digital
electronics.
Modern
computers
number
system.
use
German
In
physics,
elementar y
par ticles
were
3
Gottfried
Wilhelm
Leibniz
discovered
wrote
about
this
number
only
0s
and
1s,
in
symmetr y
he
was
studying
this
be
used
300
“Physics
not
because
physical
it
is
we
world,
only
its
years
Russell,
of
the
the
mathematics.
all
‘pure’
mathematics
will
be
used
later
.
so
much
because
we
mathematical
we
elegance
mathematical
know
but
that
Bertrand
is
or
‘p
Perhaps
would
involving
the
underlying
When
arguments
sy
beauty,
uses
through
can
about
know
so
the
little;
proper ties
discover.”
English
mathematician
and
{
philosopher
(187
rge
Boole
1970)
15–64)
Pure
ma
entirely
game
new
mathematical
disciplines,
such
as
statistics
and
theor y
.
Can
we
model
the
real
world
mathematics
because
we
mathematics
to
the
because
the
mirror
world
is
with
create
world,
What
does
this
relationship
or
sciences,
intrinsically
tell
us
between
about
the
mathematics
the
natural
and
the
natural
world?
mathematical?
Is
mathematics
invented
or
discovered?
Chapter
Calculus
with
trigonometric
functions
CHAPTER
OBJECTIVES:
6.1
T
angents
6.2
Derivative
6.3
and
rule,
Local
maximum
+
b,
integrals,
of
Kinematic
distance
Before
Y
ou
1
integral
of
Find
the
by
rules,
points,
between
and
with
a
or
boundar y
between
about
the
involving
differentiation
second
points
graphs
cos x,
inspection,
areas
problems
of
including
condition
cur ve
Use
f,
f ′
and
sum
of
inexion,
to
of
these
how
value
unit
of
displacement
with
form
determine
s,
functions
of
the
1
linear
constant
areas
cos
v
and
acceleration
the
values.
identities
exact
value
of
b
in
to
cos
4
2
11
4
=
2x
–
–
1
2cos
x
+
(2cos
x
–
=
cos
–
cos
x
for
0
≤
x
≤
3
2π.
x
=
–
cos
2
x
Solve
each
cos
x
–
1)(cos
x
1
=
+
0
1)
=
0
a
1
+
b
sin
tan
equation
2x
x
–
=
sin
for
c
1
or
cos x
sin
x
=
cos
x
=
0
1
cos
x
+
+
1
2
5
,
,
3
3
Find
the
derivative
3
3
3
Use
the
product,
quotient
and
chain
a
f
(x)
=
2x
b
f
(x)
=
x
c
f
(x )
x
e
r ules
2
to
find
ln(x
derivatives.
x
2
e.g.
Find
the
derivative
of
f
(x)
=
x
ln
x.
5
2
x
4
2
f
(x)
=
ln
x
x
ln x
d
2
f
( x )
1
(ln x )(2 x ) =
Calculus
with
x
x
f
(x )
=
x
trigonometric
functions
+
2x
ln
x
x
)
0
≤
x
2
x
2
total
sin
d
tan
2
x
a,
solve
2
cos x
cur ves,
3
7
2
x
term,
between
6
2cos
function
check
Find
c
2x
functions,
f (g(x))g ′(x)dx
the
x-axis,
velocity
Skills
to:
trigonometric
circle
trigonometric
cos
multiple,
start
know
for
Solve
real
x-axis
equations.
e.g.
a
f ″
the
the
and
graphs
composites
and
a
2
a
derivative
of
substitution
the
of
traveled
exact
functions
quotient
sin x
equations
including
minimum
revolution
you
should
their
tan x,
relationship
Anti-differentiation
volumes
6.6
and
and
integration
denite
and
cos x,
product
the
Indenite
ax
6.5
sin x,
chain
including
6.4
of
normals,
of
cos
x
≤
2π
At
The
Chocolate
Califor nia,
is
driven
bottom
In
the
by
of
vat
a
the
into
the
other
in
milk
Ghirardelli
chocolate
that
pushes
is
the
Square,
stirred
blade
San
by
back
a
Francisco,
stirrer
and
blade
forth
which
across
the
vat.
stirrer,
periodic
mechanism.
and
of
wheel
chocolate
translated
the
a
Factor y
One
end
the
linear
end
of
periodic
the
of
motion
a
rod
circular
rod
is
is
of
the
motion
blade.
attached
attached
to
to
the
of
The
a
the
wheel
diagram
wheel
or
is
shows
wheel
rod
crankshaft
i
stirrer
vat
blade
inside
the
vat.
blade
backwards
As
the
wheel
turns,
the
rod
pushes
the
stirrer
blade
between
the
and
center
of
forwards
the
wheel
across
and
the
the
vat.
stirrer
The
distance
blade
can
be
modeled
d
2
by
a
function
distance
To
find
from
in
Many
metres
the
the
this: d (θ)
like
angle
center
real
and θ
of
of
world
is
2cos θ +
=
the
rotation
the
angle
when
wheel,
you
phenomena,
of
25
rotation
the
blade
would
such
as
4 sin
heart
of
is
use
,
where
the
the
the
d
is
wheel
in
shor test
derivative
rhythms,
the
radians.
distance
of
d (θ).
y
movement
tan x
2
of
hands
on
a
clock,
the
tides
and
circular
motion,
have
periodic
sin x
behavior
–
they
follow
a
pattern
that
recurs
at
regular
intervals.
x
r
cosine
and
tangent
–
which
are
periodic
functions.
Y
ou
can
see
3r
2
r
r
r
2
2
r
r
r
2
cos x
from
In
their
this
and
graphs
chapter
tangent
order
to
that
you
will
functions
investigate
each
find
and
the
function’s
values
derivatives
integrate
behavior
of
of
sine
–2
recur.
the
and
periodic
sine,
cosine
cosine
functions,
functions
like
this
in
one.
Chapter
.
In
Derivatives
Chapter
constant
7
you
real
met
of
these
trigonometric
proper ties
of
functions
derivatives,
where c
is
a
number.
d
Constant
c
rule:
dx
d
Constant
multiple
[cf
rule:
( x )
( x ) ] cf
dx
d
Sum
or
dierence
[ f
rule:
(x )
±
g ( x )]
=
′( x )
f
g ′( x )
±
dx
d
Product
[
rule:
f
( x ) g ( x )]
( x ) g ( x )
f
g(x )
f
( x )
,
g(x )
dx
d
Quotient
f
(x )
g(x )
rule:
( x )
f
f
g ( x )
(x )
dx
g(x )
0
2
g ( x )
d
Chain
[
rule:
( g ( x ))]
f
( g ( x )) g ( x )
f
dx
Investigation:
Here
the
is
the
graph
questions
of
The
f (x)
=
derivative
sin x
for
–2π
≤
x
of
≤
sine
2π.
below.
Use
it
to
answer
f(x)
The
There
1
are
four
values
of
x
in
gradient
horizontal
f(x)
–2π
≤
x
2π
≤
gradient
of
where
the
the
tangent
line
a
is
0.
values
of
sin x
=
1
line
of
2
So
at
the
to
x
where
the
tangent
x
3r
f (x)
=
sin
x
is
equal
to
r
r
zero.
r
r
3r
lines
2
What
are
these
the
2
List
values
is
that
the
are
to
is
true
decreasing,
this
plot
and
what
sketch
Use
a
GDC
–2π
≤
x
≤
Compare
the
those
to
2π.
the
the
is
of
Be
graph
is
sign
the
GDC
is
of
of
the
plotted
in
derivative
of
in
derivative
f (x)
=
your
against
of
When
f ?
4
5
question
Make
a
f (x)
sin x.
=
Verify
of
your
What
with
Adjust
based
function
conjecture
for
chose
Calculus
2.
conjecture
values
you
in
in
the
your
on
drawing
the
do
graph
you
numerically
function
question
you
think
with
graphed
4.
trigonometric
of
f
is
f
is
When
0.
=
sin x
increasing,
f
is
derivative
of
to
x
f (x)
question
sin x
radian
on
f
graph
decreasing.
you
of
of
the
derivative
derivative
the
derivative
where
the
points
your
of
it
the
about
graph
the
sure
of
the
2π
≤
where
the
the
graph
x
sign
tr ue
and
≤
of
of
1
f ?
to
Use
make
a
f
in
the
inter val
Enter
mode.
GDC
to
the
one
the
functions
if
of
is
a
in
the
GDC
derivative
derivative
by
of
of
sine?
comparing
question
3
(sin( x ))
dx
necessar y.
the
graph:
d
you
f 1( x )
drew
horizontal
four
graph
–2π
of
about
information
possible
3
on
inter vals
increasing
what
are
derivative
equal
points
f
they?
–2
Use
to
2
and
the
the
table
function
d
In
the
investigation,
you
should
have
found
Y
ou
that
(sin x )
Now
consider
the
derivative
of
f
(x)
=
cos
can
examine
a
cos x
dx
geometric
x
of
this
justication
fact
in
the
TOK
section
If
you
translate
the
graph
of
sine
to
the
left
,
you
get
the
cosine.
So
f
(x )
cos x
sin
x
the
end
of
graph
2
of
at
this
chapter
.
f(x)
2
2
f(x)
=
r
2
cos x
f(x)
=
sin x
1
x
3r
r
r
r
2
r
r
2
–2
d
Hence,
d
(cos x )
sin
dx
dx
x
Using
d
2
cos
x
(1)
2
x
rule:
2
2
x
cos
x
d
2
cos
x
cos
sin
chain
dx
the
x
2
2
dx
1
If
you
translate
the
graph
of
cosine
to
the
left
,
you
get
a
2
reflection
of
the
graph
of
sine
in
the x-axis.
f(x)
So
r
2
f(x)
f
(x )
cos
–sin x
f(x)
r
r
2
d
conclude
that
(cos x )
cos
x
consider
the
derivative
s
f
(x )
tan x
of
f
(x)
=
tan
x.
2
We
r
r
2
2
sin x
2
r
–1
dx
Finally
x
0
3r
we
cos x
sin x
2
Therefore
=
x
=
–2
know
that
n x
,
where
cos
x
≠
0.
cos x
d
d
(tan x )
So,
⎛
sin x
⎞
=
⎜
dx
dx
⎝
⎟
cos x
⎠
cos x (cos x ) − sin
x ( − sin
Apply
x )
the
quotient
=
2
rule.
(cos x )
2
cos
2
+ sin
x
=
2
cos
x
Use
the
identity
2
1
cos
=
,
cos x
≠
2
θ
+
sin
θ
=
1
to
0
2
cos
➔
Derivatives
f
(x)
=
sin
f
(x)
=
cos
of
x
x
simplify
x
sine,
⇒
⇒
f
cos
′(x)
f
′(x)
=
and
cos
=
–
the
numerator
.
tan:
x
sin
x
1
f
(x )
tan x
f
( x )
,
cos x
0
2
cos
x
Chapter
Example
In
the
18th
Find
the
derivative
of
each
17th
and
centuries
the
function.
development
of
1
a
f
(x)
=
sin x
+
cos x
c
y
mechanical
=
devices
tan x
shifted
2
y
b
=
cos(t
the
d
f
(x)
=
sin
trigonometr y
(2x)
original
Answers
a
f
(x)
f
′(x)
sin
=
x
cos
+
x
cos
–
Take
x
sin
the
derivative
of
each
ter m.
=
cos ( t
periodic
Joseph
x
Apply
)
the
chain
rule,
where
outside
2
=
[ − sin(t
)]
derivative of
outside
with
u(t)
=
cos t
motion.
Fourier
a
mathematician
and
function
re
espect
respect
to inside function
and
2
the
derivative of
is
to
modeling
[2t ]
function
function
its
the
French
y′
to
(1768–1830),
2
y
b
of
from
connection
triangles
=
eld
3
)
inside
function
is
v (t)
=
physicist,
found
t
inside
that
with
almost
any
to t
periodic
function,
such
the
2
=
−2t
sin( t
)
Rewrite
using
rational
vibration
exponents.
of
1
y
c
as
a
violin
string
or
=
the
tan x
movement
of
the
1
= ( tan x )
pendulum
2
y′
=
1
⎛
Apply
−1(tan x )
⎜
⎝
the
chain
rule,
where
could
the
be
a
clock,
expressed
as
⎟
2
cos
x
⎠
–1
outside
1
function
is
u (x)
=
x
and
an
innite
sum
of
sine
1
= −
or
2
tan
on
⎞
2
x cos
the
2
x
sin
inside
function
is
v (x)
=
tan
x.
and
cosine
functions.
The
oscillation
x
Apply
the
chain
rule
twice.
First
3
d
f
(x )
=
sin
(2 x )
3
the
outside
function
is
u(x)
=
of
a
x
3
spring
=
( sin
=
3
(2 x ) )
and
the
inside
function
is
v(x)
and
=
movement
2
f
′( x )
( sin
(2 x ) )
sin (2x). Then
( cos
when
finding
the
of
a
the
(2 x ) ) (2 )
pendulum
derivative
of
sin (2x),
the
are
outside
2
=
6 sin
( 2 x ) cos ( 2 x )
examples
function
is
u(x)
=
sin x
and
harmonic
inside
function
is
v(x)
=
are
trigonometric
functions
A
used
to
motion?
In
questions
1
f
3
y
(x)
=
1–10,
3sin x
–
find
the
derivative
2cos x
of
each
2
y
=
4
s (t)
function.
tan (3x)
2
2
=
cos
t
sin x
2
5
f
(x )
sin
x
6
y
8
f
=
tan
x
7
y
=
cos
x
1
+
sin ( 4 x )
(x )
=
2
cos ( 2 x )
4
9
y
10
2
sin
Calculus
( x )
with
trigonometric
functions
f
(x)
=
sin (sin
x)
simple
motion.
2x.
How
Exercise
of
the
and
calculus
model
this
EXAM-STYLE
QUESTIONS
Differentiate
11
with
respect
3
tan
a
A
12
(x
to
x
4
)
cos
b
function
has
the
x
equation
y
=
sin(3x
–
4).
2
dy
Find
a
d
.
y
Find
b
2
dx
Example
Find
the
dx
equations
of
the
tangent
line
and
the
normal
line
to
the
cur ve
f
(x)
=
cos
3x
at
the
point
where
x
9
Answer
p
f
9
cos
3
9
Evaluate
find
of
tangency
=
the
point
of
to
tangenc y.
1
is
,
9
x
2
( x )
at
9
f
f
3
point
function
1
cos
The
the
3 sin
2
3 x
Find
the
derivative
of
f
and
‘Slope’
p
f
9
3 sin
3
9
3 sin
3
3
evaluate
it
at
x
to
=
find
slope
3
3
of
the
tangent
is
another
the
word
9
for
‘gradient’.
line.
2
3
2
x
at
3
The
3
is
9
2
slope
of
the
normal
line
at
The
2
x
2
is
9
to
or
3
3
nor mal
the
tangent
negative
line:
y
3
3
x
per pendicular
so
the
slopes
are
2
1
line:
y
2
3
9
x
9
Use
the
line,
y
–
y
=
1
9
point–slope
equation
for
a
2
reciprocals.
2
Normal
is
line,
9
1
Tangent
line
3
m(x
–
x
),
to
write
the
1
equations.
Exercise
In
B
questions
normal
line
1
to
and
the
2,
find
cur ve
the
at
equations
the
given
of
value
the
of
tangent
line
and
the
x.
1
f
(x )
sin x
cos x ;
f
(x )
2 tan x ;
x
2
2
x
4
Chapter
EXAM-STYLE
QUESTIONS
Most
⎛ p
The
3
point
⎞
P
the
on
the
graph
of
y
=
sin
f
(x)
⎠
gradient
=
elds
of
the
tangent
to
the
cur ve
at
can
be
Write
b
Find
cos
other
P
.
by
function.
function
is
an
An
a
function
(2x).
down
the
value
that
f
of
is
algebraic,
transcendental,
or
.
a
3
f
sciences,
and
modeled
elementary
a
the
(2x).
elementar y
Let
4
in
business
⎟
2
⎝
Find
lies
,0
⎜
phenomena
engineering ,
sum,
difference,
product,
quotient
′(x).
or
composition
of
algebraic
and
Find
c
the
equation
of
the
tangent
line
to f
x
at
transcendental
functions.
3
Algebraic
Consider
5
Find
the
the
function
value(s)
of
x
f
(x)
for
=
3
sin
which
x
the
for
0
≤
x
tangent
≤
functions
2π.
lines
●
Polynomials
●
Rational
●
Functions
to
functions
3
the
graph
of
f
are
parallel
to
the
line
y
x
4
involving
radicals
2
Transcendental
(cannot
.
More
practice
wi th
be
now
know
the
derivatives
of
d
these
n
radicals
[x
1
nx
] =
, n
≠ 1
[sin
dx
dx
d
d
x
[e
x] =
e
] =
[cos x ] =
1
d
[ln x ] =
x
,
>
sums,
and
involving
x
)
●
Logarithmic
functions
●
Exponential
functions
●
T
rigonometric
●
Inverse
functions
− sin x
dx
d
as
quotients
cos x
x
dx
products,
n
functions:
d
n
expressed
derivatives
difference,
Y
ou
functions
trigonometric
functions
1
0
[tan x ] =
,
cos x
≠
With
0
the
exception
of
the
inverse
2
dx
x
dx
cos
x
trigonometric
Using
these
facts
and
the
r ules
stated
at
the
differentiate
of
Section
variety
of
14.1,
can
find
the
derivatives
f
the
(x)
you
a
wide
any
can
now
elementar y
function.
derivative
of
each
function.
2x
a
of
almost
functions.
Example
Find
you
function,
beginning
=
4e
3
+
sin
(3x
+
2)
c
y
=
d
s(t)
cos
x
sin x
x
b
y
=
e
sin x
=
ln(sin t)
Answers
2x
f
a
(x)
=
4e
′(x)
=
4(e
+
=
8e
sin
(3x
+
2)
Use
the
constant,
multiple
and
chain
rules
to
2x
f
)(2)
+
[cos
(3x
+
2)]
(3)
dif ferentiate
the
first
dif ferentiate
the
second
ter m
and
the
chain
rule
to
2x
+
3cos
(3x
+
2)
ter m.
x
b
y
=
e
sin x
x
y ′
=
e
x
(cos x)
+
sin x
(e
)
Use
the
product
rule.
Use
the
product
rule
x
=
e
(cos x
=
cos
=
(cos x)
+
sin x)
3
y
c
x
sin x
3
sin x
3
y ′
=
(cos x)
2
(cos x)
4
=
d
cos
s (t )
=
s ′( t )
=
+
2
x
–
3cos
sin x
(3(cos x)
)
(–sin x)
x
sin
when
x
finding
Apply
sin t
Calculus
with
1
cost
(cost )
apply
the
chain
3
the
derivative
ln(sin t )
1
and
2
=
or
sin t
trigonometric
tan t
functions
the
chain
rule.
of
(cos x)
rule
Exercise
In
C
questions
1–10,
find
f
1
(x )
6 cos
2x
3
(x)
=
of
each
y
2
=
1 + cos x
x
xe
–
function.
sin x
3x
x
f
derivative
3
the
1
e
sin 2 t
4
s (t )
=
e
2
x
5
f
7
y
9
f
(x)
=
e
(sin
x
–
cos
x)
s(t)
6
=
t
tan
t
3x
=
e
(x)
cos
=
(ln
4x
x)(cos
EXAM-STYLE
x)
8
y
10
f
tan 2 x
(x)
=
ln
(cos
x)
QUESTIONS
2
11
Let
a
f
(x)
=
ln(3x
).
Write
down
f
′(x).
x
g(x )
Let
b
sin
.
Write
down
g ′(x).
2
x
2
Let
c
h( x )
ln(3 x
.
) sin
Find
h ′(x).
2
2
sin x
Given
12
f
that
(x )
cos x (1 + a cos
and
f
′( x )
find
Y
ou
a
can
and
use
+ b sin
x )
,
2
1 cos
2
x
=
2
x
(1 + cos
2
x )
b
the
first
and
second
derivatives
of
a
function
See
to
analyze
the
graph
of
a
Chapter
Section
Example
Consider
a
Find
7,
function.
7.
the
the
function
x-
and
f
(x)
=
sin x
+
cos x
for
0
b
Find
the
inter vals
on
which
f
is
increasing
c
Find
the
inter vals
on
which
f
is
concave
d
Use
the
≤
x
≤
2π.
Analyze
it
without
using
a
GDC.
y-intercepts.
information
from
par ts
a
to
c
and
up
to
decreasing
and
sketch
concave
the
and
down
graph
of
the
relative
and
the
extreme
inflexion
points.
points.
f
Answers
To
a
sin
x
+
cos
x
=
find
solve
sin
x
=
–cos
3
x
x.
x-intercept,
Use
your
set
the
function
knowledge
of
unit
equal
circle
to
0
and
values
to
x
find
7
the
solutions.
4
7
3
x-intercepts:
and
4
(0)
for
,
4
f
the
0
=
sin 0
=
0
=
1
+
+
1
y-intercept:
4
cos 0
To
x
find
=
the
y-intercept,
evaluate
the
function
when
0.
1
{
Continued
on
next
Chapter
page
b
f
(x)
f
′(x)
=
sin x
=
cos x
–
cos x
=
cos
+
x
sin x
cos x
=
′(x)
=
0
at
0
f
is
x
f
x
5
x
x
The
–
5r
4
4
maximum
point:
,
0.
and
decreasing
2r
Evaluate
cos x
– sin x
–
cos x
– sin x
=
derivative
at
x
that
relative
changes
extrema
sign.
5
and
4
find
the
maximum
and
minimum
values.
4
Find
=
f
us
2
,
–
first
tells
4
to
– sin x
the
5
Relative minimum point:
when
test
2
4
derivative
0
f
cos x
the
″(x)
Make
7
f
is
=
a
second
derivative
of
f
and
find
where
0.
sign
concave
diagram
up
when
f
is
for
f
″
f
is
″.
positive
and
concave
,
4
4
down
when
up:
x
f''(x)
4
–
0
3
down:
0
x
x
2
7
3
,
0
and
4
–
7r
4
4
2r
4
Inflexion
points:
+
3r
7
and
4
Inflexion
negative.
4
Concave
″
7
3
Concave
points
occur
when
the
,
changes
0
second
derivative
3
sign.
Evaluate
f
at
x
7
and
to
find
4
4
4
the
d
=
+
r
first
occur
′(x)
negative.
+
2
4
x
positive
is
f
4
3
′
where
′.
5
Decreasing:
=
f
find
f
4
0
″(x)
when
and
f'(x)
and
f
is
f
for
4
4
c
′
of
diagram
,
Relative
derivative
sign
5
0
a
increasing
when
4
Increasing:
the
Make
sin x
f
Find
– sin x
y-coordinates
of
the
inflexion
points.
f(x)
r
2
√2
(
)
4
1
x
0
–1
r
r
3r
4
2
4
r
5r
3r
7r
4
2
4
2r
5r
–2
, –
√2
(
)
4
Derivatives
are
useful
for
finding
both
relative
extrema
and
Absolute
absolute
extrema
on
a
closed
extrema
inter val.
sometimes
'global
Calculus
with
trigonometric
functions
called
extrema'.
are
Example
Show
a
f
(x)
how
=
ln x
to
+
use
the
sin x
on
second
0
≤
x
≤
derivative
test
to
find
the
x-coordinates
of
the
relative
extrema
of
2π
2
Find
b
the
global
extrema
of
the
function
f
(x)
=
x
+
sin (x
)
on
the
closed
inter val
0
≤
x
≤
π
Answers
Find
f
a
(x)
=
ln
x
+
sin
it
1
′( x ) =
f
the
first
derivative
and
set
x
equal
to
zero
to
find
the
critical
+ cos x
x
numbers.
Use
a
GDC
to
solve.
1
+ cos x
= 0
x
x
≈
2.04,
4.48
Find
the
second
derivative
1
f
″( x )
=
− sin x
and
evaluate
each
of
the
2
x
critical
f
″(2.04)
relative
f
≈
1.11
<
maximum
″(4.48)
relative
–
≈
0.925
⇒
at
>
minimum
0
0
at
x
=
2.07
f
⇒
x
numbers
derivative
″
>
and
=
0
f
in
the
implies
″(x)
<
0
from
the
second
a
relative
implies
first
derivative.
a
minimum
relative
maximum.
4.49
2
f
b
(x)
=
x
+
sin
(x
)
Find
the
first
derivative.
2
f
′(x)
=
1
+
2x cos (x
)
Set
it
equal
the
critical
to
zero
to
find
2
1
x
+
=
2x cos
(x
1.392,
)
=
0
2.115,
2.834
Use
a
Evaluate
f
(0)
=
f
(1.392)
≈
2.33
f
(2.115)
≈
1.14
(2.834)
≈
the
The
(π)
at
the
the
critical
the
first
numbers
is
3.82
and
GDC
the
is
is
help
and
GDCs
Exercise
D
Do
a
For
maximum
1
f
(x)
=
2
f
(x)
=
For
2 sin
points.
f
x
(x )
2,
of
x
+
+
find
cos
largest
value
is
the
global
maximum
and
the
the
minimum.
on
CD:
Alternative
for
find
any
x,
≤
0
0
the
≤
concave
points,
0
x
x
on
≤
the
TI-84
≤
to
on
the
CD.
minimum
the
given
points
and
relative
inter val.
2π
up
relative
are
FX-9860GII
2π
intervals
information
sin x ,
relative
x
Casio
1–5.
function
2x,
decreasing,
this
questions
the
cos
3–4,
minimum
Use
for
and
3 sin
questions
relative
1
points
increasing,
3
GDC
questions
from
0.
Plus
✗
of
2.71
maximum
use
each
derivative. The
demonstrations
not
endpoints
and
3.82
≈
minimum
solve.
inter val
smallest
f
f
to
0
of
f
numbers.
GDC
on
and
which
concave
maximum
sketch
a
the
function
down.
points
graph
of
Find
and
the
is
any
inflexion
function.
2
4
f
(x)
=
cos
(2x),
0
≤
x
≤
π
Chapter
EXAM-STYLE
QUESTIONS
2
5
Let
f
(x)
=
a
Show
b
f
has
0
≤
x
cos
that
one
c
Find
f
d
Find
the
6
may
Let
a
f
use
(x)
f
ii
b
f
a
≤
x
+
x
can
≤
point
coordinates
of
on
the
this
inter val
point.
and
x
of
the
inflexion
point(s)
of
f
on
the
π
≤
for
sin
be
questions
6–8.
x
expressed
in
the
form
ax
sin
x
+
b
cos
x.
b
the
equation
use
minimum
0
minimum
′(x).
Hence
ii
x
–3sin 2x
the
GDC
π
=
Solve
i
Find
0
″(x)
Find
=
coordinates
a
Find
i
′(x)
cos
″(x).
inter val
Y
ou
f
+
relative
π.
≤
2x
f
″(x)
to
points
f
′(x)
=
0
identify
and
any
for
the
0
≤
x
2π
≤
x-coordinate
relative
of
maximum
any
relative
points
of
f
for
2π
≤
2
7
Let
f
(x)
=
a
Find
f
b
Hence
x
cos
x
′(x).
2
find
inter val
8
The
≤
x
milk
pushes
Suppose
stirrer
the
that
blade
in
is
can
the
of
machine
Ghirardelli
stirred
blade
the
extrema
f
(x)
=
x
cos
x
on
the
5.
shows
Factory
chocolate
global
≤
photograph
Chocolate
that
0
the
by
back
distance
be
a
Square,
stirrer
and
by
San
chocolate
Francisco.
that
across
the
the
stirs
blade
forth
between
modeled
that
is
the
center
A
driven
bottom
of
the
in The
vat
by
of
a
of
wheel
the
wheel
vat.
and
the
function
2
d (q ) = 2 cos q
where
the
d
is
wheel
the
in
+
q
25 − 4 sin
distance
in
metres
and
θ
is
the
angle
of
rotation
of
radians.
d ′(θ).
a
Find
b
Sketch
i
a
graph
coordinates
maximum
c
i
of
of
all
d ′(θ)
for
0
≤
x-intercepts
θ
≤
and
2π,
and
relative
label
the
minimum
and
points.
Explain
how
to
use
the
graph
of
d ′(θ)
to
determine
the
d
angle
from
of
rotation
the
center
when
of
the
the
blade
wheel.
is
What
at
is
the
shor test
this
angle
distance
and
this
distance?
ii
At
which
center
of
Explain
Calculus
with
angle(s)
the
how
of
wheel
you
trigonometric
rotation
and
the
determine
functions
is
the
blade
your
distance
changing
answer.
between
the
the
fastest?
.
Y
ou
Integral
met
these
of
sine
integration
and
r ules
in
cosine
Chapter
9.
1
n
Power
x
rule:
n +1
dx
x
=
+ C,
n
≠ 1
n + 1
k dx
Constant
rule:
Constant
multiple
Sum
or
dierence
=
kx
+
C
kf (x) dx
rule:
(
rule:
f (x)
±
=
k
f (x) dx
g (x)) dx
=
f (x) dx
±
g (x) dx
1
x
Integrals
of
and
e
dx
:
=
ln x
x
e
Integral
with
+ C ,
x
> 0
x
x
linear
x
dx
=
e
+
C
composi tion:
1
f
( ax
+ b )d x
=
F ( ax
+ b) + C,
where
F
′(x)
=
f
(x).
a
These
of
integrals
sine
and
result
directly
from
the
derivatives
cosine.
Check:
➔
Integrals
of
sine
and
cosine
d
(
sin x dx
=
–cos x
+
C
cos x dx
=
sin x
+
cos x ) =
C
dx
−( −sin x ) = sin x
d
The
integrals
of
the
composition
of
sine
or
cosine
with
a
linear
(sin x ) =
cos x
dx
function
are:
1
➔
sin (ax
+ b ) dx
= -
cos (ax
+ b) + C
a
1
cos ( ax
+ b ) dx
=
sin (ax
+ b) + C
a
Y
ou
can
perhaps
f
use
the
substitution
recognize
(g (x)) g ′(x)
when
you
method
have
an
to
find
some
integral
of
integrals
the
or
form
dx.
Chapter
Example
Find
the
integrals.
a
3 sin x dx
c
e
x
b
cos (4x
d
x
x
sin
(e
–
3
) dx
6) dx
4
cos (3x
) dx
Answers
Use
3 sin x
a
dx
=
3
sin x
=
3 (–cos x)
=
–3cos x
+
+
the
constant
multiple
cos (4 x
− 6)d x
x
e
( ax
+ b ) dx
=
sin
( ax
+ b ) + C
a
(e
) dx
=
⎛ du
⎞
⎜
⎟
Recognize
dx
sin
this
as
the
for m
u dx
⎠
f(g(x))g ′(x)
=
sin
dx
and
write
u du
down
let
u
=
–cos u
=
–cos e
+
and
C
then
use
=
e
x
and
substitute
e
for
u.
x
1
3
+
C
u
⎛ du ⎞
4
4
cos
(3x
) dx
=
×
⎜
12
⎝
cos
Let
u dx
u
=
3x
3
and
then
= 12 x
⎟
dx
x
⎠
1
so
⎛ du
⎞
⎜
⎟
3
=
12
⎝
dx
x
⎠
1
=
Simplify
cos u du
and
integrate.
12
1
sin u + C
=
12
1
4
4
sin
=
(3 x
Substitute
) + C
3x
for
12
Exercise
Find
the
E
integrals
in
questions
1–10.
⎛
⎛ 1
2
1
(2cos
x
+
3sin
x) dx
2
⎜
x
+ cos
π sin(π
5
20x
x) dx
3
⎝
⎟ ⎟
4
sin(2x
6
(2x
+
3
(5x
dx
⎠ ⎠
3)dx
4
cos
⎞ ⎞
x
⎜
⎝
3
2
) dx
–
1)cos
(4x
–
4x) dx
tan( 3 x )
e
cos
dx
7
(ln x )
8
dx
2
cos
(3 x )
x
sin x
2
9
cos x sin
xdx
dx ,
10
cos x
Calculus
with
trigonometric
answer
x
e
dx
=
the
du
x
or
x
sine.
x
sin
⎝
d
integrate
1
cos
− 6) + C
4
c
then
C
sin (4 x
=
and
C
1
b
rule
dx
functions
for
cos x
>
0
u.
,
.
Simplify,
integrate
EXAM-STYLE
QUESTIONS
sin
Let
11
f
=
a
Find
b
Write
Let
12
(x)
f
f
(x)
down
=
Show
b
Hence
can
cos x.
′(x).
a
Y
ou
x
e
use
f
(x) dx
ln(cos x).
that
f
′(x)
find
the
=
–tan x.
tan x
ln(cos x) dx
Fundamental
Theorem
of
Calculus
to
evaluate
definite
See
Section
9.4.
integrals:
b
b
f
(x )
dx
=
[F
=
( x )]
F (b ) − F ( a ) ,
where
F
is
an
antiderivative
of
f.
a
a
Example
Evaluate
Then
the
check
definite
your
integral
answer
by
without
a
evaluating
GDC
the
to
get
definite
the
exact
integral
value.
on
a
GDC.
2
4
3
a
2cos x
sin( 2 x ) cos
b
dx
( 2 x )dx
0
4
Answers
4
4
2cosx dx
a
=
2
cosx dx
Apply
the
Fundamental Theorem
of
Calculus.
0
0
p
= 2 ⎡ sin x ⎤
⎣
4
⎦
0
p
⎛
= 2
⎜
sin
⎛
=
⎞
− sin 0
4
⎝
⎟
⎠
⎞
2
2⎜
0 ⎟
Use
unit
circle
values
to
evaluate.
⎟
⎜
2
⎠
⎝
=
2
Look
Using
a
at
the
investigation
in
Section
9.3
if
you
need
GDC:
to
review
how
to
enter
a
definite
integral
into
your
calculator.
4
2 cos x dx
≈ 1.41 and
since
0
2
≈ 1.41,
our
answer
is
verified.
GDC
help
on
screenshots
Plus
and
GDCs
CD:
for
Casio
are
on
Alternative
the
TI-84
FX-9860GII
the
CD.
{
Continued
on
next
Chapter
page
2
du
3
sin
b
( 2 x ) cos
Let
( 2 x )dx
u
=
cos
(2x)
and
= −2sin
(2x)
.
dx
p
4
x =
1 ⎛ du
2
⌠
=
⎞
3
u
−
⎮
⎜
1
dx
2
x =
dx
⎝
⎛ du
⎞
⎜
⎟
3
⎟
Substitute
p
⌡
⎠
2
4
dx
⎝
for
sin (2x)
and
3
u
for
cos
(2x).
⎠
u=−1
1
=
When
−
x
⎛
,
u =
cos
2
p
1
1 ⎡ 1
4
−
When
⎤
⎢
⎣
⎥
4
⎦
( ( −1)
cos
⎛ p
⎞ ⎞
⎜
⎟ ⎟
⎝
4
2
p
=
⎠ ⎠
2
=
cos
0
2
cos
1
0
Then
4
−
,
2
1
=
=
x
u
2
2
⎝
u=0
=
⎜
4
− 0
apply
the
Fundamental Theorem
of
Calculus.
)
8
1
=
−
8
2
3
Using
a
sin
GDC:
( 2 x ) cos
( 2 x )dx
=
−0 .125
Evaluate
the
definite
integral
on
your
GDC.
4
1
and
=
since
−0
125
,
our
answer
is
verified.
8
GDC
help
on
screenshots
Plus
and
GDCs
Exercise
Evaluate
Then
Casio
are
on
Alternative
the
TI-84
FX-9860GII
the
CD.
F
the
check
CD:
for
definite
your
integral
answer
by
without
evaluating
a
GDC
the
to
get
definite
the
exact
integral
on
value.
a
GDC.
π
3
cosx dx
1
(2sin x
2
+
sin 2x) dx
0
3
ln
3
2
3
2
cos
x
e
4
dx
x
x
cos (e
) dx
0
3
ln
4
Y
ou
➔
can
use
When
the
definite
the
lines
x
area
=
a
integrals
bounded
and
x
=
b
to
by
is
find
the
area
and
cur ve y
rotated
360°
=
volume.
f
(x),
about
the
the
x-axis
x-axis,
and
the
y
b
2
volume
of
the
solid
formed
π y
is
dx.
a
0
Calculus
with
trigonometric
functions
x
Example
A
por tion
a
Find
b
Write
of
the
the
graph
area
of
of
the
f (x)
=
shaded
x
sin
x
is
shown
in
the
y
diagram.
region.
f(x)
solid
down
the
formed
integral
when
the
representing
shaded
the
region
is
volume
rotated
of
=
x sin x
the
360°
A
0
about
the
Hence,
x
x-axis.
find
the
volume
of
the
solid.
Answers
Set
x sin x
a
=
of
x
=
0
x
=
0,
the
function
equal
to
0
to
find
the
x-coordinates
0
or
sin
x
=
O
and
A.
0
π
Set
π
x (sin x )dx
≈
3.14
up
the
Notice
0
definite
that
the
integral
area
of
and
this
evaluate
region
on
happens
a
to
GDC.
be
π
b
π
2
π y
Use
dx
to
set
up
the
definite
integral
and
2
b
⎡ x (sin
⎣
π
x )⎤
dx
≈ 13.8
a
⎦
0
evaluate
on
a
GDC.
y
Y
ou
can
also
find
the
area
between
two
cur ves.
Quadrant
Quadrant
2
1
b
➔
If
y
≥
y
1
for
all
x
in
a
≤
x
≤
b,
then
( y
−
1
2
y
is
)d x
the
area
2
a
between
Example
Find
the
the
two
cur ves.
x
O
Quadrant
Quadrant
3
4
area
of
the
region
in
quadrant
1
that
is
bounded
by
the
cur ves
y
=
0.4x
and
y
=
sin x
Answer
2.125
Use
Area
=
(sin (x))
−
a
GDC
to
help
sketch
a
graph
and
find
the
0.4x) dx
points
of
intersection
where
sin
x
=
0.4x.
0
b
≈
0.623
The
area
is
equal
to
( y
y
1
) dx
2
a
where
Since
a
=
0
sin x
choose
y
and
≥
=
b
0.4x
sin
x
≈
0
and
y
1
Exercise
In
1–2,
integral
1
y
=
x
2
y
=
x
≤
x
=
≤
2.125,
0.4x.
2
G
questions
definite
2.125.
for
sin
x
a
to
and
region
find
y
=
is
the
2x
–
bounded
area
6
in
of
by
the
the
given
cur ves.
Use
a
region.
quadrant
l
2
–
2
and
y
=
x
+
cos
x
Chapter
EXAM-STYLE
QUESTIONS
k
1
3
Given
that
cos xdx
and
=
f
Let
(x )
bounded
a
Find
b
Write
tan
by
the
f,
x
.
the
area
down
formed
k
,
Consider
x-axis
of
the
the
when
the
and
the
exact
value
of
k
region
the
line
x
in
=
the
first
quadrant
2.
region.
integral
the
find
2
2
0
4
0
representing
region
is
rotated
the
360°
volume
about
of
the
solid
the x-axis.
y
Hence
find
the
volume
of
the
solid.
(r
2)
2
5
The
graph
represents
the
function
f
(x)
=
a
sin
(bx).
1
a
Find
the
b
Hence
values
of
a
and
b
x
find
the
area
of
the
shaded
r
region.
–1
6
The
y
=
a
diagram
cos
y
i
x
+
=
shows
sin
cos
2x.
x
+
par t
of
the
Regions
sin
2x
A
can
graph
and
be
B
of
are
written
3r
r
2
5r
2r
3r
7r
2
r
2
–2
shaded.
as
y
y
=
cos
x(c
+
d
sin
x).
Find
the
values
of
c
and
d
2
ii
b
Hence
find
shown
in
the
the
exact
values
of
the
two
x-intercepts
1
diagram.
i
Find
the
area
of
ii
Find
the
total
A
region
A
x
0
B
c
Find
the
rotated
.
area
volume
360°
of
about
Revisiting
of
the
the
the
shaded
solid
–1
regions.
formed
when
region A
–2
is
x-axis.
linear
motion
Extension
material
Worksheet
14
-
trigonometric
Derivatives
motion
and
along
Suppose
position
that
a
s(t).
straight
an
from
function
integrals
object
an
We
used
in
kinematics
problems
involving
and
derivatives
integrals
moving
at
any
have
along
time
the
t
is
a
straight
given
following
by
line
the
and
that
its
displacement
relationships.
Remember
Displacement
function
CD:
line.
is
origin
then
are
on
More
=
that…
s (t)
Initially
⇒
at
time
0
ds
=
v (t ) =
Velocity
s′(t)
At
rest
⇒
v(t)
=
0
dt
Initially
at
rest
dv
Acceleration
⇒
a (t ) =
=
v′(t)
or
v(0)
=
0
s″(t)
dt
Moving
right
or
up
t
2
⇒
T
otal
distance
traveled
from
time t
to
=
t
Moving
t
We
will
now
modeled
by
Calculus
look
at
some
trigonometric
with
trigonometric
examples
functions.
functions
v(t)
>
0
|v (t)|dt
where
the
linear
motion
is
⇒
left
v(t)
Speed
=
<
or
down
0
|velocity|
Example
A par ticle moves along a horizontal line. The par ticle’s displacement, in
metres, from an origin O is given by s(t) = 5 – 2cos 3t
a
Find
the
par ticle’s
velocity
b
Find
the
par ticle’s
initial
c
Find
when
stopped
d
Write
the
par ticle
during
down
traveled
for
a
0
the
definite
≤
t
π
≤
is
time
and
acceleration
displacement,
moving
0
≤
t
≤
integral
seconds
to
for time t seconds.
at
any
velocity
the
right,
time
and
to
the
t
acceleration.
left
and
π
that
and
represents
use
a
the
GDC
to
total
find
distance
the
distance.
Answers
a
v(t)
a(t)
b
s(0)
v(0)
=
0
–
=
6sin 3t
=
6(cos 3t)(3)
=
18cos 3t
=
5
–
2cos (3(0))
=
5
–
2(1)
=
6sin (3(0))
=
6(0)
=
18cos (3(0))
=
18(1)
v(t)
=
s′(t)
a(t)
=
v′(t)
2(–sin 3t)(3)
=
Evaluate
each
function
at
t
=
0
3 m
–1
a(0)
=
0 m s
–2
c
v(t)
=
=
18 m s
0
6sin 3t
sin 3t
3t
The
=
=
=
0
v(t)
0
0,
when
π,
2π,
t
0,
0,
v(t)
<
0.
helpful
>
A
to
at
rest
when
par ticle
0
and
sign
moves
left
when
diagram
analyze
the
right
is
motion.
3
is
at
rest
v(t)
at
+
–
+
2
The
and
,
3
0
seconds.
r
2r
3
3
3
par ticle
moves
0
v (t)
is
0. The
,
,
par ticle
3π
2
3
The
par ticle
=
t
t
and
3
for
2
seconds
right
r
3
and
left
for
2
t
3
seconds.
3
π
d
|6sin 3t| dt
=
12 m
The
total
t
t
distance
traveled
from
time
0
t
2
to
1
is
|v(t)| dt.
Use
a
GDC
to
2
t
evaluate
the
integral.
GDC
help
on
screenshots
Plus
and
GDCs
CD:
for
Casio
are
on
Alternative
the
TI-84
FX-9860GII
the
CD.
Chapter
Example
–1
A
par ticle
moves
along
a
straight
line
so
the
par ticle
when
that
its
velocity
,
v
m s
at
time
t
seconds
is
given
by
2
v(t)
=
5sin t
cos
t
5
Find
a
the
speed
of
t
seconds.
6
b
When
t
=
0,
the
displacement,
Find
an
expression
for
s
c
Find
an
expression
for
the
in
s,
of
terms
of
the
par ticle
is
3 m.
t.
acceleration,
a,
of
the
par ticle
in
terms
of
t
Answers
a
5
v
6
5
2
5 sin
5
Velocity
cos
6
6
has
both
magnitude
and
direction,
and
speed
is
speed
=
the
magnitude
of
velocity. Therefore
2
1
5
2
3
|velocity|.
2
15
8
15
Speed
15
=
=
8
∫
5 sin t cos
t dt
=
⎮
du
⎛
⌠
2
b
1
m s
8
5
⌡
⎞
2
u
−
⎜
dt
Integrate
⎟
dt
⎝
velocity
to
get
displacement.
⎠
Using
substitution
let
u
=
cos
t,
2
5
=
du
u
∫
du
then
=
− sin t
=
sin t
dt
⎛ 1
= −5
3
⎞
+ C
u
⎜
⎝
du
⎟
3
⎠
so
−
dt
5
3
s (t ) = −
cos
t
+ C
3
5
3
3 =
−
(0 ) + C
cos
Use
the
fact
that
Use
the
product
s(0)
=
3
to
find
C.
3
5
3 =
−
(1) + C
3
14
C
=
3
5
14
3
So
s (t )
=
−
cos
t
+
3
a (t )
c
=
3
v ′( t )
2
=
5 sin t
[ 2 (cos
t )( − sin t ) ] + cos
2
=
✗
−10 sin
Exercise
H
Do
a
not
use
1
A
particle
of
rule
and
velocity.
t (5 cos t )
3
t
cos t
GDC
EXAM-STYLE
derivative
for
+ 5 cos
t
question
1–3.
QUESTION
moves
along
a
straight
line
so
that
its
displacement s
in
t
metres
from
an
origin
O
is
given
by
s (t)
a
Write
down
an
expression
for
the
b
Write
down
an
expression
for
the
Calculus
with
trigonometric
functions
=
e
sin
t
velocity
, v,
for
in
time
t
terms
acceleration, a,
in
seconds.
of
t
terms
of
t
the
chain
rule
to
find
the
2
A
par ticle
moves
displacement,
s (t)
=
1
–
2sin
Calculate
a
in
t
along
a
metres,
for
the
time
straight
from
t
velocity
when
Calculate
the
value
c
Calculate
the
displacement
is
The
origin O
par ticle’s
is
given
by
seconds.
b
velocity
line.
an
of
t
for
t
=
0
0.
<
of
t
π
<
the
when
the
par ticle
velocity
from O
is
when
zero.
the
zero.
–1
3
The
velocity
v
m s
of
a
moving
body
along
a
horizontal
line
at
sint
time
a
t
seconds
Find
i
0
ii
t
the
≤
≤
The
in
may
t
the
a
par ticle
the
body’s
initial
use
v (t)
=
is
e
at
cos
rest
t
during
the
inter val
par ticle
is
moving
left
during
the
inter val
2π
terms
object
by
2π
≤
0
c
An
when
when
Find
Y
ou
given
Find
b
s
4
≤
is
acceleration
displacement
of
s
is
a
in
4
terms
metres.
of
t
Find
an
expression
for
t
GDC
star ts
for
by
questions
moving
4–6.
from
a
fixed
point O.
Its
velocity
–1
v
m s
Let
after
d
be
t
the
seconds
is
given
displacement
a
Write
b
Calculate
down
an
the
from
integral
value
of
by
v (t)
O
which
=
4 sin
when
t
=
t
+
3cos t,
t
≥
0.
4.
represents d
d
−1
5
A
par ticle
moves
with
a
velocity
v
m s
given
by
2
⎛
v (t )
=
−( t
+ 1) sin
⎞
t
where
⎜
t
≥
0.
⎟
2
⎝
a
i
Find
ii
A
the
acceleration
par ticle
have
the
b
Find
all
c
Find
the
0
<
<
t
at
speeding
sign
at
times
1.5
when
the
1.5
seconds.
velocity
slowing
whether
time
in
time
up
and
Determine
down
the
changes
is
same
different.
slowing
⎠
down
the
and
acceleration
when
par ticle
is
the
signs
speeding
are
up
or
seconds.
inter val
0
< t
<
4
that
the
par ticle
direction.
total
distance
traveled
by
the
par ticle
during
the
time
4.
−1
6
The
velocity
,
v,
in
m s
of
a
par ticle
moving
in
a
straight
line
is
2sin t
given
0
a
≤
t
by
≤
v (t)
=
e
–
1,
where
t
is
the
time
in
seconds
for
12.
Find
the
acceleration
of
the
par ticle
at t
=1.
2sin t
b
i
ii
Sketch
a
graph
Determine
the
of
v(t)
=
value(s)
e
of
–
t,
1
for
for
0
≤
0
t
≤
≤
t
≤
12
12.
where
the
particle
−1
has
iii
At
a
velocity
time
velocity
origin
c
Find
the
t
=
to
in
of
0
5
the
m
par ticle
explain
the
s
whether
inter val
distance
is
0
≤
traveled
t
in
≤
at
or
the
origin.
not
the
Use
the
par ticle
graph
retur ns
of
to
the
12.
the
12
seconds.
Chapter
Review
exercise
✗
1
Find
the
derivative
of
3
f
a
(x)
=
cos
tan
=
(1
–
2x)
b
y
d
f
=
sin
x
2
t
c
s (t)
e
e
f
(x)
=
x
g
f
(x)
=
(ln
(x )
sin x
2
2
Find
the
cos
x
x)(sin
integral
x)
f
y
=
ln(tan
h
y
=
2
sin
x)
x
cos
x
of
3
a
(
c
sin ( 4 x
4x
)
− sin x
b
dx
cos (3 x )d x
2
sin ( 2 t
⌠
e
+ 1)d x
+ 1)
f
( 2t
⌡
+ 1)
2
sin
x
cos x
Evaluate
the
x
6 cos x
⌠
2
xe
dx
h
definite
dx
⎮
⌡
3
dx
⎮
2
cos
⌡
g
)dx
sin (ln x )
⌠
dt
⎮
x cos ( 2 x
d
2
( 2 + sin x )
integral
of
π
3
sin x dx
a
(1
b
+
sin x) dx
0
3
π
3
2
2
(sin x
c
+
cos 2x) dx
5 sin
d
0
EXAM-STYLE
4
Find
y
=
x cos x d x
0
the
QUESTIONS
equation
cos (3x
–
6)
at
of
the
the
normal
point
to
the
cur ve
with
equation
(2, 1).
5
Find
the
coordinates
of
the
point
on
the
graph
of
y
x
,
sin
2
1
0
≤
x
π,
≤
at
which
the
tangent
is
parallel
to
the
line
x
3
4
6
A
curve
with
equation y
=
f
(x)
passes
through
the
point
(0,
2).
Its
f (x)
gradient
function
is f
′(x)
=
x
–
sin
x.
Find
the
equation
of
the
curve.
4
7
The
graph
represents
the
function
f
(x)
=
p
sin(x)
+
q,
p,
q
∈
Find
2
a
the
values
b
the
area
of
of
p
the
and
q
shaded
region.
x
0
r
2
Review
1
A
to
region
find
exercise
is
the
bounded
area
of
by
the
the
given
cur ves.
Use
a
definite
region.
2
a
b
y
y
=
2cos
Calculus
x
+
2 sin x
with
cos
and
x
y
+
=
trigonometric
1,
x
=
0,
0.5x
functions
x
=
2
and
the
x-axis
integral
r
3r
2
2r
2
A
to
region
find
360°
a
y
is
the
bounded
volume
about
=
sin
the
x
by
of
the
the
given
solid
cur ves.
formed
Use
when
a
definite
the
region
integral
is
rotated
=
k,
x-axis.
and
the
x-axis
for
0
≤
x
π
≤
cos x
b
y
=
e
,
EXAM-STYLE
3
The
area
x
=
0
and
x
2π
=
QUESTIONS
under
the
cur ve
y
=
cos
x
between
x
=
0
and
x
0
where
k
,
is
0.942.
Find
the
value
of
k
2
cos
4
Let
a
s(t)
=
(5t)
2e
–
i
Find
ii
Show
iii
Hence
4.
s′ (t).
cos
that
s′′ (t)
=
(5t)
50 e
2
(sin
(5t)
–
cos
(5t)).
verify
that
s
has
a
relative
minimum
at
t
5
b
s
is
the
displacement
straight
line,
where
Find
the
total
t
to
=
=
0
t
2
CHAPTER
f
(x)
f
(x)
=
=
sin
of
x
cos
measured
traveled
a
par ticle
in
by
moving
metres
the
and
par ticle
t
is
along
in
a
seconds.
from
SUMMARY
of
trigonometric
sine,
⇒
x
is
distance
14
Derivatives
s
for
seconds.
Derivatives
●
function
f
⇒
f
cos
′(x)
=
′(x)
=
and
cos
–
functions
tan:
x
sin
x
1
f
(x )
tan x
( x )
f
,
cos x
0
2
cos
Integral
●
of
Integrals
of
sin xdx
=
cos xdx
=
sine
sine
sin x
⎮
1
sin (ax
+ b )d x
=
−
cos (ax
+ b) + C
a
1
⌠
cos ( ax
+ b )d x
sin ( ax
=
⌡
●
cosine:
+ C
⌡
⎮
cosine
+ C
⌠
●
and
and
− cos x
x
+ b) + C
a
When
the
lines
=
x
a
area
and
bounded
x
=
b
is
by
the
rotated
cur ve y
360°
=
f
about
(x),
the
the
x-axis
x-axis,
the
and
the
volume
b
2
of
the
solid
formed
πy
is
dx
a
●
If
y
1
≥
y
for
all
x
in
a
≤
b
x
≤
b,
then
(y
2
1
−
y
) dx
is
the
area
2
a
between
the
two
cur ves.
Chapter
Theory
of
knowledge
From
conjecture
The
investigation
into
the
to
derivative
of
proof
sine
in
Chapter
14
graphed
the
d
derivative
of
sin
x,
which
led
to
the
conjecture
that
(sin x)
=
cos x.
dx
This
was
tested
with
several
values
and
found
to
be
tr ue
for
these
values.
d
Does
this
prove
that
(sin x)
=
cos x?
dx
Follow
TEP
S
Here
is
a
For
these
each
steps
step,
to
find
are
you
the
derivative
using
inductive
ONE:
unit
of
or
TEP
S
circle.
QOP
=
h
radians.
As
h
using
geometr y
.
deductive
reasoning?
TWO:
approaches
length
S
sine
of
arc
QP
zero,
how
compare
does
to
the
the
length
of
Q
segment
QP?
S
P
Q
h
x
O
P
R
h
x
R
QOP
is
π
–
isosceles?
h
hy
radians?
2
y
TEP
S
i
rc
Q
?
THREE:
S
Q
π
Why
is
SOQ
equal
to
–
h
–
x?
2
F ind
a
line
segment
parallel
to
SO.
π
Hence,
why
is
OQA
also
equal
to
–
h
–
x
?
2
A
P
h
Use
OQP
and
OQA
to
explain
why
AQP
=
+
x
h
.
2
x
O
Theory
of
knowledge:
From
conjecture
to
proof
R
TEP
FOUR:
S
Why
does
QA
TEP
FIVE:
S
equal
sin (x
+
h)
–
sin x
?
d
Now
show
that
(sin x)
=
cos x.
dx
S
Q
(cos (x
+
h),
sin (x
+
h))
Explain
each
line
of
wor king:
h
+
d
x
2
sin (x
(sin x)
=
+
h)
–
sin x
lim
dx
h
h→ 0
QA
=
lim
arc QP
h→ 0
P
A
(cos x,
sin x)
QA
=
lim
QP
h→ 0
h
h
x
=
lim
[cos
+
x]
2
h→ 0
O
R
=
cos x
S
Q
“Ever y
meaningful
mathematical
h
+
statement
can
also
be
expressed
in
sin (x
language.
Many
+
h)
–
sin x
plain-language
A
statements
of
x
2
plain
mathematical
P
expressions
h
x
would
fill
several
pages,
while
to
express
O
them
in
mathematical
notation
take
as
little
as
one
line.
One
R
might
of
the
Which
type
of
reasoning
did
you
use
to
d
show
ways
to
achieve
this
that
remarkable
(sin x)
=
cos x?
Deductive
or
dx
inductive?
compression
is
to
use
symbols
to
stand
for
statements,
instructions
and
so
Explain
the
Lancelot
Hogben
The
underlying
English
for
with
an
example
of
other
type
of
reasoning.
d
Does
this
prove
that
(sin x)
=
cos x?
symbols
concepts
mathematician
credited
Give
dx
scientist
Mathematical
answer
.
(1895–1975)
English
your
on.”
of
calculus
John
introducing
the
Wallis
is
symbol
∞
innity.
Could
calculus
without
the
have
use
of
developed
mathematical
symbols?
{
)
Chapter
Probability
CHAPTER
OBJECTIVES:
Concept
5.7
distributions
of
Expected
discrete
value
random
(mean),
E(X)
variables
for
and
discrete
their
probability
distributions.
data.
Applications
5.8
Binomial
5.9
Normal
distribution
distribution
Standardization
Proper ties
Before
Y
ou
1
e.g.
the
you
should
Calculate
of
distribution
the
of
normal
normal
how
mean
Calculate
its
mean
and
variance.
cur ves.
variables
(z-values)
distribution
start
know
the
of
and
and
of
a
mean
to:
set
of
of
this
Skills
numbers
1
frequency
check
Calculate
the
distributions
x.
mean
of
of
these
frequency
x:
a
x
x
0
1
2
3
Frequency
3
6
9
2
Frequency
3
4
5
6
7
8
3
5
7
9
6
2
b
x
f
∑
x
x
(
=
0
×
3) +
(1
×
6) +
(
2
×
9
)
+
(3
×
2
12
15
17
20
3
10
15
9
2
Frequency
=
∑
10
)
f
3 + 6 + 9 + 2
Repeat
question
1
above
using
your
30
= 1
=
GDC.
5
20
2
Evaluate
⎛ n ⎞
2
Use
⎜
⎟
notation
⎛ 8 ⎞
⎛ 6 ⎞
r
a
⎜
⎟
⎛ 5 ⎞
2
⎝
e.g.
⎜
b
⎝
⎠
⎟
5
⎛ 9 ⎞
c
⎠
Evaluate
⎜
⎟
2
3
⎛ 5 ⎞
⎜
⎝
⎟
2
5!
5
=
Solve
these
equations
4
=
2 !3!
⎠
×
=
5
10
5
= 3
a
2
2
x
3
Solve
equations
x
2
5
4
e.g.
Solve
the
equation
= 3
=
b
1
0
2
x
4
9
4
= 3
4
= 3x
x
= 1
0
x
Probability
3
distributions
x
c
=
2
6
4
⎜
⎝
⎟
6
⎠
3
( 0 .3 )
6
( 0.7 )
The
the
2010
Soccer
octopus
matches
Center
correctly
between
in
after
a
of
marked
match
his
2008
and
with
were
the
placed
of
in
the
2010!
behavior
matches.
flag
produced
Germany
,
feeding
soccer
Cup
predicted
Oberhausen,
famous
series
World
Two
one
his
of
tank.
an
results
Paul
and
was
of
lived
in
to
each
national
His
12
became
used
boxes,
the
unlikely
choice
celebrity
.
out
a
of
tank
14
in
football
the
Sea
Life
inter nationally
predict
the
containing
teams
of
Paul
in
which
a
an
winners
mussel
of
and
upcoming
mussel
to
eat
Why
do
believe
first
was
inter preted
as
predicting
that
the
countr y
with
that
that
to
someone
something
(like
an
win.
octopus)
Paul
was
This
chapter
right
86%
of
the
time!
the
can
future
rationally,
will
look
at
situations
like
this
and
how
to
probability
of
an
event
if
it
were
entirely
due
to
chance.
Paul
really
was
able
to
predict
the
results
of
future
predicting
seems
to
Perhaps,
be
though,
predict
when,
deter mine
the
the
want
flag
or
would
people
illogical?
soccer
matches!
Chapter
.
➔
A
Random
random
Random
Here
is
variable
variables
are
variables
some
are
a
quantity
represented
examples
of
whose
by
random
value
capital
depends
on
chance.
letters.
variables:
A
X
=
the
number
of
sixes
obtained
when
a
dice
is
rolled
3
discrete
variable
B
=
the
number
M =
the
mass
of
babies
in
a
random
times.
does
necessarily
of
crisps
in
a
=
the
time
taken
for
a
r unner
to
complete
just
are
two
basic
types
of
random
shoe
variables:
values
sizes
students
Discrete
number
random
of
– These
variables
possible
values
(e.g.
X
have
and
B
a
finite
or
countable
possible
above).
…4,
6,
Continuous
some
random
inter val
(e.g.
–
variables
M
and
T
These
can
take
on
any
value
the
obtained
discrete
when
represent
‘the
a
dice
is
(e.g.
a
could
set
have
values
5,
of
of
5.5,
in
above).
random
variable X,
rolled
probability
4.5,
of
6.5,…).
Use
Consider
to
positive
100 m.
integer
There
need
packet
take
T
not
pregnancy
that
3
times.
the
the
Y
ou
number
number
can
of
of
write
sixes
sixes
P(X
is x’
=
random
x)
where
capital
lower
to
x
the
can
letters
variables.
case
actual
letters
for
Use
for
measured
values.
take
the
The
values
rst
random
statistician,
By
1939
Bernard
T
o
use
a
in
set
such
a
(up,
example,
1927.
of
Most
computers
numbers
20,
These
which
appearance
of
A
of
at
44,
and
are
are
took
digits
the
right,
the
62,
a
numbers
by
‘at
specialized
point,
diagonal,
by
the
on
from
before
Kendall
of
row
by
digits,
selecting
5th
an
census
operated
number
the
Tippett,
Maurice
machine
etc.)
number
,
Leonard
random’
published
star ting
the
going
English
fact
can
now
be
pseudo-random
by
a
used
to
numbers:
mathematical
generate
that
formula
is,
Probability
distri bution
possible
that
63873
90708
20025
98859
23851
27965
62394
33666
62570
64775
78428
81666
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08730
56522
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53867
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14021
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54043
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87989
72248
30500
28220
12444
71840
and
a
human.
and
the
numbers.
random
are
which
have
the
of
discrete
variables
each
78134
58303
numbers.
distributions
probability
45963
02965
backwards
they
but
73735
registers.
...
calculators
in
published
was
15th
generated
probabi li ty
list
left,
was
using
decide
random
Probability
random
Smith
down,
3.
table
100 000
list,
22,
numbers.
and
Tippett
star ting
40,
2
number
gives
➔
1,
Babington
direction
For
0,
value
each
distributions
for
of
a
discrete
the
outcome
random
random
occurs.
variable
variable
and
the
is
a
Example
Let
X
be
rolled
the
three
random
times.
variable
Tabulate
that
the
represents
probability
the
number
distribution
of
for
sixes
obtained
when
a
fair
dice
is
X
Answer
X
can
take
the
values
0,
1,
2
and
3
1
1
6
six
P(3
sixes)
=
1
1
×
6
1
×
6
1
Use
a
six
P(2
sixes)
=
six
6
1
1
1
not
5
6
1
six
6
not
5
values
to
of
5
=
6
216
P(2
sixes)
5
×
=
6
1
P ( X
=
0),
P ( X
=
1),
×
6
5
P(X
=
6
=
P ( X
1
not
5
P(1
six)
=
six
5
×
6
5
×
6
2)
and
216
six
6
the
1
six
6
5
×
×
6
diagram
216
find
6
tree
=
6
=
3)
25
=
6
216
6
1
5
six
P(2
sixes)
1
×
=
1
×
5
=
6
6
6
6
216
1
six
P(1
5
six)
=
six
not
1
5
not
6
5
5
×
×
6
6
25
=
6
216
6 1
6
six
5
six
6
P(1
six)
=
not
5
×
6
5
1
×
6
25
=
6
216
six
5
not
6
P(0
5
sixes)
=
six
5
×
6
5
×
6
125
=
6
216
6
x
0
1
2
3
Write
P(X
in
125
25
5
1
216
72
72
216
a
the
probabilities
table.
x)
=
Sometimes
Notice
that
in
the
example
the
sum
of
the
probabilities
replaced
125
5
25
+
216
➔
For
+
any
=
72
72
≤
P(X
=
P
x)
≤
–
these
variable
1
same
∑
P( X
=
x )
≤
P(X
=
x)
random
that
variable
X
has
the
probability
1
2
3
4
5
7c
5c
4c
3c
c
of
the
value
of
c
b
Find
be
between
0
distribution
means
x)
Find
a
must
1.
P( X
x
a
1
and
=
≤
1
=
always
P (X
mean
thing.
probability
The
is
P(x)
X
means
Example
x)
x
1
0
0
=
just
216
the
random
with
1
or
+
P(X
is
P (X
≥
the
that
1
the
sum
probabilities
always
4)
x)
be
will
1.
Answers
a
7c
+
20c
5c
=
+
4c
+
3c
+
c
=
1
Using
P (X
1
Solve
1
c
∑
for
=
x )
= 1
The
solutions
lots
of
of
c
examination
=
20
questions
b
P (X
≥ 4) = P (X = 4) + P (X = 5)
3
=
+
20
4
1
the
fact
that
probability
with
the
must
total
add
=
=
20
1
star t
20
5
up
to
1.
Chapter
Exercise
1
Decide
A
a
is
on
2
3
A
whether
‘the
my
age
B
is
‘the
c
C
is
‘how
d
D
the
cats
diameter
the
I
of
variable
years
the
sum
the
number
of
the
c
the
smaller
d
the
product
of
or
of
the
dice
banana
will
see
is
continuous
the
two
number
faces
has
in
distribution
when
a
next
or
person
when
the
for
first
discrete:
to
call
me
two
one
shopping’.
white
one’.
cafeteria’.
each
ordinar y
when
on
buy
the
when
when
‘1’
I
before
donuts
obtained
equal
of
six-sided
faces
sixes
next
the
probability
b
fair
of
many
a
A
completed
length
‘the
Tabulate
in
random
phone’.
b
is
each
two
two
random
dice
are
ordinar y
face,
thrown
ordinar y
ordinar y
a
‘2’
dice
dice
dice
on
variable:
are
two
are
are
thrown
thrown
thrown.
of
its
faces
A
and
a
‘3’
on
the
remaining
three
fair
dice
The
dice
is
thrown
twice.
T
is
the
random
variable
‘the
thrown’.
a
the
probability
distribution
b
the
probability
that
A
board
at
a
A
fair
S
is
If
game
time,
the
the
played
following
that
that
to
is
equally
land
on
any
of
this
dice
is
of
total
by
faces.
T
score
moving
a
is
more
than
counter S
4.
squares
forward
r ule:
thrown
once.
If
the
number
shown
is
even,
the
number
shown
on
number.
number
Write
a
is
six-sided
half
a
Find
the
4
means
total
likely
score
dice
faces.
out
shown
a
table
is
odd,
showing
S
is
the
twice
possible
values
of
S
and
the
dice.
their
probabilities.
What
b
is
counter
5
The
the
moves
random
x
P(X
=
probability
variable
1
2
1
1
3
3
x)
a
Find
the
b
Find
P(1
Exam-Style
more
value
<
X
X
3
4
c
c
of
<
that
than
has
2
the
in
a
single
go
in
the
game
the
spaces?
probability
distribution
c
4).
Question
In
6
The
probability
distribution
of
a
random
variable Y
is
given
question
by
6,
3
P(Y
=
y)
=
cy
3
P(Y
=
y)
=
cy
for
y
=
1,
2,
3
This
Given
that
c
is
a
constant,
find
the
value
of
c
is
called
probability
function
Y.
use
Y
ou
nd
can
the
various
random
Probability
distributions
a
it
probability
values
of
variable
for
to
at
the
Y
Exam-Style
7
The
Questions
random
x
variable
−1
0
X
1
2
P(X
Find
8
The
x)
=
the
2k
random
of
this
probability
distribution.
2
2
4k
value
has
6k
k
k
variable
x
X
has
the
probability
distribution
given
by
1
1
P (X
x )
k
Find
9
The
3,
the
exact
discrete
4,
5.
The
x
=
1,
2,
3,
4
and
value
of
random
X
can
distribution
take
of
X
=
0)
=
P (X
=
1)
=
P(X
=
2)
=
a
P (X
=
3)
=
P (X
=
4)
=
P(X
=
5)
=
b
P (X
≥
a
=
and
3P (X
<
are
constants.
a
Determine
the
values
b
Determine
the
probability
obser vations
discrete
these
=
The
and
from
random
only
is
the
given
values
0,
1,
2,
by
of
this
a
and
that
b
the
distribution
variables
A
and
sum
of
two
exceeds
B
are
independent
7.
independent
and
have
=
1
2
3
1
1
1
3
3
3
1
2
3
1
2
1
6
3
6
a)
B
P (B
constant.
distributions:
A
P (A
a
2)
b
The
is
k
variable
probability
2)
k
P (X
where
10
for
3
b)
random
one
variable
obser vation
C
is
the
from
sum
of
one
obser vation
from A
B
5
a
Show
that
b
Tabulate
P(C
= 3) =
18
the
probability
distribution
for C
Expectation
The
mean
value
that
or
we
expected
should
value
expect
of
for
a
X
random
over
variable
many
trials
X
of
is
the
average
the
experiment.
Expectation
is
actually
mean
of
the
the
underlying
distribution
(the
population).
The
mean
or
expected
value
of
a
random
variable X
is
is
parent
often
represented
denoted
by
It
by
μ
E (X).
Chapter
Investigation
T
wo
D,
dice
are
between
Copy
1
rolled
the
and
–
together
scores
on
complete
d
dice
0
and
the
the
1
scores
the
dice
difference,
is
noted.
probability
2
3
distribution
4
for
D.
5
10
P(D
=
d)
36
This
2
experiment
following
each
of
table
the
is
to
repeated
show
different
d
0
the
values
1
36
times.
expected
of
Copy
and
frequency
complete
of
the
obtaining
d
2
Y
ou
3
4
5
to
may
draw
nd
a
it
helpful
sample
Expected
space
diagram
like
10
frequency
ones
3
Calculate
4
The
the
original
questions
mean
of
experiment
2
and
d
3
for
0
What
do
6
What
would
we
by
would
can
just
(the
➔
you
the
=
Example
is
∑
2
of
x
be
if
3
or
each
to
=
be
value
value
Or
4
the
of
a
of
the
same
by
of
its
the
each
the
case.
random
respective
just
variable X
Therefore
variable D
probability
once).
is
x )
probability
x
from
What
expected
of
in
experiment
random
repeated
once?
value
d
were
distribution
is
Repeat
5
experiment
just
expected
conducting
P( X
the
times?
mean
mean
expected
E (X )
mean
1000
multiplying
The
Here
Or
the
equivalent
times.
notice?
the
expect
find
100
situation:
9
5
We
repeated
this
distribution.
250
frequency
times?
is
frequency
1
Expected
10
this
the
Example
0
1
2
3
1:
value
P (X
=
125
25
5
1
216
72
72
216
x)
X ?
Answer
Using
the
formula:
Use
125
⎛
E( X )
=
0
×
216
⎝
⎜
⎠
⎝
5
⎛
2
⎞
×
72
=
(X )
∑
x
P (X
=
x )
×
⎟
72
3
⎟
⎜
⎠
⎝
⎠
1
⎛
+
⎜
⎝
1
⎟
E
25 ⎞
⎛
+
⎜
+
⎞
⎞
×
⎟
216
Therefore
if
3
dice
are
rolled
a
large
⎠
number
of
times,
you
should
expect
1
E( X ) =
the
mean
number
of
sixes
to
be
0.5.
2
{
Probability
distributions
Continued
on
next
page
in
Chapter
3.
the
Using
a
GDC:
Enter
the
list
of
possible
x-values
GDC
help
on
CD:
demonstrations
in
x
and
the
set
of
Now
use
when
Use
x
P(X
=
One-Variable
finding
as
the
Frequenc y
Note
values
that
the
X1
mean
List
x)
in
p.
Statistics
of
and
a
p
not
are
=
x
=
Exercise
1
When
What
The
the
expected
to
be
a
value
value
of
of
X
X
more
defined
is
the
entering
see
Chapter
17
obtainable.
5.1
and
5.2
a
standard
by X
=
six-sided
the
expectation
square
of
of
dice,
the
let X
score
be
the
random
shown
on
the
dice.
X?
Question
z
variable
2
=
on
that
Sections
random
P (Z
CD.
B
Exam-Style
2
the
the
0.5
throwing
variable
on
List.
need
actually
TI-84
FX-9860GII
set.
data,
is
E(X)
Casio
as
data
as
and
GDCs
For
does
the
cor responding
Plus
probability
Alternative
for
3
1
1
6
6
Z
has
5
probability
7
11
x
y
distribution
1
z)
6
2
and E( Z )
= 5
3
Find x
3
A
and
‘Fibonacci
numbers:
What is
4
y
The
1,
the
dice’
2,
3,
is
5,
unbiased,
8,
expected
discrete
random
six-sided
and
labeled
with
these
13.
score
when
variable
X
the
has
dice
is
rolled?
probability
distribution
x
p( x )
for
x
=
1,
2,
3,
…,
8
36
Find
E(X).
Chapter
Exam-Style
5
For
the
given
Questions
discrete
random
P,
the
probability
distribution
is
by
⎧ kx
⎪
P( X
variable
=
x )
=
x
= 1
,
x
=
2,
3,
4,
5
⎨
⎪k (10 −
⎩
x )
6,
7,
8,
9
Find
a
6
a
the
value
Copy
for
a
of
and
the
complete,
discrete
x
P (X
b
c
7
X
What
a
Find
in
is
a
is
≤
discrete
1,
2
Ten
but
and
are
R
and
a
in
the
Calculate
c
What
balls
this
probability
distribution
can
k
take?
Give
your
answer
in
the
Q
mean
of
the
variable
which
2)
and
distribution.
can
only
take
the
three
=
0.3
that
the
mean
of
the
bag.
are
out
They
red
at
are
and
all
the
random
identical
rest
are
from
sizes
blue.
the
bag
the
the
are
balls
red
drawn
out
up
to
one.
values
most
in
drawn.
mean
a
of
the
value
likely
bag
random,
that
of
first
R
and
their
probabilities.
next
Show
a
the
at
a
the
=
number
picked
is
P(X
possible
b
Ten
k,
X:
replaced.
associated
9
k
of
4.
picked
is
∈
b
random
them
the
of
terms
variable,
2.8.
including
List
a,
E(X)
b
3
values
b,
k
1).
not
be
is
are
of
are
of
that
=
balls
Balls
Let
P(X
two
1−k
and
distribution
8
0.2
terms
known
Find
2
≤
k
in
random
1
range
form
values
It
x)
=
constant
as
but,
in
of
R
value
of
R?
question
instead,
probability
8
each
that
above.
ball
the
is
first
Again,
replaced
red
is
balls
are
before
drawn
out
the
on
the
4
second
go
is
25
b
Calculate
third
c
probability
that
the
first
red
is
drawn
after
the
go.
Derive
drawn
d
the
What
Probability
a
formula
on
is
the
the
nth
most
distributions
to
find
the
probability
go.
likely
value
of
R?
that
the
first
red
is
Exam-Style
10
An
instant
are
$0,
Question
lotter y
$2,
$20,
representing
z
$200
1000
0.2
0.05
0.001
0.0001
P(Z
E(Z)
c
How
much
=
should
of
a
are
or
ve
‘false’.
may
need
2
On
3
Bowling
4
It
5
Selachophobia
average,
look
answers
were
T
o
is
The
its
expect
to
the
ticket. Z
has
the
variable
distribution:
meaning.
to
gain
or
lose
on
average
ve
‘test’
the
at
you
ten
of
the
answer
that
elements
each
to
question
each
some
by
a
move
days
light
back
many
need
quiz
of
is
either
question.
them!
woman.
300 000
times
a
day.
Italy.
questions.
the
to
created
in
to
answer
answer
fear
How
probability
essential
binomial
answer
V inci
the
score?
guessing
the
da
is
distribution
muscles’
played
Answers
these
this
The
was
‘eye
rst
Leonardo
in
the
the
vest
the
was
good
just
pass
What
➔
a
random
prizes
distri bution
down
guess
bulletproof
took
–
Write
to
The
this
you
questions.
1
Is
inter pret
binomial
Investigation
Now
the
possible
0).
and
binomial
Definition
Y
ou
the
The
ticket?
The
‘true’
on
$2.
be
200
Determine
won
Let Z
20
Determine
Here
$1000.
for
2
b
.
and
purchased
amount
a
per
is
the
0
P (Z)
ticket
of
How
get
a
3
paint
the
Mona
book
many
you
ever y
get
you
of
would
to
to
to
Lisa’
s
lips.
ashes.
did
to
you
expect
nd
get
to
the
correct
right?
get
right
if
you
question?
correct
exactly
binomial
3
out
of
right
5.
out
of
distribution
5?
●
In
the
●
Here,
quiz
above
success
is
there
are
getting
5
trials.
the
are:
question
●
There
●
Each
is
a
fixed
number
(n)
of
has
only
two
possible
outcomes
–
●
The
or
a
to
question
In
this
of
a
success
(p)
is
getting
case
wrong.
of
success
the
probability
is
0.5
assuming
constant
you
trial
is
‘failure’.
probability
from
failure
a
●
‘success’
and
trials.
the
trial
right
obtained
ever y
answer
by
trial.
guessing.
●
Trials
are
independent
of
each
other.
●
If
you
get
question
you
the
are
the
answer
right,
more
answer
to
it
or
to
does
less
the
one
not
likely
next
mean
to
get
question
right.
Chapter
The
outcomes
probabilities
of
of
a
binomial
these
ex p e r i m e n t
o utc om es
a re
and
ca ll ed
a
the
cor respon d ing
binomial
distri bution
The
binomial
variable
➔
X
The
the
if
distri bution
the
success).
Consider
the
conditions
parameters
values
this
of
n
Any
that
(the
of
above
define
a
number
binomial
problem,
probability
describes
behavior
unique
of
you
exactly
of
a
discrete
apply
.
binomial
trials)
distribution
which
getting
the
first
two
and p
is
distribution
(the
probability
represented
met
in
heads
as X
Chapter
in
three
3:
∼
are
of
B(n,
a
p)
determine
tosses
of
a
2
biased
coin
for
which
P(head)
=
3
Y
ou
could
use
a
tree
diagram
to
help
you
answer
this
question.
2
H
3
HHH
2
H
3
1
T
HHT
H
HTH
3
H
2
2
3
1
3
T
3
T
HTT
1
3
2
H
THH
3
2
H
3
1
T
THT
H
TTH
1
3
T
3
2
3
1
T
3
T
TTT
1
3
P(two
heads
in
three
tosses)
=
P(HHT)
+
+
P(HTH)
P(THH)
We
Each
of
the
three
probabilities
are
the
often
use
a
theoretical
same.
distribution,
such
distribution,
to
as
the
binomial
2
P(HHT)
=
P(HTH)
=
P(THH)
=
2
1
4
describe
a
random
3
3
variable
27
This
that
process
occurs
is
in
called
real
life.
modeling
and
2
And
so
P(two
heads
in
three
tosses)
= 3
2
1
However,
number
What
if
you
of
trials,
you
obtaining
should
n,
were
exactly
only
is
a
tree
3
diagram
4
enables
us
tree
diagram
for
so
we
Probability
will
look
out
calculations.
27
if
9
If
the
theoretical
the
real-life
the
model
distribution
variable
matches
perfectly,
then
the
is
to
find
heads
in
the
six
probability
tosses
of
of
this
coin?
usually
not
results
of
perfect.
this
question
would
be
for
distributions
a
formula.
the
any
necessarily
However
,
case.
this
calculations
give
of
Generally
is
a
the
will
completely
real-life
the
not
accurate
situation.
too
Does
large,
carr y
3
description
The
to
small.
asked
two
use
12
this
make
them
any
less
useful?
First
note
been
met:
●
that
There
is
the
a
conditions
fixed
number
for
(n)
a
binomial
of
In
distribution
this
case
have
there
are
six
trials.
trials.
●
Each
trial
outcomes
has
–
a
two
possible
‘success’
or
Here
a
a
a
success
failure
is
is
getting
getting
a
a
head
and
tail.
‘failure’.
●
The
probability
of
a
success
(p)
2
is
The
constant
from
trial
to
probability
Trials
are
independent
of
each
the
Getting
affect
combination
of
Hs
success
and
each
is
trial.
other.
One
a
3
time
●
of
Ts
that
will
coin
a
is
head
the
tossed.
on
one
outcome
produce
2
heads
of
and
trial
the
4
will
next
not
trial.
tails
The
is
most
usual
error
HHTTTT
when
2
2
calculating
4
binomial
And
P(HHTTTT)
=
(
3
ever y
possible
probability
0.00548...)
3
729
is
to
forget
there
And
combination
of
2
Hs
and
4
Ts
will
have
are
that
r
the
there
probability
.
must
But
if
exactly
successes,
same
a
4
1
how
many
combinations
are
also
be
n
–
r
failures.
there?
⎛ n ⎞
⎜
⎝
⎟
r
represents
the
number
of
ways
of
choosing r
items
out
of
For
more
binomial
n
on
the
⎠
expansion,
items.
see
The
number
of
combinations
⎛ 6 ⎞
and
4
Ts
is
therefore
⎜
⎝
⎟
2
of
6
items
that
have
2
Chapter
6.
Hs
⎛ 6 ⎞
=
⎠
⎜
⎝
⎟
4
= 15
⎠
⎛ 6 ⎞
Y
ou
can
use
your
GDC
to
calculate
⎜
⎝
⎟
2
⎠
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
⎛ 6 ⎞
Instead,
you
could
use
the
formula
⎜
⎝
⎟
2
6!
=
Casio
are
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
6 × 5
= 15
=
2! 4 !
2
⎠
or
the
3rd
entr y
on
the
6th
row
of
Pascal’s
triangle:
Therefore
2
P(2
heads
in
6
tosses)
=
⎛ 6 ⎞
⎛
⎜
⎟
⎜
⎠
⎝
⎝
2
2
4
⎞
⎛ 1 ⎞
⎟
⎜
⎠
⎝
4
= 15
3
×
20
=
=
0. 0823
( 3 sf )
⎟
3
⎠
729
243
Chapter
Generalizing
this
method
gives
the
binomial
distribution
function:
➔
If
X
of
is
binomially
obtaining
the
r
distributed,
successes
probability
of
success
⎛ n ⎞
P( X
=
r )
=
⎝
This
is
often
r
=
r )
shor tened
⎝
X
is
r
p
=
⎜
Example
−
n
B(n,
p),
then
independent
each
trial,
the
probability
trials,
when
p
is
is
r
)
n
to
r
q
where
⎟
r
p
∼
⎠
⎛ n ⎞
P( X
n
(1
⎟
of
for
r
p
⎜
out
X
q
=
1
–
p
⎠
binomially
distributed
with
6
trials
and
a
probability
of
success
1
equal
to
at
each
attempt.
What
is
the
probability
of
5
a
exactly
c
three
four
or
successes
fewer
b
at
least
one
success?
successes?
Answers
By
hand:
You
4
a
P( X
=
4) =
⎛
⎜
⎟
⎜
⎠
⎝
⎝
⎠ ⎝
4
5
1
= 15 ×
rewrite
⎞
⎟
5
X
~
B
the
question
as
⎞
,
6,
⎜
what
is
⎟
5
⎝
⎠
16
1
⎛
If
⎟ ⎜
4
can
2
⎛ 6 ⎞ ⎛ 1 ⎞
a
P (X
=
4 )
b
P (X
≥
1)
c
P (X
≤
3)
⎠
×
625
25
48
=
⎛ n ⎞
r
3125
Use
P(X
=
r)
=
⎜
⎝
= 0.015 36
= 0.0154 (3
⎟
r
p
n
–
r
q
⎠
sf)
6
b
4
1
For
P(X
≥
1
P(X
=
1)
it
is
quicker
to
calculate
5
−
0)
than
to
calculate
4096
P(X
= 1
=
1)
+
P(X
=
2)
+
...
+
P(X
=
6)
15 625
11 529
=
15 625
= 0.738 (3 sf)
c
P (X
≤
3)
=
0.983
P(X ≤ 3) = P(X = 0) + P(X = 1) +
It
P(X = 2) + P(X = 3)
Use
the
your
GDC
Probability
distributions
this
calculation
(see
following).
{
for
is
P(X
so
easy
<
r)
read
the
carefully.
Continued
on
next
page
to
and
confuse
P(X
≤
r)
questions
GDC
Using
your
help
on
CD:
Alternative
GDC:
demonstrations
Plus
a
and
GDCs
Casio
are
on
for
the
TI-84
FX-9860GII
the
CD.
b
c
Exercise
1
X
is
C
binomially
distributed
with
4
trials
and
a
probability
1
of
success
equal
to
on
each
trial.
2
Without
a
a
P(X
=
c
P(X
≤
calculator
1)
1)
If
X
~ B
the
probability
b
P(X
<
d
P(X
≥
of
1)
1)
1 ⎞
⎛
2
determine
6,
find
⎜
to
3
significant
figures
⎟
3
⎝
In
question
and
a
P(X
=
c
P(X
≤
2)
b
P(X
<
d
P(X
≥
d
use
2)
the
you
X
is
c
binomially
distributed
with
8
of
Binompdf
2)
on
If
b
Binomcdf
2)
instead
3
2
⎠
trials
and
a
probability
of
calculator
,
are
as
calculating
cumulative
a
probability.
2
success
equal
to
at
each
attempt,
what
is
the
probability
of
7
a
exactly
c
more
5
successes
than
5
successes
b
less
d
at
than
least
5
successes
one
success?
Chapter
Example
The
probability
What
get
a
is
the
bus
that
I
get
probability
only
a
bus
that
in
to
a
work
on
working
any
mor ning
week
of
five
is
0.4.
days
I
will
twice?
Answer
By
hand:
Let
I
X
get
X
∼
be
a
the
number
B(5,
days
Can
=
you
see
why
this
is
a
binomial
situation?
0.4)
⎛ 5 ⎞
P( X
of
bus.
2)
=
⎜
⎝
⎟
2
2
3
( 0.4 )
( 0.6 )
We
require
P(X
=
2)
⎠
= 10 ×
0.16 ×
=
0.3456
=
0.346 (3 sf)
0.216
Using
your
GDC:
See
Chapter
Section
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
Example
When
were
of
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
administering
cured.
10
Casio
are
17,
5.12
The
a
dr ug
testing
it
was
program
known
that
80%
administered
the
of
people
dr ug
to
using
two
it
groups
patients.
Assume
What
is
the
probability
that
all
10
patients
were
cured
in
both
that
X
is
groups?
binomial
are
two
since
there
outcomes:
Answer
a
Let
X
be
‘the
number
of
patients
Multiply
the
success
and
cured
in
a
group
of
10’.
P(X
=
10)
is
a
cure
probabilities
and
P(X
=
a
failure
is
‘not
10)
a
cure’.
Assume
that
10
P(X
=
10)
=
0.8
=
0.10737
…
because
the
two
events
(the
patients
the
2
[P(X
=
10)]
trial
results
(0.10737…)
being
cured
in
each
group)
are
patient
=
0.0115
(3 sf)
independent.
So
for
two
groups
of
to
patient
the
probability
all
are
The
cured
xed
probability
2
is
[P(X
=
10)]
success
Probability
distributions
are
10
independent.
patients,
from
2
=
is
0.8.
of
Exercise
1
A
It
D
regular
is
tetrahedron
rolled
What
is
four
the
most
downwards?
2
The
he
shoots
the
at
a
he
hi ts
b
he
misses
factor y
produce
4
in
a
none
c
at
The
a
a
The
will
one
b
at
The
least
6
In
a
an
row
7
In
the
is
is
of
A
b
times
of
bottom
that
this
a
one
the
face
red
value
bull’s
red
face.
is
noted.
face
will
end
occurring?
eye
when
attempts
times
at
least
five
times.
making
four
be
that
a
that
the
any
component
is
type
machine
13
of
will
What
from
exactly
b
same
0.01.
components
each
will
is
the
probability
machine,
be
not
be
faulty
faulty?
telephone
If
the
lines
line
is
engaged
switchboard
lines
that
be
at
are
has
10
at
a
company
lines,
find
the
engaged
free
Nicole
most
three
hall,
it
(to
goes
that
is
probability
on
4
significant
to
bed
five
at
figures)
7:30
on
consecutive
a
given
days
day
she
is
goes
0.4.
to
days.
known
that,
in
that
a
15%
row
of
of
six
desks
desks,
are
wobbly
.
more
than
wobbly?
the
six
are
probability
the
probability
that
exactly
one
will
be
wobbly
in
a
desks?
production
Processors
packet
will
the
scores
eight
faulty
the
on
mass
defective.
a
of
0.25.
of
the
will
What
in
probability
examination
one
b
is
three
7:30
What
of
probability
machines
will
probability
at
of
and
that
half
Calculate
bed
be
two
probability
a
that
five
four
sample
probability
color
faces
0.55.
substandard
least
white
marksman
is
bull
the
number
the
a
three
Questions
switchboard
5
bull
the
has
component.
that,
that
target
the
Exam-Style
A
is
probability
a
and
likely
What
probability
Find
3
times
has
is
of
are
selected
computer
selected
at
at
random.
processors
random
Find
the
it
and
is
found
put
into
probability
that
packs
that
5%
of
are
15.
it
contain
i
three
iii
at
Two
there
defective
least
two
packets
processors
defective
are
no
ii
defective
processors
processors
selected
at
random.
Find
the
probability
that
are
i
no
ii
at
iii
no
defective
least
two
processors
defective
defective
in
either
processors
processors
in
one
packet
in
either
packet
packet
and
at
least
two
in
the
other.
Chapter
Example
A
box
red.
contains
The
How
least
rest
many
one
a
are
large
white.
flowers
red
number
car nations
Car nations
must
car nation
of
be
picked
among
are
so
them
picked
that
is
of
which
at
the
one-quarter
random
from
probability
greater
than
that
the
are
box.
there
is
at
0.95?
Answer
1
Let
X
be
the
random
variable
‘the
are
number
X
∼
P(X
of
B(n,
≥
red
red,
so
P(red)
=
0.25
4
car nations’.
0.25)
1)
=
1
–
P(X
=
1
–
(0.75)
=
0)
n
n
1 − (0.75)
We
> 0.95
require
P(X
≥
1)
>
0.95
n
0.05 > ( 0 .75)
log
and
0.05 >
0.75 <
n
least
picked
at
out
least
If
2
1%
X
∼
X
4
The
log
0.75
inequality
of
the
red
B(n,
of
the
to
least
value
of
n
is
11.
be
ensure
there
car nation
is
among
0.95.
and
a
that
in
0.2)
P(X
large
can
the
<
box
be
=
of
P(X
if
must
≥
0.0256,
fuses
taken
sample
and
1)
1)
>
are
the
be
find
n
faulty
.
What
probability
greater
0.75,
find
than
the
is
that
the
largest
there
are
0.5?
least
possible
n
probability
would
How
must
that
than
in
size
fuses
scored
5
box
0.6)
competition
she
The
E
fuses
∼
value
negative
B(n,
of
faulty
If
When
0.05
greater
sample
3
n.
log
probability
Exercise
1
for
0.05
car nations
one
is
log
> 10.4
11
the
them
inequality
you
divide
by
amount,
at
is
need
least
many
to
Anna
Find
take
once
times
probability
Probability
that
0.3.
is
at
distributions
if
least
the
greater
must
that
scores
the
an
a
penalty
number
probability
than
one
tail
in
a
hockey
attempts
that
a
goal
that
is
0.95.
unbiased
least
goal
of
coin
will
be
tossed
occur
is
at
so
that
least
0.99?
no
a
the
>
n
that
the
so
n log
At
Solve
n log 0.75
reverses.
Expectation
of
a
binomial
distribution
2
Think
of
the
example
of
the
biased
coin
where
P(H)
=
.
3
If
you
to
get
toss
a
the
coin
3
times
how
many
times
would
you
expect
head?
Intuitively
the
answer
This
same
is
2.
2
is
the
calculating 3
as
2
The
proof
of
this
3
formula
➔
For
the
binomial
expectation
The
Galton
scientist
half,
Board,
Sir
where
the
poured.
ball
falls
Each
time
a
process
heights
If
the
of
of
distribution
why
Example
A
biased
The
of
dice
sixes
bean
one
machine,
in
In
directly
this
p),
the
the
is
not
standard
on
level
syllabus.
consists
slots.
B(n,
np
arranged
is
to
hits
therefore
heaps
of
heights
fur ther
on
It
∼
where X
of
an
is
staggered
the
middle
above
the
a
device
upright
order
,
of
top
the
nail
for
board
and
top,
so
statistical
with
a
lower
there
that
evenly
is
half
a
experiments
spaced
divided
funnel
nails
into
into
a
named
driven
after
into
number
which
balls
English
its
upper
of
can
each
nail.
of
the
nails,
it
can
bounce
right
or
left
probability.
number
the
funnel
ball
This
are
=
or
Galton.
nails
The
equal
E(X)
rectangular
directly
with
X,
quincunx
Francis
evenly-spaced
be
of
distribution
(See
gives
balls
balls
of
this
of
the
is
in
rise
the
to
a
slots
sufciently
ball
heaps
will
Section
15.3).
Y
ou
binomial
at
the
large
distribution
then
the
approximate
may
in
the
bottom.
wish
to
distribution
a
normal
investigate
is.
dice
is
is
thrown
then
for
thrown
these
12
30
a
times
fur ther
and
12
the
number
times.
Find
of
the
sixes
seen
expected
is8.
number
throws.
Answer
8
X
∼
B(12,
p)
where
p
4
=
=
30
Let
X
be
‘the
number
of
sixes
seen
in
15
12
throws’.
4
E( X )
=
np
=
12
×
=
Y
ou
3 .2
may
wish
to
15
conduct
your
binominal
and
Exercise
a
A
A
fair
fair
sixes
c
A
how
close
F
coin
is
tossed
40
times.
Find
the
expected
number
of
the
results
are
expected
to
binomial
results.
heads.
b
experiment
explore
your
1
own
dice
is
rolled
40
times.
Find
the
expected
number
of
obtained.
card
is
drawn
retur ned.
repeated
13
40
of
from
these
times.
a
pack
cards
Find
the
of
are
52
cards,
labeled
expected
as
noted
and
Hear ts.
number
of
This
is
Hear ts.
Chapter
Exam-Style
X
2
is
a
mean
A
3
a
of
each
X
is
variable
choice
with
test
each
number
a
the
distribution
b
the
mean
c
the
probability
10
100
or
of
more
families
numbers
of
Number
of
Find
b
Using
two
10
15
one
X
and
~
p
B(n,
=
0.4,
questions
correct
p).
Given
find
and
answer
the
n
four
per
that
possible
question.
answers
Assume
answer.
questions
answered
correctly’
give:
X
that
purely
each
a
by
with
student
will
achieve
the
pass
mark
of
guessing.
three
children
are
found
to
have
these
girls:
girls
the
three
that
X
Frequency
a
of
of
is
has
only
guesses
‘the
such
distribution
one
student
If
4
the
multiple
for
Questions
random
0
1
2
3
13
34
40
13
probability
your
value
children,
that
from
in
a
a
a
single
baby
calculate
sample
of
the
100,
bor n
is
number
you
would
a
girl.
of
families
expect
to
with
have
girls.
Variance
of
a
binomial
distribution
The
Chapter
8
introduced
the
concept
of
the
variance
of
a
set
of
data,
proof
variance
a
measure
of
dispersion.
not
on
Level
The
the
➔
formula
formula
If
X
~
Thinking
for
the
variance
of
the
binomial
distribution
is
given
of
the
as
formula
the
is
Standard
syllabus.
in
booklet:
B(n, p)
about
then
the
Var(X)
original
=
npq
where
example
of
q
the
=
p
–
biased
1
coin
where
2
P(H)
=
,
if
you
toss
the
coin
this
will
3
times
you
expect
to
get
a
head
2
times.
3
However,
this
obviously
experiment
Using
the
many
formula
for
times
not
you
=
will
ever y
time.
sometimes
get
npq
=
3 ×
If
0,
you
1
repeat
and
3
heads.
Y
ou
variance,
2
Variance
happen
1
×
3
can
by
the
=
3
taking
square
3
The
For
the
binomial
distribution
where X
~
B (n,
of
the
variance.
expectation
of
X,
E(X)
=
expected
X,
E(X),
variance
of
X,
Var(X)
=
npq
where
q
=
1
–
distributions
the
also
mean,
p.
so
Probability
is
np
called
●
value
p)
of
●
σ,
2
general
➔
the
deviation,
root
In
nd
standard
E(X)
=
.
.
Example
In
a
large
company
,
40%
of
the
workers
travel
to
work
on
public
transpor t.
A
random
Find
on
the
sample
of
expected
public
15
workers
number
transpor t,
and
of
the
is
selected.
workers
standard
in
this
sample
that
travel
to
work
deviation.
Answer
Let
W
who
be
the
travel
to
number
work
of
on
n
workers
=
15,
p
=
0.4
public
transpor t.
W
~
B(15,
E(W )
=
Var(W )
15
=
Standard
3.6
=
0.4)
×
15
0.4
×
deviation
X
~
B
Find
1
A
the
fair
An
coin
the
b
Var(X)
c
P(X
<
in
calculate
and
the
the
standard
npq
deviation
is
square
root
of
mean
and
variance
of
X
deviation
of
the
binomial
0.6).
40
times.
number
dice
is
of
Find
the
mean
and
standard
heads.
thrown
10
times.
Let X
be
the
number
of
Find
expected
number
of
sixes
μ).
frequent
once
,
tossed
obtained.
a
A
is
of
Exam-Style
5
Standard
is
B (12,
unbiased
sixes
np
=
⎠
mean
deviation
4
=
Var(W)
3.6
⎞
⎟
4
distribution
3
=
variance
0,
⎜
⎝
2
0.6
G
⎛
If
E(W)
6
×
1.90(3 sf)
Exercise
1
=
0.4
Question
flyer
ever y
22
occasions.
a
the
5
finds
trips,
Using
expected
that
on
a
she
is
average.
binomial
number
delayed
of
One
year
model,
jour neys
at
a
par ticular
she
uses
the
air por t
air por t
on
find
that
will
be
delayed
at
that
air por t
6
b
the
variance
c
the
probability
At
the
local
can
is
r un
100
that
athletics
metres
she
club,
in
is
delayed
the
under
on
expected
13
fewer
than
number
seconds
is
4.5
of
and
4
occasions.
people
the
that
variance
3.15.
Find
13
the
probability
that
at
least
3
people
can
r un
100 m
in
under
seconds.
Chapter
Exam-Style
X
7
is
a
mean
of
the
variable
For
Find
random
the
Hence
variable
possible
calculate
The
these
P(X
foot,
a
2
Where
3
Is
4
Join
the
the
histogram
6)
the
~
and
p
variance
B (n,
n
p),
and
for
B (n,
=
of
p).
0.3
Given
that
the
find:
X.
E(X)
=
9.6
and
Var(X)
=
1.92.
p.
each
normal
50
possible
pair.
distribution
students
weight,
the
in
your
maximum
school
hand
for
span,
one
of
length
of
of
data
the
roughly
from
and
measurements
more
peak
of
7.8
X
wrist.
of
midpoints
symmetrical
If
of
histogram
the
–
height,
histogram
is
~
that
distri bution
around
circumference
Draw
Y
our
from
categories:
1
=
normal
data
X
values
Investigation
Collect
is
the
b
a
.
such
distribution
n
a
8
Question
random
of
the
the
a
measurements
tops
of
the
investigation
cur ve
around
histogram?
symmetrical?
is
is
bell-shaped
central
were
bars
of
your
probably
with
the
histograms
with
a
cur ve.
f(x)
roughly
majority
of
value.
taken,
a
histogram
plotted
and
the
O
midpoints
become
of
the
more
tops
of
the
bars
symmetrical
and
look
cur ve
joined
with
a
cur ve,
it
would
like
the
shown.
a
normal
Gaussian
normal
normal
distribution
is
probably
the
Friedrich
most
distribution
in
statistics,
since
it
is
model
for
many
naturally
These
attributes
of
include
the
people,
Gauss
cur ve
In
animals
mass-produced
and
plants,
the
old
items
from
factories.
could
approximation
also
be
of,
for
to
applied
example,
times,
complete
or
IQ
a
as
the
student
is
histor y,
it
●
the
is
piece
of
work,
scores.
work.
por trait
of
probability
Gauss
played
the
in
Gauss,
function
the
appeared
note.
an
the
French
impor tant
statisticians
and
role
in
Abraham
Pierre-Simon
De
were
Moivre
involved
in
developed
in
distribution,
the
binomial
this
was
not
1733
as
discovered
much
the
to
about
the
mean
de
Laplace
an
of
normal
the
ear ly
cur ve
approximation
although
until
his
1924
paper
mode
and
median
are
the
used
the
normal
by
cur ve
Karl
in
1783
describe
the
distribution
of
errors,
and
in
1810
same.
proved
the
distributions
Laplace
(μ)
he
Probability
the
its
mathematically
to
data
bell-shaped
symmetrical
mean,
and
(1667–1754)
Pearson.
●
used
astronomical
exam
case
cur ve
Gauss
Carl
an
on
●
analyze
10-Deutschmark
(1749–1827)
times
each
“Gaussian
The
Moivre
In
the
and
its
distribution
reaction
called
mathematician
(1777–1855).
to
Germany,
cur ve
Although
scores,
also
physical
on
even
is
German
occurring
normal
variables.
the
a
1809.
suitable
cur ve
after
normal
impor tant
Curve
distri bution
cur ve”
The
x
would
This
The
is
it
bell-shaped
The
until
then
an
Central
essential
Limit
statistical
Theorem.
theorem
called
The
characteristics
There
is
no
single
normal
of
any
cur ve,
normal
but
distribution
a fami ly of curves ,
each
one
Remember
that
mean, μ, is
the
and
defined
by
its
μ,
mean,
and
standard
the
σ
deviation,
measure
If
a
random
variable,
X,
has
a
normal
average,
standard
σ,
deviation,
➔
the
distribution
with
of
is
a
spread.
mean
2
μ
and
standard
σ,
deviation
this
is
written
X
~
N (μ,
σ
)
Note
μ
σ
and
are
called
the
parameters
of
the
that
in
the
2
distribution.
expression
X
~
N(μ,
σ
),
2
σ
The
mean
is
the
central
point
of
the
distribution
and
the
is
the
variance.
standard
Remember
deviation
describes
the
spread
of
the
distribution.
The
higher
variance
standard
deviation,
the
wider
the
normal
cur ve
will
that
the
the
is
the
standard
be.
deviation
squared.
f(x)
2
These
three
graphs
show
~
X
N (5,
2
x
),
x
1
x
2
3
1
2
~
X
N (10,
2
2
)
and
X
2
The
all
~
N (15,
2
).
3
standard
the
width
same,
μ
but
deviations
so
the
μ
<
1
are
cur ves
are
all
the
same
x
0
μ
<
2
5
10
15
20
3
2
These
three
graphs
show
X
~
N (5,
1
),
f(x)
1
X
2
~
X
N (5,
2
1
2
)
and
X
2
~
N (5,
3
).
Here
the
3
means
are
all
the
same
and
all
the
cur ves
X
2
are
centered
around
this
but
σ
σ
<
1
σ
<
2
3
X
3
so
cur ve
is
X
narrower
than
X
1
,
and
X
2
is
2
x
0
narrower
than
5
X
10
3
The
cur ves
deviations
The
No
may
but
area
matter
the
So
equal
cur ve
in
what
the
to
as
this
they
different
all
have
beneath
distribution,
and
have
1.
the
We
area
can
same
and/or
of
μ
σ
and
under
the
different
standard
characteristics.
normal
therefore
representing
normal
the
the
values
total
means
distribution
are
for
cur ve
consider
is
a
normal
always
par tial
curve
probability
the
areas
f(x)
same
under
probabilities.
distribution
we
could
find
the
probability
μ
x
P(X
<
5)
by
finding
Unfor tunately
cur ve)
to
for
the
the
the
shaded
probability
nor mal
area
on
the
function
distribution
is
(the
ver y
0
diagram.
equation
of
complicated
5
the
and
difficult
use.
2
( X )
1
f
(X )
2
2
−
e
∞
<
X
<
∞
2
It
would
cur ve!
be
too
hard
However,
for
there
us
are
to
use
other
integration
methods
we
to
find
can
areas
under
this
use.
Chapter
The
standard
normal
distribution
Note
The
μ
=
to
standard
0
σ
and
describe
normal
=
1.
the
The
distri bution
random
number
of
is
the
variable
standard
normal
is
called
deviations
distribution
Z.
It
any
uses
value
where
‘z-values’
is
P(Z
that
=
think
a)
of
having
away
the
=
no
the
P(a
Y
ou
Z
~
The
can
standard
use
N (0,
1)
Example
Given
a
no
can
line
and
area.
mean.
This
➔
Y
ou
as
width
therefore
from
0.
this
for
P(−2
d
P(Z
<
<
GDC
values
distribution
to
calculate
between
a
is
the
and
b
written Z
areas
and
~
under
hence
N (0,
the
P(a
<
that
Z
<
b)
=
P(a
≤
Z
≤
b)
=
P(a
<
Z
≤
b)
=
P(a
≤
Z
<
b)
1)
cur ve
Z
<
of
b).
that
a
your
normal
means
<
Z
~
N (0,
Z
<
1)
1),
0)
find
b
P(Z
<
e
P(|Z|
1)
>
c
P(Z
>
−1.5)
0.8)
Answers
a
P(−2
<
Z
<
1)
=
0.819
Using
the
distribution
menu
on
GDC
your
GDC,
choose
nor mCdf
and
Plus
enter
the
values
in
this
order:
help
upper
limit,
mean,
standard
deviation.
b
P(Z
<
1)
=
0.841
Enter
the
negative
lower
limit
as
a
ver y
small
number,
999
–9
c
P(Z
>
−1.5)
=
0.933
×
10
Enter
the
upper
limit
as
a
ver y
large
999
number,
9
×
10
{
Probability
distributions
Continued
on
CD:
and
Casio
Alternative
for
the
next
page
are
on
TI-84
FX-9860GII
lower
GDCs
limit,
on
demonstrations
the
CD.
f(Z)
d
P(Z
<
0)
=
Here
0.5
you
do
calculator
not
need
because
to
the
use
the
graph
is
f(Z
symmetrical
e
P(|Z|
>
0.8)
=
1 – 0.576
=
0.424
about
the
<
0)
=
0.5
mean.
z
0
|Z|
>
0.8
–0.8
<
Z
means
<
0.8
See
Chapter
Section
Exercise
1
Given
2
3
4
5
6
Find
that
<
the
Z
Z
~
<
area
1
b
between
0.5
the
a
1
b
2.4
and
area
standard
Find
the
1
b
1.75
area
standard
Given
Z
the
1.5
the
the
deviation
N (0, 1)
the
cur ve
the
the
is
the
GDC
b
P(Z
>
0.72)
d
P(Z
>
−2)
e
P(Z
≤
−0.28)
N (0, 1)
<
Z
<
1.2)
that
Z
~
N (0, 1)
P(0.2
Given
P(|Z|
a
<
use
the
P(−2
b
use
the
0.4)
Z
3)
mean
the
mean.
than:
than
≤
mean.
find
c
GDC
<
the
from
more
to
0.65)
~
<
cur ve:
from
less
<
Z
Z
mean.
P(Z
that
<
mean
below
the
P(−3
c
mean
that
below
is
a
Given
2)
deviations
above
use
<
normal
which
above
deviations
Z
deviations
cur ve
deviations
under
<
standard
standard
deviation
~
P(−2
standard
under
standard
that
2
and
standard
a
find
b
under
between
Find
N (0, 1)
1)
a
a
7
H
P(−1
a
17,
5.13.
to
to
c
1.8)
P(−1.3
≤
Z
≤
−0.3)
find
P(|Z|
b
≥
find
0.3)
GDC
P(Z
>
1.24)
f(z)
0.4
In
Question
that
Z
lies
standard
of
the
1
of
within
Exercise
one
deviations
mean
15H,
standard
of
the
you
found
deviation
mean
and
the
of
three
probability
the
mean,
standard
0.2
two
0.1
deviations
respectively
.
z
–3
Y
ou
can
see
that
most
of
the
data
for
a
normal
–2
–1
0
1
2
3
distribution
68.27%
will
lie
within
three
standard
deviations
of
the
mean.
95.45%
99.73%
Chapter
Probabilities
Clearly
,
for
however,
standard
normal
deviation
of
1).
other
ver y
few
normal
real-life
distribution
But
you
can
distributions
variables
(with
a
mean
transform
are
of
any
distributed
0
and
normal
a
like
the
standard
distribution
2
X
N ( μ,
~
normal
in
σ
)
to
the
standard
distributions
location
and
have
normal
the
same
distribution,
basic
shape
because
but
are
all
merely
shifts
spread.
2
To
transform
z-value
on
~
given
N (0,
1)
value
use
of
the
x
on
X
N ( μ,
~
σ
)
to
its
equivalent
formula
x
z
Z
any
=
Y
ou
can
then
use
your
GDC
to
find
the
required
probability
.
X
2
➔
If
X
N ( μ,
~
σ
)
then
the
transformed
random
variable Z
has
a
standard
Example
normal
distribution.
2
The
random
variable
X
~
N (10,
2
).
Find
P(9.1
<
X
<
10.3)
Answer
<
P(9.1
X
<
Draw
10.3)
a
sketch.
f(x)
P(9.1
<
x
<
10.3)
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
x
0
5
9.1
z
10
15
10.3
10
z
=
=
<
X
<
of
x.
<
10.3)
Z
<
0.15)
Enter
Check
the
values
that
reasonable
X
<10.3)
=
0.233
sketch.
value
= 0.15
P(−0.45
P(9.1<
each
2
0.45
P(9.1
10
Standardize
=
2
=
20
Probability
distributions
the
into
your
answer
when
GDC.
looks
compared
with
your
Casio
are
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
Y
ou
can
directly
the
is
also
using
the
these
GDC.
standardization
the
quickest
method
But
it
of
is
Enter
of
=
Without
most
efcient
this
to
using
this
question.
know
the
standardization.
lower
mean
and
solutions
formula,
answering
impor tant
method
Exercise
nd
limit,
10,
upper
standard
limit,
deviation
=
2.
I
2
The
1
random
variable
X
~
N (14,
5
).
Find
P(X
a
The
2
<
16)
random
P(X
b
variable
X
>
~
9)
P(9
c
N (48,
≤
X
<
12)
P(X
d
<
14)
81).
Find
P(X
a
<
52)
P(X
b
≥
42)
P(37
c
<
X
<
47)
2
The
3
random
variable
X
~
N (3.15,
0.02
).
Find
P(X
a
<
Example
Eggs
laid
is
b
P(X
≥
3.11)
c
P(3.1
<
X
<
3.15)
by
distributed,
What
3.2)
a
chicken
with
the
are
mean
probability
a
an
egg
weighs
c
an
egg
is
more
between
known
55 g
and
to
have
standard
the
mass
normally
deviation
2.5 g.
that
than
52
and
59 g
b
an
egg
is
smaller
than
53 g
54 g?
Answer
2
W
~
N
(55,
2.5
)
f(w)
Sketch
2
W
~
N(55, 2.5
Mean
)
3
×
σ
first.
=
=
55
3
×
2.5
=
7.5
0
w
45
50
55
60
65
Enter
the
value
in
your
GDC:
GDC
lower
limit,
mean
=
upper
limit,
help
Plus
and
GDCs
P(W
>
59)
=
0.0548 (3 sf)
b
P(W
<
53)
=
0.212 (3 sf)
c
P(52
<
W
<
54)
=
CD:
Casio
Alternative
for
the
TI-84
FX-9860GII
55,
standard
a
on
demonstrations
deviation
=
are
on
the
CD.
2.5
0.230 (3 sf)
Chapter
Exercise
J
Exam-Style
1
Questions
Households
groceries
in
with
distribution
what
2
is
the
less
b
more
c
between
with
a
3.5 mm
3
The
and
or
of
known
the
What
4
Packets
550 g.
the
The
more
The
with
mass
the
mass
that
The
Here
you
need
a
nominal
containing
find
the
Y
ou
value
too
cut-off
can
use
of
little
your
I
rejected.
to
wait
in
distributed
will
have
wait
breakfast
packets
What
to
of
are
Out
Dr.
0.25 mm.
smaller
of
a
than
batch
Barrett’s
with
wait
less
than
cereal
of
are
‘Flakey
with
a
propor tion
washing
powder
standard
that
a
chosen
the
of
500
mean
waiting
14
minutes
more
than
20
10
minutes?
said
to
flakes’
mean
of
of
contain
is
such
that
551.3 g,
packets
will
and
contain
is
normally
deviation
packet
chosen
of
at
distributed
20 g.
random
has
a
150
for
GDC
a
at
mass
random.
which
is
What
less
is
the
than
probability
475 g?
distribution
value
For
juice.
point
which
normally
475 g.
have
find
deviation
any
distributed
and
normal
to
are
distributed
minutes.
of
of
are
probability
.
and
patients
probability
packets
distribution,
mass?
500 g
than
normal
the
spending:
diameters
have
that
15 g.
packets
of
a
that
on
doctor.
of
stated
less
inverse
4
normally
packets
all
cumulative
to
of
Three
b
of
mean
Find
a
the
week
acceptable?
flakes’
deviation
than
a
are
per
Assuming
follows
standard
normally
production
masses
standard
5
‘Flakey
a
be
probability
propor tion
of
with
patients
the
€100
week.
4.5 mm
would
see
of
€20.
household
per
and
deviation
to
a
bolts
be
of
expenditure
of
€125
than
to
deviation
accurately
time
average
week
4 mm
many
an
week
per
and
bigger
minutes
b
of
standard
Find
a
€90
produces
length
is
per
measured
how
room
grocer y
€80
mean
spend
standard
€130
than
are
bolts
of
than
machine
Bolts
a
probability
a
A
Por tugal
ml.
The
the
to
in
the
example,
5%
of
data
a
cartons
owner
of
minimum
help
find
that
has
company
the
are
given
cartons
rejected
company
volume
this
a
fills
of
value.
a
of
juice
for
may
wish
to
carton.
The
calculator
See
has
a
function
called
Inverse
Normal
which
will
do
this.
In
Chapter
Section
examples
Z
~
we
will
retur n
N (0, 1)
Probability
distributions
to
the
standard
normal
17,
these
distribution
5.14.
Example
Given
a
that
P(Z
<
Z
a)
~
=
N (0,
1)
use
0.877
b
your
P(Z
GDC
>
a)
to
=
find
0.2
a
c
P(−a
<
Z
<
a)
=
0.42
Answers
a
f(z)
Draw
a
sketch.
0.877
a
0
z
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
P(Z
a
=
<
a)
=
1.16
Casio
are
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
0.877
(3 sf)
Notice
b
that
to
find
the
value
of
a
f(z)
f(z)
for
P(z
<
a)
=
P(Z
easily
0.8
>
a)
find
P(Z
<
a)
The
areas
a
=
=
0.2
you
can
more
for
0.8
0.8
0.2
0
P(Z
a
c
>
=
a)
=
a
0
z
0.2
P(Z
<
a)
=
a
z
0.8
0.842 (3 sf)
P(−a
<
Z
<
a)
=
0.42
either
side
of
f(z)
the
shaded
size
0.42
and
region
equal
are
the
same
to
1
(1
−
0.42)
=
0.29.
Hence
2
P(Z
–a
a
<
a)
=
1
0.29
=
0.71
z
0
a
=
0.533 (3 sf)
Chapter
Exercise
Find
1
a
<
a)
=
c
P(Z
>
a)
=
a
such
0.922
<
Z
<
a)
c
P(a
<
Z
<
−0.3)
a
such
Find
<
the
>
a)
b
P(a
b
P(|Z|
=
0.342
=
0.12
=
<
Z
<
1.6)
=
0.787
0.182
that:
Z
<
a)
values
a
P(Z
that:
P(1
P(−a
b
0.005
a
a
4
that:
P(Z
Find
3
such
a
Find
2
K
=
of
0.3
a
shown
in
these
>
a)
=
0.1096
diagrams:
b
f(z)
f(z)
0.95
0.2
z
–5
0
z
5
–3
–2
–1
0
1
2
a
a
Once
again,
however,
distributions
Example
that
are
we
not
3
are
the
more
likely
standard
to
be
normal
dealing
with
distribution.
2
Given
that
X
~
N(15, 3
)
determine
x
where
P(X
<
x)
=
0.75
sketch
to
show
Answer
f(x)
Draw
a
the
value
of
x
required.
0.75
x
0
x
15
This
question
is
best
done
on
the
GDC
help
on
CD:
demonstrations
GDC.
In
invNor m
enter
x,
Plus
standard
x
=
z
=
deviation.
You
could
also
by
first
standardizing
{
answer
the
question
15
3
Probability
distributions
and
GDCs
17.0
x
the
value
Continued
Alternative
for
the
TI-84
mean,
on
of
x.
next
page
Casio
are
on
FX-9860GII
the
CD.
P
(X
<
⎜
=
x
⎛
P
x)
15 ⎞
<
Z
= 0
⎟
3
⎝
x
0.75
75
⎠
15
= 0
6745
3
x
= 17
0
Example
Car tons
with
5%
a
of
Find
of
juice
mean
of
car tons
the
contain
are
are
it
is
to
that
and
rejected
minimum
if
such
150 ml
be
a
for
volume,
their
volumes
standard
containing
to
the
are
deviation
too
nearest
normally
of
little
ml,
that
distributed
5 ml.
juice.
a
car ton
must
accepted.
Answer
Let
V
be
the
volume
of
a
car ton.
Let
m
be
the
minimum
volume
that
2
V
~
P (V
N(150,
<
m)
5
=
)
a
car ton
must
have
to
be
accepted.
0.05
Draw
a
sketch.
f(V)
0.05
m
0
V
150
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
The
the
minimum
nearest
volume
is
142 ml
Casio
are
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
to
ml.
Chapter
Exercise
L
2
1
X
~
N(5.5, 0.2
)
and
P(X
>
a)
=
0.235
Find
the
value
of
a
1
of
2
The
mass,
M,
of
a
randomly
chosen
tin
of
dog
food
is
such
that
less
2
M
~
N(420, 10
).
the
values
are
4
than
the
rst
Find
quar tile.
the
a
first
quar tile
Exam-Style
3
contain
has
a
into
normal
An
a
percentile.
a
countr y
500 ml
must
for
bottle
have
filling
with
a
insist
that
at
all
least
bottles,
standard
mineral
that
which
bottles
amount.
puts
deviation
an
of
that
claim
‘Yummy
average
1.6 ml
of
and
Cola’
502 ml
follows
a
distribution.
inspector
What
What
b
in
machine
each
90th
Question
Regulations
to
the
b
is
the
randomly
probability
proportion
of
selects
that
bottles
it
a
bottle
will
will
of
break
contain
‘Yummy
the
Cola’.
regulations?
between
500 ml
and
505 ml?
95%
c
a
4
The
a
If
5
a
marks
standard
If
5%
by
If
b
Y
ou
of
550 g
chosen
500
the
of
the
d
at
with
of
15
of
hypermarket
standard
and
by
a
and
mean.
b ml
are
the
liquid
are a
normally
deviation
find
of
What
of
where
and
b?
distributed
25 g.
probability
that
its
570 g.
10%
in
of
an
mean
the
lettuces.
examination
of
55
marks
are
and
a
marks.
or
obtain
more,
fail
by
a
distinction
find
the
scoring
f
value
marks
of
or
d
less,
f
given
σ
a ml
the
random,
candidates
marks
(if
a
and
520 g
students
be
about
at
candidates
value
mean
between
exceeded
deviation
of
sold
between
of
also
the
is
mass
scoring
may
either
lettuce
distributed
10%
find
contain
symmetrical
mass
lies
the
normally
a
of
lettuce
Find
The
are
mean
mass
b
bottles
b
masses
with
a
of
and
cumulative
is
known)
or
probabilities
the
standard
and
asked
deviation
to
(if
find
μ
is
Extension
material
Worksheet
15
distribution
known)
or
both.
Sacks
of
loader.
Use
this
potatoes
In
a
test
it
with
was
information
mean
found
to
find
weight
that
the
5 kg
10%
of
standard
are
packed
bags
were
deviation
by
an
over
of
automatic
5.2 kg.
the
process.
Answer
Let
M
be
the
mass
of
potatoes
in
a
10%
(0.1)
of
bags
were
over
5.2 kg.
sack.
2
M
~
N(5,
σ
)
{
Probability
as
approximation
distribution
Example
-
distributions
Continued
on
next
page
on
CD:
Normal
an
to
a
binomial
P(M
>
5.2)
=
0.1
Draw
f(m)
a
sketch.
0.1
0
m
5.2
5
5.2
5
0.2
Standardize.
Z
=
0.2
P
Z
0.2
P
or
0.1
Z
0.9
From
P(Z
the
<
GDC
1.28155
. . .)
=
0.9
GDC
help
on
CD:
demonstrations
Plus
and
GDCs
Casio
are
on
Alternative
for
the
TI-84
FX-9860GII
the
CD.
0.2
= 1.28155....
σ
σ
=
0.156
Example
A
(3 sf)
manufacturer
she
is
under
5.5%
does
producing.
1.8 cm
as
too
in
big.
not
know
However
diameter.
What
is
a
It
the
the
mean
sieving
is
found
mean
and
system
that
and
standard
rejects
8%
of
standard
all
the
deviation
ball
ball
of
bearings
bearings
deviation
of
the
the
diameters
larger
are
than
rejected
ball
as
bearings
of
2.4 cm
too
ball
bearings
and
small
those
and
produced?
Answer
Let
D
be
the
diameters
of
ball
bearings
produced.
We
know
that
8%
are
too
small,
and
5.5%
are
too
2
D
~
N(μ,
σ
)
big.
P(D
<
1.8)
=
0.08
P(D
>
2.4)
=
0.055
Draw
f(d)
a
sketch.
0.08
0.055
0
1.8
2.4
1.8
2.4
and
Standardize
1.8
Z
each
value.
P
d
<
= 0.08
From
the
first
expression.
{
Continued
on
next
Chapter
page
Z
>
Z
the
second
expression.
2.4
P
From
= 0.055
or
2.4
P
<
1
= 0.945
0.005
=
0.945
From
the
GDC
we
P(Z
<
−1.40507
P(Z
<
1.59819
know
. . .)
. . .)
=
=
that
0.08
and
0.945
1.8
=
1.40507
. . .
Solve
and
simultaneously
for
μ
and
σ.
GDC
2.4
=
1.59819
. . .
help
on
CD:
demonstrations
Alternative
for
the
TI-84
Plus
and
GDCs
μ
=
2.08
σ
and
Exercise
=
Casio
are
FX-9860GII
on
the
of
σ
CD.
0.200
M
2
1
X
~
N(30, σ
2
X
~
N(μ, 4
3
X
~
N(μ,
)
and
P(X
>
40)
=
0.115.
Find
the
value
2
)
and
P(X
<
20.5)
=
0.9.
Find
the
value
of
μ
2
P(X
4
A
<
σ
41.82)
random
Exam-Style
The
mean
children
Given
=
that
0.0287,
variable
σ,
deviation
5
).
such
X
>
μ
find
is
that
P(X
58.39)
P(X
<
89)
0.0217
and
σ
and
normally
=
distributed
=
0.90
and
with
P(X
mean
<
94)
μ
=
and
standard
0.95.
Find
μ
and
σ
Questions
height
have
a
of
children
height
of
of
a
145 cm
cer tain
or
age
more.
is
Find
136 cm.
the
12%
Lamber t
Adolphe
Jacques
Quételet
of
standard
(1796–1874),
deviation
6
The
of
the
standard
deviation
of
masses
of
loaves
of
bread
is
mass
of
20 g.
Only
Flemish
scientist,
was
rst
the
loaves
weigh
less
than
500 g.
Find
the
mean
the
the
normal
loaves.
distribution
7
The
masses
of
cauliflowers
are
normally
distributed
to
1%
apply
of
a
heights.
with
to
human
mean
characteristics.
0.85 kg.
74%
of
cauliflowers
a
the
standard
deviation
b
the
percentage
of
have
of
mass
less
cauliflowers’
cauliflowers
with
than
1.1 kg.
Find:
He
masses
mass
noted
characteristics
greater
than
1 kg.
as
height,
strength
8
The
lengths
of
nails
are
normally
distributed
that
were
mean μ
with
distributed.
and
standard
than
9
A
68 cm
roll
35%
of
of
a
find
rolls
Probability
are
paper
of
normal
the
wrapping
wrapping
lengths
deviation
rolls
value
paper
over
is
of
7 cm.
3 m
2.9 m.
sold
long
Find
wrapping
distribution.
distributions
2.5%
of
the
nails
measure
more
μ
of
is
If
as
and
the
‘3 m
that
the
value
paper,
long’.
of
It
is
average
the
assuming
found
length
standard
that
that
the
of
actually
the
rolls
deviation
lengths
of
of
rolls
only
of
the
follow
such
weight,
and
normally
Exam-Style
10
It
is
suspected
score
less
Find
a
Questions
that
than
the
the
108
scores
marks
mean
and
in
on
a
the
standard
test
test,
are
normally
and
20%
deviation
of
distributed.
score
the
more
scores,
if
30%
than
154
they
are
of
students
marks.
normally
distributed.
60%
b
of
appear
scores
11
Due
a
to
students
to
are
randomly
in
chosen
given
exceeding
Exam-Style
table
than
distributed
ball
of
Find
95%
and
wool
the
of
99%
117
consistent
as
manufacturing,
that
495 m
Review
The
normally
distribution.
deviation
more
reasonably
variations
normal
1
be
score
mean
balls
have
with
be
length
idea
of
modeled
and
of
the
Does
this
that
fact
the
above?
the
can
marks.
by
in
a
standard
wool
lengths
wool
have
lengths
exceeding
490 m.
exercise
Questions
shows
the
probability
distribution
of
a
x
discrete
a
b
2
Find
In
the
by
Find
a
of
P(X
a
P(X
value
of
distribution
=
x)
value
=
of
player
−2
cx(6
a
x),
discrete
x
=
1,
Find
b
a
=
x)
The
Find the
4
A
game
The
probability
involves
other
number
Let
probability
P
is
on
be
a
Write
b
Find
biased
c
What
d
A
is
is
all
his
son
Monday
the
random
2,
3,
4,
tetrahedral
10
total
will
Exam-Style
each
then
the
score
two
2,
the
score
of
six
spinners.
4,
4.
Each
he
2
k
k
0.1
0.1
variable X
is
5.
(four-faced)
is
after
One
1
2
3
1
1
1
4
4
8
4
shown.
is
spinner
of
each
value
week
If
by
the
£10.
expect
on
values
value
of
determines
gets
boy
numbers
possible
expected
mor ning.
than
2
E(X).
possible
total
2,
of
mathematician
give
a
1
two
rolls.
numbered
is
spun
Probability
1,
once
2,
3,
and
x
4.
the
recorded.
probability
the
each
spinning
product
down
the
of
numbered
each
the
of
1
0.3
Score
dice.
0
X
of
−
c.
rolls
−1
X
k.
expected
the
game
variable
value
probability
defined
3
the
Find
The
a
random
for P
of
P
amount
him
spins
get
to
and
Otherwise
to
spinners
P?
the
getting
son
the
after
pocket
play
the
he
10
of
the
game
product
gets
£5.
weeks
of
money
is
How
to
on
greater
much
playing
the
in
game?
Question
1
5
In
a
train,
of
the
passengers
are
listening
to
music.
Five
passengers
are
chosen
3
at
random.
Find
the
probability
that
exactly
three
are
listening
to
music.
Chapter
6
When
is
that
7
a
0.1.
Let
he
X
boy
He
wins.
be
Given
that
P(X
Given
that
P(65
dice
be
Copy
b
are
is
but
the
like
eight
much
3
>
mean
a),
and
the
Find
value
P(X
>
of
If
a
1
of
or
neither
X,
a
6
is
rolled
rolled
you
the
‘the
is
will
table
you
win
of
$5.
the
dollars
won
in
a
would
you
expect
9
ii
to
gain
(or
lose)
of
the
at
songs
on
my
friends
MP3
player.
If
probability
that
I
like
at
three
songs.
probability
school
random
ii
more
the
c
How
large
one
four
than
Find
throwing
most
the
of
mean
Find
a
the
the
the
Probability
of
a
of
I
choose
least
three
sixes
twice
in
five
throws
71.
a
be
is
of
be
taken.
Find
left-handed
be
greater
the
the
probability
than
that
a)
a
group
must
a
of
of
the
that
sample
it
of
contains
10
at
people.
least
mean
0
and
variance
1.
0.85.
scores
are
of
normally
less
than
distributed
80.
scores.
more
student
with
students
have
the
score
=
in
0.95?
P(|Z|
≤
people
probability
variable
students
that
if
random
deviation
probability
left-handed.
left-handed.
to
to
of
student
is
is
sample
that
given
85%
distributions
people
normal
such
test
five
number
person
a
in
left-handed
random
standard
test
10
be
will
likely
must
value
results
Find
of
will
standardized
the
person
half
left-handed
To pass
b
of
sample
exactly
b
a
game’.
Question
i
a
you
dice.
large
with
dice
game.
random
the
The
three
the
games?
find
Find
these
play
in
b
is
on
1
songs
Z
5.
probability
three
A
E(X).
deviation
Y
ou
exactly
a
prizes
$1?
showing
number
pay
like
In
prize
a
somewhere
will
I
the
a
of
a).
that
six
wins
find
standard
probability
Find
he
number
independent,
75
find
0.954,
that
total
the
one
=
are
the
find
a
6
games
with
P(X
probability
denote
Question
songs
Exam-Style
5
the
a)
the
Let X
a
of
4
=
<
game
30%
fair,
x)
=
Exam-Style
I
if
−5
1
i
X
probability
x
How
that
65)
<
complete
distribution
c
<
thrown.
$1,
and
P(X
a
exercise
paid
What
a
at
twice.
distributed
b
Three
game
game
Assuming
a
will
2
a
the
normally
Review
1
plays
plays
than
chosen
at
65.
random
passes
the
test.
Exam-Style
7
The
lifespans
15%
of
hours.
8
The
Questions
Find
time
the
last
less
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SUMMARY
binomial
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variable
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school
probability
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at
journey
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probability
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value
of
be
consecutive
the
r
outcomes
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constant
a
‘success’
from
trial
or
to
a
‘failure’.
trial.
other.
random
variable X
successes
probability
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out
success
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for
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as
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each
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variance
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normal
distribution
with
mean μ
2
and
●
standard
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standard
deviation
normal
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this
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written
distribution
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X
2
●
If
X
∼
N(μ,
σ
)
then
the
transformed
random
variable Z
=
has
a
standard
normal
distribution.
Chapter
Theory
of
knowledge
Statistics
Social
Some
of
scientists
variables
collecting
income,
and
use
human
statistics
can
be
rates
study
measured
analyzing
bir th
to
and
data,
behavior
human
fairly
for
mor tality
behavior.
easily
example
by
population,
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and
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collect
rates.
to
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it
possible
to
reduce
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human
activity
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set
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statistical
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characteristics
of
human
behavior
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Dynamic
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courses
courses
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and
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measured?
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psychologists
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as
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example
for
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set
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experiments
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to
collect
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possible.
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type
of
when
are
being
studied.
data
one.
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the
‘Hawthorne
Effect’.
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who
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experiments
means
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carr y
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out
need
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connection
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knowledge:
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group
with
be
of
the
‘repeatable’.
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researchers
experiment
rst
conrm
experiment
Statistics
of
human
group)
should
behavior
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same
results.
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gives
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results
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repetition
prove
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theor y?
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experiment
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factors
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investigation
violence
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to
on
isolate
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child
all
other
factors
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influence
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controlled
television
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children
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of
their
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researchers
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knew
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students
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interactions
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you
questions
agree
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exercise?
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‘leading’?
children
think
today
get
too
unemployment
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Chapter
Exploration
CHAPTER
As
par t
which
This
and
of
will
a
tips
topic
.
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10
●
be
to
Mathematics
assessed
gives
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to
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get
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on
good
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your
as
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sure
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for
you
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exploration,
grade.
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hints
exploration
on
choosing
show
that
you
can
you.
spend:
time
10
assessment
suitable
your
an
exploration.
●
criteria
hours
●
topics/titles
progress
with
your
teacher
●
of
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your
your
select
and/or
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■
time
exploration,
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appropriate
collecting
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to
topic
and
organising
your
information
mathematical
Ensuring
are
own
an
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data
●
nal
suggestions
help
●
write
exploration,
making
well
to
exploration
oppor tunity
area
as
you
planning
criteria,
the
an
course,
counts
advice
star ted
About
mathematics
hours
you
and
assessment
exploration
apply
Y
ou
your
chapter
satises
OBJECTIVES:
that
derived
all
of
using
processes:
your
results
logical
deductive
reasoning
■
Ensuring
that
necessar y)
●
Demonstrating
Checking
■
Adding
and
that
terminology
proofs
coherent
(when
and
correct
mathematical
communication
■
your
are
presentation:
your
are
notation
and
consistently
diagrams,
graphs
or
correct
char ts
where
necessar y
■
Making
sure
structured
Exploration
your
and
exploration
reads
well
is
clear ly
Y
our
school
will
set
you
deadlines
for
submitting
a
draft
and
the
Ever y
final
piece
of
candidate
work.
taking
If
you
do
not
submit
an
exploration
then
you
receive
a
grade
of
“N”
SL
Mathematics
must
submit
an
exploration.
for
Mathematics
SL,
which
means
you
will
not
receive
your
IB
Ensure
that
you
diploma.
know
your
deadlines
.
Y
our
Internal
exploration
will
be
assessment
assessed
by
your
to
cri teria
teacher,
against
school’
s
and
keep
them.
given
criteria.
It
will
then
be
assessment
The
final
exter nally
moderated
by
the
IB
using
the
same
criteria.
mark
for
each
exploration
is
the
sum
of
the
scores
for
each
criterion.
These
The
maximum
possible
final
mark
is
criteria
explained
This
is
20%
of
your
final
grade
for
Mathematics
good
your
The
exploration
peers,
and
criteria
are
should
be
clear
self-explanator y
split
into
five
all
areas,
and
the
A
easily
way
to
understood
by
one
of
through.
how
A
with
E:
to
sure
Mathematical
Criterion
C
Personal
your
satises
you
understand
criteria
B
ensure
on
them.
Communication
Criterion
more
tips
exploration
Make
Criterion
in
SL.
detail,
A
are
20.
and
these
consult
presentation
them
frequently
engagement
when
Criterion
D
Reection
Criterion
E
Use
writing
your
exploration.
Criterion
A:
This
assesses
criterion
completeness
Achievement
of
of
mathematics
Communication
the
level
the
organization,
coherence,
conciseness
and
exploration.
Descriptor
The
exploration
does
not
reach
the
standard
0
described
1
by
the
descriptors
below.
The
exploration
has
some
coherence.
The
exploration
has
some
coherence
and
2
shows
The
some
organization.
exploration
is
coherent
and
is
coherent,
well
3
organized.
The
exploration
well
organized,
4
concise
and
complete.
Chapter
Y
our
To
get
A
exploration
a
good
well
mark
organized
An
A
rationale
A
description
A
conclusion
A
Y
our
Any
graphs,
A
concise
A
complete
It
is
tables
that
should
of
the
logically
‘read
is
be
one
is
an
discuss
the
explanation
exploration
developed
context
of
which
and
why
to
the
you
should
easy
of
exploration
chose
be
this
clearly
topic
identiable
follow.
well’.
diagrams
not
have
include
that
you
attached
that
one
references
exploration
from
a
in
in
as
focuses
which
all
use
should
accompany
appendices
on
the
steps
aim
are
to
the
and
avoids
clear ly
the
work
in
the
document.
irrelevancies.
explained
without
detracting
the
B:
This
assesses
criterion
are
cited
contain
a
footnotes
formula
should
end.
contain
List
any
a
from
appropriate,
as
a
i.e.,
appropriate.
mathematics
to
what
extent
mathematical
bibliography
books
Mathematical
appropriate
where
For
example,
book,
etc.,
if
put
you
the
are
using
source
of
footnote.
exploration
at
should
publication,
a
Criterion
use
is
exploration
appendix
●
aim
and
and
exploration
quote
Y
our
the
Communication
conciseness.
quote
the
you
should
should
place
essential
Y
our
of
A:
should
which
exploration
appropriate
in
which
exploration
its
Criterion
exploration
introduction
coherent
from
for
you
use,
as
any
appropriate.
websites
This
you
can
be
consult,
in
an
etc.
presentation
you
are
language
able
to:
(notation,
symbols,
terminology)
●
define
●
use
key
terms,
multiple
diagrams,
Achievement
where
forms
tables,
level
of
required
mathematical
char ts,
graphs
and
representation
such
models,
appropriate.
where
as
formulae,
Descriptor
The
exploration
does
not
reach
the
standard
described
0
by
Exploration
the
descriptors
is
some
below.
1
There
appropriate
mathematical
presentation.
2
The
mathematical
presentation
is
mostly
3
The
mathematical
presentation
is
appropriate
appropriate.
throughout.
a
Y
our
To
get
exploration
a
Y
ou
good
are
mark
expected
communicating
Y
ou
are
for
Criterion
to
use
calculators,
software,
appropriate,
Y
ou
should
dene
Y
ou
should
express
Y
ou
should
always
Variables
Do
key
where
to
scales
language
and
when
ndings.
appropriate
spreadsheets,
enhance
results
include
reasoning
use
software,
to
terms,
your
and
communication
mathematical
ideas,
choose
mathematical
as
appropriate
to
Mathematical
appropriate
mathematical
encouraged
B:
ICT
tools
such
databases,
mathematical
as
graphic
drawing
and
display
word-processing
communication.
required.
an
and
appropriate
labels
if
degree
you
use
a
of
accuracy,
graph.
when
T
ables
appropriate.
should
have
headings.
should
be
explicitly
dened.
x
not
use
use
0.028
calculator
and
not
Criterion
C:
This
assesses
criterion
exploration
and
Achievement
or
computer
For
example,
use
2
and
not
2^x;
use
×
not
*;
2.8E-2.
Personal
make
notation.
the
it
engagement
extent
your
to
which
you
engage
with
the
own.
Descriptor
level
The
exploration
does
not
reach
the
standard
0
described
There
is
by
the
descriptors
evidence
of
limited
below.
or
supercial
1
personal
There
is
engagement.
evidence
of
some
personal
of
signicant
2
engagement.
There
is
evidence
personal
3
engagement.
There
is
abundant
evidence
of
outstanding
4
personal
engagement.
Chapter
Y
our
To
get
a
Y
ou
will
exploration
good
mark
should
be
Y
ou
choose
easier
can
for
to
Criterion
a
topic
display
demonstrate
attributes
and
skills.
These
Thinking
and
Thinking
creatively
Addressing
Presenting
Asking
Looking
Considering
Exploring
working
and
D:
This
assesses
criterion
creating
historical
Criterion
that
you
are
interested
in
as
it
engagement
by
using
some
of
the
following
different
interests
ideas
making
unfamiliar
exploration
engagement.
independently
mathematical
for
engagement
include:
personal
questions,
Personal
your
personal
personal
your
for
C:
in
your
conjectures
mathematical
and
global
own
and
way
investigating
models
for
mathematical
real-wor ld
ideas
situations
perspectives
mathematics.
Reflection
how
you
review
,
analyze
and
evaluate
the
exploration.
Achievement
level
Descriptor
The
exploration
does
not
reach
the
standard
described
0
by
Y
our
To
get
a
below.
1
There
is
evidence
of
limited
2
There
is
evidence
of
meaningful
3
There
is
substantial
good
Although
Y
ou
descriptors
or
evidence
supercial
of
reection.
reection.
critical
reection.
exploration
also
the
be
can
mark
reection
found
show
Criterion
may
be
throughout
reection
Discussing
Considering
Stating
Making
Exploration
for
the
to
your
limitations
different
in
the
conclusion
exploration
of
signicance
possible
Reection
to
the
exploration,
exploration.
implications
the
links
seen
the
in
D:
of
your
elds
results
your
and/or
by
ndings
and
extensions
and/or
areas
of
to
results
your
results
mathematics.
it
may
Criterion
E:
This
assesses
criterion
Use
of
to
mathematics
what
extent
you
use
mathematics
in
your
exploration.
Achievement
Descriptor
level
The
exploration
does
not
reach
the
standard
described
0
by
1
the
descriptors
Some
relevant
Some
relevant
below.
mathematics
is
used.
mathematics
is
used.
Limited
2
understanding
Relevant
3
of
the
is
demonstrated.
mathematics
course
is
commensurate
used.
Limited
with
the
understanding
level
is
demonstrated.
Relevant
of
the
mathematics
course
is
commensurate
used.
The
with
mathematics
the
level
explored
is
4
par tially
are
correct.
knowledge
and
understanding
demonstrated.
Relevant
of
Some
the
mathematics
course
is
used.
commensurate
The
with
mathematics
the
level
explored
is
5
mostly
are
correct.
Relevant
of
Good
knowledge
and
understanding
demonstrated.
the
mathematics
course
is
used.
commensurate
The
with
mathematics
the
level
explored
is
6
correct.
Thorough
knowledge
and
understanding
are
demonstrated.
Y
our
To
get
exploration
a
Y
ou
good
are
course
the
Y
ou
If
Y
ou
you
should
the
you
for
expected
are
syllabus,
listed
mark
in
the
need
at
prior
of
only
to
produce
studying.
or
a
ensure
level
can
to
Criterion
that
the
of
that
mathematics
is
commensurate
mathematics
level
(or
you
explore
with
the
should
level
either
of
be
the
par t
of
beyond).
mathematics
involved
is
not
completely
based
on
mathematics
learning.
mathematics
get
a
is
maximum
demonstrate
mathematics
Use
work
The
similar
E:
that
not
of
commensurate
two
within
marks
your
for
wi th
this
exploration
the
level
of
the
course
cri terion.
that
you
fully
understand
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the
of
y
=
graph
2x
of
+
Using
a
y
=
0.
y
1
=
2x
+
1
(see
Example
1).
{
x-intercept,
function.
zero
draw
is
graphic
display
calculator
Continued
on
next
page
6:Analyze
menu
Graph
|
1:Zero
enter
To
find
the
bounds
The
of
zero
a
GDC
you
region
shows
need
that
a
line
to
give
includes
and
asks
the
the
lower
and
upper
zero.
you
to
set
the
lower
bound.
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the
the
left
Click
The
the
Use
the
When
The
.3
the
touchpad
and
choose
a
position
to
touchpad.
shows
another
touchpad
the
the
display
Click
using
zero.
line
and
asks
you
to
set
the
bound.
between
=
the
GDC
upper
y
line
of
the
+
move
and
contains
word
the
lower
‘zero’
in
the
a
line
so
bounds
zero,
that
the
region
contains
the
the
calculator
zero.
will
box.
touchpad.
GDC
2x
region
the
to
upper
1
displays
at
the
Finding
the
point
the
zero
of
(−0.5,
the
function
0).
gradient
(slope)
of
a
line
dy
The
correct
mathematical
notation
for
gradient
(slope)
is
,
dx
and
this
is
how
Example
Find
First
the
the
denotes
gradient.
gradient
draw
GDC
the
of
graph
y
=
of
y
2x
+
=
2x
1
+
1
(see
Example
1).
{
Continued
on
next
Chapter
page
dy
Press
menu
Press
enter
6:Analyze
Graph
|
5:
dx
Use
the
Click
The
touchpad
the
The
Move
This
=
of
the
again.
y
you
the
hand
point
confirms
+
line
at
(slope)
open-hand
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2x
selected
the
gradient
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select
a
point
on
the
line.
touchpad.
point
gradient
to
1
at
is
showing,
grasping
the
line
on
the
using
gradient
point
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together
with
the
point.
symbol
the
displayed
2.
now
ever y
now
that
is
along
that
is
click
the
(slope)
the
line
the
touchpad
point.
is
touchpad.
of
2.
equations
For
.
Solving
simultaneous
equations
graphically
solving
simultaneous
equations
To
solve
simultaneous
equations
graphically
you
draw
the
using
non-graphical
lines
and
then
find
their
point
of
intersection.
The
coordinates
point
of
Example
Use
2x
x
a
y
=
10
–
y
=
2
give
you
the
solutions x
and
rewrite
method
both
to
solve
equations
in
the
the
simultaneous
+
y
=
form
‘y
=’
10
x
–
y
=
=
10
–
2x
GDC
will
only
draw
the
y
=
2
–
x
y
=
x
–
2
are
function
a
expressed
of
different
before
x.
If
the
form,
using
explicitly, ‘y
equations
you
your
need
GDC
to
Using
a
graphic
display
calculator
to
=’
of
are
as
functions
a
written
rearrange
solve
{
graphs
2
that
y
1.5.
equations
The
2x
section
y
graphical
+
First
intersection
method,
of
see
the
a
straight
in
them
them.
Continued
on
next
page
To
draw
Open
a
the
graphs
new
The
entr y
The
default
y
=
document
line
is
10
–
and
displayed
graph
type
is
2x
and
add
at
a
the
y
=
x
–
Graphs
bottom
Function,
so
page.
of
the
2:
the
work
form
area.
‘f1(x)=’
is
displayed.
The
default
Type
The
f1(x)
Use
–
=
10
the
are
−
≤
x
≤
10
and
−6.67
≤
y
≤
6.67.
enter
displays
the
first
straight-line
graph:
2x
touchpad
cor ner
will
−10
2x
calculator
hand
This
10
axes
of
open
to
the
the
click
on
the
arrows
in
the
bottom
left-
screen.
entr y
line
again.
This
time
‘f
2(x)=’
is
displayed.
Type
The
x
–
2
and
GDC
f1(x)
=
10
f2(x)
=
x
now
–
–
press
enter
displays
both
straight-line
graphs:
2x
2
6:Analyze
menu
Graph
|
4:Intersection
Point(s)
enter
To
find
upper
The
the
bounds
GDC
bound.
to
the
of
shows
Move
position
Click
intersection
the
the
a
a
region
line
line
left
you
of
that
and
using
the
need
to
give
includes
asks
the
you
to
the
the
set
touchpad
lower
and
intersection.
the
and
lower
choose
a
intersection.
touchpad.
{
Continued
on
next
Chapter
page
The
GDC
shows
another
line
and
asks
you
to
set
the
upper
bound.
Use
the
touchpad
between
the
to
upper
move
and
the
lower
line
so
bounds
that
the
region
contains
the
intersection.
When
will
Click
The
region
the
the
The
at
contains
word
the
intersection,
‘intersection’
in
a
the
calculator
box.
touchpad.
calculator
lines
.5
the
display
the
displays
point
solution
is
x
Solving
(4,
=
the
intersection
of
the
two
straight
2).
4,
y
=
2.
simultaneous
linear
equations
Y
ou
the
be
When
solving
simultaneous
equations
in
an
examination,
you
do
not
need
do
not
written
any
method
of
solution.
Y
ou
should
simply
write
out
the
equations
correct
form
and
then
give
the
solutions.
The
GDC
will
do
all
the
use
the
linear
wor king
equation
for
any
format
in
to
the
in
to
to
par ticular
show
need
equations
solver
,
as
you.
long
as
linear,
they
that
are
is,
both
neither
2
Example
Solve
2x
x
the
y
=
10
–
y
=
2
a
Press
contains
or
order
higher
x
terms.
equations:
+
Open
equation
new
menu
document
3:Algebra
and
|
add
2:Solve
a
Calculator
Systems
of
page.
Linear
Equations…
enter
Press
Y
ou
two
will
see
this
variables,
This
Note:
solver
want
in
to
is
x
how
your
solve
equations
dialogue
and
box,
you
will
use
examinations.
a
and
more
more
showing
2
equations
and
y
In
the
linear
your
complicated
equation
project,
system
you
with
might
more
variables.
{
Using
a
graphic
display
calculator
Continued
on
next
page
enter
Type
the
arrow
and
two
keys
to
enter
giving
The
the
and
solutions
are
see
the
template
into
the
template,
x
within
GDC
in
=
will
the
4,
y
the
=
{x,
the
using
right.
the
template.
solve
form
on
the
equations,
y}.
2.
functions
Drawing
Example
will
move
the
solutions
Quadratic
.
you
equations
a
quadratic
graph
2
Draw
Open
The
a
new
is
The
form
default
of
y
=
document
line
area.
the
The
graph
entr y
work
so
the
–
and
displayed
default
‘f1(x)
axes
x
=’
are
is
2x
+
add
at
graph
3
a
the
and
display
Graphs
bottom
type
is
using
suitable
axes.
page.
of
the
Function,
displayed.
−10
≤
x
≤
10
and
−6.67
≤
y
≤
6.67.
2
Type
x
–
2x
+
The
calculator
Pan
the
For
see
axes
help
your
with
3
and
press
displays
to
get
a
the
better
enter
cur ve
view
with
of
the
the
default
axes.
cur ve.
panning ,
GDC
manual.
{
Continued
on
next
Chapter
page
Grab
the
cur ve
For
fit
x-axis
the
help
axes,
screen
with
see
and
change
it
to
make
the
quadratic
better.
changing
your
GDC
manual.
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Solving
When
show
the
for
solving
any
quadratic
quadratic
method
correct
form
of
equations
solution.
and
equations
then
in
Y
ou
give
an
examination,
should
the
simply
solutions.
you
write
The
GDC
do
out
the
will
do
you.
Example
2
Solve
3x
Press
of
2
=
0
3:Algebra
a
will
degree
|
3:Polynomial
Tools
|
1:Find
Polynomial…
see
2
(a
need
this
dialogue
quadratic
to
change
box,
showing
equation)
with
a
real
polynomial
roots.
Y
ou
of
do
anything.
enter
Press
Another
The
−
enter
Press
not
4x
menu
Roots
Y
ou
−
dialogue
general
form
box
of
opens
the
for
you
quadratic
to
enter
the
equation
equation.
is
2
a
x
+
a
2
a
x
+
a
1
,
a
2
key
move
=
3,
a
a
−4
the
coefficients
in
and
a
in
the
the
negative
dialogue
the
GDC
form
are
graphic
=
−2.
Be
sure
to
use
the
0
the
and
solutions
Using
=
enter
enter
roots
The
enter
1
to
around
Press
the
so
0
a
2
(–)
0,
a
1
Here,
=
0
and
x
{x,
=
display
values.
Use
the
tab
key
to
box.
will
solve
the
equation,
y}.
−0.387
or
calculator
x
=
1.72
(to
3 sf).
not
giving
need
to
equations
all
the
in
wor king
.
Finding
Example
a
local
minimum
or
maximum
point
2
Find
the
minimum
point
on
the
graph
of
y
=
x
–
2x
+
3
2
First
draw
Method
Y
ou
can
using
a
the
:
graph
Using
look
split
at
a
the
minimum
Look
more
around
x
=
–
2x
+
3
(see
Example
6).
table
|
a
and
9:Show
T
value
closely
the
Choose
any
Set
x
table
of
the
values
by
)
shown
at
Table
the
in
the
values
table
of
the
is
2
when
x
=
1.
function
1.
Change
Table
=
graph
ctrl
The
y
screen.
2:View
menu
of
settings
cell
in
and
the
table.
press
5:Table
menu
|
5:Edit
Settings…
Table
Star t
to
0.98
and
Table
Step
to
0.01.
enter
The
at
table
points
point
(1,
shows
around
92)
is
a
that
(1,
the
2).
local
function
We
can
has
larger
conclude
minimum
on
the
values
that
the
cur ve.
{
Continued
on
next
Chapter
page
Method
:
Press
menu
Press
enter
To
find
The
of
minimum
a
GDC
the
6:Analyze
the
bounds
Using
region
shows
a
minimum
Graph
you
that
line
|
need
2:Minimum
to
includes
and
function
give
the
asks
the
lower
and
upper
minimum.
you
to
set
the
lower
bound.
Move
the
the
left
Click
The
line
of
the
the
using
the
touchpad
and
choose
a
position
to
minimum.
touchpad.
GDC
shows
another
line
and
asks
you
to
set
the
upper
bound.
Use
the
touchpad
between
the
to
upper
move
and
the
lower
line
so
bounds
that
the
region
contains
the
minimum.
Note:
The
defined
is
being
minimum
point
you
When
between
is
Click
The
(1,
word
the
clearly
the
Make
looking
region
the
point
shown.
point.
are
the
display
minimum
In
in
this
sure
and
the
the
the
it
you
is
line
have
not
the
beyond
local
the
upper
upper
a
box
bounds.
and
the
and
The
lower
a
GDC
point
point
will
that
lies
displayed
bounds.
touchpad.
calculator
a
move
minimum,
in
displays
the
minimum
2).
Using
that
for.
contains
lower
region
screenshot
you
‘minimum’
between
the
graphic
display
calculator
point
on
the
cur ve
at
Example
2
Find
the
maximum
point
on
the
graph
of
y
=
–x
+
3x
–
4
2
First
draw
Open
a
the
new
The
entr y
The
default
graph
of
document
line
is
y
=
and
displayed
graph
type
is
−x
+
add
at
3x
a
the
−
4:
Graphs
bottom
Function,
so
page.
of
the
the
form
work
area.
‘f1(x)=’
is
displayed.
The
default
axes
are
−10
≤
x
≤
10
and
−6.67
≤
y
≤
6.67.
2
Type
−x
+
The
GDC
Pan
the
Grab
fit
For
or
values
a
and
better
change
:
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at
using
a
the
a
=
it
to
of
default
the
make
axes.
cur ve.
the
quadratic
cur ve
table
graph
split
2:View
maximum
x
view
the
see
|
1
and
x
=
a
table
of
the
screen.
T
value
and
9:Show
ctrl
when
with
manual.
menu
The
cur ve
panning
axes,
look
by
the
better.
with
GDC
can
get
x-axis
help
Method
Y
ou
to
screen
changing
your
enter
displays
axes
the
the
3x
Table
)
shown
in
the
table
is
−2
2.
{
Continued
on
next
Chapter
page
Look
x
=
1
more
and
Change
closely
x
=
the
at
the
values
of
the
function
between
2.
settings
in
the
table.
5:Table
menu
|
5:Edit
Table
Settings…
Set
Table
Press
Scroll
has
the
Star t
down
its
:
menu
Press
enter
find
The
1.0
and
table
value
Using
Table
Step
the
the
maximum
of
shows
a
a
you
can
to
see
(1.5, −1.75).
is
6:Analyze
bounds
GDC
and
at
(1.5, −1.75)
Press
upper
the
largest
point
Method
To
to
0.1.
enter
a
local
you
region
line
|
need
that
and
We
the
can
maximum
maximum
Graph
that
function
conclude
on
the
that
cur ve.
function
3:Maximum
give
the
lower
includes
to
the
maximum.
asks
you
to
set
the
and
lower
bound.
Move
the
the
left
Click
The
the
the
using
the
touchpad
and
choose
a
position
to
maximum.
touchpad.
GDC
upper
Use
line
of
shows
another
line
and
asks
you
to
set
the
bound.
the
touchpad
between
the
to
upper
move
and
the
lower
line
so
bounds
that
the
region
contains
the
maximum.
Note:
The
defined
is
being
maximum
point
you
maximum
shown.
point.
are
point
Make
looking
In
in
this
sure
the
region
that
screenshot
you
move
the
it
is
you
not
line
have
the
local
beyond
the
for.
{
Using
a
graphic
display
calculator
Continued
on
next
page
When
the
display
the
between
is
The
contains
and
the
upper
upper
maximum,
in
a
the
GDC
box
and
a
bounds.
The
point
and
lower
point
will
that
lies
displayed
bounds.
touchpad.
displays
the
maximum
point
on
the
cur ve
at
−1.75).
Exponential
.
the
‘maximum’
lower
between
the
GDC
(1.5,
word
the
clearly
Click
region
Drawing
Example
functions
an
exponential
graph
x
Draw
Open
the
a
graph
new
The
entr y
The
default
of
y
=
document
line
is
3
+
and
displayed
graph
type
is
2
add
at
a
the
Graphs
bottom
Function,
so
page.
of
the
the
form
work
‘f1(x)
area.
=’
is
displayed.
The
default
axes
are
−10
≤
x
≤
10
and
−6.67
≤
y
≤
6.67.
x
Type
y
the
3
Type
baseline
The
+
GDC
2
and
3
from
>
(Note:
=
press
x
×
the
displays
enter
to
enter
3
.
The
retur ns
you
to
exponent.)
the
cur ve
with
the
default
axes.
{
Continued
on
next
Chapter
page
Pan
the
For
axes
help
see
the
cur ve
For
with
your
Grab
fit
get
and
screen
with
better
view
of
the
cur ve.
manual.
x-axis
see
a
panning ,
GDC
the
help
axes,
to
change
it
to
make
the
exponential
better.
changing
your
GDC
manual.
.
Finding
Example
a
horizontal
asymptote
x
Find
the
horizontal
asymptote
to
the
graph
of
y
=
3
+
2
x
First
Y
ou
draw
can
using
a
Press
(or
the
look
split
the
press
The
values
x
0.
→
at
of
y
=
graph
3
+
and
2
a
(see
table
Example
of
the
10).
values
by
screen.
2:View
menu
simply
graph
of
the
|
ctrl
9:Show
T
Table
)
function
are
clearly
decreasing
as
{
Using
a
graphic
display
calculator
Continued
on
next
page
Press
The
and
hold
table
smaller,
to
shows
f1(x)
the
inspection,
you
We
actual
can
The
x
value
can
value
say
line
of
that
=
scroll
as
2
a
of
see,
table.
of
x
get
2.
at
is
→
the
values
f1(x)
f1(x)
f1(x)
is
up
the
approaches
Eventually
,
the
that
2
reaches
the
2.
bottom
On
of
closer
the
screen,
that
2.000 001 881 6...
as
x
horizontal
→
−∞.
asymptote
to
the
x
cur ve
y
=
3
+
2.
Logarithmic
.
functions
Evaluating
Example
Evaluate
logarithms
log
3.95,
ln 10.2
and
log
10
Open
a
Press
ctrl
Enter
For
new
to
base
natural
method,
document
log
the
and
open
and
the
the
base
the
that
without
.
The
y
to
Finding
inverse
y
GDC
having
values.
line
the
it
=
of
a
will
use
an
log
is
a
Calculator
page.
template.
then
possible
equal
to
e,
to
but
press
use
it
is
del
enter
the
far
same
less
time
ln
ctrl
Note
add
argument
logarithms
with
2.
5
evaluate
the
change
inverse
function
Geometrically
can
this
logarithms
of
base
with
any
base
formula.
function
be
can
found
be
by
done
interchanging
by
reflecting
the x
points
and
in
the
x
Chapter
Example
x
Show
that
the
inverse
of
the
function
y
=
10
x
is
y
=
log
x
by
reflecting
y
=
10
in
the
line
10
y
=
x
Open
First
a
new
we
recognised
plotted
Press
document
will
as
draw
the
a
7:
This
will
the
line
=
Press
menu
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both
through
Press
in
default
The
–10
&
Lines
Graphs
So
it
that
has
|
1:
page.
it
to
can
be
be
drawn
and
not
Point
enter
1
points
Points
the
&
points
the
exit
is
≤
the
entr y
(1, 1)
Lines
you
(4, 4),
|
Line
enter
4
(
and
have
drawing
line
type
which
4
both
lie
on
4:
plotted
and
draw
a
line
is
at
the
function.
bottom
Function,
so
of
the
the
work
area.
The
form
displayed.
and
calculator
x
a
x.
reflection,
enter
the
7:
to
10^x
≤
=
x
graph
“f1(x)=”
Type
add
y
them.
esc
Click
1
plot
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of
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esc
axis
and
line
function.
menu
enter
the
10
press
enter
displays
and
–6.67
the
≤
y
function
≤
with
the
default
axes,
6.67.
{
Using
a
graphic
display
calculator
Continued
on
next
page
Press
Select
the
that
is
Y
ou
it
on
the
Press
place
Use
touch
A:
moves
Click
in
the
2(x)=”
Type
is
log
the
|
to
it
is
you
of
the
have
the
pad
to
along
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the
line
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in
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|
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see
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6:
each
locus
cur ve.
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y
finished.
point
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|
the
line
select
display
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selected).
anywhere
select
then
2:
touchpad
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point
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pad
touch
calculator
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point
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menu
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when
esc
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with
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cur ve
reflected
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cur ve
menu
the
Points
highlighted
can
Press
Use
7:
menu
that
you
just
placed
x
Y
ou
line
should
y
=
see
the
x
Locus
of
of
the
the
points.
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reflection
as
the
cur ve.
at
the
bottom
of
the
work
area.
displayed.
(x)
and
press
enter
10
The
reflected
cur ve
and
the
logarithmic
function
coincide,
x
showing
that
y
=
log
x
is
inverse
of
the
function
y
=
10
10
Chapter
.3
Drawing
Example
Draw
the
a
logarithmic
graph
graph
of
y
=
2log
x
+
3.
10
Open
a
new
The
entr y
The
default
document
line
is
and
displayed
graph
type
is
add
at
a
the
Graphs
bottom
Function,
so
page.
of
the
the
work
form
“f
area.
1(x)=”
is
displayed.
The
default
Type
2log
axes
are
–10
≤
x
≤
(x
10
and
– 6.67
≤
y
≤
6.67.
enter
10
(Note:
2
logarithm.
use
the
move
The
calculator
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the
Grab
cur ve
to
axes
the
fit
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a
in
the
to
get
and
screen
graphic
a
the
the
better
change
cur ve
view
it
10
to
calculator
with
of
as
the
section
brackets
better.
display
enter
argument
beyond
displays
x-axis
the
x
and
log
ctrl
Enter
to
make
base
the
enter
the
the
of
of
+3)
default
axes.
cur ve.
the
the
template,
logarithmic
Trigonometric
.
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It
is
Degrees
in
calculator
is
TI-Nspire,
graphing
used
and
for
in
and
and
to
are
will
be
for
plane
in
are
carried
be
able
able
to
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to
default
geometrical
general
and
and
either
check
back
settings
defaults
the
calculations
out
switch
separate
geometr y
settings
be
to
three
geometr y
.
drawing
used
radians
therefore,
there
impor tant
angle
and
trigonometr y
impor tant,
radians
functions
for
is
in
which
and
to
degrees.
is
for th.
and
radians.
the
On
the
graphing
Geometr y
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for
or
general,
Normally
graphing.
graphing
mode
make:
general
figures.
degrees
the
is
only
two
refers
drawing
are
to
the
trigonometric
graphs.
Example
Change
Open
hand
Click
the
then
To
the
–
cursor
of
the
on
the
box,
add
a
or
will
to
degrees
Calculator
symbol
It
at
the
display
top
the
and
from
degrees
to
radians.
page.
right
general
angle
degrees.
and
select
setting
choose
dialogue
Graphing
and
radians
choose
either
2:Settings
degrees
or
|1:General.
radians
and
OK.
the
and
to
symbol
dialogue
change
from
screen.
radians
the
click
the
settings
document
either
in
symbol
In
new
side
mode
In
angle
a
Move
graphing,
2:Settings
box,
Angle
for
select
and
then
|
in
2:Graphs
either
click
click
degrees
on
the
&
or
Geometr y
.
radians
for
the
OK.
Chapter
.5
Drawing
Example
trigonometric
Draw
the
graph
y
of
2sin
x
a
new
document
4:Window
menu
The
entr y
line
The
default
is
and
/
Zoom
displayed
graph
type
add
is
at
6
Open
graphs
a
|
the
1
Graphs
page.
8:Zoom
bottom
Function,
so
-
Trig
of
the
the
work
form
area.
“f1(x)=”
is
displayed.
The
default
– 4.19
≤
These
with
y
are
x
mode,
axes
≤
the
basic
x-axis
y
≤
2sin
for
and
2π.
will
Type
– 6.28
axes
–2π
between
the
are
x
≤
6.28
and
4.19.
x
be
If
the
between
trigonometric
calculator
–360
and
is
in
graphs
degree
360.
6
graphing
1
and
press
enter
trig
To
enter
sin,
press
and
choose
sin
from
the
dialogue
box.
To
Pan
the
them
It
π,
enter
is
to
press
axes
to
change
also
useful
r
get
the
to
and
a
π
from
of
the
choose
better
view
the
dialogue
cur ve
and
box.
grab
view
.
change
the
x-axis
scale
to
a
multiple
of
π,
such
as
as
this
will
often
show
the
positions
of
intercepts
6
and
tur ning
points
more
clearly
.
menu
1:Window
4:Window
Settings
XScale:
Type
pi/6
Using
a
in
the
graphic
dialogue
display
box
calculator
for
XScale.
/
Zoom
|
More
complicated
functions
Follow
when
.
Solving
a
combined
quadratic
the
same
solving
equation
quadratic
See
Example
the
simultaneous
or
solving
and
a
equations
combined
exponential
Examples
4
and
equation.
17.
2
Solve
procedure
and
graphically
exponential
GDC
equation
x
−x
−
2x
+
3
=
3.2
+
4
T
o
solve
the
equation,
nd
the
point
of
intersection
2
of
the
quadratic
function
f1(x)
=
x
−
2x
+
3
with
the
−x
exponential
function
f2(x)
=
3.2
+
4.
2
To
draw
the
graphs
f1(x)
=
x
−
2x
+
3
and
−x
f 2(x)
=
Open
a
3.2
+
new
The
entr y
The
default
4:
document
line
is
and
displayed
graph
type
is
add
at
a
the
Graphs
bottom
Function,
so
page.
of
the
the
work
form
area.
‘f1(x)=’
is
displayed.
The
default
−6.67
≤
y
axes
≤
are
−10
≤
x
≤
10
and
6.67.
2
Type
The
x
−
2x
GDC
enter
displays
the
first
cur ve:
2
f 1(x)
Use
=
x
the
hand
This
−
2x
+
3
touchpad
cor ner
will
of
open
to
the
the
click
on
the
arrows
in
the
bottom
left-
screen.
entr y
line
again.
This
time
‘f 2(x)=’
is
displayed.
−x
Type
The
3.2
enter
GDC
displays
both
cur ves:
2
f 1(x)
=
x
−
2x
+
3
−x
f 2(x)
=
3.2
+
4
{
Continued
on
next
Chapter
page
Pan
the
For
axes
help
see
with
your
menu
Press
enter
find
upper
The
a
better
the
view
of
the
cur ves.
manual.
6:Analyze
Graph
intersection
bounds
GDC
get
panning ,
GDC
Press
To
to
of
a
shows
a
you
region
line
|
4:Intersection
need
that
and
to
give
includes
asks
you
to
the
the
set
Point(s)
lower
and
intersection.
the
lower
bound.
Move
the
the
left
Click
The
the
the
using
the
touchpad
and
choose
a
position
to
intersection.
touchpad.
GDC
upper
Use
line
of
shows
another
line
and
asks
you
to
set
the
bound.
the
touchpad
between
the
to
upper
move
and
the
lower
line
so
bounds
that
the
region
contains
the
intersection.
When
will
the
the
the
contains
word
the
intersection,
‘intersection’
in
a
the
calculator
box.
touchpad.
GDC
point
The
region
display
Click
The
the
displays
the
intersection
of
the
two
cur ves
at
(2.58, 4.5).
solution
is
x
=
2.58.
Modeling
.7
Using
sinusoidal
2
Note:
has
the
little
notation
algebraic
sin
regression
2
x,
cos
meaning.
2
x,
T
o
tan
enter
x,
…
is
these
a
mathematical
functions
on
the
convention
GDC,
you should
2
enter
(sin (x))
2
,
etc.
However
,
the
2
and
translate
Using
a
that
it
graphic
as
(sin (x))
display
calculator
calculator
will
conveniently
interpret
sin (
x)
Example
It
is
known
that
the
following
data
can
be
modeled
using
a
x
0
1
2
3
4
5
6
7
y
6.9
9.4
7.9
6.7
9.2
8.3
6.5
8.9
Use
sine
Open
a
new
Type
‘x’
Type
the
those
Use
regression
in
the
On
menu
Press
enter
entr y
Scatter
Enter
cell
in
and
a
3:Graph
line
plot
the
add
is
type
is
function
add
and
‘y’
the
the
to
a
in
Lists
the
x-list
to
in
model
&
cell
the
this
data.
Spreadsheet
to
first
its
cur ve.
page.
right.
column
and
second.
navigate
new
Type
around
graphs
|
displayed
names
a
and
from
y-list
keys
Press
The
first
the
Press
find
document
the
numbers
from
to
sine
page
4:Scatter
at
the
the
to
spreadsheet.
your
document.
Plot
bottom
of
the
work
area.
displayed.
of
the
lists,
x
and
y,
into
the
scatter
plot
function
key
tab
Press
and
Y
ou
your
move
from
x
to
y
del
enter
Adjust
to
window
settings
to
show
your
data
and
the
x-
y-axes.
now
Press
Select
have
to
ctrl
an
From
|
scatter
retur n
empty
Calculations
Press
a
cell
plot
to
and
the
of
against
Lists
press
&
menu
y
Spreadsheet
4:Statistics
C:Sinusoidal
Regression...
down
choose
page.
|
Stat
enter
the
drop
menus
tab
Press
x
to
‘x’
move
for
X
List
between
and
the
‘y’
for
fields.
enter
{
Continued
on
next
Chapter
page
On
screen,
regression
The
will
y
=
the
will
lists
equation
see
The
you
in
is
in
values
equation
of
to
the
of
the
1.51sin(2.00x
ctrl
see
next
–
the
to
result
the
form
a,
b,
y
c
=
retur n
to
+
the
for
x
d
sinusoidal
and
asin(bx
and
sinusoidal
0.80)
of
lists
+
y
c)
+
displayed
regression
d
and
you
separately
.
line
is
7.99
the
Graphs
page.
‹‹
Using
the
the
touchpad,
bottom
Y
ou
will
been
of
see
pasted
the
that
work
the
into
click
on
to
open
the
entr y
line
at
area.
equation
of
the
regression
line
has
f1(x).
enter
The
regression
.
line
Using
is
now
shown
on
the
transformations
graph.
to
model
Y
ou
a
quadratic
function
can
function
the
This
data
by
least
(see
Example
also
model
nding
squares
section
a
the
linear
equation
regression
5.15).
is
approximately
connected
by
a
quadratic
function.
T
ransform
x
y
−2
−1
9.1
0
0.2
1
−4.8
2
−5.9
3
−3.1
4
4.0
quadratic
nd
15.0
an
some
Find
Open
Enter
Type
Enter
a
function
a
new
the
‘x’
the
that
fits
document
data
in
the
in
two
first
x-values
and
the
and
the
keys
Using
a
graphic
display
basic
cur ve
to
equation
quadratic
to
t
data.
data.
add
a
Lists
&
Spreadsheet
page.
‘y’
first
to
in
the
column
to
cell
and
enter
navigate
a
to
its
the
right.
y-values
negative
around
the
in
the
number.
spreadsheet.
{
a
lists:
cell
in
the
(–)
Use
of
line
calculator
Continued
on
next
page
Add
a
Graphs
page
to
3:Graph
menu
your
Type
document.
|
4:Scatter
Plot
enter
The
entr y
Scatter
Enter
line
plot
the
is
type
names
displayed
is
at
the
bottom
of
the
work
area.
displayed.
of
the
lists,
x
and
y,
into
the
scatter
plot
function.
Use
the
key
tab
to
move
from
x
to
y.
enter
A:Zoom
menu
This
all
is
a
the
Y
ou
quick
to
Fit
from
choose
the
an
Window/Zoom
appropriate
scale
menu
to
show
points.
should
quadratic
The
way
–
next
recognize
that
the
points
are
in
the
shape
of
a
function.
step
is
to
enter
a
basic
quadratic
function,
2
y
=
x
,
and
manipulate
3:Graph
menu
it
to
Type
fit
|
the
points.
1:Function
enter
This
changes
the
graph
type
from
scatter
plot
to
function.
2
Type
It
is
x
clear
points,
Use
as
it
the
is
Y
ou
will
will
the
f1(x).
cur ve
the
touchpad
first
around
function
that
but
the
cur ve.
The
in
does
right
to
move
see
one
of
allow
you
to
screen
by
not
general
its
fit
any
shape
the
cursor
two
icons.
drag
the
of
to
so
the
do
it
so.
approaches
quadratic
the
function
ver tex.
or
The
second
ver tically
Use
be
or
to
allows
you
to
stretch
the
function
either
horizontally
.
position
according
to
the
the
ver tex
data
where
you
think
it
ought
to
points.
{
Continued
on
next
Chapter
page
Use
to
Make
until
The
adjust
some
you
final
have
equation
data
is:
f1(x)
=
the
a
stretch
fine
the
the
cur ve.
adjustments
good
of
of
fit
to
the
function
using
data
that
both
the
tools
points.
fits
the
2
.
2(x
−
0.75)
Using
Example
−
6.11
sliders
to
model
an
exponential
function
x
In
general,
an
exponential
function
has
the
form
y
=
ka
+
c.
x
For
this
data,
it
is
known
that
the
value
of
a
is
1.5,
so
y
=
k(1.5)
+
c.
x
−3
−2
−1
0
1
2
3
4
5
6
7
8
y
3.1
3.2
3.3
3.5
3.8
4.1
4.7
5.5
6.8
8.7
11.5
15.8
Find
the
Open
a
values
new
of
the
constants
document
and
add
k
a
and
c
Lists
&
Spreadsheet
page.
Enter
Type
the
‘x’
Enter
data
in
the
the
in
two
first
x-values
lists:
cell
in
and
the
‘y’
first
in
the
column
(–)
cell
to
and
to
its
the
enter
right.
y-values
a
negative
number.
Use
the
Add
a
keys
Graphs
Press
menu
Press
enter
The
entr y
Scatter
Enter
to
3:Graph
line
plot
the
page
is
type
names
to
navigate
your
Type
displayed
is
around
the
spreadsheet.
document.
|
4:Scatter
at
the
Plot
bottom
of
the
work
area.
displayed.
of
the
lists,
x
and
y,
into
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enter
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calculus
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Derivatives
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Example
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