STUDY GUIDE

MATH STUDIES SL
www.ib.academy
IB Academy Mathematics Studies Study Guide
Available on learn.ib.academy

Authors: Alex Barancova, Robert van den Heuvel

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INTRODUCTION

Welcome to the IB.Academy Study Guide for IB Mathematics Studies.

We are proud to present our study guides and hope that you will find them helpful. They
are the result of a collaborative undertaking between our tutors, students and teachers
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these study guides that will be continuously reviewed and improved. Should you have
any comments, feel free to contact us.

For this Mathematics Studies guide, we incorporated everything you need to know for
your final exam. The guide is broken down into chapters based on the syllabus topics
and they begin with ‘cheat sheets’ that summarise the content. This will prove especially
useful when you work on the exercises. The guide then looks into the subtopics for each
chapter, followed by our step-by-step approach and a calculator section which explains
how to use the instrument for your exam.

For more information and details on our revision courses, be sure to visit our website at
ib.academy. We hope that you will enjoy our guides and best of luck with your studies.

IB.Academy Team

3
TABLE OF CONTENTS

1. Basics 7

2. Numbers and algebra 9

3. Descriptive statistics 17

4. Logic, sets and probability 25

5. Statistical applications 35

6. Geometry and trigonometry 43

7. Mathematical models 51

8. Differentiation 61

5
 TABLE OF CONTENTS

6
BASICS 1

1.1 Notation

To begin with, it is crucial to understand some mathematical terminology that you will
hear over and over again as you work through your IB math exam. Questions might ask
you to ‘set up an equation’ or an ‘inequality’, so it is important that you know what this
means.

Equation contains an “=” sign.

e.g. − 2x − 3 = 5
−2x = 8
x = −4

Inequality contains a >, <, ≥ or ≤ sign.

e.g. −2x − 3 ≥ 5 (−2x − 3 is greater than or equal to 5).

Solve like an equation, except if you × or ÷ by a negative number,
then reverse the inequality!

−2x − 3 ≥ 5
−2x ≥ 8
x ≤ −4

0 < a < 1 means: a is between 0 and 1 (not including 0 and 1)

Absolute value |x| is the positive version of x (distance from 0).

e.g. |3| = 3
|−3| = 3
1 ≤ |x| ≤ 2 means: x is between 1 and 2 or between −2 and −1.

7
 BASICS Laws of exponents

1.2 Laws of exponents

Exponents always follow certain rules. If you are multiplying or dividing, use the
following rules to determine what happens with the powers.
.
Example

x1 = x 61 = 6
x0 = 1 70 = 1
x m · x n = x m+n 45 · 46 = 411
xm 35
= x m−n = 35−4 = 31 = 3
x n 34
€ Š2
(x m )n = x m·n 10 = 1010
5

(x · y)n = x n · y n (2 · 4)3 = 23 · 43 and (3x)4 = 34 x 4

 ‹−1
1 1 3 4
x −1 = 5−1 = and =
x 5 4 3
1 1 1
x −n = 3−5 = =
xn 35 243

1.3 Unit conversion

Units are used to measure different kinds of factors in the world; for example
temperature, weight or price are all things that can be measured in different units.
Measured values can however only be compared if they are in the same unit; so while you
may know the price of one object in EUR and of another in USD, in order to determine
which one is more expensive, you will need to convert the price of both objects into one
currency. Therefore particularly when applying mathematics to real world problems,
you will often need to convert between units.

SI units are the base units from which other units are derived. The 7 base
units are: meter, kilogram, second, ampere, kelvin, mole, candela.

e.g. the ‘meter’ is the SI unit used to measure distance; other units used
to measure distance like the centimeter (0.01 meters) or the kilometer
(100 meters) are based on the meter.

8
NUMBERS AND ALGEBRA 2
Table of contents & cheatsheet

2.1. Estimation 10 Compound interest

‹k n
r

Rounding to a FV = PV × 1 +
• decimal place 100k
• number of significant figures Where
Error approximate value − exact value = VA − VE FV Future Value
 
V −V  PV Present Value
Percentage error  A E
 × 100

 VE  r rate (%)
k compounding frequency
Standand form a × 10k , where 1 ≤ |a| < 10 and k ∈ Z n overall length of time

2.2. Sequences and series 12 Exponents

Arithmetic: +/− common difference x1 = x
x0 = 1
un = n th term = u1 + (n − 1)d
n x m · x n = x m+n
Sn = sum of n terms = 2u1 + (n − 1)d


2 xm
= x m−n
xn
with u1 = a = 1st term, d = common difference.
(x m )n = x m·n
(x · y)n = x n · y n
Geometric: ×/÷ common ratio 1
x −1 =
x
un = n th term = u1 · r n−1 1
u1 (1 − r n ) x −n =
Sn = sum of n terms = xn
(1 − r )
u1
S∞ = sum to infinity = , when −1 < r < 1
1− r
with u1 = a = 1st term, r = common ratio.

9
 NUMBERS AND ALGEBRA Estimation

2.1 Estimation

2.1.1 Rounding

In math you come across rounding almost all the time, so its important to know how to
do it accurately. The key things you need to know are:

1. which number you should be rounding;

2. whether you should round up or down.

You can round any number using these two questions:

What does the rounded digit Which digit is being rounded?

become? (2 possibillities)
The word “estimate” • If the digit is < 5, it stays the same. • A certain decimal place
without further detail
e.g. 201.78095 rounded to:
means “estimate to 1 • If the digit is ≥ 5, add +1 to the digit
significant figure”. → 2 decimal places ⇒ 201.78
When nothing is → 1 decimal place ⇒ 201.8
specified, always round
e.g. 201.78095
to 3 significant figures. Round to the nearest 10 and 10,000th
• A certain number of significant figures
Nearest 10 Rule: zeros to the left of the first
Look at the next digit → 1 non-zero digit are not significant
1 < 5 ⇒ 200
All other: numbers are significant
Nearest 10,000th (= 0.0001)
e.g. 0.0023045 rounded to:
Look at the next digit → 5
5 ≥ 5 ⇒ add +1 to 9 which carries over, → 2 significant figures ⇒ 0.0023
⇒ 201.7810. → 3 significant figures ⇒ 0.00230
→ 4 significant figures ⇒ 0.002305

10
 NUMBERS AND ALGEBRA Estimation 2

2.1.2 Errors

The error tells you by how much an estimate differed from the actual value.

This can be done by calculating the approximate value − exact value

VA − V E
 
 approximate value − exact value 
Percentage error   × 100
 
 exact value 

 
V − V 
 A E
  × 100
 VE 
.

John estimates a 119.423 cm piece of plywood to be 100 cm. What is the error?
Example

Error = VA − VE
= 100 − 119.423
= −19.423 ≈ −19.4

What is the percentage error?

 100 − 119.423 
 
Percentage error = 
  × 100
119.423 
= |−0.1626| × 100
= 0.1626 × 100 ≈ 16.3%

2.1.3 Standard form

Standard form is just a way of rewriting any number, sometimes also

referred to as ‘scientific notation’. This should be in the form a × 10k ,
where a is between 1 and 10, and k is an integer.

10 1 × 101
1000 1 × 103
3280 3.28 × 103
4582000 4.582 × 106

11
 NUMBERS AND ALGEBRA Sequences and series

2.2 Sequences and series

2.2.1 Arithmetic sequence

Arithmetic sequence the next term is the previous number + the common
difference (d ).

e.g. 2, 4, 6, 8, 10, . . . d = +2 and 2, −3, −8, −13, . . . d = −5

To find the common difference d , subtract two consecutive terms of an

arithmetic sequence from the term that follows it, i.e. u(n+1) − un .

DB 1.1 Use the following equations to calculate the n th term or the sum of n terms.
n
un = u1 + (n − 1)d Sn = 2u1 + (n − 1)d


2
with

u1 = a = 1st term d = common difference

Often the IB requires you to first find the 1st term and/or common difference.

Finding the first term u1 and the common difference d from other
terms.

In an arithmetic sequence u10 = 37 and u22 = 1. Find the common difference and the
first term.

1. Put numbers in to n th term formula 37 = u1 + 9d

1 = u1 + 21d

2. Equate formulas to find d 21d − 1 = 9d − 37

12d = −36
d = −3

3. Use d to find u1 1 − 21 · (−3) = u1

u1 = 64

12
NUMBERS AND ALGEBRA Sequences and series 2

2.2.2 Geometric sequence

Geometric sequence the next term is the previous number multiplied by the
common ratio (r ).

To find the common ratio, divide any term of an arithmetic sequence by the
second term (u2 )
term that precedes it, i.e. e.g. 2, 4, 8, 16, 32, . . . r = 2
first term (u1 )
1
and 25, 5, 1, 0.2, . . . r =
5

Use the following equations to calculate the n th term, the sum of n terms or the sum to
infinity when −1 < r < 1. DB 1.1

un = n th term Sn = sum of n terms S∞ = sum to infinity

u (1 − r n ) u
= u1 · r n−1 = 1 = 1
(1 − r ) 1− r

again with

u1 = a = 1st term r = common ratio

Similar to questions on Arithmetic sequences, you are often required to find the 1st term
and/or common ratio first.

13
 NUMBERS AND ALGEBRA GDC solvers (TI-Nspire)

2.3 GDC solvers (TI-Nspire)

There are several handy tools on your GDC which will help you answer most of the
more complicated algebra questions. You can use these in cases where you are looking to
find the roots of a quadratic equation or solve a pair of simultaneous equations.

APPS → PlYSMLT2

Solving quadratic equations

Solve 3x 2 − 4x − 2 = 0

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Degree = 2, Roots = Real, Enter values a2, a1 and a0.

Press menu , choose
3: Algebra press OK Press OK

3: Polynomial Tools
1: Find Roots of Polynomial

IB ACADEMY

so x = 1.72 or x = −0.387

14
NUMBERS AND ALGEBRA GDC solvers (TI-Nspire) 2

Solving Simultaneous equations

2x + y = 10 and x − y = 2; find the values of x and y

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Number of equations = 2. Enter the two equations

Press menu , choose ≈

3: Algebra Press OK
press enter
2: Solve System of
Linear Equations

So x = 4 and y = 2

APPS → FINANCE → TVM SOLVER

You can also use your GDC for questions dealing with money and interest rates. The
TVM Solver (“Time Value of Money”) allows you to fill in all the variables you know
and solve for the missing one.
For some questions you
might wind it simpler to
‹kn use the formula for
r

FV = PV × 1 + compound interest in
100k your data booklet!

Table 2.1: Abbreviations

Abbreviation Stands for

TVM Time Value of Money
N Number of years
I% percentage Interest rate
PV Present Value - should be negative
PMT PayMenT
FV Future Value
P/Y Payments per Year
C/Y Compounding periods per Year

15
 NUMBERS AND ALGEBRA GDC solvers (TI-Nspire)

Solving questions about compound interest

$1500 is invested at 5.25% per annum. The interest is compounded twice per year. How
much will it be worth after 6 years?

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Enter all known values Highlight cell of asked

Press menu , choose
For this example: value, in this case FV,
8: Finance ≈
N=6 (years)
1: Finance Solver press enter
I=5.25 (interest rate)
PV=-1500 (present value)
negative because
investment represents cash
outflow;
PMT=0
FV=0 (future value)
P/Y=1 (payment/yr)
C/Y=2 (compound/yr)

So FV = $2047.05

16
DESCRIPTIVE STATISTICS 3
Table of contents & cheatsheet

Definitions

Population the entire group from which statistical data is drawn (and which the statistics obtained represent).
Sample the observations actually selected from the population for a statistical test.
Random Sample a sample that is selected from the population with no bias or criteria; the observations are made at random.
Discrete finite or countable number of possible values. (e.g. money, number of people)
Continuous infinite amount of increments. (e.g. time, weight)
Note: continuous data can be presented as discrete data, e.g. if you round time to the nearest minute or weight to the nearest
kilogram.

3.1. Descriptive statistics 18

Mean the average value,

the sum of the data
x̄ =
no. of data points
Mode the value that occurs most often
Median when the data set is ordered low to high and the number of data points is:
• odd: the median is the middle value;
• even: the median is the average of the two middle values.
Range largest x-value − smallest x-value
f (x − x̄ 2 )
P
Variance σ 2 = calculator only
n
p
Standard deviation σ = variance calculator only
Grouped data: data presented as an interval. Use the midpoint as the x-value in all calculations.
Q1 first quartile = 25th percentile.
Q2 median = 50th percentile
Q3 third quartile = 75th percentile
Q3 − Q1 interquartile range (IQR) = middle 50 percent

3.2. Statistical graphs 20

Frequency the number of times an event occurs in an experiment

Cumulative frequency the sum of the frequency for a particular class and the frequencies for all the the classes below it
Histogram Cumulative frequency Box and whisker plot

lowest value highest value

Q1 Q2 Q3

Q1 Q2 Q3 Q4

17
 DESCRIPTIVE STATISTICS Descriptive statistics

3.1 Descriptive statistics

The mean, mode and median, are all ways of measuring “averages”. Depending on the
distribution of the data, the values for the mean, mode and median can differ slightly or a
lot. Therefore, the mean, mode and median are all useful for understanding your data set.

x 3 6 7 13
Example data set: 6, 3, 6, 13, 7, 7 in a table:
frequency 1 2 2 1
P
fx
P
the sum of the data x
Mean the average value, x̄ = = = P
no. of data points n f
3 + 6 + 6 + 7 + 7 + 13 1 · 3 + 2 · 6 + 2 · 7 + 1 · 13
e.g. x̄ = = =7
6 1+2+2+1

Mode the value that occurs most often (highest frequency)

e.g. The example data set has 2 modes: 6 and 7

Median the middle value when the data set is ordered low to high. Even
number of values: the median is the average of the two middle values.
1
Find for larger values as n + .
2
e.g. data set from low to high: 3, 6, 6, 7, 7, 13
6+7
median = = 6.5
2

Range largest x-value − smallest x-value

e.g. range = 13 − 3 = 10

f (x − x̄ 2 )
P
2
Variance σ = calculator only
n
p
Standard deviation σ= variance calculator only

Note on grouped data: data presented as an interval; e.g. 10–20 cm.

• Use the midpoint as the x-value in all calculations. So for 10–20 cm use
15 cm.

• For 10–20 cm, 10 is the lower boundary, 20 is the upper boundary and
the width is 20 − 10 = 10.

18
 DESCRIPTIVE STATISTICS Descriptive statistics 3

Adding a constant to all the values in a data set or multiplying the entire data set by a
constant influences the mean and standard deviation values in the following way:

Table 3.1: Adding or multiplying by a constant

adding constant k multiplying by k

mean x̄ + k k × x̄
standard deviation σ k ×σ

Q1 first quartile = 25th percentile.

The value for x so that 25% of all the data values are ≤ to it.

Q2 median = 50th percentile

Q3 third quartile = 75th percentile

Q3 − Q1 interquartile range (IQR) = middle 50 percent

.

Snow depth is measured in centimeters:

Example

30, 75, 125, 55, 60, 75, 65, 65, 45, 120, 70, 110.
Find the range, the median, the lower quartile, the upper quartile and the
interquartile range.

First always rearrange data into ascending order: 30, 45, 55, 60, 65, 65, 70, 75, 75, 110, 120, 125

1. The range:
125 − 30 = 95 cm
2. The median: there are 12 values so the median is between the 6th and 7th value.

65 + 70
= 67.5 cm
2
3. The lower quartile: there are 12 values so the lower quartile is between the 3rd
and 4th value.
55 + 60
= 57.5 cm
2
4. The upper quartile: there are 12 values so the lower quartile is between the 9th
and 10th value.
75 + 110
= 92.5 cm
2
5. The IQR
92.5 − 57.5 = 35 cm

19
 DESCRIPTIVE STATISTICS Statistical graphs

3.2 Statistical graphs

Frequency the number of times an event occurs in an experiment

Cumulative frequency the sum of the frequency for a particular class and
the frequencies for all the classes below it

Age 17 18 19 20 21
No. of students 21 45 93 61 20
Cumulative freq. 21 66 159 220 240

f
100
90 A histogram is used to display the frequency for a specific
80
70 condition. The frequencies (here: # of students) are
60 displayed on the y-axis, and the different classes of the
50
40 sample (here: age) are displayed on the x-axis. As such,
30 the differences in frequency between the different classes
20 assumed in the sample can easily be compared.
10
17 18 19 20 21 Age

cf
250 The cumulative frequency graph is used to display the
development of the frequencies as the classes of the event
200 increase. The graph is plotted by using the sum of all
frequencies for a particular class, added to the frequencies
150
for all the classes below it. The classes of the event (age)
100 are displayed on the x-axis, and the frequency is
displayed on the y-axis. The cumulative frequency graph
50 always goes upwards, because the cumulative frequency
Q1 Q2 Q3 Q4
increases as you include more classes.

17 18 19 20 21 Age

Box and whisker plots neatly summarize

the distribution of the data. It gives
information about the range, the median
and the quartiles of the data. The first
and third quartiles are at the ends of the
lowest value highest value box, the median is indicated with a
Q1 Q2 Q3 vertical line in the interior of the box,
and the maximum and minimum points
are at the ends of the whiskers.

20
 DESCRIPTIVE STATISTICS Statistical graphs 3

Outliers will be any points lower than Q1 − 1.5 × IQR and larger than
Q3 + 1.5 × IQR (IQR =interquartile range)

To identify the value of Q1 , Q2 and Q3 , it is easiest to use the cumulative frequency

graph. First, determine the percentage of the quartile in question. Second, divide the
total cumulative frequency of the graph (i.e. the total sample size) by 100 and multiply by
the corresponding percentage. Then, you will have found the frequency (y-value) at
which 25% for Q1 / 50% for Q2 / 75% for Q3 of the sample is represented. To find the
x-value, find the corresponding x-value for the previously identified y-value.
.

Using the histogram, create a cumulative frequency graph and use it to

Example

construct a box and whisker diagram.

12

10
Length (cm)

20 40 60 80 100 120
Number of fish

Write out the table for frequency and cumulative frequency.

Frequency of fish 20–30 30–40 40–50 50–60 60–70 70–80 80–90 90–100 100–110 110–120
Length of fish 2 3 5 7 11 5 6 9 1 1
Cumulative f. 2 5 10 17 28 33 39 48 49 50

21
 DESCRIPTIVE STATISTICS Statistical graphs

Plot on cumulative frequency chart. Remember to use the midpoint of the date, e.g.,

.
Example
25 for 20–30.

55
Cumulative frequency 50
45
40
35
30
25
20
15
10
5

0 25 35 45 55 65 75 85 95 105 115
Frequency of fish

Use graph to find Q1 , Q2 and Q3 .

55
50
Cumulative frequency

45
40
35
30 Q1 25% of 50 = 12.5 → 48
25 Q2 50% of 50 = 25 → 62
20
15 Q3 75% of 50 = 37.5 → 83
10
5
Q1 Q2 Q3

0 25 35 45 55 65 75 85 95 105 115
Frequency of fish

Plot box and whiskers.

20 48 62 83 120

22
DESCRIPTIVE STATISTICS GDC (TI-Nspire) 3

3.3 GDC (TI-Nspire)

To find mean, standard deviation and quartiles etc.

For the data used in the previous example showing the ages of students

IB ACADEMY IB ACADEMY IB ACADEMY

off

1: One-Variable Statistics
Press on , go to Press menu , choose
Lists and Spreadsheets. 4: Statistics
Enter x-values in L1 and, 1: Stat Calculations
if applicable, frequencies
in L2
IB ACADEMY IB ACADEMY IB ACADEMY

Enter Num of lists: 1. Enter names of columns mean = 19.06;

Press OK you used to enter your standard deviation = 1.06
x-list and frequency list etc.
and column where you
would like the solutions to
appear: a[], b[] and c[].
Press OK

23
 DESCRIPTIVE STATISTICS GDC (TI-Nspire)

24
LOGIC, SETS AND 4

PROBABILITY
Table of contents & cheatsheet

4.1. Logic 26

A proposition is any statement that can be either true or false, mathematical or not.

Propositions Negation Conjunction Disjunction Exclusive disjunction Implication

p q ¬p ¬q p∧q q∨p pÙq q⇒p
T T F F T T F T
T F F T F T T F
F T T F F T T T
F F T T F F F T
Implication ( p ⇒ q) if p, then q.
Other types of implication: Converse (q ⇒ p), Inverse (¬ p ⇒ ¬q), Contrapositive (¬q ⇒6= p).
Equivalence ( p ⇔ q) p if and only if q.
Tautology a statement that is always true.
Contradiction a statement that is always false.

4.2. Sets 27 4.3. Probability 29

Set any collection of things with a common property Sample space the list of all possible outcomes.
(capital letter, curly brackets) Event the outcomes that meet the requirement.
e.g. A = {2, 4, 6, 8} Probability for event A,
Number of ways A can happen
Number of elements in a set n(A) = 4 P (A) = .
all outcomes in the sample space
A member of a set 6 ∈ A
Conditional probability used for successive events that
An empty set ∅ come after one another. The probability of A, given
Subset a set contained in another set. P (A ∩ B)
that B has happened: P (A|B) = .
e.g. B = {4, 8} ⇒ B ⊂ A P (B)

Probability distributions
Sets can be shown in Venn diagrams.
A fair coin is tossed twice.
H T
H HH HT
N Z Q R
T TH TT
Table of probability distribuition
(x is the number of heads obtained)
Natural numbers (N)
x 0 1 2
Integers (Z) 1 1 1
Rational numbers (Q) P (X = x)
4 2 4
Real numbers (R) The sum of P(X = x) = 1.
Expected value of X E(X ) = xP(X = x) =
P
1 1 1
=0· +1· +2· =1
4 2 4

25
 LOGIC, SETS AND PROBABILITY Logic

4.1 Logic

4.1.1 Propositions

A proposition is any statement that can be either true or false, mathematical

or not.

Exclusive
2 propositions Negation Conjunction Disjunction disjunction
Bob studies Bob studies
Bob Bob Bob does Bob studies
or drinks or drinks
studies drinks not study and drinks
or both not both
p q ¬p p∧q p∨q pÙq
T T F T T F
T F F F T T
F T T F T T
F F T F F F

4.1.2 Implications

p ⇒ q is only false p q p⇒q

when p = T and
If p, then q ( p ⇒ q)
q = F. e.g. p: “you steal”, q: “you go to prison”. T T T
T F F
Implication: F T T
If you steal, then you go to prison. F F T

Contrapositive and Propositions Negation Implications Converse Inverse Contrapositive

implication are
equivalent and If x = 16 If x is If x 6= 16 If x is not
x is x is not
then x is a square then x is not a square
converse and inverse x = 16 a square x 6= 16 a square
are equivalent. a square number a square number
number number
number then x = 16 number then x 6= 16
p q ¬p ¬q p⇒q q⇒p ¬ p ⇒ ¬q ¬q ⇒ ¬ p
T T F F T T T T
T F F T F T T F
F T T F T F F T
F F T T T T T T

26
LOGIC, SETS AND PROBABILITY Sets 4

4.1.3 Equivalence (p ⇔ q)

An equivalence has an identical implication ( p ⇒ q) and converse (q ⇒ p), i.e. p if and

only if q.
e.g. p: x is even, q: x is a multiple of 2.

Tautology a statement that is always true

( p ⇒ q) ⇔ (¬q ⇒ ¬ p)

Contradiction a statement that is always false

¬( p ⇒ q) ⇔ (¬q ⇒ ¬ p)

4.2 Sets

Set any collection of things with a common property (capital letter, curly
brackets)
e.g. A = {2, 4, 6, 8}
= even numbers between 1 and 9


= x | x is even, 1 < x < 9



Number of elements in a set n(A) = 4

A member of a set 6 ∈ A

An empty set ∅

Subset a set contained in another set.

e.g. B = multiplies of 4 between 0 and 8



B = {4, 8} ⇒ B ⊂ A

27
 LOGIC, SETS AND PROBABILITY Sets

4.2.1 Venn diagrams

Sets can be shown using.

.

A room of 20 people, 11 have black hair, 6 have glasses, 2 have both.

Example

Room
Black hair Glasses

9 2 4

5
∗ When drawing venn diagrams, start from the middle.

4.2.2 Number sets

Natural numbers N = 0, 1, 2, 3 . . .

Integers Z = . . . − 3, −2, −1, 0, 1, 2, 3, . . .

Rational numbers Q; all integers and fractions

p
Real numbers R; all rational and irrational numbers (π, 2, etc.)
.

Place the following numbers on the Venn diagram:

Example

1
, −3, π, cos 120°, 2.7 × 103 , 3.4 × 10−2
4

Q R

π
Z
cos 120°
N 1
-3
2 · 7 × 103 4
3.4 × 10−2

28
 LOGIC, SETS AND PROBABILITY Probability 4

4.3 Probability

4.3.1 Single events (Venn diagrams)

Probability for single events can be expressed through venn diagrams.

Sample space the list of all possible outcomes.

Event the outcomes that meet the require-
A B ment.
Probability for event A,

Number of ways A can happen

0 P (A) =
all outcomes in the sample space

Here the shaded circle.

Imagine I have a fruit bowl containing 6 apples and 4 bananas.

.

I pick a piece of fruit.

Example

What is the probability of picking each fruit?

Apple = 0.6 Banana = 0.4

In independent events

As apples cannot be bananas this is mutually exclusive, therefore P (A∪ B) = P (A) + P (B) P (A ∩ B) =
P (A) × P (B). It will
and P (A ∩ B) = 0. It is also an exhaustive event as there is no other options apart from often be stated in
apples and bananas. If I bought some oranges the same diagram would then be not questions if events are
exhaustive (oranges will lie in the sample space). independent.

29
 LOGIC, SETS AND PROBABILITY Probability

.
Of the apples 2 are red, 2 are green and 2 are yellow.
Example
What is the probability of picking a yellow apple?

Yellow apples
A B
A: apples

B: yellow fruit

This is not mutually exclusive as both apples and bananas are yellow fruits. Here we are
interested in the intersect P (A ∩ B) of apples and yellow fruit, as a yellow apple is in both
sets P (A ∩ B) = P (A) + P (B) − P (A ∪ B).
.

What is the probability of picking an apple or a yellow fruit?

Example

A B
A: apples

B: yellow fruit

When an event is This is a union of two sets: apple and yellow fruit.
exhaustive the
probability of the union
is 1. The union of events A and B is:

• when A happens;
• when B happens;
• when both A and B happen P (A ∪ B) = P (A) + P (B) − P (A ∩ B).
.

What is the probability of not picking a yellow fruit?

Example

A B
A: apples

B: yellow fruit

This is known as the compliment of B or B 0 . B 0 = 1 − B.

Here we are interested in everything but the yellow fruit.

30
 LOGIC, SETS AND PROBABILITY Probability 4
.

What is the probability of picking an apple given I pick a yellow fruit?

Example

Yellow apples
A B
A: apples

B: yellow fruit

This is “conditional” probability in a single event. Do not use the formula in the
0.2 1
formula booklet. Here we are effectively narrowing the sample space = = .
(0.2 + 0.4) 3

You can think of it like removing the non yellow apples from the fruit bowl before
choosing.

P (A ∩ B)
Conditional probability P (A|B) = .
P (B)

4.3.2 Multiple events (tree Diagrams)

Dependent events two events are dependent if the outcome of event A

affects the outcome of event B so that the probability is changed.
Questions involving
dependent events will
often involve elements
Independent events two events are independent if the fact that A occurs that are drawn “without
does not affect the probability of B occurring. replacement”.
Remember that the
probabilities will be
Conditional probability used for successive events that come one after changing with each
another (as in tree diagrams). The probability of A, given that B has new set of branches.
P (A ∩ B)
happened: P (A|B) = .
P (B)

Probabilities for successive events can be expressed through tree diagrams. In general, if
you are dealing with a question that asks for the probability of:

• one event and another, you multiply

• one event or another, you add

31
 LOGIC, SETS AND PROBABILITY Probability

.
Two disks are randomly drawn without replacement from a stack of 4 red and 5
Example
blue disks. Draw a tree diagram for all outcomes.

The probability of drawingtwo red‹ disks can be found by multiplying both

4 3
probabilities of getting red × .
9 8
The probabilities for 1st draw 2nd draw
each event should
always add up to 1. The 3 4 3 12
probabilities describing 8 R R and R: × =
all the possible
9 8 72
outcomes should also 4 R
equal 1 (that is, the
9 5 4 5 20
probabilities that we
B R and B: × =
found by multiplying
8 9 8 72
along the individual
branches).
4 5 4 20
8 R B and R: × =
5 9 8 72
9 B
4 5 4 20
B B and B: × =
8 9 8 72

What is the probability to draw one red and one blue disk?
P (one red and one blue)
P (R) and P (B) or P (B) and P (R)

P (R) × P (B) P (B) × P (R)

It is common for
conditional probability
questions to relate to 20 20 40 5
+ = =
previous answers. 72 72 72 9

What is the probability to draw at least one red disk?

P (at least one red)
P (R and R) + P (B and R) + P (R and B) = 1 − P (B and B)
12 20 20 20 52 13
+ + = 1− = =
72 72 72 72 72 18

What is the probability of picking a blue disc given that at least one red disk is
picked?
5
P (a blue disk) 10
P (blue disk | at least one red disk) = = 9 =
P (at least one red disk) 13 13
18

32
LOGIC, SETS AND PROBABILITY Probability 4

Another way of dealing with multiple events is with a sample space diagram or a
probability distribution.

Probability distributions.

A fair coin is tossed twice, X is the number of heads obtained.

1. Draw a sample space diagram H T

H H, H H, T
T T, H T, T

2. Tabulate the probability distribution

x 0 1 2
1 1 1
P (X = x)
4 2 4

(The sum of P (X = x) always equals 1)

X : E(X )
X
3. Find the expected value of E(X ) = xP (X = x)
1 1 1
=0· +1· +2· =1
4 2 4
So if you toss a coin twice, you expect to
get heads once.

33
 LOGIC, SETS AND PROBABILITY Probability

34
STATISTICAL APPLICATIONS 5
Table of contents & cheatsheet

5.1. Normal distribution 36

Use normalcdf: for the probability that x is
Total area = 1 between any 2 values.

Use invnorm: to get an x-value for a given

probability.

Mean Standard x Expected value the value of x multiplied by

(µ) deviation probability E(x) = x · p.
(σ)

5.2. Bivariate statistics 38

Using data where two variables (x, y) are measured.

Scatter diagrams
Perfect positive No correlation Weak negative Correlation does not mean
y y y causation.

x x x

Pearsons’s correlation −1 ≤ r ≤ 1

Interpretation of r -values
r -value very weak weak moderate strong
correlation 0.00 ≤ |r | ≤ 0.25 0.25 ≤ |r | ≤ 0.50 0.50 ≤ |r | ≤ 0.75 0.75 ≤ |r | ≤ 1.00

Regression equation a mathematical model world best describe the relationship between the two measured variables; when
drawn manually, always passes through the mean point (x̄, ȳ).

5.3. Chi-squared test 41

Chi-square test Used to test independence of two variables. Using χ 2 value and/or p-value.
H0 the variables are independent (null hypothesis)
H1 the variables are not independent (alternative hypothesis)
If critical value < χ 2 or p-value < significant level (for 10% test, significant level = 0.1) reject null hypothesis.

35
 STATISTICAL APPLICATIONS Normal distribution

5.1 Normal distribution

A normal distribution is one type of probability distribution which gives a bell-shape

curve if all the values and their corresponding probabilities are plotted.

We can use normal distributions to find the probability of obtaining a certain value or a
range of values. This can be found using the area under the curve; the area under the
bell-curve between two x-values always corresponds to the probability for getting an
x-value in this range. The total area under the normal distribution is always 1; this is
because the total probability of getting any x-value adds up to 1 (or, in other words, you
are 100% certain that your x-value will lie somewhere on the x-axis below the bell-curve).

total area under the curve = 1

mean St. dv. x 0 1 z

5.1.1 Using GDC

You can use your GDC to work through questions dealing with normal distributions. In
these questions you will either need to find probabilities for given x-values or x-values
for given probabilities. In both cases, you will need to know the mean (µ) and standard
deviation (σ) for the given example.

Note: even though you will be using your GDC to find probabilities for normal
distributions, it’s always very useful to draw a diagram to indicate for yourself (and the
examiner) what area or x-value you are looking for.

Use normal cdf (lowerbound, upperbound, µ, σ): for the probability that x is
between any 2 values.
• For lower bound = −∞, use −1E99
• For upper bound = ∞, use 1E99
Use invnorm (ρ, µ, σ): to get an x-value for a given probability.

∗ Note:the calculator assumes ρ is to the left of x. When ρ is to the right of x, subtract

the output from 1 to get the final answer.

Expected value the value of x multiplied by probability.

36
STATISTICAL APPLICATIONS Normal distribution 5

To find a probability or percentage of a whole (the area under a

normal distribution curve)

The weights of pears are normally distributed with mean = 110 g and
standard deviation = 8 g.
Find the percentage of pears that weigh between 100 g and 130 g

Sketch!
Indicate:
• The mean = 110 g
• Lower bound = 100 g
• Upper bound = 130 g
• And shade the area you are looking to
find.

100 110 130 weight (g)

IB ACADEMY IB ACADEMY IB ACADEMY

Enter lower and upper Press OK

Press menu , choose
boundaries, mean (µ) and
5: Probability
standard deviation (σ).
5: Distributions
For lower bound = −∞,
2: Normal Cdf
set lower: -1E99
For upper bound = ∞,
set upper: 1E99

So 88.8% of the pears weigh between 100 g and 130 g.

37
 STATISTICAL APPLICATIONS Bivariate statistics

To find an x -value when the probability is given

The weights of pears are normally distributed with mean = 110 g and
standard deviation = 8 g. 8% of the pears weigh more than m grams. Find m.

Sketch!

8% = 0.08

110 m weight (g)

IB ACADEMY IB ACADEMY IB ACADEMY

Enter probability (Area), Press OK

Press menu
mean (µ) and standard
5: Probability
deviation (σ).
5: Distributions
The calculator assumes the
3: Inverse Normal
area is to the left of the
x-value you are looking
for.
So in this case:
area = 1 − 0.08 = 0.92

So m = 121, which means that 8% of the pears weigh more than 121 g.

5.2 Bivariate statistics

Bivariate statistics makes use of data where two different variables are measured. This
means that you can easily plot your individual measurements as (x, y) coordinates on a
scatter diagram. Analysing bivariate data allows you to asses the relationship between the
two measured variables; we describe this relationship as a correlation.

The independent variable is one you have control over and the one that you expect you
will have an effect on the other variable you are measuring - for instance time, age or
hours of sun exposure.

38
STATISTICAL APPLICATIONS Bivariate statistics 5

Scatter diagrams

Perfect positive Weak negative

correlation No correlation correlation
r =1 r =0 −1 < r < 0
y y y

x x x

5.2.1 Pearson’s correlation: −1 ≤ r ≤ 1

Besides only estimating the correlation between two variables from a scatter diagram,
you can also calculate a value that will describe it more precisely using your data. This
value is referred to as Pearson’s correlation coefficient (r ).

r = 0 means no correlation.
r ± 1 means a perfect positive/negative correlation.
Interpretation of r -values:
r −value 0 ≤ |r | ≤ 0.25 0.25 ≤ |r | ≤ 0.50 0.50 ≤ |r | ≤ 0.75 0.75 ≤ |r | ≤ 1
correlation very weak weak moderate strong

Remember that correlation does not mean causation.

Calculate by finding the regression equation on your GDC: make sure STAT DIAGNOSTICS
is turned ON (can be found when pressing MODE).

Bivariate statistics can also be used to predict a mathematical model that would best
describe the relationship between the two measured variables; this is called regression.
Here you will only have to focus on linear relationships, so only straight line graphs and
equations.

Your ‘comment’ on Pearson’s correlation always has to include two things:

1. Positive / negative and

2. Stong / moderate / weak / very weak

39
 STATISTICAL APPLICATIONS Bivariate statistics

Find Pearson’s correlation r and comment on it

The height of a plant was measured the first 8 weeks

Week x 0 1 2 3 4 5 6 7 8
Height (cm) y 23.5 25 26.5 27 28.5 31.5 34.5 36 37.5

1. Plot a scatter diagram

mean point

2. Use the mean point to draw a best fit line 0 + 1 + 2 + ... + 8

x̄ = = 3.56
9
23.5 + 25 + . . . + 37.5
ȳ = = 30
9
3. Find the equation of the regression line y = 1.83x + 22.7
Using GDC

IB ACADEMY IB ACADEMY IB ACADEMY

off

Press , got to Enter

on
Press menu
X list: A [];
“Lists and 4: Statistics
Y list: B[];
Spreadsheets” 1: Stat Calculations
1st Result Column: C[]
3: Linear Regression (mx+b)
Enter x-values in one
column (e.g A) and Press OK

y-values in another
column (e.g. B)
IB ACADEMY

So, equation of regression

line is y = 1.83x + 22.7
and Pearson’s correlation
(r -value) = 0.986
4. Comment on the result. Pearson’s correlation is r = 0.986, which
is a strong positive correlation.

40
STATISTICAL APPLICATIONS Chi-square test 5

5.3 Chi-square test

Chi-square test Used to test independence of two variables.

H0 the variables are independent (null hypothesis)

H1 the variables are not independent (alternative hypothesis)

Determine if the variables are independent by the χ2 test

Directors Managers Teachers Totals

Male 26 148 448 622

Female 6 51 1051 1108

Totals 32 199 1499 1730

2
Perform a χ test of independence at the 10% significance level to determine whether
employment grade is independent of gender.

1. State the null and alternative hypotheses H0 : gender and employment grade are
independent
H1 : gender and employment grade are
not independent

2. Calculate the table of expected e.g. expected number of male directors:

frequencies
t f1 t f2 622 32
Formula: · ·T · · 1730 = 11.5
T T 1730 1730
Directors Managers Teachers

Male 11.5 71.5 539

Female 20.5 127.5 960

3. Write down the degrees of freedom df = (2 − 1) · (3 − 1) = 2

df = (# rows - 1)(# columns - 1)

41
 STATISTICAL APPLICATIONS Chi-square test

4. Write down the chi-square value using

GDC.

IB ACADEMY IB ACADEMY IB ACADEMY

Enter data into GDC Enter dimensions of Enter the data as a matrix
matrix to fit your data. sto→

Press menu
Be sure you do not press ctrl and var

7: Matrices & Vectors include the totals, so in Give matrix a name by

1: Create this case you have a 2 × 3 typing it after the arrow
1: Matrix matrix (e.g. a)
≈
Press OK
press enter

IB ACADEMY IB ACADEMY IB ACADEMY

Enter name of Observed So Chi-square value

Press menu
Matrix (in this case a) χ 2 = 180.03, and
6: Statistics
Press OK p-value = 8.08 × 10−40
7: Statistical Tests
8: χ 2 2-way Test

42
GEOMETRY AND 6

TRIGONOMETRY
Table of contents & cheatsheet

6.1. Right triangles 44 6.3. Surface area and volume 47

Surface area the sum of the areas of all faces;

a2 = b 2 + c 2 Pythagoras
se unit2
opposite u
sin θ = en

opposite
SOH Volume amount of space it occupies;
hypotenuse p ot
hy unit3
adjacent
cos θ = CAH V = area of cross-section × height
hypotenuse
θ
opposite
tan θ = TOA adjacent
adjacent

6.3.1. Non-right angle triangles 48

c A b Remember the angles in a triangle add up to 180°.

1
Area of a triangle = ab sin C
B C 2
a

a b c Use this rule when you know:

Sine rule: = = 2 angles and a side (not between the angles)
sin A sin B sin C
or 2 sides and an angle (not between the sides).

Cosine rule: c 2 = a 2 + b 2 − 2 ab cos C Use this rule when you know:

3 sides or 2 sides and the angle between them.

43
 GEOMETRY AND TRIGONOMETRY Right triangles

6.1 Right triangles

a2 = b 2 + c 2 Pythagoras
se
opposite nu
sin θ = SOH te

opposite
hypotenuse p o
hy
adjacent
cos θ = CAH
hypotenuse θ
opposite adjacent
tan θ = TOA
adjacent

Two important triangles to memorize:

5
3 13
5

4 12

The IB loves asking questions about these special triangles which have whole numbers
for all the sides of the right triangles.

α
α = angle of elevation.
β = angle of depression.

To solve problems using Pythagoras, SOH, CAH or TOA identify what information is
given and asked. Then determine which of the equations contains all three elements and
solve for the unknown.

44
 GEOMETRY AND TRIGONOMETRY 3D applications 6

Triangle: finding an angle or the length of a side

Find c in the following triangle:

30°
12

1. Identify:
• info given • angle and adjacent
• need to find • opposite

2. pythagoras: 3x length c
tan 30° =
SOH: Θ, opp & hyp 12
CAH: Θ, adj & hyp ⇒ c = 12 × tan 30°
TOA: Θ , adj & opp = 6.92

6.2 3D applications

To find angles and the length of lines, use SOH, CAH, TOA and Pythagoras.
.
Example

Rectangular-based pyramid ABCDE with 7 cm

AD = 4 cm, C D = 3 cm, EO = 7 cm. B

A C
4 cm O
m
3c

Find the length of AC.

AC 2 = AD 2 + DC 2
= 42 + 32
= 25
p
⇒ AC = 25
= 5 cm

45
 GEOMETRY AND TRIGONOMETRY 3D applications

Find the length of AE .

.
Example
AE 2 = AD 2 + EO 2
1
(AO = AC = 2.5)
2
AE 2 = 2.52 + 72
= 55.25
p
⇒ AE = 55.25
= 7.43 cm

Find the angle AÊ C.

AÊC = 2AÊO
2.5
tan AÊO =
7 
2.5
‹
⇒ AÊO = tan−1
7
= 19.65°
⇒ AÊC = 2 × 19.65
= 39.3°

Find the angle that AE makes with the base of the pyramid.
Looking for angle E ÂO:
7
tan E ÂO =
2.5  ‹
7
⇒ E ÂO = tan−1
2.5
= 70.3°

Find the angle the base makes with E M , where M is the midpoint of C D .
Looking for angle E M̂ O:
7
tan E M̂ O =
OM
1
(O M = AD = 2 cm)
2
7
tan E M̂ O =
2
7
 ‹
⇒ E M̂ O = tan−1
2
= 74.1°

46
 GEOMETRY AND TRIGONOMETRY Surface area and volume 6

6.3 Surface area and volume

Surface area the sum of the areas of all faces cm2 .

Volume amount of space it occupies cm3 .

V = area of cross-section × height Formulas in data

booklet!
.
Example

The length of the cylindrical part of a 0.7 cm

pencil is 12.3 cm
h
13.5 cm

Write down the value of h.

h = 13.5 − 12.3
= 1.2 cm

Find the value of l .

‹2
1

l ⇒ l2 = × 0.7 + 1.22
0.7 cm

h 2
= 1.5625
⇒ l = 1.25 cm

Find the total surface area of the pencil.

SApencil = SAcylinder + SAcone + SAcircle
= 2π(0.35) · 12.3 + π(0.35)(1.25) + π(0.35)2
= 28.8 cm2

Find the volume of the pencil.

Vpencil = Vcylinder + Vcone
1
= π(0.352 ) × 12.3 + π(0.35)2 × 1.2
3
3
= 4.89 cm

47
 GEOMETRY AND TRIGONOMETRY Surface area and volume

Read the question,

does it specify if you
are looking for an acute 6.3.1 Non-right angle triangles
(less than 90°) or
obtuse (more than 90°)
angle. If not there may
To find any missing angles or side lengths in
be 2 solutions. Exam c A b
hint: Use sketches non-right angle triangles, use the cosine and sine
when working with
B C rule. Remember that the angles in the triangle
worded questions! a add up to 180°!
DB 3.6
a b c
Sine rule: = =
sin A sin B sin C
Use this rule when you know:

2 angles and a side or 2 sides and an angle

(not between the angles) (not between the sides)

a
B B
a
b
A

Cosine rule: c 2 = a 2 + b 2 − 2 a b cos C

Use this rule when you know:

3 sides or 2 sides and the angle between them

a
c C
b
b
a

1
Area of a triangle: Area = a b sin C
2
Use this rule when you know:

2 sides and the angle between them or 3 sides

first you need to use cosine rule
a to find an angle
C
b c
b

48
 GEOMETRY AND TRIGONOMETRY Surface area and volume 6

4ABC : A = 40°, B = 73°, a = 27 cm.

.
Example

Find ∠C.
∠C = 180° − 40° − 73° = 67°

Find b.
a b
=
sin A sin B
27 b
=
sin 40° sin 73°
27
b= · sin 73° = 40.169 ≈ 40.2 cm
sin 40°

Find c.
c a
=
sin C sin A
27
c= × sin 67° = 38.7 cm
sin 40°

Find the area.

1
Area = · 27 · 40 · 2 · sin 67°
2
= 499.59 ≈ 500 cm2
.
Example

m z
6k
35° x
10 km

Find z .
z 2 = 62 + 102 − 2 · 6 · 10 · cos 35°
z 2 = 37.70
z = 6.14 km

Find ∠x.
6 6.14
=
sin x sin 35°
sin x = 0.56
x = sin−1 (0.56) = 55.91°

49
 GEOMETRY AND TRIGONOMETRY Surface area and volume

50
MATHEMATICAL MODELS 7
Table of contents & cheatsheet

Definitions

Function a mathematical relationshjip where each input has a single output. It is often written as f (x) where x is the input.
Domain all possible x-values that a function can have. You can also think of this as the ‘input’ into a mathematical model.
Range all possible y-values that a function can give you. You can also think of this as the ‘output’ of a mathematical model.
Coordinates uniquely determines the position of a point, given by (x, y).

7.2. Linear 53
y = mx + c y

Where:
m = gradient (slope)
c = y-intercept

rise y − y1 x
m= = 2
run x2 − x1
For parallel lines: m1 = m2
−1
For perpendicular lines: m2 =
m1

7.3. Quadratic a>0 a<0 55

y y
y = a x2 + b x + c = 0

Axis of symmetry vertex

−b x x
x-coordinate of the vertex: x =
2a
Factorized form y = (x + p)(x + q) axis of symetry

7.4. Exponential 57
y = ka x + c y = 2x y = 2−x
y y
a>1 a<1
ax increasing decreasing
a −x decreasing increasing

Horizontal asymptote y=c

Coordinates of y-intercept (0, k + c) x x
Asymptote a line that a graph approaches
but never quite touches.

51
 MATHEMATICAL MODELS Domain & range

7.1 Domain & range

Mathematical models allows you to calculate the output that a certain input will give you.
To describe a mathematical model (or function) you therefore need to know the possible
Note: in some x and y-values that it can have; these are called the domain and the range respectively.
questions a domain will
be given to you (often
even though the
function as such could
in theory have many
Domain all possible x-values that a function can have. You can also think of
other x -value inputs).
This is relevant when
this as the ‘input’ into a mathematical model.
you have to for
example sketch the
Range all possible y-values that a function can give you. You can also think
graph; make sure that
you only draw the
of this as the ‘output’ of a mathematical model.
function for the
x -values included in
the domain.
.

1
Example

Find the domain and range for the function y =

x

Domain: x 6= 0
(all real numbers except 0)
Range: y 6= 0
(all real numbers except 0)

Find the domain and range for the function y = x 2

Domain: x ∈R
(all real numbers)
Range: y ∈ R+
(all positive real numbers)

52
MATHEMATICAL MODELS Linear 7

7.2 Linear

Linear mathematical models make straight line graphs. Two elements you need to know
to describe a linear function are its slope/gradient (how steeply it is rising or decreasing)
and its y-intercept (the y-value when the function crosses the y-axis, so when x = 0).

Straight line equation is usually written in the following form:

With
m = gradient (slope)
y = mx + c
c = y-intercept

This is useful, because this way you can read off the gradient (m) and y-intercept (c)
directly from the equation (or make a straight line equation yourself, if you know the
value of the gradient and y-intercept.)

Another form in which you may see a straight line equation is: a x + b y + c = 0.

In these cases, it is best to rearrange the equation into the y = m x + c form discussed
above. You can do this by using the rules of algebra to make y the subject of the equation.

When you are not given the value of the gradient in a question, you can find it if you
know two points that should lie on your straight line. The gradient (m) can be calculated
by substituting your two known coordinates (x1 , y1 ) and (x2 , y2 ) into the following
equation: Make sure you
substitute the y and
rise y2 − y1
= x -coordinates in the
run x2 − x1 correct order!

When you know the equation of one straight line, you can use the value of its
gradient (m) to find equations of other straight lines that are parallel or perpendicular to
it.

• Parallel lines have the same slope: m1 = m2 .

−1
• Perpendicular lines (90° angle) have: m2 = .
m1

53
 MATHEMATICAL MODELS Linear

Find the equation of the line. Then find the x -intercept.

y
7
6 •
N = (−3, 3.5) 5 M = (2, 6)
4
• 3
2
1
x
-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7 8

1. Take two points on the graph and y2 − y1

m=
substitute the values into the formula x2 − x1
6 − 3.5
= = 0.5
2 − (−3)
2. Fill in one point to find c 6 = 0.5(2)
= 1 + c , so:
c =5

3. Write down the equation y = mx + c y = 0.5x + 5

and replace m and c

4. To find the x -intercept, solve by 0 = 0.5x + 5

isolating x −5 = 0.5x
x = −10
x-intercept: (-10,0)

7.2.1 Interesection of lines

Finding a linear equation with the gradient and a point

Line L1 has gradient 5 and intersects line L2 at point A(1, 0). Find the equation of L1

1. Find slope Slope given, m=5

2. Fill in one point to find c L1 passes through (1, 0)

⇒ 0 = 5(1) + c
⇒ c = −5
⇒ y = 5x − 5

54
MATHEMATICAL MODELS Quadratic 7

Finding a linear equation with perpendicular gradient and a point

Line L2 is perpendicular to L1 . Find the equation of L2

1. Find slope L2 is perpendicular to L1 so

1
m=−
gradient
1
⇒m=−
5
2. Fill in one point to find c 1
0 = − (1) + c
5
1
⇒c=
5
1 1
⇒y =− x+
5 5

7.3 Quadratic

Functions in which one of the terms is raised to the power of 2, written in the following
form, are called quadratic:
y = a x2 + b x + c = 0

When plotted on a graph, a quadratic function always gives an upward or downward

facing U-shape – this is called a parabola. A parabola always has a vertex (the maximum
or minimum point) and an axis of symmetry. If you know the x and y
coordinate of the
vertex, the equation for
a>0 a<0 the axis of symmetry
y y will always be x = [the
x -coordinate of the
vertex]. This also works
the other way around;
the equation of the axis
of symmetry gives you
the x -coordinate of the
vertex vertex.

x x

axis of symetry

55
 MATHEMATICAL MODELS Quadratic

The equation for the axis of symmetry can be found using the following equation where
a, b and c are the corresponding numbers from your quadratic equation written in the
form y = a x 2 + b x + c:

−b
Axis of symmetry: x= = x-coordinate of vertex
2a

7.3.1 Factorization

One more thing you have to know how to find for quadratic mathematical models are its
x-intercepts. These you can find by setting your quadratic equation equal to 0. So when
a x 2 + b x + c = 0 you can solve for x to find the x-intercepts (or ‘roots’ as they are
sometimes also called). Given that quadratic equations have the shape of a parabola, they
can have up to two x-intercepts - as you can see when a quadratic equation is plotted, it
often corsses the x-axis twice.

You can find the x-intercepts using several methods. Here we show an example of
factorization, but you can also do this graphically using your GDC (See basics,
Polyrootfinder instructions).
.

Factorize x 2 − 2x − 15.
Example

x 2 − 2x − 15 = (x + a)(x + b )
a + b = −2 a = −5
⇒
a b = 15 b =3
2
x − 2x − 15 = (x − 5)(x + 3)

Find the coordinates of the x-intercepts and vertex.

x −5=0 x +3=0
or
x =5 x = −3

x-intercepts: (5, 0) and (−3, 0).

−b −(−2)
Vertex: = =1 ← x-coordinate
2a 2·1
12 − 2(1) − 15 =
1 − 2 − 15 = −16 ← y-coordinate

Vertex = (1, −16)

56
MATHEMATICAL MODELS Exponential 7

7.4 Exponential

Another type of mathematical model that you need to be familiar with is the exponential
function. An exponential function is one where the variable (x) is the power itself. An
exponential function can therefore be written in the following form:

y = ka x + c

Asymptote a line that a graph approaches but never quite touches.

In questions dealing with exponential functions, you will need to know how to describe
the following three things: their asymptotes, y-intercepts and whether they are
increasing or decreasing. When an exponential function is written in the form described
above (y = ka x + c), you can use its different parts to find these.

Where is the asymptote?

The horizontal asymptote is at y = c.

Where is the y -intercept?

y-intercept has coordinates (0, k + c).

Is the function increasing or decreasing?

a>1 a<1
ax increasing decreasing
a −x decreasing increasing

57
 MATHEMATICAL MODELS Exponential

.
Example
y = 2x y = 3 · (2 x )
y y

x x
Use GDC to sketch
more complicated
functions.
y = 2−x − 3 y = −2−x + 1
y y

x x

−2

58
MATHEMATICAL MODELS Intersection 7

7.5 Intersection

When functions intersect the x and y-values are equal, so at the point of intersection
f (x) = g (x).

To find the intersection point(s) of two functions.

1
f (x) = x − 2 and g (x) = −x 2 + 4
2

IB ACADEMY IB ACADEMY IB ACADEMY

Plot both functions Find the intersection: Approach the intersection

off
you are trying to find with
Press on , go to Press menu
the cursor and click once
“Graph” 8: Geometry you near it. Repeat for any
Enter the two functions: 1: Points & lines other intersections.
1
f1(x)= x − 2, 3: Intersection Point(s)
2
press tab to input
f2(x)= −x 2 + 4

In this case the intersection points are (−1.68, 1.19) and (2.41, −1.81).
Note: if you can’t find 8: Geometry make sure you use graph mode instead of the
scratchpad, otherwise please update your N-spire to the latest version.

59
 MATHEMATICAL MODELS Intersection

60
DIFFERENTIATION 8
Table of contents & cheatsheet

8.1. Polynomials 62

Polynomial a mathematical expression or function that contains several terms often raised to different powers
dy
When y = f (x) = a x n then the derivative is = f 0 (x) = nax n−1 .
dx

Derivative of a constant (number) 0

Derivative of a sum sum of derivatives.
dy
When y = a x n + b x m , = na n−1 + mb x m−1
dx

8.2. Tangent/Normal 63

Tangent a straight line that touches a curve at one single point. At that point, the gradient of the curve = the gradient of
the tangent.
−1
Normal a straight line that is perpendicular to the tangent line. Slope of normal =
slope of tangent

8.3. Turning points 65

Turning points occur when a function has a local maximun or local minimum. At these points f 0 (x) = 0.
y
D
•
•A C•

f 0 (x) f (x) is

B A − decreasing
• B 0 at local minimum
x
C + increasing
D 0 at local maximum

61
 DIFFERENTIATION Polynomials

8.1 Polynomials

As you have learnt in the section on functions, a straight line graph has a gradient. This
gradient describes the rate at which the graph is changing and thanks to it we can tell
how steep the line will be. In fact gradients can be found for any function - the special
thing about linear functions is that their gradient is always the same (given by m in
y = m x + c). For polynomial functions the gradient is always changing. This is where
calculus comes in handy; we can use differentiation to derive a function using which we
can find the gradient for any value of x.

Using the following steps, you can find the derivative function ( f 0 (x)) for any
polynomial function ( f (x)).

Polynomial a mathematical expression or function that contains several

terms often raised to different powers

2 1
e.g. y = 3x 2 , y = 121x 5 + 7x 3 + x or y = 4x 3 + 2x 3

dy
Principles y = f (x) = a x n ⇒ = f 0 (x) = na x n−1 .
dx

The (original) function is described by y or f (x), the derivative

dy
(gradient) function is described by or f 0 (x).
dx

Derivative of a constant (number) 0

e.g. For f (x) = 5, f 0 (x) = 0

Derivative of a sum sum of derivatives.

If a function you are looking to differentiate is made up of several

summed parts, find the derivatives for each part separately and
then add them together again.

e.g. f (x) = a x n and g (x) = b x m

f 0 (x) + g 0 (x) = na x n−1 + m b x m−1

When differentiating it is important to rewrite the polynomial function into a form that
is easy to differentiate. Practically this means that you may need to use the laws of
exponents before or after differentiation to simplify the function.

62
 DIFFERENTIATION Tangent/Normal 8

5
For example, y = seems difficult to differentiate, but using the laws of exponents we
x3
5
know that y = = 5x −3 . Having the equation in this form allows you to apply the
x3
same rules again to differentiate.
.

f (x) f 0 (x)
Example

5 −→ 0

x2 −→ 2 · 1x 2−1 = 2x

4x 3 −→ 3 · 4x 3−1 = 12x 2

3x 5 − 2x 2 −→ 5 · 3x 5−1 − 2 · 2x 2−1 = 15x 4 − 4x

2 −8
= 2x −4 −→ (−4) · 2x −4−1 = −8x −5 =
x4 x5
2 6
3x 4 − +3 −→ 4 · 3x 4−1 − 3 · (−2)x −3−1 + 0 = 12x 3 +
x3 x4

8.2 Tangent/Normal

Tangent a straight line that touches a curve at one single point. At that
point, the gradient of the curve is equal to the gradient of the tangent.

Normal a straight line that is perpendicular to the tangent line:

−1
slope of normal =
slope of tangent

For any questions with tangent and/or normal lines, use the steps described in the
following example.

63
 DIFFERENTIATION Tangent/Normal

Finding the linear function of the tangent.

Let f (x) = x 3 . Find the equation of the tangent at x =2

1. Find the derivative and fill in value of x to f 0 (x) = 3x 2

Steps 1, 2 and 4 are determine slope of tangent
identical for the f 0 (2) = 3 · 22 = 12
equation of the tangent
and normal 2. Determine the y value f (x) = 23 = 8

3. Plug the slope m and the y value in 8 = 12x + c

y = mx + c

4. Fill in the value for x to find c 8 = 12(2) + c

c = −16
eq. of tangent: y = 12x − 16

Finding the linear function of the normal.

Let f (x) = x 3 . Find the equation of the normal at x =2

Steps 1, 2 and 4 are

1. f 0 (2) = 12
identical for the
equation of the tangent
2. f (x) = 8
and normal
3. Determine the slope of the normal −1
m=
−1 12
m= and plug it and the
slope tangent 1
8=− x+c
y -value into y = mx + c 12
4. Fill in the value for x to find c 1
8=− (2) + c
12
49
c=
6
1 49
eq. of normal: y =− x+
12 6

64
DIFFERENTIATION Turning points 8

To find the gradient of a function for any value of x.

f (x) = 5x 3 − 2x 2 + x. Find the gradient of f (x) at x = 3.

IB ACADEMY IB ACADEMY IB ACADEMY

Enter the variable used in Type in your function

Press menu
your function (x) and the ≈

4: Calculus press enter

value of x that you want to
1: Numerical Derivative
find. Keep the settings on
at a Point
1st Derivative
Press OK

In this case, f 0 (3) = 124

8.3 Turning points

Turning points are when a graph shows a local maximum (top) or minimum (dip). This
occurs when the derivative f 0 (x) = 0.

Use the graph (GDC) to see whether a turning point is a maximum or minimum.

y
D
•
•A C•
f 0 (x) f (x) is

B
A − decreasing
•
x B 0 at local minimum
C + increasing
D 0 at local maximum

65
 DIFFERENTIATION Optimization

8.4 Optimization

As we saw in the previous section, differentiation is useful for identifying maximum and
minimum points of different functions. We can apply this knowledge to many real life
problems in which we may seek to find maximum or minimum values; this is referred to
as optimization.
Note: The most
important thing to
remember is that at a
maximum or minimum Determine the max/min value with certain constraints
point f 0 (x) = 0. So,
often if a question asks
a maximum/minimum The sum of the height h and base x of a triangle is 40 cm. Find an expression for the
value of something, like area in terms of x , hence find the maximum area of the triangle.
in this example,
differentiation may well 1. First write expression(s) for constraints x + h = 40
be a useful way to
approach it.
followed by an expression for the actual h = 40 − x
calculation. Combine two expressions so
1
that you are left with one variable. A= xh
2
1
= x(40 − x)
2
1
= − x 2 + 20x
2
2. Differentiate the expression dA
= −x + 20
dx
3. The derivative = 0, solve for x −x + 20 = 0
x = 20

4. Plug the x value into the original function 1

A = − (20)2 + 20(20)
2
= −200 + 400
= 200 cm2

66