KSSM_Form_4_Additional_Mathematics_Notes

FORM 4

LAIZHIJUN
LIM JUN HAO
OOIHIAN GEE
OOIMINGYANG
SIAW JIAQI

FORM 4

LAIZHIJUN
LIM JUN HAO
OOIHIAN GEE
OOIMINGYANG
SIAW JIAQI

Project Leader: Lim Jun Hao
Associate Editor: Lai Zhi Jun
Associate Project Editor: Siaw Jia Qi
Senior Managing Editor: Lim Jun Hao
Production Coordinator: Ooi Ming Yang
Cover Design: Ooi Hian Gee
Cover image: Pixabay

About the cover: The cover images comprise of dictionary, notes and an ancient clock.
The illustrations mean well-management of time and notes-taking habit can help us
in achieving our goals. We shall also always make inquiry to the right person
whenever we have any doubts during learning process.
A collaboration work by Faculty of Science, University of Malaya Applied
Mathematics undergraduate students.

Form 4 Additional Mathematics / Lim Jun Hao, Lai Zhi Jun, Ooi Hian Gee, Ooi Ming
Yang, Siaw Jia Qi
Font used: Times New Roman
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stored in a retrieval system, or transmitted, in any form or by any means, electronic,
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at [email protected].

i

Table of Contents

Chapter 1 Functions
1.1 Functions ................................................................................................................................ 1
1.2 Composite Functions.............................................................................................................. 7
1.3 Inverse Functions ............................................................................................................... 12
Summative Exercises ................................................................................................................ 18

Chapter 2 Quadratic Functions
2.1 Quadratic Equations and Inequalities ............................................................................... 19
2.2 Types of Roots of Quadratic Equations ............................................................................. 24
2.3 Quadratic Functions ........................................................................................................... 26
Summative Exercises ............................................................................................................... 38

Chapter 3 Systems of Equations
3.1 Systems of Linear Equations in Three Variables .............................................................. 39
3.2 Simultaneous Equations Involving One Linear Equation & One Non-Linear Equation . 45
Summative Exercises ............................................................................................................... 49

Chapter 4 Indices, Surds and Logarithms
4.1 Laws of Indices.................................................................................................................... 50
4.2 Laws of Surds ..................................................................................................................... 52
4.3 Laws of Logarithms ........................................................................................................... 55
4.4 Applications of Indices, Surds and Logarithms ................................................................ 62
Summative Exercises ............................................................................................................... 63

Chapter 5 Progressions
5.1 Arithmetic Progressions ..................................................................................................... 64
5.2 Geometric Progressions ..................................................................................................... 71
Summative Exercises ............................................................................................................... 82

Solutions

References

ii

CHAPTER 1 FUNCTIONS

Chapter 1 Functions

1.1 Functions

1.1.1 What is a function?

Let’s take an example.
If the radius of a circle is 2 cm, then the circumference, C of the circle is C = 2πr = 4π cm.
If the radius of a circle is 4 cm, then the circumference, C of the circle is C = 2πr = 8π cm.
If the radius of a circle is 5 cm, then the circumference, C of the circle is C = 2πr = 10π cm.

From this example, we know that the radius of a circle will affect the circumference of the circle;
therefore, the circumference of a circle is a function of radius, such that C = 2πr.

Look at the graph of y = x − 2 on the right. The relation (8,6)
between the value of 3 on the x-axis and the value of 1 on the
y-axis can be written as 3 → 1. (5,3)
(3,1)
We can say that 3 is mapped to 1, and likewise for 5
→ 3 and 8 → 6. Every point (x, y) on the line corresponds to
the mapping of x → y where the values of x on the x-axis are
mapped to the values of y on the y-axis.

A relation from x-axis to y-axis is called a function if
each element in the x-axis is related to exactly one element in
the y-axis. Hence, this is a function since each element from
X is related to only one element in Y.

A relation can be represented by an arrow diagram as shown below.

31
53
86

Set X Set Y

1. f is a function from set X to set Y .

2. The element in set X is known as object, whereas the element in set Y is known as image.
Note that the term “preimage” is in common usage in math instead of the term “object”.

3. Example, the element 3 in set X and the element 1 in set Y correspond to each other whereby 3 is the
object of 1 and 1 is the image of 3.

4. Any element x in set X that is mapped to one element y in set Y by a function is written in function
notation as ∶ → or ( ) = , where x is the object and y is the image.

5. A function maps an object to an image only. Therefore, the relation is called a function if the relation

is either one-to-one relation or many- to-one relation.

1

CHAPTER 1 FUNCTIONS

EXAMPLE 1

Which of the following is a function? Explain.

(a) Solution:
12 8 This is a function since each object is mapped to

20 only one image although the element 1 has no
6 object.
8
This relation is known as one-to-one function.
12

1

(b) a 5 Solution:
b This is a function since each object has only one
c
4 image although the element 10 has no object.
(c) This relation is known as many-to-one relation.
9
1 10

(d) 3 Solution:
2 This is not a function since one object has two
4
-3 images. (Does not satisfy the condition of a
function)

1 This relation is known as one-to-many relation.
-1

Solution:
1 This is not a function since each object has more
2 than one image. (Does not satisfy the condition of

a function)
4 This relation is known as many-to-many relation.

2

CHAPTER 1 FUNCTIONS
To determine whether a graph represents a function or not, we use vertical line test.

The vertical line intersects more than The vertical line intersects only 1
1 point of the graph. point of the graph.
Hence the graph does not represent a Hence the graph represents a function.
function.

1.1.2 Introducing asymptotes

The asymptote of a curve is a line such that the distance between the line and the curve approaches zero
as one or both of the x or y coordinates tend to infinity. However, the asymptote of a curve will not touch
the curve.

y From the graph, we know that:
1. When the x-value approaches 3 from the left
( 3− ), the corresponding y-value approaches

negative infinity (−ꚙ).

Graph of = 1 2. When the x-value approaches 3 from the right
−3 ( 3+ ), the corresponding y-value approaches

infinity (ꚙ).

x

Function is undefined when denominator = 0,

Vertical asymptote x=3 because division by zero makes an operation
undefined.
− 3 = 0

= 3

The function is undefined at x = 3, therefore the graph will not touch the line x = 3.
Is there a way to calculate the vertical asymptote?
Take note that the function is undefined when denominator is 0.
In this example, the denominator of the function is x−3. We will have the function undefined when
x−3 = 0, therefore x = 3.

3

CHAPTER 1 FUNCTIONS

1.1.3 Introducing the absolute-value function

y
The absolute value |x| is expressed as:

x when x ≥ 0
|x| = −x when x < 0

Graph of y=|x| For instance, |1| = 1; |−1| = 1.
x

1.1.4 How to determine the domain and the range of a discrete function?

Let’s take back Example 1(a). 1. The domain of a function is the set of all possible input
a) values.

12 8 2. The codomain of a function is the set into which all of the
output function is forced to fall.

20 6 3. The range of a function is the set of all possible output
values.

8 12 In this example,
Domain = {12, 20, 8}
Set X 1 Codomain = {8, 6, 12, 1}
Domain Set Y Range = {8, 6, 12}
Codomain

1.1.5 How to determine the domain and the range of a continuous function?

y
( ) =

x Range

Domain
In this example, the domain is −5 ≤ ≤ 6 whereas the range is −1 ≤ ( ) ≤ 1.

4

CHAPTER 1 FUNCTIONS

EXAMPLE 2

Sketch the graph of f : x → |2x–4| in the domain −3 ≤ ≤ 3.

Solution:

Step 1: You need to have 4 points in order to construct a graph of an x |2 − 4|
absolute-value function. -3 10
The four points are: 04
(a) vertex, 20
(b) y-intercept, 32
(c) left endpoint of the graph,

(d) right endpoint of the graph.

Step 2: You need to plot the points onto a graph paper and indicate the x-axis and y-axis.

Step 3: Connect the points in the shape of V with vertex as the turning point.

The vertex can be obtained by equating the (-3,10)
algebraic expression inside the modulus
sign with 0 and solving it. (0,4)
In this example, ( ) = |2 − 4|
2x – 4 = 0 (3,2)

2x = 4 (2,0)
x=2
when x = 2,
f(2) = |2(2) – 4| = 0
Hence the coordinate of the vertex is (2,0).

This is the vertex.

5

CHAPTER 1 FUNCTIONS

1.1.6 How to determine the image of a function when its domain is given and vice
versa?

Note: For a function, for example f : x → 2x + 2,
if x = 1, then f (1) = 2(1) + 2 = 4;
if x = 2, then f (2) = 2(2) + 2 = 6;
if x = u, then f (u) = 2(u) + 2 = 2u + 2.

EXAMPLE 3 (b) When f (x) = 4
2x + 2 = 4
A function f is defined by f : x → 2x + 2, find 2x = 2
(a) the image of 2, x=1
(b) the object that have the image 4.
Hence the object that has the image 4 is 1.
Solution:
(a) f : x → 2x + 2

f (x) = 2x + 2
When x = 2,

f (2) = 2(2) + 2
=6

Hence, the image of 2 is 6.

EXAMPLE 4 2 + for all values of x, except x = b, where a is a constant.
− 2
A function f is defined by f : x →
(a) State the value of b.

(b) Find the value of a given that the value 5 is mapped to itself under f.

Solution: (b) 5 is mapped to itself, therefore
(a) Function is undefined when: (5) = 5

denominator = 0 2(5) +
x–2=0 5−2 =5
x=2 10 + = 5(3)

It is given that function is undefined = 5
when x = b, therefore b = 2.

6

CHAPTER 1 FUNCTIONS

1.2 Composite Functions

1.2.1 What are composite functions?

f f (x) g gf (x)
x

For example, let x be a tree. In order to produce furniture, the tree (x) undergoes cutting process (function

f) to obtain wood (f (x)). The wood (f (x)), is then manufactured (Function g) to produce furniture (gf (x)).

fg(x) is read as “f composed with g of x” . fg(x) = f[g(x)].

gf(x) is read as “g composed with f of x” . gf(x) = g[f(x)].

Note: ( ) = ( ).

1.2.2 How to determine composite functions?

EXAMPLE 5

Two functions f and g are defined by f (x) = x + 1 and g (x) = 2. Find the following composite
functions:
(a) 2 ( ) 2 ( ) ( )

Solution: f (x) = x + 1 (c) ( ) = [ ( )]
(a) 2( ) = ( ) f (x + 1) = (x + 1) + 1 = ( 2)
= 2 + 1
= [ ( )]
(d) ( ) = [ ( )]
= ( + 1)
= ( + 1) + 1 = ( + 1)
= ( + 1)2
= + 2 = 2 + 2 + 1
(b) 2( ) = ( )

= [ ( )]
= ( 2)
= ( 2)2
= 4

7

CHAPTER 1 FUNCTIONS

1.2.3 How to determine the objects or the images of composite functions?
EXAMPLE 6

Two functions f and g are defined by ( ) = + 2 and ( ) = 2, find ( ). Then, find the value
of (2).

Solution: ( ) = + 2 (2) = 22 + 4(2) + 4
( ) = [ ( )] = 16

= ( + 2)
= ( + 2)2
= 2 + 4 + 4

EXAMPLE 7

Two functions f and g are defined by f (x) = x − 2 and g (x) = 2 + 2. Find the values of x when
gf (x) = 6.

Solution: ( ) = 6
( ) = [ ( )] 2 − 4 + 6 = 6

= ( − 2) 2 − 4 = 0
= ( − 2)2 + 2 ( )( − 4) = 0
= ( 2 − 4 + 4) + 2
= 2 − 4 + 6 = 0, = 4

Tip: The word “values”
indicates that there will
be at least 2 values of x

8

CHAPTER 1 FUNCTIONS

1.2.4 How to determine a function when the composite function and one of the
functions are given?

EXAMPLE 8

(a) A function f is defined as ∶ → + 2. Find the function g such that ∶ → 3 + 4.
(b) A function f is defined as ∶ → + 2. Find the function g such that ∶ → + 5.

Solution: (b) (Case where the function determined is
(a) (Case where the function determined is situated “outside” the composite function)
( ) = + 5
situated “inside” the composite function) ( + 2) = + 5
∶ → 3 + 4 ( ) = ( − 2) + 5
( ) = 3 + 4 ( ) = + 3
( ) = ( ) + 2 ( ) = + 3
3 + 4 = ( ) + 2
How to change
( ) = 3 + 2 ( + ) to ( )?

Note: For a function, for example f : x → 2x + 2, + 2 =
if x = 1, then f (1) = 2(1) + 2 = 4; = − 2
if x = 2, then f (2) = 2(2) + 2 = 6;
if x = u, then f (u) = 2(u) + 2 = 2u + 2.

9

CHAPTER 1 FUNCTIONS

1.2.5 How to determine the domain of a composite function?

If ƒ and g are functions, the composite function (“ƒ composed with g”) is defined by ƒ(g(x)).
The domain of this composite function consists of the numbers x in the domain of g for which g(x) lies
in the domain of ƒ.
Similarly, if ƒ and g are functions, the composite function (“g composed with f”) is defined by g(f(x)).
The domain of this composite function consists of the numbers x in the domain of f for which f(x) lies in
the domain of g.

EXAMPLE 9

Two functions f and g are defined by ( ) = √ and ( ) = − 2. Find ( ) and its domain.

Solution:
( ) = ( − 2)

= √ − 2

Notice that ( ) = − 2 is defined for all real x but ( ) belongs to the domain of ƒ only if x ≥ 0.
The domain of the composite function ( ) = √ − 2 is ≥ 2 .

− 2 ≥ 0

1.2.6 Further examples and applications on composite functions

EXAMPLE 10

A function f is defined as : → 1 , ≠ 0. Find 2, 4, 5 and 20 .


Solution: 4( ) = ( 2)2( )
( ) = 1
= 2[ 2( )] 2( ) =
= 2( )

2( ) = ( ) =

1 5( ) = [ 4( )]
= ( )
= ( )
1
=1 1
=
20 ( ) = 4 4 4 4 4( )
=

=

10

CHAPTER 1 FUNCTIONS

EXAMPLE 11

In a grocery store, the discounted price, q of a product A is denoted as ( ) = 1 , where x is the
2

original price of the product. Upon checking out, the product will be charged service tax, p and is

denoted by ( ) = 1.2 . How much does product A costs if the original price of the product is

RM20?

Solution:

pq

The question is to find (20).

( ) = 1 (20) = 0.6(20)
(2 ) = 12
1
= 1.2 (2 ) Therefore product A costs RM12.

= 0.6

11

CHAPTER 1 FUNCTIONS

1.3 Inverse Functions

1.3.1 What are inverse functions?

Let’s take an example of ice and water.

f

Water Ice

−1

Water undergoes freezing process to become ice. Freezing process is denoted by the function f.

Ice undergoes melting process to become water again. Melting process is the inverse of freezing process
and is denoted by the inverse function −1.

1.3.2 What are the characteristics of an inverse function?

1) A function f that maps set X to set Y has an inverse function, −1 if f is a one-to-one function.
(Inverse function, −1 is valid when each element in set Y is mapped onto one and only one element in

set X and the function f is valid)
2) The inverse of an inverse function gives back the original function.

Example − ( ) = and − ( ) =

3) If two functions f and g are inverses of each other:

a) The domain of f = the range of g.
b) The domain of g = the range of f.

c) The graph of function f is the reflection of graph of function g across the line y = x.
4) If (a,b) lies on the graph f , where a and b are real numbers, then (b,a) lies on the graph − since

graph f is the reflection of graph −1 at line y = x.

Note: − ( ) ≠
( )

1.3.3 How to determine whether the function is one-to-one?

A function ƒ(x) is one-to-one on a domain D if ( 1) ≠ ( 2) whenever 1 ≠ 2 in D.
For example, ( ) = 3 is one-to-one on any values of x in the domain D because 13 ≠ 23 whenever 1 ≠ 2
in D.

( )
( ) = 3

( 2) x

1 2
( 1)

12

CHAPTER 1 FUNCTIONS

1.3.4 How to determine whether the graph of function has an inverse function?

We use horizontal line test to test whether the graph of function has an inverse function.

y = 2 y
= + 2

x x

= 2 is a function. = + 2 is a function.
However, it does not have an inverse It has an inverse function since horizontal
function since horizontal line cuts the line cuts the graph of function at only 1
graph of function at least 2 points. point.

EXAMPLE 12

Given that (3) = 5 and (1) = 3. Determine −1(5) and −1(3).

Solution:
(3) = 5 therefore −1(5) = 3
(1) = 3 therefore −1(3) = 1

EXAMPLE 13

Given that : → 1 , ≠ 0. Find −1(3).


Solution: ( ) =
= −1(3) −1( ) =

( ) = 3

( ) = 1


13

CHAPTER 1 FUNCTIONS

To determine whether two functions f and g are inverses of each other, we need to make sure both the
domain for ( ) and ( ) are the same.

EXAMPLE 14

Determine whether the function ( ) = − 3 has an inverse of ( ) = − 3 + 2 .
+ 2 − 1

Solution:
Check the domain of the function ( ).
Function is undefined when denominator = 0
+ 2 = 0

= −2
Therefore the domain of the function ( ) is < −2 and > −2.

Check the domain of the function ( ).
Function is undefined when denominator = 0
− 1 = 0

= 1
Therefore the domain of the function ( ) is < 1 and > 1.

Check the domain of the function ( ).

Notice that ( ) 3 + 2 is defined for all real x except =1 but ( ) belongs to the domain of
= − − 1
ƒ only if ≠ −2.

The domain of the composite function ( ) = is < −2, −2 < < 1, > 1.

Check the domain of the function ( ).

Notice that ( ) = − 3 is defined for all real x except = −2 but ( ) belongs to the domain of
+ 2

only if ≠ 1.

The domain of the composite function ( ) = is < −2, −2 < < 1, > 1.

14

CHAPTER 1 FUNCTIONS

( ) = − 3
( + 2)

= − 3 + 2 ( − 23)
+

− 3 − 1
+ 2

−3( + 2) − 2( − 3) − 3 − ( + 2)
= + 2 ÷ + 2

−5 + 2
= + 2 × −5
=

Since ( ) = ( ) = and both the domain for ( ) and ( ) are the same, therefore

( ) = − 3 + 2 is an inverse of ( ) = − 3 .
− 1 + 2

EXAMPLE 15

A function f is defined as : → 2 + 1 for the domain 0 ≤ ≤ 3.

(a) Sketch the graphs of and −1 on the same graph.

(b) State the domain and image of and −1.

Solution: (3,10) y=x
(a) x 0 1 2 3 f −1 (10,3)

y 1 2 5 10

The graph of f is a part of the quadratic curve y = 2 + 1.

The graph of −1 is a reflection of the graph f by the line y=x

(b) Domain of f is 0 ≤ ≤ 3.

Range of f is 1 ≤ ( ) ≤ 10.

Domain of −1 is 1 ≤ ≤ 10. (0,1)

Range of −1 is 0 ≤ −1( ) ≤ 3 (1,0)

15

CHAPTER 1 FUNCTIONS

1.3.5 How to determine the inverse function?

There are a few steps to determine the inverse function:
Step 1: Change the function from = ( ) to = ( ).
Step 2: Write x as −1( ).
Step 3: Substitute the variable y with x.

EXAMPLE 16

If ( ) = 3 + 1, find −1( ) and hence find −1(4).

Solution:

( ) = 3 + 1

= 3 + 1

3 = − 1 Step 1: Change the function from = ( ) to = ( ).

− 1 Step 2: Write x as −1( ).
= 3 Step 3: Substitute the variable y with x.
− 1
−1( ) = 3

−1( ) = − 1
3

−1(4) = 4 − 1
3

=1

Always do silly mistakes?

You can check the validity of the inverse function by using the characteristic of inverse function.
You must make sure that − ( ) = and − ( ) = .

1.3.6 Further examples on inverse functions

Inverse functions do apply in our everyday life.

1) For example, you are running a marathon. You travelled 10 miles per hour constantly, and want to
know how far you have gone in x hours. This could be represented by the function ( ) = 10 . Now if

I know I have travelled x miles, then how long I have been travelling for? This could be represented by

the inverse function of −1( ) = 10 .

2) For example, 3 Malaysian Ringgit (MYR) is equivalent to 1 Singaporean Dollar (SGD). The currency

from x SGD to MYR could be represented by the function ( ) = 3 . Now if I have x MYR and I

want to convert it to SGD, how much is it? This could be represented by the inverse function of

−1( ) = 3 .

16

CHAPTER 1 FUNCTIONS

Further Exploration

Functions (Part I)

https://youtu.be/oCf4-G6wojY

Functions (Part 2)

https://youtu.be/g7RZ5aV7RG4

Composite Functions (Part 1)

https://youtu.be/TIdFFbM2MNA

Composite Functions (Part 2)

https://youtu.be/SpK4wsd_W6g

Composite Functions (Part 3)

https://youtu.be/uZTuBdNYFxQ

Inverse Functions

https://youtu.be/LsXrVOdAgOI

17

CHAPTER 1 FUNCTIONS

Summative Exercises

1.1 Functions 4) 7cm
25cm
1) A function f is defined by : → 2 + 4. State
(a) the image of 4,
(b) the object that have the image 4.

2) A function f is defined by f : x → 2 + 3 for all
− 4

values of x, except x = q and p is a constant. The diagram shows a container having the shape
of a cylinder. It had 1540 3 of water and is
(a) State the value of q.

(b) Find the value of p given that the value 2 is being filled with water at a constant rate. The

mapped to itself under f. height of the water in the container increases at a

constant rate of 1 per second as water was

poured in. The radius and height of the container
3) Sketch the graph of : → | + 3| in the domain 22
−4 ≤ ≤ 1. Hence, state the range of the is 7 and 50 respectively.[ = 7]
function.
(a) Write the function for the height h of water
1. 2 Composite Functions
after t seconds.
1) Given the functions ( ) = 2 + 1 and (b) Write the function for the volume V of the
( ) = 2 2 + 5. Find ( ). Hence, find the
value of x when ( ) = 2. water in terms of h, height of water.
(c) Find the function ℎ( ).
(d) Hence, find the volume of water in 3 after

10 seconds.

2) Given the function ∶ ⟶ 1 + , ≠ 1. Find 1.3 Inverse Functions
(a) 2( ),
1 − 2
1) Given the functions : → − 2 , ≠ 2 and
(b) 3( ),
: → 3 − . Find
(c) 6( ). (a) −1
(b) −1
3) Roland got lost from Robin Hood, his dad in the (c) −1 −1
woods. Robin Hood is determining the area in (d) ( )−1
which to search. Roland can walk at an estimated (e) ( )−1
5 miles per hour. [ ℎ ] Hence, find the function which is the same as
(a) Write the function ( ) which would be the −1 −1 .
distance Roland could walk in x hours.
(b) Write the function ( ) which would give the 2) Given the function : → − 4.
potential search area given Roland walked r (a) Find the expressions for 2 and −1.
miles. (b) Show that ( −1)2 = ( 2)−1.
(c) Find the composite function of area for Robin (c) What is the value of x when
Hood to search for him. −1 ( ) = −1(5 + 2).

18

CHAPTER 2 QUADRATIC FUNCTIONS

CHAPTER 2 Quadratic Functions

2.1 Quadratic Equations and Inequalities

2.1.1 How to solve quadratic equations?

• By using Completing The Square method

EXAMPLE 1

By using the completing the square method to solve the following equation.
2 # + 7 + 6 = 0.

2 # + 7 + 6 = 0 Factorise the equation by 2 so that the
coefficient of # becomes 1
# + 7 + 3 = 0
2

# + 7 = −3 Move the constant term to the RHS
2 of the equation

# + 7 + +,722-.# = −3 + +,722-.# Add the term
2 89:;<<=9#=;>? :< @A# to
the both RHS and LHS
# + 7 + ,74-# = −3 + 49
2 16 of the equation

, + 74-# = 1 ( + )# = # + 2 + #
16

+ 7 = ±3116
4

= − 3 = −2
2

Hence, the solutions of the equation 2 # + 7 + 6 = 0 are − 7 and −2 .
#



19

CHAPTER 2 QUADRATIC FUNCTIONS

• Formula method

= − ± I # − 4 is used to solve a quadratic equation # + + = 0
2

EXAMPLE 2

By using the formula to solve 2 # + 7 + 6 = 0.

Compare the given equation 2 # + 7 + 6 = 0 with the general form
equation # + + = 0. Hence, = 2, = 7 and = 6.

−7 ± J(7)# − 4(2)(6)
= 2(2)

= −7 ± I1
4

= − 3 = −2
2

Hence, the solutions of the equation 2 # + 7 + 6 = 0 are − 7 and −2.
#

2.1.2 How to form quadratic equations from given roots?

Assume the given roots of a quadratic equation are and then we calculate the sum of roots and the
product of roots.

Sum of roots = + = − K
L

Product of roots = = 9
L

Therefore, the quadratic equation with the given roots a and b can be written as:
# − M S + M S = 0

20

CHAPTER 2 QUADRATIC FUNCTIONS

EXAMPLE 3

Form a quadratic equation where the roots are 5 and -7.
= 5 = −7
Sum of roots = 5 + (−7) = −2
Product of roots = 5 × (−7) = −35
# − M S + M S = 0
# − (−2) + (−35) = 0
# + 4 − 35 = 0

EXAMPLE 4

Assume a and b are the roots of the quadratic equation 2 # − 3 − 5 = 0,
form a quadratic equation with the following roots:

(a) 2 , 2

2 # − 3 − 5 = 0 where = 2, = −3 = −5

+ = − = 3
2

= = − 5
2

Sum of roots: Product of roots:

2 + 2 = 2M + S (2 )M2 S = 4

= 2 87#A = 4 8− Z#A
=3 = −10

Thus, the quadratic equation with the roots 2a and 2b is
# − 3 − 10 = 0.

21

CHAPTER 2 QUADRATIC FUNCTIONS

2.1.4 How to solve quadratic inequalities?

• By using graph sketching method
• Number line method
• Table method

EXAMPLE 5

Find the range of values of for the following quadratic inequalities
# − 11 + 24 ≥ 0 by using graph sketching, number line and table method.
Graph sketching method:
# − 11 + 24 ≥ 0
( − 8)( − 3) ≥ 0
When ( − 8)( − 3) = 0, = 8 = 3
Thus, the graph will intersect the -axis at the point = 8 = 3. Since
( − 8)( − 3) ≥ 0, hence the values of is determined on the graph above
the -axis. Therefore, the range of values of is ≤ 3 ≥ 8.


= ( − 8)( − 3)

++


38
-

22

CHAPTER 2 QUADRATIC FUNCTIONS

Number line method: Test point 5: Test point 9:
Test point 2: (5 − 8)(5 − 3) ≤ 0 (9 − 8)(9 − 3) ≥ 0
(2 − 8)(2 − 3) ≥ 0

+- +

≤ 3 3 3 ≤ ≤ 8 8 ≥ 8

Since ( − 8)( − 3) ≥ 0, then the range of values of is determined on the
positive part of the number line. Hence, the range of values of is

≤ 3 ≥ 8.

Table method: ≤ 3 Range of ≥ 8
− 3 ≤ ≤ 8 +
( − 3) − +
( − 8) + +
( − 8)( − 3) + −
−

Since ( − 8)( − 3) ≥ 0, then the range of values of is determined on
the positive part of the number line. Hence, the range of values of is

≤ 3 ≥ 8.

23

CHAPTER 2 QUADRATIC FUNCTIONS

2.2 Types of Roots of Quadratic Equations

2.2.1 What are the types of roots of quadratic equations and the value of
discriminant?

What is a discriminant? A discriminant is the part under the square root in the quadratic formula

# − 4 . The discriminant will tell us whether there are two solutions, one solution or no solution in
the quadratic equation.

= − ± I # − 4 discriminant
2

In general:

• # − 4 > 0, the equation has two different solutions or real roots.
• # − 4 < 0, the equation has no real roots.
• # − 4 = 0, the equation has one solution.

EXAMPLE 6

Determine the type of roots for the quadratic equation 6 # + 10 − 1 = 0.
= 6, = 10 = −1
# − 4 = (10)# − 4(6)(−1)

= 124
# − 4 > 0
Thus, the quadratic equation 6 # + 10 − 1 = 0 has two different real roots.

2.2.2 How to solve the problems involving types of roots of quadratic
equations?

The discriminant besides that telling us the types of roots of the quadratic equation, it can be
used to:

• Find the value of a variable in the quadratic equation
• Derive a relation

24

CHAPTER 2 QUADRATIC FUNCTIONS

EXAMPLE 7

Find the value of a variable in the quadratic equation.

The quadratic equation 4 # + 2 + 9 = , where k is a constant, it has two
real roots. Find the possible values for k.

4 # + 2 + 9 =

4 # + (2 − ) + 9 = 0

= 4 , = 2 − , = 9

Since it has two real roots, we use # − 4 > 0.

# − 4 > 0 Has two real roots.

(2 − )# − 4(4)(9) > 0

4 − 8 + # − 144 > 0

# − 8 − 140 > 0

( − 14)( + 10) > 0

= 14 = −10

EXAMPLE 8

Derive a relation.
The quadratic equation # − 6 + 11 = 0 has one real root, express in
terms of .
# − 6 + 11 = 0
= 1 , = −6 , = 11
Since it has only one real root, we use # − 4 = 0

# − 4 = 0
(−6 )# − 4(1)(11 ) = 0

25

CHAPTER 2 QUADRATIC FUNCTIONS

36 # − 44 = 0
36 # = 44

# = bb
c

= ±Jbcb

2.3 Quadratic Functions

2.3.1 What are the effect of changes on the shape and position of the quadratic
equation ( ) = # + + when the values of , change?

EXAMPLE 9

The diagram shows the sketch of graph for ( ) = 2 # + 4 + 5 where

= 2, = 4 = 5. Resketch the shape and position of the graph when
the following values changes.

( )
( ) = 2 # + 4 + 5

5
3

1

26

CHAPTER 2 QUADRATIC FUNCTIONS

(a) The values of become 1 and 6
( )
( ) = 2 # + 4 + 5
( ) = 6 # + 4 + 5
5 ( ) = # + 4 + 5
3


1

(b) The values of become 3 and 7
( )
( ) = 2 # + 4 + 7
( ) = 2 # + 4 + 5
( ) = 2 # + 4 + 3
5
3
1

27

CHAPTER 2 QUADRATIC FUNCTIONS

2.3.2 What is the relationship between the graph of quadratic function and type
of roots?

# − 4 Types of roots and position > 0 < 0
# − 4 > 0 of graph

• Two real roots
• Two intersection b b

points on -axis

a a

# − 4 = 0 • One real root = b

• One intersection
point on -axis

= b

# − 4 < 0 • No real root

• Does not intersect on
-axis



EXAMPLE 10

Determine types of roots of quadratic function ( ) = 6 # + 10 − 1 when
( ) = 0. Hence, sketch the graph generally.
( ) = 6 # + 10 − 1
= 6, = 10, = −1
# − 4 = 10# − 4(6)(−1)

= 124 (> 0)
Thus, the quadratic function ( ) = 6 # + 10 − 1 has two real roots and it
has to intersect two points on -axis. Since > 0, the graph of ( ) is a
parabola which passes through the minimum point.



28

CHAPTER 2 QUADRATIC FUNCTIONS

EXAMPLE 11

Find the values of when the -axis is tangent to the graph of a quadratic
function ( ) = ( + 1) # + 4( − 2) + 2 .
Tangent to the graph = One real root
Since it has only one real root, we use # − 4 = 0.
= + 1, = 4 − 8, = 2
# − 4 = 0
(4 − 8)# − 4( + 1)(2 ) = 0
16 # − 64 + 64 − 8 # − 8 = 0
8 # − 72 + 64 = 0
# − 9 + 8 = 0
( − 1)( − 8) = 0
= 1 = 8

2.3.3 What is the relation exists between the vertex form of a quadratic function
( ) = ( − ℎ)# + with the other form of quadratic functions (general form

and intercept form)?

What is a vertex form? A vertex form is a quadratic function given by ( ) = ( − ℎ)# + , where
, ℎ are constants. (ℎ , ) is the vertex and it is symmetrical about the -axis if = ℎ.

There are two form of quadratic functions:

• General form: ( ) = # + + where , are constants. A vertex is at point

,− K , 8− #KLA- and it is said to be symmetric about the -axis when = − #KL.
#L

• Intercept form: ( ) = ( − )( − ) where , are constants. are the

roots or we called it as -intercepts for ( ). A vertex is at the point ,fg#h , 8fg#hA- and it is
said to be symmetric about the -axis when = fg#h.

29

CHAPTER 2 QUADRATIC FUNCTIONS

Completing Vertex Form Expansion
the Square ( ) = ( − ℎ)# +
Factorisation
General Form or Formula
( ) = # + +

Expansion

Intercept Form
( ) = ( − )( − )

EXAMPLE 12

Express quadratic function ( ) = 2 # + 9 + 10 in the intercept form,
( ) = ( − )( − ) where , are constants and > . Hence,
state the values of , .

( ) = 2 # + 9 + 10 General form
( ) = (2 + 5)( + 2)

( ) = 2( + 25)( + 2) Intercept form

= 2, = − 5 = −2
2

30

CHAPTER 2 QUADRATIC FUNCTIONS

EXAMPLE 13

Express quadratic function ( ) = 2 # + 9 + 10 in the form of
( ) = ( − ℎ)# + where , ℎ are constants. Hence determine the
value of , ℎ .

( ) = 2 # + 9 + 10

= 2 j # + 9 + 5k
2

= 2 ⎡ + 9 + +,292-.# − +,292-.# + ⎤
⎢ # 2 5⎥
⎢
⎥
⎣⎦

= 2 rj + 94k# − 116s

= 2 j + 49k# − 1
8

Thus = 2, ℎ = − c = − b
t u

31

CHAPTER 2 QUADRATIC FUNCTIONS

2.3.4 What are the effect of changes of , ℎ on the shape and position of
graph for ( ) = ( − ℎ)# + ?

EXAMPLE 14

The diagram shows the sketch of graph for ( ) = 2( + 2)# + 3 where =
2, ℎ = −2 = 3. Resketch the shape and position of the graph when the
following values changes.

( )
( ) = 2( + 2)# + 3

11


3
−2

(a) The values of change to b and 6.
#

( ) = 6( + 2)# + 3 ( )
( ) = 2( + 2)# + 3

( ) = 1 ( + 2)# + 3
2



32

CHAPTER 2 QUADRATIC FUNCTIONS

(b) The values of ℎ change to -6 and 4.

( ) = 2( + 6)# + 3 ( )

( ) = 2( + 2)# + 3

( ) = 2( − 4)# + 3


−6 4

(c) The values of k change to 1 and 8.

( )
( ) = 2( + 2)# + 3
( ) = 2( + 2)# + 3
16
( ) = 2( + 2)# + 3

9
8

1


33

CHAPTER 2 QUADRATIC FUNCTIONS

2.3.5 How to sketch the graphs of quadratic functions?

Determine the shape
of graph by

identifying the
value of .

Determine the
position of graph by
finding the value of

discriminant.

Determine the
vertex.

Determine the
intersection point on

the -axis by
solving ( ) = 0.

Determine the -
intercept by solving

(0).

Plot the points on
Cartesian plane and

draw a parabola
which is

symmetrical at the
vertical line passing
through the vertex.

34

CHAPTER 2 QUADRATIC FUNCTIONS

EXAMPLE 15

Sketch the graph of quadratic function ( ) = −3 # + + 1.

Since < 0, so ( ) has a maximum point.

# − 4 = (1)# − 4(−3)(1)

= 13 (> 0)

Since # − 4 > 0, the ( ) has two real roots which it intersects the

-axis at two points.

( ) = −3 # + + 1

= −3 8 # − @ − b7A
7

= −3 r # − @ + j8v#xwAk# − j8v#xwAk# − 7bs
7

= −3 y8 − zbA# − b7z7{

= −3 8 − zbA# + b7
b#

Hence, the maximum point is 8bz , bb7#A and the axis of symmetry is = zb.

( ) = 0

−3 # + + 1 = 0

8 − bvz√b7A 8 − bgz√b7A = 0

= bv√b7 ≈ −0.434 = bg√b7 ≈ 0.768
z z

Thus, the intersection points on the -axis are = −0.434 and = 0.768.

(0) = −3(0)# + 0 + 1 = 1

The graph intersect at -axis at (0,1).

35

CHAPTER 2 QUADRATIC FUNCTIONS

( ) 1
6
=

1

−0.434 0.768

( ) = −3 # + + 1

2.3.6 How to solve problems of quadratic functions?

EXAMPLE 16

Alvin dives into the pool at a distance of 6 meters from the surface of pool.
The height of the Alvin is given by ℎ( ) = 4 + 8 − 5 # where ℎ is his
height in meters and is the time in seconds.

ℎ( ) = 4 + 8 − 5 # where = −5, = 8 = 4

(a) Find the maximum height achieved by Alvin.

= −
2
8
= − 2(−5)

= 0.8

Then substitute = 0.8 into ℎ( )

ℎ(0.8) = 4 + 8(0.8) − 5(0.8)#

= 7.2

Thus, the maximum height reached by Alvin is 7.2 meters at 0.8
seconds.

36

CHAPTER 2 QUADRATIC FUNCTIONS
(b) Calculate the time for Alvin to reach the surface of the pool.
ℎ( ) = 0
−5 # + 8 + 4 = 0
5 # − 8 − 4 = 0
(5 + 2)( − 2) = 0
= 2 = − Z# (ignored)
Thus, the time for Alvin to reach the surface of pool is 2 seconds.

37

CHAPTER 2 QUADRATIC FUNCTIONS

Summative Exercises

2.1 Quadratic Equations and Inequalities

1. Solve the quadratic equation 3 ( − 5) = 2 − 1 without using calculator and give your
answer in three decimal places.

2. Form an equation with roots 5 and 9.
3. If are the roots of the quadratic equation 2( − 5)# = 4( + 7), form an equation

with the roots of 2 2 .
4. Given that the roots of the quadratic equation 2 # + ( + 1) + = 2 are -3 and #b. Find

the value of .
5. Find the range of values of for the quadratic inequalities # − 4 + 3 > 0.

2.2 Types of Roots of Quadratic Equations

1. Determine the types of roots for the following quadratic equation:
(a) ( − 2) = 5
(b) ( + 5) = 2 − 14

2. The quadratic equation # + 2ℎ + 4 = where ℎ is a constant, has one real root. Find all
the possible values for ℎ.

3. Given the equation # + 5 = 4 has only one real root, express in terms of .
4. The quadratic equation # + + = 0 where are positive integers, has a

discriminant of 16 and − = −4. Find the possible value for .

2.3 Quadratic Functions

1. Given the quadratic function ( ) = − # + + 6 where = −1, = 1 = 6 .

Sketch the graph generally when the values changes.

(a) The value of changes to − b
#

(b) The value of b changes to −1

(c) The value of changes to 1

2. Determine the types of roots for the quadratic function ( ) = 2 # + − 5 when ( ) =

0. Thus, sketch the graph generally at the -axis.

3. Given the quadratic function ( ) = # + 4 − 6 where is a constant, has no real

roots, find the range of values for .

4. Find the range of values of if the given quadratic function has two real roots,

( ) = 5 # − ( + 4) − 2.

38

CHAPTER 3 SYSTEMS OF EQUATIONS

Chapter 3 Systems of Equations

3.1 Systems of Linear Equations in Three Variables

3.1.1 What is a system of linear equations?

A system of linear equation is the existence of two or more linear equations which contain the same
set of variables.

3.1.2 The General Form of a linear equation

A linear equation in two variables can be written in the form Ax + By = C, where A, B and C are
constant.
The General Form of a linear equation in three variables is
Ax + By + Cz = D

Note that:  A, B, C and D are constants
 A, B and C are not equal to zero
 D can be zero or non-zero

3.1.3 Determining systems of linear equation in three variables

A linear equation is an equation where the power of the variable is 1. If the power of any variables in
the system of equation is more than 1, then it is NOT a system of linear equation.

EXAMPLE 1

Describe whether the following equations are systems of linear equations in three variables or not.
(a) x + 2y + 3z = 29

3x + 4y + z3 = 61
2x – 3y – z = –12
Solution: No, a linear equation is an equation where the power of the variable is 1, however there is an
equation in which the highest power of the variable is 3.

(b) 3a + b – c = 4
7b + 2c – 4a = 41
2(a + 6b) – 4c = 36

Solution: Yes, a linear equation is an equation where the power of the variable is 1. Since, all three
equations have three variables with power 1, therefore it is a system of linear equations in three
variables.

39

CHAPTER 3 SYSTEMS OF EQUATIONS

3.1.4 The Graphing of System of linear equations in Three Variables

In systems of linear equation in three variables, the equations are represented as three-dimensional
planes in Cartesian space. Each plane might be orientated at any angle.

A system of linear equations in three variables has three axes: x-axis, y-axis and z-axis. The three
linear equations will form a plane on each axis, therefore a 3D plane is formed.

3.1.4.1 Why each linear equation in three variables will form a plane on each axis?

Let’s set the coefficients of all but one of the variables to zero by

using the general form of a linear equation in three variables: Ax + By + Cz = D z=3
Step 1: Begin by setting A and B to zero and C to one.

0x + 0y + 1z = D

This eliminates all the variables except z which becomes z = D

Step 2: Since D is a constant, if we set D to 3. We obtain z = 3

Since both x and y are free to take any value, the graph of this equation consists of every point in 3D
space where z=3 while x-coordinate and y-coordinate are any real numbers.

The graph of this equation is therefore a horizontal plane three units above the origin.

Step 3: Repeat step 1 by setting A and C to zero and B to one.

0x + 1y + 0z = D

This eliminates all the variables except y which becomes y = D

Step 4: Repeat step 2. Since D is a constant, if we set D to 2. We obtain y = 3

Since both x and z are free to take any value, the graph of this equation consists of

every point in 3D space where y=2 while x-coordinate and z-coordinate are any real numbers. y=2
The graph of this equation is therefore a vertical plane two units to the right of the origin.

The same argument can be applied by setting all variables with any value.
Depending upon the values of A, B, C and D, the plane may lie in any position and orientation. Each
plane can be positioned anywhere in space relative to other planes.
Since each plane graphically represents the set of solutions to one to three equation, the points where
all three planes simultaneously intersect correspond to the solutions which simultaneously satisfy all
three equations.
These points therefore correspond to the solutions of the system.

40

CHAPTER 3 SYSTEMS OF EQUATIONS
To illustrate, below shows a system of linear equation in three variables.
2x + y + z = 20
x + y + 2z = 22
3x + 2y + 2z = 35
These equations formed a system of linear equations.
By graphing, each equation can be displayed in a 3D plane.
Therefore, three planes are formed.

3.1.5 Three Cases when Solving a System of linear equations

There are three possible results when solving a system of linear equations in three variables.
(a) The 3 planes intersect at only one point. The system is consistent and has only one solution.
(b) The 3 planes intersect along a line or a plane. The system is consistent and infinitely many
solutions.
(c) The 3 planes do not intersect or intersect with no common points. The system is inconsistent
and no solution.

Hence, the system of linear equation in three variables can have a single unique solution, no solution
or infinite number of solutions depending on the way in which the three planes are oriented.
Note that:  When there is no solution the system is called "inconsistent".

 One or infinitely many solutions the system is called "consistent”

3.1.6 Methods to solve the system of linear equations

• Substitution Method
• Elimination Method

41

CHAPTER 3 SYSTEMS OF EQUATIONS

EXAMPLE 2

Solve the following system of linear equations using the substitution method.
2x + y + z = 20
x + y + 2z = 22
3x + 2y + 2z = 35

Solution:

2x + y + z = 20 … ○1
… ○2
x + y + 2z = 22 … ○3
… ○4
3x + 2y + 2z = 35

From ○1 , z = 20 – 2x – y

Substitute ○4 into ○2

x + y + 2(20 – 2x – y) = 22

x + y + 40 – 4x – 2y = 22

–3x – y = –18 … ○5
y = –3x + 18

3x + 2(–3x + 18) + 2[20 – 2x – (-3x+18)] = 35 Substitute equation ○4
and ○5 into equation ○3

3x – 6x + 36 + 2(2 + x) = 35

x=5 Substitute x = 5
y = –3(5) + 18 into equation ○5

=3

z = 20 –2(5) – 3 Substitute x = 5 and
y = 5 into equation ○4

=7
Thus, x = 5, y = 3 and z = 7 are the solutions to this system of linear equation.

42

CHAPTER 3 SYSTEMS OF EQUATIONS

EXAMPLE 3

Solve the following system of linear equations using the elimination method.
x –2y + z = –8
x + 3y – 2z = 41
3x + 2y + 2z = 214

Solution:

Choose any two equations.

x – 2y + z = –8 … ○1

x + 3y – 2z = 41 … ○2

Since the coefficient of x in ○1 and ○2 are equal

Eliminate the variable x by subtracting ○1 from ○2

○2 –○1 : 5y – 3z = 49 … ○4

Choose another two sets of equation

5y – 3z = 49 … ○4

–8y + z = –238 … ○6

Multiple equation ○6 with 3 so that the coefficient of x is equal to ○4

○6 ×3: –24y + 3z = –714 ... ○7

Eliminate the variable y by adding ○7 with ○4

○4 +○7 : –19y = –665

y = 35

5(35) – 3z = 49

–3z = –126

z = 42

x – 2(35) + 42 = –8

x = 20

Thus, x = 20, y = 35 and z = 42 are the solutions to this system of linear equations.

43

EXAMPLE 4

Solve the following system of linear equations.
x – 2y – 4z = 3
4x – 8y – 16z = –12
x – 4z = 3

Solution:

x – 2y – 4z = 3 … ○1

4x – 8y – 16z = –12 … ○2

x – 4z = 3 … ○3

Since equation ○3 only has two variables, elimination method is used to eliminate the variable x in

equations ○1 and ○2 . … ○4 Multiple ○1 with 4 so that the
○1 × 4: 4x – 8y – 16 z = 12
○4 –○2 : 0x + 0y + 0z = 24 coefficient of x are equal in
equation ○1 and ○2

0 = 24

Hence, we obtain 0 = 24. This is a contradiction as 0 ≠ 24. Therefore, the system of linear equation has

no solution.

EXAMPLE 5

Solve the following system of linear equations.
3x + 2y + z = 3
2x + y + z = 0
6x + 2y + 4z = –6

Solution: … ○1
3x + 2y + z = 3

2x + y + z = 0 … ○2

6x + 2y + 4z = –6 … ○3

○1 ×2: 6x + 4y + 2z = 6 … ○4 Multiple ○1 with 2
○4 –○3 : 2y – 2z = 12 … ○5 Multiple ○2 with 3
○2 ×3: 6x + 3y + 3z = 0 … ○6

○6–○3 : y – z = 6 … ○7 Multiple ○7 with 2
○7 ×2: 2y – 2z = 12 … ○8

○8 –○5 : 0y + 0z = 0

0=0

Since 0=0, the system of linear equations has infinite number of solutions.

44

3.2 Simultaneous Equations Involving One Linear Equation
and One Non-Linear Equation

3.2.1 What is non-linear equation?

A non-linear equation is an equation that does not form a straight line, but a curve. A non-linear
equation has the variable with power more than or equal to two.

3.2.2 Difference between linear equations and non-linear equations

Comparison between linear equations and non-linear equations:

Linear Equations Non-Linear Equations
The graph forms a straight line It does not form a straight line, but forms a curve
It has only one degree It has more than or equal to two degree
(The power of variable = 1) (The power of variable ≥ 2)
All these equations form a straight line in XY It forms a curve and if the value of the degree
plane. These lines can be extended to any increases, the curvature of the graph increases.
direction but in a straight line form.
Examples Examples:
• x2 + y2 = 1
• 6y + 2x + 8 = 0 • x2 + 12xy + y2 = 0
• 5y = 9x • x2 + x + 4 = 0
• 4x + 8 = 9y

3.2.3 Methods to solve these Simultaneous Equations

The methods to solve the simultaneous equations involving one linear equation and one non-linear
equation are

• Solving by Substitution
• Solving by elimination
• Graphical representation method

45

EXAMPLE 6

Solve the following simultaneous equations using substitution method.
y–2=x
y = x2

Solution:

Let y–2=x … ○1

y = x2 … ○2

From ○1 : y=x+2 … ○3

x + 2 = x2 Compare equation
x2 –x – 2 = 0 ○2 and ○3
(x – 2)(x + 1) = 0

x = 2 or x = –1

y=2+2 or y = –1 + 2

=4 =1

Thus, the solutions to these simultaneous equations are x= 2, y = 4 and x = –1, y = 1.

EXAMPLE 7

Solve the following simultaneous equations using elimination method.

x + 2y = 3

x² + 3xy = 10

Solution: … ○1
x + 2y = 3

x² + 3xy = 10 … ○2

Multiple ○1 with –3x

○1 ×–3x: –3x2 – 6xy = –9x … ○3

○2 × 2: 2x2 + 6xy = 20 … ○4 Multiple equation ○2 with 2

○3 + ○4 : –x2 = –9x + 20

x2 – 9x + 20 = 0

(x – 5)(x – 4) = 0

x = 5 or x = 4

5 + 2y = 3 or 4 + 2y = 3 Substitute x= 5 and

y = –1 y = – 1 x = 4 into equation ○1
2

When x = 5, y = –1

When x = 4, y = – 1
2

Thus, the solutions to these simultaneous equations are x = 5, y = –1 and x = 4, y = – 1 .
2

46