Page 2 - Spotlight A+ Form 4 & 5 Mathematics KSSM

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CHAPTER
1 Quadratic Functions and
Equations in One Variable
SMART concept map
SMART SCOPE • Identify and describe the characteristics of quadratic expressions in Page 3
Important Learning Standards
one variable.
• Recognise quadratic function as many-to-one relation, hence, The entire content of the
describe the characteristics of quadratic functions. 3 chapter is summarised in the
Contains Learning Standards 1.1 Quadratic • Investigate and make generalisation about the effect of changing the 4
values of a, b and c on graphs of quadratic functions, f (x) = ax 2 + bx + c.
Functions and
Equations • Form quadratic functions based on situations, and hence relate to 5 form of a concept map.
the quadratic equations.
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(LS) that need to be achieved • Explain the meaning of roots of a quadratic equation. 5
• Determine the roots of a quadratic equation by factorisation method. 6
in each chapter. • Sketch graphs of quadratic functions. 6
• Solve problems involving quadratic equations. 7 1 CHAP.
Concept 4 Form
2
Words
• Effect of change/ Kesan perubahan
• Horizontal line test/ Ujian garis mengufuk One variable, x Power of x is a whole number Highest power of x is 2
• Many-to-one relation/ Hubungan banyak kepada satu
• Maximum point/ Titik maksimum
• Method of factorisation/ Kaedah pemfaktoran
• Minimum point/ Titik minimum Many-to-one relations Characteristics of quadratic
Form • Quadratic equation/ Persamaan kuadratik expressions ax 2 + bx + c Life situations
• Quadratic function/ Fungsi kuadratik
4 Mathematics Chapter 1 Quadratic Functions and Equations in One Variable
• Rate of change/ Kadar perubahan are identify based on
• Root/ Punca Determine the roots of a quadratic equation
CHAP. Example 9 • Real root/ Punca nyata identify Quadratic Functions and Equations form Quadratic equation ax 2 + bx + c = 0
1 • Variable/ Pemboleh ubah Quadratic functions in One Variable Mathematics Chapter 1 Quadratic Functions and Equations in One Variable
by factorisation method
Determine whether each of the following values
of x is the root of the quadratic equation Solve quadratic equations by the method of describe sketch meaning determine
x 2 + x – 6 = 0. factorisation.
(a) x = 1 (b) x = 2 (c) x = –3 1 Write the quadratic equation in the form
Solution: ax 2 + bx + c = 0. Characteristics based on Quadratic graphs f (x) = ax 2 + bx + c Roots of quadratic equations
(a) When x = 1, x 2 + x – 6 = 1 2 + 1 – 6 2 Factorise ax 2 + bx + c = 0 in the form
(mx + p)(nx + q) = 0.
= –4 ≠ 0 3 State mx + p = 0 or nx + q = 0. 1
∴ x = 1 is not the root of the equation investigate the effect of
x 2 + x – 6 = 0. 4 Solve the two linear equations in 3 to obtain Curved- Axis of symmetry of graph change of a, b and c
(b) When x = 2, x 2 + x – 6 = 2 2 + 2 – 6 x = – p —– or x = – q —–. n shape is parallel to the y-axis
m
= 0
∴ x = 2 is the root of the equation Example 10 Maximum or c = 0 b = c = 0 b = 0
x 2 + x – 6 = 0. minimum point
(c) When x = –3, x 2 + x – 6 = (–3) 2 + (–3) – 6 Determine the roots of the following quadratic a . 0 a , 0 a , 0 a . 0 a , 0 a . 0
equations by the method of factorisation.
= 9 – 3 – 6 (a) x 2 + 5x = 14 y y y y y y
= 0 (b) (3x + 2)(x – 1) = 3x + 13
∴ x = –3 is the root of the equation O x O x x
x 2 + x – 6 = 0. Solution: x x O x
Try question 9 in Formative Zone 1.1 (a) x 2 + 5x = 14 x +7 +7x O O O
x 2 + 5x – 14 = 0 x –2 –2x
(x + 7)(x – 2) = 0 x 2 –14 +5x
BRILLIANT Tips x + 7 = 0 or or x – 2 = 0
x = –7
x = 2
y (b) (3x + 2)(x – 1) = 3x + 13
y = x 2 + x – 6 The graph y = x 2 + x – 6 3x 2 – 3x + 2x – 2 = 3x + 13
3x 2 – 4x – 15 = 0
cuts the x-axis at x = –3 (3x + 5)(x – 3) = 0 3x +5 +5x
–3 O 2 x and x = 2. Therefore, the x –3 –9x
roots of the quadratic 3x 2 –15 –4x
equation x 2 + x – 6 = 0 are 3x + 5 = 0 or x – 3 = 0
the x-intercepts of the x = – 5 — or x = 3 Form
graph y = x 2 + x – 6.
brilliant tips Sketch graphs of quadratic functions Try question 10 in Formative Zone 1.1 1.1 Quadratic Functions and (c) The expression 6t 2 + pt – 9 contains two CHAP.
4
3
Mathematics

Chapter 1 Quadratic Functions and Equations in One Variable
Graphs of quadratic functions of the form Equations variables p and t. Therefore, 6t 2 + pt – 9 is not 1
a quadratic expression in one variable.
1 —
Identify and describe the characteristics of (d) The expression y – 7y 2 contains one variable 1 —
y = a(x + m) 2
y = a(px + m)(qx + n)
y = ax 2 + bx
y = ax 2 + c
Useful tips to help (a) a . 0 (a) a . 0 (a) a . 0 quadratic expressions in one variable y. However, the power of y in the term 7y 2 is
(a) a . 0
1 —
not a whole number. Therefore, y – 7y 2 is not
y
y y y 1. Quadratic expression in one variable is an a quadratic expression in one variable.
algebraic expression of the form ax 2 + bx + c, a, b
and c are constants, a ≠ 0 and x is a variable.
students solve c x x 2. Characteristics of quadratic expressions in one Try question 2 in Formative Zone 1.1
x
m
O
n
O x O b – – a O –m variable: – –– p – –– q
(b) a , 0
problems in the (b) a , 0 y (b) a , 0 y (b) a , 0 y • Expressions contain only one variable. Spotlight portal
• The power of the variable is a whole number.
y
c O –m x • The highest power of the variable is 2. Quadratic equations in one variable
related subtopics. O x O b – – a x BRILLIANT – –– q n Tips https://bit.ly/3d1ZKeM
x
m O

– –– p
The variable x in quadratic expressions can also be
represented by other alphabet letters.
6 1.1.5 1.1.6 1.1.7 BRILLIANT Tips Scan the QR code to
If p represents a constant, then 6t 2 + pt – 9 is a
Example 1 quadratic expression in one variable, t.
4a 2 + b + 3 r 2 — – 2r –h 2 + 8h – 2 3t 2 + 5 — t browse website or video
2
Identify quadratic expressions in one variable Recognise quadratic function as many-
from the list of expressions above. Form to-one relation, hence, describe the
Chapter 2 Number Bases Mathematics 4 characteristics of quadratic functions related to the subtopics
Solution:
r 2
Example 12 2. Numbers in base two can be converted to 1. Quadratic function is a many-to-one relation. learned.
— – 2r, –h 2 + 8h – 2
2
numbers in base eight and vice versa based on
2. Characteristics of quadratic functions
calculator Convert the table above. Number in CHAP. • The graph has a curved shape.
Try question 1 in Formative Zone 1.1
(a) 1011101 2 to a number in base eight,
(b) 53 8 to a number in base two.
• It has a maximum point or a minimum point.
Example 2
• The axis of symmetry of the graph is parallel
base two
Solution:
(a) 1011101 2 = 1 × 2 6 + 0 × 2 5 + 1 × 2 4 + 1 × 2 3 Divide the digits of Replace each digit of the 2 to the y-axis.
Determine whether each of the following
expressions is a quadratic expression in one
+ 1 × 2 2 + 0 × 2 1 + 1 × 2 0 the number in base number in base eight
variable. Give your justification.
= 64 + 0 + 16 + 8 + 4 + 0 + 1 two into groups of with the corresponding BRILLIANT Tips
(a) 2m 2 – 9m + 5 three digits in base two.
= 93 10 8 93 Remainder three digits from
the right to the left.
(b) 5x 3 – 2x + 10 Ignore any zero in front
Explains how to use a = 135 8 8 8 11 … 5 3 Subsequently, replace two. Graphs of quadratic functions y = ax 2 + bx + c
(c) 6t 2 + pt – 9 of the number in base
the three digits with
1 …
0 … 1 the corresponding 1 —
digit in base eight.
(d) y – 7y 2
scientific calculator in (b) 53 8 = 5 × 8 1 + 3 × 8 0 2 43 Remainder Solution: Number in a . 0 a , 0 Form
= 40 + 3
base eight
Maximum point
= 43 10 2 21 … 1 (a) The expression 2m 2 – 9m + 5 contains one Chapter 1 Quadratic Functions and Equations in One Variable Mathematics 4
variable m and the highest power of m is
= 101011 2 2 2 10 … 1 0 2. Therefore, 2m 2 – 9m + 5 is a quadratic Minimum point CHAP.
Example 14
5 …
mathematics calculations. 2 2 … 1 Perform computations involving addition Sketch the graph for each of the following Sketch the graph for each of the following 1
Example 11
expression in one variable.
and subtraction of numbers in various bases
(b) The expression 5x 3 – 2x + 10 contains one
2 1 … 0 variable x. However, the highest power of x is Axis of symmetry Axis of symmetry quadratic functions. Mark the points where the
quadratic functions.
3. Therefore, 5x 3 – 2x + 10 is not a quadratic
0 … 1 1. Addition of number bases in various bases: (a) y = x 2 – 2 graph cuts the x-axis and the y-axis.
expression in one variable.
Alternative Method (a) Numbers in base two (b) y = –x 2 + 4 (a) y = (x – 1)(x – 3)
+ 0 2 1 2 (b) y = –2x 2 – 11x – 14
(a) 1011101 2 = 001 011 101 2 Solution:
0 2 0 2 1 2 (a) (b) y Solution:
= 1 3 5 8 y y = x 2 – 2
(b) 53 8 = 5 3 8 1.1.1 1 2 1.1.2 10 2 4 y = –x 2 3 + 4 (a) y y = (x – 1)(x – 3)
1 2
= 101 011 2 (b) Numbers in base three x 3
Calculator O O x
+ 0 3 1 3 2 3 –2
(a) Press:  MODE MODE 3 BIN x
0 3 0 3 1 3 2 3 O 1 3
2 1 0 1 1 1 0 1 = OCT Try question 11 in Formative Zone 1.1
1 3 1 3 2 3 10 3 (b) y = –2x 2 – 11x – 14
(b) Press:  MODE MODE 3 OCT
2 3 2 3 10 3 11 3 Example 12 = –(2x 2 + 11x + 14)
2 5 3 = BIN (c) Numbers in base four = –(2x + 7)(x + 2) y
Form Sketch the graph for each of the following
+
Try question 12 in Formative Zone 2.1 4 Mathematics Chapter 2 Number Bases quadratic functions. Mark the points where the 7 – – –2 O x
0 4
3 4
2 4
1 4
graph cuts the x-axis.
0 4 0 4 1 4 2 4 3 4 (a) y = x 2 + 3x (b) y = –2x 2 + 7x 2
–14
2 2
1 4• • • •
• • • •
• • • •
BRILLIANT Tips • • • • • • • • • • • • • • • • • • • • 1 4 2 4 2 4 3 4 10 4 3 4 10 4 11 4 9 × 8 + 3 (b) Place value 2 5 1 2 4 1 2 3 0 (a) Solution: 2 1 1 2 0 y 0 (b) y y = –2x 2 – 11x – 14
× 1
0
• • • •
Digit
• • • •
• • • •
2 4 • • • •
1. Table shows the digits for number in base eight • • • • • • • • • • • • • • • • 3 4 10 4 11 4 12 4 y = x 2 + 3x y = –2x 2 + 7x Try question 14 in Formative Zone 1.1
3 4
that correspond to the three digits for number • • • • • • • • ••• The place value of the underlined digit 0 = 2 3 x
The value of digit 0 = 0 × 2 3
in base two. CHAP. (d) Numbers in base five = 0 x O 7 – 2 Solve problems involving quadratic equations
9 eights and
Number 2 + 0 5 3 ones are 2 5 3 5 4 5 (c) Place value 7 2 7 1 7 0 –3 O
1 5
regrouped as
in base 0 1 2 3 4 5 6 7 0 5 0 5 1 sixty-fours, 2 5 3 5 4 5 Example 15
1 5
eight Digit 4 1 Try question 12 in Formative Zone 1.1
0
3 5
Number 1 5 1 5 1 eights and 3 4 5 10 5 In the diagram, PQRS is a rectangular plot of land.
2 5
ones.
The shaded region that is planted with brinjol has
Example 13
in base 000 001 010 011 100 101 110 111 2 5 2 5 3 5 4 5 10 5 11 5 The place value of the underlined digit 4 = 7 2 an area of 388 m 2 .
two The value of digit 4 = 4 × 7 2
3 5 4 5 10 5 11 5 12 5 = 196 Sketch the graph for each of the following S 30 m R
1 × 64
• • • •
• • • • • • • • • • • • • • • • • • • • 3 5 • • • • 4 5 10 5 11 5 12 5 + 1 × 8 13 5 (d) Place value 6 4 6 3 quadratic functions. Label its maximum point or x m
4 5 • • • •
6 2
6 1
• • • • • • • • • • • • • • • • • • • • • • • • + 3 × 1 Digit 2 0 minimum point. 6 0 N 20 m
• • • •
• • • •
5
1
(a) y = (x + 2) 2 1
(b) y = –2x 2 + 4x – 2
2.1.2 2.1.3 • • • • • • • • ••• 25 The place value of the underlined digit 2 = 6 4 P M
Solution:
The value of digit 2 = 2 × 6 4
75 = 1 × 64 + 1 × 8 + 3 × 1 = 2 592 (a) y = (x + 2) 2 y Determine the value of x. x m Q
Number in base eight to represent 75 is 113 8 . Try question 5 in Formative Zone 2.1 4 Solution:
Try question 4 in Formative Zone 2.1 (–2, 0) O x S 30 m R
tAgging ‘Try question ... 2. The place value for each digit of a number in Diagram shows number base blocks representing O y (1, 0) • x x m N 20 m
Example 6
(b) y = –2x 2 + 4x – 2
base a is a times greater than the place value of
= –2(x 2 – 2x + 1)
the digit on its right-hand side.
= –2(x – 1) 2
For example, the place values for each digit in a number in a certain base. Determine the number –2 (20 – x) m P (30 – x) m M x m Q
and represent it in terms of number value.
the number 32014 5 are shown in the following
in Formative Zone ...’ table. Place value 5 4 5 3 5 2 5 1 5 0 (a) (b) Try question 13 in Formative Zone 1.1 y = –2x 2 + 4x – 2 Area of the shaded region = 388 m 2 example
Digit 3 2 0 1 4
3. The value of a digit in the number x a is
determined by multiplying the digit with the Solution: 1.1.7 1.1.8 7
place value of its corresponding digit.
The tagging is located at the end Example 5 (a) Place value 6 1 6 0 Examples with complete
Digit 1 4
Based on the table of place values, determine
of the example guides the student the value of the underlined digit in each of the 14 6 = 1 × 6 1 + 4 × 6 0 solutions to enhance students’
= 6 + 4
following numbers.
(a) 1502 8 (b) 110010 2 = 10 10
(c) 410 7 (d) 20511 6 (b) Place value 5 2 5 1 5 0
to answer the corresponding Solution: Digit 3 2 2 understanding of the chapters
(a) Place value 8 3 8 2 8 1 8 0 322 5 = 3 × 5 2 + 2 × 5 1 + 2 × 5 0
questions in Formative Zone. Digit 1 5 0 2 = 75 + 10 + 2 learned.
= 87 10
The place value of the underlined digit 5 = 8 2 Try question 6 in Formative Zone 2.1
The value of digit 5 = 5 × 8 2
= 320
22 2.1.1
iv
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