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Stand SW 100
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Drawing the graph of
a quadratic function?
8
7
LoS
y
Drawing quadratic graphs of
the form y = ax2 + bx + c
Equation
of Line of
symmetry
is x = 1
6
5
Example 1.
4
y = x2 - 2x - 8
3
2
x
1
-3
-2
-1
0
-1
-2
-3
-4
1
2
3
4
5
x
-3 -2 -1 0 1
-7
-8
-9
3 4
5
x2
9
4
1
0
1
4
9
16
25
-2x
6
4
2
0
-2
-4
-6
-8
-10
-8
-8
-8
-8
-8
-8
-8
-8
-8
-8
y
7
0 -5 -8 -9 -8 -5
0
7
-5
-6
2
Minimum point
at (1, -9)
Look at graphs of
some trig functions?
The
Trigonometric
Ratios for any
angle
90o
-360 -270
180o
0o
-360
-180 -90
-270 -180
-90
-450o -360o -270o -180o -90o
0
0
0o
90
180
270 360
90
180
270
90o 180o
270o 360o 450o
270o
1
sinx + circle
0o
-1
90o
180o
270o
360
360o

y = f(x)
f(x) = cosx
f(x) = cos2x
f(x) = cos3x
f(x) = cos ½ x
2
1
x
-360
-270
-180
-90
0
-1
-2
90
180
270
360
Introducing addition
of fractions with
different
denominators?
2
3
1
4
+
8
12
=
+
3
12
11
12
Multiples of
3 and 4
3
4
6
8
9
12
12
16
15
20
12 is the LCM
Probability for
Dependent Events
Conditional Probability: Dependent Events
When events are not independent, the outcome of earlier
events affects the outcome of later events. This happens in
situations when the objects selected are not replaced.
Conditional Probability: Dependent Events
A box of chocolates contains twelve chocolates of three different types.
There are 3 strawberry, 4 caramel and 5 milk chocolates in the box. Sam
chooses a chocolate at random and eats it. Jenny then does the same.
Calculate the probability that they both choose a strawberry chocolate.
P(strawberry and strawberry) =
3/12 x
Conditional Probability: Dependent Events
A box of chocolates contains twelve chocolates of three different types.
There are 3 strawberry, 4 caramel and 5 milk chocolates in the box. Sam
chooses a chocolate at random and eats it. Jenny then does the same.
Calculate the probability that they both choose a strawberry chocolate.
P(strawberry and strawberry) =
3/12 x 2/11 = 6/132 (1/22)
Enlarge on object?
Enlargements from a Given Point
Centre of Enlargement
To enlarge the kite by
scale factor x3 from the
point shown.
B
A
Object
C
B/
1. Draw the ray lines
through vertices.
D
C/
A/
Image
No Grid 2
D/
2. Mark off x3 distances
along lines from C of E.
3. Draw and label image.
Do some Loci?
Loci (Dogs and Goats)
Q2
Billy the goat is tethered by a 15m long chain to a tree at A. Nanny the goat is
tethered to the corner of a shed at B by a 12 m rope. Draw the boundary locus
for both goats and shade the region that they can both occupy.
Scale:1cm = 3m
Wall
A
Shed
B
Wall
1. Draw arc of circle of radius 5 cm
2. Draw ¾ circle of radius 4 cm
3. Draw a ¼ circle of radius 1 cm
4. Shade in the required region.
Investigate some
Properties of
Pascal’s Triangle
Pascal’s
Triangle
1
1
1
1
Blaisé Pascal
(1623-1662)
1
1
1
1
1
1
1
1
1. Complete the rest of the triangle.
1
7
8
9
6
2
3
4
5
1
1
6
10 10
Counting/Natural Numbers
1
3
4
1
5
15 20 15
1
6
21 35 35 21
Triangular Numbers
1
1
7
28 56 70 56 28
36 84 126 126 84 36
1
8
9
Tetrahedral Numbers
1
10 45 120 210 252 210 120 45 10
1
Pascal’s
1
66 220 495Triangle
792 924 792 495 220 66 12 1
11 55 165 330 462 462 330 165 55 11
12
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13
1
Pyramid
Numbers
(square
base)
Add the numbers shown
along each of the
shallow diagonals to find
Leonardo
of Pisa
another
well known
- 1250
sequence1180
of numbers.
The Fibonacci Sequence
1
1
1
1
1
2
3
1
1
3
1
2
3
5
1
1
1
The sequence first appears as a
recreational maths problem
about the growth in population
of rabbits in book 3 of his
famous work, Liber – abaci (the
book of the calculator).
10 45 120 210 252 210 120 45 10
1
Fibonacci
1
66 220
495 792 924 792 495 220 66 12 1
Sequence
11 55 165 330 462 462 330 165 55 11
12
13
21 34 55 89 144 233 377
Fibonacci travelled
1 4 6 4 1
extensively throughout
1 5 10 10 5 1
the Middle East and
elsewhere. He strongly
1 6 15 20 15 6 1
recommended that
Europeans adopt the
1 7 21 35 35 21 7 1
Indo-Arabic system of
numerals including the 1 8 28 56 70 56 28 8 1
use of a symbol for
1 9 36 84 126 126 84 36 9 1
zero “zephirum”
1
8
1 13 78 286 715 1287 1716 1716 1287 715 286 78 13
1
National Lottery Jackpot? Row 0
12
7
31
49C
6
39
16
4
9
46
34
2
29
42
5
There are 13 983 816 ways of
choosing 6 balls from a set of
49. So buying a single ticket
means that the probability of a
win is 1/13 983 816
10
28
25
1
13
43
36
24
3
45
15
33
38
30
17
20
11
49
21
49 balls choose 6
Choose 6
35
14
37
19
22
44
40
32
47
8
26
23
6
41
18
27
Row 49
13 983 816
48
The Theorem of
Pythagoras?
A 3rd Pythagorean Triple
In a right-angled triangle,
the square on the
hypotenuse is equal to the
sum of the squares on the
other two sides.
49
625
7, 24, 25
25
7
24
576
72+ 242= 252
49 + 576 = 625
The Theorem of Pythagoras: A Visual Demonstration
In a right-angled triangle,
the square on the
hypotenuse is equal to the
sum of the squares on the
other two sides.
Henry Perigal
(1801 – 1898)
Perigal’s Dissection
Gravestone
Inscription
Draw 2 lines through the centre of the middle square, parallel to the sides of the large square
This divides the middle square into 4 congruent quadrilaterals
These quadrilaterals + small square fit exactly into the large square
Look at one of the 6 proofs
of the Theorem from the
Pythagorean Treasury.
President James Garfield’s Proof(1876)
To prove that a2 + b2 = c2
We first need to show that the angle between
angle x and angle y is a right angle.
•This angle is 90o since x + y = 90o (angle sum of a
triangle) and angles on a straight line add to 180o 
Area of trapezium
Draw line:The boundary shape is a trapezium
= ½ (a + b)(a + b) = ½ (a2 +2ab + b2)
Area of trapezium is also equal to the
areas of the 3 right-angled triangles.
yo
= ½ ab + ½ ab + ½ c2
c
xo
a
b
c
 ½ (a2 +2ab + b2) = ½ ab + ½ ab + ½ c2
 a2 +2ab + b2 = 2ab + c2
yo
b
So
 a2 + b2 = c2
xo
QED
a
Take 1 identical copy of this right-angled triangle and arrange like so.
Sample some material
from the Golden section
presentation.
THE GOLDEN SECTION
Constructing a Golden Rectangle.
1. Construct a square and the
perpendicular bisector of a side
to find its midpoint p.
2. Extend the sides as shown.
M
L
Q
3. Set compass to
length PM and draw
an arc as shown.
1
O
P
LQRO is a Golden Rectangle.
N
R
4. Construct a
perpendicular QR.
THE GOLDEN SECTION
"Geometry has two great treasures: one is the Theorem of
Pythagoras, and the other the division of a line into extreme
and mean ratio; the first we may compare to a measure of
gold, the second we may name a precious jewel."
Johannes Kepler
1571- 1630
Or just simply ride
your bike!
Wheels in Motion
The Cycloid
It’s true! The point at the bottom
of a moving wheel is not moving!
Wheel
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