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Drawing the graph of a quadratic
function?
This is a short demo that auto-runs.
0 1 2 3 4-1-2
1
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5
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7
8
-1
-2
-3
-4
-5
-6
-7
-8
5
y
-3
-9
x
y = x2 - 2x - 8
x -3 -2 -1 0 1 2 3 4 5
x2
-2x
-8
y
9 4 1 0 1 4 9 16
25
6 4 2 0 -2 -4 -6 -8 -10
-8 -8 -8 -8 -8 -8 -8 -8 -8
7 0 -5 -8 -9 -8 -5 0 7
LoSEquation of Line of symmetry is x = 1
Drawing quadratic graphs of the form y = ax2 + bx + c
Example 1.
Minimum point
at (1, -9)
Look at graphs of some trig functions?
sinx + circle90o 180o
0o 270o
1
-1
The Trigonometric Ratios for any
angle
0 90 180 360270-90-180-270-360
0 90 180 270-90-180-270-360 360
450o0o 90o 180o 270o 360o-90o-180o-270o-360o-450o
360o
0o
90o
180o
270o
x
270 36090-360 180
y = f(x)
0-90-180-270
1
-1
2
-2
f(x) = cosx f(x) = cos2x f(x) = cos3x f(x) = cos ½ x
Introducing addition of fractions with
different denominators?
23
14
+
+
=1112 2015
1612
129
86
43
Multiples of 3 and 4
12 is the LCM
812
Equivalent312
Equivalent
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.
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.
Conditional Probability: Dependent Events
P(strawberry and strawberry) = 3/12 x
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.
Conditional Probability: Dependent Events
P(strawberry and strawberry) = 3/12 x 2/11 = 6/132 (1/22)
Enlarge on object?
D
To enlarge the kite by scale factor x3 from the
point shown.
Centre of Enlargement
ObjectA
B
C
Image
A/
B/
C/
D/
Enlargements from a Given Point
1. Draw the ray lines through vertices.
2. Mark off x3 distances along lines from C of E.
3. Draw and label image.
No Grid 2
Do some Loci?
Q2Loci (Dogs and Goats)
Scale:1cm = 3m
Shed
Wall
Wall
A
B
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.
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.
Investigate some Properties of
Pascal’s Triangle
Pascal’s Triangle
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
55 10 10
6156 15 20
7 72121 35 35
88 285628 56 70
9 36 84 126 126 84 36 9
10 45 210 252 210 120 45120 10
11 55 330 462 462 330 165165 55 11
12 66 495 792 924 792 495220 220 66 12
171613 78 286 715 1287 1716 1287 715 286 78 13
Pascal’s Triangle
1. Complete the rest of the triangle.
Blaisé Pascal (1623-1662)
Counting/Natural Numbers
Triangular Numbers
Tetrahedral Numbers
Pyramid Numbers (square base)
Fibonacci Sequence
1716
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
2
3 3
4 6 4
55 10 10
6156 15 20
7 72121 35 35
88 285628 56 70
9 36 84 126 126 84 36 9
10 45 210 252 210 120 45120 10
11 55 330 462 462 330 165165 55 11
12 66 495 792 924 792 495220 220 66 12
13 78 286 715 1287 1716 1287 715 286 78 13
Add the numbers shown along each of the shallow diagonals to find another well known sequence of numbers.
1 2 3 8 13
21 55 89
1 5
34 144 233 377
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).
Fibonacci travelled extensively throughout the Middle East and elsewhere. He strongly recommended that Europeans adopt the Indo-Arabic system of numerals including the use of a symbol for zero “zephirum”
The Fibonacci Sequence
Leonardo of Pisa 1180 - 1250
8
7
65
49
3
2
1
27
26
2524
23
28
22
21
20
29
17
16
15
1413
18
12 11
10
1937
36
35
34
3338
32
31 30
39
4746
45
44
43
48
42
41
40
49
National Lottery Jackpot? 49 balls choose 6
Choose 6
Row 49 13 983 816
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
49C6
Row 0
The Theorem of Pythagoras?
625
576
49
72+ 242
=
252
49 + 576 = 625
7
24
25
A 3rd Pythagorean Triple
7, 24, 25
In a right-angled triangle, the square on
the hypotenuse is equal to the sum of the
squares on the other two sides.
Perigal’s Dissection
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.
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
Henry Perigal
(1801 – 1898)
Gravestone Inscription
Look at one of the 6 proofs of the Theorem from the Pythagorean
Treasury.
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
Take 1 identical copy of this right-angled triangle and arrange like so.
Area of 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.
= ½ ab + ½ ab + ½ c2
So
½ (a2 +2ab + b2) = ½ ab + ½ ab + ½ c2
a2 +2ab + b2 = 2ab + c2
a2 + b2 = c2 QED
a
b
cxo
yo
a
bc
xo
yo
Draw line:The boundary shape is a trapezium
To prove that a2 + b2 = c2
President James Garfield’s Proof(1876)
Sample some material from the Golden section
presentation.
Constructing a Golden Rectangle.
1. Construct a square and the perpendicular bisector of a side to find its midpoint p.
3. Set compass to length PM and draw an arc as shown.
2. Extend the sides as shown.
L M
4. Construct a perpendicular QR.
O N
Q
THE GOLDEN SECTION
LQRO is a Golden Rectangle.
1
P R
THE GOLDEN SECTION
Johannes Kepler 1571- 1630
"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."
Or just simply ride your bike!
Wheels in Motion
Wheel
The Cycloid
It’s true! The point at the bottom of a moving wheel is not moving!
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Stand SW 100
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