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KING KHALID
ISLAMIC COLLEGE
OF VICTORIA
Proudly presents…
Mathematical Methods
CONTENTS: (Click on topic you want)
1. Factorisation
2. Graphing
3. Differentiation
4. Integration
Mathematical Methods
CONTENTS: (Click on topic you want)
2. Graphing
Quadratics
Cubics
Quartics
Circular functions
Exponential
Logarithmic
Domain and Range
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MAIN MENU
Factorisation
(1.01)
12x3 + 8x2 – 4x = 4x (3x2 + 2x – 1)
1. Common factors
2. Difference of Perfect Squares
a2 – b2 = (a + b)(a – b)
3. Perfect Squares
a2 + 2ab + b2 = (a + b)2
a2 - 2ab + b2 = (a - b)2
4. Sum and Product
x2 + (a + b)x + ab = (x + a)(x + b)
5. Completion of Perfect Squares
eg
x2 – 6x + 3 can’t be factorised by 1, 2, 3, or 4, so
= x2 – 6x + 9
- 9 + 3 Completion of square
= (x – 3)2
-
(6 )2
= (x – 3 + 6)(x –3 - 6)
Difference of squares
Factorisation

(1.01)
Cubic expressions
1. Sum of Cubes
(a + b)3 = (a + b)(a2 – ab + b2)
2. Difference of Cubes
(a - b)3 = (a - b)(a2 + ab + b2)
Factorisation

(1.01)
Factor Theorem
For a polynomial in x, P(x) ,
if (x + a) is a factor,
then P(-a) = 0
eg Find the factors of x3 + 4x2 – 7x - 10
Try (x - 1): Test P(1) = (1)3 + 4(1)2 – 7(1) – 10 = -12
(x – 1) is not a factor
Try (x + 1): Test P(-1) = (-1)3 + 4(-1)2 – 7(-1) – 10 = 0 (x + 1) is a factor
Try (x – 2): Test P(2) = (2)3 + 4(2)2 – 7(2) – 10 = 0
(x - 2) is a factor
If (x + 1) and (x – 2) are factors then (x + 5) must be the third factor (1 x –2 x 5 = -10)
Remainder Theorem

If P(-a) does not equal zero, the number left
is the remainder of P(x) divided by (x + a)
Division of a polynomial
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CONTENTS
(1.01)
If (x – 3) is a factor of 3x3 – 7x2 – 9x + 9, find the other factor.
3x2 + 2x - 3
)
(x – 3) 3x3 – 7x2 – 9x + 9
3x3 – 9x2
+2x2 - 9x
2x2 – 6x
- 3x + 9
-3x + 9
No remainder
3x3 – 7x2 – 9x + 9 = (x – 3)(3x2 +2x – 3)
Graphing

All graphs follow a standard set of transformation rules:
1. Reflections
Negative before the function, y = - f(x) reflects vertically (top for bottom).
Negative inside the function, y = f(-x) reflects horizontally (left for right).
2. Dilations
Number (>1) before the function, y = a.f(x), dilates (stretches) vertically.
Number (>1) inside the function, y = f(ax), dilates (compresses) horizontally.
3. Translations
Number added to or subtracted from the function, slides up or down.
Number added or subtracted inside the function (inside the brackets),
slides the graph horizontally in the opposite direction.
Parabolas
y=
y
5
x2
3
1
-4
-2
2
-1
-3
-5
4
x
Parabolas
y
- Vertical translation
5
y = x2 + 2
3
1
-4
-2
2
-1
-3
-5
4
x
Parabolas
y
- Vertical translation
5
y = x2 - 3
3
1
-4
-2
2
-1
-3
-5
4
x
Parabolas
y
- Vertical reflection
5
y = - x2
3
1
-4
-2
2
-1
-3
-5
4
x
Parabolas
y
- Vertical translation
- Vertical reflection
5
3
y = 5 - x2
1
-4
-2
2
-1
-3
-5
4
x
Parabolas
y
- Horizontal translation
5
y = (x – 4)2
3
1
-4
-2
2
-1
-3
-5
4
x
Parabolas
y
- Horizontal translation
5
y = (x + 3)2
3
1
-4
-2
2
-1
-3
-5
4
x
Parabolas
y
- Horizontal translation
5
- Vertical reflection
3
y = - (x – 1)2
1
-4
-2
2
-1
-3
-5
4
x
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Parabolas
y
Vertical and horizontal
translation
5
3
y = (x –
2)2
1
-3
Turning point form
-4
-2
2
-1
-3
Turning point at (2, -3)
-5
4
x
Cubics
y
5
y = x3
3
1
Three factors
are the same, so
Point of inflection
occurs at (0, 0)
-4
-2
2
-1
-3
-5
4
x
Cubics
y
5
y = (x – 3)3 + 1
3
Turning point form
1
Three factors
are the same, so
Point of inflection
occurs at (3, 1)
-4
-2
2
-1
-3
-5
4
x
Cubics
y
What is the equation to this curve?
y = (x + 1)(x - 2)(x - 4)
6
2
-4
-2
2
-2
-6
4
x
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CONTENTS
Cubics
y
What is the equation to this curve?
5
3
y = -(x + 3)(x - 1)2
1
or
-4
y=–
x3
-
x2
+ 5x - 3
-2
2
-1
-3
4
x
Quartics
y
What is the equation to this curve?
6
y = x(x + 3)(x - 3)(x –5)
2
or
-4
-2
2
-2
y = x4 – 5x3 - 9x2 + 45x
-6
4
x
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CONTENTS
Quartics
y
What is the equation to this curve?
y = x(x + 3)(x - 3)(x –5)
6
2
or
y = x4 – 5x3 - 9x2 + 45x
-4
-2
2
-2
-6
4
x
TRIGONOMETRIC
functions
(Circular Functions)
Circular functions
y
y = a sin nx
a
a – amplitude
(the height)
0

n
Period, T =
-a
(the length)
2
n
2
n
x
Circular functions
y
y = 3sin2x + 2
5
Amplitude, a = 3
Vertical Translation = +2
x
0
-5
Period, T = 
Circular functions
y
What is
the
equation
to this
curve?
1
y = sin x
0

-1
2
x
Circular functions
y
What is
the
equation
to this
curve?
1
0

y = cos x
-1
2
x
Circular functions
y
What is
the
equation
to this
curve?
5
0
y = tan x

-5
2
x
Circular functions
y
What is
the
equation
to this
curve?
4
0

y = 4 sin x
-4
2
x
Circular functions
What is
the
equation
to this
curve?
y
5
y = 5 sin 2x + 2
0

-5
x
Circular functions
л
2
л-θ
л
θ–л
or
θ+л
θ
Only
Sin
is positive
All
functions are
1 positive
2
3 4 Only
Only
Tan
Cos
is positive
is positive
3л
2
0, 2л
2л - θ
Circular functions
What is
the
equation
to this
curve?
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CONTENTS
y
5
y = 5 sin 2x + 2
0

-5
x
Exponential Equations
y = ax
(1.02)
1
To solve an exponential equation, make the bases the same.
eg 25x = 125
Becomes
52x = 53
Now, equate indices:
2x = 3
and so,
x = 1.5
Exponential Equations
(1.02)
To solve an exponential equation when
you cannot find common bases.
eg
73x = 66
No common base, so
take log10 of both sides: log10 73x = log10 66
and so,
3x.log10 7 = log10 66
By rearrangement,
3x = log10 66
log10 7
and, x = 0.718
Exponential Equations
eg
becomes
or
(1.02)
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CONTENTS
9x - 4  3x + 3 = 0
(32)x + 4(3x) + 3 = 0
(3x)2 + 4(3x) + 3 = 0 Notice the quadratic?
To make solving easier, let 3x = a
 a2 + 4a + 3 = 0
and (a - 1)(a - 3) = 0
 a = 1, 3
Since a = 3x ,
3x = 1 or 3x = 3

x = 0 or 1
LOGARITHMS
Logarithms
The logarithm is the inverse operation of the exponential.
eg
If ax = b then logab = x
So, 53 = 125 is equivalent to log5 125 = 3
If log4 x = 3
then x = 64
(because 43 = x
and  x = 64)
Logarithmic functions
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y
What is
the
equation
to this
curve?
No, think about
it a bit more!

1
(5, 1)
0
1
5
What does the point
(5, 1) tell you?
When y = 1,
x = 5, so…
y = log5 x
-1
x
Domain and Range
Domain means all of the x-values for which the relation exists.
If you substitute an x-value into the equation and you get
an answer for y, then that x-value is part of the domain.
eg for the expression y =
x
,
if x = 4 then y = 4 = 2
so x = 4 is part of the domain.
However, if x = - 4 then y =
4
= undefined
so x = - 4 is not part of the domain.
The domain for y =
x , is x  0 or [0, )
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Domain and Range
Range means all of the y-values for which the relation exists.
If you substitute an x-value into the equation and you get
an answer for y, then that y-value is part of the domain.
eg for the expression y =
x
,
if x = 4 then y = 4 = 2
so y = 2 is part of the range.
Since it is impossible to get a real solution for y =
x
which is negative, y can only have non-negative values.
The range for y =
x , is y  0 or [0, )
Inverse Functions
Gradient
Differentiation is a technique for finding the gradient
of any point on a curved line.
Finding the gradient of a straight line is easy
because the gradient is the same at every part of the line.
y
y2
Gradient of a straight line,
y2  y1
rise
m=
=
x2  x1
run
You only need to know the coordinates of
two points on the line, (x1, y1) and (x2, y2)
(x1,y1)
.
x1
(x2,y2)
.
rise
y1
run
x2
x
Average Gradient
Average gradient of a curved line is just as easy because it is the same
as the gradient of the straight line between the two points
y2  y1
Average Gradient, m =
x2  x1
y
B(x2,y2)
You only need to know the coordinates of two
points on the curve, (x1, y1) and (x2, y2)
A(x1,y1)
x
or (using f(x) instead of y)
Average Gradient, m =
f ( x  h)  f ( x )
( x  h)  x
f ( x  h)  f ( x )
Therefore, m =
h
You only need to know the coordinates of two
points on the curve,
(x, f(x)) and ((x+h), f(x+h))
f(x)
B(x+h,f(x+h))
A(x,f(x))
x
x+h
x
Differentiation
is a technique for finding the gradient
at one point on a curved line.
Gradient of a straight line between two points is given by:
m=
y
x
=
f ( x  h)  f ( x )
where h is the distance between
h
the x coordinates.
However, to find the gradient at one point, the distance, h,
between the x-coordinates has to become zero.
The problem here is that we end
up with zero in the denominator.
y
To get around this we must simplify the
numerator so that we can cancel the h
from the denominator
Mathematically, this is written as:
dy
lim f ( x  h)  f ( x)
m = dx = f’(x) =
h 0
h
A.
x
h
h
h
x
Differentiation
is a technique for finding the gradient
at one point on a curved line.
The wonderful thing about differentiation
is that there is a rule to make things easier:
If y = axn
then dy = naxn-1
dx
If f(x) = axn
and
then f΄(x) =naxn-1
If y = x3 + 5x2 - 3x + 7
dy
Then
= 3x2 + 10x - 3
dx
eg
Gradients
Differentiation is a technique for finding the gradient of a curve.
The derivative is also called the Gradient Function
- it gives the gradient at any point on the curve.
eg. f(x) = x2 - 4x + 7
f΄(x) = 2x – 4
(gradient function)
y
The
The gradient
gradient at
at xx =
= 231
f΄(2)
f΄(3)
2(3)
f΄(1) =
= 2(2)
2(1) –– 44
m==-022
m
x
Finding the coordinates of a point
where the gradient is given
eg What are the coordinates of the point on the curve
y = x2 + 3x – 7 where the gradient is equal to 5?
1. Differentiate
y΄ = 2x + 3
2. Equate to m
2x + 3 = 5
3. Solve for x
2x = 2
x =1
4. Substitute into the original equation
y = (1)2 + 3(1) - 7
y=-3
The gradient of
y = x2 + 3x – 7
is equal to 5
at (1, - 3)
Finding the coordinates of a
turning point (stationary point)
Horizontal turning points have a gradient of zero
eg What are the coordinates of the stationary points
on the curve y = x3 + 3x2 – 9x ?
1. Differentiate
y΄ = 3x2 + 6x – 9
2. Equate to zero
3x2 + 6x – 9 = 0 (turning point)
3. Solve for x
x2 + 2x - 3 = 0
(x + 3)(x – 1) = 0
x = 1, -3
4. Substitute into the original equation
At x = 1, y = (1)3 + 3(1)2 – 9(1)
y=-5
There are turning
points at (1, -5)
and at (-3, 27)
At x = -3, y = (-3)3 + 3(-3)2 – 9(-3)
y = 27
Antidifferentiation
– the indefinite integral
Antidifferentiation “undoes” differentiation.
It turns the derivative back into the original function.

axn
dx =
ax n 1
n 1
+c
The indefinite integral reverses the differentiation
but c must always be added to make up for the constant
which disappears during differentiation.
eg 
(5x3
–
3x2)
dx
5x4
=
4
=
5x4
4
-
3x 3
3
+c
- x3 + c
The Indefinite Integral – yields
yields the
a function
original with
an
function
unknown
when constant,
you have been
c. given the derivative.
eg.The
Findvalue
the equation
thefound
function
whose the
of c canto2be
by knowing
gradient
is
given
by
6x
4x
–
3,
given
that
the
coordinates
of a point
the original
curve.
graph
of the function
passeson
through
the point
(2, 3)
Gradient means
dy
dx
so, the original function, y =
dy
 dx .dx
y = (6x2 - 4x – 3) dx
and hence, y = 2x3 - 2x2 – 3x + c
Now, since this function passes through the point (2, 3),
we can say: 3 = 2(2)3 - 2(2)2 – 3(2) + c (Substitute into x and y)
Now, solving for c, c = 3 – (16 – 8 – 6) = 1
And so, y = 2x3 - 2x2 – 3x + 1
The Indefinite Integral – yields the original
function when you have been given the derivative.
Try this: If f(x) = 3x2 + 8x and f(-3) = 2,
find f(x).
f(x) =  f (x) dx =  f(3x2 + 8x) dx
= x3 + 4x2 + c
If f(-3) = 2,
then (-3)3 + 4(-3)2 + c = 2
c = 2 +27 –36 = - 7
So, f(x) = x3 + 4x2 - 7
Definite Integral
The Definite Integral gives the integral between two limits.
Area
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