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Unit 3 Notes
Mathematical Methods
Foundational Knowledge
Created by
Triumph Tutoring
1
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2
Polynomials
Basic Polynomials
This section covers concepts from Units 1 and 2 that are considered presumed knowledge for
Unit 3 Mathematical Methods. Without a good prior knowledge of basic Unit 1 and 2 concepts,
it will be difficult to build on skills required in Units 3 and 4.
A polynomial in x, for example, is an expression that consists of terms which have non-negative
integer powers of x only. That is, an expression must only have positive whole number powers
of x to be classed as a polynomial.
1
eg. x4 − 3x2 + 1 is a polynomial whereas x 2 − 2x is not.
The ‘degree’ of a polynomial refers to the highest power of x.
eg. p(x) = x4 + 5x3 + 1, the degree of p(x) is 4.
Example Question
If p(x) = x5 + ax4 + 3x2 + bx − 1, p(1) = 9 and p(2) = 81, find the values of a and b.
Division and Factorisation of Polynomials
Division of polynomials must be completed in order to factorise cubic or higher order graphs.
When one polynomial p(x) is divided by another d(x), the result can be expressed as
r(x)
p(x)
= q(x) +
d(x)
d(x)
where q(x) is the quotient, r(x) is the remainder and d(x) is the divisor.
At this stage, the divisor will always be a simple linear expression. The linear expression is
usually found by using the remainder theorem, or through analysis to find an x-intercept.
eg. using the remainder theorem find a factor of p(x) = x3 + 2x2 − 5x − 6
1. Look at factors of the constant − 6: 1, 2, 3, 6
2. Substitute the factors in until the expression equals 0.
p(1) = (1)3 + 2(1)2 − 5(1) − 6
p(1) = 1 + 2 − 5 − 6
p(1) = −8 6= 0
p(−1) = (−1)3 + 2(−1)2 − 5(−1) − 6
p(−1) = −1 + 2 + 5 − 6
p(−1) = 0
Therefore, x + 1 is a factor.
3
eg. divide x4 − 3x3 + 2x2 − 8 by x + 2
Image sampled from Maths Quest 10
If the remainder is 0 (which can be verified using the remainder theorem), then the new expression for the original function can be represented by: quotient×divisor.
If the quotient is not a linear expression, then it can likely be factorised again (if it is a quadratic
you may use simpler methods to achieve this).
Note: not all functions can be factorised into linear terms. Often times this will be due to a
quadratic factor of the function. If this occurs, you can verify this by using the discriminant.
The discriminant of ax2 + bx + c can be found using the following formula:
∆ = b2 − 4ac
If ∆ < 0, then we know that there are no solutions (or that it cannot be factorised further).
4
Linear Graphs
The gradient of a straight line joining two points is m =
y2 − y1
, also known as rise over run.
x2 − x1
The general equation of a straight line is y = mx+c. m represents the gradient and c represents
the vertical translation (also the y-intercept).
The equation of a straight line passing through the point (x1 , y1 ) and having a gradient of m
is: y − y1 = m(x − x1 ).
TIP: use this equation when evaluating straight lines, instead of y = mx + c.
Parallel lines have the same gradient.
Perpendicular lines, however, have gradients such that m1 × m2 = −1 (or m1 =
−1
).
m2
Domain and Range
The domain of a function is the set of values of x for which the function is defined.
The range of a function is the set of values of y for which the function is defined.
Notation
If the domain or range is unrestricted it can be denoted as R (all real numbers) or (−∞, ∞).
Circular brackets ( or ) are used to denote that the limiting number is not included.
eg. (0, ∞) means ’from zero to infinity’ not including 0
Square brackets [ or ] are used to denote that the limiting number is included.
eg. [0, ∞) means ’from zero to infinity’ including 0
Quadratics
The general form of quadratic functions is y = ax2 + bx + c. For a > 0, the graph has a
minimum value (smile). For a < 0, the graph has a maximum value (mountain). Both of these
types are called turning points.
The y-intercept is c. (Thinking: at the y intercept x=0. If x=0 all x terms equal zero, therefore
only c is left).
The equation of the axis of symmetry and the x-value of the turning point is x =
The x-intercepts are found by solving the equation ax2 + bx + c = 0.
You can factorise and then use the null factor law or use the quadratic formula.
Quadratic formula: x =
−b ±
√
b2 − 4ac
2a
Note: Always check solvability using the discriminant.
−b
2a
5
The discriminant (∆) is the value underneath the square root sign in the quadratic formula.
Discriminant: ∆ = b2 − 4ac.
If ∆ > 0, there are two solutions to the equation and hence two x-intercepts on the graph.
If ∆ = 0, the two solutions are equal and hence there is only one x-intercept on the graph.
That is, the graph has a turning point on the axis.
If ∆ < 0, there are no solutions to the equation and hence no x-intercepts on the graph.
The power form of quadratic functions looks like y = a(x − b)2 + c, where
a = dilation factor. Dilates the graph in the y direction.
b = horizontal translation. Translates the graph horizontally to either the left or right.
c = vertical translation. Translates the graph vertically upwards or downwards.
The turning point of the graph occurs at (b, c)
To change from the general form to power form we need to complete the square.
eg. By completing the square, factorise y = x2 + 10x + 3 into turning point form.
y = x2 + 10x + 3
1. Half the coefficient of x, square the result, add and subtract it like the example.
y = (x2 + 10x + 25) + 3 − 25
2. Factorise the first three expressions.
y = (x + 5)2 − 22
Now that we have it in power form we know that:
The graph is translated 5 units left
The graph is translated 22 units down
The turning point is at (−5, −22)
25 y
20
15
10
5
−15
−10
−5
(−5, −22)
−5
−10
−15
−20
−25
x
5
10
15
6
Cubics
Cubic functions are polynomials of degree 3.
The general form of a cubic function is y = ax3 + bx2 + cx + d.
The power form of a cubic function is very similar to that of a quadratic function: y =
a(x − b)3 + c.
The translations of a cubic graph are the same as a quadratic:
a = dilation factor. Dilates the graph in the y direction.
b = horizontal translation. Translates the graph horizontally to either the left or right.
c = vertical translation. Translates the graph vertically upwards or downwards.
The point of inflection of the graph occurs at (b, c)
For example, the graph of y = 2(x − 3)3 + 5 is the graph of y = x3 translated 3 units right, 5
units up and dilated by a factor of 2 from the y-axis.
25 y
20
15
10
5
−15
−10
−5
−5
−10
−15
−20
−25
(3, 5)
5
x
10
15
7
Exponential and Logarithmic Equations
The index laws
A number in index form has two parts, the base and the index, power, exponent or logarithm.
A number in index form is represented like this:
ax where a is our base and x is known as the index, power or exponent
The index laws are as follows
ax × ay = ax+y
ax ÷ ay = ax−y
(ax )y = axy
a0 = 1 (a 6= 0)
a−x = a1x (a 6= 0)
√
1
a x = x a (x 6= 0)
√
x
y
a y = ax (y 6= 0)
(ab)x = ax bx
x
( ab )x = abx (b 6= 0)
The logarithm laws
Logarithms are an alternative way of expressing a number in index form. If y = ax , then
loga (y) = x.
For example, 23 = 8 can be written as log2 (8) = 3.
The logarithm laws are as follows:
since ax 6= 0, loga (0) is undefined
a0 = 1 → loga (1) = 0
a1 = a → loga (a) = 1
loga (m) + loga (n) = loga (mn)
)
loga (m) − loga (n) = loga ( m
n
p
loga (m ) = p loga (m)
8
Practicing exponential equations
(2x2 y 3 )3 × 3(xy 4 )2
6x4 × 2xy 4
Remove the brackets by multiplying out the indices
(2x2 y 3 )3 × 3(xy 4 )2
23 x6 y 9 × 3x2 y 8
=
6x4 × 2xy 4
6x4 × 2xy 4
Add the indices of x and add the indices of y. Simplify 23 to 8 and
multiply the whole numbers together.
23 x6 y 9 × 3x2 y 8
24x8 y 17
=
6x4 × 2xy 4
12x5 y 4
Subtract the indices of x and y. Divide the whole numbers.
24x8 y 17
= 2x3 y 13
12x5 y 4
eg. simplify
1.
2.
2.
3.
eg. write 32−0.4 in simplest form
1. Write as a fraction with a positive index
1
32−0.4 = 0.4
32
2. Change 0.4 into a fraction and simplify
1
1
= 2
0.4
32
32 5
3. Rewrite using the index laws
1
1
2 = √
5
322
32 5
4. Simplify
1
1
1
√
= 2 =
5
2
4
322
eg. solve 32x − 12 × 3x + 27 = 0 for x
1. Write 32x as (3x )2
(3x )2 − 12 × 3x + 27 = 0
2. Let 3x = a to create a simper quadratic equation to solve
a2 − 12a + 27 = 0 where a = 3x
3. Factorise and solve for a
(a − 3)(a − 9) = 0
a − 3 = 0 or a − 9 = 0
a = 3 or a = 9
4. Substitute back a = 3x and solve for x
3x = 3 or 3x = 9
3x = 31 or 3x = 32
∴ x = 1 or x = 2
9
Practicing logarithmic equations
eg. solve log5 (x − 1) = 2 for x
1. Rewrite using the law ax = y → loga (y) = x
52 = x − 1
2. Solve for x
25 = x − 1
∴ x = 26
eg. solve log10 (x) + log10 (x − 3) = log10 (4) for x
1. Simplify the left hand side using the law loga (m) + loga (n) = loga (mn)
log10 (x(x − 3)) = log10 (4)
2. Equate the logs
x(x − 3) = 4
3. Expand and rearrange the equation
x2 − 3x − 4 = 0
4. Solve the equation
(x − 4)(x + 1) = 0
x = 4 or x = −1
∴ x = 4 (as x > 0)
Euler’s number
Euler’s number (e) is a number that is very important, used in problems involving natural
growth and natural decay. It is very similar to π in that it is irrational and has to be approximated: e = 2.7183.
The number e can be use like any other number. The same index and logarithm laws apply
to it. However, certain calculus rules and calculations can only be performed on indices and
d
(loge (x) = x1 . This cannot be performed with any other
logarithms involving e. For example, dx
base logarithm.
eg. Write 2 loge (x) + 1 = loge (y) with y as the subject
1. Use the laws loga (m)p = p loga (m) and loga (a) = 1 to make each term a log
loge (x2 ) + loge (e) = loge (y)
2. Use the law loga (m) + loga (n) = loga (mn) to create one logarithm on each side
loge (ex2 ) = loge (y)
3. Equate the ’insides’ of the logs
y = exx
10
Unit 1 & 2 Trigonometry Revision
Radians and the Unit Circle
Angles are measured in degrees or radians.
One radian is written as 1c
π c = 180◦
Degrees → Radians: multiply angle by
eg. 60◦ ×
60π
π
π
=
=
180
180
3
Radians → Degrees: multiply angle by
eg.
π
180
180
π
π 180
180π
×
=
= 30◦
6
π
6π
The unit circle is a circle with a radius of one
unit. This circle can be used to define
radians, trigonometric ratios and other relationships.
The unit circle has a circumference of 2π units,
and the angle at the centre of the circle is radians, or 360◦ .
As shown in the first unit circle to the right:
- the horizontal distance is defined as cos(θ)
- the vertical distance is defined as sin(θ)
- the point the diagonal hits on the line ST is
defined as tan(θ)
According to the ’CAST’ notation on the
second unit circle to the right:
- Cos(θ) is positive in the fourth quadrant
- All are positive in the first quadrant
- Sin(θ) is positive in the second quadrant
- Tan(θ) is positive in the third quadrant
T
y
tan(θ)
sin(θ)
θ
x
cos(θ)
S
y
π−θ
θ
S
A
x
T
π+θ
C
2π − θ
11
Exact Values and Symmetry
In a right-angled triangle like the one to the
right, we know that
sin(θ) =
O
H
cos(θ) =
A
H
tan(θ) =
O
A
H
O
θ
A
With this in mind, we can use the following triangles to create a table of values to use.
√
2
30◦
2
1
45◦
√
2
3
60◦
1
1
1
According to the above triangles, the following table is created.
30◦
sin
1
2
cos
√
3
2
tan
√
3
3
45◦
60◦
√
√
3
2
2
2
√
2
2
1
2
1
√
3
a
Try to memorise this table. Practicing questions that involve these values will help cement the
knowledge and really speed up your equation-solving skills.
12
The following two tables define complementary angles (angles which add up to 90◦ ) as part of
the unit circle.
First Quadrant
Second Quadrant
sin( π2 − θ) = cos(θ)
sin( π2 + θ) = cos(θ)
cos( π2 − θ) = sin(θ)
cos( π2 + θ) = − sin(θ)
a
In the first two quadrants, we can see that the complementary angles revolve around
to translate between the quadrants.
Third Quadrant
Fourth Quadrant
sin( 3π
− θ) = − cos(θ)
2
sin( 3π
+ θ) = − cos(θ)
2
cos( 3π
− θ) = − sin(θ)
2
cos( 3π
+ θ) = − sin(θ)
2
a
In the last two quadrants, we can see that the complementary angles revolve around
to translate between the quadrants.
3π
2
eg. find the exact value for sin(120◦ ) without a calculator
Method 1
1. Rewrite the angle in terms of 180◦
120◦ = (180 − 60)◦
2. Identify the ”base angle”
120◦ = (180 − 60)◦
base angle: 60◦
3. Identify the quadrant and whether the value will be positive or negative
2nd Quadrant (S of CAST) - sin(120◦ ) will be positive
4. Reference the√exact values table
sin(60◦ ) = 23
5. Consolidate the information
from steps 3 and 4
√
3
◦
◦
sin(120 ) = sin(60 ) = 2
Method 2
1. Reference the complementary angles table
sin(120◦ ) = sin(90◦ + 30◦ ) = cos(30◦ )
2. Reference the √exact values table
cos(30◦ ) = 23
∴ sin(120◦ ) =
√
3
2
This method can be followed to find the exact values of many different angles.
π
2
(90◦ )
(270◦ )
13
Pythagorean Formula
One of the most important formulas that you need to follow is this:
(sin(θ))2 + (cos(θ))2 = 1
This formula is derived from Pythagoras’ Theorem a2 + b2 = c2 . As seen in the unit circle on
page 9, a triangle is created with the centre angle. This triangle has a hypotenuse of 1 unit,
and a = cos(θ) and b = sin(θ).
Solutions to trigonometric equations
From the general equation equation sin(x) = a, we can find an infinite number of solutions.
√
An example of this general equation is sin(x) =
3
.
2
A solution of this equation is x = 60◦ or π3 , because sin(60◦ ) =
We also know that sin(π − π3 ) =
√
3
2
or sin( π3 ) =
√
3
.
2
√
3
,
2
because sin is positive in the second quadrant.
Therefore, for this particular equation there are two solutions between 0 and 2π. These solutions
are π3 and 2π
.
3
Sin is negative in quadrants three and four, so there are no solutions in those areas. We can go
’around’ the unit circle as many time as we need to in order to find more solutions based on
the required domain.
For example, since sin(2π + π3 ) =
, 7π
, 8π
.
solutions: π3 , 2π
3
3
3
√
3
2
and sin(3π − π3 ) =
√
√
3
2
in the domain [0, 4π] there are 4
eg. Find all solutions to the equation cos(x) = − 22 in the domain [0, 2π].
1. Write out the √equation.
cos(x) = − 22
2. Find the base angle by looking at the exact values table (ignore the negative)
base angle: π4
3. Identify the quadrants in which cos is negative
2nd and 3rd Quadrants (ST from CAST)
4. Find the angle in these quadrants using the formulas on page 9
Second Quadrant: x = π − π4 = 3π
4
π
5π
Third Quadrant: x = π + 4 = 4
∴ x = 3π
, 5π
4
4
14
Plotting circular functions
Now that we know how to find the solutions to an equation, we can combine this with a few
tricks to plot the entire graph on a cartesian plane.
To begin with, we need to recognise a normal sin, cos and tan graph.
1
y
π
3π
2
2π
π
2
−1
3π x
5π
2
one period
1
y
π
3π
2
2π
π
2
−1
3π
x
5π
2
one period
y
2
1
π
3π
2
2π
x
π
2
−1
−2
one period
These basic shaped graphs can be modified with transformations and dilations in the same way
that any other graph can be.
The basic form these translations take is:
y = a sin(b(x − c)) + d
y = a cos(b(x − c)) + d
y = a tan(b(x − c)) + d
where:
a affects the amplitude of the graph
b affects the period
c affects the horizontal translation (in the x-direction)
d affects the vertical translation (in the y-direction)
The important translations to note to begin with for circular function graphing is a and b
(amplitude and period respectively).
15
The amplitude (a) affects how high and low the graph goes (the range). For example, a standard
sin (and cos) graph’s range is [−1, 1]. If we graph 3 sin(x) we are changing the graphs range,
stretching the graph in the y-axis as seen below.
3
y
2
1
−1
π
2π
π
2
3π
2
3π x
5π
2
−2
−3
As we can now see, changing the amplitude to 3 adjusted the highest and lowest points of the
graph.
The period affects how often the shape of a graph repeats itself. The period of a standard sin
graph is 2π as this is one full revolution of the unit circle. Changing b will change this domain,
and thus have a big effect on the results.
The period of a sin or cos graph is
The period of a tan graph is πb .
2π
.
b
We can use the period to know where the important points on our graph are. This includes
maximums, minimums and inflection points. We know a graph has 4 of these important points
every period. Therefore a graph with a period of π2 will have either an inflection, maximum or
minimum every π8 units.
y
( π8 , 1)
( 5π
, 1)
8
( π4 , 0)
( 3π
, 0)
4
( π2 ,
( 3π
, −1)
8
0)
x
(π, 0)
( 7π
, −1)
8