1. science
2. mathematics
3. algebra

1
HOW TO THINK
ABOUT
PRELIMINARY
MATHEMATICS
JW Hick, 2013
2
Contents
Preliminary Mathematics
1. Algebra and Surds
2. Equations and Inequalities
3. Plane Geometry
4. Trigonometry
5. Coordinate Geometry
6. Functions and Relations
7. Locus
8. The Quadratic Function
9. Introduction to Calculus
10. Summary
7
17
24
28
39
42
47
51
57
63
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Algebra and Surds
Basic operations with Algebra
It is important to think of (addition and subtraction) as one set of rules, and (multiply and
divide) as another separate set of rules.
Addition and subtraction we need “like terms”, i.e. group all the “same letter stuff”
together. Note a same “block of letters” is a like term. Example:
3x + 7xy – 2x + 3xy =
x + 10xy
Multiplication and division you do not need like terms. You “merge” it all into one
answer. The powers of the letters can change in multiply and divide questions. If you
multiply you add powers of the same letter, if you divide you subtract powers of the same
letter. Example:
2
5
3
3x × 4x y =
7
3
12x y
Index laws
When we multiply we add the indices, when we divide we subtract the indices.
Example:
5
6
7
5
3
2x y × 3x y = 6x
2
12
-1
y
7
20xy ÷ 4y x = 5x y
2
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If we have a power to a power, then we multiply the indices
Example:



3
3
9
2x y 
3
= 8x y
Two final, and very important index laws are the fractional index and the negative index.
These will be used extensively throughout differentiation and integration in HSC
Mathematics. It is important to understand them.
Fractional indices:
3
5
x =
5
x
3
“The number on the bottom goes out the front.”
Negative indices:
x
-4
=
1
x
4
“When the power is negative, put a 1 on top of it and make it positive.”
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Substitution
Substitution is simply “replacing letters with numbers” then simplifying or solving the
expression.
For example if the substitution was L = 3, then everywhere we see an L, we replace it
with the number 3. Example: Solve when x = 5 and y = 1
e = 2xy + x
e = 2×5×1 + 5
e = 15
“We replaced the x with 5 and the y with 1 and ended up with e equals 15.”
Bracket Expansions
To expand brackets multiply terms from the first bracket with terms from the following
brackets, one at a time.
EG (X+Y)(A+B) = XA + XB + YA +YB
“We multiplied the X into the second bracket, then the Y into the second bracket.”
Note: Perfect squares are in the form:
2
2
2
(a ± b) = a ± 2ab + b
It is not critical to remember this as you can just do normal bracket expansion, but it can
come in handy.
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Algebraic Fractions
“There are one set of rules for plus and minus, and another set of rules for times and
divide.”
When we plus or minus fractions, we need a “common denominator”. To get this we can
multiply the top and bottom of the fraction by any number we choose. What you do to the
top, you must do to the bottom. Example:
2x + 3y =
y
4x
2
2
2
2
8x + 3y =
4xy
4xy
8x + 3y
4xy
When we times or divide we do not need a common denominator. We simply multiply
both the numerators together, and multiply both the denominators together. If it is a
fraction division we “invert and multiply”, which means flip the second fraction upside
down then multiply. Example:






4 2
4
2

3x  y × i =
3
2
4x






3
3x  y ÷ 4x =
i
2i
4 2
3x  y  i
=
3
8x
2
3xy  i
8
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Remember, when you are manipulating the fraction to find a common denominator, we
are not changing the fraction. What we do to the top, we must do the same thing to the
bottom. EG double the top, double the bottom.
Factorising
There are many ways to factorise, depending on the nature of the question.
(i)
Highest common factor: “Just divide out the highest common factor.”
(ii)
Four term expressions: “Split into two groups, take out the highest common
factors, then take out the brackets.”
(iii)
x
Quadratic Trinomials: “What multiplies to the coefficient of
and the constant term?” Split the x term up into these two numbers and treat
as a four term expression.
(iv)
Difference of squares: “When you have something squared – something
4
2
 2 2
squared.” EG x – 49 is  x  – 7 .
(v)
Sum and difference of cubics: “Basically learn the formula, the sign in the
cubic will be the same as the sign in the two term bracket.”
2
3
3

2
2
3
3

2
2
x + y = (x + y)  x – xy + y
x – y = (x – y)  x + xy + y




Note: We can only “cancel” in fractions when we have stuff being multiplied on the
whole numerator and the whole denominator, IE a common factor. That is why we
factorise, so we can cancel down.
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Example: Factorise the following quadratic equation to solve for x.
2
6x + 9x – 6 = 0

2

2

3  2x + 3x – 2  = 0

3  2x + 4x – x – 2  = 0
3[2x(x + 2) – 1(x + 2)] = 0
3(2x – 1)(x + 2) = 0
Hence x is equal to 0.5 or -2
Example: Factorise the following
16
4y



8
– 25 =

8

2y – 5   2y + 5 
Here we have “something squared minus something squared” so we can use a difference
of two squares.
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Basic Operations with Surds
The surd rules are very similar to algebra rules. The main new ideas are ‘simplifying
surds’ and ‘rationalising the denominator’, other than that it is an extension of the
algebraic laws.
Addition and Subtraction Involving Surds
“We can only add or subtract like surds.’
Example:
3 +3 2 –3 3 +4 2 =7 2 –2 3
Multiplication and Division Involving Surds
“Pretend the square root isn’t there and multiply and divide as per normal.”
Example 1:
2 5 × 3 10 = 6 50
Example 2:
4 18 ÷ 3 2 = 4 9
3
“Just multiply/divide the numbers out the front and multiply/divide the numbers under
the square root sign.”
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Simplifying Surds
“When you multiply surds you merge the two multiples into the one surd. You can also
go in the opposite direction and break a surd up into its multiples.”
Example: 3 × 7 = 21
Hence, 21 = 3 × 7
The idea is to break a surd up, into “perfect square” multiples, as these can just be written
as a normal number, without a square root sign.
Example: Simplify 3 75 + 2 27
3 × 25 × 3 + 2 × 9 × 3 =
3×5× 3 +2×3× 3 =
15 3 + 6 3 =
21 3
Here we broke up 75 and 27 using “useful factors”. The square root of 25 turned into the
integer 5, and the square root of 9 turned into the integer 3. When we simplified we
noticed that we had “like surds” and could add them together.
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Rationalising the Denominator
Basically this means “manipulate the fraction so there is a rational number on the
denominator, not a surd on the denominator.” There are two main types.
Rationalising the denominator when there is a single surd on the denominator
In this case, we simply multiply the top and bottom by the surd on the bottom.
Example: Rationalise the denominator in
3 5
3
3 5
3
×
3
3 15
=
3
=
9
3 15 =
3
15
Here we multiplied the numerator and the denominator by the surd on the denominator.
Notice that we produced a whole number in the denominator, hence it is now rational.
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Rationalising the denominator when there are two terms on the denominator
In this case we will multiply the top and bottom by the “conjugate” of the denominator,
which basically means “change the sign of the second term”.
Example: Rationalise the denominator in
2 3
5 +2
2 3
5 –2
×
5 +2
2 3







5 – 2 


=
5 + 2   5 – 2 
2 15 – 4 3



=
5 –2
5
2


– (2)
2
=
2 15 – 4 3 =
5–4
2 15 – 4 3
When we multiply the numerator and denominator by the “conjugate” we will create a
difference of two squares in the denominator. You can just expand the brackets normally
if you want. Either way you will generate an integer in the denominator.
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Equations and Inequalities
In Algebra we need to “simplify” stuff, in Equations we need to “find the value of x”.
Linear equations
“Do the opposite to get the unknown letter by itself”.
When we have variables on both sides of the equation we can make it disappear off one
of the sides by adding or subtracting it away.
When equations involve fractions you often need to multiply every term by a number
which will eliminate all the fractions. This number is called the lowest common multiple.
Example: Solve the following linear equation
3x + 6(x + 1) = 8x + 1
4
3x + 6x + 6 = 8x + 1
4
3x + 5 = 2x
4
3x + 20 = 8x
20 = 5x
x=4
We started by expanding the 6 into the brackets. Then we subtracted 6x from both sides
to make the +6x disappear, and we subtracted 1 from both sides to make the +1 disappear.
To get rid of the fraction we multiplied everything by 4. To finish we subtract 3x from
both sides, then divide by 4 to get x by itself.
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Quadratic equations
A quadratic equation is an equation where the highest power of x is 2. You can solve it
by factorizing, using the formula or completing the square.
Before you can solve a quadratic equation, you need to get it into the general format, ie
get everything on one side equal to zero. There are three potential methods to solve
quadratic equations, which are underlined below.
Factorising is the easiest way to solve, provided there are integer answers. Try and
factorise first, and if you can’t think of the two numbers to split the x term into, then just
use the quadratic formula, as you probably have non integer answers.
Quadratic formula
2
x = – b± b – 4ac
2a
To use the quadratic formula you just have to substitute the values in for a, b and c and
solve. This will work on every quadratic equation.
Completing the square: “halve it, and square it”, this is the number you will add to both
sides of the equation to create a perfect square. You perform this procedure on the
coefficient of the x term.
Example: Solve the following quadratic equation
2
4x + x = 5
2
4x + x – 5 = 0
2
4x + 5x – 4x – 5 = 0
x(4x + 5) – 1(4x + 5) = 0
(x – 1)(4x + 5) = 0
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Hence x equals 1 or -0.8
In this example we could find integers to break the middle term up into, hence we used
the factorising method. If we couldn’t think of the numbers then we could just use the
quadratic formula, with a = 4, b = 1, c = -5.
Quadratic Inequalities
This follows very closely from quadratic equations. Basically solve the quadratic
equation, and graph the solution so you can easily use a test point to see what the solution
region is. This process also holds true for higher degree polynomials.
2
Example: Solve 2x + x – 6 > 0
2
2x + 4x – 3x – 6 > 0
2x(x + 2) – 3(x + 2) > 0
(2x – 3)(x + 2) > 0
so x = 3 and x = -2 are the roots. Let's graph this and test for the solution region.
2
We have solved the quadratic equation, now it is the “graph and test” part. The question
is asking “where is the parabola greater than (above) zero (the x-axis). So let’s graph and
see what values of x give us “the parabola above the x-axis.”
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Hence x < -2 and x > 1.5 are the solutions, as for these values of x, the parabola is above
the x-axis.
Note: You don’t have to draw in the y-axis in quadratic inequalities, we only care about
the x-axis and the two roots of the equation. Once you have that you have everything you
need to draw the parabola and solve the question. Just be aware of whether your parabola
is a positive or negative parabola.
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Absolute values
The easiest way to think about absolute values is “the stuff inside could equal the positive
or negative answer.”
This works for both equalities and inequalities.
Example: Solve the following inequality.
|7x + 3| < – 11
This means that it is true for both the positive and negative version of what is inside the
absolute value. Hence,
– (7x + 3) < – 11
7x + 3 < – 11
OR
x< –2
7x + 3 > 11
x>8
7
Simultaneous equations
Basically we now have two unknown variables and two equations, as opposed to linear
equations, which have one unknown variable and one equation.
There are 2 main methods to solve simultaneous equations, substitution and elimination
method. Unless there is an obvious elimination then just use substitution.
Substitution – “Get a letter by itself, then in the equation you haven’t used yet, replace
that letter with the stuff it is equal to.”
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Example: Solve the following equations simultaneously
3x – 2y = 5 7x + 3y = 4 x = 5 + 2y
3
7 × (5 + 2y) + 3y = 4
3
35 + 14y + 9y = 12
23y = -23
y = -1
hence x = 1
We solved this by getting x by itself from (1), and then replacing x in the second equation
with the stuff it was equal to. Solve this new equation, get your answer for y and then use
that value of y in either equation to find the value of x.
Elimination – “Add or subtract the two equations so that one of the 2 letters will
disappear.”
Example: Solve the following simultaneous equations
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3x – 2y = 5 7x + 3y = 4 9x – 6y = 15 14x + 6y = 8 (3) + (4) g ives us
23x = 23
hence x = 1 y = -1
What I did here was triple the first equation and double the second equation. This had the
effect of creating a negative 6y in the first equation and a positive 6y in the second
equation. When I added both the equations together the y variable was eliminated.
The graphical representation of solving simultaneous equations is where the two lines or
curves will intersect each other on the Cartesian plane.
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Plane Geometry
Angle definitions
Complementary angles – Angles which add to 90 degrees
Supplementary angles – Angles which add to 180 degrees
Revolution – Angles which add to 360 degrees
Adjacent angles – Angles which are next to each other
Summary of main theorems
•
Vertically opposite angles are equal.
•
Angles on a straight line add to 180 degrees, angles at a point add to 360 degrees.
•
Angle sum of a triangle is 180 degrees. (N-2)*180 to determine the angles sum of
any polygon.
•
Base angles of an isosceles triangle are equal, and the sides opposite the equal
angles are also equal.
•
The three parallel line theorems. Alternate angles are equal, Corresponding angles
are equal, Co interior angles add to 180 degrees.
If you need to prove lines are parallel, the best way to do this is to prove that the parallel
line theorems hold true in that specific scenario. EG I have shown that alternate angles
are equal, hence the lines must be parallel.
Ratio of intercept theorem – A family of parallel lines will cut transversals in a given
ratio. Basically use ratios to find unknown lengths.
Pythagoras Theorem is also something that may be used in conjunction with other
theorems.
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Congruent triangles
Most of the theorems you learn will lead into using them to prove triangles are either
congruent or similar.
Congruent triangles have exactly the same shape and size as each other. There are 4 tests
to prove that two triangles are congruent to each other.
“Basically to prove triangles are congruent we need three facts that match up between the
triangles. We need to then compare our three facts to see if we have satisfied one of the
tests.”
1. RHS – Right angle, hypotenuse, side
2. SAS – Side, angle, side
3. AAS – Angle, angle, side
4. SSS – Side, side, side
Once we have proved triangles are congruent we know that all the sides and angles in one
triangle are identical to all the sides and angles in the other triangle. This is phrased as:
“Corresponding sides and angles are equal in congruent triangles.”
A common trick is for triangles to both share one same line as a side, so obviously this
side will be equal on both triangles.
Example: Given AB is parallel to DE, prove that ∆ A B C ≡ ∆ D E C
In ∆ A B C and ∆ D E C
BC = EC (given)
<ACB = < DCE (vertically opposite)
<ABC = < DEC (alternate angles on parallel lines)
Therefore ∆ A B C ≡ ∆ D E C (AAS)
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Similar triangles
Similar triangles have the same shape, but are bigger or smaller versions of each other.
The difference in size is called the “scale factor”. EG if the scale factor is 2, then double
all the side lengths to get from the small triangle to the big one.
There are four tests, which are slightly different to the congruence tests:
1. Equiangular (all angles in one triangle are equal to all the angles in the other
triangle).
2. All three sides of one triangle are in proportion to the three sides of another
triangle. (Common scale factor between all the corresponding sides)
3. Two pairs of corresponding sides are in proportion and the angle included is equal.
Similar triangle questions are usually in two parts. First you need to prove the triangles
are similar, then you need to use the fact that similar triangles have corresponding sides
in proportion to solve a ratio based problem to find an unknown length.
Example: Prove ABC ||| ADE, hence find the value of x
In ∆ A B C and ∆ A D E
<A is common
<ABC=<ADE (alternate angles on parallel lines)
Therefore ∆ A B C ||| ∆ A D E (equiangular)
Now that we have proved they are similar, we can use the ratios to find the value of
unknown sides.
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Big : Small =
8:5=
12 : x
Hence
x = 12
5 8
x = 15
2
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Trigonometry
Right angles triangles
When we have right angled triangles we can use SOHCAHTOA to find unknown angle
or side lengths. We can also use Pythagoras’ Theorem.
When we are finding an unknown side length, write out the Sin, Cos or Tan ratio and if
the unknown is in the numerator we will multiply, if it is in the denominator we will
divide.
When finding an unknown angle, remember to use the inverse function on your calculator.
Example: Find the unknown side length in the following triangle.
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sinθ =  O 
 H
sin 25° = 6
x
x × sin 25° = 6
x=
6
sin 25°
x = 14.2 (1d p)
Example: Find the unknown angle in the following triangle.
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cosθ =  A 
 H
cosθ =  7.4 
 11.3 
-1
θ = cos  7.4 
 11.3 
θ = 49°
(to the nearest degree)
60 minutes = 1 degree
60 seconds = 1 minute
The angle of elevation is the angle, “going upwards” that is made with a horizontal line,
for example the ground, etc. The angle of depression is the angle “going downwards”
made with a horizontal line. This is normally not “inside” the triangle you draw in your
diagram.
Here alpha is the angle of elevation, and beta is the angle of depression.
Bearings can be measured in True Bearings, or in Compass Bearings. True bearings are
measured out of 360 degrees, starting at North and moving clockwise. Compass Bearings
start with a North or South, then how far you need to move in the East or West direction.
A common theme in more difficult bearing questions is the fact that “alternate angles are
equal on parallel lines.”
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Reciprocal functions
The third letter rule; Cosec, Secant and Cot are the reciprocal functions. The third letter
tells you which function it is the reciprocal of. For example the third letter of “Cot” is t,
hence it is the reciprocal function of Tan.
“The reciprocal functions simply are the inverse of their corresponding functions.”
cotθ given that
Example: Find
If sinθ = 3
4
then we can use Pythagoras’ to construct the entire triangle. Doing so will
reveal a 3, 4,
tanθ =
sinθ = 3
4
7
triangle. Hence,
3
7
hence
cotθ = 7
3
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Graphs
Sin and Cosine are periodic every 360 degrees. That is when they start to repeat
themselves again. Tan is periodic every 180 degrees. Sine and Cosine have an amplitude
of 1.
Cosine starts at (0,1), the other two functions go through the origin.
y = sinθ ( red)
y = cosθ ( blue)
y = tanθ ( green)
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Exact ratios
When angles are 30, 45 or 60 degrees we can use exact fractional answers. These are our
“exact” angles. When you see the word “exact” in a trigonometric question it usually is
implying that you need to use these exact triangles.
Remember in a triangle “The biggest angle is opposite the biggest side”. This is useful
when trying to remember how to draw the exact triangles.
Example: Find the exact value of the following.
2
2
sin 45° + cos 45° =
2
2
 1  + 1  =
 
 
 2
 2
1+1=1
2 2
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1
“we can replace sin 45° and cos 45° with
2 by using the exact triangle.
Angles of any magnitude
This is basically using ASTC to reduce angles over 90 degrees back to their equivalent
acute angle expression.
1.
2.
3.
4.
Draw the right angles triangle
Find the acute angle made with the x-axis. This is your corresponding acute angle.
What quadrant are we in? Will my ratio be positive or negative?
Usually use the exact triangles for an exact answer.
The main idea is “what acute angle is made with the x-axis”, then all you need is “is the
answer positive or negative”, and that depends on the quadrant you are in.
Example:
sin 330° = sin(360° – 30°)
= – sin(30°)
= –1
2
“The 330 degree line make a 30 degree acute angle with the x-axis, hence our
corresponding acute angle is 30 degrees. We are in the fourth quadrant, hence our answer
will be negative. As our angle of 30 degrees corresponds with an angle on our exact
triangles, we can use them, rather than typing it into our calculator.”
As the graphs of sin, cos and tan keep repeating themselves forever, we have to restrict
the domain of our answers. Remember we can always plus or subtract 360 degrees from
our angles and have the same answer.
For example
sin(540°) = sin(180°)
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So just solve for sin(180) as this will give you the same answer.
Trigonometric Equations
Basically this is the reverse process of angles of any magnitude. You will get a numerical
value and have to find the corresponding angles.
“It all comes back to finding the acute angle made with the x-axis, everything flows from
this.”
1. Find the acute angle answer
2. Apply this acute angle to the relevant quadrants to get all the required answers.
Example: Solve the following for the given domain.
cos( θ) = 0.5
for
-180° < θ < 180°
cos( θ) = 0.5
-1
θ = cos (0.5)
θ = 60°
Hence Q 1 ⇒ 60°
Q 4 ⇒ 360° – 60° = 300°
We were going to get answers in quadrants 1 and 4 as we were solving for positive 0.5.
However, one of our answers lies outside the given domain. We can always add or
subtract 360 degrees from any of our answers, so let’s do that to make sure both answers
are in the required domain.
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θ1 = 60°
θ2 = – 60°
Trigonometric Identities
This is basically about remembering some facts, and using them to simply an expression
or prove that the left hand side of an equation equals the right hand side.
cosθ = sin(90° – θ)
tanθ = sinθ
cosθ
2
2
(sinθ) + (cosθ) = 1
As Sin and Cos are equal if the angles add together to give 90 degrees, so to are Cot and
Tan, and Sec and Cosec. This is derived from manipulating the original equation.
Everything else can be derived from these main ones. For example with the third identity,
2
2
if you divide all the terms by sin (θ) you will derive the cot (θ) identity and if you
2
2
divide all the terms by cos ( θ) you derive the tan ( θ) identity.
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Example: Prove the following identity.
cosθ = secθ + tanθ
1 – sinθ
RHS =
1 + sinθ
cosθ cosθ
= 1 + sinθ
cosθ
=
=
cosθ(1 + sinθ)
2
(cosθ)
cosθ(1 + sinθ)
1 – sinθ
2
=
cosθ(1 + sinθ)
(1 + sinθ)(1 – sinθ)
=
cosθ
1 – sinθ
Hence we have proved RHS = LHS so we are done.
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Non right angled trigonometry
For right angled triangles we can use SOHCAHTOA. For non right angles triangles we
use the Sine Rule and the Cosine Rule. Basically try and use the Sine rule, if you can’t
then try the cosine rule. They will both give the same answer anyway. This non right
angled section is basically “pick the formula and substitute correctly”.
Sine Rule – “Match up opposites”
Cosine Rule – “Based on Pythagoras’ Theorem”
Sine Rule
a = b
SinA SinB
This also equals c divided by SinC but you only ever have to use two of them, not the
entire three. Just match up the opposites.
Note: You can inverse the equation which can be handy if you want the unknown value
on the numerator. Just makes solving the equation a little easier.
Cosine Rule
2
2
2
a = b + c – 2bc cos A
There is an “angle” version as well but that is this same formula rearranged so there is no
real point of remembering that as well.
The key with the Cosine formula is noting that the side “a” is the one opposite to the
angle you are using “A”. The order you choose “b” and “c” are of no consequence in this
formula.
Area of a triangle
Area = 1 abSinC
2
“Multiply the sides, half the answer and multiply by Sin of the angle in between the
sides.”
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Coordinate Geometry
Basic formulae
Midpoint
M = x1 + x2
 2
y1 + y2 
2 


This is simply the averaging the x coordinates and the y coordinates.
Distance
2
d = (x1 – x2) + (y1 – y2)
2
This formula is based on Pythagoras’ Theorem. “What is the difference between the x’s?
Square it, what is the difference between the y’s? Square it. Add the results and take the
square root.
Gradient
m = y2 – y1
x2 – x1
How far we rise between the two points, divided by how far we run across between the
two points.
A positive gradient slopes upward to the right, a negative gradient slopes upward to the
left.
This whole chapter is basically remembering the formulae and substituting points into
them.
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Gradient intercept form of a line and General form of a line
A straight line may be written in either “gradient intercept” form or in “general form”.
Gradient intercept form: y=mx+b
The number in front of x, denoted by m, will be the gradient of the line. The constant, b,
is where the line cuts the y-axis.
General form: ax+by+c=0
Get all the terms on one side, equal to zero and make sure all the coefficients are integers.
The point gradient form is a more useful form to have a line in, as this form tells us useful
information quite quickly. There is no real benefit of having a line in general format in
regards to reading information off of it.
Note: To find the y-intercept, make x = 0, to find the x-intercept, make y = 0.
Note: If a point lies on the line that means you can substitute it into the equation and “it
will work out” for example you will get 10 = 10 when you evaluate it.
Point gradient formula
This is possibly the most useful formula in this chapter. You will use it extensively in
calculus also.
m(x1 – x) = y1 – y
Whenever you see the phrase “find the equation of the (line) (tangent) (normal)” then you
should be thinking about this equation.
If you have the gradient, and one point on the line then you can find the equation of a
straight line by using this formula.
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Parallel and perpendicular lines
If lines are parallel then they have the same gradient. If they are perpendicular then the
gradients are negative reciprocals, i.e. they multiply to give -1.
Example: If the gradient of line 1 is 4, the gradient of a perpendicular line would be
–  1 
 4
Intersection of 2 lines
To find the point of intersection of 2 lines, solve the equations simultaneously. There is
another method but learning this offers no real advantage unless you are asked to solve
for the point of intersection by using the “k” method and you don’t have a choice.
Perpendicular distance from a point to a line
The distance formula you learned previously was between 2 points. This formula is
between a point and a line. It is basically just substitution.
d=
|ax1 + by1 + c|
2
a +b
2
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Functions and Relations
Domain and range
The domain is the restrictions on the x-values, the range is the restriction on the y-values.
You can remember the basic shapes of graphs to try and recall the domain and range of
specific questions, however it is easier to do the following:
1. Make x the subject, can y be any possible value? This is the domain.
2. Make y the subject, can x be any possible value? This is the range.
The only things that will restrict domain and range include “not getting a zero in a
denominator” and “not getting a negative under a square root”.
Functions and relations
A function is a graph which is known as “one to one”. For every x value there is only one
corresponding y value. Think of a straight line, or a parabola as an example.
A relation is a graph which is not “one to one”. Think of a circle.
If a horizontal line can cut a graph in more than one spot, then it is not a function.
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Odd and even functions
Even functions are reflected by the y-axis. Think of a parabola with vertex on the origin.
Odd functions are reflected in the line x=y. Think of a cubic.
The blue function is an even function, the red function is an odd function. It is possible
for a function to be neither odd nor even.
Algebraic tests for odd and even functions:
Even functions – If we replace x with (-x) and simplify we end up with the same thing we
started with.
Odd functions – If we replace x with (-x) and simplify, we end up with the negative
version of what we started with.
Example: Prove the following function is an odd function.
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3
f(x) = x + x
3
f( – x) = ( – x) + ( – x)
3
f( – x) = – x – x

3

f( – x) = -1 ×  x + x 
f( – x) = -1 × [(f)x)]
Hence f(x) is odd as substituting (-x) gives us -1 times the original function.
Basic curves
You should know the basic shape of:
1.
2.
3.
4.
5.
6.
Straight line with positive and negative gradients
Positive and negative parabolas
Cubic
Circles
Modulus functions
Semi circles
These basic graphs can be shifted up and down, left and right by adding or subtracting to
the x or y value. If you replace (x) with (x+1) you will shift the graph 1 unit to the left. If
you replace (y) with (y+1) you will shift the graph down one unit. The opposite applies
for -1, etc.
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Regions and inequalities
A region is generated from an inequality, rather than an equality which generates a line or
a curve.
Basically to graph a region, pretend the inequality is an equal sign to graph the curve,
noting whether you need a dotted or a solid line. Once you have the curve then use a test
point to see which side of the curve to shade in. The easiest test point is (0,0), unless this
point lies on the curve.
2
Example: Shade the region given by the following. y < 3x – 5
2
Basically what we do is graph y = 3x – 5 and then use a test point. The point (0,0)
satisfies the above inequality, so we shade the region containing (0,0).
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Modulus functions
The absolute value functions will generate a modulus function on the Cartesian plane.
This is a straight line which will “bounce” off the x-axis, hence forming a V like shape.
The following is a graph of y = |x+2|
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Locus
Definition of a locus
A locus is a set of points, like a graph which satisfy some condition such as always being
a certain distance from some point, etc. There are three main types, straight lines, circles
and parabolas.
We call the locus point P(x,y), which represents any point on the locus graph.
The straight line as a locus
You will get a straight line locus when you require a locus which is equidistant from two
points. You solve it as distance formula = distance formula.
There are slight variations to these type of questions. You may get the distance is in the
ratio of 2:5 for example. Same idea, but 5*PA = 2*PB and progress with the distance
formulae as per normal. You will get a curve in these scenarios.
Example: Find the locus, P(x,y), such that it is equidistant from A(1,4) and B(5, 8)
dpa = dpb
2
2
2
(x – 1) + (y – 4) = (x – 5) + (y – 8)
2
2
2
2
2
x – 2x + 1 + y – 8y + 16 = x – 10x + 25 + y – 16y + 64
-2x – 8y + 17 = -10x – 16y + 89
8x + 8y – 72 = 0
x+y–9=0
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The circle as a locus
The circle is a locus as any point on the circle is always the same distance away from the
centre of the circle.
There are two main type of questions, identifying the key properties of a circle, such as
the centre and radius, or completing the square to put a circle back into general format so
that you can read the key properties off it.
The general format of a circle with centre (a,b) and radius r is
2
2
( x – a) + ( y – b) = r
2
“You subtract away the centre coordinates, and all this is equal to the radius squared.”
Example: Find the centre and radius of the following circle.
2
2
x – 8x + y + 6y = 23



2


2

x – 4x + 4  +  y + 6y + 9  = 23 + 4 + 9
2
2
( x – 2) + ( y + 3) = 36
Hence the centre is (2,-3) and radius is 6
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The parabola as a locus
The parabola is the main locus we focus on. Any point on the parabola is equidistant
from the focus to the directrix. The focus is always on the inside of the parabola and the
directrix is always outside the parabola.
The main idea is to find “a”, the focal length. The distance from the vertex to the directrix
is the same as the vertex to the focus. That distance is “a”, the focal length.
“The distance from the red dot (focus) to the black line (directrix) is always the same
length. This is known as the focal length, represented by the letter “a” in the equation
2
x = 4ay ”.
Shifted parabolas
The same idea as circles, subtract away the new vertex point from the x and y coordinates.
Where the vertex is at the coordinate ( x1 y1) .
2
(x – x1) = 4a(y – y1)
To solve most problems it is just a case of drawing the picture with the primary goal to
find the value of “a”. Once you have “a” then it is normally pretty easy.
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Example: Find the equation of the parabola with vertex (2,4) and focus (2, 6)
Hence the jump up from the vertex to the focus is 2 units, so a = 2. We know the vertex
and the parabola is not a negative (upside down) version so we can write the equation in
general form.
2
( x – 2) = 8( y – 4)
Note: If you swap the x’s and the y’s then you will have a sideways parabola. The same
rules apply, just treat x’s like y’s and y’s like x’s. The focal length concept still remains
intact.
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Quadratic Functions
Quadratic functions
A quadratic function is in the form:
2
y = ax + bx + c
When a quadratic is graphed, it forms a parabola. If the leading coefficient is negative the
parabola will be upside down.
Axis of symmetry
x= –b
2a
This gives you the equation of the vertical line which cuts the parabola exactly in half. If
you want to find the vertex then substitute this x value into the equation to find the
corresponding y value.
The discriminant
The discriminant is a very useful tool. It is the stuff under the square root sign in the
quadratic formula. It tells us how many times a parabola crosses the x-axis, as well as
some information about those points where it crosses the x-axis, known as the roots of the
equation.
2
∆ = b – 4ac
If the discriminant is a positive number, then there are 2 real roots. Ie the parabola crosses
the x-axis twice. If the number is a perfect square then the roots are “real and rational”, if
the number is not a perfect square then the roots are “real and irrational.
If the discriminant equals zero, then there is one root, known as “equal” roots, meaning
both the roots are equal to the same number. This means the parabola will glance the xaxis in one spot.
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If the discriminant is a negative number that means there are no real roots. This can also
be described as “imaginary” roots. This means that the parabola does not cross the x-axis.
If the leading coefficient is a positive number then the parabola is “positive definite”,
meaning it is always above the x-axis. Same idea for “negative definite”.
It is important to understand the terms that are used. They include: real, imaginary,
rational, irrational, equal and no real roots.
Example: Find the values of “a” if there are no real roots in the following quadratic.
2
2x + x – a = 0
2
∆=1 –4×2× –a
∆ = 1 + 8a
Hence 1 + 8a < 0
a< –1
8
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Sum and product of roots
The roots are the numbers which make the quadratic equal zero. They are also the points
where the parabola cuts the x-axis. Let’s call these roots alpha and beta.
The sum of the roots
α+β= –b
a
The product of the roots
αβ = c
a
This is basically substitution, however there is often a bit of algebraic manipulation
required.
2

Example: Find the value of  α + β
2


in the following quadratic
2
3x + 2x + 5 = 0
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2
3x + 2x + 5 = 0
hence
α+β = –b
a
α+β = –2
3
αβ = c
a
αβ = 5
3
hence
2
2
2
2
2
=  – 2  – 2 × 5
3
 3
2
2
= – 26
9



α +β



α +β



α +β






2
= ( α + β) – 2αβ
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Reducible quadratics
Some equations are similar to quadratic equations and can be treated the same after a
little bit of algebraic manipulation. Remember x
2
just means “something squared”.
If it helps you can do a substitution to make it a quadratic, solve it and then substitute
back as your final step.
Example:
x
16
8
– 4x + 3 = 0
8
For simplicity, let u = x then,
2
u – 4u + 3 = 0
( u – 3)( u – 1) = 0
hence



8

8

x – 3   x – 1  = 0
Note: we could continue this process on both the brackets, as we have “something
squared – something squared” again. It depends if you want to factorise over the rational
numbers or over integers.
Equivalent quadratics
If two quadratics are equivalent then all the corresponding coefficients are equal. For
example, all the stuff in front of x
on the other quadratic.
2
on one quadratic equals all the stuff in front of x
2
In harder examples you will probably end up with some simultaneous equations to solve
for the unknown coefficients.
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Example: Find the values of a, b and c given the following,
2
2
a( x + 1) + b( x – 2) + c ≡ 3x + 7x
2
2
ax + 2ax + a + bx – 2b + c ≡ 3x + 7x
2
2
ax + (2a + b) x + ( a – 2b + c) ≡ 3x + 7x
hence
a = 3 b = 1 c = -1
a has to be 3, so then you have 6 + b = 7 gives b equals 1, then the constant term is zero
so 3 – 2 + c = 0 gives c = -1
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Introduction to Calculus
Limits
There are three main types of limit problems you will be faced with in HSC Mathematics.
The first is the simple type where you just substitute the value into the expression.
The next type is when you try to substitute the value into the expression, but realize doing
so will make the denominator of a fraction equal zero, which we can’t have. Invariably
you will be required to factorise, and cancel down, after which you can substitute the
value into the expression without creating a zero in the denominator.
The third type is the limit to infinity. In this case you will be required to “divide every
term by the highest power of x”. Then you apply the limit to infinity to all the terms,
noting any term with an x in the denominator will approach (become) zero.
Example:
2
lim
3x – 9x
2
x → 3
lim
x → 3
=
x –9
3x( x – 3) =
( x + 3)( x – 3)
lim  3x  =
 x + 3
x → 3
9= 3
6 2
Here we couldn’t substitute in 3 as it would have made the denominator zero, but if we
factorise and cancel down then we can avoid this problem.
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Example:
3
4x + 3x + 4
lim
3
x → ∞
2x + 4x
4x
3
3x
+
3
x
lim
x
3
2x
x → ∞
x
4+
+
3
3
2
+
x
lim
x → ∞
x
4x
2+4
x
=
4
+
x
3
2
3
2
=
3
4
x
3
=
lim 4 = 2
2
x → ∞
“We have to divide everything by the highest power of x, then anything with an x in the
denominator will approach zero as x approaches infinity. The constant terms will
determine your answer.”
Differentiation by first principles
In the past, we have found the gradient of a straight line by using the rise on run
technique. We will now find the gradient of curves, more specifically, at certain points on
a curve. The gradient is often different at every point on a curve, so we have to be
specific as to which x value we are calculating the gradient at now.
dy
dx
=
lim f(x + h) – f(x)
h
h → 0
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Note: The gradient at a certain point on a curve is equivalent to the gradient of a tangent
at that point. I.e. a line which glances the curve at that point has the same gradient as the
curve at that point.
2
Example: Differentiate from first principles f( x) = 2x + x
lim f( x + h) – f( x) =
h
h → 0
2

2



lim 2( x + h) + ( x + h) –  2x + x  =
h
h → 0
2
2
2
lim 2x + 4xh + 2h + x + h – 2x – x =
h
h → 0
lim h(4x + 2h + 1) =
h
h → 0
lim 4x + 2h + 1 =
h → 0
4x + 1
Hence dy = 4x + 1
dx
“To differentiate by first principles you replace all the x’s with (x+h), and subtract away
the original function. You divide this answer by h, which should cancel down if you
factorise the numerator.”
Note: Students find the “replacing x with (x+h)” as the hardest part. Everywhere you see
an x, replace it with (x+h).
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Differentiation
The simple way to differentiate is to “bring the power down and subtract one from it”.
When you differentiate a function, you now have the gradient function, also known as the
first derivative. This gradient function will tell you the value of the gradient at any x
value you substitute into it.
If a question gives you the derivative and you need to find the corresponding point on the
curve, follow your normal procedures, however now you will have an equation to solve
for an unknown coefficient.
Example: Find the first derivative of the following.
3
y = 13x + 9x + 4
dy = 39x2 + 9
dx
The three rules
The normal process of differentiation is to “bring the power down and subtract one from
it”, and to do this separately to all the terms in the expression. However there are 3 cases
where it is a bit more complex and you need to use certain rules.
Product rule
“The derivative of the first function multiplied by the second function + the derivative of
the second function multiplied by the first function.”
Think “take turns and add them”.
Example:
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2
y = 3x × (2x + 7)
dy = 6x(2x + 7) + 3x2 × 2
dx
dy = 12x2 + 42x + 6x2
dx
dy = = 18x2 + 42x
dx
Chain rule
“Bring the power down from the bracket, subtract 1 from it and multiply by the derivative
of what is in the brackets.”
Example:

2
4
y =  3x + x 
dy = 4  3x2 + x  3 × 6x


dx
dy = 24x  3x2 + x  3


dx
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Quotient rule
Use the formula for the quotient rule. Remember whatever is on the denominator stays
there and gets squared. It is also the first term on the numerator.
Example:
4
y=
3x
(2x + 3)
3
4
dy = (2x + 3) × 12x – 2 × 3x
2
dx
(2x + 3)
3
dy = 12x (2x + 3) – 6x
2
dx
(2x + 3)
4
Note: sometimes you will get questions where you need to use a combination of the rules
at the same time.
Finding the equation of tangents and normals
The idea of differentiating is to be able to find the gradient at certain points on the curve.
If we can find the gradient, and we have that certain point, then we can use the point
gradient formula to find the equation of the tangent, or the normal at that point.
Note: the gradient of the normal will be the negative reciprocal of the gradient of the
tangent. Just follow the normal process to find the gradient of the tangent, then as the last
step, take the negative inverse of it and you have the gradient of the normal.
1. Differentiate the function
2. Substitute the value of x into the gradient function to find the gradient at that
point.
3. Use this gradient (or the negative reciprocal for normals), with the point gradient
formula to find the equation of the tangent or normal.
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3
Example: Find the equation of the normal to the curve y = 4x + x where x = 2.
dy = 12x2 + 1
dx
when x = 2
dy = 49 ⇒ gradient is – 1
dx
49
when x = 2
y = 34
m(x1 – x) = (y1 – y)
– 1 (2 – x) = 34 – y
49
x – 2 = 1666 – 49y
x + 49y – 1668 = 0
Note: you may have to substitute the value of x into the original function to find the
corresponding value of y to use in the point gradient formula. You need both an x and y
value to use this formula.
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Summary
This provides you with an overview of all the important concepts. You should read over
each of these and then see if you can recall all of the formulae or principles. After you do
so you can look back in the book to check if you were correct. This tests your ability to
recall information, in much the same way as test does.
Algebra and Surds
•
•
•
•
•
•
•
•
Basic operations with Algebra
Index laws (including negative and fractional indices)
Substitution
Bracket Expansions
Algebraic Fractions
Factorising (5 methods)
Basic Operations with Surds
Rationalising the Denominator
Equations and Inequalities
•
•
•
•
•
Linear equations
Quadratic equations
Quadratic Inequalities
Absolute values
Simultaneous equations
Plane Geometry
•
•
•
•
Angle definitions
Summary of main theorems
Congruent triangles (4 tests)
Similar triangles (3 tests)
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Trigonometry
•
•
•
•
•
•
•
•
Right angles triangles
Reciprocal functions
Graphs
Exact ratios
Angles of any magnitude
Trigonometric Equations
Trigonometric Identities
Non right angled trigonometry
Coordinate Geometry
•
•
•
•
•
•
Basic formulae
Gradient intercept form of a line and General form of a line
Point gradient formula
Parallel and perpendicular lines
Intersection of 2 lines
Perpendicular distance from a point to a line
Functions and Relations
•
•
•
•
•
•
Domain and range
Functions and relations
Odd and even functions
Basic curves
Regions and inequalities
Modulus functions
Locus
•
•
•
•
Definition of a locus
The straight line as a locus
The circle as a locus
The parabola as a locus
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The Quadratic Function
•
•
•
•
•
Quadratic functions
The discriminant
Sum and product of roots
Reducible quadratics
Equivalent quadratics
Introduction to Calculus
•
•
•
•
•
Quadratic functions
The discriminant
Sum and product of roots
Reducible quadratics
Equivalent quadratics
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