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Mathematics Teaching Self-evaluation Tools

You are viewing a limited version of the NCETMâ€™s self-evaluation tools. Any answers you save during this session will be removed after seven days. Log in or sign up to view, and use, the full version of the tools.

Here you can see a summary of the areas in which you are confident and those in which you are less confident; there are some ideas and suggestions which may help you in your professional learning.

Showing all next steps for the selected topic.Click on a question to show more information.

the words expression, formula, equation and identity?(show/hide all)

What this might look like in the classroom

Problem
In the diagram below, x, y and z are the lengths of the sides of the cuboid.
Use the diagram to find an (i) expression, (ii) formula, (iii) equation

Possible solutions (there are many)
2y + 2z is an expression with two terms that represents the perimeter of the shaded rectangle. This can be written as the formula P = 2y + 2z
where P is the perimeter.

What would be the formula for the perimeter of the other faces you can see?

A formula for the volume of this shape (V) would be V = xyz (where xyz means x times y times z).

What would the value of the formula be if x = 3cm, y = 4cm and z = 8cm? (V = 3 × 4 × 8 = 96cm^{3})

If we know that V = 60cm^{3} and that y = 3cm and z = 2cm then we know that 6x = 60. This is an equation, and we can work out that for this to be true there is only one possible value for x which is 10cm

Show that if the volume was 90cm^{3} and y and z were both 3cm, then x would again be 10cm

*****

Terms either side of an identity sign are always equal regardless of the value of the variables. A simple example of this would be that 4a + 2 = a + a + a + a + 2. We could take this further by saying that
4a + 2 = a + a + a + a + 2 = 4(a+1) − 2 = 2(2a − 1) + 4

Taking this mathematics further

Write down as many formulae as you can remember. Which formulae have 3 variables, 4 variables etc?

Find out what is meant by dependent and independent variables.

Making connections

Learners will begin to develop algebraic reasoning without even realising that that is what they are doing. The ability to pattern−spot and generalise in simple cases is well−established in the primary phase, and a clear understanding of the technical vocabulary helps ensure learners do not get confused as the process becomes more formal.

These ideas form a basis on which to build algebraic processes. The next stage would be to generate expressions, manipulate formulae and solve equations.

The equation in the example is a linear equation but the principle is the same for higher order equations e.g. those with terms in x^{2}, x^{3} etc. although there is then usually more than one value that works.

Learners sometimes find it difficult to understand the difference between an expression and an equation.

the nth term of an arithmetic sequence and the nth term of a geometric sequence?(show/hide all)

What this might look like in the classroom

Problem A sequence starts with the first two terms 2 and 4. How could the sequence continue?

Possible response (there are plenty more though)
If we have the first two numbers in a sequence then it can be the beginning of an arithmetic sequence: in this case 2, 4, 6, 8 (add 2), or a
geometric sequence 2, 4, 8, 16 (multiply by 2)

The nth term of the arithmetic sequence is T(n) = 2n. It is quite easy to spot this just by looking (by inspection).This means that, for example, the 30th term will be 60.

The nth term of the geometric sequence is T(n) = 2n. The 30th term, for example, will be 1073741824.

Show that the difference between the 6th term of an arithmetic sequence starting 1, 5 and the 6th term of a geometric sequence starting 1, 5 is 3104

Taking this mathematics further

If the sequence is 2, 5, 8, 11
A finite series would be 2+5+8+11.
We could write the sum as

We can add up these terms to get 26 but if there were many more terms we would need a formula.

If an arithmetic series has first term a and common difference d, then the sum of the first n terms of the progression is

1

2

n(2a + ( n − 1 )d).

Making connections

An understanding of sequences is an important step in generalising. Many using and applying problems can be generalised to an arithmetic sequence or a geometric sequence.

These basic sequences help to develop an understanding of the more complex sequences; e.g. consecutive squares found in A and AS level curriculum.

how to identify and cancel common algebraic factors in rational expressions?(show/hide all)

What this might look like in the classroom

Question:
Simplify

Answer
You can factorise x^{2} + 3x + 2 into (x + 1)(x + 2)
and x^{2} + 4x + 4 as (x + 2)(x + 2) so

Taking this mathematics further

Research some of the real−life applications of rational functions; for example: click here.

Making connections

Factorising rational expressions is another step towards confidence in manipulating algebraic expressions.
You will meet such expressions again when solving more complex equations and graphing functions

Find the formulae sheet for a GCSE higher examination. Can you rearrange the formulae to make each of the variables the subject in turn?

Making connections

Rearranging formulae is a development of the process of solving equations e.g.
x + 8 = 25, as now the numbers are replaced by algebraic terms e.g. x + 2g = h^{2}

The complexity of the problem and the need to rearrange increases as the work progresses to KS5 but the principle remains the same.

situations in which pupils can generate formulae?(show/hide all)

What this might look like in the classroom

Question
A savings account offers 3% interest on the first £2000 and 4% on the remainder.
Find a formula for the interest, I, on £n if n is more than 2000

Answer
I = 0.03 × 2000 + 0.04(n − 2000)
So
I = 60 + 0.04n − 80
So
I = 0.04n − 20

Taking this mathematics further

Consider contract versus pay−as− you go tariffs. Generate formulae which can then be used to evaluate the cost effectiveness of any given tariff for a likely usage.

Research some accessible examples of famous mathematicians and interesting formulae that they developed; for example, Heron’s formula for the area of a triangle.

There are (apparently) 12 conceptually different ways of deriving a formula for the area of an isosceles trapezium. For example, two different dissections of an isosceles trapezium result in
A=

1

2

(a+b)h

And

Can you find the other ten?

Making connections

It is useful to get learners to consider how often they actually work with algebraic formula without realising it (e.g. mobile phone bills).

Learners use formulae to find area, volumes, etc.

Generating formulae becomes modelling as the mathematics curriculum progresses.
.

know how to find the gradient of lines given by equations of the form y = mx + c?(show/hide all)

What this might look like in the classroom

Finding the gradient of lines directly from the graph will help to develop the connection between the equation of the line and the gradient. Care needs to be taken with scales.

Question Find the gradient of the lines
4y = 6x + 1
3y − 9x + 2 = 0
Answer
4y = 6x + 1
y = 1.5x + 0.25
so the gradient is 1.5

3y − 9x + 2 = 0
3y = 9x − 2
y = 3x −

2

3

so the gradient is 3

Taking this mathematics further

By drawing the tangent to the curve it is possible to find the gradient of the curve at any given point. By undertaking this task and tabulating values it is possible to see that for y = x^{2} the gradient is always twice the value of x at the point. This is the beginning of calculus. Try it yourself.

Does it make a difference to the gradient function if your original graph is of y = x^{2} + 7 ?

Making connections

Learners will be familiar with graphs as a means of displaying information but at this stage they are beginning to consider the link with linear equations.
This will lead on to measuring the gradient of second/third order functions at various points and at KS5 to the use of calculus.

The effect on the equation of transposing the graph will be studied.

find the gradients of parallel lines and lines perpendicular to them?(show/hide all)

What this might look like in the classroom

Investigation
Using dynamic geometry software find a line perpendicular to each of these lines:
y = 2x
y = 5x + 2
y =

1

2

x

y =

2

3

x + 1
Write down the gradient of the original four lines, and of your four new lines. What do you notice? Test this idea with some lines of your own.

Solution
The gradients are:
2, 5,

1

2

and

2

3

And
−

1

2

, −

1

5

, −2 and −

3

2

The product of the gradients of perpendicular lines is always −1.

Taking this mathematics further

Research some of the history of algebraic graphing; why are coordinates properly known as ‘Cartesian coordinates’.

What other coordinate systems exist and what is the reason for their existence?

Making connections

Learners will be familiar with the terms ‘parallel’ and ‘perpendicular’, and should have had plenty of opportunity to investigate features of algebraic graphs.

The relationship between the value of the gradients is used in Key Stage 5 to find equations of tangents to curves.

For example if you know that the gradient of a curve at x = 5 is 2 then the equation of the tangent at the this point through (1,3) can be found.

y = mx + c
y = 2x + c
3 = 2(1) + c
1 = c

y = 2x+1

What is the y co−ordinate of the point where this tangent touches the curve?

know how to deduce and graph inverse functions?(show/hide all)

What this might look like in the classroom

Question 1 Find the inverse of the following functions:
y = 4x
y =

1

2

x
y = 5x + 2

Solution 1
y =

1

4

x
y = 2x

A common approach to solving these problems is to rearrange each function to find x in terms of y, and then to swap y and x around.

Question 2
Using tracing paper overlaid on a coordinate grid, plot each of the functions in question 1 and their inverses. Now overlay each graph and its inverse on the graph of y = x. What do you notice?

Solution 2
The graph of an inverse function is a reflection in the line y = x of the graph of the original function.

A dynamic geometry package could be used if available.

Taking this mathematics further

What are the potential problems with the inverse of the function y = x^{2}?

A self−inversing function is one where the inverse of the function is the same as the function itself. Think of two examples of self−inversing functions.

Making connections

The concept of writing a function as a function machine is useful for finding inverses; by following the machine from output to input and ‘undoing’ each operation, the inverse function can be found quite easily in most cases.

Learners should be familiar with the idea of plotting graphs both using pen and paper, and using graphing software.

Some learners will be confident reflecting in y = x others may need tracing paper to help with sketches.

find the exact solution of a pair of linear simultaneous equations by eliminating one variable?(show/hide all)

What this might look like in the classroom

Question
Pairs of simultaneous equations may be generated when more than one set of information is provided.
For example
3 drinks and 5 doughnuts cost £4.50
2 drinks and 1 doughnut
cost £2.30
Generate and solve the equations
Answer
If x is the cost of a drink and y is the cost of a doughnut
Then
3x + 5y = 450
2x + y = 230
Multiplying the second equation by 5 gives
10x + 5y = 1150
but 3x + 5y = 450
so subtracting the equations gives
7x = 700 x = 100 pence

substituting this solution into 2x + y = 230
gives 200 + y = 230
y = 30 pence

Taking this mathematics further

Simultaneous equations can have any number of unknowns, and solutions can be achieved (in many cases but not all) provided that there are as many equations as there are unknowns.

Some open ended tasks may result in equations with 3 or even 4 variables. The process is the same but some approaches are more efficient than others.
Have a look here.

Making connections

Take the time to explore the language: each pair of the two equations in question will have an infinity of solutions, but the word ‘simultaneous’ means ‘at the same time’. When considered together, there is a solution which fits both of the equations ‘at the same time’. For this reason it is always worth exploring the graphical representation of the simultaneous equations from the start; the visual aid of having two straight lines that intersect at one point, which just happens to give the solution, is very powerful. Dynamic graph drawing software (e.g. Autograph) should be used if possible.

Solving simultaneous equations using an algebraic method relies on a secure ability to manipulate basic algebraic expressions, substitute into expressions and solve linear equations.

The graphical representation of the simultaneous equations also makes it very clear why some simultaneous equations have no solutions, and some have an infinity of solutions

At the higher end of KS4 some learners will experience solving simultaneous equations when one of the equations is linear and the other is not. In these cases substitution is usually the best approach.

Matchless Starting with a challenge: This can be used as an introduction to simultaneous equations and their solution, encouraging learners to realise what they already know.

National Strategies- Algebra

Framework for mathematics – learning objectives and supplementary examples

find an approximate solution of a pair of linear simultaneous equations by graphical methods?(show/hide all)

What this might look like in the classroom

Question
3 drinks and 5 doughnuts cost £4.50
2 drinks and 1 doughnut
cost £2.30
Generate a pair of simultaneous equations and use a graphical method to find approximate solutions for the cost of each item
Answer
If x is the cost of a drink and y is the cost of a doughnut

Then
3x + 5y = 4.5
2x + y = 2.3

3x + 5y = 4.5
When x = 0 y = 0.9 y = 0 x = 1.5

2x + y = 2.3
When x = 0 y = 2.3 y = 0 x = 1.15

x = 1.00 y = 0.30

Taking this mathematics further

How does an understanding of this concept show that a trigonometric equation has an infinite number of solutions? Does this explain why, at AS/A2 Level, a range of values is always specified when solving trigonometrical equations?

Making connections

Learners need to be confident at plotting algebraic graphs

The principle of moving between and algebraic and graphical form to solve a problem is a powerful one. The solution of 2x − 1 = 7 can be seen by the intersection of the lines y = 2x − 1 and y = 7. If this is understood then the link to simultaneous equations is clear.

If this concept is understood then there is a clear link to the solution of a quadratic equation such as x^{2} − 3x + 2 = 0.

solve exactly, by elimination, one linear and one quadratic equation?(show/hide all)

What this might look like in the classroom

Question
Solve y = x^{2} + 3x + 1
and y = x + 4 simultaneously
Answer x^{2} + 3x + 1 = x + 4 (substitute) x^{2} + 2x − 3 = 0 (simplify)
(x − 1)(x + 3) = 0 (factorise − possible in this case)
So either x − 1 = 0 or x + 3 = 0
So x = 1 or x = −3
Now x = 1 ⇒ y = 5 (substituting back in to either of the original equations)
or x = −3 ⇒ y = 1

Taking this mathematics further

Research the ‘real−life’ situations which involve solving a pair of simultaneous equations where one is linear and one is quadratic. Prepare a list which you could share with colleagues and learners.

Making connections

Two linear simultaneous equations can be solved by substitution, and if learners are given the chance to experiment with carefully chosen sets of equations they might well develop this approach themselves; for example y = x + 4 and 3x + y = 28. If they are confident with the algebraic manipulation they can then apply this approach when one of the equations is not linear.

Encourage learners to consider the graphical representation of simultaneous equations in this case too. The (probably quadratic) graph makes it clear why there might very well be two pairs of solutions.

solve quadratic equations by factorisation?(show/hide all)

What this might look like in the classroom

Question
Solve 2x^{2} + x − 3 = 0

Answer
2x^{2} + x − 3 = 0
(x −1)(2x + 3) = 0 x = 1 or x = −1.5

Taking this mathematics further

Physics problems frequently require that a quadratic equation be set up and solved. Try this one:

A runner has half the kinetic energy that his younger brother has. The runner speeds up by 1.3 m/s, at which point he has the same kinetic energy as his brother. If the runner's mass is twice as large as his brother's mass, what were the original speeds of both the learner and his brother?

Can you solve it? (It doesn't factorise)

Making connections

Encourage learners to consider what ‘factorise’ means, in terms of factorising a number such as 12.

Ensure that learners are aware that a simple equation such as x^{2} = 9 has two solutions.

Learners should see factorising a quadratic expression as the reverse of expanding two sets of linear expressions.

Consider the graphical representation. If the equations y = 2x^{2} + x − 3 and y = 0 are plotted, the points of intersection of the graphs indicate the solutions to the quadratic equation 2x^{2} + x − 3 = 0.

Ensure that learners appreciate the fact that not all quadratic equations can be neatly factorised and that other approaches are necessary on occasions

solve quadratic equations by completing the square?(show/hide all)

What this might look like in the classroom

Question Solve x^{2} − 10x − 47 = 0
by completing the square
Answer
The first two terms determine that the equation must be rewritten as
(x − 5)^{2} − 25 − 47 = 0
(x − 5)^{2} − 72 = 0
(x − 5)^{2} = 72 x − 5 = ± √72 x = 5 ± 6√2

Taking this mathematics further

Program a spreadsheet to carry out the ‘completing the square’ process on any equation you enter.

Try completing the square to derive the formula for solving a quadratic equation of the form ax^{2} + bx + c = 0. It is remarkably satisfying − and is likely to amaze learners, even if they don’t entirely keep up with every step along the way!

Making connections

Learners will have had experience of multiplying out brackets at KS3 and of factorising quadratic equations.

Investigate the approach of completing the square: by considering squaring linear expressions such as x + 3, most learners will soon recognise that the coefficient of x in the resulting quadratic expression is always double the constant term in the linear expression. Therefore, completing the square is the reverse of this (hence the halving), and a correction has to be made (hence the subtraction of the square of half the coefficient of x)

To achieve exact solutions the answer will be given in surd form: an idea developed further at KS5.
Ask your learners what they think the advantages are of leaving the answer in surd form rather than using a calculator to evaluate.

solve quadratic equations using the quadratic formula?(show/hide all)

What this might look like in the classroom

Question
Solve x^{2} − 5x − 3 = 0

Answer using the quadratic formula
a = 1, b = −5, c = −3

or

x = −0.54 or 5.54 to 2 d.p.

Taking this mathematics further

Try programming a spreadsheet to carry out the ‘quadratic solving formula’ on any equation you enter.

There will be some equations which learners will declare to have no solutions as the number under the root sign is negative.
Once the concept of imaginary numbers has been considered then complex roots to such equations may be obtained.
The nature of the roots depends on the value of the discriminant. The discriminant (b^{2} − 4ac) is so called as it discriminates between the three cases of no real solutions, two equal real solutions and two real solutions.

Making connections

Learners need to be aware that there are three approaches they should draw from in order to solve quadratic equations. Factorising is likely to be the preferred method for many, but only works in certain cases.

Check for the common errors when multiplying, dividing and squaring with negative numbers

The formula is derived by completing the square on the general quadratic equation ax^{2} + bx + c = 0.

how to check the answers to questions that involve solving quadratic equations?(show/hide all)

What this might look like in the classroom

Question
Jo is not sure whether the solutions to x^{2} − 5x − 14 = 0 are x = 7 and x = −2, or x = 7 and x = 2
How could Jo check?
Answer
Substituting x = 7 into x^{2} −5x − 14 gives 49 − 35 − 14 = 0, so x = 7 is correct.

Substituting x = 2 into x^{2} − 5x − 14 gives 4 − 10 − 14 ( 0, so x = 2 is not a solution.

Substituting x = −2 into x^{2} − 5x − 14 gives 4 + 10 − 14 = 0, so x = −2 is correct.

So x = 7 and x = −2 are the solutions.

Problem
What mistake has been made here?

(x + 1)(x + 3) = 8
So x + 1 = 8
or x + 3 = 8
so x = 7 or x = 5

Taking this mathematics further

Program a spreadsheet to substitute any number into a specified quadratic equation and therefore quickly check the solutions generated.

Making connections

Encouraging learners to check their answers should be something that is well established already. However, unless actively encouraged to do so, many learners may opt out of the process.

examples of real−life relationships that are directly proportional and can relate these relationships to graphical representations?(show/hide all)

What this might look like in the classroom

Question 1
Which of these pairs of variables are directly proportional to each other?

The diameter of a circle and the circumference of a circle

Cost of diesel purchased and amount of diesel purchased

The diameter of a circle and the area of a circle

The perimeter of a rectangle and the area of a rectangle

Amount in pounds and amount in euros

Answer 1
a), b) and e and directly proportional to each other

Question 2 What can you say about a graph of each of the pairs of variables a), b) and e) in question 1?

Answer 2
Each of the graphs will be linear, will pass through the origin, and will have a gradient that is equivalent to the ‘multiplier’ between the two variables.

Taking this mathematics further

While the area of a circle is not directly proportional to the radius of a circle, it is directly proportional to square of the radius of a circle. You could plot a graph of A against r2 and find that the graph is linear. (What would the gradient be?) Can you think of similar examples where one quantity is directly proportional to the square of another?

Find out about the British physicist Robert Hooke and his law about springs

Find out about the Golden Proportion

Making connections

Conversion graphs are an example of a graph of quantities that are directly proportional to each other that learners will be familiar with.

Learners should be able to identify sets of numbers that are directly proportional to each other; see the example from the SNS materials below:

examples of real−life relationships that are in inverse proportion and can relate these relationships to graphical representations?(show/hide all)

What this might look like in the classroom

Question 1
Which of these pairs of variables are directly proportional to each other?

Given some sandwiches, the number of sandwiches each guest can receive, and the number of guests who turn up

Given a length in a rectangle, the perimeter of a rectangle and the area of a rectangle

Given some milk, the number of containers needed and the capacity of each of the containers.

Given an exchange rate, the amount in pounds and the amount in dollars

Given the distance in a cycling race, the average speed of a cyclist and the time taken to complete the race.

Answer 1 a), c) and e and inversely proportional to each other

Question 2 What can you say about a graph of each of the pairs of variables a), c) and e) in question 1?

Answer 2
Each of the graphs will be a reciprocal one. It is only necessary to consider the first quadrant in these real−life scenarios.

Taking this mathematics further

Quantities which are inversely proportional, or which are related by an inverse square law, are surprisingly common in the world around us; e.g. acoustics, electromagnetic radiation. See what examples you can find.

Making connections

Learners should be encouraged to see the relationship between this and direct proportion, and also the ‘directly proportional to the square of’ and ‘inversely proportional to the square of’ relationships.

Learners should also be able to use algebraic methods to find a missing value in a proportional relationship. That missing value might be a value in a set given the multiplier, or it might be the multiplier given the proportional set.

The conjecture that the sum of the squares of the shorter sides equals the square of the hypotenuse (in a right−angled triangle) was well−known before the time of Pythagoras. Indeed, the 3, 4, 5 relationship was used to measure right angles accurately. But Pythagoras was the first person to actually prove it was true for all right−angled triangles. In doing so he simultaneously promoted the notion of mathematical proof as important. It is for this reason that the theorem is named after him.

Research the mathematician Pythagoras. Other than his famous theorem, what other contributions did he make (consider music, fractions, proof, …)?

Making connections

Clearly, an understanding of Pythagoras’ Theorem is needed in order to apply it in a context such as this one.

Pythagoras’ Theorem can be developed into finding lengths in three dimensional shapes, and correspondingly it can be used to find lengths between points given as coordinates in 3D.

Pythagoras theorem as a means of finding length will be developed into finding the magnitude of a vector: the magnitude of the vector (

find points that divide a line in a given ratio, using the properties of similar triangles?(show/hide all)

What this might look like in the classroom

Question A triangle has vertices A(−2, 3), B(4, 1) C (2, −5).

Find the mid point of AB

Find the point which divides AC in the ratio 1:3

Answer The x coordinate of the midpoint of AB is (−2 + 4) ÷ 2

The y coordinate of the midpoint of AB is (3 + 1) ÷ 2

Therefore the midpoint of AB is (1, 2)

The point which divides the line in the ratio 1:3 must be one quarter of the way along the line from A. Mark this point on a diagram (as below) with L.

To move from the x coordinate of A to the x coordinate of C, move 2 − −2 = 4 units. Therefore, to move from the x coordinate of A to the x coordinate of L, move

1

4

of 4 units = 1 unit. So the x coordinate of L is −1.

To move from the y coordinate of A to the y coordinate of C, move −5 − 3 = −8 units (8 units down). Therefore, to move from the y coordinate of A to the y coordinate of L, move

1

4

of −8 units = −2 units. So the y coordinate of L is 1.

Taking this mathematics further

Find out about the golden ratio with its links to Nature, Art and Architecture.

Making connections

This problem can be solved simply by drawing an accurate diagram, but the numbers might not always be so kind. Learners should develop an understanding of the process through straightforward examples so that it can then be applied in more complex situations.

Knowledge of similarity, built on an understanding of enlargement provides the basis for this approach.

Learners could also be encouraged to develop a purely numerical approach to solving such a problem which does not rely on knowledge of similar triangles.

explain and exemplify the use of the inequality symbols?(show/hide all)

What this might look like in the classroom

Question List all the possible integer solutions if 0 < x < 7

Answer x = 1, 2, 3, 4, 5, or 6

Taking this mathematics further

Find out about the application of inequalities in linear programming. This process in developed in Decision Mathematics A Level modules.

Making connections

Children are introduced to the inequality symbols as early as KS1, and by KS2 they are expected to be able to use them correctly with negative numbers in simple cases.

Through KS3 learners will develop this understanding, and some will solve inequalities in one variable representing solution sets on a number line.

Learners will progress to solve inequalities in two variables representing solution sets on a pair of axes.

solve simple linear inequalities in one variable?(show/hide all)

What this might look like in the classroom

Activity 1:
Match the cards to their correct place on the grid:

Solution 1:
The solution is not unique! Some of the cards go in more than one place, and misconceptions are addressed. Make sure that learners justify their answers!

Taking this mathematics further

What if the inequalities were not written in base 10?
For example:

x + 10 < 100 in Base 2
or
x − 20 < 12 in Base 3

Making connections

Learners will be familiar with solving equations. The mathematical process of solving inequalities is very similar but learners may feel less secure when the answer is expressed as a range.

The one thing to be particularly careful with is the main difference between solving equations and solving inequalities: if you multiply or divide both sides by a negative number, the sign swaps direction. Explore the reason for this with learners. However, it can usually be avoided. If faced with −2x > 4, you could just add 2x to both sides and continue as normal.

solve simple linear inequalities in two variables?(show/hide all)

What this might look like in the classroom

Question Draw the graph of y = 2x + 3
and shade the region where y < 2x+3

Taking this mathematics further

Companies (especially manufacturing), routinely use linear programming approaches to maximise profits. Find out more about this application of inequalities.

Making connections

This consolidates the work on linear graphs, but through testing ether side of the line the learners decides where the inequality holds. The problem becomes more complex as more variables and more conditions, are introduced.

Ensure that in the case of the inequality symbols, ‘<’ and ‘>’, a dotted line is used to represent the fact that the boundary is not included. In the case of the inequality symbols, ‘≥’ and ‘≤’, a solid line is used.

recognise and understand the key features of cubic graphs, the reciprocal graph and simple exponential graphs?(show/hide all)

What this might look like in the classroom

Question 1
What is the same / different about the graphs of y = x^{3}, y =

3

x

and y = 3?

Answer 1
The only real similarities are that the functions contain the same number and variables, and that all the graphs are non−linear.

y =

3

x

is a cubic graph, and will demonstrate the ‘s−shaped’ curve

is a reciprocal graph. It will have rotational symmetry about the origin, and will extend to infinity in both horizontal and vertical directions without ever touching the axes.

y = 3^{x} is an exponential graph (the variable is the exponent). It will have an increasingly steep curve as x increases.

Taking this mathematics further

Consider the example of exponential decay:
The Half Life of any substance is the amount of time that passes such that the amount of the substance is halved. Examples would be of half life of radioactive substances, or caffeine presence within the brain. The half life of coffee is somewhere between 3 and 7 hours.

Making connections

Learners will be familiar with the properties of linear and quadratic graphs, and the process of plotting the graphs.

To plot the graphs, learners will need to be able to substitute positive and negative numbers into formulae. Care should be taken with substituting x = −1 into y = x^{3} for example.

This is a ‘pre− calculus’ point where an appreciation of the more complex curves is developed.
At this point in the KS4 curriculum the vocabulary to describe graphs will extend to include maximum, minimum, asymptote, etc.

recognise and understand the key features of the basic trigonometric functions y = sin x, y = cos x and y = tan x?(show/hide all)

What this might look like in the classroom

Question
Here are three statements we could make about the sine and cosine functions:

The maximum value is 1 and the minimum value is −1.

The cycle repeats itself every 360 degrees.

It is a smooth curve

What extra information do you need know to be able to sketch the curve?

Answer
For example, we need to know:
what the value of the function is at x = 0
where the maximum / minimum points are

Taking this mathematics further

Find out about the multitude of examples of waves that can be described by the trigonometrical functions. Share this with your learners.
The displacement, d, of an object moving with Simple Harmonic motion is given by:
d = R sin wt
where R is the radius of the rotating object and w is the angular velocity of the object.
This is just one of numerous real world situations which can be modelled by a trigonometric graph.

Making connections

Learners are likely to have an initial view of trigonometry that relates it only to right−angled triangles. This might make the consideration of angles greater than 180^{o} a tricky one to start with. The trigonometrical functions are ‘circular functions’, which explains there periodic nature and there relationship to right angled triangles.
Angles can be measured in radians rather than degrees and the trigonometric graphs then become useful to solve equations such as x = sinx at Key Stage 5.
Look up how to convert degrees to radians.

apply simple transformations to graphs of y = f(x)?(show/hide all)

What this might look like in the classroom

Create 6 cards using ideas from BBC bitesize
1
Multiplying the angle by a number

2
Changes the length of the part of the graph which repeats itself.

3
Multiplying the trigonometric ratio by a number

4
Changes the maximums and minimums of the graph (i.e. how far the graph reaches up and how far below it reaches down )

5
Adding on a number

6
Moves the whole graph (without changing its shape) up (if the number is positive) or down (if the number is negative).

Learners work in groups on one of the trigonometric curves and decide how the cards should be pared.

Taking this mathematics further

Functions such as y = sin x can undergo a process of transformation so that the curve obtained is a match (or near match) to the real world data.

1822, Joseph Fourier, a French mathematician, discovered that sinusoidal waves can be used to describe most periodic waveform. The process is named Fourier analysis, and is a useful analytical tool in the study of waves, heat flow etc.

Making connections

Dynamic graphing software (e.g. Autograph), is a tremendous tool to aid understanding of this concept: try entering a function as ‘y = x^{2 }+ a’ and use the constant controller to vary the value of a and observe the effect.

When working on quadratics learners will have acquired an understanding of how the equation and the curve are related; e.g. the difference between a sketch of y = x^{2} and
y = x^{2} − 8
Use simple examples of quadratic curves as an introduction to more complex functions. Consideration of what happens to particular coordinates can help − particularly with the multiplicative transformations, which can appear similar unless looked at in detail.

why the quadratic relation x² + y² = r² represents a circle with radius r, centred on the origin?(show/hide all)

What this might look like in the classroom

Question
Find the equation of a circle centre (0,0) radius 4.

Answer
Draw the circle, centred on the origin with radius 4. Pick any point on the circumference. From this point, drop a vertical line to the x axis. From this you can create a right−angled triangle with hypotenuse 4. Therefore, by Pythagoras’ theorem x² + y² = 4² or x² + y² = 16. (think about the coordinates of the point on the circle − they are x and y). The circle is actually the locus of all points 4 units from the origin.

Taking this mathematics further

Research circles which are not centred on the origin.

What if the centre (of the circle x^{2} + y^{2} = 16) was moved to (2,3)? By considering transformations of any graphs it can be shown that in this case, the equation of the circle becomes
(x − 2)² + (y − 3)² = 4²
Find out about conics; e.g. ellipses. How can you construct an ellipse?

Making connections

The connection with locus and Pythagoras’ theorem is necessary in order to understand why the equation describes a circle.

Take the opportunity to construct the situation described in ‘What this might look like in the classroom’ with dynamic graphing software. You should be able to drag the point around the circumference of the circle, which in turn demonstrates the repeated right−angled triangles that occur

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