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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.

For the following questions, select the statement which most accurately matches your level of confidence (1 is not confident and 4 is very confident) or choose from the alternatives detailed in the question. You do not have to answer all questions. Your answers will be saved so you can exit and come back to your self-evaluation at any time. Click Save and Results to view the next steps for questions you have answered.

1.

How confident are you that you understand the distinction between:

1

2

3

4

a.

the words expression, formula, equation and identity?

An algebraic expression is formed from letter symbols and numbers, combined with operation signs and brackets. Each part of an expression is called a term. In the expression 3n + 5 the terms are 3n and 5.

A formula is an equation linking sets of physical variables. For example, in the formula v = u + at, has 4 variables v, u, a and t related by the formula. If the values of three variables are known, the fourth value can be calculated.

An equation is a mathematical statement showing that two expressions have equal value. The expressions are linked with the symbol =. For example, in the equation 5x + 4 = 2x + 31, x is a particular unknown number for which the expressions on either side of the equation have equal value. Solving the equation simply means finding the value of x for which this occurs (in this case, x = 9).

In an identity, the expressions on each side of an equation always take the same value, whatever numbers are substituted for the letters; the expressions are said to be identically equal. The expressions are linked with the symbol ≡. For example, 4(a + 1) ≡ 4a + 4 is an identity, because the expressions 4(a + 1) and 4a + 4 always have the same value, whatever value a takes.

Additional User Example

Expression

5x +3

Identity

3x+9 = 3(x+3)

Formula

P=2l +2w

2.

How confident are you that you can find:

1

2

3

4

a.

the nth term of an arithmetic sequence and the nth term of a geometric sequence?

A sequence is a series of shapes or numbers which follow a rule.

An arithmetic sequence is a sequence of numbers with the rule ‘add a fixed number’. The number is the difference between consecutive terms.

For example, in the sequence 5, 8, 11, 14, 17, …, the rule is ‘add 3’. The nth term is 3n + 2. The 3 in 3n + 2 is the difference between consecutive terms. The +2 in 3n + 2 is the term that would come before the 1st term, sometimes called the zero term.

In general, for an arithmetic sequence with first term a and difference d, the nth term is given by:

T_{n} = a + (n – 1)d

A visual example of an arithmetic sequence:

Write down the number of dots in pattern 4. Find an expression, in terms of n, for the nth term in the sequence.
Does any pattern in the sequence contain 245 dots?

To create the next pattern requires adding 4 more dots, giving 19 dots, and the nth pattern will have 4n + 3 dots in it.
If a pattern has 245 dots then the equation 4n + 3 = 245 will have an integer solution. n =

245 – 3

4

= 60.5 shows that no pattern in the sequence will have 245 dots in it.

A geometric sequence is a sequence of numbers with the rule ‘multiply by a fixed number’. The number, called the common ratio, is the ratio between successive terms.

For example, in the sequence 24, 36, 54, 81, 121.5, …, the rule is ‘multiply by 1.5’. The nth term is 24 × 1.5^{n-1}.

In general, for a geometric sequence with first term a and common ratio r, the nth term is given by:

T_{n} = ar^{n-1}

In both cases the rule for finding the nth term involves applying the term-to-term rule exactly n – 1 times.

3.

How confident are you that you know :

1

2

3

4

a.

different ways of factorising algebraic expressions?

Factorising is the reverse process to expanding brackets.

Removing the highest common factor

Expanding 6(2a + 5) gives 12a + 30. So factorising 12a + 30 gives 6(2a + 5). 6 is the highest common factor of 12 and 30. 2(6a + 15) and 3(4a + 10) are also factorisations of 12a + 5) but do not take out the highest common factor.

Factorising by grouping

Expanding (a + b)(c + d) gives ac + ad + bc + bd. So factorising ac + ad + bc + bd gives (a + b)(c + d). In the expression ac + ad + bc + bd, pairs of terms with a common factor are grouped together and factorised to give a(c + d) + b(c + d) These two terms have a common factor (c + d). Removing this common factor gives (c + d)(a + b), which can also be written as (a + b)(c + d).

Factorising the difference of two squares

Expanding (a + b)(a – b) gives a^{2} – ab + ab – b^{2} = a^{2} – b^{2}. So factorising a^{2} – b^{2} gives (a + b)(a – b).

Factorising expressions of the form x^{2} + bx + c

Since (x + p)(x + q) = x^{2} + (p + q)x + pq, to factorise x^{2} + 6x + 8, we need to find two numbers whose product is +8 and whose sum is +6. The two numbers are +2 and +4.

Factorising expressions of the form ax^{2} + bx + c

For example 2x^{2} – 5x – 12. The rationale is rather more difficult, but if you find two numbers whose product is ac (here ac = -24), you can split the middle term into two terms and factorise by grouping: so here the two numbers are -8 and +3.

how to solve a linear equation with fractional coefficients?

In algebra, the expression (x + 3) ÷ 4 is usually written as:

x + 3

4

To solve a linear equation with fractional coefficients, the fractions can be changed to equivalent fractions with a common denominator.

For example:

Solve

Here, each term could be changed to have a denominator of 12 since 12 is the lowest common multiple of 2, 3 and 4. However, since the purpose of this is to then multiply through by the common denominator to remove all the fractions, it is usual to just multiply both sides by 12 without introducing the equivalent fractions.

Then cancel

or

Expand the brackets.

12x – 6 – 4x + 20 = 15

Simplify the left side.

8x + 14 = 15

Subtract 14 from both sides.

8x = 1

Divide both sides by 8.

x =

1

8

c.

how to identify and cancel common algebraic factors in rational expressions?

For example:

If we needed to simplify

provided x ≠ 1.

d.

how to change the subject of a formula where the subject appears twice or where a power of the subject appears?

Here are examples of changing the subject of a formula.

If the target variable appears more than once, you will need to take it out as a common factor. Then dividing will leave the target variable on its own.

Make s the subject of this formula: 2s – t = 3t – s Rearranging using the rules of algebra, 3s = 4t or s =

4

3

t.

Make u the subject of this formula:s =

v^{2} - u^{2}

t

Rearranging using the rules of algebra, u^{2} = v^{2} – st, so u = √v^{2} - st

4.

How confident are you that you can think of:

1

2

3

4

a.

situations in which pupils can generate formulae?

Many real-life situations involve two or more variables depending on one another. Some have simple relationships which can be expressed algebraically. Here are some examples.

Concert tickets cost £45 each. For booking over the Internet there is a credit card surcharge of 3%, plus postage, packing and insurance at £4.99 per order. Write a formula to give the total cost of ordering n tickets using a credit card.

In a pathology laboratory, batches of blood are used for experiments. From each batch, 25 ml are retained for reference and the remainder is divided (as far as possible) into 5 ml samples. Write a formula to show how many samples (S) can be made from a batch of V ml of blood.

A 1 m square has identical isosceles right-angled triangles cut from each corner. The shorter sides of the triangles have length s. Write a formula for the area (A) of the resulting octagon in terms of s.

5.

How confident are you that you can:

1

2

3

4

a.

know how to find the gradient of lines given by equations of the form y = mx + c?

In general, the line y = mx + c has interceptc on the y-axis, since it crosses the y-axis at (0, c). For example, this diagram shows the line y = 3x + 1.5. It crosses the y-axis at the point (0, 1.5). The intercept on the y-axis is +1.5.

The steepness of the slope or gradient of the line is found by dividing the vertical distance by the horizontal distance:

vertical distance

horizontal distance

=

3 units

1 unit

= 3

The gradient of the line y = 3x + 1.5 is 3, and in general the gradient of the line y = mx + c is m.

A line that slopes up from left to right has a positive gradient.
A line that slopes down from left to right has a negative gradient.

b.

find the gradients of parallel lines and lines perpendicular to them?

Parallel lines
Parallel lines have the same gradients.

The three lines y = 3x + 3, y = 3x – 2 and y = 3x – 5 are parallel since m = 3 in each case.

Perpendicular lines
If a line has gradient m, all lines perpendicular to it have gradient -

1

m

.

So two lines are perpendicular if the product of their gradients is –1 (except in the special case of the horizontal-vertical pair).

c.

know how to deduce and graph inverse functions?

An inverse of a function reverses the effect of that function. The inverse of function f is conventionally written f ^{–1}.

The inverse of ‘divide by 2’ is ‘multiply by 2’. So if f(x) =

1

2

x then f ^{-1}(x) = 2x.

The function f(x) = 5x^{2} – 3 represents the operation ‘square, multiply by 5 and subtract 3’. The inverse operation is ‘add 3, divide by 5 and take the square root’. So, in this case,
.

This diagram shows the graph y = 2x - 3 and its inverse y =

1

2

(x + 3).

Note: the two lines are symmetrical about the line y = x.

Many pupils find sketching the graph of y = √x difficult, because the graph is always increasing, but at a decreasing rate. Because it is the inverse function of y = x^{2} for x > 0 the graph will be the reflection of the the graph of y = x^{2} in the line y = x.

6.

How confident are you that you know how to

1

2

3

4

a.

find the exact solution of a pair of linear simultaneous equations by eliminating one variable?

Two linear equations that apply at the same time to two given variables are called simultaneous linear equations. The solution to the simultaneous equations is the pair of values for the variables that satisfies both equations.

To get the same number of ys in each equation, multiply (1) by 2.

6x + 4y = 14 (3)

Add (3) and (2).

11x = 22

Divide both sides by 11.

x = 2

Substitute x = 2 in (1).

6 + 2y = 7

Subtract 6 from both sides.

2y = 1

Divide both sides by 2.

y =

1

2

Where the two equations have the same sign it will be necessary to subtract the equations (or multiply through one of them by a negative to be able to add).

To get the same number of ys in each equation,
multiply (1) by 2
multiply (2) by 3

8x + 6y = 18 (3)
15x + 6y = 39 (4)

Subtract (3) from (4)

7x = 21

Divide both sides by 7

x = 3

Substitute x = 3 in (1)

12 + 3y = 9

Subtract 12 from both sides

3y = –3

Divide both sides by 2

y = –1

b.

find an approximate solution of a pair of linear simultaneous equations by graphical methods?

The graphical solution to simultaneous equations is a point where the lines representing the equations intersect. The diagram shows the graphs of the lines:

2y + x = 10 y = 2x – 7

Reading from the graph, the point of intersection is approximately (4.9, 2.6). So the approximate solution to the pair of equations is x = 4.9, y = 2.6.

The solution can be checked by substituting the values in the original equations. More accurate graphical solutions can be obtained by increasing the scale around the point of intersection.

7.

How confident are you that you know how to::

1

2

3

4

a.

generate points and plot graphs of quadratic functions?

A quadratic expression is of the form ax^{2} + bx + c where a, b and c are numbers, and a ≠ 0.

The graphs of all quadratics are parabolas, meaning it will always have a vertical axis of symmetry and it will turn smoothly – very often the table of values will not include the actual turning point (unless it happens to be an integer) so pupils need to know they must continue the graph below (or above if it has a negative x^{2} term) the lowest (highest) plotted value(s).

To draw the graph of a quadratic function such as y = x^{2} + 2x – 2 a table of values can be used.

x

-4

-3

-2

-1

0

1

2

x^{2}

16

9

4

1

0

1

4

2x

-8

-6

-4

-2

0

2

4

–2

-2

-2

-2

-2

-2

-2

-2

y

6

1

-2

-3

-2

1

6

Here the graph is relatively easy to draw because the minimum point (–1, –3) is plotted – we know it is the minimum because of the symmetry of the other points.

For y = 2x^{2} – x – 3 a table of values gives.

The lowest point of this graph where it turns is called the minimum point. It will be just to the right of x = 0, and just below where the graph cuts the axis at x = 0 (you know it is to the right because the value at x = 1 is lower than the value at x = -1).

The estimated values of x are –1 and 1.5, which are the approximate solutions of the equation 2x^{2} – x – 3 = 0.

b.

find approximate solutions of a quadratic equation from the graph of the corresponding quadratic function?

In general, the approximate solutions of a quadratic equation can be found from the graph of the corresponding function, or from the graph of a quadratic function and a linear function.

The points of intersection of the graphs of y = px^{2} + qx + r and the line y = mx + c can be used to solve the equation px^{2} + qx + r = mx + c.

For example, to solve x^{2} + x + 2 = 0 from the graph of y = x^{2}, draw the graph of y = x + 2 and read off the x-coordinates of the points where the graphs cross.

In this case the solutions are exact (x^{2} – x – 2 = (x + 1)(x – 2)) illustrating how the technique works, but it can be used to find approximate solutions where the solutions are not exact.

A quadratic graph can be used to solve other quadratic equations by drawing another straight line on the graph, but you need to find the appropriate straight line to draw.

For example, if x^{2} – 3x + 2 = 0 has been drawn you can solve x^{2} – 4x + 1 = 0 by writing it as x^{2} – 3x + 2 = x + 1, then drawing y = x + 1 and reading off the x values off the points of intersection.

The approximate solutions to x^{2} – 4x + 1 = 0 are x = 0.3 and x = 3.7.

c.

solve exactly, by elimination, one linear and one quadratic equation?

By drawing the graph on a graph plotter, or solving the equation by the quadratic formula find that
x = 0.893 or x = -2.115
if x = 0.893, y = 4.071
if x = -2.115, y = -20.765

First, look for a complete square which will deal with the x^{2} and x terms.

So we can rewrite the initial equation as:

So the roots are and .

f.

solve quadratic equations using the quadratic formula?

The roots of a quadratic equation written in the form ax^{2} + bx + c = 0 can be found using the quadratic formula:

Solve y = 3x^{2} – 4x – 5.

Here, a = 3, b = –4 and c = –5.

Substituting into the quadratic formula:

x = 2.12 or x = -0.79 (to 2 d.p.)

g.

how to check the answers to questions that involve solving quadratic equations?

Pupils should be encouraged to routinely check the answers to problems that involve finding the solutions to equations by substituting the solution(s) into the original equation(s).

For example, a pupil has x = 2 and x = 4 as the solutions to x^{2} + 2x – 8 = 0. They check by substitution:
If x = 2, x^{2} + 2x – 8 = 4 + 4 – 8 = 0 check
If x = 4, x^{2} + 2x – 8 = 16 + 8 – 8 0 error!

If one is right and the other wrong (as here) it suggests that the factorisation may be correct but there is a mistake (usually with negatives) in extracting the solution from one of the brackets: x^{2} + 2x – 8 = (x + 4)(x – 2) and the solutions should be -4 and 2.

Where a bracket has ax rather than just x, pupils may forget they need to divide by a, so in solving 2x^{2} – 5x – 3 = 0, the factorisation is (2x + 1)(x – 3) = 0 and the pupil may give x = – 1 and 3 as the solutions instead of x = –

1

2

and 3.

If both are wrong then it suggests the pupil may have not been careful enough in finding the two numbers giving the sum and product, for example they will often factorise x^{2} – 5x – 6 = (x – 3)(x – 2) and get ‘solutions’ of 2 and 3. Checking, both of these are wrong – and it is not just a problem of signs as it was previously.

8.

How confident are you that you are able to:

1

2

3

4

a.

a range of graphs modeling real situations and can interpret them?

For example, this line graph shows telephone charges for a fixed charge of £15 plus 10p per minute for calls.

Real-life graphs may be sketched rather than plotted accurately. Here are some examples.

This graph shows the length of a chocolate bar as a child nibbles at it.

This graph shows the temperature of water in a pan of boiling water as frozen peas are added to it.

This graph shows the depth of water in a bowl as the washing up is done.

This graph shows the depth of water in a conical container which is being filled at a constant rate.

b.

a range of formulae from mathematics and other subjects for pupils to investigate, derive or change the subject?

A formula is an equation linking sets of physical variables. Here are some relevant formulae from mathematics and other subjects.

density =

mass

volume

average speed =

distance

time

wage earned = hours worked × wage per hour

A =

1

2

bh where A is the area of a triangle, b is the length of the base and h is the perpendicular height

C = πd where C is the circumference of a circle and d the diameter

k = 1.6m where k is distance in kilometres and m is distance in miles

T = 40w + 20 where T is the time in minutes to roast a chicken and w is the mass of the chicken in kilograms

V = IR where V is voltage, I is current and R is resistance

F =

9

5

C + 32 and C =

5

9

(F – 32) where F is the temperature in degrees Fahrenheit and C is the temperature in degrees Celsius

P = 2(l +

A

l

) where P is the perimeter of a rectangle, A the area and l the length of one side

F = kx where F is the force in a spring and x is the extension of the spring

The subject of a formula appears just once and only on the left-hand side of the formula but it can be rearranged to make a different variable the subject.

Rearrange the formula T = 40w + 20 to make w the subject.

The formula is: T = 40w + 20 Subtract 20 from both sides. T – 20 = 40w Divide both sides by 40.

T - 20

40

= w This can be written as: w =

T - 20

40

.

The temperature C in degrees Celsius can be found from the temperature F in degrees Fahrenheit by the formula C =

5

9

(F – 32). Make F the subject of the formula.

C =

5

9

(F – 32)

9

5

C = F – 32 F =

9

5

C + 32.

c.

examples of real−life relationships that are directly proportional and can relate these relationships to graphical representations?

When c = 0, the equation y = mx + c becomes y = mx, where m is constant. We say that y is directly proportional to x. The conventional notation is: y α x or y = kx
All graphs of the form y = kx pass through the origin. The gradient of the graph is given by the constant k.
Here are some real-life examples of directly proportional relationships.

The cost of petrol C is directly proportional to the number n of litres of petrol that the motorist buys, so C α n or C = kn, where k is the cost per litre of petrol.

When a car is travelling at constant speed s, the distance travelled d is proportional to the time taken t: d α t or d = st.

The circumference C of a circle is directly proportional to the radius r, so C α r or C = 2πr.

The cost C of using a pay-as-you-go mobile phone is directly proportional to the number of minutes t of calls made, so C α t or C = kt, where k is the cost of calls per minute.

The volume V of paint required is directly proportional to the area a of wall to be painted, so V α a or V = ka, where k is the volume of paint needed to paint one square unit of the wall.

d.

examples of real−life relationships that are in inverse proportion and can relate these relationships to graphical representations?

When y =

k

x

(where k is constant) we say that y is inversely proportional to x. We can write y α

1

x

.

The graph of y =

k

x

is a hyperbola:

Here are some real-life examples of inverse proportion relationships.

The time t taken to drive a fixed distance d at constant speed s is inversely proportional to s, so t α

1

s

or t =

d

s

.

The number of canisters, n, required to hold V litres of radioactive liquid is inversely proportional to the capacity c of each canister, so n α

1

c

or n =

V

c

.

The charge per person, c, sharing the cost H of hiring a coach is inversely proportional to the number of people, n, so c α

1

n

or c =

H

n

.

9.

How confident are you that you know how to:

1

2

3

4

a.

find the length of the line segment AB, given the coordinates of A and B?

For example, if A is the point (3, 2) and B is the point (7, 6), we can find the length of AB by completing a right-angled triangle with AB as the hypotenuse.

Use Pythagoras’ theorem to calculate AB.
AB^{2} = 4^{2} + 4^{2} = 32, so AB = 4√2.

If A is the point (-3, 4) and B is the point (4, -1), care needs to be taken to get the horizontal and vertical distances correctly.

Use Pythagoras’ theorem to calculate AB. AB2 = 7^{2} + 5^{2} = 74, so AB = √74 = 8.60 (2 dp).

b.

find points that divide a line in a given ratio, using the properties of similar triangles?

For example, if A is the point (3, 2) and B is the point (15, 8), what are the coordinates of points M and N that divide AB in the ratios 3 : 2 : 1?

Let points P, P1 and P2 form three similar right-angled triangles APB, AP_{1}M and AP_{2}N.

Consider the x-coordinates.
M and N divide AB in the ratio 3 : 2 : 1 so AM : AN : AB = 3 : 5 : 6.
Using corresponding sides of the similar triangles, this means that AP_{1} : AP_{2} : AP = 3 : 5 : 6.
AP has length 15 – 3 = 12.
So AP_{1} =

3

6

× 12 = 6
and AP_{2} =

5

6

× 12 = 10.
So the x-coordinate of M is 3 + 6 = 9 and that of N is 3 + 10 = 13.

Consider the y-coordinates.
PB = 8 – 2 = 6 and MP1 : NP2 : BP = 3 : 5 : 6.
So MP1 =

3

6

× 6 = 3
and NP1 =

5

6

× 6 = 5.

So the y-coordinate of M is 2 + 3 = 5 and that of N is 2 + 5 = 7.

So M is (9, 5) and N is (13, 7).

10.

How confidently can you:

1

2

3

4

a.

explain and exemplify the use of the inequality symbols?

The inequality signs have the following meanings:

< means less than, e.g. 7 < 7.01;

≤ means less than or equal to, e.g. the number of days I will visit a shop next week ≤ 7;

> means greater than, e.g.

7

15

>

7

16

;

≥ means greater than or equal to, e.g. h ≥ 1.5 m where h is the height required to ride the Dark Canyon rollercoaster.

Negative numbers are difficult for many students working with inequalities because –4 < –3 i.e. a numerically larger negative number is less than. However –4 < 5 since any negative number is less than any positive number.

b.

solve simple linear inequalities in one variable?

Solutions to linear inequalities where the variable is restricted to integers are often shown as sets, like this.

x > 7 ÞxÎ {8, 9, 10, 11, …}
15 > x ≥ 11 ÞxÎ {14, 13, 12, 11}

Solutions to inequalities where the variable is a real number are often shown using a number line as follows.

Linear inequalities can be solved in the same way as equations with one important exception – where the inequality is multiplied or divided by a negative number, the sense of the inequality must be reversed.

Dividing through by a negative can be avoided if the terms in x are collected on the side of the inequality where there will be a positive number of them.

3x + 7 ≤ 5x + 8

⇒ –1 ≤ 2x

OR

⇒ –2x ≤ 1

⇒ –0.5 ≤ x

⇒ x ≥ –0.5

⇒ x ≥ –0.5

Here is an example:

4

(x + 2)

The area of this rectangle is numerically greater than the perimeter. Write down an inequality in y, and solve it.

Area = 4(x + 2)= 4x + 8.
Perimeter = 2(4 + x + 2) = 2x + 12.
So 4x + 8 > 2x + 12 giving x > 21.

Explain why it is not possible to find a value of x which satisfies both 4x + 7 < 27 and x + 7 > 13

Solving these inequalities gives x < 5 for the first and x > 6 for the second, and there is no value of x for which both these statements is true.

c.

solve simple linear inequalities in two variables?

Here are examples of solving simple linear inequalities in two variables.

Graph the solution to y ≤ 2x – 1.
First consider the graph of y = 2x – 1.

For all points on the line, y = 2x – 1.
For all points above the line, y > 2x – 1.
For all points below the line, y < 2x – 1.
So the ‘solution set’ for y ≤ 2x – 1 is all the points on and below the line. This is shown by a continuous line on the graph with shading in the region below it.

y ≤ 2x – 1

For a strict inequality of the type > or <, the border line which is not part of the solution set is conventionally shown dashed. So y > 5 – x is drawn thus:

y > 5 – x

d.

solve several linear inequalities in two variables and find the solution set?

Here is an example of solving several linear inequalities in two variables.

Show the solution set for these three simultaneous inequalities: y < x y > –2 y ≤ 10 – x

Consider the graphs of the corresponding equations.

Consider each inequality in turn and shade the region where all three overlap.

A different approach is to shade the region excluded by each linear inequality in turn, and the required region will then be the part of the graph without any shading. It is then important to clearly identify the unshaded region as the solution.

11.

How confidently are you able to:

1

2

3

4

a.

recognise and understand the key features of cubic graphs, the reciprocal graph and simple exponential graphs?

Here are examples of cubic graphs.

y = x^{3}

y = x^{3} – 2x^{2} + 3

y = 3x – x^{3}

The reciprocal graph y =

1

x

is a hyperbola. It is undefined when x = 0.

The reciprocal graph y =

1

x – 3

is a hyperbola which has been translated by 3, and is undefined when x = 3.

The function y = k^{x}, where k is a constant, is called an exponential function. For example, the graph of y = 3^{x} looks like this:

The graph of y = (

1

2

)^{x} looks like this:

b.

recognise and understand the key features of the basic trigonometric functions y = sin x, y = cos x and y = tan x?

The graph of y = sin x is a wave function which is periodic (360°) and oscillates between –1 and 1, going through the origin moving in the positive direction.

The graph of y = cos x is a wave function which is periodic (360°) and oscillates between –1 and 1, going through the point (0, 1) i.e. when x = 0 it is at a maximum.

The graph of y = tan x is a periodic function (180°) which moves from –∞ to +∞, going through the origin.

c.

apply simple transformations to graphs of y = f(x)?

The transformation f(x) + a translates the graph y = f(x) by a units parallel to the y-axis.

The transformation = f(x + a) translates the graph y = f(x) by (–a) units parallel to the x-axis.

The transformation a f(x) stretches the graph y = f(x) parallel to the y-axis by scale factor a.

The transformation f(ax) stretches the graph y = f(x) parallel to the x-axis by scale factor

1

a

.

Rearrange x^{2} + 6x – 1 into the form (x + a)^{2} + b, where a and b are integers. In the graph of x^{2} + 6x – 1, write down the coordinates of the minimum point, A, and the point, B, where the curve cuts the y-axis.

x^{2} + 6x – 1 = (x + 3)^{2} – 10 so the minimum point A must be (-3, –10), and B will be (0, –1).

Combinations of transformations can be used to identify the important features of graphs for example writing y = 2x^{2} – 12x + 14
as y = 2(x – 3)^{2} – 4
shows the graph will be the quadratic y = x^{2} stretched by a factor in the y direction, and then translated by 3 to the right and 4 down, and will look like this:

Two special cases are worth noting explicitly.

f(–x) is a reflection of the graph in the y axis:

–f(x) is a reflection of the graph in the x axis:

12.

How confident are you in knowing

1

2

3

4

a.

why the quadratic relation x² + y² = r² represents a circle with radius r, centred on the origin?

Using Pythagoras from the origin to the general point (x, y) it can be seen that the pairs of coordinates that satisfy the equation x^{2} + y^{2} = r^{2} all lie on a circle radius centred on the origin. This is the quadratic relation that gives a circle.

For example, this is the graph of the quadratic relation x^{2} + y^{2} = 25.

b.

why two simultaneous equations representing a straight line and a circle can have 0, 1 or 2 points of solution?

The equation of a straight line and the equation of a circle can be combined to produce a quadratic in either x or y. The number of points of intersection depends on the number of solutions of this quadratic equation.

A straight line can cut a circle in 0, 1 or 2 points.
In the diagram below:

Line A does not touch or intersect the circle. So the simultaneous equations for the circle and line A have no points of solution (the discriminant of the quadratic will be negative).

Line B touches the circle at just one point (it is a tangent to the circle). So the simultaneous equations for the circle and line A have one point of solution (the discriminant of the quadratic will be zero, giving a single repeated root).

Line C intersects the circle in two points. So the simultaneous equations for the circle and line C have two points of solution (the discriminant of the quadratic will be strictly positive, giving two distinct values).

No straight line can intersect the circle in more than two points.