Mini-Lesson

Sequences, Functions & Graphs

6 missions. Guess, learn, then check. Your place is saved.

Mission 1Chessboard: 1 grain, then double each square. How many on square 64?
Mission 2Rise 1 for every 12 along: which of three ramps is legal?
Mission 3Four graphs, four equations: can you match them without plotting?
Mission 4A machine doubles your number, then adds 3. How do you undo it?
Mission 5Two runners' distance–time graphs cross. What does the crossing mean?
Mission 640 m of fence and a wall: which rectangle encloses the most area?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

Chessboard: 1 grain, then double each square. How many on square 64?

Make a guess: Chessboard: 1 grain, then double each square. How many on square 64?

Algebra · sequences

Two kinds of rule

A term-to-term rule gets each term from the one before: 5, 9, 13, 17, … is "add 4".
A position-to-term rule (the nth term) gets any term from its position $n$ n: here $4 n + 1$ 4n + 1.
Substitute $n$ n to find a term: the 10th term is $4 \times 10 + 1 = 41$ 4 × 10 + 1 = 41.
The chessboard has term-to-term rule "double".

Worked example Algebra · sequences

Finding the nth term of a linear sequence

Question

Find the nth term of 7, 10, 13, 16, … and its 100th term. Is 100 a term?

1
Find the common difference: $10 - 7 = 3$ 10 − 7 = 3, so the nth term starts $3 n$ 3n
2
Compare with $3 n$ 3n (3, 6, 9, 12): each term is 4 more, so add 4
3
Write the rule: nth term $= 3 n + 4$ = 3n + 4
4
Substitute $n = 100$ n = 100: $3 \times 100 + 4 = 304$ 3 × 100 + 4 = 304
5
Solve $3 n + 4 = 100$ 3n + 4 = 100: $n = 32$ n = 32, a whole number, so 100 is the 32nd term

Answer

nth term

3 n + 4

; 100th term 304; 100 is a term (the 32nd).

⚠️ Watch out

the nth term is not "

n + 3

". The difference 3 multiplies

n

.

Calculate

?The sequence 40, 37, 34, 31, … is linear. Find its 20th term.

Calculate

?Exam-style. Pattern

n

is a row of

n

hexagons made from matchsticks. Pattern 1 uses 6 sticks, pattern 2 uses 11 and pattern 3 uses 16. Which pattern uses exactly 101 matchsticks?

Algebra · sequences

Special sequences

Square numbers $n^{2}$ n²: 1, 4, 9, 16, … and cube numbers $n^{3}$ n³: 1, 8, 27, 64, ….
Triangular numbers: 1, 3, 6, 10, 15, … (add 2, then 3, then 4, …).
A geometric sequence multiplies by the same number $r$ r each time: 3, 6, 12, 24 ( $r = 2$ r = 2). The chessboard 1, 2, 4, 8, … is geometric: square $n$ n holds $2^{n - 1}$ 2ⁿ⁻¹.

Sort it

?Sort each sequence.

Tap a statement, then tap the group it belongs in. Then press Check.

Algebra · sequences

Arithmetic series (Higher)

An arithmetic sequence has first term $a$ a and common difference $d$ d: nth term $= a + (n - 1) d$ = a + (n−1)d.
If the 3rd term is 11 and the 7th term is 23, then $4 d = 12$ 4d = 12, so $d = 3$ d = 3 and $a = 5$ a = 5.
The sum of the first $n$ n terms is

S_{n} = \frac{n}{2} (2 a + (n - 1) d)

Calculate

?Find the sum of the first 40 terms of

5 + 9 + 13 + 17 + \dots

Mission check

?On the chessboard, square 1 has 1 grain and each square doubles the one before. How many grains are on square 64?

Mission 2

Rise 1 for every 12 along: which of three ramps is legal?

Make a guess: Rise 1 for every 12 along: which of three ramps is legal?

Algebra · straight-line graphs

Gradient measures steepness

Gradient = change in $y$ y ÷ change in $x$ x ("rise over run").

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Uphill from left to right: positive. Downhill: negative. Horizontal: 0.
The bigger the gradient (ignoring its sign), the steeper the line.
"Rise 1 for every 12 along" is a gradient of $\frac{1}{12}$ 1/12.
Foundation: read the gradient from a graph. Higher: calculate it from two given coordinates, as in the worked example below.

Algebra · straight-line graphs

The equation $y = m x + c$ y = mx + c

In $y = m x + c$ y = mx + c, $m$ m is the gradient and $c$ c is the $y$ y-intercept: the line crosses the $y$ y-axis at $(0, c)$ (0, c).
Rearrange to $y = \dots$ y = … first: $2 y = 6 x - 4$ 2y = 6x − 4 gives $y = 3 x - 2$ y = 3x − 2, so $m = 3$ m = 3 and $c = - 2$ c = −2.
To plot: make a table of values (three or more $x$ x values), plot the points, rule the line.
$y = k$ y = k is a horizontal line; $x = k$ x = k is a vertical line.
Lines may be given as $a x + b y = c$ ax + by = c: $3 x + 2 y = 5$ 3x + 2y = 5 rearranges to $y = - 1.5 x + 2.5$ y = −1.5x + 2.5.

Quick check

?What is the gradient of the line

2 y = 12 - 3 x

?

Worked example Algebra · straight-line graphs

Equation of the line through two points

Question

Find the equation of the line through

(2, 7)

and

(5, 13)

.

1
Find the gradient: $m = \frac{13 - 7}{5 - 2} = \frac{6}{3} = 2$ m = (13 − 7)/(5 − 2) = 6/3 = 2
2
Substitute one point into $y = 2 x + c$ y = 2x + c: $7 = 2 \times 2 + c$ 7 = 2 × 2 + c
3
Solve for $c$ c: $c = 3$ c = 3
4
Check with the other point: $2 \times 5 + 3 = 13$ 2 × 5 + 3 = 13 ✓

Answer

y = 2 x + 3

⚠️ Watch out

subtract the coordinates in the same order on the top and the bottom, or the sign of the gradient flips.

Calculate

?A taxi fare is a fixed charge plus a set amount per mile, so fare against miles is a straight line. A 3-mile trip costs £8.40 and an 8-mile trip costs £15.40. Find the fixed charge. Give your answer as a decimal.

£

Algebra · straight-line graphs

Parallel and perpendicular lines

Parallel lines have equal gradients. The line parallel to $y = 4 x - 1$ y = 4x − 1 through $(1, - 3)$ (1, −3): $- 3 = 4 \times 1 + c$ −3 = 4 × 1 + c, so $y = 4 x - 7$ y = 4x − 7.
Higher: perpendicular lines have gradients that multiply to $- 1$ −1. Use the negative reciprocal: gradient 2 becomes $- \frac{1}{2}$ −1/2.
Higher: $2 y = 3 x + 1$ 2y = 3x + 1 has gradient $\frac{3}{2}$ 3/2, so a perpendicular line has gradient $- \frac{2}{3}$ −2/3.
For CCEA (Unit M3) and Edexcel IGCSE, parallel lines are Higher tier content too.

Calculate

?(Higher) Line

L

has equation

y = 2 x + 5

. Line

M

is perpendicular to

L

and passes through

(4, 1)

. Find the

y

-intercept of line

M

.

Algebra · straight-line graphs

Midpoint and length of a line segment

The midpoint averages the coordinates:

(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

The length comes from Pythagoras: $(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$ √((x₂ − x₁)² + (y₂ − y₁)²).
For $A (1, 2)$ A(1, 2) and $B (7, 10)$ B(7, 10): the midpoint is $(4, 6)$ (4, 6) and the length is $6^{2} + 8^{2} = 10$ √(6² + 8²) = 10.

Calculate

?Point

P

is

(- 3, 8)

and point

Q

is

(9, - 2)

. Find the length of

PQ

. Give your answer to 3 significant figures.

Mission check

?The rule: no steeper than a rise of 1 for every 12 along. Ramp P rises 0.6 m over 9 m, ramp Q is the line

y = 0.1 x

, ramp R rises 0.25 m over 3 m. Which ramps are legal?

Mission 3

Four graphs, four equations: can you match them without plotting?

Make a guess: Four graphs, four equations: can you match them without plotting?

Algebra · graphs of functions

Lines and parabolas

A linear graph $y = m x + c$ y=mx+c is a straight line: gradient $m$ m, crossing the $y$ y-axis at $(0, c)$ (0, c).
A quadratic graph $y = a x^{2} + b x + c$ y=ax²+bx+c is a parabola: U-shaped if $a > 0$ a>0, ∩-shaped if $a < 0$ a<0.
The roots are where the curve crosses the $x$ x-axis ( $y = 0$ y=0). The $y$ y-intercept is at $(0, c)$ (0, c).
The turning point lies on the line of symmetry, halfway between the roots.

Worked example Algebra · graphs of functions

Reading a quadratic graph

Question

The graph of

y = x^{2} - 2 x - 3

is drawn. Find its roots, its turning point, and use it to solve

x^{2} - 2 x - 3 = 5

.

1
Read where the curve crosses the $x$ x-axis: $x = - 1$ x = −1 and $x = 3$ x = 3
2
The line of symmetry is halfway between: $x = \frac{- 1 + 3}{2} = 1$ x = (−1+3)/2 = 1
3
Substitute $x = 1$ x = 1: $y = 1 - 2 - 3 = - 4$ y = 1 − 2 − 3 = −4, so the turning point is $(1, - 4)$ (1, −4)
4
Draw the line $y = 5$ y = 5 and read where it meets the curve: $x = - 2$ x = −2 and $x = 4$ x = 4

Answer

roots

x = - 1

and

x = 3

; turning point

(1, - 4)

; solutions

x = - 2

or

x = 4

⚠️ Watch out

to solve

x^{2} - 2 x - 3 = 5

you read off at

y = 5

, not at

y = 0

.

Calculate

?The graph of

y = x^{2} + 2 x - 8

is drawn for

- 6 \leq x \leq 4

. Use the graph to solve

x^{2} + 2 x - 8 = 7

. Enter the positive solution.

Algebra · graphs of functions

Cubics and reciprocals

A cubic $y = a x^{3} + \dots$ y=ax³+… is S-shaped: rising left to right if $a > 0$ a>0, falling if $a < 0$ a<0. $y = x^{3} - 4 x$ y = x³ − 4x has roots $- 2$ −2, 0 and 2.
The reciprocal graph $y = \frac{a}{x}$ y=a/x has two branches that never touch the axes (the asymptotes); $x \neq = 0$ x ≠ 0.
To match an equation to a graph: check the highest power of $x$ x and its sign, then the intercepts.
On your course cubic and reciprocal graphs are Higher tier.

Sort it

?Sort each equation by the shape of its graph.

Tap a statement, then tap the group it belongs in. Then press Check.

Calculate

?Higher. The graphs of

y = \frac{6}{x}

and

y = x + 1

are drawn on the same axes. Use the graphs to find the

x

-coordinate of the intersection point in the first quadrant.

Algebra · graphs of functions

Trigonometric graphs (Higher)

$y = sin x$ y = sin x and $y = cos x$ y = cos x are waves between $- 1$ −1 and 1 that repeat every $36 0^{\circ}$ 360°.
$sin x$ sin x starts at 0 and peaks at $x = 9 0^{\circ}$ x = 90°; $cos x$ cos x starts at 1 and is $- 1$ −1 at $x = 18 0^{\circ}$ x = 180°.
$y = tan x$ y = tan x repeats every $18 0^{\circ}$ 180° and has asymptotes at $x = 9 0^{\circ}$ x = 90°, $27 0^{\circ}$ 270°, ….
Symmetry gives more solutions: $sin x = sin (18 0^{\circ} - x)$ sin x = sin (180° − x).
You need these graphs for angles of any size, including negative angles.

Quick check

?You know that

sin 3 0^{\circ} = 0.5

. Use the graph of

y = sin x

to find the other angle between

0^{\circ}

and

36 0^{\circ}

with

sin x = 0.5

.

Mission check

?Four graphs, four equations. Graph P has two separate branches and never meets either axis. Which equation matches it, and why?

Mission 4

A machine doubles your number, then adds 3. How do you undo it?

Make a guess: A machine doubles your number, then adds 3. How do you undo it?

Algebra · functions

Function notation (Higher)

A function turns each input into one output. $f (x) = 3 x - 5$ f(x) = 3x − 5 means "multiply by 3, then subtract 5".
$f (4)$ f(4) means the output when the input is 4: $f (4) = 3 \times 4 - 5 = 7$ f(4) = 3 × 4 − 5 = 7.
Solving $f (x) = 13$ f(x) = 13 means finding the input: $3 x - 5 = 13$ 3x − 5 = 13, so $x = 6$ x = 6.
Composite function: $f g (x)$ fg(x) means $f (g (x))$ f(g(x)). Do $g$ g first, then $f$ f.
You may also see $f : x \mapsto 3 x - 5$ f: x mapsto 3x − 5, read " $f$ f maps $x$ x to $3 x - 5$ 3x − 5".

Worked example Algebra · functions

Composite functions (Higher)

Question

f (x) = 2 x + 3

and

g (x) = x^{2}

. Find

f g (3)

,

g f (3)

and an expression for

f g (x)

.

1
For $f g (3)$ fg(3), apply $g$ g first: $g (3) = 9$ g(3) = 9
2
Then apply $f$ f: $f (9) = 2 \times 9 + 3 = 21$ f(9) = 2 × 9 + 3 = 21
3
For $g f (3)$ gf(3), apply $f$ f first: $f (3) = 9$ f(3) = 9, then $g (9) = 81$ g(9) = 81
4
For $f g (x)$ fg(x), put $g (x)$ g(x) into $f$ f: $f (x^{2}) = 2 x^{2} + 3$ f(x²) = 2x² + 3

Answer

f g (3) = 21

,

g f (3) = 81

,

f g (x) = 2 x^{2} + 3

⚠️ Watch out

f g (x)

is not

f (x) \times g (x)

, and the order matters:

f g (3) \neq = g f (3)

.

Calculate

?

f (x) = 5 - x

and

g (x) = x^{2}

. Work out

f g (- 3)

.

Algebra · functions

Inverse functions (Higher)

The inverse function $f^{- 1}$ f⁻¹ undoes $f$ f: if $f (5) = 13$ f(5) = 13 then $f^{- 1} (13) = 5$ f⁻¹(13) = 5.
Undo the steps in reverse order, like taking off shoes before socks.
Algebra method: write $y = f (x)$ y = f(x), rearrange to make $x$ x the subject, then write it in terms of $x$ x.

y = 2 x + 3 \Rightarrow x = \frac{y - 3}{2} \Rightarrow f^{- 1} (x) = \frac{x - 3}{2}

Check: $f^{- 1} (f (x)) = x$ f⁻¹(f(x)) = x.

Quick check

?

f (x) = \frac{x + 4}{5}

. Which is

f^{- 1} (x)

?

Calculate

?Exam-style.

f (x) = 2 x - 1

and

g (x) = x^{2} + 1

. Solve

g f (x) = 26

. Enter the positive solution.

Algebra · functions

Domain and range (Higher)

The domain is the set of inputs allowed; the range is the set of outputs.
Exclude any input that would mean dividing by zero: $f (x) = \frac{1}{x - 2}$ f(x) = 1/(x − 2) needs $x \neq = 2$ x ≠ 2.
Exclude any input that would mean the square root of a negative: $g (x) = x - 3$ g(x) = √(x − 3) needs $x \geq 3$ x ≥ 3.
$h (x) = x^{2} + 1$ h(x) = x² + 1 for all $x$ x has range $h (x) \geq 1$ h(x) ≥ 1, because $x^{2}$ x² is never negative.

Algebra · functions

Transforming graphs (Higher)

$y = f (x) + a$ y = f(x) + a: translation $(0 a)$ ( 0 ; a ), up by $a$ a.
$y = f (x + a)$ y = f(x + a): translation $(- a 0)$ ( −a ; 0 ), left by $a$ a. So $f (x - 3)$ f(x − 3) moves the graph right 3: you need $x = 3$ x = 3 to get the old $f (0)$ f(0).
$y = - f (x)$ y = −f(x): reflection in the $x$ x-axis. $y = f (- x)$ y = f(−x): reflection in the $y$ y-axis.
$y = a f (x)$ y = af(x): stretch parallel to the $y$ y-axis, scale factor $a$ a. $y = f (a x)$ y = f(ax): stretch parallel to the $x$ x-axis, scale factor $\frac{1}{a}$ 1/a.

Quick check

?The graph of

y = f (x)

has a turning point at

(2, - 3)

. Where is the turning point of

y = f (x - 4) + 1

?

Mission check

?A machine doubles your number, then adds 3. Which machine undoes it?

Mission 5

Two runners' distance–time graphs cross. What does the crossing mean?

Make a guess: Two runners' distance–time graphs cross. What does the crossing mean?

Algebra · real-life graphs

Distance–time graphs

The gradient of a distance–time graph is the speed: distance ÷ time.
Steeper means faster. A horizontal line means stopped.
Where two graphs cross, the two objects are at the same place at the same time. If the lines do not have the same gradient there, one is overtaking the other.
Use the units on the axes: metres per minute, km/h, m/s.

Worked example Algebra · real-life graphs

Reading two runners' graphs

Question

Runner A runs 2000 m in 10 minutes at a steady pace. Runner B covers 1200 m in the first 4 minutes. Find each speed for the first 4 minutes, in m/s, and say who is faster at the start.

1
Gradient of B's first section: $\frac{1200}{4} = 300$ 1200/4 = 300 m per minute
2
Convert to m/s: $\frac{1200}{4 \times 60} = 5$ 1200/(4 × 60) = 5 m/s
3
Gradient of A's line: $\frac{2000}{10 \times 60} = 3.33$ 2000/(10 × 60) = 3.33 m/s (3 s.f.)
4
Compare: B's line is steeper at the start

Answer

B runs at 5 m/s and A at 3.33 m/s, so B is faster at the start.

⚠️ Watch out

change minutes to seconds before dividing, or the answer is in metres per minute.

Calculate

?Exam-style. A cyclist rides 15 km in 45 minutes, stops for 15 minutes, then rides 15 km back home in 30 minutes. Find the average speed for the whole journey.

km/h

Algebra · real-life graphs

Conversion graphs and rates

A conversion graph is a straight line through $(0, 0)$ (0, 0): the two quantities are in direct proportion.
The gradient is the rate: kilometres per mile, euros per pound.
A line that starts above $(0, 0)$ (0, 0) has a fixed charge. A plumber charging £40 call-out plus £30 per hour has intercept 40 and gradient 30 (£ per hour).
Read across from one axis to the line, then down (or across) to the other axis.

Calculate

?The conversion graph is a straight line through

(0, 0)

and (50 miles, 80 km). Use it to convert 28 miles to kilometres.

km

Algebra · real-life graphs

Speed–time graphs

The gradient is the acceleration, in m/s². On this graph, $\frac{12}{4} = 3$ 12/4 = 3 m/s² for the first 4 s.
A horizontal line means constant speed; a downward slope means deceleration.
The area under the graph is the distance travelled: split it into triangles, rectangles and trapezia.

Calculate

?Exam-style. A car speeds up from rest to 12 m/s in 4 s, travels at 12 m/s for 6 s, then slows steadily to rest in 3 s. Find the total distance travelled.

m

Algebra · real-life graphs

Gradient of a curve (Higher)

On a curve the gradient changes. Draw a tangent: a straight line that just touches the curve at the point.
The gradient of the tangent is the instantaneous rate of change there, e.g. the speed at that moment on a distance–time graph.
Pick two points far apart on the tangent: gradient = change in $y$ y ÷ change in $x$ x.

Calculate

?Higher. A tangent is drawn to a distance–time curve at

t = 3

s. It passes through (1, 2) and (5, 14), with time in seconds and distance in metres. Estimate the speed at

t = 3

s.

m/s

Mission check

?Two runners start together. Their distance–time graphs cross at (8 minutes, 1600 m). Runner B's line is steeper at the start. What does the crossing mean?

Mission 6

40 m of fence and a wall: which rectangle encloses the most area?

Make a guess: 40 m of fence and a wall: which rectangle encloses the most area?

Algebra · differentiation

The gradient function (Higher)

A curve's gradient changes along it. The derivative $\frac{d y}{d x}$ dy/dx gives the gradient of the tangent at any $x$ x.
Rule: if $y = a x^{n}$ y = axⁿ, then

\frac{d y}{d x} = na x^{n - 1}

Multiply by the power, then reduce the power by 1: $5 x^{3}$ 5x³ gives $15 x^{2}$ 15x², $7 x$ 7x gives $7$ 7, a constant gives $0$ 0.
Differentiate a sum term by term. Substitute an $x$ x value to get the gradient at that point.
$n$ n can be any integer: $y = \frac{3}{x} = 3 x^{- 1}$ y = 3/x = 3x⁻¹ gives $\frac{d y}{d x} = - 3 x^{- 2} = - \frac{3}{x ^{2}}$ dy/dx = −3x⁻² = −3/x².

Worked example Algebra · differentiation

Gradient at a point (Higher)

Question

Find the gradient of the curve

y = 4 x^{3} - 5 x^{2} + 7

at the point where

x = 2

.

1
Differentiate each term: $4 x^{3} \to 12 x^{2}$ 4x³ → 12x², $- 5 x^{2} \to - 10 x$ −5x² → −10x, $7 \to 0$ 7 → 0
2
Write the derivative: $\frac{d y}{d x} = 12 x^{2} - 10 x$ dy/dx = 12x² − 10x
3
Substitute $x = 2$ x = 2: $12 \times 4 - 10 \times 2 = 48 - 20$ 12 × 4 − 10 × 2 = 48 − 20

Answer

the gradient is 28

⚠️ Watch out

substitute into

\frac{d y}{d x}

, not into

y

. Putting

x = 2

into

y

gives the height of the curve (19), not its gradient.

Quick check

?Find

\frac{d y}{d x}

when

y = 2 x^{5} - 3 x

.

Calculate

?Find the gradient of the curve

y = x^{3} - 6 x + \frac{8}{x}

at the point where

x = 2

.

Algebra · differentiation

Stationary points (Higher)

At a stationary point (turning point) the gradient is zero: solve $\frac{d y}{d x} = 0$ dy/dx = 0, then find $y$ y.
Maximum: the gradient goes positive, zero, negative. Minimum: negative, zero, positive.
Decide which is which from the shape of the graph: a cubic with positive $x^{3}$ x³ term rises, falls, then rises, so its left turning point is the maximum.
Edexcel IGCSE expects you to use the general shape of the graph.

Worked example Algebra · differentiation

Finding and classifying turning points (Higher)

Question

Find the turning points of

y = x^{3} - 12 x + 5

and say which is a maximum.

1
Differentiate: $\frac{d y}{d x} = 3 x^{2} - 12$ dy/dx = 3x² − 12
2
Set it to zero: $3 x^{2} = 12$ 3x² = 12, so $x^{2} = 4$ x² = 4 and $x = 2$ x = 2 or $x = - 2$ x = −2
3
Find each $y$ y: $x = - 2$ x = −2 gives $- 8 + 24 + 5 = 21$ −8 + 24 + 5 = 21; $x = 2$ x = 2 gives $8 - 24 + 5 = - 11$ 8 − 24 + 5 = −11
4
Use the shape: a positive cubic rises, falls, rises, so the left point is the maximum

Answer

maximum at

(- 2, 21)

, minimum at

(2, - 11)

⚠️ Watch out

x^{2} = 4

has two solutions,

\pm 2

. Missing

- 2

loses a turning point.

Calculate

?A ball is thrown upwards. Its height after

t

seconds is

h = 20 t - 5 t^{2}

metres. Use differentiation to find the greatest height the ball reaches.

m

Algebra · differentiation

Calculus and motion (Higher)

For a particle with displacement $s$ s at time $t$ t: velocity $v = \frac{d s}{d t}$ v = ds/dt and acceleration $a = \frac{d v}{d t}$ a = dv/dt.
The particle is at rest when $v = 0$ v = 0.
If $s = 2 t^{3} - 15 t^{2} + 36 t$ s = 2t³ − 15t² + 36t, then $v = 6 t^{2} - 30 t + 36 = 6 (t - 2) (t - 3)$ v = 6t² − 30t + 36 = 6(t − 2)(t − 3), so it is at rest when $t = 2$ t = 2 and $t = 3$ t = 3.

Calculate

?A particle moves in a straight line. Its displacement from

O

after

t

seconds is

s = 2 t^{3} - 15 t^{2} + 36 t

metres. Find its acceleration when

t = 4

.

m/s²

Mission check

?40 m of fence makes three sides of a rectangular pen; a wall makes the fourth side. Which pen encloses the most area?

Recap

Big ideas

A linear sequence adds the same amount each time; its nth term is (difference) $\times n$ × n + adjustment.
A geometric sequence multiplies by the same amount each time, so it grows much faster.
Higher: an arithmetic sequence has nth term $a + (n - 1) d$ a + (n−1)d and sum $S_{n} = \frac{n}{2} (2 a + (n - 1) d)$ Sₙ = n/2(2a + (n−1)d).
Gradient = change in $y$ y ÷ change in $x$ x; in $y = m x + c$ y = mx + c, $m$ m is the gradient and $(0, c)$ (0, c) the $y$ y-intercept.
To find a line's equation: gradient first, then substitute a point to find $c$ c.
Parallel lines share a gradient; perpendicular gradients multiply to $- 1$ −1 (Higher).
Midpoint: average the coordinates. Length: Pythagoras on the changes in $x$ x and $y$ y.
The highest power of $x$ x and its sign set the shape: line, parabola, S-shaped cubic, or two-branch reciprocal.
Roots are where $y = 0$ y = 0; to solve $f (x) = k$ f(x) = k from a graph, read off where the line $y = k$ y = k meets the curve.
Higher/Extended: $sin x$ sin x and $cos x$ cos x repeat every $36 0^{\circ}$ 360°, $tan x$ tan x every $18 0^{\circ}$ 180°; use symmetry for second solutions.
$f g (x)$ fg(x) means do $g$ g first, then $f$ f; order matters.
To find $f^{- 1} (x)$ f⁻¹(x), undo the steps in reverse order, or rearrange $y = f (x)$ y = f(x) to make $x$ x the subject.
$f (x) + a$ f(x) + a moves the graph up $a$ a; $f (x + a)$ f(x + a) moves it left $a$ a; $- f (x)$ −f(x) and $f (- x)$ f(−x) reflect in the $x$ x- and $y$ y-axes.
The domain excludes inputs that would divide by zero or take the square root of a negative.
Gradient of a distance–time graph = speed; steeper means faster; a crossing means the same place at the same time.
A straight-line graph's gradient is a rate (km per mile, £ per hour); its intercept is a fixed amount.
On a velocity–time (speed–time) graph, gradient = acceleration and area = distance.
Higher/Extended: the tangent's gradient gives the rate of change at an instant.
If $y = a x^{n}$ y = axⁿ, then $\frac{d y}{d x} = na x^{n - 1}$ dy/dx = naxⁿ⁻¹; substitute $x$ x into $\frac{d y}{d x}$ dy/dx for the gradient at a point.
Stationary points: solve $\frac{d y}{d x} = 0$ dy/dx = 0, find $y$ y, then decide maximum or minimum.
To maximise a quantity, write it in one variable, differentiate and set the derivative to zero.
$v = \frac{d s}{d t}$ v = ds/dt and $a = \frac{d v}{d t}$ a = dv/dt.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Sequences, Functions & Graphs for Edexcel International GCSE Maths A (4MA1). 🎉

Mission 1Chessboard: 1 grain, then double each square. How many on square 64?
Mission 2Rise 1 for every 12 along: which of three ramps is legal?
Mission 3Four graphs, four equations: can you match them without plotting?
Mission 4A machine doubles your number, then adds 3. How do you undo it?
Mission 5Two runners' distance–time graphs cross. What does the crossing mean?
Mission 640 m of fence and a wall: which rectangle encloses the most area?

Your stars: 0 / 0

Next: test yourself in the Verify stage.

📣 Smashed it? Share your score

Challenge a mate to beat your stars, or show a parent how you got on.

→ Back to IGCSE Mathematics · Edexcel

Specification points and sources for this lesson are listed at the bottom of the page.

Sequences, Functions & Graphs

Chessboard: 1 grain, then double each square. How many on square 64?

Two kinds of rule

Finding the nth term of a linear sequence

Special sequences

Arithmetic series (Higher)

Rise 1 for every 12 along: which of three ramps is legal?

Gradient measures steepness

The equation y=mx+cy = mx + c

Equation of the line through two points

Parallel and perpendicular lines

Midpoint and length of a line segment

Four graphs, four equations: can you match them without plotting?

Lines and parabolas

Reading a quadratic graph

Cubics and reciprocals

Trigonometric graphs (Higher)

A machine doubles your number, then adds 3. How do you undo it?

Function notation (Higher)

Composite functions (Higher)

Inverse functions (Higher)

Domain and range (Higher)

Transforming graphs (Higher)

Two runners' distance–time graphs cross. What does the crossing mean?

Distance–time graphs

Reading two runners' graphs

Conversion graphs and rates

Speed–time graphs

Gradient of a curve (Higher)

40 m of fence and a wall: which rectangle encloses the most area?

The gradient function (Higher)

Gradient at a point (Higher)

Stationary points (Higher)

Finding and classifying turning points (Higher)

Calculus and motion (Higher)

Big ideas

Mini-lesson complete!

📣 Smashed it? Share your score

The equation $y = m x + c$ y = mx + c