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IGCSE Mathematics 0580 · Chapter 5 notes · ~18 min read

Sequences, functions & graphs

Spot patterns and write the nth term, use function machines, sketch and read every graph shape, and meet the gradient of a curve and differentiation, all with clear pictures.

nth termFunctions & inverseStraight-line & curve graphsGradient of curvesDifferentiation
RSby Riyaz Sir
mentorriyaz.in
No tag = Core (everyone)EXTENDED Extended onlyhighlight = comes up again and againTest yourself at the end

In 30 seconds

  • A sequence follows a rule. For a linear sequence the nth term is dn + c, where d is the common difference. A function is a machine: inverse = undo it, composite = do one then the other.
  • Graphs have shapes: straight lines, U-shaped parabolas, S-shaped cubics and the two-branch reciprocal. Gradient on a distance-time graph is speed, and on a speed-time graph it is acceleration. The area under a speed-time graph is distance.
  • Extended: quadratic sequences, function notation, tangents and differentiation to find gradients and turning points.

By the end you can

  • Continue sequences and describe the term-to-term rule.
  • Find the nth term of a linear sequence and test whether a number is in it.
  • Recognise and sketch straight-line, quadratic, cubic and reciprocal graphs, and plot graphs from tables of values.
  • Solve equations by reading intersections from graphs.
  • Draw and interpret distance-time and speed-time graphs.
  • Find the nth term of quadratic and exponential sequences. EXT
  • Use function notation, composite and inverse functions. EXT
  • Estimate the gradient of a curve with a tangent, and differentiate powers of x. EXT
  • Find turning points and apply differentiation to problems. EXT

1Sequences

Types of sequence

Arithmetic3+47+411+415+419add 4 each timeGeometric2× 36× 318× 354× 3162multiply by 3Square numbers1+34+59+716+925n²: 1², 2², 3², …Triangular1+23+36+410+515add 2, 3, 4, …Fibonacci1+01+12+13+25add the last two
Find the difference or the ratio between neighbouring terms. That tells you which kind of sequence you have.

Term — one number in a sequence. The first term is T(1), the second T(2), and the nth term is T(n).

Term-to-term rule — how to get from one term to the next, for example "add 4".

nth term — a formula that gives any term directly from its position n.

  • Arithmetic (linear): same difference each time. Geometric: same multiplier. Square numbers: 1, 4, 9, 16, … Cube numbers: 1, 8, 27, 64, … Triangular: 1, 3, 6, 10, …
  • Fibonacci-type: each term is the sum of the two before: 1, 1, 2, 3, 5, 8, 13, …
Worked example

Write the next two terms of 2, 6, 18, 54, …

  1. 6 ÷ 2 = 3, 18 ÷ 6 = 3, so the rule is × 3
  2. 54 × 3 = 162
  3. 162 × 3 = 486

nth term of a linear sequence

position n1234term581114+3+3+33n (3 × n)36912term − 3n2222the difference is 3, so start with 3n, then add 2: nth term = 3n + 2
The difference tells you the number in front of n. Then adjust with a constant.
nth term = dn + c
d = common difference. c = (first term) − d
  • Method: (1) find the common difference d; (2) write dn; (3) compare with the sequence and add or subtract the number that fixes it.
  • Is a number in the sequence? Set the nth term equal to the number and solve for n. If n is a whole number, it is a term.
Worked example

Find the nth term of 20, 17, 14, 11, …

  1. common difference d = −3, so start with −3n
  2. n = 1: −3(1) = −3, but the term is 20, so add 23
  3. nth term = −3n + 23 (or 23 − 3n)
  4. check n = 4: −12 + 23 = 11 ✓
Worked example

Find the 50th term of 4n − 1, and decide whether 150 is a term.

  1. 50th term: 4 × 50 − 1 = 199
  2. set 4n − 1 = 150, so 4n = 151
  3. n = 37.75, not a whole number
  4. 150 is not a term
Common mistakeWriting the nth term as 3n + 5 for the sequence 5, 8, 11. Check n = 1: 3 + 5 = 8, not 5. The constant is first term − d: 5 − 3 = 2.

Quadratic and exponential sequencesEXTENDED

position n1234sequence2714231st difference5792nd difference22constant: so an n² term2 ÷ 2 = 1, so n²14916sequence − n²13571, 3, 5, 7 is 2n − 1, so the nth term = n² + 2n − 1
A constant second difference means the nth term has n² in it. The number in front of n² is half the second difference.
  • EXT Quadratic: the second difference is constant. The coefficient of n² is half the second difference. Subtract the n² terms, then find the linear nth term of what is left.
  • EXT Exponential (geometric): nth term = a × rn−1, where a is the first term and r the common ratio. 3, 6, 12, 24, … is 3 × 2n−1.
Worked example

EXT Find the nth term of 4, 10, 18, 28, …

  1. first differences: 6, 8, 10
  2. second differences: 2, 2, so a = 2 ÷ 2 = 1: n²
  3. n² gives 1, 4, 9, 16
  4. sequence − n²: 3, 6, 9, 12 = 3n
  5. nth term = n² + 3n
  6. check n = 3: 9 + 9 = 18 ✓
Worked example

EXT Find the nth term of 2, 6, 18, 54, …

  1. ratio = 3 and first term = 2
  2. nth term = 2 × 3n−1
  3. check n = 4: 2 × 3³ = 54 ✓

2Functions EXT

Function notation, composite and inverse

f(x) = 2x + 3: the machinex× 2+ 3f(x)4811f(x) = 2x + 3inverse f⁻¹(x): undo in reverse ordery− 3÷ 2x1184f⁻¹(x) =(x − 3) ÷ 2composite fg(x): g FIRST, then fxg: x²f: 2x + 3fg(x)3921fg(x) = f(g(x))= 2x² + 3
A function is a machine: input x, output f(x). Composite functions apply two machines in order, and the inverse undoes the machine.

Function — a rule that gives exactly one output for every input. f(x) = 2x + 3 means "double x, then add 3".

Composite function — fg(x) means g first, then f: fg(x) = f(g(x)).

Inverse function — f⁻¹ undoes f. It takes the output back to the input.

  • EXT f(4) means "put x = 4 into f": f(4) = 2(4) + 3 = 11.
  • EXT To find f⁻¹: write y = f(x), make x the subject, then swap y back to x. (See rearranging formulae.)
Worked example

EXT f(x) = 3x − 2. Find f(4) and solve f(x) = 13.

  1. f(4) = 3(4) − 2 = 10
  2. 3x − 2 = 13
  3. 3x = 15, so x = 5
Worked example

EXT f(x) = 2x + 3 and g(x) = x². Find fg(x) and gf(x).

  1. fg(x) = f(x²) = 2x² + 3 (square first, then double and add 3)
  2. gf(x) = g(2x + 3) = (2x + 3)²
  3. = 4x² + 12x + 9
  4. the order matters: fg ≠ gf
Worked example

EXT f(x) = x + 53. Find f⁻¹(x).

  1. y = x + 53
  2. multiply by 3: 3y = x + 5
  3. x = 3y − 5
  4. swap: f⁻¹(x) = 3x − 5

The graph of an inverse

xy-2-1123456-2-11234560ff⁻¹y = xf(x) = 2x + 1f⁻¹(x) = (x − 1) ÷ 2The graph of f⁻¹ is thereflection of f in theline y = x.Swap the coordinates:(0, 1) on f ↔ (1, 0) on f⁻¹(2, 5) on f ↔ (5, 2) on f⁻¹
The graph of the inverse is the mirror image of the original in the line y = x. Every point (a, b) becomes (b, a).
  • EXT Reflect the graph of f in the line y = x to get the graph of f⁻¹. Swap the coordinates of every point.

3Graphs

Shapes of graphs

y = ax + bstraight liney = x²U-shaped parabolay = −x²n-shaped parabolay = x³S-shaped cubicy = 1/xhyperbolay = 2xexponential (Ext)
Recognise the shape from the equation. A positive x² gives a U, a negative x² gives an n.
EquationShapeKey facts
y = ax + bstraight linegradient a, y-intercept b
y = ax² + bx + cparabolaU if a > 0, n if a < 0
y = ax³ + …cubicS-shape, up to two turning points
y = k/xhyperbolatwo branches, never touches the axes
y = aˣ (Extended)exponentialpasses (0, 1), never below the x-axis
  • Plot from a table: substitute each x to get y, plot the points, and join with a smooth curve (not straight ruler lines).
  • Check the shape: a parabola is symmetrical, so y-values repeat on either side of the turning point.
Worked example

Complete a table for y = x² − 3x, for x = −1 to 4, and find the turning point.

  1. x = −1: 1 + 3 = 4
  2. x = 0: 0
  3. x = 1: 1 − 3 = −2
  4. x = 2: 4 − 6 = −2
  5. x = 3: 9 − 9 = 0
  6. x = 4: 16 − 12 = 4
  7. symmetrical about x = 1.5, so the turning point is at (1.5, −2.25)

Solving equations from graphs

xy-2-11234-5-4-3-2-11234560−1.43.4y = 2Solve x² − 2x − 5 = 0same as x² − 2x − 3 = 21. draw y = x² − 2x − 32. draw the line y = 23. read the x-values wherethey cross:x ≈ −1.4 or x ≈ 3.4
To solve f(x) = a number, draw the horizontal line y = that number and read the x-values where it meets the curve.
  • Solve f(x) = k: draw the line y = k. The x-coordinates of the crossing points are the solutions.
  • Solve f(x) = g(x): draw both graphs. The x-coordinates of the crossings are the solutions. For example, y = x² − 2x − 3 meets the line y = x + 1 at x = −1 and x = 4.
  • Read to the accuracy of your graph: "x ≈ −1.4" means about.
Worked example

Use the graph of y = x² − 2x − 3 to solve x² − 2x − 3 = 2.

  1. draw the horizontal line y = 2
  2. it crosses the curve at two points
  3. read the x-values: x ≈ −1.4 and x ≈ 3.4
  4. (this solves x² − 2x − 5 = 0)

Distance-time graphs

km123102030400time (h)1231 out: 30 km in 1 h30 ÷ 1 = 30 km/h2 flat: stopped for 0.5 h3 back: 30 km in 1.5 h30 ÷ 1.5 = 20 km/hgradient = speedsteeper line = fasterflat line = not moving
Gradient means speed. A steeper line is a faster journey, and a flat line means stopped.
  • Gradient = distance ÷ time = speed. Flat line: not moving. Return journey: the line goes down.
  • Average speed = total distance ÷ total time (including any stops).
Worked example

Use the graph. Find the speed on the return journey and the average speed for the whole trip.

  1. return: 30 km in 1.5 h
  2. 30 ÷ 1.5 = 20 km/h
  3. total distance = 30 + 30 = 60 km, total time = 3 h
  4. average speed = 60 ÷ 3 = 20 km/h
Worked example

A cyclist rides 12 km in 40 minutes. Find the speed in km/h.

  1. 40 min = 4060 h = 23 h
  2. 12 ÷ 23 = 12 × 32
  3. = 18 km/h

Speed-time graphs

m/s1020304048121620240time (s)ABCA: speeding upgradient = 20 ÷ 10= 2 m/s²B: constant 20 m/sC: slowing downgradient = −2 m/s²area = distance½(20 + 40) × 20= 600 m
Gradient means acceleration. The area under the graph is the distance travelled.
acceleration = change in speedtime taken
= gradient of a speed-time graph. Negative = deceleration (slowing down)
  • Gradient = acceleration (m/s²). Area under the graph = distance. A flat line means constant speed.
  • EXT For a curved graph: the gradient at a point is the gradient of a tangent; the area can be estimated with trapezia.
Worked example

Use the graph to find the acceleration in the first 10 s and the total distance.

  1. acceleration = 2010 = 2 m/s²
  2. total distance = area of the trapezium
  3. = ½ × (20 + 40) × 20
  4. = 600 m
Worked example

EXT Speeds are 0, 6, 10 and 12 m/s at times 0, 2, 4 and 6 s. Estimate the distance using trapezia.

  1. strip 1: ½ × 2 × (0 + 6) = 6
  2. strip 2: ½ × 2 × (6 + 10) = 16
  3. strip 3: ½ × 2 × (10 + 12) = 22
  4. total: 6 + 16 + 22 = 44 m

4Gradient of curves and differentiation EXT

The gradient of a curve

Tangent and radiusTwo tangents from a pointtangentradius90°a tangent meets the radiusat exactly 90°ABPOtwo tangents from a pointare equal: PA = PB(triangle OAP and OBP are congruent: RHS)
The gradient of a curve changes from point to point. At a point it equals the gradient of the tangent there.

Tangent — a straight line that touches a curve at one point and has the same direction as the curve there.

  • EXT To estimate: draw a tangent with a ruler, make a right-angled triangle, then gradient = rise ÷ run.
  • EXT A positive gradient means the curve is rising; negative, falling; zero, flat (a turning point).

Differentiation

y = axⁿ → dy/dx = n × a × xn−1(a and n are numbers, x is the letter)1. multiply by the power2. reduce the power by 1y = x⁴dy/dx = 4x³y = 3x²dy/dx = 6xy = 5xdy/dx = 5y = 7dy/dx = 0y = 2x³ − 6x + 1dy/dx = 6x² − 6y = 1/x = x⁻¹dy/dx = −x⁻² = −1/x²y = √x = x½dy/dx = ½x−½last row: Extended
Multiply by the power, then reduce the power by one. A number on its own differentiates to zero.

Derivative — the gradient function, dy/dx. It gives the gradient of the curve at any x.

  • EXT Differentiate each term separately. Rewrite roots and fractions as powers first: 1x = x⁻¹ and √x = x½.
Worked example

EXT Differentiate y = 5x³ − 2x² + 7x − 4.

  1. 5x³ → 5 × 3 × x² = 15x²
  2. −2x² → −2 × 2 × x = −4x
  3. 7x → 7
  4. −4 → 0
  5. dy/dx = 15x² − 4x + 7
Worked example

EXT Find the gradient of y = x² + 3x at x = 2.

  1. dy/dx = 2x + 3
  2. when x = 2: 2(2) + 3 = 7
Worked example

EXT Differentiate y = 4x².

  1. rewrite: y = 4x⁻²
  2. multiply by the power: 4 × (−2) = −8
  3. reduce the power: x⁻³
  4. dy/dx = −8x⁻³ = −8x³
Worked example

EXT Find the equation of the tangent to y = x² at the point (2, 4).

  1. dy/dx = 2x, so the gradient at x = 2 is 4
  2. y − 4 = 4(x − 2)
  3. y − 4 = 4x − 8
  4. y = 4x − 4

Turning points

xy-2-112345-30-20-10100max (0, 5)min (4, −27)gradient > 0gradient < 0gradient > 0y = x³ − 6x² + 5dy/dx = 3x² − 12xTurning points: dy/dx = 03x(x − 4) = 0x = 0 or x = 4x = 0: y = 5 (maximum)x = 4: y = −27 (minimum)
At a turning point the gradient is zero. Solve dy/dx = 0, then find y, then decide whether it is a maximum or minimum.
  • EXT Method: (1) find dy/dx; (2) set dy/dx = 0 and solve; (3) find y for each x; (4) decide maximum or minimum using the gradient either side, or the second derivative: positive → minimum, negative → maximum.
Worked example

EXT Find and classify the turning points of y = x³ − 6x² + 5.

  1. dy/dx = 3x² − 12x = 3x(x − 4)
  2. dy/dx = 0: x = 0 or x = 4
  3. y(0) = 5, y(4) = 64 − 96 + 5 = −27
  4. second derivative = 6x − 12: at x = 0 it is −12 (negative), so (0, 5) is a maximum
  5. at x = 4 it is +12 (positive), so (4, −27) is a minimum
Worked example

EXT A farmer has 40 m of fencing for a rectangle with width x. Find the maximum area.

  1. length = 20 − x, so area A = x(20 − x) = 20x − x²
  2. dA/dx = 20 − 2x = 0, so x = 10
  3. A = 10 × 10 = 100 m² (a maximum, because the x² term is negative)

5Test yourself

Test yourself — cover the answers first

Quick check

  1. Write the next two terms of 2, 6, 18, 54.
  2. Find the nth term of 7, 11, 15, 19.
  3. Find the nth term of 20, 17, 14, 11.
  4. Is 100 a term of the sequence 3n + 1?
  5. What is the next term of 1, 1, 2, 3, 5, 8?
  6. Complete y = x² − 3x when x = 2.
  7. What does a flat line on a distance-time graph mean?
  8. What does the area under a speed-time graph represent?
  9. EXT Find the nth term of 3, 6, 11, 18.
  10. EXT f(x) = 5x − 1. Find f(3).
  11. EXT f(x) = 2x + 1 and g(x) = x − 4. Find fg(x).
  12. EXT Find the inverse of f(x) = x + 53.
  13. EXT Differentiate y = 3x⁴ − 5x.
  14. EXT Differentiate y = 6x².
  15. EXT Find the gradient of y = x² + 3x at x = 2.
  16. EXT Find the turning point of y = x² − 4x + 1.

Exam-style

  1. A sequence begins 5, 9, 13, 17. Find the nth term and decide whether 99 is a term. [3]
  2. Complete the table for y = x² − 2x − 1 for x = −1, 0, 1, 2, 3. [2]
  3. A car accelerates from rest to 24 m/s in 8 s, travels at 24 m/s for 20 s, then brakes to rest in 6 s. Find the initial acceleration and the total distance. [4]
  4. EXT f(x) = 3x + 2 and g(x) = x². Find fg(2) and f⁻¹(x). [4]
  5. EXT Find the nth term of 2, 5, 10, 17, 26. [3]
  6. EXT Find the coordinates of the turning points of y = x³ − 3x + 2 and decide whether each is a maximum or minimum. [5]
  7. EXT Find the equation of the tangent to y = x² at the point where x = 3. [3]
  8. EXT Speeds are 0, 6, 10 and 12 m/s at 0, 2, 4 and 6 s. Estimate the distance using trapezia. [3]
Show answers
  1. 162, 486
  2. 4n + 3
  3. 23 − 3n
  4. Yes (n = 33)
  5. 13
  6. −2
  7. the object is stationary
  8. the distance travelled
  9. n² + 2
  10. 14
  11. 2x − 7
  12. f⁻¹(x) = 3x − 5
  13. 12x³ − 5
  14. −12x⁻³
  15. 7
  16. (2, −3), a minimum
  17. d = 4, so nth term = 4n + 1
    4n + 1 = 99 gives n = 24.5
    not a whole number, so 99 is not a term
  18. x = −1: 1 + 2 − 1 = 2
    x = 0: −1
    x = 1: 1 − 2 − 1 = −2
    x = 2: 4 − 4 − 1 = −1
    x = 3: 9 − 6 − 1 = 2
  19. acceleration = 24 ÷ 8 = 3 m/s²
    total time = 34 s, parallel sides 34 and 20
    area = ½ × (34 + 20) × 24 = 648 m
  20. g(2) = 4, then f(4) = 14
    y = 3x + 2, so x = (y − 2) ÷ 3
    f⁻¹(x) = (x − 2) ÷ 3
  21. first differences 3, 5, 7, 9, second difference 2, so n²
    2 − 1 = 1, 5 − 4 = 1, 10 − 9 = 1, …
    nth term = n² + 1
  22. dy/dx = 3x² − 3 = 0, so x = 1 or x = −1
    y(1) = 0, y(−1) = 4
    second derivative 6x: at x = 1 it is 6 (positive): minimum at (1, 0)
    at x = −1 it is −6 (negative): maximum at (−1, 4)
  23. dy/dx = 2x = 6 at x = 3
    y = 9 at x = 3
    y − 9 = 6(x − 3), so y = 6x − 9
  24. ½ × 2 × (0 + 6) = 6
    ½ × 2 × (6 + 10) = 16
    ½ × 2 × (10 + 12) = 22
    total = 44 m
Practise this chapter
SequencesFunctionsGradient & equation of a lineDifferentiation
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