Pictures first. Formulas boxed. Every exam keyword highlighted.
HOW TO STUDY WITH THIS
Read the 💡 box the whole topic in 1 line
Study the diagram cover it, redraw it
Learn the yellow formulas & keywords
Redo the worked example without looking
THE CHAPTERS
⛰️
A1
Quadratic Functions
Complete the square → turning point.
🔍
A2
Equations & Inequalities
b² − 4ac decides it.
√
A3
Surds
Rationalise with the conjugate.
🧮
A4
Polynomials
f(a) = 0 → (x − a) is a factor.
🧩
A4
Partial Fractions
Split by the denominator type.
🎯
A5
Binomial Theorem
General term ⁿCᵣ aⁿ⁻ʳ bʳ.
📈
A6
Exponentials & Logs
log = power.
〰️
G1
Trig Functions & Graphs
ASTC + amplitude, period.
🔁
G1
Trig Identities & Equations
Prove one side only.
⭕
G2
Coordinate Geometry & Circles
m₁m₂ = −1.
📏
G2
Linear Law
Make it Y = mX + c.
🖨 PRINTCtrl+P → Save as PDF · Scale 100% · Margins None · Background graphics ON
A Math · Sec 3 · SEC K341 (G3)
⛰️
A1
Quadratic Functions
max/min, always positive/negative, models
BIG IDEAComplete the square: y = a(x − h)² + k → turning point (h, k). a > 0: minimum k. a < 0: maximum k.
y = 2x² − 8x + 3 = 2(x − 2)² − 5
Completing the square — steps
① take out a from the x terms ② halve the x coefficient, square it, add & subtract ③ tidy up 2x² − 8x + 3 = 2(x² − 4x) + 3 = 2[(x − 2)² − 4] + 3 = 2(x − 2)² − 5
Condition
For y = ax² + bx + c
always positive (above x-axis)
a > 0 and b² − 4ac < 0
always negative (below x-axis)
a < 0 and b² − 4ac < 0
As a model
ball height h = −5t² + 20t + 1 → h = −5(t − 2)² + 21 max height 21 m at t = 2 s
⚠ TRAPWith a ≠ 1, factor a out before halving. The min/max value is k; it occurs at x = h.
🔎 WHY IT WORKSWhy completing the square finds the min: (x − h)² is a square, so it is never negative — its smallest value is 0, when x = h. So a(x − h)² + k is smallest (a > 0) when the bracket is 0, giving y = k. That is why the turning point is (h, k).
🧠 Worked example — follow the thinking
Find the range of k for which kx² + 4x + k is always positive.
Step
Why I do this
Need a > 0 → k > 0
💭 "Always positive" means the graph is a ∪ that never touches the x-axis. A ∪ needs a positive x² coefficient.
Need D < 0: 4² − 4(k)(k) < 0
💭 "Never touches the x-axis" = no real roots = discriminant negative.
16 − 4k² < 0 → k² > 4 → k < −2 or k > 2
💭 Solve like a quadratic inequality: k² > 4 means k is further than 2 from 0.
Combine with k > 0 → k > 2
💭 Both conditions must be true, so take the overlap.
BIG IDEAThe discriminant D = b² − 4ac tells you how many real roots — and whether a line meets a curve.
D > 0 · 2 real roots
D = 0 · 2 equal roots
D < 0 · no real roots
Line & curve
Sub line into curve, then
meet at 2 points
D > 0
tangent (touch once)
D = 0
do not meet
D < 0
real roots / meets
D ≥ 0
Quadratic inequality
x² − x − 6 < 0 → (x − 3)(x + 2) < 0 sketch ∪ → below axis between roots −2 < x < 3 > 0 would give x < −2 or x > 3
< 0 → between the roots
Simultaneous (one linear)
x + y = 5, x² + y² = 13 → y = 5 − x x² + (5 − x)² = 13 → x² − 5x + 6 = 0 → (2, 3) and (3, 2)
🔎 WHY IT WORKSWhy b² − 4ac counts the roots: in the formula x = [−b ± √(b² − 4ac)]/2a, the ± gives two answers only if the square root is a real non-zero number. If b² − 4ac > 0 → 2 roots; = 0 → the ± adds nothing → 1 (equal) root; < 0 → you cannot square-root a negative → no real roots. For a line and a curve, substituting gives a quadratic whose roots are the x-values where they meet.
🧠 Worked example — follow the thinking
Find m if y = mx + 1 is a tangent to y = x² + 2x + 2.
Step
Why I do this
x² + 2x + 2 = mx + 1
💭 Where they meet, the y-values are equal, so set the two equations equal.
x² + (2 − m)x + 1 = 0
💭 Bring everything to one side to get a quadratic in x. Group the x terms: coefficient is (2 − m).
Tangent → D = 0: (2 − m)² − 4(1)(1) = 0
💭 A tangent touches at exactly ONE point = one (repeated) root = discriminant 0.
square both sides → solve → CHECK every answer squaring can create fake roots
⚠ TRAP√a + √b ≠ √(a + b). And (2√3)² = 4 × 3 = 12, not 6.
🔎 WHY IT WORKSWhy the conjugate removes the surd: (a + √b)(a − √b) = a² − (√b)² = a² − b — this is the difference of two squares, and squaring √b leaves a whole number. So multiplying top AND bottom by the conjugate clears the surd from the denominator without changing the value (you are multiplying by 1).
🧠 Worked example — follow the thinking
Solve √(x + 5) = x − 1.
Step
Why I do this
x + 5 = (x − 1)² = x² − 2x + 1
💭 Square both sides to remove the square root. Square the WHOLE right side, not each term.
x² − 3x − 4 = 0 → (x − 4)(x + 1) = 0
💭 Rearrange to = 0 and factorise.
x = 4 or x = −1
💭 These are only candidates — squaring can create fake answers because (−2)² = 2² too.
Check x = −1: √4 = 2 but −1 − 1 = −2 ✗ · x = 4: √9 = 3 = 3 ✓
💭 √ always means the POSITIVE root, so x = −1 fails. x = 4 only.
A Math · Sec 3 · SEC K341 (G3)
🧮
A4
Polynomials
remainder & factor theorems, cubics, a³ ± b³
BIG IDEARemainder theorem: f(x) ÷ (x − a) leaves remainder f(a). If f(a) = 0, then (x − a) is a factor.
Remainder & factor theorems
divisor (x − a) → remainder f(a) divisor (ax − b) → remainder f(b/a) f(a) = 0 ⇔ (x − a) is a factor
Identities (memorise — not given)
a³ + b³ = (a + b)(a² − ab + b²) a³ − b³ = (a − b)(a² + ab + b²) 8x³ − 27 = (2x − 3)(4x² + 6x + 9)
Solving a cubic
Try x = ±1, ±2, ±3 … (factors of the constant) until f(a) = 0
Divide (long division or compare coefficients) → quadratic
(x − 3)(2x² + bx + c) = 2x³ − 3x² − 11x + 6 x²: b − 6 = −3 → b = 3 const: −3c = 6 → c = −2
🔎 WHY IT WORKSWhy the remainder theorem works: division always gives f(x) = (x − a) × Q(x) + R. Put x = a: the bracket becomes 0, so the whole first part vanishes and f(a) = R. If R = 0, then (x − a) divides exactly — it is a factor.
🧠 Worked example — follow the thinking
f(x) = 2x³ − 3x² − 11x + 6. Solve f(x) = 0.
Step
Why I do this
Try f(1) = −6, f(−1) = 18, f(3) = 0 ✓
💭 Only try factors of the constant (±1, ±2, ±3, ±6) — a root must divide 6. f(3) = 0 means (x − 3) is a factor.
f(x) = (x − 3)(2x² + bx + c)
💭 If (x − 3) is a factor, the rest must be a quadratic. 2x² so the x³ terms match; −3c = 6 so c = −2.
Compare x²: −3 = b − 6 → b = 3
💭 Expand mentally: x² terms come from x·bx and −3·2x². Match with −3x².
2x² + 3x − 2 = (2x − 1)(x + 2) → x = 3, ½, −2
💭 Factorise the quadratic and set each factor = 0.
A Math · Sec 3 · SEC K341 (G3)
🧩
A4
Partial Fractions
3 denominator types, improper fractions
BIG IDEASplit one fraction into simpler ones. The denominator decides the form. Top degree must be lower than bottom (else divide first).
Denominator
Write as
(ax + b)(cx + d)
A/(ax + b) + B/(cx + d)
(ax + b)(cx + d)²
A/(ax + b) + B/(cx + d) + C/(cx + d)²
(ax + b)(x² + c²)
A/(ax + b) + (Bx + C)/(x² + c²)
Linear factors (cover-up)
(5x + 1)/((x − 1)(x + 1)) x = 1: A = 6/2 = 3 x = −1: B = −4/−2 = 2 → 3/(x − 1) + 2/(x + 1)
Repeated factor
1/((x − 1)(x + 1)²) x = 1: A = ¼ · x = −1: C = −½ compare x²: 0 = A + B → B = −¼ check x = 0: −¼ − ¼ − ½ = −1 ✓
Improper fraction
top degree ≥ bottom → long divide first (x² + 1)/(x² − 1) = 1 + 2/(x² − 1) = 1 + 1/(x − 1) − 1/(x + 1)
⚠ TRAPA repeated factor needs both B/(x + 1) and C/(x + 1)². A quadratic factor x² + c² needs Bx + C on top.
🔎 WHY IT WORKSWhy cover-up works: partial fractions reverse adding fractions. Multiply both sides by the denominator and you get an identity true for EVERY x — so choose x-values that make brackets 0: each one "kills" all terms except one, leaving one unknown at a time.
⚠ TRAPKeep the sign and coefficient inside b: (2 − x/2) means b = −x/2, so bʳ = (−½)ʳ xʳ. Not tested: greatest term, properties of coefficients.
🔎 WHY IT WORKSWhy ⁿCᵣ appears: (a + b)ⁿ = (a + b)(a + b)…(a + b). To make the term with bʳ you pick b from r of the n brackets and a from the rest. The number of ways to choose those r brackets is ⁿCᵣ — that is the coefficient.
🧠 Worked example — follow the thinking
Find the coefficient of x³ in (2 − x/2)⁸.
Step
Why I do this
Tᵣ₊₁ = ⁸Cᵣ (2)⁸⁻ʳ (−x/2)ʳ
💭 Write the general term with a = 2 and b = −x/2 (keep the minus inside b).
Power of x is r → need r = 3
💭 Only the b part contains x, and it is raised to r.
⁸C₃ × 2⁵ × (−½)³
💭 Substitute r = 3. (−x/2)³ = (−½)³x³.
56 × 32 × (−⅛) = −224
💭 Odd power of a negative stays negative.
A Math · Sec 3 · SEC K341 (G3)
📈
A6
Exponential & Logarithmic Functions
graphs, laws, change of base, equations, models
BIG IDEAA log is a power: y = aˣ ⇔ x = logₐ y. lg = log₁₀, ln = logₑ.
reflections in y = x · (0, 1) and (1, 0)
Law
Rule
product
log(xy) = log x + log y
quotient
log(x/y) = log x − log y
power
log xⁿ = n log x
change base
logₐ b = log b / log a
specials
logₐ a = 1 · logₐ 1 = 0
Solve exponentials
take lg/ln of both sides 2ˣ⁺¹ = 5ˣ → (x + 1) lg 2 = x lg 5 x = lg 2 / (lg 5 − lg 2) = 0.756 disguised quadratic: 4ˣ − 3(2ˣ) − 4 = 0 → let u = 2ˣ
Solve logs
combine into one log, then remove it log₂x + log₂(x − 2) = 3 x(x − 2) = 8 → x = 4 (reject −2)
📈 As a model
N = 500e⁰·²ᵗ → when N = 2000: e⁰·²ᵗ = 4 t = ln 4 ÷ 0.2 = 6.93 N₀ = value at t = 0
⚠ TRAPlog(x + y) ≠ log x + log y. You can only take logs of positive numbers — reject answers that make a log argument ≤ 0.
🔎 WHY IT WORKSWhy the log laws work: a log just asks "what power?" — log₂8 = 3 because 2³ = 8. Multiplying numbers ADDS their powers (2³ × 2⁴ = 2⁷), so log(xy) = log x + log y. Dividing subtracts powers, and a power multiplies them — every log law is an index law in disguise.
🧠 Worked example — follow the thinking
Solve lg(x + 3) − lg(x − 1) = lg 5.
Step
Why I do this
lg[(x + 3)/(x − 1)] = lg 5
💭 Quotient law: subtracting logs = log of a division. Now one log on each side.
(x + 3)/(x − 1) = 5
💭 If lg A = lg B then A = B (log is one-to-one).
x + 3 = 5x − 5 → x = 2
💭 Cross-multiply and solve.
Check: x + 3 = 5 > 0 and x − 1 = 1 > 0 ✓
💭 You can only take logs of positive numbers — always check.
A Math · Sec 3 · SEC K341 (G3)
🧭
G1
Trig for Any Angle
unit circle → ASTC → basic angle method
BIG IDEAFor ANY angle, a point on a circle of radius 1 is (cos θ, sin θ). So cos is the x-coordinate and sin is the y-coordinate — that is the whole reason for ASTC.
rotate anticlockwise from the +x-axis
Read the signs off the circle
Q1 (0–90°): x +, y + → All positive Q2 (90–180°): x −, y + → only Sin positive Q3 (180–270°): x −, y − → only Tan positive (− ÷ − = +) Q4 (270–360°): x +, y − → only Cos positive
So ASTC is not magic — it is just the signs of x and y.
🔎 WHY IT WORKSSymmetry: the point for 180° − θ is the mirror image of θ in the y-axis — same height (same sin), opposite x (cos flips sign). That is why sin(180° − θ) = sin θ. Same idea gives 180° + θ (both flip) and 360° − θ (only y flips).
Exact values — derive them, don't just memorise
half a square: sin 45° = 1/√2 = √2/2, tan 45° = 1
half an equilateral △ (side 2): sin 30° = ½, cos 30° = √3/2
🧠 Worked example — follow the thinking
Solve sin x = −0.5 for 0° ≤ x ≤ 360°.
Step
Why I do this
Basic angle α = sin⁻¹(0.5) = 30°
💭 Always use the POSITIVE value to get the acute basic angle. The sign only tells me which quadrants.
sin is negative → Q3 and Q4
💭 From the circle: y is negative below the x-axis, i.e. Q3 and Q4.
Q3: 180° + 30° = 210°
💭 Q3 angles are 180° plus the basic angle.
Q4: 360° − 30° = 330°
💭 Q4 angles are 360° minus the basic angle.
x = 210°, 330°
💭 Check with calculator: sin 210° = −0.5 ✓
A Math · Sec 3 · SEC K341 (G3)
〰️
G1
Trig Graphs — what a, b, c do
sketching y = a sin bx + c, a cos bx + c, a tan bx
BIG IDEAIn y = a sin bx + c: a stretches up/down, b squashes sideways (more waves), c lifts the whole graph.
a = 3: taller · amplitude 3
b = 2: 2 waves in 360° · period 180°
c = 2: lifted up 2
🔎 WHY IT WORKSWhy period = 360°/b: sin bx repeats when bx goes through 360°, so x only needs to go through 360° ÷ b. For sin(x/b), x must go through 360° × b. Max = c + |a|, min = c − |a| because sin only ranges −1 to 1.
🧠 Worked example — follow the thinking
Sketch y = 2 cos 2x − 1 for 0° ≤ x ≤ 360°.
Step
Why I do this
Amplitude = 2, period = 360° ÷ 2 = 180°, shift c = −1
💭 Read a, b, c first — they fix the shape before plotting anything.
Max = −1 + 2 = 1, min = −1 − 2 = −3
💭 cos 2x goes from −1 to 1, so ×2 then −1 gives −3 to 1.
Cos starts at its MAX: (0, 1)
💭 cos 0 = 1, so a cosine graph starts at the top (sine starts at the middle).
Quarter-period = 45°: points at 0, 45, 90, 135, 180 → max, mid, min, mid, max
💭 Every wave has 4 equal quarters: max → middle → min → middle → max.
Repeat for 180° to 360°, join smoothly
💭 Period 180° means 2 full waves fit in 0–360°.
y = 2 cos 2x − 1 · dashed line = centre y = −1
⚠ TRAPb must change the period, not the height. A negative a flips the graph: y = −2 sin x starts going down.
A Math · Sec 3 · SEC K341 (G3)
🔁
G1
Trig Identities — how to use them
turning trig into algebra, proving
BIG IDEAIdentities let you swap one trig function for another so an equation has only ONE function — then it becomes algebra.
🔎 WHY IT WORKSWhere sin² + cos² = 1 comes from: on the unit circle the point (cos θ, sin θ) is 1 unit from O, so by Pythagoras cos²θ + sin²θ = 1. Divide every term by cos²θ → 1 + tan²θ = sec²θ. Divide by sin²θ → cot²θ + 1 = cosec²θ. That is all three given identities from one picture.
If the equation has…
Swap using
To get only…
sin² and cos
sin² = 1 − cos²
cos
cos² and sin
cos² = 1 − sin²
sin
tan² and sec
tan² = sec² − 1
sec
cot² and cosec
cot² = cosec² − 1
cosec
sin and cos (no squares)
divide by cos → tan
tan
🧠 Worked example — follow the thinking
Solve tan²x + sec x = 1 for 0° ≤ x ≤ 360°.
Step
Why I do this
tan²x = sec²x − 1
💭 The equation has tan² AND sec — swap tan² so only sec is left.
sec²x − 1 + sec x = 1 → sec²x + sec x − 2 = 0
💭 Now it is a quadratic in sec x. Let u = sec x if it helps: u² + u − 2 = 0.
(sec x + 2)(sec x − 1) = 0
💭 Factorise like normal algebra.
sec x = 1 → cos x = 1 → x = 0°, 360°
💭 sec = 1/cos, so flip it: always convert sec, cosec, cot back to cos, sin, tan to use the calculator.
sec x = −2 → cos x = −½ → basic 60°, Q2 & Q3 → x = 120°, 240°
💭 Work on ONE side only. Pick the side with more to simplify.
1 − cos 2x = 1 − (1 − 2sin²x) = 2sin²x
💭 There are 3 forms of cos 2x — choose 1 − 2sin²x because it CANCELS the 1.
sin 2x = 2 sin x cos x
💭 Double-angle formula from the sheet.
LHS = 2sin²x / (2 sin x cos x) = sin x / cos x = tan x = RHS ∎
💭 Cancel 2 sin x top and bottom, then sin/cos = tan.
A Math · Sec 3 · SEC K341 (G3)
🔀
G1
Compound & Double Angle — how to use them
exact values, given ratios, equations
BIG IDEACompound-angle formulas break a "weird" angle into two nice angles. Double-angle is just compound with A = B.
🔎 WHY IT WORKSDouble angle comes free: put B = A in sin(A + B) = sin A cos B + cos A sin B → sin 2A = 2 sin A cos A. Put B = A in cos(A + B) → cos 2A = cos²A − sin²A, then use sin² + cos² = 1 to get the other two forms. You don't need to memorise them separately — they're on the sheet anyway.
Which cos 2A form?
Use when
cos²A − sin²A
both sin and cos appear
2cos²A − 1
the rest of the equation is in cos
1 − 2sin²A
the rest of the equation is in sin, or there is a "1 −" to cancel
🧠 Worked example — follow the thinking
Find the exact value of sin 75°.
Step
Why I do this
75° = 45° + 30°
💭 Split into two angles whose exact values I know.
sin(45° + 30°) = sin 45° cos 30° + cos 45° sin 30°
💭 Compound formula from the sheet, with A = 45°, B = 30°.
= (√2/2)(√3/2) + (√2/2)(½)
💭 Substitute exact values (from the special triangles).
= (√6 + √2)/4
💭 Combine over 4. Calculator check: 0.966 ✓
draw a right △ to find the missing ratio
🧠 Use a triangle
sin A = 3/5, cos B = 5/13, A and B acute. Find sin(A + B).
Step
Why I do this
cos A = 4/5, sin B = 12/13
💭 Draw the triangles: 3-4-5 and 5-12-13 (Pythagoras).
sin(A + B) = (3/5)(5/13) + (4/5)(12/13)
💭 Substitute into the compound formula.
= 15/65 + 48/65 = 63/65
💭 If A were obtuse, cos A would be −4/5 (Q2) — always check the quadrant.
🧠 Worked example — follow the thinking
Solve sin 2x = cos x for 0° ≤ x ≤ 360°.
Step
Why I do this
2 sin x cos x = cos x
💭 Expand sin 2x so every term is in x, not 2x.
2 sin x cos x − cos x = 0 → cos x(2 sin x − 1) = 0
💭 Move everything to one side and FACTORISE — never divide by cos x (you would lose cos x = 0).
cos x = 0 → x = 90°, 270°
💭 Read straight off the graph of cos.
sin x = ½ → x = 30°, 150°
💭 Basic 30°, sin + → Q1, Q2.
x = 30°, 90°, 150°, 270°
💭 4 answers. Dividing by cos x would have lost 90° and 270°.
A Math · Sec 3 · SEC K341 (G3)
🌊
G1
R-Formula — how and why
a sin x + b cos x as one wave
BIG IDEAa sin x + b cos x is secretly ONE sine wave. Rewrite it as R sin(x + α) — then max, min and solving become easy.
R = √(a² + b²), tan α = b/a
🔎 WHY IT WORKSWhy it works: expand R sin(x + α) = R sin x cos α + R cos x sin α. Match with 3 sin x + 4 cos x: R cos α = 3 and R sin α = 4. That is a right triangle with sides 3 and 4 → R = 5 (Pythagoras), tan α = 4/3 (opp/adj).
Expression
Write as
R cos α = …, R sin α = …
a sin x + b cos x
R sin(x + α)
a, b
a sin x − b cos x
R sin(x − α)
a, b
a cos x + b sin x
R cos(x − α)
a, b
a cos x − b sin x
R cos(x + α)
a, b
🧠 Worked example — follow the thinking
Solve 3 sin x + 4 cos x = 2 for 0° ≤ x ≤ 360°.
Step
Why I do this
3 sin x + 4 cos x = 5 sin(x + 53.13°)
💭 R = √(9 + 16) = 5, α = tan⁻¹(4/3) = 53.13°. Keep 2 d.p. for α in working.
5 sin(x + 53.13°) = 2 → sin(x + 53.13°) = 0.4
💭 Now it is a single sine — solve for the whole bracket first.
Range: 53.13° ≤ x + 53.13° ≤ 413.13°
💭 Add α to BOTH ends of the interval — the bracket has a different range from x.
list vertices anticlockwise, repeat the first: ½ |(x₁y₂ + x₂y₃ + …) − (y₁x₂ + y₂x₃ + …)| (0,0), (4,0), (5,3), (1,4) → ½|32 − 3| = 14.5
(x − a)² + (y − b)² = r²
General form
x² + y² + 2gx + 2fy + c = 0 centre (−g, −f) · radius √(g² + f² − c) x² + y² − 6x + 4y − 12 = 0 → centre (3, −2), r = √(9 + 4 + 12) = 5 tangent ⊥ radius at the point of contact
⚠ TRAPCoefficients of x² and y² must be 1 before reading g, f, c — divide through first. Problems with two circles are not tested.
🔎 WHY IT WORKSWhy m₁m₂ = −1: turning a line 90° swaps the "rise" and "run" and flips one sign: gradient a/b becomes −b/a. Multiply them: (a/b)(−b/a) = −1. A circle equation is just Pythagoras: every point (x, y) is distance r from the centre (a, b), so (x − a)² + (y − b)² = r².
🧠 Worked example — follow the thinking
A(1, 2), B(5, −4). Find the perpendicular bisector of AB.
Step
Why I do this
Midpoint = ((1 + 5)/2, (2 − 4)/2) = (3, −1)
💭 A bisector passes through the MIDDLE of AB.
m_AB = (−4 − 2)/(5 − 1) = −3/2
💭 Gradient = change in y ÷ change in x.
Perpendicular m = 2/3
💭 Flip and change sign so that m₁m₂ = −1.
y + 1 = ⅔(x − 3) → y = ⅔x − 3
💭 Use y − y₁ = m(x − x₁) with the midpoint.
A Math · Sec 3 · SEC K341 (G3)
📏
G2
Linear Law
converting to Y = mX + c
BIG IDEATurn a curved relationship into a straight line Y = mX + c, read m and c from the graph, then find the constants.
Given
Rearrange
Y
X
m
c
y = axⁿ
lg y = n lg x + lg a
lg y
lg x
n
lg a
y = kbˣ
lg y = (lg b)x + lg k
lg y
x
lg b
lg k
y = ax² + bx
y/x = ax + b
y/x
x
a
b
y = a/x + b
y = a(1/x) + b
y
1/x
a
b
straight line → read gradient & intercept
Steps
① rearrange to Y = mX + c — Y and X must contain only x, y (no unknown constants) ② make a table of X, Y ③ plot, draw line of best fit ④ m = gradient, c = Y-intercept ⑤ solve for the constants
⚠ TRAPThe Y-axis intercept is only "c" if the X-axis starts at 0. If not, use c = Y − mX with a point on the line.
🔎 WHY IT WORKSWhy taking lg makes a straight line: y = axⁿ is a curve. Take lg of both sides: lg y = lg a + n lg x. If you call Y = lg y and X = lg x, this is Y = nX + lg a — exactly y = mx + c. So the gradient is n and the intercept is lg a.
🧠 Worked example — follow the thinking
y = axⁿ. The graph of lg y against lg x passes through (0.2, 1.1) and (0.8, 2.3). Find a and n.
Step
Why I do this
lg y = n lg x + lg a
💭 Take lg of both sides and use the log laws to make it Y = mX + c.
n = gradient = (2.3 − 1.1)/(0.8 − 0.2) = 2
💭 The gradient of the straight line IS n.
lg a = 1.1 − 2(0.2) = 0.7
💭 Use c = Y − mX with a point (safe even if the axis does not start at 0).