IB Math AA HL — All Flashcards

a × 10ᵏ, with 1 ≤ a < 10 and k a whole number. Example: 4.53 × 10⁴.

1 ≤ a < 10 (at least 1, less than 10). So 7 × 10³ ✓ but 12 × 10³ ✗.

Positive. Example: 52 000 = 5.2 × 10⁴.

Negative. Example: 0.0007 = 7 × 10⁻⁴.

Count how many places the point moves to leave one non-zero digit in front. Left → positive, right → negative.

7.3 × 10⁴.

6.1 × 10⁻⁵ — the ᴇ symbol means × 10.

The coefficient 45.3 is not between 1 and 10. Correct: 4.53 × 10⁷.

Add them. Example: 10³ × 10⁴ = 10⁷.

Subtract them. Example: 10⁸ ÷ 10³ = 10⁵.

Multiply the exponents. Example: (10⁴)² = 10⁸.

Multiply the coefficients, add the powers of ten, then re-normalise. Example: (3×10⁴)(2×10³) = 6×10⁷.

4 × 10⁶ — square the coefficient (2² = 4) and double the exponent.

(3×10²)³ = 27×10⁶ = 2.7 × 10⁷ cm³.

5 × 10⁻⁴ — 0.5 = 5 × 10⁻¹, so subtract 1 from the exponent.

2.7 × 10⁷ — 27 = 2.7 × 10¹, so add 1 to the exponent.

Order matters — the position of each object counts, so ABC and CBA are different. Count by filling positions and multiplying.

Draw one box per position, write how many choices go in each box, then multiply. Fill the most restricted box first.

The first box has 9 choices (1–9, not 0); fill the rest from the remaining digits, then multiply.

n! = n × (n − 1) × … × 1. Example: 9 people in a line = 9! = 362880.

ⁿPᵣ = n!/(n − r)! — the number of ways to arrange r objects out of n in a definite order.

It multiplies r numbers counting down from n: n × (n − 1) × … (r factors). e.g. ⁸P₃ = 8 × 7 × 6 = 336.

All n → n!. Just r of them, in order → ⁿPᵣ = n!/(n − r)!.

Arrangements — the position matters, so use n! (all) or ⁿPᵣ (some), filling positions and multiplying.

MATH, arrow right to PRB, then 2: nPr. Type n, choose nPr, type r, ENTER.

n! — arranging all n in order (since n!/(n − n)! = n!/0! = n!).

Does order matter? Arrange/rank/roles → yes (ⁿPᵣ); choose/team/committee → no (ⁿCᵣ). Then check for restrictions.

Block method — glue them into one item, arrange, then multiply by the internal orders (k!).

Total − together (for a pair), or the gap method (seat the others, drop the rest into separate gaps).

Fix that member in place, then choose the rest from those remaining.

Add up the allowed cases, or use total − unwanted (the complement).

Divide n! by the factorial of each repeated letter: n! ÷ (p! q! …).

Choose which of the moves go 'up' (or right): ⁿCᵣ.

Because the wording sets the type. Medals (distinct roles) → ⁿPᵣ; a committee (no roles) → ⁿCᵣ.

When n is negative or a fraction (not a positive whole number) — the terms never stop.

1 + nx + n(n−1)/2! x² + n(n−1)(n−2)/3! x³ + … , valid for |x| < 1.

Multiply by the next falling factor of n on top and divide by the next factorial: n, then n(n−1)/2!, then n(n−1)(n−2)/3!.

Replace every x in the formula with kx, then simplify — remember to square and cube the k.

1 − x + x².

1 + 6x + 27x².

n.

n(n − 1)/2.

T_(r+1) = ⁿCᵣ aⁿ⁻ʳ bʳ — choose r of the b's.

Use the general term; set the power of x equal to k to fix r, then compute ⁿCᵣ aⁿ⁻ʳ bʳ.

Write the coefficient as an expression in the unknown, set it equal to the given number, and solve.

ⁿC₂ a².

⁵C₃ × 2² = 10 × 4 = 40.

10a² = 40 ⇒ a² = 4 ⇒ a = 2.

Forgetting to raise the inside coefficient to the power r — (2x)² = 4x², not 2x².

The coefficient counts the ⁿCᵣ ways that term is formed — the powers alone aren't enough.

When |x| < 1 — only then do the terms shrink so the series settles to a value.

|kx| < 1, i.e. |x| < 1/|k|.

Choose x so the bracket equals the target number (x should be small), then evaluate the first few terms.

x = 0.02: 1 + 0.01 − 0.00005 = 1.00995.

Only then do the powers of x get smaller, so the later terms fade and the sum converges.

|2x| < 1, so |x| < ½.

x = 0.04 (since 1 + x = 1.04).

Yes — the later (ignored) terms are even tinier, so the first few terms are more accurate.

When you pick a GROUP and the order inside doesn't matter — a team, a committee, a sample. {A,B} is the same as {B,A}.

Top: r numbers counting down from n. Bottom: r! = r × (r − 1) × … × 1. e.g. ⁵C₂ = (5 × 4)/(2 × 1) = 10.

Picking r in order counts each group r! times (its r! orderings). A group has no order, so divide those out.

Ask: does the order matter? Order matters → arrangement (ⁿPᵣ). Order doesn't → selection (ⁿCᵣ).

ⁿCᵣ = n!/(r!(n − r)!) — the number of ways to choose r objects from n with order ignored.

(8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56.

MATH, arrow right to PRB, then 3: nCr. Type n, choose nCr, type r, ENTER.

ⁿCᵣ — these are unordered groups, so order doesn't matter.

Choose — the increasing order is fixed, so each set of digits gives exactly one number. Count with ⁿCᵣ.

Once you pick the items, there is only ONE way to put them in increasing order — nothing is left to arrange, so order doesn't add anything.

Choose 4 from 1–9 (0 can't appear): ⁹C₄ = 126.

'increasing order', 'decreasing order', 'alphabetical order' — all fix the order, so use ⁿCᵣ.

In increasing order 0 would have to come first, but a number can't start with 0. So choose from 1–9.

Choose 3 of the 7 letters: ⁷C₃ = 35 (alphabetical order is automatic).

You counted the orderings, but the order is forced (one per set). Divide out — i.e. use ⁿCᵣ instead.

No — both fix the order, so both are ⁿCᵣ. Each chosen set has one increasing and one decreasing arrangement.

Put that person in first (they use one place), then choose the rest from the people who are left.

Captain in, choose 3 from the remaining 9: ⁹C₃ = 84.

That person is unavailable, so just choose from everyone else.

Choose all 4 from the other 9: ⁹C₄ = 126.

An ordinary selection (ⁿCᵣ) on the remaining people for the remaining places.

Both drop by 1: one fewer place to fill AND one fewer person to choose from.

The number of people drops by 1, but the number of places stays the same.

'must include', 'always plays', 'the chair is on the committee', or 'must not be chosen / refuses'.

Choose from each group separately, then multiply (AND → ×). e.g. 2 men and 2 women = ⁵C₂ × ⁶C₂.

Each way of choosing the first group can pair with each way of choosing the second — that's the multiplication principle.

Add the cases: exactly 2 + exactly 3 + … Each case = choose that many women AND the rest from the other group.

When there are many 'at least' cases but few to exclude — e.g. 'at least 1' = total − (none).

The other 2 must be women, so multiply ⁵C₂ (men) × ⁶C₂ (women) — the numbers add to 4.

Exactly 0 girls + exactly 1 girl, added together.

Adding the two ⁿCᵣ values instead of multiplying them (it's AND, not OR).

Split the numbers into remainder groups (mod 3), then count the choices that make the total work — choosing from each group, multiplying and adding cases.

Glue them into one block, arrange the blocks, then multiply by the ways to arrange inside the block.

k! — the ways to arrange the k people inside the block.

Treat AB as one block. Arrange the (n − 1) things, then × 2 for the AB / BA orders inside.

'Together' allows both internal orders (× 2). 'Immediately after' fixes the order (× 1).

One fewer for each extra person in the block — a block of k among n total leaves (n − k + 1) things to arrange.

Glue the 3: arrange 5 things (5!) × 3! inside = 120 × 6 = 720.

The order inside the block is fixed (only one way), so there's nothing extra to multiply by.

'together', 'next to each other', 'side by side', 'consecutive', 'immediately after/before'.

Total − together. Count all arrangements, then subtract the ones where the two ARE together (block method).

Gap method: arrange the other items first; they make gaps; place the restricted items into different gaps.

n + 1 gaps — one between each pair plus one at each end.

They go into different gaps and the items are distinct, so the order in which gaps are filled matters → ⁿPᵣ.

5! − (4! × 2) = 120 − 48 = 72.

Boys 4! = 24, 5 gaps, girls ⁵P₂ = 20: 24 × 20 = 480.

A single pair not together → total − together. A whole group with none adjacent → the gap method.

'not next to', 'not adjacent', 'apart', 'at least one seat between', 'not sharing a boundary'.

A route is an arrangement of right and up moves. Choose which moves go 'up' (or right) with ⁿCᵣ.

Make a right moves and b up moves (a + b total); choose the b up moves: ⁽ᵃ⁺ᵇ⁾Cᵦ.

Divide n! by the factorial of each repeated letter: n! ÷ (p! q! …).

Swapping two identical letters gives the same word, so plain n! counts each arrangement several times; dividing removes the duplicates.

6 letters with A×3, N×2: 6! ÷ (3! 2!) = 720 ÷ 12 = 60.

7 moves, choose 3 up: ⁷C₃ = 35.

A grid route is a word made of two letters (R and U), so both use the same arrangement idea.

11 letters with S×4, I×4, P×2: 11! ÷ (4! 4! 2!) = 34650.

ⁿP₂ = n(n − 1). Set n(n − 1) = k, rearrange to a quadratic, solve, and keep the positive whole-number root.

ⁿC₂ = n(n − 1)/2. Multiply by 2 to get n(n − 1) = 2k, solve the quadratic, keep the positive root.

n counts objects, so it must be a positive whole number — a negative root has no meaning.

n(n − 1).

n(n − 1)/2.

a + b = n, because ⁿCᵣ = ⁿCₙ₋ᵣ. e.g. ⁿC₃ = ⁿC₇ ⇒ n = 10.

n(n − 1) = 42 ⇒ (n − 7)(n + 6) = 0 ⇒ n = 7.

n(n − 1) = 56 ⇒ (n − 8)(n + 7) = 0 ⇒ n = 8.

Write it as A/(x − a) + B/(x − b), then find A and B.

Clear the fractions, then substitute the x that makes one bracket zero — it deletes a term and leaves a single unknown.

x = a — the root of its own bracket — which zeroes the OTHER term and isolates A.

Multiply both sides by the whole denominator to clear the fractions.

2/(x + 1) + 3/(x − 2).

2/(x − 1) − 1/(x + 3).

When the bottom is two DIFFERENT linear factors, e.g. (x − 1)(x + 3).

Add the two fractions back over a common denominator — you should recover the original fraction.

Factorise it first into (x − a)(x − b), then split as usual.

When the top's highest power is as big as (or bigger than) the bottom's. You must divide first.

Divide to peel off a whole part, leaving a proper fraction; then split the proper part into partial fractions.

1 + (1/2)/(x − 1) − (1/2)/(x + 1) — divide first since x² = (x² − 1) + 1.

(x + 3)(x − 2).

(x − 2)(x + 2) — difference of two squares.

Equate coefficients: expand the right side and match the x-terms and the constant terms, then solve.

No — same degree top and bottom (top-heavy). Divide first: it's 1 + 2/(x² − 1).

The imaginary unit, defined by i² = −1 (so i = √(−1)).

z = a + bi, where a is the real part and b is the imaginary part.

Combine the real parts together and the imaginary parts together — they don't mix.

Expand like two brackets (FOIL), then replace every i² with −1 and collect terms.

4 − 3i.

3 − 12i + 2i − 8i² = 3 − 10i + 8 = 11 − 10i.

i, −1, −i, 1 — then the cycle repeats.

i² = −1; it's the step that turns a multiplication of complex numbers back into a + bi form.

z* = a − bi — flip the sign of the imaginary part (real part unchanged).

a² + b², which is always a real number (= |z|²).

Multiply top and bottom by the conjugate of the bottom, making the denominator real, then write as a + bi.

4 + 7i.

Because z × z* = a² + b² is real, so it clears the i from the denominator.

Multiply by (2 + 3i)/(2 + 3i): (7 + 17i)/13 = 7/13 + (17/13)i.

The mirror image of z in the real axis (same real part, opposite imaginary part).

4 + 25 = 29.

The plane for complex numbers: real part across (horizontal), imaginary part up (vertical).

At the point (a, b).

The distance from the origin to z; |z| = √(a² + b²).

It's Pythagoras — the point (a, b) is at horizontal distance a and vertical distance b from the origin.

√(9 + 16) = √25 = 5.

√(25 + 144) = √169 = 13.

No — it's a distance, so |z| ≥ 0.

It decides up (b > 0) or down (b < 0) on the diagram; the modulus is unaffected because b is squared.

z = r(cosθ + i sinθ) = r cisθ, where r = |z| (modulus) and θ = arg z (argument).

r = √(a² + b²); θ = arctan(b/a), then adjust for the quadrant.

a = r cosθ and b = r sinθ, then write a + bi.

arctan(b/a) only returns quadrant 1 or 4 angles; quadrant 2 or 3 points need ±π added.

r = 2, θ = π/3, so 2 cis(π/3).

r = 2; quadrant 2 so θ = π − π/6 = 5π/6; 2 cis(5π/6).

a = 2cos30° = √3, b = 2sin30° = 1, so √3 + i.

The angle the line from 0 to z makes with the positive real axis.

Multiply the moduli and add the arguments: r₁r₂ cis(θ₁ + θ₂).

Divide the moduli and subtract the arguments: (r₁/r₂) cis(θ₁ − θ₂).

Scales the point by r and rotates it by θ about the origin.

Rotates 90° anticlockwise (i = cis(π/2), modulus 1).

6 cis(5π/12) — moduli 2×3 = 6, arguments π/6 + π/4 = 5π/12.

3 cis(5π/12) — moduli 12÷4 = 3, arguments 2π/3 − π/4 = 5π/12.

Adding the moduli or multiplying the arguments. It's moduli × and arguments +.

Multiplying just scales and rotates, so it needs only one multiply and one add — no FOIL.

z = r e^(iθ), with r the modulus and θ the argument in radians.

e^(iθ) = cosθ + i sinθ — it links the exponential form to the polar form.

Multiply the moduli and ADD the exponents: r₁r₂ e^(i(θ₁+θ₂)) (an index law).

Divide the moduli and SUBTRACT the exponents: (r₁/r₂) e^(i(θ₁−θ₂)).

r = 4√2, θ = π/4, so 4√2 e^(iπ/4).

−1 (so e^(iπ) + 1 = 0, Euler's identity).

Cartesian a + bi, polar r cisθ, exponential r e^(iθ) — all the same number.

Because e^(iθ) = cosθ + i sinθ; the i is essential — e^(θ) would be a real exponential.

Its conjugate a − bi — complex roots come in conjugate pairs.

z² − 2az + (a² + b²) (middle term −sum, constant = product).

Write the conjugate, form their real quadratic, divide it out, then solve what's left.

The proof relies on conjugating the whole equation; with real coefficients the equation is unchanged, forcing the conjugate to be a root.

2 − i.

z² − 2z + 5 (sum 2, product 5).

No — complex roots come in pairs, so a cubic has either 3 real roots or 1 real + a conjugate pair.

1 + 2i, 1 − 2i, 1.

(r cisθ)ⁿ = rⁿ cis(nθ) — power the modulus, multiply the argument by n.

Convert to polar form, apply De Moivre (rⁿ, nθ), then convert back if needed.

De Moivre only applies to polar/exponential form; powering a + bi directly means a messy binomial expansion.

(√2 cis(π/4))⁸ = (√2)⁸ cis(2π) = 16.

(2 cis(π/6))⁶ = 2⁶ cis(π) = −64.

(r e^(iθ))ⁿ = rⁿ e^(inθ) — same idea via index laws.

Multiplying r by n instead of powering it (it's rⁿ), or forgetting to multiply the angle by n.

(√2 cis(−π/4))⁴ = 4 cis(−π) = −4.

Exactly n distinct nth-roots.

On a circle of radius R^(1/n), equally spaced 2π/n apart (a regular n-gon).

z_k = R^(1/n) cis((φ + 2πk)/n) for k = 0, 1, …, n − 1.

Keep adding 2π/n to the argument until you have n of them.

1, cis(2π/3) = −½ + (√3/2)i, cis(4π/3) = −½ − (√3/2)i.

2, −1 + √3 i, −1 − √3 i (modulus 2, spaced 120°).

R^(1/n), where R is the modulus of the original number.

They share the same modulus (so lie on a circle) and are equally spaced 2π/n apart.

Base case (n = 1), assume true for n = k, prove true for n = k + 1, conclude true for all n.

Knock the first domino (base case) and show each knocks the next (k ⇒ k + 1), so they all fall (all n).

The assumption (the result for n = k) — that's the link that proves n = k + 1.

State the conclusion: true for n = 1 and 'true for k ⇒ true for k + 1', so true for all n ∈ ℤ⁺.

n = 1: LHS = 1, RHS = 1(2)/2 = 1 ✓.

Rewrite the (k+1) expression so the assumption (e.g. 6ᵏ − 1 = 5m) appears, then factor out the divisor.

Without a true starting case, the chain k ⇒ k + 1 never gets going — nothing is ever shown true.

Take the statement as given for one (unspecified) value k, so you can use it to prove the next case.

Assume the statement is false, derive something impossible (a contradiction), so the assumption is wrong and the statement is true.

The negation (opposite) of what you want to prove.

That the assumption (the opposite) is false — so the original statement is true.

Assume √2 = p/q in lowest terms; show p² = 2q² makes both p and q even, contradicting 'no common factor'.

Assume n is odd (n = 2k+1); then n² = 2(2k²+2k)+1 is odd, contradicting n² even.

Assuming r + x is rational forces x = (r + x) − r to be rational, contradicting x irrational.

'Assume, for contradiction, that … [the negation].'

No — it's the goal; it shows the assumption can't hold.

A single case where a 'for all' statement fails — enough to prove the statement false.

Just one.

Small numbers (0, 1), negatives, fractions, and edge cases.

2 — it's prime and even.

a = 2, b = −2: 4 = 4 but 2 ≠ −2.

x = ½: ¼ < ½.

√2 + (−√2) = 0, which is rational.

That it meets the statement's condition (hypothesis) but breaks the conclusion.

Eliminate one variable to get two equations in two unknowns, solve those, then back-substitute for the third.

Add or subtract two equations (scaling one first if needed) so that variable cancels.

Back-substitute the known values into one of the original equations.

Use the simultaneous-equation solver (PlySmlt2) or enter the coefficient matrix and use rref.

Multiply an equation by a constant first so a variable's coefficients are equal (or opposite).

Substitute it into the equation you didn't use last; all three should hold.

x = 1, y = 2, z = 3.

It leaves two equations in the same two unknowns, which you can then solve as a 2×2 system.

Exactly one, none, or infinitely many — never a finite number greater than one.

An equation was redundant → infinitely many solutions (planes meet in a line).

The equations contradict each other → no solution (inconsistent).

The three planes meet at a single point.

The three planes meet along a common line (or coincide).

The planes have no point common to all three (e.g. they form a triangular prism).

Eliminate to two copies of the same left side, then set their right-hand sides equal.

No — only 0, 1, or infinitely many.

A sequence where each term differs from the previous one by a constant amount, the common difference d. Example: 4, 7, 10, 13 has d = 3.

The constant gap between consecutive terms: d = uₙ − uₙ₋₁. Subtract any term from the next. Example: in 9, 5, 1 the difference is d = −4.

Both use uₙ = u₁ + (n − 1)d. Want a VALUE (u₂₀)? Put n = 20 and compute. Want a POSITION ('which term = 100')? Set the formula equal to that value and solve for n. Spot value-vs-position first.

You start at u₁ and add d only on each step after the first, so reaching the nth term takes (n − 1) steps. Example: u₅ = u₁ + 4d.

Divide the difference by the number of steps between them: (41 − 17) ÷ (7 − 3) = 24 ÷ 4 = 6.

Set uₙ = u₁ + (n − 1)d equal to the value and solve for n. Example: 4 + (n−1)5 = 99 ⇒ n = 20.

Substitute n = 1 for the first term (u₁ = 16); the coefficient of n is the common difference (d = −4).

When the differences are equal: u₂ − u₁ = u₃ − u₂. The middle term is the average of its neighbours.

Set u₂ − u₁ = u₃ − u₂ and solve: (k + 1) = (3k − 5) ⇒ k = 3.

A sequence is the list of terms (3, 7, 11, …); a series is their sum (3 + 7 + 11 + …).

Spot u₁ first. If you want a TERM ('the 6th row'), it's n − 1 jumps (row 1 = 0 jumps). If u₁ is a starting amount and you want the value 'after n years/steps', it's n jumps. Always ask: how many times do I add d to get there?

Go straight to Sₙ = (n/2)(2u₁ + (n − 1)d): S₂₀ = 10(14 + 19×4) = 900. Trap: do not waste time finding u₂₀ first — the u₁-and-d form needs only what you are given.

It is a TOTAL, so use the sum: set Sₙ > 500 and solve for n, then round UP to the next whole number (on Paper 2, scan the GDC table of Sₙ). Spot 'total/altogether' ⇒ Sₙ, not uₙ.

Know u₁ and d → use (n/2)(2u₁ + (n − 1)d). Know u₁ and the last term → use (n/2)(u₁ + uₙ).

u₁ = S₁ — substitute n = 1 into the sum. Example: Sₙ = 2n² + 3n ⇒ u₁ = 5.

uₙ = Sₙ − Sₙ₋₁ — the running total up to n minus the running total up to n − 1.

Its sum is a quadratic in n with no constant term (Sₙ = an² + bn). The common difference is 2a.

Use S₅ = (5/2)(u₁ + u₅): 70 = (5/2)(u₁ + 20) ⇒ u₁ + 20 = 28 ⇒ u₁ = 8.

Pairing the first and last terms gives a constant total u₁ + uₙ, and there are n/2 such pairs, so Sₙ = (n/2)(u₁ + uₙ).

u₆ = S₆ − S₅ gives a term; combined with the sum formula you can solve for u₁ and d.

S₈ = (8/2)(10 + 45) = 4 × 55 = 220.

uₙ is a single term (the nth one); Sₙ is the total of the first n terms: Sₙ = u₁ + u₂ + … + uₙ.

Add up. Substitute the index from the lower limit to the upper limit into the expression and sum the results. Example: Σ r=1→4 of (2r+1) = 3+5+7+9 = 24.

The lower limit (below Σ) is where the index starts; the upper limit (above Σ) is where it stops. Both endpoints are included.

Upper limit − lower limit + 1. Example: Σ r=2→9 has 9 − 2 + 1 = 8 terms.

When the summand is linear in the index (like 3r + 2). The common difference equals the coefficient of the index.

First term 5, last term 32, n = 10; then S = (10/2)(5 + 32) = 185.

u₁ = the summand at the lower limit; d = the coefficient of the index. Then use the arithmetic sum formula.

Counting terms as (upper − lower) and forgetting the +1, or assuming the index starts at 1.

On Paper 2 use sum(seq(expression, index, lower, upper)). On Paper 1 you must use the arithmetic sum formula by hand.

Look for a quantity that changes by the same amount each step (a fixed raise, a fixed number per row). Then u₁ = the start and d = the constant change.

u₁ = 20 and d = 4. The nth value is uₙ = 20 + (n − 1)4.

At the last term that is still positive or zero — find where uₙ = 0. Adding later negative terms only shrinks the total.

Solve uₙ = 0 for n, then evaluate Sₙ at that position. Example: u₁ = 48, d = −3 ⇒ u₁₇ = 0 ⇒ S₁₇ = 408.

Graph Sₙ or scan a table of Sₙ and read off the largest value; the peak is at the term where uₙ = 0.

Set up an inequality with uₙ, solve for n, then round to the next whole number (n must be a positive integer).

90 − 7(n − 1) < 20 ⇒ n > 11 ⇒ n = 12; u₁₂ = 13.

A total or 'altogether' is a sum, so use Sₙ. A single 'nth' value is a term uₙ.

n counts terms (rows, years, balls), which only come in whole numbers; round a decimal n to the appropriate integer and check.

A sequence where each term is the previous one multiplied by a constant, the common ratio r. Example: 2, 6, 18, 54 has r = 3.

The constant multiplier: r = uₙ ÷ uₙ₋₁. Divide any term by the one before. Example: 12 ÷ 4 = 3.

'by 8%' multiplies ⇒ geometric, r = 1.08, uₙ = u₁rⁿ⁻¹. 'by 8' adds ⇒ arithmetic, d = 8. Words like percent / ratio / times / doubles signal geometric; a fixed amount signals arithmetic.

You start at u₁ and multiply by r only on each step after the first — (n − 1) times. Example: u₅ = u₁ r⁴.

Divide the values, then take the (steps)-th root: 48 ÷ 6 = 8 over 3 steps, so r = ∛8 = 2.

When the number of steps between the two terms is even, r = ±(root of the value-ratio). An odd number of steps gives a unique r.

When the ratios are equal: u₂/u₁ = u₃/u₂, i.e. u₂² = u₁u₃ (middle squared = product of neighbours).

k² = 4 × 25 = 100, so k = 10.

Arithmetic ADDS the same d each step; geometric MULTIPLIES by the same r. 3, 6, 9 is arithmetic; 3, 6, 12 is geometric.

u₆ = 5 × 2⁵ = 160.

Because consecutive ratios are equal: u₂/u₁ = u₃/u₂. Cross-multiplying gives u₂² = u₁u₃.

Up to two — solve the quadratic and report both. Use any stated condition (e.g. all terms positive) to choose between them.

Count the ×r jumps from the start — that count is the power. The nth TERM is n − 1 jumps (term 1 = 0 jumps); 'after n bounces / years' is n jumps (the start is counted). E.g. dropped 6 m, after the 4th bounce = 6(½)⁴ = 0.375.

Look for a TOTAL — 'sum of', 'altogether', 'total saved' over terms that multiply by a constant ⇒ Sₙ = u₁(rⁿ − 1)/(r − 1). A single value ('the 8th term', 'value after 8 years') ⇒ uₙ = u₁rⁿ⁻¹.

Either works. Use (rⁿ − 1)/(r − 1) when r > 1 and (1 − rⁿ)/(1 − r) when 0 < r < 1 to keep the numbers positive.

u₁, r and n. If r isn't given, find it first from two terms.

u₁ = 3, r = 2: S₅ = 3(2⁵ − 1)/(2 − 1) = 3 × 31 = 93.

Set Sₙ > target and solve for n; on Paper 2 scan the GDC table of Sₙ and round up to the next whole number.

If r = 1 every term equals u₁, so the sum is just n × u₁ (and the formula would divide by zero).

S₄ = 6(1 − 0.5⁴)/(1 − 0.5) = 6(0.9375)/0.5 = 11.25.

Compare with u₁(rⁿ − 1)/(r − 1): r = 3 and u₁/(3 − 1) = 2 ⇒ u₁ = 4.

Use sum(seq(u₁ r^(x−1), x, 1, n)) or read a table of Sₙ. Round n up for 'smallest n' questions.

Substitute u₁ and r into Sₙ = u₁(rⁿ − 1)/(r − 1) and simplify until it matches. E.g. u₁ = 4, r = 3 → 4(3ⁿ − 1)/2 = 2(3ⁿ − 1).

Distance = the first drop + 2 × (sum of the rebound heights). The drop counts once; every rebound is travelled up and down. Use the finite geometric sum Sₙ for the rebound heights.

Each period the balance multiplies by r = 1 + (rate as a decimal). After n periods: balance = start × rⁿ.

Growth: r = 1 + x/100. Decay: r = 1 − x/100. E.g. 6% growth → 1.06; 15% decay → 0.85.

Solve rⁿ = 2 (logs) or use the GDC/TVM solver; round n up to the next whole period.

2000 × 1.06ⁿ > 4000 → 1.06ⁿ > 2 → n ≈ 11.9 → 12 years.

Enter I% = rate, PV = −start, PMT = 0, FV = target, P/Y = C/Y = periods per year, then solve for N. Money out is negative.

Depreciation is decay: r = 1 − rate (0 < r < 1), so the value shrinks by a fixed percentage each period.

Compound multiplies the growing balance by r each period (geometric); simple adds a fixed amount each period (arithmetic).

r = 0.85; 20 000 × 0.85⁴ ≈ $10 440.

Put them into FV = PV(1 + r/(100k))^(kn) and solve for n — take logs, or on Paper 2 scan the GDC table for when the balance first reaches FV. Spot which letter is the unknown before substituting.

The number of compounding periods per year: annual k = 1, half-yearly 2, quarterly 4, monthly 12.

Divide the annual rate by k and raise to (k × n) periods: FV = PV(1 + r/(100k))^(kn).

5000(1 + 0.04/4)^(4×3) = 5000(1.01)¹² ≈ $5634.13.

Compound multiplies the growing balance by (1 + rate) each period (geometric); simple adds a fixed amount each year (arithmetic).

Write the one-year amount as PV(1 + x)⁴ for quarterly (the power = the number of periods in the year; x = the per-period rate), expand with the binomial theorem, and substitute the small x.

Yes — for the same nominal rate, monthly beats quarterly beats annual, because interest compounds sooner.

Interest = FV − PV (the growth above the amount invested).

1 + r/(100k) — one plus the per-period rate as a decimal.

Compound decay — a value loses a fixed percentage each year, multiplying by r = 1 − rate (0 < r < 1).

Set PV × rⁿ < X with r = 1 − rate, then solve for n (logs or the GDC table) and round UP to the next whole year. It is 'first below', so you need the smallest whole n.

24 000 × 0.88⁵ ≈ $12 666.

1 − b as a percent. E.g. V = 5000(0.92)ᵗ loses 8% a year.

Growth multiplies by 1 + rate (> 1); depreciation multiplies by 1 − rate (< 1).

It keeps a fixed percentage each year, so it shrinks geometrically but never actually hits 0.

The yearly multiplier: 90% is kept, so 10% is lost each year.

2000 × 0.7² = 2000 × 0.49 = $980.

N (periods), I% (annual rate as a %), PV (present value), PMT (regular payment), FV (future value), P/Y and C/Y (periods per year).

Money you pay out (invest) is negative; money you receive is positive.

The compounding frequency: 1 annually, 2 half-yearly, 4 quarterly, 12 monthly. N = years × that frequency.

Enter N, PV (negative), PMT = 0, FV, P/Y = C/Y; leave I% blank and solve.

Leave N blank, fill I%, PV (negative), PMT = 0, FV, P/Y = C/Y; solve, then round N up (and ÷ frequency for years).

Months — divide by 12 to get years.

A part-period hasn't reached the target yet, so you need the next whole period.

On Paper 2 for any compound-interest problem — especially finding the rate or the time, which are awkward by hand.

The power you raise a to, to get b. a^x = b ⇔ x = log_a b. Example: log₂ 8 = 3 because 2³ = 8.

Keep the base: log_a b = x. The log equals the exponent.

a^x = b. The log value x is the exponent of the base a.

log₁₀ x — base 10.

log_e x — the natural logarithm, base e ≈ 2.718.

Taking log_a undoes raising a to a power: log_a(a^x) = x, and a^(log_a b) = b.

1, because e¹ = e.

log₅ 125 = 3.

Ask 'a to what power gives b?' — write b as a power of a; the exponent is the answer. E.g. log₃ 81 = 4 since 3⁴ = 81.

0 — because a⁰ = 1.

1 — because a¹ = a.

1/16 = 2⁻⁴, so log₂(1/16) = −4 (a reciprocal gives a negative power).

3 = 9^(1/2), so log₉ 3 = ½ (a root gives a fractional power).

−log_a b — a reciprocal flips the sign.

Type log or ln on the GDC; for other bases use the change-of-base rule (topic 1.7).

8^(1/3) = 2, so log₈ 2 = ⅓.

A chain of justified steps from what you know to the result. The marks are the reasons, not the final answer.

Start from the given (or one side) and work toward the target. Never start from the answer and work backwards.

= (equation) is true for particular values you solve for; ≡ (identity) is true for ALL values.

Even = 2k, odd = 2k + 1, where k is an integer.

n, n + 1, n + 2, … — start from n and add 1 each time.

(2a + 1) + (2b + 1) = 2a + 2b + 2 = 2(a + b + 1), which is even. Use different letters a, b.

Manipulate it until you can take out a factor of k: write it as k × (an integer).

It is a product of two consecutive integers, and one of any two consecutive integers is even.

Reusing one letter (e.g. 2k + 1 twice) forces the two numbers to be equal, which breaks a general proof.

Reach the target exactly, then conclude in words — "… = 2m, which is even, so …". That sentence is often the last mark.

3: n + (n + 1) + (n + 2) = 3(n + 1).

n, n + 1, n + 2 — they go up by 1. Use one starting letter.

n + (n+1) + (n+2) = 3n + 3 = 3(n + 1) = 3 × a whole number.

One of any two consecutive integers is even, and an even factor makes the product even.

Take out a factor of k: write it as k × (a whole number).

Show it always leaves the same remainder: k × (whole) + r with r ≠ 0.

No — the sum is 3(n² + 2n + 1) + 2, so it always leaves remainder 2.

3 × the middle integer (3n + 3 = 3(n + 1)).

No — examples never prove 'for all'. Use algebra (n, n + 1, …) and reason generally.

Name them with one letter (n, n + 1, n + 2), then add or multiply.

= (equation) is true for particular values you solve for; ≡ (identity) is true for ALL values, which you prove.

Transform ONE side (the busier one) step by step into the other. Never move terms across the ≡.

One value only checks that single case; an identity must hold for every x.

The busier side — the one with brackets/powers to expand or fractions to combine.

Expand all brackets, then collect like terms until it matches the target.

a² − 2ab + b² — don't forget the middle term −2ab.

Put the side over a common denominator, combine the numerators, then simplify/factor.

x(x + 1) — the product of the two distinct denominators.

Use the identity/result you just proved — don't re-derive it from scratch.

Expand: (x² + 6x + 9) − (x² − 6x + 9) = 12x. The x² and 9 cancel.

aᵐ × aⁿ = aᵐ⁺ⁿ (multiply→add); aᵐ ÷ aⁿ = aᵐ⁻ⁿ (divide→subtract); (aᵐ)ⁿ = aᵐⁿ (power of a power→multiply).

a⁰ = 1 for any a ≠ 0. Example: 7⁰ = 1.

A reciprocal: a⁻ⁿ = 1/aⁿ. Example: 2⁻³ = 1/8.

A root: a^(1/n) = ⁿ√a, and a^(m/n) = (ⁿ√a)ᵐ. Example: 8^(2/3) = (∛8)² = 4.

Cube root first (∛27 = 3), then square: 3² = 9.

√x = x^(1/2), and the reciprocal flips the sign: 1/√x = x^(−1/2).

Raise both sides to the reciprocal 3/2: a = 4^(3/2) = (√4)³ = 8.

No — the laws need the SAME base. 2³ × 3² = 8 × 9 = 72 must be done directly.

It is a quadratic in disguise: a²ˣ = (aˣ)², so substitute y = aˣ, solve the quadratic, then solve back for x.

Because y = aˣ and a power is always positive — only positive y can give a real x. Discard zero or negative roots.

Coefficients go UP as powers, + becomes ×, − becomes ÷: 2 log x + log y − log z = log(x²y/z). To go the other way (one log → several), read the laws right-to-left. Spot whether they want 'a single log' or 'expanded'.

No — that is the #1 trap. The laws only act on products, quotients and powers, never on a sum.

log_a 1 = 0 (since a⁰ = 1) and log_a a = 1 (since a¹ = a).

It brings an exponent down to the front as a coefficient: log_a(xᵐ) = m log_a x. Example: log 8 = log 2³ = 3 log 2.

2 ln 3 = ln 9; then ln 6 + ln 9 − ln 2 = ln(6 × 9 ÷ 2) = ln 27.

24 = 2³ × 3, so log 24 = 3 log 2 + log 3 = 3p + q.

Change of base: log₂ 50 = (log 50)/(log 2) ≈ 5.64 (any base b works: log_a x = (log_b x)/(log_b a)). Use it whenever the base is not 10 or e, or to combine logs of different bases.

Change to base 2: log₂32 ÷ log₂8 = 5 ÷ 3 = 5/3.

Change to base 10: log 8 ÷ log 3 = 3 log 2 ÷ log 3 = 3p/q.

Product then power law: log₂8 + log₂x³ = 3 + 3 log₂ x.

Write both sides as powers of the same base, then equate the exponents. E.g. 4ˣ = 8 → 2²ˣ = 2³ → 2x = 3 → x = 3/2.

Take logs of both sides; the power law brings x down: x log a = log b, so x = log b / log a.

The power law: log aˣ = x log a turns the unknown exponent into a coefficient you can divide out.

x log 5 = log 20, so x = log 20 / log 5 = log₅ 20 (≈ 1.86).

Convert to exponential form: expr = aᶜ, then solve. E.g. log₃(x − 1) = 2 → x − 1 = 9 → x = 10.

e⁰ = x² − 16, so x² = 17 and x = ±√17 (both keep the argument positive).

Combine them into a single log with the product or quotient law, then convert to exponential form and solve.

A logarithm needs a positive argument, so reject any solution that makes an argument ≤ 0.

k⁴ = 81, so k = ⁴√81 = 3 (take the positive root).

Enter each side as Y₁ and Y₂, graph, then use 2nd → TRACE → 5: intersect. Check for more than one crossing.

Only when |r| < 1 — the terms shrink toward 0. Then S∞ = u₁/(1 − r).

Write it as a GP: 0.47 + 0.0047 + … with u₁ = 0.47 and r = 0.01. Then S∞ = u₁/(1 − r) = 0.47/0.99 = 47/99. Each repeating block is the previous one × 0.01.

If |r| ≥ 1 the terms don't approach zero, so the running total grows without limit — there is no finite sum.

r = 8/12 = ⅔, so S∞ = 12/(1 − ⅔) = 12/(⅓) = 36.

25 = 10/(1 − r) → 1 − r = 0.4 → r = 0.6.

u₁ = S∞(1 − r) = 40 × 0.8 = 32.

Putting r in the denominator instead of (1 − r), or using S∞ when |r| ≥ 1.

State that |r| ≥ 1, so the terms do not approach zero and the total is unbounded.

The gap is S∞ − Sₙ = u₁rⁿ/(1 − r). Set it below the tolerance and solve (GDC table or logs), rounding n up.

Use the finite sum with n = 2m: S₂ₘ = u₁(r²ᵐ − 1)/(r − 1). The power law r²ᵐ = (r²)ᵐ usually simplifies it (e.g. 3²ᵐ = 9ᵐ).

Total = drop + 2 × S∞ of the rebound heights, where the rebounds are a GP with first term (drop × r). For r = ½ this is 3 × the drop.

Start and end every row with 1; each inside number is the sum of the two directly above it. Rows: 1 / 1 1 / 1 2 1 / 1 3 3 1.

The coefficients of (a + b)ⁿ. E.g. row 3 (1, 3, 3, 1) → (a + b)³ = a³ + 3a²b + 3ab² + b³.

Take r = 3 factors counting down from 8 on top, r! on the bottom: ⁸C₃ = (8×7×6)/(3×2×1) = 56. The big factorials cancel — never expand them in full. (nCr = n!/(r!(n − r)!).)

5!/(2! 3!) = (5 × 4)/(2 × 1) = 10.

Type n, then MATH → ▶ (PRB) → 3: nCr, then r, then ENTER. E.g. 10 nCr 4 = 210.

Small n (about ≤ 6) and you want the WHOLE expansion → Pascal's triangle is fastest. Large n, or you only need ONE term/coefficient → use nCr (the general term nCr·aⁿ⁻ʳbʳ) and skip the rest.

n + 1 terms. E.g. (x + 2)⁹ has 10 terms.

The power of a falls from n to 0; the power of b rises from 0 to n; in every term the two powers sum to n.

Both equal 1 (the first and last coefficient of every row). The row is symmetric: nCr = ⁿCₙ₋ᵣ.

nCr is the entry in row n, position r (counting from 0). Row 5 = ⁵C₀, ⁵C₁, …, ⁵C₅ = 1, 5, 10, 10, 5, 1.

Multiply each coefficient nCr by a falling power of a and a rising power of b: aⁿ + ⁿC₁aⁿ⁻¹b + … + bⁿ.

Coeffs 1, 4, 6, 4, 1 with rising powers of 3: x⁴ + 12x³ + 54x² + 108x + 81.

Raise the WHOLE term: (3x)² = 9x², not 3x². Always bracket the term before squaring/cubing.

Use −b as the second term; even powers come out +, odd powers −. So the signs alternate: + − + − …

1 + 4(−2x) + 6(−2x)² + 4(−2x)³ + (−2x)⁴ = 1 − 8x + 24x² − 32x³ + 16x⁴.

Take r = 0, 1, 2, 3 in turn (the lowest powers of x) and stop — no need for the whole expansion.

1 + ¹⁰C₁x + ¹⁰C₂x² = 1 + 10x + 45x².

(1 + rate)ⁿ is the compound-interest factor; expanding it gives the value (or a quick approximation) by hand.

The two powers in every term sum to n, and there are n + 1 terms in total.

Two errors — the sign and the 3. Here a = 3x (not x) and b = −2 (not 2). Correct: ⁵C₃ (3x)²(−2)³ = 10 × 9 × (−8) x² = −720x². Always raise the WHOLE bracket term — number, variable and sign — to its power. (General term: nCr·aⁿ⁻ʳbʳ.)

Use the general term nCr aⁿ⁻ʳ bʳ; set the exponent of x equal to the power you want, solve for r, then compute that one term.

Write the general term, find the r that gives that power of x, and compute the coefficient (raising the whole coefficient/sign to the power).

The power of x is 0. Set the exponent of x to 0, solve for r, then compute that term.

Write that coefficient via the general term, set it equal to the given value, and solve. E.g. (x+k)⁷ coeff x⁵ = 63 → 21k² = 63 → k = ±√3.

An even power of the unknown (e.g. k²) gives two values. Check for a restriction like 'k > 0' before keeping both.

Use the simplest coefficient: ⁿC₂ = n(n − 1)/2 gives a quadratic in n; solve for the positive integer.

Use ⁿC₁k = (x coefficient) and ⁿC₂k² = (x² coefficient); eliminate k and solve for n, then k.

r = 2: ⁶C₂(2x)⁴(−3)² = 15 × 16 × 9 = 2160.

Form both equations and divide one by the other to eliminate a variable.

m = (y₂ − y₁)/(x₂ − x₁) = rise ÷ run. Subtract the coordinates in the same order top and bottom.

m > 0 uphill, m < 0 downhill, m = 0 horizontal (y = c), vertical lines (x = a) have no gradient.

Gradient–intercept y = mx + c; point–gradient y − y₁ = m(x − x₁); general ax + by + d = 0.

m is the gradient; c is the y-intercept (where the line crosses the y-axis).

Rearrange to y = mx + c — the gradient is m = −a/b.

Put m into y = mx + c, then substitute the point to find c. (Or use point–gradient form y − y₁ = m(x − x₁).)

Find the gradient m = (y₂ − y₁)/(x₂ − x₁) first, then substitute one point into y = mx + c to find c.

Set x = 0 (or, in y = mx + c, read off c).

Set y = 0 and solve for x.

Vertical: x = a (gradient undefined). Horizontal: y = b (gradient 0).

When they have the same gradient: m₁ = m₂ (with different y-intercepts).

Use the SAME gradient, put it into y = mx + c, then substitute the point to find c.

When their gradients multiply to −1: m₁m₂ = −1, i.e. m₂ = −1/m₁.

Take the negative reciprocal — flip the fraction and change the sign. E.g. ⅔ → −3/2.

Write 5 as 5/1; the perpendicular gradient is −1/5.

Use the negative-reciprocal gradient, put it into y = mx + c, then substitute the point to find c.

The line perpendicular to the tangent at a point; its gradient is −1/(tangent gradient). Used in calculus.

They are perpendicular, but a vertical line (x = a) has no gradient, so the product rule can't be applied — state it separately.

The line through the midpoint of a segment, perpendicular to it (negative-reciprocal gradient).

1) Midpoint of the endpoints. 2) Gradient of the segment, then its negative reciprocal. 3) Substitute the midpoint into y = mx + c.

((x₁ + x₂)/2, (y₁ + y₂)/2) — average each coordinate.

The negative reciprocal of the segment's gradient (flip the fraction and change the sign).

The midpoint of the two endpoints — not either endpoint.

It is equidistant from the two endpoints (the same distance from A as from B).

Midpoint (3, 5); m_AB = 3/2 → bisector gradient −2/3; y = −2/3 x + 7.

Build y = mx + c first, then clear fractions and move everything to one side: ax + by + d = 0.

Find the value(s) of x that make both sides equal.

Rearrange so one side is 0, then find the roots.

You can lose the solution x = 0; factor instead.

The x-intercepts (zeros) of y = f(x).

The x-coordinates of the intersection points of the two graphs.

Graph both sides and use the GDC intersect tool (no neat algebra).

Paper 1: analytic (algebra). Paper 2: graphical / GDC.

Substitute it back — both sides should be equal.

The 'zero' tool (or read the x-intercepts).

Translates the graph up by k (down if k < 0) — an outside change, as expected.

Translates the graph RIGHT by a — inside changes move the opposite way.

To get the same output, x must be 3 bigger, so the graph sits 3 to the right.

Left 2 (inside +2 is the opposite of its sign).

Top a (right), bottom b (up).

(5, 6) — right 2, up 1.

Yes — every feature slides by the same vector.

Inside f acts on x (left/right, opposite sign); outside f acts on y (up/down, as written).

Stretches the graph vertically by factor a (every y ×a).

Stretches the graph horizontally by factor 1/b (the reciprocal).

Squash horizontally by factor 1/2 (the graph narrows).

Reflects the graph in the x-axis (y-coordinates flip sign).

Reflects the graph in the y-axis (x-coordinates flip sign).

(2, 15) — multiply y by 3.

(−2, 5) — negate x.

The x-intercepts (their y is 0, so 0 × a = 0).

Outside the function affects y; inside affects x (reciprocal for stretch, opposite for shift/flip).

A vertical stretch by a, then a translation k up — both on the y-values.

Multiply by 2 first (stretch), then subtract 1 (translate).

Stretch before translate: ×2 then −1, not −1 then ×2.

(1, 12) — multiply y by 3, x unchanged.

(3, 8) — right 1, up 5.

A translation 2 right and 5 up (vector (2, 5)).

A reflection in the x-axis, then a translation 4 up.

The inside ones: stretch by 1/b, then translate right h.

Translation / stretch (scale factor) / reflection (in which axis), with direction and amount.

The remainder when P(x) is divided by (x − a) is P(a).

(x − a) is a factor of P(x) if and only if P(a) = 0.

Substitute: the remainder is P(a) — no long division needed.

It gives an equation (P(value) = 0 for a factor, or = remainder); solve the equations together.

Both P(a) = 0 and P′(a) = 0 (a is a repeated root).

P(2) = 8 − 8 + 10 − 1 = 9.

P(1) = 1 − 6 + 11 − 6 = 0, so yes.

x = −2 (the root of x + 2).

Sum = −b/a, product = c/a.

Sum = −aₙ₋₁/aₙ; product = (−1)ⁿ a₀/aₙ.

Sum = −b/a; product = −d/a (the (−1)³ makes it negative).

It reads the symmetric functions of the roots straight off the coefficients — no need to find the roots.

Sum = 6/2 = 3, product = 1/2.

Sum = 4, product = −6.

Yes — it's (−1)ⁿ a₀/aₙ, so + for even degree, − for odd.

Sum = k = 5.

Find one root with the factor theorem, divide it out, then factorise/solve the resulting quadratic.

Small integers — factors of the constant term (±1, ±2, …).

The conjugate a − bi.

Use the sum of roots (−b/a): subtract the known roots from it.

x = 1, 3, −2 (factorises as (x − 1)(x − 3)(x + 2)).

1 − i (conjugate) and 2 (from sum of roots = 4).

It sets the end behaviour: +xⁿ rises to the right; the parity of n sets the left end.

1 or 3 (complex roots come in pairs, and degree 3 is odd).

Where the denominator = 0 (and the numerator isn't also 0 there).

Compare degrees: top < bottom → y = 0; equal → y = (ratio of leading coefficients).

The horizontal asymptote is y = 0.

y = (leading coefficient of top) ÷ (leading coefficient of bottom).

x = 3 and x = −2 (from (x − 3)(x + 2) = 0).

y = 2 (equal degrees, 4/2).

There's a hole at x = a, not a vertical asymptote (the factor cancels).

There's a slant (oblique) asymptote instead of a horizontal one.

When the numerator's degree is exactly one more than the denominator's.

Divide top by bottom; the quotient line y = mx + c is the asymptote (the remainder term → 0).

Divide: x + 1 + 2/(x − 1), so y = x + 1.

x-intercepts (top = 0), y-intercept (x = 0), vertical asymptotes (bottom = 0), horizontal/slant asymptote, then fit the branches.

Yes — e.g. (x² + 1)/(x − 1) has vertical x = 1 and slant y = x + 1.

y = 2x − 3 (the quotient of the division).

It can cross it (asymptotes are about behaviour as x → ±∞), unlike never crossing a vertical one.

The asymptotes as dashed lines, then the intercepts.

f(−x) = f(x) — symmetric in the y-axis (e.g. x², cos x).

f(−x) = −f(x) — symmetric about the origin (e.g. x³, sin x).

Compute f(−x): if it equals f(x) it's even; if it equals −f(x) it's odd; otherwise neither.

0 — the halves cancel.

2 × ∫₀ᵃ f(x) dx.

Odd: f(−x) = −x³ + 4x = −(x³ − 4x) = −f(x).

Only even powers (and a constant); odd polynomials have only odd powers.

Only f(x) = 0.

When it's one-to-one (each output comes from exactly one input — passes the horizontal-line test).

Restrict its domain to a stretch where it IS one-to-one, then invert.

Write y = f(x), make x the subject, then swap x and y.

One that is its own inverse: f(f(x)) = x, so f⁻¹ = f; its graph is symmetric in y = x.

f(x) = 1/x and f(x) = a − x.

The domain of f⁻¹ is the range of f, and the range of f⁻¹ is the domain of f.

[0, π] — where cos is one-to-one (the arccos domain).

On x ≥ 0, f⁻¹(x) = √x.

Find the intersections (f = g); they bound the intervals where f is on or above g.

Move everything to one side, find zeros and undefined points, then do a sign analysis on a number line.

The denominator could be negative, which would flip the inequality direction.

Outside its roots (x < smaller or x > larger); negative between them.

Roots of x² = x + 2 are −1, 2; the parabola is below the line between them: −1 ≤ x ≤ 2.

Critical points 1 (zero), −2 (undefined); sign analysis gives x < −2 or x ≥ 1.

Yes for ≤/≥ — except any x that makes a denominator zero (always excluded).

Where the expression is zero (numerator = 0) or undefined (denominator = 0).

Reflect any part below the x-axis up above it; leave the rest unchanged.

Keep the graph for x ≥ 0 and reflect it across the y-axis (the left half is discarded).

Take reciprocals of the heights: zeros of f → vertical asymptotes; large f → near 0; max ↔ min; sign kept.

Wherever f(x) = 0.

Yes — it's symmetric in the y-axis (an even function).

A V with vertex (2, 0); minimum value 0.

It becomes a minimum of 1/f (with the same sign).

|f(x)| reflects below-axis parts up; f(|x|) mirrors the right half across the y-axis.

Set inside = +c and inside = −c (provided c ≥ 0), and solve both.

−a < x < a (a band around 0).

x < −a or x > a (everything outside the band).

No — a modulus is always ≥ 0, so |…| = (negative) has no solution.

2x − 1 = ±5 ⇒ x = 3 or x = −2.

−3 < x − 2 < 3 ⇒ −1 < x < 5.

f = g or f = −g (or square both sides).

2x + 1 ≥ 4 or ≤ −4 ⇒ x ≥ 3/2 or x ≤ −5/2.

The output of function f for input x. f(3) means substitute x = 3 into the rule.

No — it's 'f of x', the function applied to x. The brackets hold the input.

Each input gives exactly ONE output. (Different inputs may share an output.)

Replace every x with a (in brackets for negatives/expressions), then simplify.

(−3)² − 4(−3) = 9 + 12 = 21.

Set the rule equal to k and solve for x (output → input).

Yes — e.g. f(x) = x² gives f(2) = f(−2) = 4. So f(x) = k may have several solutions.

Go up from x = a to the curve, then across to the y-axis.

Read across from y = k to the curve, then down to the x-axis (there may be several x).

Substitute the whole expression: 3(2a) − 5 = 6a − 5.

The set of all allowed inputs (x-values).

The set of all possible outputs (y-values).

How far the graph extends left ↔ right (the x-extent).

How far the graph extends down ↕ up (the y-extent).

No dividing by zero (denominator ≠ 0) and no even root of a negative (under √ ≥ 0). Also a log argument must be > 0.

x ≠ 3 — the denominator can't be zero.

x ≥ 2 — what's under the root must be ≥ 0 (0 is allowed).

y ≥ k — the vertex (h, k) is the minimum.

y > 0 — always positive, approaching but never reaching 0.

All real numbers, x ∈ ℝ.

It undoes f: if f(a) = b then f⁻¹(b) = a. Inputs and outputs swap.

No — f⁻¹ is the inverse function (reverses f), not the reciprocal.

y = x. Each point (a, b) on f becomes (b, a) on f⁻¹.

Write y = f(x), swap x and y, then solve for y — that's f⁻¹(x).

Swap: x = 2y + 3 ⇒ y = (x − 3)/2, so f⁻¹(x) = (x − 3)/2.

They swap: domain of f⁻¹ = range of f; range of f⁻¹ = domain of f.

On the line y = x — solve f(x) = x to find where.

Pick a point: f(a) = b should give f⁻¹(b) = a. Or check f(f⁻¹(x)) = x.

Its domain is f's range, which can be limited (e.g. √x has range y ≥ 0, so its inverse x² is restricted to x ≥ 0).

It maps to itself — which is why f and f⁻¹ meet on y = x.

A sketch shows the correct SHAPE plus the KEY FEATURES labelled — not every point plotted exactly.

Axis intercepts, turning points (max/min), asymptotes, and correct end-behaviour — each labelled.

Set x = 0.

Set y = 0 and solve (factor, formula, or GDC).

From the leading coefficient: a > 0 opens up (minimum), a < 0 opens down (maximum).

A line the curve approaches but doesn't reach — drawn as a dashed guide line.

Where the denominator equals zero.

Graph it on the GDC, then transfer the shape to paper with intercepts, turning points and asymptotes labelled (values read off the GDC).

Not exact, but key points must be in the right relative positions and labelled with their values.

Vertical asymptote x = 2, horizontal asymptote y = 1, two branches approaching them.

Intercepts, maximum/minimum points, asymptotes, increasing/decreasing intervals, symmetry, and behaviour as x → ±∞.

An x-intercept — a value of x where f(x) = 0 (also called a root).

y-intercept: set x = 0. x-intercept (zero/root): set y = 0.

Point = coordinates (a, b); value = the y-coordinate b; 'where' = the x-coordinate a. Read the question.

Local = highest in its neighbourhood; global = highest over the whole graph.

A line x = a the curve shoots toward (±∞) — where a denominator is 0.

The value y approaches as x → ±∞ (the curve levels off).

As x increases, y increases — the graph goes up from left to right.

Decreasing for x < 0, increasing for x > 0 — it turns at the vertex (x = 0).

Graph it on the GDC and use the maximum/minimum tool to read the coordinates.

It lies on both curves, so f(x) = g(x) there; the shared value is the y-coordinate.

Set f(x) = g(x), bring everything to one side, solve for x, then substitute back for y.

Graph both functions on the GDC and use the intersect tool — once per crossing.

The horizontal line y = k.

The x-intercepts (zeros) — where the graph meets the x-axis.

Usually not — substitute each x back into a function to get the y-coordinate of the point.

Yes — e.g. a line can cut a parabola twice; find every crossing.

x² + 1 = 2x + 1 ⇒ x² − 2x = 0 ⇒ x = 0 or 2 ⇒ (0, 1) and (2, 5).

x³ − 2x = 1 (i.e. x³ − 2x − 1 = 0) — the intersection IS the solution.

f(g(x)) — apply the inner function g first, then f. Read right-to-left.

Not in general — order matters; the inner function changes the result.

Compute g(a) first, then put that value into f. Work inside-out.

Replace every x in f with the whole expression g(x) (in brackets), then simplify.

f(x²) = 2x² + 1.

g(2x + 1) = (2x + 1)² = 4x² + 4x + 1.

Form the composite expression, set it equal to k, and solve for x.

Compose with the unknown f, then match coefficients to the given result.

Doing the functions in the wrong order, or forgetting brackets when substituting (e.g. (2x+1)²).

Write y = f(x), swap x and y, solve for y. (Geometrically, reflect in y = x.)

Swap x and y, multiply up to clear the fraction, gather the y-terms, factor out y, then divide.

x = 5y − 2 ⇒ y = (x + 2)/5, so f⁻¹(x) = (x + 2)/5.

f(f⁻¹(x)) = x (and f⁻¹(f(x)) = x) — they undo each other.

They swap: domain of f⁻¹ = range of f; range of f⁻¹ = domain of f.

x² isn't one-to-one over all x; restricting to x ≥ 0 makes it invertible, giving f⁻¹(x) = √x.

Swap, multiply up, gather y: f⁻¹(x) = (3x + 1)/(x − 2).

No — f⁻¹ is the inverse function; 1/f is the reciprocal.

Go up from x = a to the curve, then across to the y-axis.

Read f(a), then read f of that result — two read-offs, inside first.

Start at y = b on the y-axis, go across to the curve, then down to the x-axis.

Reflect the graph in the line y = x; every point (a, b) becomes (b, a).

They swap: a y-intercept (0, k) becomes an x-intercept (k, 0), and vice versa.

Standard ax²+bx+c, factored a(x−p)(x−q), vertex a(x−h)²+k.

a > 0 opens up (minimum); a < 0 opens down (maximum).

It's c — the constant term (set x = 0).

Set each bracket to zero: a(x − p)(x − q) gives x = p and x = q.

The turning point (h, k) and the max/min value k.

x = 4 and x = −1 (watch the sign on (x + 1)).

Exactly midway between the two x-intercepts.

Direction (sign of a), x-intercepts, y-intercept, and the vertex.

Factored form, a(x − p)(x − q).

x = −b/(2a) for y = ax² + bx + c — also midway between the x-intercepts.

Vertex form a(x − h)² + k, which shows the vertex (h, k) directly.

Halve b, square it for the bracket (x + b/2)², then adjust the constant to keep it equal.

It's k, reached at x = h: minimum if a > 0, maximum if a < 0.

(x − 3)² + 2, so the vertex is (3, 2).

(h, k) — note (x − 3)² means h = +3.

Write y = a(x − h)² + k, substitute the point to find a.

Value = k (a number); point = (h, k) (coordinates).

x = −(−6)/(2·1) = 3.

It's k in vertex form a(x − h)² + k: minimum if a > 0, maximum if a < 0, at x = h.

y ≥ k, where k is the vertex's y-value (the minimum).

y ≤ k, where k is the vertex's y-value (the maximum).

y ≥ 3 — opens up, vertex (2, 3).

Find the vertex (x = −b/(2a), then substitute), check the opening, then range is y ≥ k or y ≤ k.

k (the y-value of the vertex). h is just where it occurs.

Set it to = 0, write as two brackets, then set each bracket to zero.

x = (−b ± √(b² − 4ac)) / (2a), for ax² + bx + c = 0.

Rearrange the equation so one side is 0.

Square-root both sides with ±: x − h = ±√n, then solve.

A square has two roots; dropping ± loses a solution.

The quadratic formula.

On Paper 2 — use the equation solver or read the x-intercepts.

(x − 2)(x − 3) = 0 ⇒ x = 2 or 3.

From ax² + bx + c = 0 (set to zero first); keep their signs.

Δ = b² − 4ac — the expression under the root in the quadratic formula.

Two distinct real roots; the graph cuts the x-axis twice.

One repeated real root; the graph touches the x-axis (tangent).

No real roots; the graph misses the x-axis.

Set line = curve, form a quadratic = 0, then set Δ = 0 (one intersection).

Δ = 0.

Δ < 0.

Δ = k² − 36 = 0 ⇒ k = 6.

No — just compute Δ and read its sign.

Rearrange to one side, then find the roots (solve = 0).

Between the roots.

Outside the roots: x < p or x > q.

Between the roots (the reverse of an upward one).

Two inequalities joined by 'or': x < p or x > q.

A single chain: p ≤ x ≤ q (or p < x < q).

Use ≤/≥ (closed) when equality is included; </> (open) when not.

Reverse the inequality sign.

Roots −2, 3; upward → outside: x < −2 or x > 3.

A hyperbola with two branches (top-right and bottom-left), hugging both axes.

Domain x ≠ 0, range y ≠ 0.

x = 0 (vertical) and y = 0 (horizontal).

None — it never reaches either axis.

Vertical x = h, horizontal y = k.

Right by 3, so the vertical asymptote is x = 3.

Raises the horizontal asymptote to y = k.

Draw the asymptotes, find any intercepts, then draw the two branches hugging the asymptotes.

Division by zero is undefined — that's the vertical asymptote.

Where the denominator is zero: cx + d = 0.

y = a/c — the ratio of the leading coefficients.

Where the numerator = 0 (a fraction is zero only when its top is zero).

At x = 0: y = b/d.

x = 4 (denominator zero).

y = 2 (leading coefficients 2/1).

Dividing top and bottom by x, the b and d terms vanish, leaving a/c.

One — the linear denominator has a single zero.

Draw the asymptotes, plot the x- and y-intercepts, then draw the two branches.

(0, 1), because a⁰ = 1.

Horizontal asymptote y = 0; range y > 0 (always positive).

a > 1 grows; 0 < a < 1 decays. Both pass through (0, 1).

It lifts to y = c (the curve levels off at c, not 0).

At x = 0: k·1 + c = k + c.

The initial value (at t = 0).

The per-period factor: b > 1 growth, b < 1 decay.

Solve A₀·bᵗ = target with logs, or graph and use intersect on the GDC.

No — it's always positive, never reaching the x-axis.

(1, 0), because logₐ1 = 0.

Vertical asymptote x = 0; domain x > 0.

y = aˣ — they're reflections in y = x.

All real numbers (y ∈ ℝ).

x = 0 isn't in the domain (log 0 is undefined).

No — it keeps increasing (slowly) forever.

x = h (the inside must be positive: x > h).

x > 2.

They swap (inverses): (0,1)↔(1,0), asymptote y=0↔x=0, domains/ranges swap.

d = √((x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²) — Pythagoras with a z-term.

((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2) — average each coordinate.

Distance squares the gaps and roots; midpoint averages the coordinates.

No — each gap is squared, so the sign disappears.

B = 2M − A (each coordinate).

√(4 + 9 + 36) = √49 = 7.

√(l² + w² + h²).

Average the two endpoints — you should recover the midpoint.

V = ⁴⁄₃πr³.

A = 4πr².

V = ⅓πr²h — one third of the cylinder.

πrl, where l is the slant height = √(r² + h²).

V = πr²h; A (closed) = 2πr² + 2πrh.

⅓ × base area × perpendicular height.

Add the volumes of the parts (subtract for a hole).

Don't count the join between two pieces; only exposed faces.

⁴⁄₃πr³ = 36π ⇒ r³ = 27 ⇒ r = 3.

Spot a right-angled triangle inside the solid and use SOH-CAH-TOA.

The angle between the line and its projection (shadow) on the plane.