IB Math AI HL — All Flashcards

It is the front number in a × 10ⁿ. In valid standard form, it must be at least 1 but smaller than 10.

1. The coefficient must be at least 1 but smaller than 10. 2. The exponent must be an integer.

Move the decimal so only the first non-zero digit stays before it. Count how many places it moved left. That count becomes the positive exponent.

5.84 × 10⁶. The decimal moves 6 places left, so the exponent is +6.

Move the decimal so only the first non-zero digit stays before it. Count how many places it moved right. That count becomes the negative exponent.

5.2 × 10⁻⁴. The decimal moves 4 places right to make 5.2, so the exponent is −4.

Check that the coefficient is at least 1 but smaller than 10, the exponent sign matches the size of the number, and the question asks for standard form rather than ordinary form.

7.39 × 10⁷. Move the decimal 7 places left to get a coefficient between 1 and 10.

1.205 × 10¹⁰. Move the decimal 10 places left and keep all significant digits in the coefficient.

8.4 × 10⁻⁵. Move the decimal 5 places right to make 8.4, so the exponent is −5.

3.02 × 10⁻⁷. Move the decimal 7 places right so the coefficient is between 1 and 10.

9.6 × 10⁻³. Move the decimal 3 places right to get 9.6, so the exponent is −3.

0.0072. Move the decimal 3 places left.

30 600. Move the decimal 4 places right.

Positive exponent -> bigger than 1. Negative exponent -> smaller than 1.

Because 0.48 is less than 1, so the coefficient is not valid.

4.8 × 10⁶

Because 31.5 is greater than 10. Rewrite it as 3.15 × 10⁵.

Check coefficient, exponent sign, and the form asked for.

Because 0.6 is smaller than 1. Rewrite it as 6.0 × 10⁷.

4 700 000. Move the decimal 6 places right.

It will be a large number greater than 1.

2.46 × 10⁵

Multiply coefficients. Add exponents.

Check the coefficient is between 1 and 10.

Coefficient too small. Correct form: 6.0 × 10⁷.

1.56 × 10⁷

(a × 10ᵐ) ÷ (b × 10ⁿ) = (a ÷ b) × 10^(m-n). Divide the coefficients and subtract the exponents, then re-normalise if needed.

The powers of 10 must match first. Rewrite one number so both terms use the same power of 10, then add or subtract the coefficients.

3.0 x 10^4 = 0.3 x 10^5. Then 1.8 x 10^5 - 0.3 x 10^5 = 1.5 x 10^5.

Use matching powers first. Keep both terms as x 10^4, then subtract the coefficients.

3.0 × 10⁴ = 0.3 × 10⁵. Then 1.8 × 10⁵ - 0.3 × 10⁵ = 1.5 × 10⁵.

Because the coefficient 0.5 is less than 1. Re-normalise it to 5.0 × 10¹.

Because the coefficient is less than 1. Re-normalise it to 5 x 10^1.

Because the coefficient 0.5 is less than 1. Re-normalise it to 5.0 x 10^1.

2.46 × 10⁵, because the coefficient must be between 1 and 10 in valid standard form.

Rewrite to valid standard form by making the coefficient between 1 and 10, and adjust the exponent to keep the same value.

2.46 x 10^5, because the coefficient must be between 1 and 10 in valid standard form.

Multiply coefficients. Add exponents.

Divide coefficients. Subtract exponents.

Match the powers first.

5 × 10¹

4.8 × 10⁶

Coefficient, sign, requested form.

5.08 × 10⁻⁴ = 0.000508

E means × 10^(the number after E). So 3.7E9 means 3.7 × 10⁹, and 5.1E-4 means 5.1 × 10⁻⁴.

3.7 × 10⁹

(a) 6.4 × 10⁻³ (b) 0.0064

No. Writing E notation earns zero marks. You must write 1.25 × 10¹¹.

Step 1: Read the number before E → that is your coefficient a. Step 2: Read the number after E → that is your exponent n. Then write a × 10ⁿ.

A reciprocal: x⁻ⁿ = 1/xⁿ. It does NOT make the value negative.

1 — anything (non-zero) to the power 0 is 1.

The n-th root of x: x^(1/n) = ⁿ√x. So x^(1/2) = √x, x^(1/3) = ³√x.

ⁿ√(xᵐ) = (ⁿ√x)ᵐ — the bottom n is the root, the top m is the power.

Cube root of 27 is 3, then square: 3² = 9.

x^(1/2). (And ³√x = x^(1/3), 1/√x = x^(−1/2).)

Reciprocal of 16^(3/4): 4th root of 16 is 2, cube it (8), reciprocal = 1/8.

x^(1/2) / x² = x^(1/2 − 2) = x^(−3/2).

S∞ = u₁/(1 − r), the finite total of all (infinitely many) terms — valid only when |r| < 1.

Only when −1 < r < 1 (|r| < 1). The terms must shrink toward 0. If |r| ≥ 1, no sum to infinity exists.

When |r| < 1 each term is a fraction of the last, so the terms shrink toward 0 fast enough that the running total closes in on a fixed value.

S∞ = 12/(1 − 0.5) = 12/0.5 = 24.

Rearrange S∞ = u₁/(1 − r): 1 − r = u₁/S∞, so r = 1 − u₁/S∞.

1 − r = 24/36 = 2/3, so r = 1/3 (and |1/3| < 1, so it converges).

Add the first drop, then count each rebound height TWICE (up and down): total = drop + 2 × (rebound series).

Interpreting it in context (e.g. the steady drug level or long-run total) and commenting on whether the model is realistic.

i is 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 (both real numbers).

Combine the real parts together and the imaginary parts together (treat i like a letter).

Expand the brackets, then replace every i² with −1 and collect like terms.

z* = a − bi (flip the sign of the imaginary part). z·z* = a² + b² is real, which lets you divide.

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

It plots complex numbers; z = a + bi is the point (a, b) — real across, imaginary up.

|z| = √(a² + b²), the distance from the origin to (a, b) on the Argand diagram.

r = |z| = √(a² + b²) — its distance from the origin on the Argand diagram.

The angle θ from the positive real axis (anticlockwise positive); fix the quadrant by sketching the point.

z = r(cos θ + i sin θ) = r cis θ = r e^(iθ).

Multiply the moduli and add the arguments: z₁z₂ = r₁r₂ e^(i(θ₁+θ₂)).

Divide the moduli and subtract the arguments: z₁/z₂ = (r₁/r₂) e^(i(θ₁−θ₂)).

r = √(1+3) = 2, θ = arctan(√3) = π/3, so z = 2 e^(iπ/3).

4(cos π/6 + i sin π/6) = 4(√3/2 + i/2) = 2√3 + 2i.

It scales (stretches) by r and rotates by angle θ — that's why complex numbers model rotations and phase shifts.

(r cis θ)ⁿ = rⁿ cis(nθ) = rⁿ e^(inθ): power the modulus, multiply the argument by n.

r = √2, θ = π/4; (√2)⁸ cis(8·π/4) = 16 cis(2π) = 16.

Real when nθ is a multiple of π; positive real when nθ is a multiple of 2π.

Z = R + iX, where R is resistance and X is reactance; |Z| is the total opposition and arg Z is the phase angle.

They add as complex numbers: Z_total = Z₁ + Z₂ + …

Represent each as a phasor A e^(iφ), add the phasors, then read off the resultant amplitude (modulus) and phase (argument).

|Z| = √(3² + 4²) = √25 = 5 Ω.

r = 2, θ = −π/3; z⁵ = 32 cis(−5π/3) = 32 cis(π/3) = 16 + 16√3 i.

Rows × columns, rows first. E.g. 2 rows and 3 columns is a 2 × 3 matrix.

Only when they have the same order; then you add matching entries.

Multiply every entry by the scalar (e.g. 2A doubles each entry).

When A's columns = B's rows (inner numbers match). The result is (rows of A) × (columns of B) — the outer numbers.

Slide along a row of A and down a column of B, multiply pair-by-pair, and add.

I = ((1, 0), (0, 1)); AI = IA = A, so it leaves a matrix unchanged (like ×1).

det A = ad − bc. If it equals 0, A has no inverse (singular).

A⁻¹ = 1/(ad − bc) · ((d, −b), (−c, a)): swap diagonal, negate off-diagonal, divide by det.

A non-zero direction v that A only stretches: Av = λv (same direction, scaled by the eigenvalue λ).

Solve the characteristic equation det(A − λI) = 0 for λ (subtract λ down the diagonal, set the determinant to zero).

Solve (A − λI)v = 0; the rows give one line, so pick the simplest non-zero whole-number vector on it.

P has the eigenvectors as its columns; D has the matching eigenvalues on its diagonal (same order).

Aⁿ = PDⁿP⁻¹, and Dⁿ just raises each diagonal eigenvalue to the power n.

Its eigenvector is the steady state — the long-run mix the matrix leaves unchanged (rescale so the entries sum to the total).

It is multiplied by λⁿ → 0, so that part fades away, leaving only the λ = 1 piece.

They are exactly the entries on its main diagonal.

A sequence with the same difference each time.

Subtract one term from the next term.

uₙ = u₁ + (n − 1)d

30

Sequence = list. Series = sum.

Sₙ = (n/2) × (2u₁ + (n − 1)d)

2 + 4 + 6 + 8 = 20

It is a short way to write a sum.

Simple interest adds the same amount each time.

Write uₙ = u₁ + (n−1)d for each term. Subtract one equation from the other — u₁ cancels, leaving d.

One value -> nth term. Total -> sum formula.

Both equations contain u₁. Subtracting cancels u₁ so only d remains.

Real data can be close to arithmetic without being exact.

Yes, because the common difference is 60.

n = 12. Always round up — you need the first whole term that passes the threshold.

d = $2 400. Eq(1): u₁ + 7d = 43 200. Eq(2): u₁ + 2d = 31 200. Subtract: 5d = 12 000.

|r| < 1 — the terms must be getting smaller toward zero.

S∞ = u₁ ÷ (1 − r). Only valid when |r| < 1.

No. r = 6 ÷ 3 = 2. |r| = 2 ≥ 1, so S∞ does not exist.

r = 0.5. |r| = 0.5 < 1 ✓. S∞ = 10 ÷ (1 − 0.5) = 20.

u₁ = S∞ × (1 − r) = 30 × (1 − 0.4) = 30 × 0.6 = 18.

1 − r = u₁ ÷ S∞ = 12 ÷ 20 = 0.6, so r = 0.4.

Yes. |r| = |−0.6| = 0.6 < 1 ✓. Negative r is fine — |r| strips the sign.

State: |r| < 1 ✓. IB mark schemes award this step — you earn the method mark even if the final answer is wrong.

log(xy) = log x + log y; log(x/y) = log x − log y; log(xⁿ) = n log x.

aʸ = x — 'the power of a that gives x'. Logs and powers are inverse operations.

logₐ x = log x / log a = ln x / ln a (number on top, old base on the bottom).

Take logs both sides: x log a = log b, so x = log b / log a (the doubling-time / half-life move).

Rewrite in index form: N = aᵏ. Then check the inside of the log is positive.

logₐ 1 = 0 (a⁰ = 1) and logₐ a = 1 (a¹ = a).

Logs are only defined for positive inputs; reject any solution that makes the inside ≤ 0, even if the algebra allows it.

log(P²Q / 5) — power law first, then product (add) and quotient (subtract).

The gradient measures the steepness and direction of a line — how much y changes for every 1 unit increase in x. Positive gradient → rises left to right. Negative gradient → falls left to right. Zero gradient → horizontal line.

Gradient = rise ÷ run = 8 ÷ 2 = 4. The line goes up by 4 for every 1 unit to the right. This is a positive, fairly steep gradient.

The line falls steeply — for every 1 unit moved right, y drops by 5. Steepness = |−5| = 5 (compare using absolute value). The negative sign means it slopes downward from left to right.

Wrong — steepness uses absolute value: |−4| = 4 > |3| = 3. The line with gradient −4 is steeper. Never compare signed gradient values to decide steepness — always compare |m₁| and |m₂|.

m = (y₂ − y₁) / (x₂ − x₁) The y-change (rise) goes on top. The x-change (run) goes on the bottom. Use the same pair order for both: subtract in the same direction.

m = (9 − 1) / (7 − 3) = 8 / 4 = 2. y increased and x increased → positive gradient makes sense. ✓

m = (−1 − 5) / (4 − (−2)) = −6 / 6 = −1. Key step: 4 − (−2) = 4 + 2 = 6. Subtracting a negative flips the sign.

They have swapped Δy and Δx. The gradient formula is m = Δy/Δx, not Δx/Δy. Fix: always write the formula first — m = (y₂ − y₁)/(x₂ − x₁) — before substituting numbers.

The y-intercept is the point where the line crosses the y-axis — the value of y when x = 0. In y = mx + c, the y-intercept is c, the constant term. Example: y = 4x − 7 has y-intercept = −7, so it crosses at (0, −7).

m is the gradient — it is the coefficient of x. c is the y-intercept — it is the constant term. Example: y = −2x + 9 → gradient = −2, y-intercept = 9.

Gradient m = −3. y-intercept c = 7. Coordinates of y-intercept: (0, 7).

The equation is not in y = mx + c order. Rewrite: y = −3x + 5. Gradient m = −3, y-intercept c = 5. Always rearrange into y = mx + c form before reading off m and c.

Horizontal line: gradient = 0 (no rise — Δy = 0). Vertical line: gradient is undefined — Δx = 0, so we would divide by zero.

Compare the absolute values of their gradients. The line with the larger |m| is steeper. Example: |−5| = 5 > |2| = 2, so y = −5x is steeper than y = 2x.

y-intercepts: A → c = 1, B → c = −5. Line A crosses higher. Steepness: |−3| = 3 vs |4| = 4. Line B is steeper. Two different comparisons — do them separately.

They dropped the negative sign. The gradient is m = −1/3 (negative, because it is − times 1/3). The y-intercept is 9. Read the coefficient of x including its sign.

y = mx + c m = gradient (slope), c = y-intercept. This form directly shows both key features of the line.

Substitute directly into y = mx + c: y = 5x − 3. The gradient goes with x; the y-intercept is the constant.

Gradient m = −1/2. y-intercept c = 6. The line starts high on the y-axis and falls gently — it goes down 1 for every 2 units to the right.

IB always requires a full equation, not just the values of m and c. Write: y = 3x + 7. The equation must start with "y =" and show both m and c in the correct form.

1. Write y = mx + c with the known gradient m. 2. Substitute the coordinates of the given point for x and y. 3. Solve for c. 4. Write the full equation with both m and c.

y = 3x + c. Substitute (2, 8): 8 = 3(2) + c → 8 = 6 + c → c = 2. Equation: y = 3x + 2.

y = −2x + c. Substitute (−1, 5): 5 = −2(−1) + c → 5 = 2 + c → c = 3. Equation: y = −2x + 3. Check: plug in x = −1: y = −2(−1) + 3 = 5 ✓

They left m as a letter instead of replacing it with the actual gradient value. If gradient = 2 and c = 4, the equation must be: y = 2x + 4. Always replace m with its value in the final answer.

Step 1: Calculate the gradient using m = (y₂ − y₁)/(x₂ − x₁). Step 2: Use one point and the gradient to find c (substitute into y = mx + c).

m = (10 − 4)/(3 − 1) = 6/2 = 3. y = 3x + c. Use (1, 4): 4 = 3(1) + c → c = 1. Equation: y = 3x + 1.

m = (5 − (−3))/(4 − 0) = 8/4 = 2. y-intercept: when x = 0, y = −3, so c = −3 directly. Equation: y = 2x − 3. Shortcut: if one point is the y-intercept (x = 0), c = that y-value immediately.

They must use one of the given points to substitute into y = 4x + c and solve for c. The equation y = 4x only works if the line passes through the origin — that must be verified, not assumed.

ax + by + d = 0 (sometimes written ax + by = c). All terms are moved to one side, leaving zero on the other. IB accepts both y = mx + c and general form unless the question specifies which.

Move all terms to the left: 3x − y − 5 = 0. Or equivalently: −3x + y + 5 = 0 (both are valid; IB usually wants positive leading coefficient).

Rearrange: y = 2x + 8. Gradient m = 2, y-intercept c = 8.

IB requires the specific form asked for. Leaving it as y = 2x − 4 does not match ax + by + d = 0. Correct: 2x − y − 4 = 0. Always re-read what form the question requires before writing the final answer.

Two lines are parallel if and only if they have the same gradient. They never intersect (unless they are the same line). Example: y = 3x + 2 and y = 3x − 7 are parallel — both have m = 3.

Any line parallel to L₁ also has gradient m. The gradient is the same — only the y-intercept (c) can differ.

Yes — both have gradient m = −2. They are different lines (different y-intercepts: 5 and −3), so they are parallel, not the same line.

No — gradients are +2 and −2. These are different gradients, so the lines are not parallel. They intersect at (0, 1). Similar equations do not mean parallel lines — the gradient values must match exactly.

Two lines are perpendicular if the product of their gradients equals −1: m₁ × m₂ = −1. This means the gradients are negative reciprocals of each other.

The perpendicular gradient is −1/m (flip the fraction and change the sign). Examples: m = 3 → m⊥ = −1/3 m = −2/5 → m⊥ = 5/2 m = 4 → m⊥ = −1/4

m⊥ = −1 / (−3/4) = 4/3. Rule: flip the fraction (4/3) and change the sign. Starting negative → perpendicular is positive. Check: (−3/4) × (4/3) = −12/12 = −1 ✓

They only changed the sign but did not take the reciprocal. The perpendicular gradient is −1/5 (flip to 1/5, then negate). "Negative reciprocal" means both steps: flip AND change sign.

1. Identify the gradient: m = 4 (same as the original line — parallel). 2. Substitute into y = 4x + c using (3, 7): 7 = 4(3) + c → c = −5. 3. Equation: y = 4x − 5.

m⊥ = −1/2. y = −(1/2)x + c. Use (4, 1): 1 = −(1/2)(4) + c → 1 = −2 + c → c = 3. Equation: y = −(1/2)x + 3.

m⊥ = 1/3 (negative reciprocal of −3). The line passes through (0, 5), so c = 5 directly (it is the y-intercept). Equation of L₂: y = (1/3)x + 5.

Their answer will be a parallel line, not a perpendicular one — a completely different type of answer. Always find m⊥ = −1/m first, before substituting the given point to find c.

The perpendicular bisector is a line that: 1. Passes through the midpoint of AB. 2. Is perpendicular to AB (i.e. meets AB at a right angle). Every point on the perpendicular bisector is equidistant from A and B.

1. The midpoint of AB — the perpendicular bisector passes through this point. 2. The perpendicular gradient — find the gradient of AB first, then take the negative reciprocal.

Midpoint M = ((2+6)/2, (4+8)/2) = (4, 6). Gradient of AB: m = (8−4)/(6−2) = 4/4 = 1. So m⊥ = −1. y = −x + c. Use (4, 6): 6 = −4 + c → c = 10. Perpendicular bisector: y = −x + 10.

The line will pass through the wrong point — it will be perpendicular to AB but not at the midpoint. The perpendicular bisector must pass through the midpoint, not through A or B. Always substitute the midpoint to find c.

A linear model describes a situation where the output increases or decreases at a constant rate as the input changes. It has the form y = mx + c. Use it when: the rate of change is constant (e.g. fixed cost per unit, steady temperature drop).

C = 2.5d + 4. Gradient m = 2.50 (cost per km). y-intercept c = 4 (fixed booking fee — the cost when d = 0).

Model: C = 0.15t + 10. When t = 40: C = 0.15(40) + 10 = 6 + 10 = $16.

They have swapped m and c. m (gradient) = the rate — the amount added per unit (per km, per hour, etc.). c (y-intercept) = the fixed starting value — the value when the variable equals 0.

The gradient is the rate of change — how much y changes for each 1-unit increase in x. Examples: • m = 3 km/h → speed of 3 km per hour. • m = −50 → value decreases by 50 per unit. Always state the units when interpreting.

The y-intercept is the initial value — the value of y when x = 0. Examples: • c = 200 → 200 items in stock at the start. • c = 15 → the temperature was 15°C at time 0. It is the starting point before any change occurs.

Gradient m = 8: the cost increases by $8 per hour. y-intercept c = 25: the initial cost (before any time passes) is $25 — a fixed/setup fee.

IB requires contextual interpretation — the gradient must be described in terms of the variables in the problem. For example: "The cost increases by $50 per kilogram." Just stating the number "50" earns no credit for an interpretation question.

1. The rate of change (→ this becomes m). 2. An initial value or a specific data point (→ this lets you find c). If two data points are given, find m first using the gradient formula, then find c.

V = −60t + 800. m = −60 (rate of decrease — negative because draining). c = 800 (starting volume at t = 0).

m = (300 − 180)/(7 − 3) = 120/4 = 30. C = 30d + c. Use (3, 180): 180 = 30(3) + c → c = 90. Model: C = 30d + 90 (daily rate $30, fixed fee $90).

A decrease means a negative gradient: m = −3. When a quantity is falling or decreasing, the gradient must be negative. Always check the direction of change before assigning the sign to m.

Substitute the given input value for x into the model equation and calculate y. Example: If C = 12t + 30 and t = 4, then C = 12(4) + 30 = 78.

Interpolation: predicting within the range of the original data — generally reliable. Extrapolation: predicting outside the range of the original data — less reliable; the model may not hold. IB questions often award 1 mark for commenting on reliability.

Set P = 0: 0 = −3t + 120 → t = 40 years. This is extrapolation (t = 40 is beyond the data range of 0–20) — the prediction is less reliable.

No — IB always requires a reason for reliability judgements. A correct answer gives: (a) whether it is interpolation or extrapolation, and (b) a reason (e.g. "within the data range" or "outside the data range — the trend may not continue").

Log-log: plot log y against log x. It's straight, with gradient n and intercept log a.

Semi-log: plot log y against x. It's straight, with gradient log b and intercept log a.

log y = log a + n·log x is LINEAR in log x, so a curved power law becomes a straight line you can read.

The power n.

log b (so b = 10^gradient).

The intercept is log a, so a = 10^(intercept).

Linearise both ways and compare R² — the fit with R² closer to 1 is straighter, so that's the model.

a = 10² = 100; b = 10^0.3 ≈ 2.00, so y ≈ 100·2ˣ.

A function is a rule that assigns exactly one output to each input. Every input (x-value) maps to one and only one output (y-value). Example: f maps every temperature in °C to a temperature in °F — one input, one output.

First mapping (1→5, 2→7, 3→5): YES, this is a function. Two inputs (1 and 3) share the same output — that is allowed. Second mapping (1→5, 1→9): NOT a function. Input 1 maps to two different outputs — that breaks the rule.

Example: "Country → Capital city." Each country has exactly one capital — every input (country) maps to exactly one output (capital). Non-example: "Person → Friend" — a person can have many friends, so one input maps to many outputs.

Yes — this is perfectly fine and does NOT stop something from being a function. What is NOT allowed: one input mapping to two different outputs. Example: f(2) = 5 and f(3) = 5 is fine. But f(2) = 5 and f(2) = 9 means it is not a function.

f(x) is the output of the function f when the input is x. Read it as "f of x." f is the name of the function. x is the input. f(x) is the corresponding output. Example: if f(x) = 2x + 1, then f(3) = 7.

f(x) = 4x − 3. Replace y with f(x). The name "f" is conventional but any letter works (g, h, p, etc.). Both y = 4x − 3 and f(x) = 4x − 3 describe the same rule.

g(t): apply the same rule but with t as the input → g(t) = t² + 1. g(a + 1): replace every x with (a + 1) → g(a + 1) = (a + 1)² + 1. The letter inside the bracket is always the input — substitute it everywhere x appears.

f(x) is not multiplication — the parentheses here mean "function of," not "times." f(x) = 4x + 2 does not mean f × x = 4x + 2. f is the function name; f(x) is the output value when the input is x.

Substitute a for every x in the function rule, then simplify. Example: f(x) = 3x + 5. Find f(4). Replace x with 4: f(4) = 3(4) + 5 = 12 + 5 = 17.

f(3) = 2(3) − 7 = 6 − 7 = −1. f(0) = 2(0) − 7 = 0 − 7 = −7. f(0) gives the y-intercept of the function.

Replace x with −2: h(−2) = (−2)² − 4(−2) + 1 = 4 + 8 + 1 = 13. Key: (−2)² = 4 (positive). −4(−2) = +8 (negative times negative = positive).

The error is in −4². When substituting a negative number, use brackets: (−4)² = +16. Without brackets: −4² = −16 (squaring only 4, then negating — wrong). Correct: f(−4) = (−4)² + 3 = 16 + 3 = 19.

The vertical line test: draw (or imagine) any vertical line through a graph. If every vertical line crosses the graph at most once → the graph represents a function. If any vertical line crosses the graph more than once → it is NOT a function (one x has two y-values).

No — a vertical line through the centre of the circle crosses it twice (two y-values for one x). Since one input (x) gives two outputs (y), the circle fails the vertical line test and is not a function.

Yes — every vertical line crosses the V-shape exactly once. Although the V looks like two lines meeting at a point, each x-value still gives exactly one y-value. y = |x| passes the vertical line test and is a function.

No — this describes a one-to-one function (injective), not just any function. The vertical line test only checks if each x gives at most one y. It is fine for two different x-values to produce the same y (many-to-one is still a function).

The domain is the set of all valid input values (x-values) for which the function is defined. Example: f(x) = √x has domain x ≥ 0 because you cannot take the square root of a negative number.

1. Division by zero — values of x that make the denominator = 0 must be excluded. Example: f(x) = 1/(x − 3) → x ≠ 3. 2. Square root of a negative — the expression inside √ must be ≥ 0. Example: f(x) = √(x + 4) → x ≥ −4.

The expression inside √ must be ≥ 0: x − 5 ≥ 0 → x ≥ 5. Domain: x ≥ 5 (or [5, ∞) in interval notation). At x = 5: f(5) = √0 = 0 ✓. At x = 4: f(4) = √(−1) — undefined ✗.

The denominator is x² − 9 = (x − 3)(x + 3). This equals zero when x = 3 or x = −3. The domain excludes x = 3 and x = −3, not x = 9. Always set the denominator equal to 0 and solve — do not guess.

The range is the set of all possible output values (y-values) that the function can produce. Example: f(x) = x² has range y ≥ 0 because squaring any real number gives a non-negative result.

Squaring any real number always gives a non-negative result: (−3)² = 9, 0² = 0. The output can never be negative. So no matter what x you input, f(x) ≥ 0. The minimum value is 0 (at x = 0); the function grows without limit as x → ±∞.

Since x² ≥ 0, we have x² + 3 ≥ 3. Range: g(x) ≥ 3 (or [3, ∞)). The minimum value is 3, reached at x = 0: g(0) = 0 + 3 = 3.

The square root function only outputs non-negative values: √x ≥ 0 for all x ≥ 0. Correct range: f(x) ≥ 0 (or [0, ∞)). The function cannot produce negative outputs — √9 = 3, not ±3.

Look at the graph horizontally — the domain is the set of x-values covered by the graph. Find the leftmost and rightmost x-values. Filled circle (●) = endpoint included. Open circle (○) = endpoint not included.

Look at the graph vertically — the range is the set of y-values covered by the graph. Find the lowest and highest y-values reached by the graph. A filled dot means that y-value is included; an open dot means it is excluded.

Domain: −2 ≤ x ≤ 6. Range: −3 ≤ y ≤ 8 (or −3 ≤ f(x) ≤ 8). IB also accepts interval notation: domain [−2, 6], range [−3, 8].

Domain → x-axis (horizontal). Range → y-axis (vertical). Memory trick: "D for domain, D for direction left-right (x-axis)." Domain = span of x-values; range = span of y-values.

A restricted domain limits the valid inputs to a practical range — not all mathematical values make sense. Examples: • Time t: must be t ≥ 0 (time cannot be negative). • Number of items n: must be a positive integer (you cannot buy half an item). • Distance d: must be d ≥ 0.

Domain: 0 ≤ t ≤ 15. t ≥ 0: time cannot be negative. t ≤ 15: V(15) = 1200 − 80(15) = 0 — the pool is empty; the model stops being valid.

No — x = 11 is outside the domain [2, 10]. The function is not defined for x = 11; the output is meaningless in this context. Always check inputs are within the stated domain before calculating.

n must be a non-negative integer (you cannot sell −3.7 units). A more appropriate domain is n ∈ {0, 1, 2, 3, ...} or n ≥ 0 with n ∈ ℤ. IB context questions often award a mark for recognising this restriction.

It tells you that when the input is x, the output is y — i.e. f(x) = y. The x-axis shows inputs; the y-axis shows outputs.

f(3) = 7. Read the y-value at x = 3 directly from the graph.

Locate x = 4 on the horizontal axis, go straight up to the curve, then read across to the y-axis. That y-value is f(4).

f(0) = −5. The point (0, −5) is on the graph, so when x = 0 the output is −5.

Shape of the curve, any x- and y-intercepts, turning points (if present), and asymptotes (if relevant). Label key values. Accuracy matters less than the correct shape and labelled features.

Straight line → linear. U-shape → quadratic. J-curve → exponential. Wave → sinusoidal.

y-intercept at (0, 6). Gradient = −2: from (0, 6), go right 1 and down 2 to reach (1, 4). Draw a straight line through both points and label the y-intercept.

a < 0. The parabola has a maximum (peak) at the vertex. If a > 0 it opens upward with a minimum.

Go to x = 2 on the horizontal axis, read straight up to the curve, then across to the y-axis. Write the y-value you find. "Write down" means no working is needed.

Draw a horizontal line at y = 5. Where it meets the curve, read straight down to the x-axis. There may be more than one solution.

The function has two x-intercepts (zeros/roots) at x = −1 and x = 3. The curve crosses the x-axis at those points.

Printed graphs have limited precision. As long as your reading is within 0.2 of the true value, the mark is awarded. Always read as carefully as possible.

Exponential: approaches a horizontal asymptote (y → 0 as x → −∞), never crosses the x-axis (if a > 0). Quadratic: has a vertex (turning point), usually has two x-intercepts, is symmetric.

A horizontal asymptote at y = 4. The curve gets arbitrarily close but never equals 4.

A cubic polynomial or a sinusoidal function. A quadratic has only one turning point; two suggests a higher-degree polynomial or a periodic function.

y → ∞. The graph grows without bound — steeper and steeper. As x → −∞, y → 0 (horizontal asymptote).

x-intercept: where the graph crosses the x-axis — this is where y = 0. y-intercept: where the graph crosses the y-axis — this is where x = 0.

No. A function produces exactly one output for x = 0, so there is exactly one y-intercept. However, a function can have zero, one, or many x-intercepts.

The curve stays entirely above or below the x-axis — its output is never zero.

All three refer to the values of x where f(x) = 0 — i.e. where the graph meets the x-axis. They are the same thing.

Substitute x = 0 into the function and calculate the output. The y-intercept is at the point (0, f(0)).

f(0) = 0 − 0 + 7 = 7. y-intercept is (0, 7).

f(0) = 5 · 2⁰ = 5 · 1 = 5. y-intercept is (0, 5). For any exponential y = a · bˣ, the y-intercept is always (0, a).

When x = 0: y = m(0) + c = c. So the line always meets the y-axis at the constant term.

Set f(x) = 0 and solve. Each solution is an x-intercept (root/zero).

Set x² − x − 6 = 0. Factor: (x − 3)(x + 2) = 0. So x = 3 or x = −2. x-intercepts are (3, 0) and (−2, 0).

The x-values where f(x) = 0, typically as coordinates: e.g. (−2, 0) and (3, 0), or just x = −2 and x = 3. Using the GDC Zero function is fine.

No real x-intercepts — the parabola is entirely above or below the x-axis. The equation has no real solutions.

Times when h = 0 — i.e. when the ball is on the ground: t = 0 (launch) and t = 4 (lands). x-intercepts are times, not heights.

P(0) = 800. The y-intercept is the initial population of 800 (at time t = 0).

State what the y-intercept value represents using the context's real-world units and language. E.g. "800 is the initial population at the start of the study."

The fixed cost of 400 — even if n = 0 units are produced, the cost is still 400 (overhead/startup cost).

The range of x and y values displayed on screen. Set using Xmin, Xmax, Ymin, Ymax. If the window is wrong, key features of the graph will be off-screen.

The turning points are outside the default window. Zoom out — increase the x and y range (e.g. −15 to 15). Use ZoomFit or adjust Ymin/Ymax manually.

Key features (intercepts, turning points, asymptotes) may be off-screen in the default window. Missing them leads to incomplete or wrong answers.

Automatically adjusts the y-window to show all points of the graph within the current x-range. Use it when the default window shows nothing useful.

Graph the function. Use 2nd → Calc → Zero (TI-84). Set a left bound and right bound on either side of each zero. The GDC gives the exact x-value.

Graph both functions. Use 2nd → Calc → Intersect (TI-84). Move the cursor near the intersection and press Enter three times. The GDC gives both x and y coordinates.

The y-coordinate. Substitute x = 2.31 into either equation, or read y from the GDC screen. IB expects both coordinates: e.g. (2.31, 5.62).

Graph h(x) = f(x) − g(x) and find its zeros using the Zero function. Where h(x) = 0 is exactly where f(x) = g(x).

Graph f(x). Use 2nd → Calc → Maximum. Set a left bound before the peak and a right bound after it. The GDC returns both x and y coordinates of the maximum.

Both the x and y coordinates as a pair: e.g. (2, −3). Never write only the x-value — that loses the second mark.

Use Maximum for the peak and Minimum for the trough — run them separately with appropriate bounds around each turning point.

12. The maximum value is the y-coordinate of the turning point, not the x-coordinate.

Write both coordinates clearly: x = 3.46, y = 8.92 (3 s.f. unless told otherwise). Or write the coordinate pair (3.46, 8.92).

No — you must state the GDC result clearly (coordinates, equation, etc.) but no algebraic working is needed. Always write what you found, not how the GDC found it.

Paper 2 only. Paper 1 is the non-calculator paper. No GDC allowed on Paper 1.

3 significant figures (3 s.f.), unless the question specifies otherwise. Using more decimal places is not wrong but messy; using fewer can cost marks.

A point where the function value is higher than all nearby values — the graph has a peak there. The function increases up to it and decreases after it.

Maximum point: both coordinates, e.g. (2, 9). Maximum value: just the y-value, e.g. 9. IB questions ask for either — read carefully.

The gradient is zero at every turning point. The tangent line is horizontal there.

Yes — local max/min are only local (in a neighbourhood). The global maximum is the highest point overall, which may be different from any local maximum.

Local maximum at (3, 8). The x-coordinate is 3 and the maximum value is 8. State both.

A coordinate pair: e.g. (−1, −5). Both x and y must be stated. Writing only x = −1 loses the second mark.

1. The minimum value is the y-coordinate. The point (−2, 1) tells you the minimum occurs at x = −2, and the minimum value is 1.

Look for a trough — a point where the graph changes from decreasing (falling) to increasing (rising). The curve dips down then comes back up.

1. Graph f(x) with an appropriate window. 2. Press 2nd → Calc → Maximum. 3. Move left of the peak: press Enter (left bound). 4. Move right of the peak: press Enter (right bound). 5. Press Enter again (guess). GDC shows coordinates.

x = 2.72 (3 s.f.). Then substitute into f to find y, e.g. y = f(2.72). State both coordinates.

IB markschemes award separate marks for each coordinate. Writing only x earns 0 marks for the y-coordinate. Always give both.

The local minimum. Run GDC Minimum with bounds around the other turning point to get its coordinates too.

After 3 seconds the ball reaches its highest point of 36 m above the ground.

Maximum profit of 8000 occurs when 500 units are produced. Producing more or fewer reduces profit.

State what the x-value represents (e.g. time, units) and what the y-value represents (e.g. height, profit) using the context's specific units. E.g. "After 3 hours, temperature reaches its peak of 36°C."

At n = 10 units, profit is at its lowest. The business loses the most money at this production level, and should either produce fewer or more units.

f is increasing on an interval if the output rises as you move left to right: whenever x₁ < x₂, we have f(x₁) < f(x₂). The graph goes upward.

The graph moves downward as you read from left to right — outputs fall as inputs increase.

Increasing — the function rises up to the maximum, then begins decreasing after it.

Inequalities (e.g. 1 < x < 4) and interval notation (e.g. (1, 4)) are both accepted. Write whichever matches the question's phrasing.

f is increasing on −2 < x < 1 (or [−2, 1]).

Increasing: x < 2 and x > 5. Decreasing: 2 < x < 5.

An inequality or interval notation including both endpoints. E.g. 2 ≤ x ≤ 5 or [2, 5]. The interval must refer to x-values (inputs), not y-values.

For x < 0. The parabola falls from left toward x = 0, then rises for x > 0. The minimum is at (0, 0).

"Increasing at a point" is meaningless. Increasing is a property of an interval, not a single point. Write "f is increasing for x > 3" or "f is increasing on (1, 3)".

The answer should be an interval of x-values, not y-values. Correct: e.g. "1 < x < 3." The y-values (4 to 9) are outputs, not the interval.

IB accepts both x < 2 and x ≤ 2 for the increasing interval up to a maximum. Either strict or inclusive inequalities are fine unless the question specifies.

Increasing everywhere — gradient is 3 > 0, so the output always rises as x increases. No turning points.

The temperature rises during the first 5 hours.

An interval of t-values (the input variable), e.g. "3 < t < 8 hours." Not y-values. Use the same variable as the context.

n = 200 is where the profit function has its local maximum — the production level giving the greatest profit.

State whether f is increasing or decreasing, and whether it approaches a fixed value (asymptote) or continues without bound. E.g. "f is decreasing and approaches y = 3."

A horizontal line y = k that the graph approaches as x → ∞ or x → −∞, but (usually) never reaches or crosses.

Exponential: y = a · bˣ. As x → −∞ (for b > 1) or x → ∞ (for 0 < b < 1), the output approaches 0.

Write it as a full equation: e.g. y = 3. Not just "3" — the y = must be included.

As x gets very large (or very negative), the output of f gets arbitrarily close to the asymptote value — but the curve never quite touches that line.

y = 5. As x → −∞, 3 · 2ˣ → 0, so f(x) → 5. The +5 shifts the asymptote up from y = 0 to y = 5.

Range is f(x) > 4. The function always stays above y = 4 (never equals it), so 4 is excluded from the range.

Horizontal asymptote y = 10. As x → ∞, 100 · 0.5ˣ → 0, so f(x) → 10 from above.

The function never reaches the asymptote value, so that value is excluded from the range. E.g. if asymptote y = 3 and function approaches from above, range is f(x) > 3.

A vertical line x = a where the function is undefined and its output grows to ±∞ as x approaches a from either side.

At x = 3 — the denominator is zero there, so the function is undefined. The graph blows up to ±∞ near x = 3.

x-intercept: the curve actually touches or crosses y = 0. Asymptote y = 0: the curve approaches y = 0 but never reaches it.

Set denominator = 0: 2x + 4 = 0 → x = −2. Vertical asymptote at x = −2.

How f(x) behaves as x → ∞ or x → −∞ — whether it grows, falls, or approaches a limiting value (asymptote).

As x → ∞, 0.5ˣ → 0, so f(x) → 0. The graph approaches the asymptote y = 0 from above and decreases toward it.

f(x) → ∞ as x → ∞. There is no horizontal asymptote — the function grows forever.

State whether f increases, decreases, or approaches a fixed value. If it approaches a value, give the equation of the asymptote. Use context language if relevant.

1. Constant rate of change — each unit increase in x produces the same change in y. 2. The graph is a straight line.

When the data shows a constant rate of change — equal steps in x produce equal steps in y. A scatter plot that looks like a straight line suggests a linear model.

5n: cost increases by 5 per unit produced (variable cost, the gradient). 200: fixed cost regardless of production level (the y-intercept).

Yes — constant speed means equal distance in equal time intervals. Distance = 80t is linear with gradient 80.

1. Calculate gradient: m = (y₂ − y₁)/(x₂ − x₁). 2. Use y = mx + c with one point to find c. 3. Write the model.

T = −2.5(12) + 80 = −30 + 80 = 50.

m = (20 − 60)/8 = −5. Using (0, 60): T = −5t + 60.

The full equation in y = mx + c form, with numerical values for m and c, using the variables named in the context.

The population increases by 4.5 people per year.

The initial weight is 50 kg — the weight at the start (d = 0), before any distance has been covered.

State: the numerical value, the units, and what it means for the specific context. E.g. "The water level rises by 3 cm per hour."

The quantity is decreasing at a constant rate as the input variable increases.

The model gives reliable, meaningful predictions for x-values within the range of the original data (interpolation). Outside this range, the model may break down.

Check if x = 50 is within the data range. If yes: "Yes — x = 50 is within the data range so the estimate is reliable (interpolation)." If no: "Less reliable — x = 50 is outside the data range (extrapolation)."

Physically extreme or impossible values signal model breakdown — this is extrapolation far beyond the data range. Real temperatures may not follow this pattern at t = 100.

Interpolation: predicting within the data range — generally reliable. Extrapolation: predicting outside the range — less reliable, the pattern may not continue.

A parabola — a symmetric U-shape. Opens upward (∪) if a > 0, downward (∩) if a < 0.

A ball thrown upward: h(t) = −5t² + 20t + 3. Height rises, reaches a maximum, then falls — the parabolic path of projectile motion.

Linear: constant rate of change, straight line. Quadratic: changing rate of change, has a maximum or minimum turning point (vertex), curved graph.

Revenue increases, reaches a maximum at the vertex (optimal price), then decreases. There is one best price for maximum revenue.

x = −b/(2a). The y-coordinate is found by substituting this x back into the equation.

x = −(−8)/(2·2) = 2. y = 2(4) − 8(2) + 3 = 8 − 16 + 3 = −5. Vertex at (2, −5).

x = −(−6)/(2·1) = 3. f(3) = 9 − 18 + 11 = 2. Minimum value is 2 (at x = 3).

If a > 0 (parabola opens up), the vertex is a minimum. If a < 0 (parabola opens down), the vertex is a maximum.

x = 3 is the x-coordinate of the vertex, not the maximum value. The maximum value is f(3) = −9 + 18 − 5 = 4.

The formula is x = −b/(2a). Forgetting the negative gives the wrong x-value and hence the wrong vertex.

No — if a > 0 the parabola opens upward and only has a minimum. Only quadratics with a < 0 have a maximum.

h(t) = 0. Set −5t² + 20t = 0 → −5t(t − 4) = 0 → t = 0 or t = 4. Ball is on the ground at t = 0 and t = 4.

t = −20/(2·−5) = 2. h(2) = −5(4) + 40 + 1 = 21. Maximum height = 21.

n = −10/(2·−1) = 5. Maximum profit at n = 5 units.

Find x-intercepts (set P = 0, solve). P is positive between the roots if a < 0, or outside them if a > 0.

R = 0 at p = 0 and p = 40. These are the prices at which revenue is zero: free (no payment) or so expensive no one buys.

y = a · bˣ. a = initial value (y-intercept at x = 0). b = growth/decay factor per unit of x.

500 = initial value (at x = 0). 1.06 = growth factor — 6% growth per unit of x.

Growth — the output increases as x increases. The greater b is above 1, the faster the growth.

Decay — the output decreases as x increases. The closer b is to 0, the faster the decay.

P(t) = 4000 · 1.05ᵗ. Initial value a = 4000, growth factor b = 1 + 0.05 = 1.05.

Q(t) = 200 · 0.5ᵗ. Initial value a = 200, decay factor b = 0.5.

Substitute both points to get two equations. Divide one by the other to eliminate a and solve for b. Then substitute b back to find a.

P = 3000 · 1.04⁵ = 3000 · 1.2167 ≈ 3650.

y = 5 · 1.03 · x is linear, not exponential. In an exponential model, x must be the exponent: y = 5 · 1.03ˣ.

b is the growth factor, not the rate. b = 1 + rate = 1 + 0.08 = 1.08. Using b = 8 would give wildly wrong values.

No — a · bˣ is always positive when a > 0 and b > 0. A negative result always means a calculation error.

Check the ratio of successive y-values: if y₂/y₁ is approximately constant, the data is exponential.

y = 0. As x → −∞, 2ˣ → 0, so the whole expression approaches 0 from above. The x-axis is the asymptote.

P → 0. The substance/quantity decays toward zero but never fully disappears (according to the model).

y = 0. Write as a full equation. The growth model approaches 0 as x → −∞.

The model predicts the quantity approaches zero but never reaches it. In reality, the quantity may reach zero (e.g. a substance fully decays). The model is a mathematical idealisation.

y = axⁿ, where a is a constant and n is any real-number power.

Area of circle: A = πr² (power 2). Distance under gravity: s = 5t² (power 2). Surface area ∝ length² for similar shapes.

Power model: x is the base, n is a fixed exponent. Exponential: x is the exponent, b is a fixed base. Very different shapes for large x.

y increases by a factor of 2² = 4. Power models scale multiplicatively: doubling x multiplies y by 2ⁿ.

y = 2x³ is a power model — x is the base. y = 2 · 3ˣ is exponential — x is the exponent.

Exponential always eventually grows faster than any power model. Even x¹⁰⁰ is eventually overtaken by 2ˣ.

No — exponential y = a · bˣ passes through (0, a), not the origin (unless a = 0). A power model with n > 0 passes through (0, 0).

Exponential — x is in the exponent. Base 0.7 means decay. It is NOT a power model.

Power regression (PwrReg on TI-84). Returns a and b for y = axᵇ.

y = 3.25x^0.812 (all values to 3 s.f.).

When the scatter plot shows a curved relationship (not straight), the data passes near the origin, and a straight line clearly doesn't fit the pattern.

Very strong fit — 97% of variation in y is explained by the power model. It is a very good fit for the data.

Mass grows slightly faster than the square of diameter. Doubling d multiplies M by 2^2.1 ≈ 4.3.

The model was built from data in a limited range. Using it for x well beyond that range is extrapolation — the pattern may not continue and the model may give unrealistic values.

y = 2 · 4^1.5 = 2 · 8 = 16.

The model is not valid for that input. Negative length, mass, or similar quantities are physically impossible. Either the input is outside the valid domain or the model breaks down.

f(t) = a sin(bt + c) + d. a = amplitude (half the range). Period = 2π/b. c = phase shift. d = midline (vertical shift).

Amplitude = 3. It is the coefficient of sin — the distance from the midline to the maximum or minimum.

Period = 2π ÷ (π/6) = 2π × 6/π = 12.

Midline y = 10. Amplitude = 4, so f oscillates between 10 − 4 = 6 and 10 + 4 = 14.

Amplitude = (max − min) / 2. Midline = (max + min) / 2.

Amplitude = (18 − 4)/2 = 7. Midline = (18 + 4)/2 = 11.

Midline = (8 + 24)/2 = 16°C. Amplitude = (24 − 8)/2 = 8°C.

2π/b = 24 → b = 2π/24 = π/12.

Amplitude = (max − min)/2, not the maximum value alone. If min = 4, amplitude = (18 − 4)/2 = 7, not 18.

Period: how long one complete cycle takes (in time units, e.g. hours). Frequency: cycles per unit time = 1/period.

b is not the period — it is a parameter inside the argument. Period = 2π/b. For b = 2, period = π, not 2.

Maximum = midline + amplitude = 12 + 5 = 17. The midline d is not the maximum.

f(6) = 7 sin(π · 6/12) + 15 = 7 sin(π/2) + 15 = 7(1) + 15 = 22.

h(3) = 3 sin(π/2) + 5 = 3(1) + 5 = 8 m.

Either a calculation error, or the model is being used outside its valid range. A sinusoidal model never exceeds amplitude + midline.

8 sin(πt/12) + 12 = 20 → sin(πt/12) = 1 → πt/12 = π/2 → t = 6 hours.

Linear: y = mx + c. Quadratic: y = ax² + bx + c. Exponential: y = a · bˣ. Power: y = axⁿ. Sinusoidal: y = a sin(bx + c) + d.

Exponential — constant percentage growth = constant ratio between successive values = exponential model.

Sinusoidal (trigonometric). Tides, temperature cycles, sound waves — any periodic real-world quantity.

Linear. A straight-line scatter plot is the defining sign of a linear model.

Power (y = axⁿ) or exponential (y = a · bˣ). The near-origin hint favours power. Compare R² after fitting both.

Quadratic — single turning point, symmetric parabola shape.

Sinusoidal — regular repeating pattern = periodic = trigonometric model.

Name the model type AND give one clear reason based on the shape or context. E.g. "Exponential, because the data shows a constant ratio between successive values."

Power (y = axⁿ): may pass through origin, no horizontal asymptote to the right. Exponential (y = a · bˣ): never passes through origin, has horizontal asymptote y = 0 as x → −∞.

12/3 = 4 and 48/12 = 4. Constant ratio → exponential model.

Exponential — higher R² means it explains more of the variation. Choose the model with the higher R².

State that the data shows a constant multiplicative (percentage) growth rate, not a constant additive change — which matches exponential, not linear.

Exponential — doubling at a constant time interval means a constant ratio between values, which is the defining feature of exponential models.

Quadratic — the path is a parabola. It has one turning point and is not periodic (doesn't repeat).

Sinusoidal — regular repeating cycle with constant period.

Power model: F = av², where n = 2.

1. Enter x data in L1, y data in L2. 2. Stat → Calc → LinReg(ax+b). 3. Note a and b from output. 4. Write the equation y = ax + b.

The best-fit equation parameters (a, b, etc.) and the correlation coefficient r (or R² for non-linear).

The full equation with all parameters to 3 s.f. E.g. y = 2.35x + 4.18. Include what regression type you used if asked.

Substitute x = 10 into the regression equation and calculate. Show the substitution clearly.

Exponential (ExpReg) and power (PwrReg). Run both and compare R² values.

Sinusoidal regression (SinReg on TI-84).

Use the exponential model — much higher R² means far better fit.

Exponential regression (ExpReg). Constant ratio is the hallmark of exponential growth/decay.

y = 2.35 · 0.812ˣ (all values to 3 s.f.).

No — just state the values clearly. "From GDC: a = 2.35, b = 0.812." No algebraic working is needed.

y = 3.7(5) − 12.4 = 18.5 − 12.4 = 6.1.

IB expects 3 significant figures unless specified. Using fewer can cause errors in later parts; IB may not award accuracy marks if rounding is too severe.

Very strong positive linear correlation. The model fits the data extremely well.

r: Pearson correlation coefficient, ranges from −1 to 1, linear regression only. R²: coefficient of determination, ranges 0 to 1, applies to all regression types. R² = r² for linear.

The model has a moderate fit (R² = 0.72 — 72% of variation is explained). Predictions may not be highly reliable.

Perfect fit — every data point lies exactly on the regression curve. All predicted values match observed values exactly.

Using a model to predict a value for an input that is within the range of the original data. Generally reliable.

Using a model to predict a value for an input that is outside the range of the original data. Less reliable — the pattern may not continue.

Extrapolation — 2025 is beyond the end of the data range.

Interpolation — we stay within the range where the model was built and validated. Extrapolation assumes the pattern continues, which may not hold in new conditions.

"Yes, the estimate is reliable as the value x = [n] is within the data range (interpolation)."

"The estimate is less reliable as the value x = [n] is outside the data range (extrapolation). The model may not hold beyond the collected data."

The model breaks down for large t — populations cannot be negative. The model is only valid within the original data range.

Conditions change over time (resources, policy, environment). The model was built on past data and assumes the same pattern continues indefinitely.

The range of input values for which the model produces meaningful, realistic outputs — usually the range of the original data.

Negative height is physically impossible — the ball has already hit the ground. The model is only valid for 0 ≤ t ≤ 4 (while airborne).

Ask: Is the output physically possible? Is the input within the data range? Does the result make sense in the context (correct units, realistic magnitude)?

One reason the model may not be perfectly accurate, e.g. "The model assumes constant growth rate, but this may not hold over long periods as conditions change."

Yes/No + one reason referencing whether the input is within or outside the data range (interpolation vs extrapolation).

"The estimate is reliable as t = 8 is within the data range (interpolation)."

"The estimate is less reliable as t = 15 is outside the data range (extrapolation). The model may not hold beyond the collected data."

"The model assumes exponential growth continues indefinitely, but in reality growth may slow due to limited resources or carrying capacity."

f(g(x)): apply the inner function g first, then the outer function f.

g — the one written closest to x (read right-to-left).

Substitute the whole expression g(x) into f wherever f's input appears, then simplify.

Usually no — order matters, so they are generally different functions.

g(5)=2, then f(2)=2(2)+1=5.

(f∘g)(x)=(x+1)²; (g∘f)(x)=x²+1. They differ.

x must be allowed into g, AND g(x) must be a legal input for f.

√(x−4) needs x−4 ≥ 0, so x ≥ 4.

It undoes f: if f maps a→b, then f⁻¹ maps b→a. So f⁻¹(f(x)) = x.

Write y = f(x), swap x and y, then solve for y.

The input that gives output b — i.e. solve f(x) = b.

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

f⁻¹ is the reflection of f in the line y = x; each (a,b) becomes (b,a).

Only when f is one-to-one (passes a horizontal line test); otherwise restrict the domain.

They swap: the range of f becomes the domain of f⁻¹.

y = 4x+3 → x = 4y+3 → f⁻¹(x) = (x−3)/4.

Translates it UP by b (b outside f changes the height; negative b moves it down).

Translates RIGHT by a. The input must be a units larger to reproduce an old height, so features move to bigger x.

Vertical stretch by factor p (every height ×p). Points on the x-axis don't move.

Horizontal stretch by factor 1/q (width scaled by the reciprocal). q > 1 squashes toward the y-axis.

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

Reflects the graph in the y-axis (left↔right).

To (x + a, p·y + b): inside shifts x, outside scales then shifts y.

y = 5 — the outside +5 lifts the whole graph, asymptote included.

Match the behaviour: steady change → linear; one peak → quadratic; constant multiplier each step → exponential; fast-then-levels-off at a ceiling → logistic; repeats/oscillates → sinusoidal; different rule on each interval → piecewise.

It grows fast at first but LEVELS OFF at a finite ceiling (carrying capacity). A plain exponential never flattens.

L is the carrying capacity (long-run ceiling): as t → ∞, e^{−kt} → 0 so y → L.

Put t = 0: e⁰ = 1, so y(0) = L/(1 + C).

k is the starting value (at t = 0); r is the continuous growth (r > 0) or decay (r < 0) rate.

a = amplitude (half the swing), d = midline (average level), period = 2π/b, c = horizontal shift.

One linear rule up to the threshold, then a second linear rule (fixed cost so far + new lower rate × extra units) beyond it.

Interpret each parameter in context AND comment on validity (where the model is reliable vs where extrapolation is unsafe).

d = sqrt((x2-x1)^2 + (y2-y1)^2)

M = ((x1+x2)/2, (y1+y2)/2)

When points have x, y, and z coordinates.

Mixing up subtraction order before squaring and arithmetic slips.

d = √[(x₂−x₁)² + (y₂−y₁)² + (z₂−z₁)²] — the 2D formula plus a z-term.

M = ((x₁+x₂)/2, (y₁+y₂)/2, (z₁+z₂)/2) — average all three coordinates.

The line from one corner of a cuboid to the opposite corner — found with the 3D distance formula.

√(4+9+36) = √49 = 7.

B = 2M − A (double the midpoint, subtract the known end), coordinate by coordinate.

Adding the gaps before squaring: √(6+4+3) instead of √(36+16+9). Square each gap first.

Volume = cross-sectional area x length

V = pi r^2 h

Surface area measures outside covering; volume measures inside space.

Material needed to cover an object.

A magnitude (size) and a direction. A scalar has size only (e.g. temperature, price).

Pythagoras on its components: |v| = √(v₁² + v₂²) in 2D, √(v₁² + v₂² + v₃²) in 3D.

√(6² + 8²) = √100 = 10.

Component by component: add (or subtract) the matching entries.

Multiplies every component by k — it stretches/shrinks the vector, and flips its direction if k is negative.

Divide v by its own magnitude: v̂ = v / |v|. It keeps the direction but has length 1.

The arrow from the origin O to a point: the position vector of A is a = OA.

AB = b − a ('finish minus start') — the displacement from A to B.

a is the position vector of a point ON the line, d is the direction the line points, and λ is a number that slides you along it (λ = 0 gives a, λ = 1 gives a + d).

d (the vector multiplied by λ) is the direction; a (the constant part) is a point on the line.

d = B − A (finish minus start). Any scalar multiple of it is also a valid direction.

r = A + λ(B − A) — start at A, walk in the direction B − A.

Substitute that number for λ and add component by component: r = a + λd.

Set a + λd = P, solve ONE component for λ, then check the SAME λ works in every other component. One λ fits all → on the line; otherwise → off it.

No — different start points a (any point on the line) and any scalar multiple of d describe the same line.

At λ = ½ in r = A + λ(B − A): a half-step of the direction from A.

r(t) = r₀ + t·v — the start position r₀ plus the velocity v added on t times.

v is the vector multiplying t — the coefficient of t in each coordinate.

speed = |v| = √(vₓ² + v_y²) — the length of the velocity vector (one positive number).

Velocity is a vector (direction + rate); speed is a single positive number, the magnitude of velocity.

√(6² + (−8)²) = √100 = 10.

(4 + 12, 12 − 16) = (16, −4).

Distance = speed × t (the path is a straight line, direction constant).

The speed — a single positive number (the magnitude of the velocity vector), not a vector.

They must be at the SAME position at the SAME value of t — one t satisfies every coordinate of r_A(t) = r_B(t).

Two objects can pass through the same point at different times; a collision needs the same point at the same moment.

Set r_A(t) = r_B(t), solve one coordinate for t, then CHECK that t in the other coordinate(s). If it fits all, they collide.

d(t) = |r_B(t) − r_A(t)| = √((Δx)² + (Δy)²), the length of the displacement between them.

Find the MINIMUM of d(t) — graph d(t) on the GDC and read the lowest point (x = time, y = least distance).

d(t) is least exactly where d(t)² is least (both positive, square root is increasing), and d² has no awkward square root.

Its x-coordinate is the time of closest approach; its y-coordinate is the least distance between the objects.

Interpret in context — compare it to the required safety/separation distance and state whether the rule is met.

Multiply matching components and add: v·w = v₁w₁ + v₂w₂ + v₃w₃. The result is a single number.

A number (scalar) — that's why it's called the scalar product.

v·w = |v||w| cos θ, so cos θ = (v·w)/(|v||w|).

θ = cos⁻¹[ (v·w)/(|v||w|) ] — dot product over the product of the magnitudes.

The angle is obtuse (between 90° and 180°), because cos θ is negative.

They are perpendicular exactly when v·w = 0.

(1)(3) + (2)(0) + (−2)(1) = 3 + 0 − 2 = 1.

AB and AC (both pointing OUT from A); then cos A = (AB·AC)/(|AB||AC|).

A VECTOR perpendicular to both v and w (the dot product gives a scalar).

v×w = (v₂w₃ − v₃w₂, v₃w₁ − v₁w₃, v₁w₂ − v₂w₁).

|v×w| = |v||w| sin θ = the area of the parallelogram with sides v and w.

½|AB×AC| — two sides from the same vertex, crossed, length halved.

It must be perpendicular to both inputs: v·(v×w) = 0 and w·(v×w) = 0.

Dot → a scalar (number); cross → a vector.

No — w×v = −(v×w) (opposite direction); and v×v = 0.

When v and w are parallel (sin θ = 0) — they span no area.

Vertices (the dots/points) and edges (the lines joining pairs of vertices).

The number of edge-ends meeting at that vertex (a loop counts as 2).

The sum of all vertex degrees equals 2 × (number of edges): Σ deg(v) = 2E.

Because the total degree Σ deg = 2E is always even, the odd degrees must pair up to keep the sum even.

Every pair of vertices is joined; it has n(n − 1)/2 edges and every vertex has degree n − 1.

A connected graph with no cycles; it has exactly n − 1 edges.

The vertices split into two groups, with edges only between the groups (never within a group).

Trail = no repeated edge (may revisit a vertex); path = no repeated vertex; cycle = a path that returns to its start.

A square matrix where rows and columns are the vertices (same order); entry (i,j) = the number of edges from vertex i to vertex j.

An undirected graph gives a SYMMETRIC matrix (edges go both ways); a directed graph is usually not symmetric.

The number of walks of length n (n edges, repeats allowed) from vertex i to vertex j.

No — length = the number of EDGES used. Use the weighted/distance matrix for actual distance.

On the DIAGONAL — entry (i,i) of Aⁿ counts walks of length n that start and end at vertex i.

Replace each 1 in the adjacency matrix with the weight (distance/cost/time) of that edge; keep 0 (or blank/∞) where there is no edge.

Enter A, compute A⁴, then read the entry in row X, column Y.

The degree of vertex i (the number of edges meeting it) — 'out and straight back'.

The set of edges connecting every vertex with the least total weight and no cycle. On n vertices it has exactly n − 1 edges.

Exactly n − 1, and it contains no cycle.

Sort all edges from cheapest to dearest; add them one at a time, skipping any edge that would create a cycle, until all vertices are connected.

Start at any vertex; repeatedly add the cheapest edge that joins a new (un-included) vertex to the tree, until all vertices are included.

They always give the same total weight (and the same tree when all edge weights are distinct).

The shortest path (least total weight) from a start vertex to a target vertex — not the whole connecting tree.

new label = min(old label, permanent value of current vertex + edge to this vertex). The smallest temporary label becomes permanent next.

MST (Prim/Kruskal): connect ALL vertices cheaply. Dijkstra: shortest route between TWO specific vertices.

The shortest CLOSED walk that uses every EDGE at least once and returns to the start.

When every vertex has even degree (an Eulerian circuit exists); then the answer is just the sum of all edges.

Pair the odd-degree vertices so the total of the shortest paths between the pairs is least; add that to the sum of all edges.

Sum of all edges + minimum pairing of the odd-degree vertices' shortest paths.

The shortest CLOSED route visiting every VERTEX exactly once and returning to the start (a Hamiltonian cycle).

Nearest-neighbour algorithm: from the start, always go to the nearest unvisited vertex, then return to start. Its length is an upper bound.

Delete a vertex; find the MST of the rest; add the two cheapest edges from the deleted vertex back.

Route inspection covers every EDGE; travelling salesman visits every VERTEX. Pick by what must be covered.

sin = opp/hyp, cos = adj/hyp, tan = opp/adj

When angle is unknown and side ratio is known.

The side directly across from angle theta.

Using wrong ratio due to side mislabelling.

When you have AAS, ASA, or SSA triangle data.

When you have SAS or SSS triangle data.

a^2 = b^2 + c^2 - 2bc cos A

SSA data can produce two possible triangles.

Angle measured upward from horizontal line of sight.

Angle measured downward from horizontal line of sight.

They often form alternate interior angles in parallel-line setup.

Using vertical line as reference instead of horizontal.

Sketch and isolate a right triangle in 3D shape.

Use Pythagoras twice or 3D distance formula.

Need plan view + elevation view for full geometry.

Using wrong triangle for angle asked.

s = r theta

s = (theta/360) * 2pi r

Many formulas become simpler and less error-prone.

Distance traveled along circular path, not straight line.

A = (1/2) r^2 theta

A = (theta/360) * pi r^2

P = 2r + arc length

Using degrees formula with radians (or vice versa).

Set equations equal (or solve simultaneous equations).

Same gradient m, different intercept c.

Same gradient and same intercept.

Break-even point or equal-cost point in models.

Line through midpoint of AB and perpendicular to AB.

If gradient is m, perpendicular gradient is -1/m.

Bisector must pass through midpoint of original segment.

Voronoi boundaries are perpendicular bisectors between sites.

Set of points closer to one site than any other site.

Using perpendicular bisectors of neighbouring sites.

Point equidistant from 3 or more sites.

Compare distances from the point to each site.

Only local neighbouring cells around insertion region change.

Usually at a Voronoi vertex.

Service zones (hospitals, towers, warehouses).

State which zones change and justify using nearest-distance logic.

The angle at the centre of a circle subtended by an arc equal in length to the radius (≈ 57.3°).

Multiply by π/180 (since π rad = 180°).

Multiply by 180/π.

2π radians = 360° (and π radians = 180°).

l = rθ, with θ in radians.

A = ½r²θ, with θ in radians.

60 × π/180 = π/3 radians.

The formulas come from the fraction θ/(2π) of the circle; the 2π only cancels cleanly when θ is measured in radians.

(cos θ, sin θ) — cos is the x-coordinate (across), sin is the y-coordinate (up).

They're coordinates of a point on a circle of radius 1, so neither can exceed the radius.

sin²θ + cos²θ = 1 (it's Pythagoras applied to the unit-circle point).

cos 60° = ½, sin 60° = √3⁄2.

Both equal √2⁄2 (≈ 0.707).

tan θ = sin θ / cos θ.

sin²θ = 1 − 0.36 = 0.64, so sin θ = 0.8 (positive in Quadrant 1).

Graph both sides over the interval, use 'intersect' to read EVERY crossing, then keep only solutions inside the interval.

Write the point as a column vector and multiply: matrix on the left, point on the right. (a b; c d)(x; y) = (ax+by; cx+dy).

(cos θ −sin θ; sin θ cos θ).

(k 0; 0 k) — multiplies every distance from O by k.

(cos 2θ sin 2θ; sin 2θ −cos 2θ).

(0 −1; 1 0)(4; 0) = (0, 4).

A rotation of 180° about the origin: (x, y) → (−x, −y).

Transform each vertex (multiply each corner's column vector), then re-join the images.

A reflection in the x-axis (it flips the sign of y only).

Multiply BA (B on the left): image = B(Ax) = (BA)x. The right-most matrix acts first.

Yes — matrix multiplication is not commutative, so BA ≠ AB in general (different transformations).

ad − bc.

New area = |det| × old area; |det| is the area scale factor.

The transformation reverses orientation — the shape is reflected (flipped over).

Area scale factor 0: the plane collapses onto a line/point, so there is no inverse (it can't be undone).

|−4| × 6 = 24 (and the shape is flipped).

det(BA) = det B × det A.

The population is every individual of interest; the sample is the subset you actually collect data from. A good sample mirrors the population.

Every member of the population has an equal chance of selection (e.g. names drawn from a hat, or GDC random numbers).

Order the population, choose a random start, then pick every kᵗʰ member down the list.

(group size ÷ population size) × sample size. Each stratum is sampled in proportion to its size.

Both target groups in proportion, but stratified picks members randomly within each group, while quota lets the interviewer choose — so quota is non-random and can be biased.

It surveys whoever is easiest to reach, so the sample is usually unrepresentative — it tends to over- or under-represent certain people, giving biased estimates.

Consistency — repeating the measurement gives (almost) the same result each time (small random error).

It measures what it is supposed to measure, with no systematic bias. A measure can be reliable yet still invalid (consistently wrong).

Look at the SHAPE of the scatter and pick the family that matches (exponential = constant % change, power = scaling/flattening, quadratic = rise-then-fall, sinusoidal = repeating). Then compare R² between candidates.

How much of the variation in the data the model explains. R² = 1 is a perfect fit; closer to 1 is better; near 0 is poor.

Fit both on the GDC and compare R² — the model with the higher R² (closer to 1) fits better. Also check the shape and context make sense.

A residual is data − model (y − ŷ) for one point. SSres = Σ(y − ŷ)² is the sum of squared residuals; regression minimises it, and a smaller SSres gives a higher R².

So positive and negative residuals don't cancel out, and larger misses are penalised more heavily.

y = k·aˣ (base form) or y = k·eʳˣ (natural form). a > 1 (or r > 0) = growth; 0 < a < 1 (or r < 0) = decay.

Interpolation = predicting inside the data range (safer). Extrapolation = predicting outside it (riskier — the model may not hold).

No — R² only measures fit to the EXISTING data. Predictions far beyond the range (extrapolation) can be unreliable even when R² is near 1.

Multiply each value by its probability and add them: E(X) = Σ x·P(X = x). It's the long-run average.

Var(X) = E(X²) − [E(X)]² — the mean of the squares minus the square of the mean. SD = √Var(X).

No — e.g. the expected number of heads in one flip is 0.5, even though you only ever see 0 or 1.

aE(X) + b — the scale multiplies and the shift b is added on.

a²Var(X). The shift b drops out completely; the scale a enters SQUARED.

|a|·SD(X). The standard deviation multiplies by |a| and is unaffected by the shift b.

Adding a constant slides every value equally, so the gaps between values (the spread) are unchanged.

Enter values in L1 and probabilities in L2, run 1-Var Stats with L1 as data and L2 as frequencies: x̄ = E(X), σ = SD.

E(X + Y) = E(X) + E(Y); E(X − Y) = E(X) − E(Y). Means take the sign.

Var(X + Y) = Var(X − Y) = Var(X) + Var(Y). Variances ALWAYS add, even for a difference.

No — add the VARIANCES (square the SDs), then square-root: SD(X±Y) = √(SD(X)² + SD(Y)²).

Mean = nμ; Variance = nσ² (so SD = σ√n).

Var(nX) = n²σ² (one copy scaled up); Var(sum of n independent copies) = nσ² (separate items partly cancel).

The sample mean x̄ — it's unbiased as is.

sₙ₋₁² = Σ(x − x̄)²/(n − 1) — divide by n − 1, the GDC's Sx² (not σx² which uses ÷n).

sₙ₋₁² = [n/(n − 1)]·sₙ² — scale the biased variance up by n/(n − 1).

For a large sample size n, X̄ is approximately Normal — even if the population itself is not Normal.

It equals the population mean μ (averaging does not shift the centre).

σ/√n — the population standard deviation divided by the square root of the sample size.

X̄ ≈ N(μ, σ²/n), i.e. Normal with mean μ and standard error σ/√n.

Averaging cancels out highs and lows; dividing σ by √n means larger samples give steadier (less variable) means.

Four times as much — because √n must double, and √4 = 2.

When the population is already Normal — then no large-sample approximation is needed.

A single value uses σ; the mean (or total) of n values uses the standard error σ/√n.

A range of believable values for the true population mean, built as x̄ ± margin of error. A 95% interval is produced by a method that captures the true mean about 95% of the time.

x̄ ± t*·s_{n-1}/√n, where t* comes from the t-distribution with df = n − 1.

df = n − 1 (one less than the sample size). The GDC's t-interval applies this automatically.

The unbiased estimate s_{n-1} (the GDC's 'Sx', dividing by n − 1), because the true population σ is unknown and must be estimated from the sample.

Use the t-interval menu: enter x̄, s_{n-1} and n (or the raw data) and the confidence level (C-Level), then read off the interval (a, b).

It makes the interval NARROWER: a larger n increases √n in the denominator, so the margin t*·s/√n shrinks — a more precise estimate.

It makes the interval WIDER: a higher confidence level uses a larger t*, so the margin grows — you cast a wider net to be more sure of trapping μ.

If the claimed value lies INSIDE the interval it is plausible (consistent with the data); if it lies OUTSIDE, the data give evidence against the claim.

For X ~ Po(m): P(X = x) = e^(−m)·mˣ/x!, for x = 0, 1, 2, …

The mean (average) number of events in the fixed interval.

They are EQUAL: mean = variance = m, so σ = √m.

poissonpdf(m, x) — the probability of exactly x events.

poissoncdf(m, x) — the probability of at most x events.

Use the complement: P(X ≥ k) = 1 − poissoncdf(m, k − 1).

Also Poisson: X + Y ~ Po(m₁ + m₂) — the means add.

Events occur independently, at a constant average rate, singly (not in clumps), with no fixed upper limit on the count.

The probability of getting data at least this extreme if H₀ were true. Small p = the data are surprising under H₀.

p < α → reject H₀; p ≥ α → fail to reject H₀ (α = significance level).

z-test when the population σ is known; t-test when σ is unknown (you only have the sample sd) — the usual case.

One-sample: compare one group's mean to a fixed claimed value. Two-sample: compare two independent groups' means.

One-tailed for a directional H₁ (μ < or μ >); two-tailed for H₁: μ ≠ (a difference either way).

When the same subjects are measured twice (before/after). Test the mean of the differences d: H₀: μ_d = 0.

A test can only fail to find evidence against H₀ — absence of evidence isn't proof. Say 'do not reject H₀'.

In the in-context conclusion — naming the real quantities (bottles, runners…), not just 'reject H₀'.

Rejecting H₀ when H₀ is actually true (a false alarm — concluding there's an effect when there isn't).

Failing to reject H₀ when H₀ is actually false (a miss — concluding there's no effect when there is one).

It equals the significance level α (e.g. 0.05). Computed from the H₀ distribution / critical region.

β = P(not rejecting H₀ | the alternative is true). You need a SPECIFIC alternative value, and you use that distribution.

The set of extreme outcomes for which you reject H₀. α = P(landing in it when H₀ is true).

α comes from the H₀ distribution; β comes from the alternative (H₁) distribution.

Lowering α shrinks the critical region, which makes a Type II error more likely (β rises), and vice versa.

P(X ≥ 15 | p = 0.5) = 1 − P(X ≤ 14) ≈ 0.0207.

The probability of moving FROM state j INTO state i in one step.

Everyone who starts in that state (column) must end up in some state, so the probabilities over all destinations total 1.

Multiply the transition matrix by the current state vector: s₁ = T s₀.

sₙ = Tⁿ s₀ — raise T to the power n, then multiply by the start vector.

Column A is 0.7 in the A-row and 0.3 in the B-row (column = where you start, row = where you end).

Store T and s₀, then compute T^n and the product T^n × s₀.

No — the chance of A→B need not equal B→A, so it is generally not symmetric.

Interpret the numbers IN CONTEXT — say which group/café/patch they describe and round sensibly.

As x gets closer to a (from both sides), f(x) gets closer and closer to L. The limit does not depend on f(a).

Direct substitution: replace x with a. E.g. lim_(x → 3)(2x+1) = 2(3)+1 = 7.

Factor and cancel the common factor, then substitute. E.g. (x^2-4)/(x-2) = x+2, so the limit at x=2 is 4.

lim_(x → a^-): approach from the LEFT (values below a). lim_(x → a^+): approach from the RIGHT (values above a).

Only when both one-sided limits exist AND are equal: lim_(x → a^-) f(x) = lim_(x → a^+) f(x).

YES. The limit only depends on values near a, not AT a. Example: (x^2-4)/(x-2) is undefined at x=2 but the limit is 4.

lim_(x → 5) f(x) = 8. Read from the table: both sides converge to the same value.