IB Math AI SL SL — 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.

4 700 000. Move the decimal 6 places right.

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

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

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

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.

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

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

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.

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

3.7 × 10⁹

(a) 6.4 × 10⁻³ (b) 0.0064

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

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 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.

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

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

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

Real data can be close to arithmetic without being exact.

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

Yes, because the common difference is 60.

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

A sequence is geometric if you multiply by the same number each step. That fixed multiplier is the common ratio r.

Divide any term by the term before it: r = uₙ₊₁ ÷ uₙ.

uₙ = u₁ · rⁿ⁻¹

It shrinks. When 0 < r < 1, each term is a fraction of the one before it.

The signs alternate. For example, r = −2 gives 4, −8, 16, −32, ...

u₁ = 3 and r = 2, because each term is multiplied by 2.

n is the position number of the term. uₙ is the value of that term.

Match the exponents: n − 1 = 7, so n = 8.

A sequence is the list of terms. A series is what you get when you add those terms together.

Sₙ = a(1 − rⁿ) ÷ (1 − r), for r ≠ 1.

a is the first term of the geometric sequence.

Use the sum formula when the question wants the total of several terms, not just one term.

They find only one term instead of adding the terms. If the question asks for the total, use Sₙ.

a = 5 and r = 2.

It adds repeated growth amounts together, so it is useful when the question wants a running total, not just the latest value.

No. The denominator becomes 0. If r = 1, every term is the same, so Sₙ = n × a.

Look for the same percentage change each period. Constant percentage change means geometric.

r = 1 + p/100

r = 1 − p/100

r = 1 − 15/100 = 0.85

The number of periods that have passed. It is the number of times you multiply by r.

Round up. You need the first whole period where the threshold has actually been crossed.

Because the amount added is fixed each year. It is arithmetic, not geometric.

Give the value with sensible rounding, units, and a short sentence in context.

|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.

FV = PV × (1 + r/(100k))^(kn)

Enter r as 5. The formula already divides by 100.

k is the number of compounding periods per year. For example: 1 yearly, 4 quarterly, 12 monthly.

kn is the total number of compounding periods.

Because the money is leaving your pocket when you invest it. TVM uses cash-flow signs.

Yearly: 1. Quarterly: 4. Monthly: 12.

9 full years. Round up because 8 full years is not enough.

Simple interest adds the same amount each period, so it is arithmetic. Compound interest multiplies by the same factor each period, so it is geometric.

1.07. Keep 100% and add 7%.

0.88. Keep 88% of the value each period.

A = P(1 + r/100)^n

A = P(1 - r/100)^n

Because 0.10 is the amount lost, not the amount kept. The correct multiplier is 0.90.

V = 400(0.95)^n

A sensible rounded value and a short sentence in context.

The number of percentage-change periods.

The advertised annual percentage rate before compounding frequency is taken into account.

k is the number of compounding periods per year.

k = 12.

The total number of compounding periods.

Monthly compounding, because interest is added more often.

1% per month.

Because different compounding frequencies create different effective yearly growth.

Write a decision sentence explaining which option is better and why.

It means better for the criterion in the question, such as a larger final balance or lower total cost.

No. The starting amount and compounding frequency can change the final result.

The final values that answer the question, not just the deposits or rates separately.

It gives no mathematical reason.

The second one, because it gives a numerical contextual reason.

Option B is better because it gives the larger final balance.

A comparison and a clear decision sentence.

Because unsupported claims like “better” or “more” usually do not earn full marks.

The total number of periods.

N = 60.

Because it is money leaving your pocket at the start.

N.

C/Y = 12, and often P/Y = 12 if there are monthly periods.

The final value should usually be larger than the starting value if the rate is positive.

Because N must count total periods, so it should be 72.

When the arithmetic is messy or the question asks for an unknown like N, PV, or I%.

a^m × a^n = a^(m+n).

a^m ÷ a^n = a^(m-n).

a^(-n) = 1/a^n. It means reciprocal, not negative answer.

x^6. Multiply the powers: 3 × 2 = 6.

It means a^c = b. A logarithm gives the exponent needed on the base.

3, because 2^3 = 8.

log_10 10000 = 4.

It usually means log base 10.

log_a(xy) = log_a x + log_a y.

log_a(x/y) = log_a x - log_a y.

log_a(x^n) = n log_a x.

Because the product law works for multiplication, not addition.

Rewrite 16 as 2^4, then set exponents equal: x + 1 = 4, so x = 3.

Use logs: x = log(20)/log(3).

x = 32, because x = 2^5.

The input must be positive: x > 0.

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").

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.

A composite function applies one function to the output of another. f(g(x)): first apply g to x, then apply f to the result. Notation: (f ∘ g)(x) = f(g(x)) — read "f of g of x."

(f ∘ g)(x) = f(g(x)). g is applied first (the inner function), then f is applied to the result (the outer function). Think of it like nested brackets — work from the inside out.

Step 1: g(x) = 3x (the inner function). Step 2: f(g(x)) = f(3x) = (3x) + 2 = 3x + 2. Substitute g(x) = 3x wherever x appears in f.

Composition (f ∘ g) is not multiplication. f(g(x)) means "substitute g(x) into f" — apply one function to the output of the other. f(x) × g(x) means multiply the two outputs — a completely different operation.

Step 1: Calculate the inner function first — find g(a). Step 2: Substitute that result into f — find f(g(a)). Always work inside out: inner function first, outer function second.

Step 1: g(3) = 3² = 9. Step 2: f(g(3)) = f(9) = 2(9) + 1 = 19.

Step 1: f(5) = 5 − 4 = 1. Step 2: g(f(5)) = g(1) = 3(1) + 2 = 5. Note: this asks for g(f(5)), so f is applied first, then g.

They applied the functions in the wrong order. For f(g(4)): compute the inner function g(4) first, then substitute into f. The function written on the right (inside the bracket) is always applied first.

Step 1: Write out g(x). Step 2: Substitute g(x) into f — replace every x in f(x) with the expression g(x). Step 3: Simplify if possible.

g(x) = x². f(g(x)) = f(x²) = 2(x²) + 3 = 2x² + 3.

f(x) = x − 1. g(f(x)) = g(x − 1) = 3(x − 1) = 3x − 3.

They applied the wrong rule. f(x) = (x + 1)² means: take the input, add 1, then square. f(g(x)) = (g(x) + 1)² — substitute g(x) for x throughout. Always replace every x in f with the full expression g(x), including inside brackets.

No — in general f(g(x)) ≠ g(f(x)). Counterexample: f(x) = x + 1, g(x) = x². f(g(x)) = x² + 1. g(f(x)) = (x + 1)² = x² + 2x + 1. These are different.

f(g(2)): g(2) = 5, then f(5) = 25. g(f(2)): f(2) = 4, then g(4) = 7. f(g(2)) = 25 ≠ g(f(2)) = 7. The order of composition matters.

f and g are inverse functions of each other: g = f⁻¹ (and f = g⁻¹). Each function "undoes" the other. Example: f(x) = 2x + 1 and g(x) = (x − 1)/2 satisfy f(g(x)) = x and g(f(x)) = x.

Read carefully: g(f(x)) means "f is inside g" — apply f first, then g. Memory check: the function closest to x (written on the right) is always applied first. In g(f(x)): f is closer to x → f goes first → then g.

f⁻¹ undoes the effect of f — it reverses the mapping. If f maps a → b, then f⁻¹ maps b → a. Together: f⁻¹(f(x)) = x and f(f⁻¹(x)) = x.

f(f⁻¹(x)) = x (applying f after f⁻¹ gives back x). f⁻¹(f(x)) = x (applying f⁻¹ after f gives back x). Both compositions return the original input — they cancel each other out.

f⁻¹ reverses the mapping: f⁻¹(8) = 3 and f⁻¹(12) = 5. No formula needed — just swap the input and output of f.

f⁻¹(x) is the inverse function, not the reciprocal. 1/f(x) means "1 divided by the output of f" — a completely different thing. The −1 in f⁻¹ is function notation for "inverse," not an exponent.

1. Write y = f(x). 2. Swap x and y (write x = f(y)). 3. Rearrange to make y the subject. 4. Replace y with f⁻¹(x).

y = 4x − 7. Swap: x = 4y − 7. Rearrange: x + 7 = 4y → y = (x + 7)/4. f⁻¹(x) = (x + 7)/4.

y = (2x + 1)/3. Swap: x = (2y + 1)/3. Rearrange: 3x = 2y + 1 → 2y = 3x − 1 → y = (3x − 1)/2. f⁻¹(x) = (3x − 1)/2.

They will get x = (expression in y), not y = (expression in x). The swap is essential — it converts the input-output relationship. Without swapping, the result is not expressed as f⁻¹(x).

An inverse only exists if f is one-to-one (each output comes from exactly one input). Example: f(x) = x² over all ℝ is not one-to-one — f(3) = f(−3) = 9, so the inverse would give two outputs. Restricting to x ≥ 0 makes it one-to-one: f⁻¹(x) = √x.

y = x². Swap: x = y². Rearrange: y = √x (take positive root since original domain x ≥ 0). f⁻¹(x) = √x, domain x ≥ 0.

The domain of f⁻¹ equals the range of f. The range of f⁻¹ equals the domain of f. The inverse swaps domain and range — inputs become outputs and vice versa.

Without a domain restriction, f(x) = x² is not one-to-one — the inverse is not unique. The full answer must be: f⁻¹(x) = √x for x ≥ 0. IB questions typically award a separate mark for correctly stating the domain of f⁻¹.

The graph of f⁻¹ is the reflection of the graph of f in the line y = x. Every point (a, b) on f maps to the point (b, a) on f⁻¹ — x and y coordinates are swapped.

(7, 2) and (4, −1). The inverse swaps x and y — every (a, b) on f becomes (b, a) on f⁻¹.

At any intersection point, f(x) = f⁻¹(x). These points also lie on the line y = x (since they satisfy f(x) = x at the intersection in the most common case). Note: f and f⁻¹ can intersect off the line y = x too, but they always cross y = x when they intersect.

The correct reflection is over the line y = x, not the x-axis. Reflecting over the x-axis would flip the graph vertically — that gives −f(x), not f⁻¹(x). The line y = x is the mirror that swaps x and y coordinates.

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."

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.

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.

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.

Entire group studied

Subset selected for study

Equal probability selection

Saves time, cost, resources

Non-random selection excludes members

Population: 2000 students. Sample: 200 random

Sample accurately reflects population

Expects random selection to minimize bias

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.

Factor: x^2 - 16 = (x-4)(x+4). Cancel (x-4). Substitute x=4: 4+4 = 8. The limit is 8.

f is increasing if f'(x) > 0 for all x in that interval. As x gets bigger, f(x) gets bigger — the graph goes UP.

f is decreasing if f'(x) < 0 for all x in that interval. As x gets bigger, f(x) gets smaller — the graph goes DOWN.

1) Find f'(x). 2) Solve f'(x) = 0 — these are the critical x-values. 3) Test a value in each interval: if f'(x) > 0, increasing; if f'(x) < 0, decreasing.

f'(x) = 2x − 4. Critical point: x = 2. For x < 2: f'(x) < 0 → DECREASING. For x > 2: f'(x) > 0 → INCREASING.

It marks the boundary between increasing and decreasing. At that point, the function is momentarily flat — it is a critical (stationary) point.

Draw a number line. Mark critical x-values. Pick one test x in each interval, evaluate f'(x). Label each interval + (increasing) or − (decreasing).

f'(x) = −2x + 6. Substitute x = 2: f'(2) = 2 > 0. Yes — f is increasing at x = 2.

The function is increasing for all x. It never turns around. Example: f(x) = x³ has f'(x) = 3x² ≥ 0 but is still overall increasing.

f′(x) is the gradient function. It gives the gradient of the curve y = f(x) at any x-value. Substitute a number into f′(x) to get the gradient at that point.

dy/dx is "the derivative of y with respect to x". It is exactly the same thing as f′(x). Both notations appear in IB papers.

f′(x) > 0 when the curve is increasing (rising left to right). f′(x) < 0 when decreasing. f′(x) = 0 at a local maximum or minimum.

f′(3) = 0. At any local maximum (or minimum), the tangent is horizontal, so the gradient is zero.

A straight line has the same gradient everywhere. For y = mx + c, the gradient is always m. Only curves have a different gradient at each point.

The volume is decreasing at a rate of 5 litres per unit time. The negative sign means the function is falling. Always include units in your interpretation.

f(a) is the y-value (height) of the curve at x = a.\nf′(a) is the gradient (steepness) of the curve at x = a.\nThey are completely different quantities.

Yes. f(x) and f′(x) are independent. A curve can be at any height while being momentarily flat — for example, at the top of a hill.

d/dx[axⁿ] = naxⁿ⁻¹. Multiply the coefficient by the power, then reduce the power by one.

f′(x) = 20x³. (Multiply 5 by 4 = 20, reduce power from 4 to 3.)

0. The derivative of any constant is zero.

−7. The derivative of ax is a. Here a = −7.

f′(x) = 9x² − 4x + 1. Apply the power rule to each term. The constant −9 disappears. The linear x term gives 1.

Expand: y = 4x² − x. Then differentiate: dy/dx = 8x − 1. You cannot apply the power rule inside a product without expanding.

dy/dx = 6x² − 1. At x = 2: 6(4) − 1 = 23.

f(3) = 9 is the y-value of the curve at x = 3. f′(3) = 6 is the gradient of the curve at x = 3. Different quantities with different meanings.

y − y₁ = m(x − x₁), where m is the gradient and (x₁, y₁) is the point of tangency.

1. Differentiate f(x) to get f′(x).\n2. Substitute x₁ into f′(x) to get the gradient m.\n3. Write y − y₁ = m(x − x₁) and simplify.

dy/dx = 2x. At x = 3: m = 6.

dy/dx = 2x → m = 4. y₁ = 5. Tangent: y − 5 = 4(x − 2) → y = 4x − 3.

Because f(x) gives y-values (heights). f′(x) gives gradients. You need the y-coordinate of the point of tangency — that comes from the original function.

Set f′(x) = given gradient and solve for x. There may be one or two solutions. Find y at each solution using f(x).

f′(x) = 3x². m = 3. f(−1) = −1 → point (−1, −1). Tangent: y + 1 = 3(x + 1) → y = 3x + 2.

The tangent is the best linear approximation to the curve at that point. It has exactly the same gradient as the curve at that point — but the curve will curve away from the tangent for x-values further away.

m_tangent × m_normal = −1, so m_normal = −1/m_tangent. The normal is perpendicular to the tangent.

m_n = −1/5.

m_n = −1/(−3) = 1/3. Two negatives cancel.

dy/dx = 2x − 2. m_t = 4. m_n = −1/4.

dy/dx = 2x → m_t = 6 → m_n = −1/6. Normal: y − 9 = −(1/6)(x − 3) → y = −(1/6)x + 19/2.

The normal is vertical: a line of the form x = x₁. You cannot divide −1 by zero.

True. Both lines pass through the point of tangency (x₁, y₁). They differ only in their gradients.

Using the tangent gradient (from f′) directly as the normal gradient, without applying m_n = −1/m_t. Always take the negative reciprocal.

"Integrate with respect to x." The integral symbol ∫ paired with dx means find the antiderivative — the reverse of differentiation.

∫xⁿ dx = xⁿ⁺¹/(n+1) + C, provided n ≠ −1. Add 1 to the power, divide by the new power, add C.

Because constants disappear when you differentiate. Infinitely many functions have the same derivative — +C represents all of them.

x⁴ − 3x² + 2x + C. Integrate each term: 4·x⁴/4 = x⁴, 6·x²/2 = 3x², 2·x = 2x.

Expand the brackets first: x(x+3) = x² + 3x. Then integrate: x³/3 + 3x²/2 + C.

(2/3)x^(3/2) + C. Add 1: 1/2 + 1 = 3/2. Divide by 3/2: divide by 3/2 = multiply by 2/3.

Differentiate your answer. If you get back the original integrand, your integral is correct.

Rewrite: x²/x − 3/x = x − 3x⁻¹. Integrate: x²/2 − 3ln|x| + C.

An integral with limits [a, b] that gives a specific number — the signed area between the curve and the x-axis from x = a to x = b.

∫[a to b] f(x) dx = F(b) − F(a), where F is any antiderivative of f.

F(x) = x². F(3) − F(1) = 9 − 1 = 8.

A negative number. The integral gives signed area — negative when the curve is below the x-axis. For total area, take the absolute value.

1) Find intersections: solve f(x) = g(x) to get limits a and b. 2) Identify the top function. 3) Integrate [f(x) − g(x)] from a to b.

∫[0 to 2] (x²+1) dx = [x³/3 + x] from 0 to 2 = (8/3 + 2) − 0 = 14/3 ≈ 4.67 square units.

Use your GDC. But always write the integral notation first (e.g., ∫[a to b] f(x) dx = ...). Marks are given for the setup, not just the answer.

∫[0 to 1] (x − x²) dx = [x²/2 − x³/3] from 0 to 1 = 1/2 − 1/3 = 1/6 square units.

A specific point (x₀, y₀) that the function passes through. Used to find the exact value of the constant C.

Step 1: Integrate → f(x) = 2x² − x + C. Step 2: f(2) = 8 − 2 + C = 5 → C = −1. Answer: f(x) = 2x² − x − 1.

Integrate: f(x) = 3x² + C. Substitute (0, 4): 3(0) + C = 4 → C = 4. So f(x) = 3x² + 4.

Integrate: s(t) = t³ + C. Use s(0) = 5: C = 5. So s(t) = t³ + 5.

Two initial conditions — one for each integration, since each introduces a new constant (C₁ and C₂).

v(t) = 10t + 3 (use v(0)=3 → C₁=3). s(t) = 5t² + 3t + C₂. Use s(0)=1 → C₂=1. s(t) = 5t² + 3t + 1.

General solution: f(x) + C (all possible solutions). Particular solution: the specific function once C is found using an initial condition.

Integrate: y = x³ + x² + C. Use (1, 10): 1 + 1 + C = 10 → C = 8. So y = x³ + x² + 8.

A point where f'(x) = 0. The tangent is horizontal — the function momentarily stops increasing or decreasing.

A stationary point where the function changes from INCREASING to DECREASING. f'(x) goes from + to −. The point is the highest nearby.

A stationary point where the function changes from DECREASING to INCREASING. f'(x) goes from − to +. The point is the lowest nearby.

1) Find f'(x). 2) Solve f'(x) = 0. 3) Use a sign diagram: if + then − → local max; if − then + → local min. 4) Find the y-value using f(x).

f'(x) = 3x² − 3 = 3(x−1)(x+1). Critical points: x = 1 and x = −1. Sign: +,−,+ → x=−1 local max, x=1 local min. y values: f(−1)=2, f(1)=−2.

An inflection point is where concavity changes. It is only a stationary point if f'(x) = 0 there too (a "saddle point" like x=0 on y=x³).

f'(x) = 6x² − 6x = 6x(x−1). Critical x: 0 and 1. Sign diagram: + before x=0, − between 0 and 1. So x=0 is local max. f(0) = 0.

At a critical point where f'(x)=0: if f''(x) < 0 → local max; if f''(x) > 0 → local min; if f''(x) = 0 → inconclusive, use sign diagram.

Finding the maximum or minimum value of a quantity. You use derivatives to locate stationary points, then determine if it is a max or min.

1) Write an expression for the quantity to optimise. 2) Express it in terms of ONE variable (use a constraint). 3) Differentiate and set f'(x) = 0. 4) Solve and classify (max or min). 5) State the answer with units.

Use a sign diagram of f'(x), OR check the endpoints. In closed-interval problems, also evaluate f at the endpoints.

Let width = x. Then length = 80 − 2x. Area A = x(80−2x) = 80x − 2x². A' = 80 − 4x = 0 → x = 20. A = 20 × 40 = 800 m².

A rule that links two or more variables. You use it to eliminate one variable so you can write everything in terms of one unknown.

R'(x) = 40 − 2x = 0 → x = 20. R'(20) = −2 < 0 → local max. Max revenue = R(20) = 40(20)−400 = 400.

1) The optimal VALUE of x. 2) The optimal value of the quantity (max area, min cost, etc.). 3) Confirmation it is a max or min (sign diagram or second derivative). 4) Units if the problem has them.

C' = 4x − 12 = 0 → x = 3. C'(3) = 4 > 0 → local min. Min cost = 2(9) − 12(3) + 20 = 18 − 36 + 20 = 2.

A ≈ (h/2)(y₀ + 2y₁ + 2y₂ + ... + 2yₙ₋₁ + yₙ), where h = (b − a)/n and yᵢ = f(a + i·h).

h is the step width — the horizontal width of each trapezoid strip. h = (b − a) / n.

Because each interior vertical line is shared by two adjacent trapezoids — it counts as a side of both.

h = 1. y₀ = 0, y₁ = 1, y₂ = 4. A ≈ (1/2)(0 + 2×1 + 4) = 0.5 × 6 = 3. (Exact = 8/3 ≈ 2.67)

Overestimate. The trapezoids sit above the curve, so the total estimated area is larger than the actual area.

Underestimate. The trapezoids fall below the curve, so the estimated area is smaller than the actual area.

1. Calculate h = (b−a)/n. 2. List all x-values: a, a+h, a+2h, ..., b. 3. Calculate yᵢ = f(xᵢ) for each. 4. Apply: A ≈ (h/2)(y₀ + 2y₁ + ... + yₙ).

When the function is linear (a straight line). Trapezoids perfectly fit straight-line sections with no gap or overlap.