IB MYP Mathematics: Functions and models: linear, quadratic, exponential and rational (MYP Mathematics) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – the IB holds copyright in its own papers. Use these alongside the official past papers available through your school or the IB store.

These questions practise functions and models from the thinking with models framework of the IB Middle Years Programme Mathematics guide (first assessment 2022). Questions 1 to 8 use standard content. Questions 9 to 11 are marked (Extended) and question 12 is marked (Enrichment), following the framework’s own labels. The mark values are for self-marking only. Read the study guide first, or the revision notes for a quick refresher. The course page is MYP Mathematics and the checklist tracks coverage. Show every step, as the criteria reward clear reasoning.

Questions

1. The function f is defined by f(x) = 4 - 3x.

(a) Find f(-2). [1] (b) Solve f(x) = -11. [2]

2. The mapping x → x² is applied to the inputs -2, -1, 0, 1, 2.

(a) Explain why this is a function. [1] (b) Is it one-to-one or many-to-one? Give a reason. [1] (c) List the set of outputs. [1]

3. Line A passes through (-1, 2) and (3, 10).

(a) Find the equation of line A in the form y = mx + c. [3] (b) Line B is perpendicular to line A and passes through (4, 1). Find its equation. [3] (c) Find the coordinates where the two lines meet. [2]

4. (a) Solve the simultaneous equations 3x - y = 7 and x + 2y = 7. [3] (b) Plan A costs 8 dollars plus 0.10 dollars per minute of calls. Plan B costs 20 dollars plus 0.04 dollars per minute. Write each cost as a function of m minutes, state what the intersection of their graphs means, and find it. [4]

5. The function f is given by f(x) = x² - 4x - 12.

(a) Factorise f(x) and state the x-intercepts. [2] (b) Find the coordinates of the vertex. [2] (c) Write f(x) in the form (x - h)² + k. [2]

6. A ball is kicked so that its height in metres after t seconds is h = -5t² + 15t.

(a) Find when the ball returns to the ground. [2] (b) Find the greatest height reached and when. [3] (c) Give one limitation of this model. [1]

7. The area of algae on a pond is modelled by A = 40 × 1.5ᵈ square metres, where d is the number of days.

(a) State the area at the start. [1] (b) Find the area after 4 days. [2] (c) Find the first whole day on which the area exceeds 1000 square metres. [2] (d) Give a reason why the model may stop being valid. [1]

8. Consider g(x) = 2ˣ - 4.

(a) Write down the equation of the horizontal asymptote. [1] (b) Find the y-intercept. [1] (c) Find the x-intercept. [2]

9. (Extended) Let f(x) = (3x - 2)/(x + 4).

(a) State the equation of each asymptote of the graph. [2] (b) Find the x-intercept and the y-intercept. [2] (c) State the domain and the range. [2]

10. (Extended) The function g is given by g(x) = -3(x + 2)² + 5.

(a) Describe the transformations that take the graph of y = x² to the graph of g, in order. [3] (b) Expand g(x) in the form ax² + bx + c. [2] (c) State the range of g. [1]

11. (Extended)

(a) A cubic is given by f(x) = (x + 1)(x - 2)(x - 4). Find its x-intercepts and y-intercept, and describe what happens to f(x) as x becomes very large and positive, and very large and negative. [4] (b) For y = log₃ x, state the equation of the vertical asymptote and the coordinates of the points where x = 1 and x = 3. [2]

12. (Enrichment) Let f(x) = 3x - 1 and g(x) = x².

(a) Find g∘f(2) and f∘g(2). [2] (b) Find the inverse function f⁻¹(x). [2] (c) Solve 2ˣ = 50, giving your answer to 3 significant figures. [2]

Answers

1. (a) f(-2) = 4 - 3(-2) = 10 [1] (b) 4 - 3x = -11 gives -3x = -15 [1], so x = 5 [1]. [3] Examiner insight: Substituting a negative number is the usual slip, so write the bracket first. Checking your answer back in f(x) takes seconds and catches sign errors.

2. (a) Each input gives exactly one output. [1] (b) Many-to-one, because -2 and 2 both map to 4 (and -1 and 1 both map to 1). [1] (c) {0, 1, 4} [1] [3] Examiner insight: A reason should point to specific inputs. List a repeated output only once when asked for the set of outputs.

3. (a) m = (10 - 2)/(3 - (-1)) = 8/4 = 2 [1]. Using (-1, 2): 2 = -2 + c [1], so y = 2x + 4 [1]. (b) Perpendicular gradient = -1/2 [1]. Using (4, 1): 1 = -2 + c [1], so y = -x/2 + 3 [1]. (c) 2x + 4 = -x/2 + 3 gives 2.5x = -1, so x = -0.4 [1]; y = 2(-0.4) + 4 = 3.2, so the lines meet at (-0.4, 3.2) [1]. [8] Examiner insight: State the gradient before finding c so your method is visible. Give both coordinates in part (c), not just x.

4. (a) From the first equation y = 3x - 7 [1]. Then x + 2(3x - 7) = 7 gives 7x = 21 [1], so x = 3, y = 2 [1]. (b) C_A = 8 + 0.10m and C_B = 20 + 0.04m [1]. The intersection is the number of minutes at which both plans cost the same [1]. 8 + 0.1m = 20 + 0.04m gives 0.06m = 12, so m = 200 [1]. Cost = 8 + 20 = 28 dollars at 200 minutes [1]. [7] Examiner insight: Both coordinates matter in a modelling question: the minutes and the cost. Say what the intersection means in words, not just where it is.

5. (a) (x - 6)(x + 2) [1]; x-intercepts 6 and -2 [1]. (b) The axis is halfway: x = (6 + (-2))/2 = 2 [1]; f(2) = 4 - 8 - 12 = -16, so vertex (2, -16) [1]. (c) Completing the square: x² - 4x = (x - 2)² - 4 [1], so (x - 2)² - 16 [1]. [6] Examiner insight: Part (c) can be checked against part (b): the vertex form must show the same vertex. Expand your answer to confirm it matches the original.

6. (a) Set h = 0: -5t(t - 3) = 0 [1], so t = 3 seconds [1] (t = 0 is the kick). (b) The axis is halfway between 0 and 3: t = 1.5 [1]; h = -5(2.25) + 15(1.5) [1] = 11.25 m [1]. (c) For example: it ignores air resistance, or it gives negative heights after t = 3, which have no meaning. [1] [6] Examiner insight: Quote the limitation in terms of the situation. Naming a general weakness without linking it to the ball earns less than a specific one.

7. (a) At d = 0 the area is 40 m² [1]. (b) 40 × 1.5⁴ [1] = 40 × 5.0625 = 202.5 m² [1]. (c) Day 7: 40 × 1.5⁷ = 683.4, which is below 1000 [1]. Day 8: 40 × 1.5⁸ = 1025.2, which is above, so day 8 [1]. (d) For example: the pond has a limited size, so area cannot grow without limit. [1] [6] Examiner insight: Show the two values either side of 1000 so the reader sees why day 8 is the first. A reason for part (d) should be tied to the pond, not just say the model is unrealistic.

8. (a) y = -4 [1] (b) g(0) = 1 - 4 = -3 [1] (c) 2ˣ - 4 = 0 gives 2ˣ = 4 [1], so x = 2 [1]. [4] Examiner insight: The -4 shifts the asymptote down from y = 0, so a bare y = 0 is wrong. Remember that the y-intercept is found by substituting x = 0 into the full rule, including the shift.

9. (a) x + 4 = 0 gives x = -4 [1]; the horizontal asymptote is y = 3 [1]. (b) 3x - 2 = 0 gives x-intercept 2/3 [1]; f(0) = -2/4 = -1/2 [1]. (c) Domain: x ≠ -4 [1]; range: f(x) ≠ 3 [1]. [6] Examiner insight: The horizontal asymptote is the ratio of the x coefficients, 3/1. Write asymptotes as equations of lines, not as bare numbers.

10. (a) Translate 2 units left [1], dilate vertically by factor 3 and reflect in the x-axis [1], then translate 5 units up [1]. (b) g(x) = -3(x² + 4x + 4) + 5 [1] = -3x² - 12x - 7 [1]. (c) The vertex is (-2, 5) and the graph opens downward, so g(x) ≤ 5 [1]. [6] Examiner insight: The order of transformations should match the order in which you would evaluate the rule. State the direction of each translation, not just its size.

11. (a) x-intercepts at x = -1, 2, 4 [1]; f(0) = (1)(-2)(-4) = 8 [1]. As x becomes very large and positive, f(x) becomes very large and positive [1]; as x becomes very large and negative, f(x) becomes very large and negative [1]. (b) The vertical asymptote is x = 0 [1]; the points are (1, 0) and (3, 1) [1]. [6] Examiner insight: Describe both ends of a cubic separately. For a logarithm, remember that the graph exists only for positive x.

12. (a) f(2) = 5, so g∘f(2) = g(5) = 25 [1]; g(2) = 4, so f∘g(2) = f(4) = 11 [1]. (b) y = 3x - 1 gives x = 3y - 1 after swapping [1], so f⁻¹(x) = (x + 1)/3 [1]. (c) x = log₂ 50 [1] = 5.6439, so x = 5.64 [1]. [6] Examiner insight: In g∘f the function nearest the input acts first. Round only at the end, and state the rounding you used.

Where marks are usually lost

• Substituting a negative input without brackets, so the sign is lost.
• Giving only x for an intersection when both coordinates are needed.
• Forgetting the minus sign when finding a perpendicular gradient.
• Reading the vertex of a vertex-form quadratic with the wrong sign for h.
• Using y = 0 as the asymptote of an exponential that has been shifted.
• Mixing up the vertical and horizontal asymptotes of a rational function.
• Stopping at a number in a modelling question without its unit and its meaning in the situation.
• Naming a limitation of a model without connecting it to the situation.
• Applying the functions in a composite in the wrong order.

Next steps

• Review the revision notes and the study guide.
• Go to the MYP Mathematics course page and tick items off on the checklist.
• Try all free 10-minute diagnostics to find your next weak area.
• Next, book a free trial class.

Official syllabus

Source: Mathematics guide of the Middle Years Programme, International Baccalaureate Organization; first assessment 2022. Section used: the framework for thinking with models.