IB DP Mathematics: Analysis and Approaches – Straight lines, functions, inverses and quadratics 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.

This practice set covers straight lines, functions, inverses and quadratics in IB Diploma Programme Mathematics: Analysis and Approaches. It is aligned to the IB Mathematics: analysis and approaches guide, first assessment 2021, syllabus sections 2.1–2.7, which are common content for SL and HL. It follows the IB guide for first assessment 2021, which remains the examined syllabus until the new course is first assessed in May 2029, so it applies to the May and November 2026, 2027 and 2028 sessions.

Paper 1 allows no technology; Paper 2 requires it. Questions are labelled “(calculator-free)” or “(calculator allowed)” to match. Give answers exactly or to 3 significant figures unless a question says otherwise.

Learn the methods in the study guide and the revision notes. The Functions strand overview has a separate short set. The IB DP Maths AA course hub and the printable syllabus checklist show where this unit sits.

Questions

1. (calculator-free) Find the equation of the line through P(1, −4) that is perpendicular to 3x − 6y + 5 = 0. Give your answer in the form ax + by + d = 0, where a, b, d ∈ ℤ. [4]

2. (calculator-free) f(x) = 5 − √(8 − 2x). Write down the largest possible domain of f and the range of f. [2]

3. (calculator-free) f(x) = 3x + 1 and g(x) = x² − 2.

(a) Find (f ∘ g)(x). [2] (b) Solve (g ∘ f)(x) = 14. [3]

4. (calculator-free) h(x) = (x + 2)² − 1, x ≥ −2.

(a) Find h⁻¹(x) and state its domain. [4] (b) Hence solve h(x) = 24. [1]

5. (calculator-free) f(x) = −x² + 6x − 5.

(a) Write f(x) in the form a(x − h)² + k. [2] (b) Write down the coordinates of the vertex and the equation of the axis of symmetry. [2] (c) Write f(x) in the form a(x − p)(x − q) and write down the x-intercepts. [2] (d) Sketch the graph of y = f(x), labelling the vertex and all axis intercepts. [2]

6. (calculator-free) Solve 2x² + 3x − 9 ≤ 0. [3]

7. (calculator-free) Find the set of values of k for which (k + 1)x² − 4x + (k − 2) = 0 has two distinct real roots. [5]

8. (calculator allowed) f(x) = x³ − 5x + 2.

(a) Find the zeros of f. [2] (b) Find the coordinates of the local maximum and local minimum points. [2] (c) Find the coordinates of the points where y = f(x) meets y = x + 2. [2]

9. (calculator allowed) A temperature of x °C is x × 1.8 + 32 in °F, so F(x) = 1.8x + 32.

(a) Find F(−12). [1] (b) Find F⁻¹(x). [2] (c) Find F⁻¹(98.6) and say what it means. [1] (d) Find the temperature that has the same value on both scales. [2]

10. (calculator allowed) A ball is thrown upwards. Its height in metres after t seconds is h(t) = −4.9t² + 14t + 1.5, until it lands.

(a) Write down the height from which the ball is thrown. [1] (b) Find the maximum height and the time at which it occurs. [3] (c) Find the time at which the ball hits the ground. [2] (d) Find the length of time for which the ball is more than 8 m above the ground. [3]

11. (calculator-free) The curve C has equation y = x² + 3. The line L has equation y = mx − 1.

(a) Show that the x-coordinates of any intersection points satisfy x² − mx + 4 = 0. [2] (b) Find the values of m for which L is a tangent to C. [3] (c) For the positive value of m, find the point of contact. [2] (d) Find the equation of the line through this point perpendicular to L, in the form ax + by + d = 0. [3] (e) Find the values of m for which L meets C at two distinct points. [2]

12. (calculator-free) The graph of a quadratic function f crosses the x-axis at (−1, 0) and (4, 0) and passes through (2, −12).

(a) Find f(x) in the form ax² + bx + c. [3] (b) Find the coordinates of the vertex. [2]

Answers

1. 6y = 3x + 5, so the gradient of the given line is 1/2 [1]. Perpendicular gradient = −2 [1]. y + 4 = −2(x − 1) [1], so 2x + y + 2 = 0 [1]. [4] Examiner insight: the last mark is for the requested form; y = −2x − 2 is correct but loses the accuracy mark when integer ax + by + d = 0 is asked for.

2. Domain: 8 − 2x ≥ 0, so x ≤ 4 [1]. Since √(8 − 2x) ≥ 0, range f(x) ≤ 5 [1]. [2] Examiner insight: write the range in terms of f(x) (or y), not x; “x ≤ 5” for a range is marked wrong.

3. (a) f(x² − 2) = 3(x² − 2) + 1 [1] = 3x² − 5 [1]. (b) (3x + 1)² − 2 = 14 [1], so 3x + 1 = ±4 [1], giving x = 1 or x = −5/3 [1]. [5] Examiner insight: dropping the negative square root loses the final accuracy mark; “±” must appear in your working.

4. (a) y = (x + 2)² − 1, so x + 2 = √(y + 1) [1]; positive root since x ≥ −2 [1]. h⁻¹(x) = −2 + √(x + 1) [1], domain x ≥ −1 (the range of h) [1]. (b) x = h⁻¹(24) = −2 + √25 = 3 [1]. [5] Examiner insight: leaving “±√” in h⁻¹(x) means it is not a function, and the accuracy mark is lost even if the algebra is otherwise right.

5. (a) −(x² − 6x) − 5 = −[(x − 3)² − 9] − 5 [1] = −(x − 3)² + 4 [1]. (b) Vertex (3, 4) [1]; axis x = 3 [1]. (c) −(x − 1)(x − 5) [1]; x-intercepts (1, 0) and (5, 0) [1]. (d) Concave-down parabola with the correct shape [1]; vertex (3, 4), x-intercepts (1, 0) and (5, 0), y-intercept (0, −5) all labelled [1]. [8] Examiner insight: the axis of symmetry must be given as an equation, x = 3; “3” alone does not earn the mark.

6. (2x − 3)(x + 3) ≤ 0 [1]. Critical values x = −3 and x = 3/2 [1]. The parabola opens upwards, so it is below the axis between the roots: −3 ≤ x ≤ 3/2 [1]. [3] Examiner insight: using < instead of ≤ at the end points loses the final mark even when the critical values are right.

7. Δ = 16 − 4(k + 1)(k − 2) [1] = −4k² + 4k + 24 = −4(k − 3)(k + 2) [1]. Two distinct roots need Δ > 0 [1], so −2 < k < 3 [1]. When k = −1 the equation is linear (−4x − 3 = 0), so −2 < k < 3, k ≠ −1 [1]. [5] Examiner insight: the final accuracy mark needs both the interval and k ≠ −1; −2 < k < 3 on its own is incomplete.

8. (a) From the GDC, x = −2.41, 0.414 [1] and x = 2 [1]. (b) Local maximum (−1.29, 6.30) [1]; local minimum (1.29, −2.30) [1]. (c) (−2.45, −0.449), (0, 2) [1] and (2.45, 4.45) [1]. [6] Examiner insight: GDC answers must be given to 3 significant figures; writing −2.4 or 6.3 loses the accuracy mark. Exact zeros (2 and −1 ± √2) are also accepted.

9. (a) F(−12) = 10.4 °F [1]. (b) y = 1.8x + 32 gives x = (y − 32)/1.8 [1], so F⁻¹(x) = (x − 32)/1.8 [1]. (c) F⁻¹(98.6) = 37: 98.6 °F is 37 °C [1]. (d) 1.8x + 32 = x [1], so x = −40 (−40 °C = −40 °F) [1]. [6] Examiner insight: in (c) the mark needs both the value and its meaning in context; a bare 37 does not score.

10. (a) h(0) = 1.5 m [1]. (b) Vertex at t = −14/(2 × −4.9) (or GDC maximum) [1]: t = 1.43 s [1], height 11.5 m [1]. (c) Solve h(t) = 0 [1]: t = 2.96 s (reject the negative root) [1]. (d) Solve −4.9t² + 14t + 1.5 = 8 [1]: t = 0.583 and t = 2.27 [1]. Time above 8 m = 1.69 s [1]. [9] Examiner insight: in (d) subtract unrounded times; 2.27 − 0.583 gives 1.687, which rounds correctly here, but early rounding can shift the third figure and cost the final mark.

11. (a) x² + 3 = mx − 1 [1], so x² − mx + 4 = 0 [1]. (b) Tangent when Δ = 0 [1]: m² − 16 = 0 [1], so m = 4 or m = −4 [1]. (c) x² − 4x + 4 = 0, (x − 2)² = 0, x = 2 [1]; point (2, 7) [1]. (d) Gradient −1/4 [1]; y − 7 = −(1/4)(x − 2) [1]; x + 4y − 30 = 0 [1]. (e) Δ > 0: m² > 16 [1], so m < −4 or m > 4 [1]. [12] Examiner insight: “show that” in (a) needs the equating step written out; stating only the final equation earns nothing.

12. (a) f(x) = a(x + 1)(x − 4) [1]. f(2) = a(3)(−2) = −12, so a = 2 [1]. f(x) = 2x² − 6x − 8 [1]. (b) Axis x = (−1 + 4)/2 = 3/2 [1]; f(3/2) = 2(5/2)(−5/2) = −25/2, vertex (3/2, −25/2) [1]. [5] Examiner insight: the method mark in (a) is for using the factorised form with the given roots; you can still earn it, and follow-through in (b), if a is wrong.

Where marks are usually lost

• Leaving a line in y = mx + c when ax + by + d = 0 with integers is asked for.
• Taking the gradient of 3x − 6y + 5 = 0 as 3 or −6 instead of rearranging.
• Writing a range in terms of x, or with the wrong inequality direction.
• Composing in the wrong order, or losing the negative root when solving (3x + 1)² = 16.
• Keeping ± in an inverse on a restricted domain, or not stating the domain of the inverse.
• Giving an axis of symmetry as a number instead of an equation.
• Missing the exclusion of the value of k that makes the x² coefficient zero.
• Rounding GDC values to 2 significant figures, or rounding early before subtracting.
• Not interpreting a value in context when the question asks what it means.
• Omitting the equating step in a “show that”.

Next steps

• Revision notes for this unit
• Study guide for this unit
• IB DP Maths AA course hub
• Printable syllabus checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

International Baccalaureate Organization, Diploma Programme, Mathematics: analysis and approaches guide, first assessment 2021 (published February 2019, updated November 2020), syllabus sections SL 2.1–2.7.