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IB Maths: Functions and Graphs – Practice Paper 2 (10 Questions)
IB Maths: Functions and Graphs – Practice Paper 2 (10 Questions)
Paper 2 adds equations and curves: solving back from a function value, vertex form, the asymptote of an exponential, completing the square and restricted inverses.
Quick answer Ten items, about twenty-five minutes. Items 01 to 04 cover substitution, domain, solving f of x and vertex form; items 05 to 07 cover composites, inverses and the exponential curve; items 08 and 09 are completing the square and a restricted inverse; item 10 redoes a double restriction.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 questions |
| Suggested time | 25 minutes |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: IB Maths: Functions and Graphs – Tutor Lesson Plan |
| Level | IB SL |
02 Questions (10 items)
- FoundationGiven f(x) = 5 – 2x, find f(-1).
- FoundationState the domain of f(x) = the square root of x minus 4.
- Foundationf(x) = 3x + 1. Given f(x) = 16, find x.
- FoundationGive the vertex and the axis of symmetry of y = (x – 2) squared plus 7.
- Intermediatef(x) = x + 3 and g(x) = x squared minus 1. Find g(f(2)).
- Intermediatef(x) = (x + 5) / 2. Find the inverse and check it at 1.
- IntermediateState the horizontal asymptote and the y-intercept of y = 2 to the power x.
- ChallengeComplete the square for y = x squared plus 2x minus 3 and give the minimum value.
- Challengef(x) = 1 / (x – 2) + 3. Find the inverse and its domain.
- Error redoA student gives only x at least 2 for the domain of root x minus 2 over x minus 5. What is missing?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | f(-1) = 7 | Five minus two times minus one is five plus two, that is seven. |
| 2 | x is at least 4 | An even root may not hold a negative, so x minus four is at least zero. |
| 3 | x = 5 | Solve three x plus one equals sixteen, so three x is fifteen and x is five. |
| 4 | vertex (2, 7), axis x = 2 | In vertex form the numbers inside the bracket and at the end give the vertex directly. |
| 5 | g(f(2)) = 24 | f of two is five, and five squared minus one is twenty-four. |
| 6 | inverse is 2x – 5, and at 1 it gives -3 | Rearranging gives x equal to two y minus five, and f of minus three returns one. |
| 7 | asymptote y = 0, intercept (0, 1) | The curve stays positive and tends to zero, and any number to the power zero is one. |
| 8 | y = (x + 1) squared minus 4, minimum -4 | Adding and removing one completes the square, and the squared term is never below zero. |
| 9 | inverse is 2 plus 1 over x minus 3, x not equal to 3 | Isolate the fraction and invert; the excluded value three in the range becomes the excluded value in the new domain. |
| 10 | x at least 2 and x not equal to 5 | The root needs x at least two and the denominator forbids five, and both must hold together. |
04 What to do after marking
- After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
- Repeat the same item two days later; move to the next paper only when both attempts are correct.
- Parents only need the answer and note columns; there is no need to explain every item.
Questions and answers are written originally by the TutorZone team and are not taken from any school, publisher or examination authority paper. Every value is checked by computer.
