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S4 Quadratic Functions, Paper 3: transformations and applied maxima
S4 Quadratic Functions, Paper 3: transformations and applied maxima
Vertex form shows every shift at a glance, and applied questions are answered at the vertex.
Focus Read the opening direction and vertex, then use completing the square or the vertex formula for maxima.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 4 · Intermediate 3 · Challenge 2 · Error redo 1 |
| Chapter | Chapter: S4 Maths: Quadratic Functions and Graphs |
| Level | S4 |
02 Questions (10 items)
- FoundationState the vertex and minimum of y = (x − 5)².
- FoundationHow is the graph of y = x² + 2 obtained from y = x², and where is its vertex?
- FoundationFind the maximum and the vertex of y = −(x + 2)² − 1.
- FoundationFind the intercepts of y = x² − 4x.
- IntermediateComplete the square for y = x² − 6x + 11 and give the vertex.
- IntermediateComplete the square for y = 2x² + 8x + 5 and state the vertex.
- IntermediateThe graph of y = x² − 2x + k touches the x-axis. Find k.
- ChallengeA ball follows h = −5t² + 30t + 1 metres. Find the time and height at its highest point.
- ChallengeA 40 m fence makes a rectangle against a wall on one side. Find the greatest area.
- Error redoError: a student says y = (x − 2)² + 5 is least at x = −2. What is wrong?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | Vertex (5, 0), minimum 0 | The bracket subtracts 5, so the vertex sits at 5 and the least value is zero. |
| 2 | Shifted up 2 units, vertex (0, 2) | Adding a constant lifts the whole curve without changing its shape. |
| 3 | Maximum −1, vertex (−2, −1) | The minus sign opens the curve downwards, so the largest value is minus 1. |
| 4 | x-intercepts 0 and 4, y-intercept 0 | Putting x to zero gives zero, and x(x minus 4) gives the two x-intercepts. |
| 5 | y = (x − 3)² + 2, vertex (3, 2), minimum 2 | Adding and subtracting 9 completes the square, leaving 2. |
| 6 | y = 2(x + 2)² − 3, vertex (−2, −3) | Take out 2 first, then add and subtract 4 inside the bracket to get minus 3. |
| 7 | k = 1 | Touching the axis means the discriminant 4 minus 4k is zero, so k is 1. |
| 8 | t = 3 s, height 46 m | The vertex is at t equals 30 over 10, namely 3, and substituting gives 46. |
| 9 | 200 m², 10 m by 20 m | With width x the length is 40 minus 2x, giving area 200 when x is 10. |
| 10 | The minimum is at x = 2 | The bracket is x minus 2, so it vanishes at 2 rather than minus 2. |
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.
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