S4 Maths quadratic graphs paper 4: ten common slips taken apart
- 2026-11-10
S4 Maths quadratic graphs paper 4: ten common slips taken apart
Ten items target the slips seen most often earlier: intercept order, discriminant arithmetic, the axis sign, repeated roots and always-positive conditions.
Direct answers Items 01 to 03 rebuild the y-intercept, a discriminant and the opening direction, 04 to 06 a repeated root, the axis sign and an equation from two roots, 07 to 09 an outside-the-roots set, the twin always-positive conditions and tangency for m, and 10 a missing negative branch.
01 Paper details (check answers afterwards)
| Item | Detail |
|---|---|
| Questions | 10 |
| Suggested time | 25 |
| Difficulty mix | Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1 |
| Chapter | Chapter: S4 Maths Quadratic Graphs |
| Level | S4 |
02 Questions (10 items)
- FoundationWhere does the graph of y = x^2 – 6x + 5 meet the y-axis?
- FoundationFind the discriminant of x^2 – 5x + 6 = 0 and state the number of real roots.
- FoundationGive the opening direction and the vertex of y = -2(x – 1)^2 + 8.
- IntermediateSolve x^2 + 10x + 25 = 0.
- IntermediateFind the axis of symmetry of y = x^2 + 6x + 1.
- IntermediateA quadratic with leading coefficient 1 has roots 3 and 5. Write the equation.
- ChallengeSolve the inequality x^2 – 2x – 3 > 0.
- ChallengeThe inequality kx^2 + 4x + 2 > 0 holds for every real x. Find the range of k.
- ChallengeThe curve y = x^2 – 4x + m touches the x-axis. Find m.
- Error redoError drill: solve x^2 <= 16. A student wrote x <= 4. What is the correct set?
03 Answers and one-line notes
| # | Answer | One-line note |
|---|---|---|
| 1 | (0, 5) | Putting x = 0 gives y = 5, so the point is (0, 5) and not the reversed (5, 0). |
| 2 | Discriminant = 1, so there are two distinct real roots | The discriminant is 25 – 24 = 1, which is positive, so the two roots are distinct. |
| 3 | It opens downward and the vertex is (1, 8) | The coefficient of x^2 is -2, so the curve opens downward, and the vertex form gives the vertex (1, 8) directly. |
| 4 | x = -5 (a repeated root) | The left side is (x + 5)^2, so the only root is -5 and not plus or minus 5. |
| 5 | x = -3 | Completing the square gives (x + 3)^2 – 8, so the axis is x = -3 and the sign flips. |
| 6 | x^2 – 8x + 15 = 0 | The sum is 8 and the product 15, so the equation is x^2 – 8x + 15 = 0. |
| 7 | x < -1 or x > 3 | Factorising gives (x – 3)(x + 1) > 0, and the curve opens upward, so it is positive outside the roots. |
| 8 | k > 2 | It needs k > 0 together with discriminant 16 – 8k < 0, and the two conditions give k > 2. |
| 9 | m = 4 | Touching means a zero discriminant: 16 – 4m = 0, hence m = 4. |
| 10 | -4 <= x <= 4 | The inequality means |x| <= 4, so both directions are needed; writing only x <= 4 wrongly keeps values such as -3. |
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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