| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Session | Specimen |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Complete square then find vertex/turning point |
| Difficulty | Moderate -0.8 This is a straightforward completing-the-square question with direct application to find the vertex. It requires only standard algebraic manipulation and recognition that the vertex form immediately gives the minimum point coordinates—both are routine C1 skills with no problem-solving or insight required. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(x^2 - 8x + 3 = (x-4)^2 - 13\), i.e. \(a = -4, b = -13\) | B1 M1 A1 | For \((x-4)^2\) seen, or statement \(a = -4\); For use of (implied) relation \(a^2 + b = 3\); For correct value of \(b\) stated or implied |
| 3 | ||
| (ii) Minimum point is \((4, -13)\) | B1 √ B1 √ | For \(x\)-coordinate equal to their \((-a)\); For \(y\)-coordinate equal to their \(b\) |
| 2 |
(i) $x^2 - 8x + 3 = (x-4)^2 - 13$, i.e. $a = -4, b = -13$ | B1 M1 A1 | For $(x-4)^2$ seen, or statement $a = -4$; For use of (implied) relation $a^2 + b = 3$; For correct value of $b$ stated or implied
| | 3 |
(ii) Minimum point is $(4, -13)$ | B1 √ B1 √ | For $x$-coordinate equal to their $(-a)$; For $y$-coordinate equal to their $b$
| | 2 |
---2 (i) Express $x ^ { 2 } - 8 x + 3$ in the form $( x + a ) ^ { 2 } + b$.\\
(ii) Hence write down the coordinates of the minimum point on the graph of $y = x ^ { 2 } - 8 x + 3$.
\hfill \mbox{\textit{OCR C1 Q2 [5]}}