| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Transformations of quadratic graphs |
| Difficulty | Moderate -0.8 This is a straightforward C1 question testing completing the square (routine algebraic manipulation), reading vertex coordinates from completed square form, and applying a translation vector. All parts follow standard procedures with no problem-solving or novel insight required, making it easier than average but not trivial due to the multi-step translation in part (c). |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\left(x + \frac{3}{2}\right)^2 - \frac{9}{4} + 2\) | M1 | Correct method completing the square |
| \(\left(x + \frac{3}{2}\right)^2 - \frac{1}{4}\) | A1 | Both \(p = \frac{3}{2}\) and \(q = -\frac{1}{4}\) correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Vertex \(= \left(-\frac{3}{2}, -\frac{1}{4}\right)\) | B1 B1 | B1 for each coordinate; follow through from part (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = -\frac{3}{2}\) | B1 | Follow through from their \(p\) value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Replace \(x\) with \((x-2)\) and \(y\) with \((y-4)\) | M1 | Correct translation method |
| \(y - 4 = (x-2)^2 + 3(x-2) + 2\) | A1 | Correct substitution |
| \(y = x^2 - x + 2\) | A1 | Correct final equation in required form |
## Question 5:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\left(x + \frac{3}{2}\right)^2 - \frac{9}{4} + 2$ | M1 | Correct method completing the square |
| $\left(x + \frac{3}{2}\right)^2 - \frac{1}{4}$ | A1 | Both $p = \frac{3}{2}$ and $q = -\frac{1}{4}$ correct |
**Part (b)(i):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Vertex $= \left(-\frac{3}{2}, -\frac{1}{4}\right)$ | B1 B1 | B1 for each coordinate; follow through from part (a) |
**Part (b)(ii):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = -\frac{3}{2}$ | B1 | Follow through from their $p$ value |
**Part (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Replace $x$ with $(x-2)$ and $y$ with $(y-4)$ | M1 | Correct translation method |
| $y - 4 = (x-2)^2 + 3(x-2) + 2$ | A1 | Correct substitution |
| $y = x^2 - x + 2$ | A1 | Correct final equation in required form |5
\begin{enumerate}[label=(\alph*)]
\item Express $x ^ { 2 } + 3 x + 2$ in the form $( x + p ) ^ { 2 } + q$, where $p$ and $q$ are rational numbers.
\item A curve has equation $y = x ^ { 2 } + 3 x + 2$.
\begin{enumerate}[label=(\roman*)]
\item Use the result from part (a) to write down the coordinates of the vertex of the curve.
\item State the equation of the line of symmetry of the curve.
\end{enumerate}\item The curve with equation $y = x ^ { 2 } + 3 x + 2$ is translated by the vector $\left[ \begin{array} { l } 2 \\ 4 \end{array} \right]$. Find the equation of the resulting curve in the form $y = x ^ { 2 } + b x + c$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2015 Q5 [8]}}