| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Transformations of quadratic graphs |
| Difficulty | Easy -1.2 This is a routine completing-the-square question with standard follow-up parts requiring direct reading from the completed square form. All parts are textbook exercises with no problem-solving or novel insight required—significantly easier than average A-level questions. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| \((x + 2.5)^2\) | B1 | \(p = \frac{5}{2}\) |
| \(q = 7 - \text{'their } p^2\) | M1 | unsimplified attempt at \(q = 7 - \text{'their } p^2\): \(q = 7 - \frac{25}{4} = \frac{3}{4}\) |
| \((x + 2.5)^2 + 0.75\) | A1 | mark their final line as their answer |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = -\text{'their } p\) or \(y = \text{'their } q\) | M1 | or \(x = -\frac{5}{2}\) cao found using calculus |
| \(\left(-\frac{5}{2}, \frac{3}{4}\right)\) | A1 cao | condone correct coordinates stated: \(x = -2.5, y = 0.75\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = -\frac{5}{2}\) | B1 \(\checkmark\) | correct or ft "\(x = -\text{'their } p\)" |
| Answer | Marks |
|---|---|
| B1 | \(y\) intercept \(= 7\) stated or seen in table as \(y = 7\) when \(x = 0\) or 7 marked as intercept on \(y\)-axis (any graph) |
| M1 | \(\cup\) shape |
| A1 | vertex above \(x\)-axis in correct quadrant and parabola extending beyond \(y\)-axis into first quadrant |
| Answer | Marks | Guidance |
|---|---|---|
| Translation through \(\begin{bmatrix} -\frac{5}{2} \\ \frac{3}{4} \end{bmatrix}\) | E1 | and no other transformation |
| M1 | ft either 'their' \(-p\) or 'their' \(q\) or one component correct for M1 | |
| A1 cao | both components correct for A1; may describe in words or use a vector |
**4(a)**
$(x + 2.5)^2$ | B1 | $p = \frac{5}{2}$
$q = 7 - \text{'their } p^2$ | M1 | unsimplified attempt at $q = 7 - \text{'their } p^2$: $q = 7 - \frac{25}{4} = \frac{3}{4}$
$(x + 2.5)^2 + 0.75$ | A1 | mark their final line as their answer
**4(b)(i)**
$x = -\text{'their } p$ or $y = \text{'their } q$ | M1 | or $x = -\frac{5}{2}$ cao found using calculus
$\left(-\frac{5}{2}, \frac{3}{4}\right)$ | A1 cao | condone correct coordinates stated: $x = -2.5, y = 0.75$
**4(b)(ii)**
$x = -\frac{5}{2}$ | B1 $\checkmark$ | correct or ft "$x = -\text{'their } p$"
**4(b)(iii)**
| B1 | $y$ intercept $= 7$ stated or seen in table as $y = 7$ when $x = 0$ or 7 marked as intercept on $y$-axis (any graph)
| M1 | $\cup$ shape
| A1 | vertex above $x$-axis in correct quadrant and parabola extending beyond $y$-axis into first quadrant
**4(c)**
Translation through $\begin{bmatrix} -\frac{5}{2} \\ \frac{3}{4} \end{bmatrix}$ | E1 | and no other transformation
| M1 | ft either 'their' $-p$ or 'their' $q$ or one component correct for M1
| A1 cao | both components correct for A1; may describe in words or use a vector
---4
\begin{enumerate}[label=(\alph*)]
\item Express $x ^ { 2 } + 5 x + 7$ in the form $( x + p ) ^ { 2 } + q$, where $p$ and $q$ are rational numbers.\\
(3 marks)
\item A curve has equation $y = x ^ { 2 } + 5 x + 7$.
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of the vertex of the curve.
\item State the equation of the line of symmetry of the curve.
\item Sketch the curve, stating the value of the intercept on the $y$-axis.
\end{enumerate}\item Describe the geometrical transformation that maps the graph of $y = x ^ { 2 }$ onto the graph of $y = x ^ { 2 } + 5 x + 7$.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2011 Q4 [12]}}