| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Complete the square |
| Difficulty | Moderate -0.8 This is a routine C1 question testing completing the square, sketching, and transformations—all standard textbook exercises requiring only procedural knowledge. The multi-part structure guides students through each step with no problem-solving or novel insight required, making it easier than average. |
| 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 |
|---|---|---|
| \((x - 3)^2\) | M1 | or \(p = 3\) seen |
| \((x - 3)^2 + 2\) | A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \((x - 3)^2 = -2\) | M1 | FT their positive value of \(q\); not use of discriminant for graphical approach see below to see if SC1 can be awarded |
| No (real) square root of \(-2\) therefore equation has no real solutions | A1 cso | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(x =\) 'their' \(p\) or \(y =\) 'their' \(q\) | M1 | or \(x = 3\) found using calculus |
| Vertex is at \((3, 2)\) | A1 cao | 2 marks |
| Answer | Marks |
|---|---|
| B1 | \(y\) intercept \(= 11\) stated or marked on \(y\)-axis (as \(y\) intercept of any graph) |
| M1 | ∪ shape (generous) |
| Answer | Marks | Guidance |
|---|---|---|
| A1 | 3 marks | above \(x\)-axis, vertex in first quadrant crossing \(y\)-axis into second quadrant |
| Translation | E1 | and no other transformation |
| through \(\begin{bmatrix} -3 \\ -2 \end{bmatrix}\) | M1 | FT negative of BOTH 'their' vertex coords |
| A1 | 3 marks | both components correct for A1; may describe in words or use a column vector |
**4(a)(i)**
$(x - 3)^2$ | M1 | or $p = 3$ seen
$(x - 3)^2 + 2$ | A1 | 2 marks |
**4(a)(ii)**
$(x - 3)^2 = -2$ | M1 | FT their positive value of $q$; not use of discriminant for graphical approach see below to see if SC1 can be awarded
No (real) square root of $-2$ therefore equation has no real solutions | A1 cso | 2 marks |
**4(b)(i)**
$x =$ 'their' $p$ or $y =$ 'their' $q$ | M1 | or $x = 3$ found using calculus
Vertex is at $(3, 2)$ | A1 cao | 2 marks |
**4(b)(ii)**
| B1 | $y$ intercept $= 11$ stated or marked on $y$-axis (as $y$ intercept of any graph)
| M1 | ∪ shape (generous)
**4(b)(iii)**
| A1 | 3 marks | above $x$-axis, vertex in first quadrant crossing $y$-axis into second quadrant
Translation | E1 | and no other transformation
through $\begin{bmatrix} -3 \\ -2 \end{bmatrix}$ | M1 | FT negative of BOTH 'their' vertex coords
| A1 | 3 marks | both components **correct for A1**; may describe in words or use a column vector
---4
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Express $x ^ { 2 } - 6 x + 11$ in the form $( x - p ) ^ { 2 } + q$.
\item Use the result from part (a)(i) to show that the equation $x ^ { 2 } - 6 x + 11 = 0$ has no real solutions.
\end{enumerate}\item A curve has equation $y = x ^ { 2 } - 6 x + 11$.
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of the vertex of the curve.
\item Sketch the curve, indicating the value of $y$ where the curve crosses the $y$-axis.
\item Describe the geometrical transformation that maps the curve with equation $y = x ^ { 2 } - 6 x + 11$ onto the curve with equation $y = x ^ { 2 }$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA C1 2013 Q4 [12]}}