| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch transformed curve from description |
| Difficulty | Moderate -0.8 This is a straightforward transformation question requiring application of standard rules: vertical translation shifts points and asymptote down by 2, reflection in y-axis negates x-coordinates. Both transformations are routine P1 content with no problem-solving or novel insight required, making it easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape, translated downward | B1 | Same shape as original but translated vertically downwards. No coordinates considered here. Look for minimum left of \(y\)-axis and maximum to the right |
| Horizontal asymptote \(y = -1\) labelled | B1 | Must be below \(x\)-axis. Intention must be for curve to be asymptotic at both ends. Be tolerant of asymptote not being exactly same level at either end |
| Correct maximum \(\left(\frac{3}{2}, 0\right)\) and minimum \((-1, -2)\) labelled | B1 | Must be in correct positions. \(\left(\frac{3}{2}, 0\right)\) on positive \(x\)-axis, \((-1,-2)\) in quadrant 3. Coordinates may be in body of script but must match points marked on graph |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct shape, reflected in \(y\)-axis | B1 | Intention for asymptotes at same level. Image \(Q'\) a maximum to left of \(y\)-axis in quadrant 2, \(P'\) a minimum on positive \(x\)-axis |
| Horizontal asymptote \(y = 1\) labelled | B1 | Must be above \(x\)-axis. Curve must be asymptotic at both ends at same height/level |
| Correct maximum \(\left(-\frac{3}{2}, 2\right)\) and minimum \(P'(1, 0)\) labelled | B1 | Must be in correct positions |
# Question 4(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape, translated downward | B1 | Same shape as original but translated vertically downwards. No coordinates considered here. Look for minimum left of $y$-axis and maximum to the right |
| Horizontal asymptote $y = -1$ labelled | B1 | Must be below $x$-axis. Intention must be for curve to be asymptotic at both ends. Be tolerant of asymptote not being exactly same level at either end |
| Correct maximum $\left(\frac{3}{2}, 0\right)$ and minimum $(-1, -2)$ labelled | B1 | Must be in correct positions. $\left(\frac{3}{2}, 0\right)$ on positive $x$-axis, $(-1,-2)$ in quadrant 3. Coordinates may be in body of script but must match points marked on graph |
# Question 4(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct shape, reflected in $y$-axis | B1 | Intention for asymptotes at same level. Image $Q'$ a maximum to left of $y$-axis in quadrant 2, $P'$ a minimum on positive $x$-axis |
| Horizontal asymptote $y = 1$ labelled | B1 | Must be above $x$-axis. Curve must be asymptotic at both ends at same height/level |
| Correct maximum $\left(-\frac{3}{2}, 2\right)$ and minimum $P'(1, 0)$ labelled | B1 | Must be in correct positions |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3cf69966-e825-4ff0-a6e8-c5dfdc92c53f-08_604_1207_251_370}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a sketch of a curve with equation $y = \mathrm { f } ( x )$\\
The curve has a minimum at $P ( - 1,0 )$ and a maximum at $Q \left( \frac { 3 } { 2 } , 2 \right)$\\
The line with equation $y = 1$ is the only asymptote to the curve.
On separate diagrams sketch the curves with equation\\
(i) $y = \mathrm { f } ( x ) - 2$\\
(ii) $y = \mathrm { f } ( - x )$
On each sketch you must clearly state
\begin{itemize}
\item the coordinates of the maximum and minimum points
\item the equation of the asymptote
\end{itemize}
\hfill \mbox{\textit{Edexcel P1 2022 Q4 [6]}}