Edexcel C1 2014 January — Question 4 4 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Year2014
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeSingle transformation sketches
DifficultyModerate -0.8 This is a straightforward C1 transformation question requiring only recall of standard rules: horizontal translation shifts x-coordinates (0,3)→(-4,3) and (4,2)→(0,2), and vertical stretch multiplies y-coordinates (0,3)→(0,6) and (4,2)→(4,4). No problem-solving or conceptual insight needed, just direct application of memorized transformation rules.
Spec1.02w Graph transformations: simple transformations of f(x)

4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{6081d81b-51d2-4140-9834-71ef7fd700b0-08_835_777_118_596} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} Figure 1 shows a sketch of a curve with equation \(y = \mathrm { f } ( x )\). The curve crosses the \(y\)-axis at \(( 0,3 )\) and has a minimum at \(P ( 4,2 )\). On separate diagrams, sketch the curve with equation
  1. \(y = \mathrm { f } ( x + 4 )\),
  2. \(y = 2 \mathrm { f } ( x )\). On each diagram, show clearly the coordinates of the minimum point and any point of intersection with the \(y\)-axis.

AnswerMarks Guidance
(a)Horizontal translation of \(\pm 4\) M1
Minimum point on the \(y\)-axis at \((0,2)\)A1 The shape remains unchanged and has a minimum at \((0,2)\). Condone U shaped curves.
(b)Correct "shape" with \(P'\) adapted M1
\(y\) intercept \((0,6)\) and \(P'(4,4)\)A1 Correct shape, condoning U shapes with the \(y\) intercept at \((0, 6)\) and \(P' = (4,4)\). The coordinates of the points may appear in the text or besides the diagram. This is acceptable but if they contradict the diagram, the diagram takes precedence.
(a) | Horizontal translation of $\pm 4$ | M1 | A horizontal translation of $\pm 4$. The $y$ coordinate of $P$ remains unchanged at 2. Look for $P' = (0,2)$ or $(8,2)$. Condone U shaped curves.

| Minimum point on the $y$-axis at $(0,2)$ | A1 | The shape remains unchanged and has a minimum at $(0,2)$. Condone U shaped curves.

(b) | Correct "shape" with $P'$ adapted | M1 | The curve remains in quadrant 1 and quadrant 2 with the minimum in quadrant 1. The shape must be correct. Condone U shaped curves. $P'$ must have been adapted. The mark cannot be scored for drawing the original curve with $P' = (4,2)$.

| $y$ intercept $(0,6)$ and $P'(4,4)$ | A1 | Correct shape, condoning U shapes with the $y$ intercept at $(0, 6)$ and $P' = (4,4)$. The coordinates of the points may appear in the text or besides the diagram. This is acceptable but if they contradict the diagram, the diagram takes precedence.

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4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{6081d81b-51d2-4140-9834-71ef7fd700b0-08_835_777_118_596}
\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 crosses the $y$-axis at $( 0,3 )$ and has a minimum at $P ( 4,2 )$.

On separate diagrams, sketch the curve with equation
\begin{enumerate}[label=(\alph*)]
\item $y = \mathrm { f } ( x + 4 )$,
\item $y = 2 \mathrm { f } ( x )$.

On each diagram, show clearly the coordinates of the minimum point and any point of intersection with the $y$-axis.
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1 2014 Q4 [4]}}