Edexcel M2 2014 January — Question 4 4 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2014
SessionJanuary
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFunction Transformations
TypeMultiple separate transformations (sketch-based, standard transformations)
DifficultyModerate -0.8 This is a straightforward application of standard function transformation rules (horizontal translation and vertical stretch) with clearly given key points. Students only need to recall and apply the transformation formulas to two specific points, requiring minimal problem-solving beyond direct rule application.
Spec1.02w Graph transformations: simple transformations of f(x)

\includegraphics{figure_1} Figure 1 shows a sketch of a curve with equation \(y = f(x)\). 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. \(y = f(x + 4)\), [2]
  2. \(y = 2f(x)\). [2]
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 \(\pm4\)M1 A horizontal translation of \(\pm4\). 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
(2 marks)
(b) Correct "shape" with \(P'\) adaptedM1 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.
(2 marks)
(4 marks)
(a) Horizontal translation of $\pm4$ | M1 | A horizontal translation of $\pm4$. 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

| | **(2 marks)**

(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.

| | **(2 marks)**
| | **(4 marks)**

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\includegraphics{figure_1}

Figure 1 shows a sketch of a curve with equation $y = f(x)$.

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 $y = f(x + 4)$, [2]
\item $y = 2f(x)$. [2]
\end{enumerate}

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

\hfill \mbox{\textit{Edexcel M2 2014 Q4 [4]}}