| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2023 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Combined stretch and translation |
| Difficulty | Moderate -0.3 This is a straightforward multi-part question on completing the square and applying transformations. Part (a) is routine (completing the square or differentiation), parts (b) and (c) require systematic application of transformation rules but involve no novel insight—slightly easier than a typical A-level question due to the mechanical nature of the transformations. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02w Graph transformations: simple transformations of f(x)1.02x Combinations of transformations: multiple transformations |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a) | d ( x2 −8x+5 ) =0 2x−8 = 0 | |
| dx | M1 | dy |
| Answer | Marks | Guidance |
|---|---|---|
| y=(x−4)2 −11 | M1 | y=(x−4)2k |
| Answer | Marks | Guidance |
|---|---|---|
| 2a 2 | M1 | |
| x=4,y=-11 | A1 | 8 64−20 |
| Answer | Marks | Guidance |
|---|---|---|
| 6(b) | x=(their x value from a )+4 [=8] | B1 FT |
| Answer | Marks | Guidance |
|---|---|---|
| y= (their y value from a)2 +1 [−21] | B1 FT | Can be from putting x = 8 in the equation of the transformed |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | If B0B0 scored, SC B1 for sight of (4,−22) . | |
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 6(c) | 2 ( x2 −8x+5 ) or 2{(x−4)2 −11} | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| (x−4)2 −8(x−4)+5 +1 or (x−4−4)2 −their11 +1 | M1 | For the x translation, each x becomes (x−4). |
| M1 | For the y translation of +1. | |
| y=2x2 −32x+107 or a=2, b=−32, c=107 | A1 | Evidence to support their answer may be in (b) but answer |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 6:
--- 6(a) ---
6(a) | d ( x2 −8x+5 ) =0 2x−8 = 0
dx | M1 | dy
Correct differentiation of x2 and equating their to 0.
dx
Alternative method 1 for first mark of Question 6(a)
y=(x−4)2 −11 | M1 | y=(x−4)2k
Attempt to complete the square as far as .
Alternative method 2 for first mark of Question 6(a)
−b 8
x= =
2a 2 | M1
x=4,y=-11 | A1 | 8 64−20
Answers from x= leading to x=4 11
2
scores M0A0
2
--- 6(b) ---
6(b) | x=(their x value from a )+4 [=8] | B1 FT | Can be from finding the equation of the transformed curve,
dy
differentiating and putting =0.
dx
y= (their y value from a)2 +1 [−21] | B1 FT | Can be from putting x = 8 in the equation of the transformed
curve.
2 | If B0B0 scored, SC B1 for sight of (4,−22) .
Question | Answer | Marks | Guidance
--- 6(c) ---
6(c) | 2 ( x2 −8x+5 ) or 2{(x−4)2 −11} | B1 | Can be implied if both transformations done together:
( )
2 (x−4)2 −8(x−4)+5 +1 OE.
( )
(x−4)2 −8(x−4)+5 +1 or (x−4−4)2 −their11 +1 | M1 | For the x translation, each x becomes (x−4).
M1 | For the y translation of +1.
y=2x2 −32x+107 or a=2, b=−32, c=107 | A1 | Evidence to support their answer may be in (b) but answer
must be seen in (c).
4
Question | Answer | Marks | GuidanceThe equation of a curve is $y = x^2 - 8x + 5$.
\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the minimum point of the curve. [2]
\end{enumerate}
The curve is stretched by a factor of 2 parallel to the $y$-axis and then translated by $\begin{pmatrix} 4 \\ 1 \end{pmatrix}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the coordinates of the minimum point of the transformed curve. [2]
\item Find the equation of the transformed curve. Give the answer in the form $y = ax^2 + bx + c$, where $a$, $b$ and $c$ are integers to be found. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2023 Q6 [8]}}