AQA AS Paper 1 2023 June — Question 6 6 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2023
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCompleting the square and sketching
TypeTransformations of quadratic graphs
DifficultyModerate -0.8 This is a straightforward AS-level question testing completing the square (routine algebraic manipulation), reading the minimum from vertex form (direct recall), and applying a stretch transformation (standard procedure). All parts are textbook exercises requiring no problem-solving insight, making it easier than average.
Spec1.02e Complete the square: quadratic polynomials and turning points1.02w Graph transformations: simple transformations of f(x)
  1. The curve \(C_1\) has equation \(y = 2x^2 - 20x + 42\) Express the equation of \(C_1\) in the form $$y = a(x - h)^2 + c$$ where \(a\), \(b\) and \(c\) are integers. [3 marks]
  2. Write down the coordinates of the minimum point of \(C_1\) [1 mark]
  3. The curve \(C_1\) is mapped onto the curve \(C_2\) by a stretch in the \(y\)-direction. The minimum point of \(C_2\) is at \((5, -4)\) Find the equation of \(C_2\) [2 marks]
Question 6:
AnswerMarks Guidance
6(a)Obtains a = 2 1.1b
y = 2(x2 – 10x + 25 – 4)
y = 2((x – 5)2 – 4)
y = 2(x – 5)2 – 8
AnswerMarks Guidance
Obtains b = 51.1b B1
Obtains c = –81.1b B1
Subtotal3
QMarking instructions AO
AnswerMarks
6(b)Obtains correct coordinates of
their minimum point.
FT their b and c
AnswerMarks Guidance
Condone missing brackets.1.1b B1F
Subtotal1
QMarking instructions AO
AnswerMarks
6(c)1
Uses a stretch scale factor of
2
−4
FTtheir ,do not FT c =  4
c
AnswerMarks Guidance
PI by correct answer3.1a M1
s c a l e f a c t o r = =
− 8 2
y = 2(x – 5)2 – 8
1
y =  2 ( x – 5 ) 2 – 8 
2
y = ( x – 5 ) 2 – 4
y = x2 – 10x + 21
Deduces their correct equation
using their vertical stretch factor.
ACF
FT their c , do not FT c=4
AnswerMarks Guidance
ISW2.2a A1F
Subtotal2
Question 6 Total6
QMarking instructions AO 
Question 6:
--- 6(a) ---
6(a) | Obtains a = 2 | 1.1b | B1 | y = 2(x2 – 10x + 21)
y = 2(x2 – 10x + 25 – 4)
y = 2((x – 5)2 – 4)
y = 2(x – 5)2 – 8
Obtains b = 5 | 1.1b | B1
Obtains c = –8 | 1.1b | B1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b) ---
6(b) | Obtains correct coordinates of
their minimum point.
FT their b and c
Condone missing brackets. | 1.1b | B1F | (5, –8)
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 6(c) ---
6(c) | 1
Uses a stretch scale factor of
2
−4
FTtheir ,do not FT c =  4
c
PI by correct answer | 3.1a | M1 | − 4 1
s c a l e f a c t o r = =
− 8 2
y = 2(x – 5)2 – 8
1
y =  2 ( x – 5 ) 2 – 8 
2
y = ( x – 5 ) 2 – 4
y = x2 – 10x + 21
Deduces their correct equation
using their vertical stretch factor.
ACF
FT their c , do not FT c=4
ISW | 2.2a | A1F
Subtotal | 2
Question 6 Total | 6
Q | Marking instructions | AO | Marks | Typical solution
\begin{enumerate}[label=(\alph*)]
\item The curve $C_1$ has equation $y = 2x^2 - 20x + 42$

Express the equation of $C_1$ in the form
$$y = a(x - h)^2 + c$$
where $a$, $b$ and $c$ are integers.
[3 marks]

\item Write down the coordinates of the minimum point of $C_1$
[1 mark]

\item The curve $C_1$ is mapped onto the curve $C_2$ by a stretch in the $y$-direction.

The minimum point of $C_2$ is at $(5, -4)$

Find the equation of $C_2$
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1 2023 Q6 [6]}}
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Q1 1 Q2 1 Q3 3 Q4 5 Q5 7 Q6 6 Q7 5 Q8 7 Q9 3 Q10 8 Q11 7 Q12 1 Q13 1 Q14 4 Q15 4 Q16 7 Q17 4 Q18 6