AQA Paper 3 2023 June — Question 6 9 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2023
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeVertical translation of cubic with factorisation
DifficultyStandard +0.3 This is a multi-part question requiring standard A-level techniques: sketching a cubic (routine), applying factor theorem (direct substitution), identifying a vertical translation (basic transformation), and using the discriminant condition for no real roots (standard reasoning linking graph behavior to algebraic properties). While part (b)(iii) requires connecting multiple parts, the logic is straightforward—the cubic touches the x-axis once, so the quadratic factor has no real roots, giving b²<4ac. All steps are well-practiced techniques with no novel insight required, making this slightly easier than average.
Spec1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x)
  1. Sketch the curve with equation $$y = x^2(2x + a)$$ where \(a > 0\) [3 marks] \includegraphics{figure_6a}
  2. The polynomial \(p(x)\) is given by $$p(x) = x^2(2x + a) + 36$$
    1. It is given that \(x + 3\) is a factor of \(p(x)\) Use the factor theorem to show \(a = 2\) [2 marks]
    2. State the transformation which maps the curve with equation $$y = x^2(2x + 2)$$ onto the curve with equation $$y = x^2(2x + 2) + 36$$ [2 marks]
    3. The polynomial \(x^2(2x + 2) + 36\) can be written as \((x + 3)(2x^2 + bx + c)\) Without finding the values of \(b\) and \(c\), use your answers to parts (a) and (b)(ii) to explain why $$b^2 < 8c$$ [2 marks]
Question 6:
AnswerMarks Guidance
6(a)Draws cubic curve in the correct
orientation1.1a M1
Deduces minimum or maximum
AnswerMarks Guidance
at (0,0) on their curve2.2a M1
Draws a fully correct cubic curve
a
with x-intercept at − shown on
2
AnswerMarks Guidance
the curve2.2a A1
Subtotal3
QMarking instructions AO
AnswerMarks
6(b)(i)Substitutes x=−3 into p(x)
Condone missing bracket for
(−3) 2
Must see an expression in terms
AnswerMarks Guidance
of a1.1a M1
−54+9a+36=0
9a−18=0
a =2
Completes reasoned argument
with at least one correct
intermediate step and no error
seen to show a = 2 AG
Must set an expression for
p(−3) = 0
Condone recovery of missing
bracket for (−3) 2 to get 9
Do not condone any other
AnswerMarks Guidance
missing bracket2.1 R1
Subtotal2
QMarking instructions AO
AnswerMarks
6(b)(ii)States ‘translation’ or ‘translate’
or ‘translated’
Must not have other
transformation other than
AnswerMarks Guidance
translation1.1b B1
Translation  
36
 0 
States the vector  or 36j
AnswerMarks Guidance
361.1b B1
Subtotal2
QMarking instructions AO
AnswerMarks
6(b)(iii)Explains that the translated
graph only has one real solution
or only has a root at −3
AnswerMarks Guidance
Condone missing ‘real’2.4 E1
one real solution.
b 2 −4ac<0
Hence b 2 −4×2×c<0
b 2 <8c
Deduces that the discriminant of
2x 2 +bx+cmust be negative
and shows the required result
Do not allow the use of a = 2
with reference to part (b)(i)
Allow b 2 −8c<0 following
2−4ac
AnswerMarks Guidance
from b seen2.2a E1
Subtotal2
Question 6 Total9
QMarking instructions AO 
Question 6:
--- 6(a) ---
6(a) | Draws cubic curve in the correct
orientation | 1.1a | M1
Deduces minimum or maximum
at (0,0) on their curve | 2.2a | M1
Draws a fully correct cubic curve
a
with x-intercept at − shown on
2
the curve | 2.2a | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b)(i) ---
6(b)(i) | Substitutes x=−3 into p(x)
Condone missing bracket for
(−3) 2
Must see an expression in terms
of a | 1.1a | M1 | (−3 )2( 2×−3+a )+36=0
−54+9a+36=0
9a−18=0
a =2
Completes reasoned argument
with at least one correct
intermediate step and no error
seen to show a = 2 AG
Must set an expression for
p(−3) = 0
Condone recovery of missing
bracket for (−3) 2 to get 9
Do not condone any other
missing bracket | 2.1 | R1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b)(ii) ---
6(b)(ii) | States ‘translation’ or ‘translate’
or ‘translated’
Must not have other
transformation other than
translation | 1.1b | B1 |  0 
Translation  
36
 0 
States the vector  or 36j
36 | 1.1b | B1
Subtotal | 2
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b)(iii) ---
6(b)(iii) | Explains that the translated
graph only has one real solution
or only has a root at −3
Condone missing ‘real’ | 2.4 | E1 | The translated graph will only have
one real solution.
b 2 −4ac<0
Hence b 2 −4×2×c<0
b 2 <8c
Deduces that the discriminant of
2x 2 +bx+cmust be negative
and shows the required result
Do not allow the use of a = 2
with reference to part (b)(i)
Allow b 2 −8c<0 following
2−4ac
from b seen | 2.2a | E1
Subtotal | 2
Question 6 Total | 9
Q | Marking instructions | AO | Marks | Typical solution
\begin{enumerate}[label=(\alph*)]
\item Sketch the curve with equation
$$y = x^2(2x + a)$$
where $a > 0$

[3 marks]

\includegraphics{figure_6a}

\item The polynomial $p(x)$ is given by
$$p(x) = x^2(2x + a) + 36$$

\begin{enumerate}[label=(\roman*)]
\item It is given that $x + 3$ is a factor of $p(x)$

Use the factor theorem to show $a = 2$

[2 marks]

\item State the transformation which maps the curve with equation
$$y = x^2(2x + 2)$$
onto the curve with equation
$$y = x^2(2x + 2) + 36$$

[2 marks]

\item The polynomial $x^2(2x + 2) + 36$ can be written as $(x + 3)(2x^2 + bx + c)$

Without finding the values of $b$ and $c$, use your answers to parts (a) and (b)(ii) to explain why
$$b^2 < 8c$$

[2 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2023 Q6 [9]}}
This paper (17 questions)
View full paper
Q1 1 Q2 1 Q3 1 Q4 5 Q5 3 Q6 9 Q7 14 Q8 7 Q9 12 Q10 1 Q11 1 Q12 8 Q13 4 Q14 10 Q15 11 Q16 9 Q17 6