Edexcel P1 2024 January — Question 9 10 marks

Exam BoardEdexcel
ModuleP1 (Pure Mathematics 1)
Year2024
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeFinding quadratic from vertex information
DifficultyModerate -0.8 This is a straightforward multi-part question testing standard techniques: (a) sketching from given information, (b) finding a quadratic from vertex form using three known points, and (c) solving simultaneous equations. All parts use routine methods with no novel insight required, making it easier than average for A-level.
Spec1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem

  1. The curve \(C _ { 1 }\) has equation \(y = \mathrm { f } ( x )\).
Given that
  • \(\mathrm { f } ( x )\) is a quadratic expression
  • \(C _ { 1 }\) has a maximum turning point at \(( 2,20 )\)
  • \(C _ { 1 }\) passes through the origin
    1. sketch a graph of \(C _ { 1 }\) showing the coordinates of any points where \(C _ { 1 }\) cuts the coordinate axes,
    2. find an expression for \(\mathrm { f } ( x )\).
The curve \(C _ { 2 }\) has equation \(y = x \left( x ^ { 2 } - 4 \right)\) Curve \(C _ { 1 }\) and \(C _ { 2 }\) meet at the origin, and at the points \(P\) and \(Q\) Given that the \(x\) coordinate of the point \(P\) is negative,
  • using algebra and showing all stages of your working, find the coordinates of \(P\)

  • Question 9:
    Part (a):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    Correct \(\cap\) shape passing through \((0,0)\), maximum on rhs of \(y\)-axisB1 \(\cap\) shaped quadratic through origin with maximum on rhs of \(y\)-axis; appears in quadrants 3, 1 and 4; ignore extra graphs superimposed
    Intersection at \((4, 0)\)B1 Allow 4 marked on positive \(x\)-axis; \((0,4)\) is B0; condone graphs that just meet \(x\)-axis at 4
    (2 marks)
    Part (b):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    Attempts form of equation e.g. \(y = Ax(x-4)\) or \(y = 20 \pm C(x-2)^2\)M1 Condone \(A,C=1\); also allow \(y=ax^2+bx+c\) with attempt to use all three coordinates
    Full attempt to find equation e.g. \(20 = A\times2(2-4) \Rightarrow A = \ldots\) or \(0 = 20+C(4-2)^2 \Rightarrow C=\ldots\)dM1 Full attempt at finding values of \(A\) or \(C\) or \(a,b,c\)
    \(y = -5x(x-4)\), \(y = 20-5(x-2)^2\) o.e.A1 Allow \(f(x)\leftrightarrow y\); allow \(5x(4-x)\); ISW after correct answer
    (3 marks)
    Part (c):
    AnswerMarks Guidance
    Answer/WorkingMark Guidance
    Sets \(x(x^2-4) = ``{-5x(x-4)}''\)M1 Form of quadratic must be correct (pass through \((0,0)\)); condoning slips
    \(x^3+5x^2-24x=0 \Rightarrow x\left(x^2+5x-24\right)=0\)dM1 Multiplies out to form \(px^3+qx^2+rx=0\), factorises/cancels \(x\)
    \((x+8)(x-3)=0 \Rightarrow x\) coordinate of \(P\) is \(-8\)ddM1, A1 Solves resulting quadratic; chooses \(-8\); reject or ignore \(x=0,3\)
    \(P = (-8, -480)\)A1 May be given separately as \(x=\ldots, y=\ldots\)
    (5 marks)
    (10 marks total) All stages of working must be shown
    # Question 9:
    
    ## Part (a):
    
    | Answer/Working | Mark | Guidance |
    |---|---|---|
    | Correct $\cap$ shape passing through $(0,0)$, maximum on rhs of $y$-axis | B1 | $\cap$ shaped quadratic through origin with maximum on rhs of $y$-axis; appears in quadrants 3, 1 and 4; ignore extra graphs superimposed |
    | Intersection at $(4, 0)$ | B1 | Allow 4 marked on positive $x$-axis; $(0,4)$ is B0; condone graphs that just meet $x$-axis at 4 |
    | **(2 marks)** | | |
    
    ## Part (b):
    
    | Answer/Working | Mark | Guidance |
    |---|---|---|
    | Attempts form of equation e.g. $y = Ax(x-4)$ or $y = 20 \pm C(x-2)^2$ | M1 | Condone $A,C=1$; also allow $y=ax^2+bx+c$ with attempt to use all three coordinates |
    | Full attempt to find equation e.g. $20 = A\times2(2-4) \Rightarrow A = \ldots$ or $0 = 20+C(4-2)^2 \Rightarrow C=\ldots$ | dM1 | Full attempt at finding values of $A$ or $C$ or $a,b,c$ |
    | $y = -5x(x-4)$, $y = 20-5(x-2)^2$ o.e. | A1 | Allow $f(x)\leftrightarrow y$; allow $5x(4-x)$; ISW after correct answer |
    | **(3 marks)** | | |
    
    ## Part (c):
    
    | Answer/Working | Mark | Guidance |
    |---|---|---|
    | Sets $x(x^2-4) = ``{-5x(x-4)}''$ | M1 | Form of quadratic must be correct (pass through $(0,0)$); condoning slips |
    | $x^3+5x^2-24x=0 \Rightarrow x\left(x^2+5x-24\right)=0$ | dM1 | Multiplies out to form $px^3+qx^2+rx=0$, factorises/cancels $x$ |
    | $(x+8)(x-3)=0 \Rightarrow x$ coordinate of $P$ is $-8$ | ddM1, A1 | Solves resulting quadratic; **chooses** $-8$; reject or ignore $x=0,3$ |
    | $P = (-8, -480)$ | A1 | May be given separately as $x=\ldots, y=\ldots$ |
    | **(5 marks)** | | |
    | **(10 marks total)** | | All stages of working must be shown |
    \begin{enumerate}
      \item The curve $C _ { 1 }$ has equation $y = \mathrm { f } ( x )$.
    \end{enumerate}
    
    Given that
    
    \begin{itemize}
      \item $\mathrm { f } ( x )$ is a quadratic expression
      \item $C _ { 1 }$ has a maximum turning point at $( 2,20 )$
      \item $C _ { 1 }$ passes through the origin\\
    (a) sketch a graph of $C _ { 1 }$ showing the coordinates of any points where $C _ { 1 }$ cuts the coordinate axes,\\
    (b) find an expression for $\mathrm { f } ( x )$.
    \end{itemize}
    
    The curve $C _ { 2 }$ has equation $y = x \left( x ^ { 2 } - 4 \right)$\\
    Curve $C _ { 1 }$ and $C _ { 2 }$ meet at the origin, and at the points $P$ and $Q$\\
    Given that the $x$ coordinate of the point $P$ is negative,\\
    (c) using algebra and showing all stages of your working, find the coordinates of $P$
    
    \hfill \mbox{\textit{Edexcel P1 2024 Q9 [10]}}