Edexcel C1 — Question 10 11 marks

Exam BoardEdexcel
ModuleC1 (Core Mathematics 1)
Marks11
PaperDownload PDF ↗
TopicTangents, normals and gradients
TypeTangent meets curve/axis — further geometry
DifficultyModerate -0.8 This is a straightforward C1 differentiation question requiring routine techniques: substituting to verify a point lies on a curve, finding dy/dx and evaluating at a point for the tangent equation, then solving dy/dx = m to find another point with parallel tangent. All steps are standard textbook exercises with no problem-solving insight required, making it easier than average but not trivial due to the cubic differentiation and multi-part structure.
Spec1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations

The curve \(C\) has equation \(y = \frac{1}{3}x^3 - 4x^2 + 8x + 3\). The point \(P\) has coordinates \((3, 0)\).
  1. Show that \(P\) lies on \(C\). [1]
  2. Find the equation of the tangent to \(C\) at \(P\), giving your answer in the form \(y = mx + c\), where \(m\) and \(c\) are constants. [5]
Another point \(Q\) also lies on \(C\). The tangent to \(C\) at \(Q\) is parallel to the tangent to \(C\) at \(P\).
  1. Find the coordinates of \(Q\). [5]

The curve $C$ has equation $y = \frac{1}{3}x^3 - 4x^2 + 8x + 3$.

The point $P$ has coordinates $(3, 0)$.

\begin{enumerate}[label=(\alph*)]
\item Show that $P$ lies on $C$. [1]
\item Find the equation of the tangent to $C$ at $P$, giving your answer in the form $y = mx + c$, where $m$ and $c$ are constants. [5]
\end{enumerate}

Another point $Q$ also lies on $C$. The tangent to $C$ at $Q$ is parallel to the tangent to $C$ at $P$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the coordinates of $Q$. [5]
\end{enumerate}

\hfill \mbox{\textit{Edexcel C1  Q10 [11]}}