Edexcel Paper 1 2021 October — Question 2 4 marks

Exam BoardEdexcel
ModulePaper 1 (Paper 1)
Year2021
SessionOctober
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCompleting the square and sketching
TypeComplete square then find vertex/turning point
DifficultyEasy -1.8 This is a routine completing-the-square exercise with direct reading of vertex coordinates—purely procedural with no problem-solving required. The completing the square is straightforward (coefficient of x² is 1), and finding P and Q requires only substitution and reading from completed square form. This is easier than typical A-level questions, being foundational GCSE/AS-level material.
Spec1.02e Complete the square: quadratic polynomials and turning points1.04e Sequences: nth term and recurrence relations

  1. Given that
$$\mathrm { f } ( x ) = x ^ { 2 } - 4 x + 5 \quad x \in \mathbb { R }$$
  1. express \(\mathrm { f } ( x )\) in the form \(( x + a ) ^ { 2 } + b\) where \(a\) and \(b\) are integers to be found. The curve with equation \(y = \mathrm { f } ( x )\)
    • meets the \(y\)-axis at the point \(P\)
    • has a minimum turning point at the point \(Q\)
    • Write down
      1. the coordinates of \(P\)
      2. the coordinates of \(Q\)

Question 2:
Part (a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(f(x) = (x-2)^2 \pm \ldots\)M1 Achieves \((x-2)^2 \pm \ldots\) or states \(a = -2\)
\(f(x) = (x-2)^2 + 1\)A1 Correct expression; ISW after sight of this. Condone \(a=-2, b=1\). Condone \((x-2)^2+1=0\)
(2 marks)
Part (b)(i):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(P = (0, 5)\)B1 Correct coordinates for \(P\). Allow \(x=0, y=5\)
(1 mark)
Part (b)(ii):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(Q = (2, 1)\)B1ft Correct coordinates for \(Q\). Allow \(x=2, y=1\). Follow through from part (a): allow \((-a, b)\) where \(a, b\) numeric
(1 mark)
## Question 2:

### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $f(x) = (x-2)^2 \pm \ldots$ | M1 | Achieves $(x-2)^2 \pm \ldots$ or states $a = -2$ |
| $f(x) = (x-2)^2 + 1$ | A1 | Correct expression; ISW after sight of this. Condone $a=-2, b=1$. Condone $(x-2)^2+1=0$ |
| **(2 marks)** | | |

### Part (b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P = (0, 5)$ | B1 | Correct coordinates for $P$. Allow $x=0, y=5$ |
| **(1 mark)** | | |

### Part (b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $Q = (2, 1)$ | B1ft | Correct coordinates for $Q$. Allow $x=2, y=1$. Follow through from part (a): allow $(-a, b)$ where $a, b$ numeric |
| **(1 mark)** | | |

---
\begin{enumerate}
  \item Given that
\end{enumerate}

$$\mathrm { f } ( x ) = x ^ { 2 } - 4 x + 5 \quad x \in \mathbb { R }$$

(a) express $\mathrm { f } ( x )$ in the form $( x + a ) ^ { 2 } + b$ where $a$ and $b$ are integers to be found.

The curve with equation $y = \mathrm { f } ( x )$

\begin{itemize}
  \item meets the $y$-axis at the point $P$
  \item has a minimum turning point at the point $Q$\\
(b) Write down\\
(i) the coordinates of $P$\\
(ii) the coordinates of $Q$
\end{itemize}

\hfill \mbox{\textit{Edexcel Paper 1 2021 Q2 [4]}}