| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Finding quadratic from vertex information |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing standard P1 techniques: finding a quadratic from vertex form (routine substitution), finding a perpendicular line (basic coordinate geometry), and writing inequalities from a diagram. All parts are textbook exercises requiring only direct application of formulas with no problem-solving insight needed. |
| Spec | 1.02e Complete the square: quadratic polynomials and turning points1.02f Solve quadratic equations: including in a function of unknown1.02i Represent inequalities: graphically on coordinate plane1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| E.g. \(12 - a(x+2)^2\) or \(a(x-1)(x+5)\) or \(y = ax^2 + bx + c \Rightarrow 4a - 2b + c = 12\) and \(25a - 5b + c = 0\) | M1A1 | For knowing a quadratic form and using two correct pieces of information |
| E.g. \(0 = 12 - a(-5+2)^2 \Rightarrow a = ...\) or \(12 = a(-2-1)(-2+5) \Rightarrow a = ...\) or \(2a(-2)+b = 0, 25a-5b+c=0, 4a-2b+c=12 \Rightarrow a=..., b=..., c=...\) | dM1 | Full method for finding \(f(x)\); dependent on previous M1 |
| \(12 - \frac{4}{3}(x+2)^2\) or \(-\frac{4}{3}(x-1)(x+5)\) or \(-\frac{4}{3}x^2 - \frac{16}{3}x + \frac{20}{3}\) | A1 | Correct equation/expression; accept any equivalent form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Gradient of \(l_2\) is \(\frac{-5}{4}\) | M1 | Applies perpendicular condition to find gradient of \(l_2\) |
| Equation of \(l_2\) is \(y = {-\frac{5}{4}}(x+5)\) | M1 | Uses changed gradient with \((-5,0)\); must proceed to \(c = ...\) if using \(y = mx + c\) |
| \(y = -\frac{5}{4}x - \frac{25}{4}\) | A1 | Correct equation; must be in form \(y = mx + c\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For two of: \(y \geqslant -\frac{5}{4}x - \frac{25}{4}\); \(y \geqslant \frac{4}{5}x\); \(y \leqslant -\frac{4}{3}x^2 - \frac{16}{3}x + \frac{20}{3}\) (or strict inequalities) | M1 | FT on answers to (a) and (b); use of \(R\) instead of \(y\) is M0 A0 |
| All three inequalities correct (or with strict inequalities, consistently) | A1ft | Follow through their answers to (a) and (b); set notation accepted |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| E.g. $12 - a(x+2)^2$ or $a(x-1)(x+5)$ or $y = ax^2 + bx + c \Rightarrow 4a - 2b + c = 12$ and $25a - 5b + c = 0$ | M1A1 | For knowing a quadratic form and using two correct pieces of information |
| E.g. $0 = 12 - a(-5+2)^2 \Rightarrow a = ...$ or $12 = a(-2-1)(-2+5) \Rightarrow a = ...$ or $2a(-2)+b = 0, 25a-5b+c=0, 4a-2b+c=12 \Rightarrow a=..., b=..., c=...$ | dM1 | Full method for finding $f(x)$; dependent on previous M1 |
| $12 - \frac{4}{3}(x+2)^2$ or $-\frac{4}{3}(x-1)(x+5)$ or $-\frac{4}{3}x^2 - \frac{16}{3}x + \frac{20}{3}$ | A1 | Correct equation/expression; accept any equivalent form |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Gradient of $l_2$ is $\frac{-5}{4}$ | M1 | Applies perpendicular condition to find gradient of $l_2$ |
| Equation of $l_2$ is $y = {-\frac{5}{4}}(x+5)$ | M1 | Uses changed gradient with $(-5,0)$; must proceed to $c = ...$ if using $y = mx + c$ |
| $y = -\frac{5}{4}x - \frac{25}{4}$ | A1 | Correct equation; must be in form $y = mx + c$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For two of: $y \geqslant -\frac{5}{4}x - \frac{25}{4}$; $y \geqslant \frac{4}{5}x$; $y \leqslant -\frac{4}{3}x^2 - \frac{16}{3}x + \frac{20}{3}$ (or strict inequalities) | M1 | FT on answers to (a) and (b); use of $R$ instead of $y$ is M0 A0 |
| All three inequalities correct (or with strict inequalities, consistently) | A1ft | Follow through their answers to (a) and (b); set notation accepted |
---\begin{enumerate}
\item The curve $C$ has equation $y = \mathrm { f } ( x )$
\end{enumerate}
Given that
\begin{itemize}
\item $\mathrm { f } ( x )$ is a quadratic expression
\item the maximum turning point on $C$ has coordinates $( - 2,12 )$
\item $C$ cuts the negative $x$-axis at - 5\\
(a) find $\mathrm { f } ( x )$
\end{itemize}
The line $l _ { 1 }$ has equation $y = \frac { 4 } { 5 } x$
Given that the line $l _ { 2 }$ is perpendicular to $l _ { 1 }$ and passes through $( - 5,0 )$\\
(b) find an equation for $l _ { 2 }$, writing your answer in the form $y = m x + c$ where $m$ and $c$ are constants to be found.\\
(3)
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3cf69966-e825-4ff0-a6e8-c5dfdc92c53f-10_983_712_1126_616}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of the curve $C$ and the lines $l _ { 1 }$ and $l _ { 2 }$\\
(c) Define the region $R$, shown shaded in Figure 2, using inequalities.
\hfill \mbox{\textit{Edexcel P1 2022 Q5 [9]}}