| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2021 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Completing square from standard form |
| Difficulty | Moderate -0.8 This is a straightforward multi-part question testing routine algebraic skills. Part (a) is standard completing the square with a coefficient, part (b) requires simple reading of the vertex form, and part (c) uses factor form of a cubic with one additional constraint. All techniques are textbook exercises requiring minimal problem-solving insight. |
| Spec | 1.02d Quadratic functions: graphs and discriminant conditions1.02e Complete the square: quadratic polynomials and turning points1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3x^2+6x+9 = 3(x\pm\ldots)^2\pm\ldots\), states \(a=3\) | B1 | Or states \(a=3\) |
| \(3x^2+6x+9 = 3(x+1)^2\pm\ldots\), states \(a=3\) and \(b=1\) | M1 | Deals correctly with first two terms; may use \(a(x+b)^2+c = ax^2+2abx+ab^2+c\) |
| \(3x^2+6x+9 = 3(x+1)^2+6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((-1, 6)\) | B1ft | Follow through from \((-b, c)\) in part (a); condone lack of brackets; accept \(x=-1\), \(y=6\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = \alpha(x+4)(x+2)(x-3)\) | B1 | Identifies three factors \((x+4)(x+2)(x-3)\) |
| \(6 = \alpha(-1+4)(-1+2)(-1-3)\) | M1 | Correct method to find scale factor using minimum point from (b) |
| \(\alpha = -\frac{1}{2}\) | A1 | |
| \(y = -\frac{1}{2}(x+4)(x+2)(x-3) \Rightarrow y = \ldots x^3+\ldots x^2+\ldots x+\ldots\) | M1 | Attempts to expand to achieve \(x^3+\ldots\pm 24\) |
| \(A=-\frac{1}{2},\ B=-\frac{3}{2},\ C=5,\ D=12\) | A1 | Accept \(y=-\frac{1}{2}x^3-\frac{3}{2}x^2+5x+12\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Three correct equations using \(x=-4,-2,3\) with \(y=0\): \(-64A+16B-4C+D=0\); \(-8A+4B-2C+D=0\); \(27A+9B+3C+D=0\) | B1 | |
| Fourth equation \(-A+B-C+D=6\) using point \(P\) | M1 | |
| One of \(A=-\frac{1}{2}\), \(B=-\frac{3}{2}\), \(C=5\), \(D=12\) correct | A1 | |
| Fully solves simultaneous equations | M1 | May use matrices or elimination; calculator acceptable |
| \(A=-\frac{1}{2},\ B=-\frac{3}{2},\ C=5,\ D=12\) | A1 | Accept \(y=-\frac{1}{2}x^3-\frac{3}{2}x^2+5x+12\) |
# Question 8:
## Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $3x^2+6x+9 = 3(x\pm\ldots)^2\pm\ldots$, states $a=3$ | B1 | Or states $a=3$ |
| $3x^2+6x+9 = 3(x+1)^2\pm\ldots$, states $a=3$ and $b=1$ | M1 | Deals correctly with first two terms; may use $a(x+b)^2+c = ax^2+2abx+ab^2+c$ |
| $3x^2+6x+9 = 3(x+1)^2+6$ | A1 | |
**(3 marks)**
## Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(-1, 6)$ | B1ft | Follow through from $(-b, c)$ in part (a); condone lack of brackets; accept $x=-1$, $y=6$ |
**(1 mark)**
## Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = \alpha(x+4)(x+2)(x-3)$ | B1 | Identifies three factors $(x+4)(x+2)(x-3)$ |
| $6 = \alpha(-1+4)(-1+2)(-1-3)$ | M1 | Correct method to find scale factor using minimum point from (b) |
| $\alpha = -\frac{1}{2}$ | A1 | |
| $y = -\frac{1}{2}(x+4)(x+2)(x-3) \Rightarrow y = \ldots x^3+\ldots x^2+\ldots x+\ldots$ | M1 | Attempts to expand to achieve $x^3+\ldots\pm 24$ |
| $A=-\frac{1}{2},\ B=-\frac{3}{2},\ C=5,\ D=12$ | A1 | Accept $y=-\frac{1}{2}x^3-\frac{3}{2}x^2+5x+12$ |
**(5 marks)**
**Alternative (c) via simultaneous equations:**
| Answer | Mark | Guidance |
|--------|------|----------|
| Three correct equations using $x=-4,-2,3$ with $y=0$: $-64A+16B-4C+D=0$; $-8A+4B-2C+D=0$; $27A+9B+3C+D=0$ | B1 | |
| Fourth equation $-A+B-C+D=6$ using point $P$ | M1 | |
| One of $A=-\frac{1}{2}$, $B=-\frac{3}{2}$, $C=5$, $D=12$ correct | A1 | |
| Fully solves simultaneous equations | M1 | May use matrices or elimination; calculator acceptable |
| $A=-\frac{1}{2},\ B=-\frac{3}{2},\ C=5,\ D=12$ | A1 | Accept $y=-\frac{1}{2}x^3-\frac{3}{2}x^2+5x+12$ |
**(9 marks total)**8. The curve $C _ { 1 }$ has equation
$$y = 3 x ^ { 2 } + 6 x + 9$$
\begin{enumerate}[label=(\alph*)]
\item Write $3 x ^ { 2 } + 6 x + 9$ in the form
$$a ( x + b ) ^ { 2 } + c$$
where $a$, $b$ and $c$ are constants to be found.
The point $P$ is the minimum point of $C _ { 1 }$
\item Deduce the coordinates of $P$.
A different curve $C _ { 2 }$ has equation
$$y = A x ^ { 3 } + B x ^ { 2 } + C x + D$$
where $A$, $B$, $C$ and $D$ are constants.
Given that $C _ { 2 }$
\begin{itemize}
\item passes through $P$
\item intersects the $x$-axis at $- 4 , - 2$ and 3
\item find, making your method clear, the values of $A , B , C$ and $D$.\\
\includegraphics[max width=\textwidth, alt={}, center]{877d03f2-d62c-4060-bdd2-f0d5dfbe6203-27_2644_1840_118_111}
\end{itemize}
\includegraphics[max width=\textwidth, alt={}, center]{877d03f2-d62c-4060-bdd2-f0d5dfbe6203-29_2646_1838_121_116}
\end{enumerate}
\hfill \mbox{\textit{Edexcel P1 2021 Q8 [9]}}