| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Factorise then sketch |
| Difficulty | Moderate -0.8 This is a straightforward C1 algebraic manipulation and sketching question. Part (a) requires routine expansion and collection of terms, part (b) is simple factorisation once the common factor x is identified, and part (c) is a standard cubic sketch with roots already found. All steps are mechanical with no problem-solving insight required, making it easier than average. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials |
| Answer | Marks | Guidance |
|---|---|---|
| \(f(x)=x[(x-6)(x-2)+3]\) or \(x^3-6x^2-2x^2+12x+3x=x(\ldots)\) | M1 | For a correct method to get the factor of \(x\). \(x(\) as printed is the minimum |
| \(f(x)=x(x^2-8x+15)\), \(b=-8\) or \(c=15\) | A1 | \(-8\) comes from \(-6-2\) and must be coefficient of \(x\), and 15 from \(6\times2+3\) and must have no \(x\)s |
| both and \(a=1\) | A1 | (3). For \(a=1\), \(b=-8\) and \(c=15\). Must have \(x(x^2-8x+15)\) |
| Answer | Marks | Guidance |
|---|---|---|
| \((x^2-8x+15)=(x-5)(x-3)\) | M1 | For attempt to factorise their 3TQ from part (a) |
| \(f(x)=x(x-5)(x-3)\) | A1 | (2). For all 3 terms correct. They must include the \(x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Cubic curve, correct shape (bottom-left to top-right, two turning points) | B1 | For correct shape |
| Crossing at their 3 or their 5 | B1f.t. | For crossing at their 3 or their 5 indicated on graph or in text |
| Both their 3 and their 5, and \((0,0)\) by implication | B1f.t. | (3). If graph passes through \((0,0)\) [needn't be marked] and both their 3 and their 5 |
## Question 9:
### Part (a):
| $f(x)=x[(x-6)(x-2)+3]$ or $x^3-6x^2-2x^2+12x+3x=x(\ldots)$ | M1 | For a correct method to get the factor of $x$. $x($ as printed is the minimum |
| $f(x)=x(x^2-8x+15)$, $b=-8$ or $c=15$ | A1 | $-8$ comes from $-6-2$ and must be coefficient of $x$, and 15 from $6\times2+3$ and must have no $x$s |
| both and $a=1$ | A1 | (3). For $a=1$, $b=-8$ and $c=15$. Must have $x(x^2-8x+15)$ |
### Part (b):
| $(x^2-8x+15)=(x-5)(x-3)$ | M1 | For attempt to factorise their 3TQ from part (a) |
| $f(x)=x(x-5)(x-3)$ | A1 | (2). For all 3 terms correct. They must include the $x$ |
### Part (c):
| Cubic curve, correct shape (bottom-left to top-right, two turning points) | B1 | For correct shape |
| Crossing at their 3 or their 5 | B1f.t. | For crossing at their 3 or their 5 indicated on graph or in text |
| Both their 3 and their 5, and $(0,0)$ by implication | B1f.t. | (3). If graph passes through $(0,0)$ [needn't be marked] and both their 3 and their 5 |
---9. Given that $\mathrm { f } ( x ) = \left( x ^ { 2 } - 6 x \right) ( x - 2 ) + 3 x$,
\begin{enumerate}[label=(\alph*)]
\item express $\mathrm { f } ( x )$ in the form $x \left( a x ^ { 2 } + b x + c \right)$, where $a$, $b$ and $c$ are constants.
\item Hence factorise $\mathrm { f } ( x )$ completely.
\item Sketch the graph of $y = \mathrm { f } ( x )$, showing the coordinates of each point at which the graph meets the axes.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2006 Q9 [8]}}