| Exam Board | OCR |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Sketch transformations from algebraic function |
| Difficulty | Moderate -0.3 This is a straightforward C1 question involving routine algebraic expansion to verify factorization, basic curve sketching from factored form, and standard transformations (horizontal translation and reflection). All techniques are standard textbook exercises requiring no problem-solving insight, though the multi-part nature and transformation work elevates it slightly above pure recall. |
| Spec | 1.02j Manipulate polynomials: expanding, factorising, division, factor theorem1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{LHS} = (x+1)(x^2-7x+12)\) | ||
| \(= x^3 - 7x^2 + 12x + x^2 - 7x + 12\) | M1 | Expanding |
| \(= x^3 - 6x^2 + 5x + 12 = \text{RHS}\) | A1 |
| Answer | Marks |
|---|---|
| Graph with correct shape, \(y\)-intercept \((0,12)\), roots \((-1,0)\), \((3,0)\), \((4,0)\) | B3 |
| Answer | Marks |
|---|---|
| Graph showing translation, roots \((-4,0)\), \((0,0)\), \((1,0)\) | B2 |
| Answer | Marks |
|---|---|
| Graph showing reflection, \(y\)-intercept \((0,12)\), roots \((-4,0)\), \((-3,0)\), \((1,0)\) | B2 |
# Question 8:
## Part (a):
| $\text{LHS} = (x+1)(x^2-7x+12)$ | | |
|---|---|---|
| $= x^3 - 7x^2 + 12x + x^2 - 7x + 12$ | M1 | Expanding |
| $= x^3 - 6x^2 + 5x + 12 = \text{RHS}$ | A1 | |
## Part (b):
| Graph with correct shape, $y$-intercept $(0,12)$, roots $(-1,0)$, $(3,0)$, $(4,0)$ | B3 | |
|---|---|---|
## Part (c)(i):
| Graph showing translation, roots $(-4,0)$, $(0,0)$, $(1,0)$ | B2 | |
|---|---|---|
## Part (c)(ii):
| Graph showing reflection, $y$-intercept $(0,12)$, roots $(-4,0)$, $(-3,0)$, $(1,0)$ | B2 | |
|---|---|---|
---8.
$$f ( x ) = x ^ { 3 } - 6 x ^ { 2 } + 5 x + 12$$
\begin{enumerate}[label=(\alph*)]
\item Show that
$$( x + 1 ) ( x - 3 ) ( x - 4 ) \equiv x ^ { 3 } - 6 x ^ { 2 } + 5 x + 12 .$$
\item Sketch the curve $y = \mathrm { f } ( x )$, showing the coordinates of any points of intersection with the coordinate axes.
\item Showing the coordinates of any points of intersection with the coordinate axes, sketch on separate diagrams the curves
\begin{enumerate}[label=(\roman*)]
\item $\quad y = \mathrm { f } ( x + 3 )$,
\item $y = \mathrm { f } ( - x )$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR C1 Q8 [9]}}