OCR C1 — Question 9 10 marks

Exam BoardOCR
ModuleC1 (Core Mathematics 1)
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCurve Sketching
TypeMultiple transformations in sequence
DifficultyStandard +0.3 This is a multi-part question covering standard C1 topics: solving a quadratic, sketching a parabola, applying a horizontal stretch transformation, and finding the equation after a translation. All parts use routine techniques with no novel problem-solving required, though the multiple parts and transformation work make it slightly above average difficulty for C1.
Spec1.02f Solve quadratic equations: including in a function of unknown1.02n Sketch curves: simple equations including polynomials1.02w Graph transformations: simple transformations of f(x)

9. \(f ( x ) = 2 x ^ { 2 } + 3 x - 2\).
  1. Solve the equation \(\mathrm { f } ( x ) = 0\).
  2. Sketch the curve with equation \(y = \mathrm { f } ( x )\), showing the coordinates of any points of intersection with the coordinate axes.
  3. Find the coordinates of the points where the curve with equation \(y = \mathrm { f } \left( \frac { 1 } { 2 } x \right)\) crosses the coordinate axes. When the graph of \(y = \mathrm { f } ( x )\) is translated by 1 unit in the positive \(x\)-direction it maps onto the graph with equation \(y = a x ^ { 2 } + b x + c\), where \(a , b\) and \(c\) are constants.
  4. Find the values of \(a , b\) and \(c\).

AnswerMarks Guidance
(i) \((2x-1)(x+2) = 0\)M1
\(x = -2, \frac{1}{2}\)A1
(ii) Sketch showing curve passing through \((-2,0)\), \((\frac{1}{2}, 0)\), and \((0,-2)\) with correct shapeB2
(iii) \((0,-2)\), \((-4,0)\), \((1,0)\)B1 M1 A1
(iv) \(f(x-1) = 2(x-1)^2 + 3(x-1) - 2\)M1 A1
\(= 2x^2 - x - 3\)A1
\(\therefore a = 2, b = -1, c = -3\)A1 (10)
**(i)** $(2x-1)(x+2) = 0$ | M1 |
$x = -2, \frac{1}{2}$ | A1 |

**(ii)** Sketch showing curve passing through $(-2,0)$, $(\frac{1}{2}, 0)$, and $(0,-2)$ with correct shape | B2 |

**(iii)** $(0,-2)$, $(-4,0)$, $(1,0)$ | B1 M1 A1 |

**(iv)** $f(x-1) = 2(x-1)^2 + 3(x-1) - 2$ | M1 A1 |
$= 2x^2 - x - 3$ | A1 |
$\therefore a = 2, b = -1, c = -3$ | A1 | **(10)**

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9. $f ( x ) = 2 x ^ { 2 } + 3 x - 2$.\\
(i) Solve the equation $\mathrm { f } ( x ) = 0$.\\
(ii) Sketch the curve with equation $y = \mathrm { f } ( x )$, showing the coordinates of any points of intersection with the coordinate axes.\\
(iii) Find the coordinates of the points where the curve with equation $y = \mathrm { f } \left( \frac { 1 } { 2 } x \right)$ crosses the coordinate axes.

When the graph of $y = \mathrm { f } ( x )$ is translated by 1 unit in the positive $x$-direction it maps onto the graph with equation $y = a x ^ { 2 } + b x + c$, where $a , b$ and $c$ are constants.\\
(iv) Find the values of $a , b$ and $c$.\\

\hfill \mbox{\textit{OCR C1  Q9 [10]}}