**(i)** 
| B1 | +ve cubic with 3 distinct roots | For first B1, left end of curve must finish below x axis and right end must end above x axis. Allow slight wrong curvature at one end but not both ends. No cusp at either turning point. No straight lines drawn with a ruler. Condone (0, 3) as maximum point.

| B1 | (0, 3) labelled or indicated on y-axis |

| B1, 3 | $(-3, 0), (\frac{1}{2}, 0)$ and $(1, 0)$ labelled or indicated on x-axis and no other x-intercepts | To gain second and third B marks, there must be an attempt at a curve, not just points on axes. Final B1 can be awarded for a negative cubic.

**(ii)** $2x^3 + 5x - 3 = x + 2x - 3, 2x^3 - 3x + 1$ | B1 | Obtain one quadratic factor (can be unsimplified) |

$(2x^2 + 5x - 3)(x - 1)$ | M1 | Attempt to multiply a quadratic by a linear factor |

$2x^3 + 5x^2 - 3x + 2x - 3$ | A1 | | Alternative for first 3 marks: Attempt to expand all 3 brackets with an appropriate number of terms (including an x² term) M1 | Expansion with at most 1 incorrect term A1 | Correct, answer (can be unsimplified) A1 | Allow if done in part(i) please insert AG.

$\frac{dy}{dx} = 6x^2 + 6x - 8$ | M1 | Attempt to differentiate (one non-zero term correct) |

When $x = 1$, gradient $= 4$ | A1, 6 | Fully correct expression www | Confirms gradient = 4 at x = 1 **AG |

**(iii)** Gradient of $l = 4$ | B1 | May be embedded in equation of line |

On curve, when $x = -2$, $y = 15$ | B1 | Correct y coordinate |

$y - 15 = 4(x + 2)$ | M1 | Correct equation of line using their values |

$y = 4x + 23$ | A1, 4 | Correct answer in correct form |

**(iv)** Attempt to find gradient of curve when $x = -2$ | M1 | Substitute $x = -2$ into their $\frac{dy}{dx}$ |

$6(-2)^2 + 6(-2) - 8 = 4$ | A1 | Obtain gradient of 4 CWO |

So line is a tangent | A1, 3, 16 | Correct conclusion | Alternatives 1) Equates equation of l to equation of curve and attempts to divide resulting cubic by (x + 2) M1 Obtains (x + 2)² (2x - 5) (=0) A1 Concludes repeated root implies tangent at x = -2 A1 2) Equates their gradient function to 4 and uses correct method to solve the resulting quadratic M1 Obtains (x + 2)(x - 1) = 0 oe A1 Correctly concludes gradient = 4 when x = -2 A1

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# Allocation of Method Mark for Solving a Quadratic

e.g. $2x^2 - 5x - 18 = 0$

**1) If the candidate attempts to solve by factorisation, their attempt when expanded must produce the correct quadratic term and one other correct term (with correct sign):**

| $(2x + 2)(x - 9) = 0$ | M1 | $2x^2$ and $-18$ obtained from expansion |
| $(2x + 3)(x - 4) = 0$ | M1 | $2x^2$ and $-5x$ obtained from expansion |
| $(2x - 9)(x - 2) = 0$ | M0 | only $2x^2$ term correct |

**2) If the candidate attempts to solve by using the formula**

a) If the formula is quoted incorrectly then **M0**.

b) If the formula is quoted correctly then one **sign slip is permitted**. Substituting the wrong numerical value for a or b or c scores **M0**

| $\frac{-5 \pm \sqrt{(-5)^2 - 4 \times 2 \times -18}}{2 \times 2}$ | M1 | (minus sign incorrect at start of formula) |
| $\frac{5 \pm \sqrt{(-5)^2 - 4 \times 2 \times 18}}{2 \times 2}$ | M1 | (18 for c instead of -18) |
| $\frac{-5 \pm \sqrt{(-5)^2 - 4 \times 2 \times 18}}{2 \times 2}$ | M0 | (2 sign errors: initial sign and c incorrect) |
| $\frac{5 \pm \sqrt{(-5)^2 - 4 \times 2 \times -18}}{2 \times 5}$ | M0 | (2b on the denominator) |

**Notes** – for equations such as $2x^2 - 5x - 18 = 0$, then $b^2 = 5^2$ would be condoned in the discriminant and would not be counted as a sign error. Repeating the sign error for a in both occurrences in the formula would be two sign errors and score M0.

c) If the formula is not quoted at all, substitution must be completely correct to earn the **M1**

**3) If the candidate attempts to complete the square, they must get to the "square root stage" involving ±; we are looking for evidence that the candidate knows a quadratic has two solutions!**

| $2x^2 - 5x - 18 = 0$ |
| $2(x^2 - \frac{5}{2}x) - 18 = 0$ |
| $2[(x - \frac{5}{4})^2 - \frac{25}{16}] - 18 = 0$ |
| $(x - \frac{5}{4})^2 = \frac{169}{16}$ |
| $x - \frac{5}{4} = \pm\sqrt{\frac{169}{16}}$ | This is where the M1 is awarded – arithmetical errors may be condoned provided $x - \frac{5}{4}$ seen or implied |

If a candidate makes repeated attempts (e.g. fails to factorise and then tries the formula), mark only what you consider to be their last full attempt.