## Question 6:

### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \int(6x^2 - 20x + 6)\,dx = 2x^3 - 10x^2 + 6x + c$ | M1 | Attempt to integrate. Do not allow if multiplied by $x$ |
| $(2, 6)$ on curve so $6 = 2\times2^3 - 10\times2^2 + 6\times2 + c \quad [c=18]$ | M1 | |
| So $y = 2x^3 - 10x^2 + 6x + 18$ | A1 | Complete equation must be seen |

### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = 2(x+1)(x^2-6x+9)$ or $2(x^2-2x-3)(x-3)$ | M1 | o.e. Attempt to remove all brackets leading to a cubic expression. A quadratic factor must be seen |
| So $y = 2x^3 - 10x^2 + 6x + 18$ | E1 | Correct expression found |
| **Alternative:** Division of $2x^3-10x^2+6x+18$ by $(x+1)$ or $(x-3)$ and attempt to factorise quadratic factor | M1 | Leading to a product of linear factors. A quadratic factor must be seen |
| $2(x+1)(x-3)^2$ | E1 | Correct expression found |
| **Alternative:** roots of $y=0$ are $-1$ and $3$ [repeated], so $(x+1)$ and $(x-3)$ are factors; $(x-3)$ is a repeated factor and leading coefficient is 2 so $y=2(x+1)(x-3)^2$ | M1, E1 | Method using factor theorem with roots found by calculator or substitution in their (a). Fully explained |

### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct shape | B1 | Correct shape |
| $y$-intercept at 18 labelled | B1 | |
| Curve crosses $x$-axis at $(-1, 0)$ and touches at $(3, 0)$ | B1 | |

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