## Question 9(a):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Forms equation $\frac{x(x+3)}{x+4} = k$ | M1 (3.1a) | Or differentiates using quotient rule |
| Rearranges to quadratic in $x$: $x^2 + (3-k)x - 4k = 0$; or obtains $f'(x) = \frac{(x+4)(2x+3)-(x^2+3x)(1)}{(x+4)^2}$ | M1 (1.1a) | |
| Explains discriminant of quadratic $< 0$; or explains vertical asymptote means minimum is higher than maximum and turning points lie on different branches | E1 (2.4) | |
| Forms quadratic in $k$ from discriminant: $(3-k)^2 - 4(1)(-4k) < 0$, giving $k^2 + 10k + 9 < 0$, $(k+9)(k+1) < 0$; or equates $f'(x)$ numerator to zero and solves | M1 (1.1a) | |
| Concludes $-9 < k < -1$, so $f(x)$ does not take values in interval $(-9,-1)$ | R1 (2.1) | Condone $-9 < k < -1$ |

## Question 9(b):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Substitutes $k = -9$ or $k = -1$ into $y = f(x)$ and forms quadratic in $x$; or differentiates using quotient rule and equates $f'(x)$ or numerator to 0 | M1 (1.1a) | |
| Stationary points $(-6, -9)$ and $(-2, -1)$ | A1F (1.1b) | Both coordinates of both points required |

## Question 9(c):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Divides numerator by $x+4$: $f(x) = \frac{x^2+3x}{x+4} = \frac{(x+4)(x-1)+4}{x+4} = x - 1 + \frac{4}{x+4}$ | M1 (3.1a) | Obtains $f(x) = x + \cdots$ |
| Asymptote is $y = x - 1$ | A1 (1.1b) | |

## Question 9(d):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Draws curve asymptotic to $x = -4$ or $y = x-1$ | B1F (1.1b) | |
| Draws curve with two branches asymptotic to their asymptotes | B1F (1.1b) | |
| Draws one branch in correct position | M1 (1.1b) | |
| Fully correct sketch with roots at $-3$ and $0$, correct oblique asymptote, showing $(-6,-9)$ and $(-2,-1)$ | A1F (2.2a) | FT their oblique asymptote |