(a)

Vertical asymptote occurs where the denominator is zero:
\( 2x + 6 = 0 \implies x = -3 \).
Equation: \( x = -3 \) (A1, N1).

[1 mark]

(b)

Graph crosses x-axis when \( f(x) = 0 \):
\( \frac{x^2 – 14x + 24}{2x + 6} = 0 \implies x^2 – 14x + 24 = 0 \).
Solve: \( x = \frac{14 \pm \sqrt{196 – 96}}{2} = \frac{14 \pm 10}{2} \).
\( x = 12 \), \( x = 2 \).
Coordinates: \( (2, 0) \), \( (12, 0) \) (M1 for equation, A1 for coordinates, N2).

[2 marks]

(c)

Perform polynomial long division for \( \frac{x^2 – 14x + 24}{2x + 6} \):
Divide \( x^2 \) by \( 2x \): \( \frac{x}{2} \).
\( \frac{x}{2} (2x + 6) = x^2 + 3x \).
Subtract: \( (x^2 – 14x + 24) – (x^2 + 3x) = -17x + 24 \).
Divide \( -17x \) by \( 2x \): \( -\frac{17}{2} \).
\( -\frac{17}{2} (2x + 6) = -17x – 51 \).
Subtract: \( (-17x + 24) – (-17x – 51) = 75 \).
Thus: \( f(x) = \frac{x}{2} – \frac{17}{2} + \frac{75}{2x + 6} \).
As \( x \to \pm \infty \), \( \frac{75}{2x + 6} \to 0 \), so oblique asymptote is \( y = \frac{x}{2} – \frac{17}{2} \).
\( a = \frac{1}{2} \), \( b = -\frac{17}{2} \) (M1 for division setup, A1 for quotient, A1 for remainder, A1 for a, b, N4).

[4 marks]

(d)

Sketch for \( -50 \leq x \leq 50 \):
– Vertical asymptote at \( x = -3 \).
– Oblique asymptote: \( y = \frac{x}{2} – \frac{17}{2} \).
– X-intercepts: \( (2, 0) \), \( (12, 0) \).
– Y-intercept: \( f(0) = \frac{24}{6} = 4 \), so \( (0, 4) \).
Behavior: As \( x \to -3^- \), \( f(x) \to -\infty \); as \( x \to -3^+ \), \( f(x) \to \infty \). As \( x \to \pm \infty \), \( f(x) \to \frac{x}{2} – \frac{17}{2} \). Local minimum and maximum at \( y = -5 \pm 2\sqrt{10} \). (A1 for asymptotes, A1 for intercepts, A1 for shape, A1 for critical points, N4).

[4 marks]

(e)

Find range of \( f(x) = \frac{x^2 – 14x + 24}{2x + 6} \):
Rewrite: \( f(x) = \frac{x}{2} – \frac{17}{2} + \frac{75}{2x + 6} \).
Let \( y = f(x) \):
\( y(2x + 6) = x^2 – 14x + 24 \implies x^2 – 14x – 2xy – 6y + 24 = 0 \).
Quadratic in \( x \): \( x^2 – (14 + 2y)x + (24 – 6y) = 0 \).
Discriminant: \( \Delta = (14 + 2y)^2 – 4(24 – 6y) = 4y^2 + 56y + 196 – 96 + 24y = 4y^2 + 80y + 100 \).
For real \( x \), \( \Delta \geq 0 \): \( 4(y^2 + 20y + 25) \geq 0 \), always true.
Simplify: \( y^2 + 20y + 100 = (y + 10)^2 + 25 \geq 25 \).
Critical points: Derivative: \( f'(x) = \frac{(2x – 14)(2x + 6) – (x^2 – 14x + 24)(2)}{(2x + 6)^2} = \frac{2x^2 + 12x – 132}{(2x + 6)^2} \).
Solve \( 2x^2 + 12x – 132 = 0 \implies x^2 + 6x – 66 = 0 \).
\( x = \frac{-6 \pm \sqrt{36 + 264}}{2} = -3 \pm 5\sqrt{3} \).
Evaluate \( f \): At \( x = -3 \pm 5\sqrt{3} \), \( f(x) = -5 \pm 2\sqrt{10} \).
Range: \( y \leq -5 – 2\sqrt{10} \) or \( y \geq -5 + 2\sqrt{10} \) (M1 for setup, M1 for critical points, A1 for range, A1 for exact form, N4).

[4 marks]

(f)

Solve \( f(x) > x \):
\( \frac{x^2 – 14x + 24}{2x + 6} > x \implies \frac{x^2 – 14x + 24 – x(2x + 6)}{2x + 6} > 0 \implies \frac{-x^2 – 20x + 24}{2x + 6} > 0 \).
Numerator: \( -x^2 – 20x + 24 = -(x^2 + 20x – 24) \).
Solve \( x^2 + 20x – 24 = 0 \): \( x = \frac{-20 \pm \sqrt{400 + 96}}{2} = -10 \pm 2\sqrt{31} \).
Roots: \( x = -10 \pm 2\sqrt{31} \).
Denominator zero: \( x = -3 \).
Critical points: \( x = -10 + 2\sqrt{31} \), \( x = -3 \), \( x = -10 – 2\sqrt{31} \).
Sign analysis:
– \( x < -10 – 2\sqrt{31} \): Negative numerator, positive denominator, negative.
– \( -10 – 2\sqrt{31} < x < -3 \): Positive numerator, positive denominator, positive.
– \( -3 < x < -10 + 2\sqrt{31} \): Positive numerator, negative denominator, negative.
– \( x > -10 + 2\sqrt{31} \): Negative numerator, negative denominator, positive.
Solution: \( -10 – 2\sqrt{31} < x < -3 \) or \( x > -10 + 2\sqrt{31} \) (M1 for setup, M1 for critical points, A1 for sign analysis, A1 for solution, N4).

[4 marks]

Total [19 marks]