(a) Vertical Asymptote:

The vertical asymptote occurs where the denominator is zero:
\( 2x + 6 = 0 \) ⇒ \( x = -3 \)
Answer: \( \boxed{x = -3} \)

(b) x-intercepts:

Solve \( f(x) = 0 \):
\( x^2 – 14x + 24 = 0 \) ⇒ \( (x-2)(x-12) = 0 \)
\( x = 2 \) or \( x = 12 \)
Answer: \( \boxed{(2, 0)} \) and \( \boxed{(12, 0)} \)

(c) Oblique Asymptote:

Perform polynomial long division:
\( \frac{x^2 – 14x + 24}{2x + 6} = \frac{x}{2} – \frac{17}{2} + \frac{75}{2x + 6} \)
As \( x \to \pm\infty \), the term \( \frac{75}{2x + 6} \to 0 \)
Thus, the oblique asymptote is \( y = \frac{x}{2} – \frac{17}{2} \)
Answer: \( a = \boxed{\frac{1}{2}} \), \( b = \boxed{-\frac{17}{2}} \)

(d) Graph Sketch:

Key features to include:
– Vertical asymptote at \( x = -3 \)
– Oblique asymptote \( y = \frac{x}{2} – \frac{17}{2} \)
– x-intercepts at (2, 0) and (12, 0)
– y-intercept at (0, 4)

(e) Range of f:

Find critical points by solving \( f'(x) = 0 \):
\( f'(x) = \frac{2(x^2 + 6x – 66)}{(2x + 6)^2} = 0 \) ⇒ \( x \approx -11.66, 5.66 \)
Evaluate \( f \) at critical points:
\( f(-11.66) \approx -18.7 \), \( f(5.66) \approx -1.34 \)
Answer: \( \boxed{y \leq -18.7} \) or \( \boxed{y \geq -1.34} \)

(f) Inequality Solution:

Solve \( \frac{x^2 – 14x + 24}{2x + 6} > x \):
Rearrange: \( \frac{-x^2 – 20x + 24}{2x + 6} > 0 \)
Critical points: \( x \approx -21.1, 1.14 \) (numerator), \( x = -3 \) (denominator)
Test intervals:
Answer: \( \boxed{x < -21.1} \) or \( \boxed{-3 < x < 1.14} \)