(a) The vertical asymptote occurs where the denominator is zero: \(x = 2\). For the oblique asymptote, divide the numerator by the denominator: \(x^2 + 2x - 15 = (x - 2)(x + 4) - 7\), giving \(y = x + 4\).

(b) Differentiate \(y\) to find stationary points: \(\frac{dy}{dx} = \frac{(x - 2)(2x + 2) - (x^2 + 2x - 15)}{(x - 2)^2} = \frac{x^2 - 4x + 11}{(x - 2)^2}\). The quadratic \(x^2 - 4x + 11 = 0\) has no real roots (discriminant \(4^2 - 4 \times 1 \times 11 = -28 < 0\)), so no stationary points.

(c) The curve intersects the y-axis at \(x = 0\): \(y = \frac{0^2 + 2 \times 0 - 15}{0 - 2} = 7.5\). Intersects the x-axis where \(y = 0\): \(x^2 + 2x - 15 = 0\), giving roots \(x = -5\) and \(x = 3\).

(d) Sketch the absolute value function based on the sketch from (c), reflecting the negative parts above the x-axis.

(e) Solve \(\left| \frac{2x^2 + 4x - 30}{x - 2} \right| < 15\). Critical points from \(2x^2 + 4x - 30 = 15(x - 2)\) and \(2x^2 + 4x - 30 = -15(x - 2)\) give \(x = 0, \frac{11}{2}, -12, \frac{5}{2}\). Test intervals to find \(-12 < x < 0\) or \(\frac{5}{2} < x < \frac{11}{2}\).