(a) The vertical asymptote occurs where the denominator is zero, so \(x = 3\). For the oblique asymptote, perform polynomial long division on \(\frac{x^2 + 2x + 1}{x - 3}\):

\(y = \frac{(x-3)(x+5) + 16}{x-3} = x + 5 + \frac{16}{x-3}\)

Thus, the oblique asymptote is \(y = x + 5\).

(b) Differentiate \(y = \frac{x^2 + 2x + 1}{x - 3}\) to find turning points:

\(\frac{dy}{dx} = 1 - \frac{16}{(x-3)^2}\)

Set \(\frac{dy}{dx} = 0\):

\(1 - \frac{16}{(x-3)^2} = 0 \Rightarrow (x-3)^2 = 16\)

Solving gives \(x = -1\) and \(x = 7\). Substitute back to find \(y\)-coordinates: \((-1, 0)\) and \((7, 16)\).

(c) Sketch the curve using the asymptotes and turning points found.

(d) For \(y = \left| \frac{x^2 + 2x + 1}{x - 3} \right|\), reflect the negative parts of the curve above the x-axis. For \(y^2 = \frac{x^2 + 2x + 1}{x - 3}\), consider both positive and negative square roots, sketching both branches.