(a) To find the vertical asymptote, set the denominator equal to zero: \(x + 1 = 0\), giving \(x = -1\).

For the oblique asymptote, perform polynomial long division on \(\frac{x^2 + x + 9}{x + 1}\):

\(y = x + \frac{9}{x + 1}\), so the oblique asymptote is \(y = x\).

(b) To find stationary points, differentiate \(y = \frac{x^2 + x + 9}{x + 1}\) and set \(\frac{dy}{dx} = 0\).

\(\frac{dy}{dx} = 1 - \frac{9}{(x+1)^2} = 0 \Rightarrow (x+1)^2 = 9\).

Solving gives \(x + 1 = 3\) or \(x + 1 = -3\), so \(x = 2\) or \(x = -4\).

Substitute back to find \(y\):

For \(x = 2\), \(y = \frac{2^2 + 2 + 9}{2 + 1} = 5\), giving point \((2, 5)\).

For \(x = -4\), \(y = \frac{(-4)^2 - 4 + 9}{-4 + 1} = -7\), giving point \((-4, -7)\).