Answer: \(A=-3\), \(B=-4\); the stationary point is \(\left(-\frac43,-3e^6\right)\).

Use differentiation to find the gradient information needed. Stationary points occur when the derivative is zero, and tangents or normals are found from the gradient at the given point.

\(y=\frac{e^{-3x+2}}{x+1}\).

Differentiate using the quotient rule:

\(\frac{dy}{dx}=\frac{(x+1)(-3e^{-3x+2})-e^{-3x+2}}{(x+1)^2}\).

Factorising the numerator gives

\(\frac{dy}{dx}=\frac{e^{-3x+2}(-3x-4)}{(x+1)^2}\).

So \(A=-3\) and \(B=-4\).

For a stationary point, \(\frac{dy}{dx}=0\). Since \(e^{-3x+2}\ne0\), and \((x+1)^2\ne0\) for \(x\lt -1\), the only possible zero comes from

\(-3x-4=0\).

Hence \(x=-\frac43\). This is the only stationary point.

Substitute \(x=-\frac43\):

\(y=\frac{e^{-3(-4/3)+2}}{-4/3+1}=\frac{e^6}{-1/3}=-3e^6\).

This gives the required answer: \(A=-3\), \(B=-4\); the stationary point is \(\left(-\frac43,-3e^6\right)\).