The original stationary point is at \((p, q)\).

The transformation applied to the curve is \(y = -3f\left(\frac{1}{4}(x + 8)\right)\).

This transformation involves:

• A horizontal translation of \(-8\) units (due to \(x + 8\)).
• A horizontal scaling by a factor of \(4\) (due to \(\frac{1}{4}x\)).
• A vertical scaling by a factor of \(-3\) (due to \(-3f(x)\)).

For the x-coordinate:

Start with the horizontal translation: \(x' = x + 8\), so \(x = x' - 8\).

Apply the horizontal scaling: \(x'' = 4x\), so \(x'' = 4(x' - 8) = 4p - 8\).

For the y-coordinate:

Apply the vertical scaling: \(y'' = -3y\), so \(y'' = -3q\).

Thus, the coordinates of the stationary point for the transformed curve are \((4p - 8, -3q)\).