Answer: \(h=5\); \(x_A=2+2\sqrt5\); angle \(=65.9^\circ\) or \(1.15\text{ rad}\).

Start by taking the key information from the graph or diagram, such as intercepts, turning points, amplitudes, periods, or intersection points. Then use the relevant algebra to justify the result.

The curve is \(y=x-\frac{x^2}{4}\).

Differentiate:

\(\frac{\mathrm dy}{\mathrm dx}=1-\frac{x}{2}.\)

At the maximum point, \(1-\frac{x}{2}=0\), so \(x=2\). Then

\(y=2-\frac{2^2}{4}=1.\)

The perpendicular distance from \(y=1\) to the line \(y=-4\) is \(h=5\).

For point \(A\), set the curve equal to \(-4\):

\(-4=x-\frac{x^2}{4}.\)

Multiplying by 4 gives \(x^2-4x-16=0\).

Using the quadratic formula,

\(x=\frac{4\pm\sqrt{16+64}}{2}=2\pm2\sqrt5.\)

Point \(A\) is the right-hand intersection, so \(x_A=2+2\sqrt5\).

The gradient of the tangent at \(A\) is

\(1-\frac{2+2\sqrt5}{2}=-\sqrt5.\)

The acute angle with the horizontal line \(y=-4\) therefore satisfies

\(\tan\theta=\sqrt5.\)

Thus \(\theta=\tan^{-1}\sqrt5=65.9^\circ\), or \(1.15\text{ rad}\).

This gives the required answer: \(h=5\); \(x_A=2+2\sqrt5\); angle \(=65.9^\circ\) or \(1.15\text{ rad}\).