The figure below shows part of the curve whose equation is given by y = (6−x)(x+2) and the line y = mx + c. (a) Find the coordinates of the point A and B. (b) Determine the equation of the straight line y = mx + c. (c) Find the area between the curve y = (6−x)(x+2) and the line y = mx + c.

Substitute $x=0$ into the curve equation:

So, the coordinates of A are $(0, 12)$.

Set $y=0$ in the curve equation:

This gives two possible x-values: $6-x=0 \implies x=6$ or $x+2=0 \implies x=-2$.
From the graph, point B is the positive x-intercept.
So, the coordinates of B are $(6, 0)$.
The coordinates of A and B are $\boxed{A(0, 12) \text{ and } B(6, 0)}$.

The slope $m$ is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Since point A is $(0, 12)$, the y-intercept $c$ is 12.
Substitute $m=-2$ and $c=12$ into $y=mx+c$.
The equation of the straight line is $\boxed{y = -2x + 12}$.

The area between two curves $f(x)$ and $g(x)$ from $x=a$ to $x=b$ is $\int_{a}^{b} (f(x) - g(x)) \,dx$, where $f(x)$ is the upper curve.
Let $f(x) = -x^2 + 4x + 12$ (curve) and $g(x) = -2x + 12$ (line).
In the interval $[0, 6]$, the curve is above the line.
The difference between the curve and the line is:

The area is given by the integral:

Integrate term by term:

The area between the curve and the line is $\boxed{36 \text{ units}^2}$.