Consider the graphs of k(x) = ax2 and p(x) = mx + c. Determine the range of k, the domain of p, the equation of AD (given gradient is 1), and the equation of k if B is equidistant from A to C.

The line $p(x) = mx + c$ passes through points A, B, C, and D. The gradient of AD (which is the gradient of line $p$) is given as $m=1$.
Point C is $(0, -6)$. This is the y-intercept of the line.

The equation of the line is $p(x) = mx + c$.
Given $m=1$ and $c=-6$.

The equation of AD is $\boxed{p(x) = x - 6}$.

Point B is equidistant from A(6,0) and C(0,-6). This means B is the midpoint of the line segment AC.
Using the midpoint formula $B = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$:

Since B(3,-3) lies on the parabola $k(x) = ax^2$, substitute these coordinates into the equation:

The equation of $k$ is $\boxed{k(x) = -\frac{1}{3}x^2}$.

From 4.3, $p(x) = x - 6$.
From 4.4, $k(x) = -\frac{1}{3}x^2$.

Multiply by 3 to clear the fraction:

Rearrange into standard quadratic form $ax^2 + bx + c = 0$:

Factor the quadratic equation:

The possible x-coordinates for the intersection points are $x = -6$ or $x = 3$.

We know from 4.4 that point B has an x-coordinate of 3. Therefore, point D must have an x-coordinate of -6.
Substitute $x=-6$ into the equation of the line $p(x) = x - 6$ to find the y-coordinate of D:

The coordinates of D are $\boxed{(-6, -12)}$.

A = (6,0)
C = (0,-6)
O = (0,0) (the origin)

Length of OA:

Length of OC:

Length of AC:

Since $OA = OC = 6$, two sides of the triangle are equal in length.
Therefore, $\triangle ACO$ is an isosceles triangle.