The image shows graphs of the functions k(x) = ax2 and p(x) = mx + c. a) Determine the range of k. b) Determine the domain of p. c) Given that the gradient of AD is 1, determine the equation of AD in

The graph $k(x) = ax^2$ is a parabola with its vertex at the origin $(0,0)$. From the diagram, the parabola opens downwards.

Since the parabola opens downwards from the vertex $(0,0)$, the maximum y-value is 0, and all other y-values are less than or equal to 0.
The range of k is $\boxed{y \le 0}$.

The graph p is a straight line, given by $p(x) = mx + c$.

A straight line extends infinitely in both the positive and negative x-directions.
The domain of p is $\boxed{x \in \mathbb{R}}$.

The gradient of line AD (which is line p) is given as $m = 1$.
The line p passes through point C(0, -6). This is the y-intercept, so $c = -6$.

Substitute $m=1$ and $c=-6$ into the equation $p(x) = mx + c$:

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

Let B be $(x_B, y_B)$. A is $(6,0)$ and C is $(0,-6)$.
Since B is equidistant from A and C, $BA^2 = BC^2$.

From 4.3, the equation of line p is $y = x - 6$.
Substitute $y_B = -x_B$ into the line equation:

Now find $y_B$:

So, the coordinates of B are $(3, -3)$.

Point B(3, -3) lies on the parabola $k(