Find the x-intercepts, axis of symmetry, range of f(x) = -x2 - x + 6, and the equation of g(x) passing through a point.

To find the x-intercepts, set $f(x) = 0$:
\begin{align*} -x^2 - x + 6 &= 0 \ x^2 + x - 6 &= 0 \ (x+3)(x-2) &= 0 \end{align*}
This gives two x-intercepts:

The x-intercepts are $\boxed{(-3; 0) \text{ and } (2; 0)}$.

The axis of symmetry for a parabola $f(x) = ax^2 + bx + c$ is given by the formula $x = -\frac{b}{2a}$.
For $f(x) = -x^2 - x + 6$, we have $a = -1$ and $b = -1$.

The equation of the axis of symmetry is $\boxed{x = -\frac{1}{2}}$.

Since $a = -1 < 0$, the parabola opens downwards, meaning it has a maximum value.
The maximum value occurs at the axis of symmetry, $x = -\frac{1}{2}$.
Substitute $x = -\frac{1}{2}$ into $f(x)$:
\begin{align*} f\left(-\frac{1}{2}\right) &= -\left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right) + 6 \ &= -\frac{1}{4} + \frac{1}{2} + 6 \ &= -\frac{1}{4} + \frac{2}{4} + \frac{24}{4} \ &= \frac{25}{4} \end{align*}
The maximum y-value is $\frac{25}{4}$. Since the parabola opens downwards, the range is all y-values less than or equal to this maximum.
The range of $f$ is $\boxed{y \le \frac{25}{4}}$.

The line $g(x)$ passes through point A(-2; 4).
From the graph, D is the y-intercept of $f(x)$. To find D, set $x=0$ in $f(x)$:

So, point D is (0; 6).
Now, calculate the gradient $m$ of the line passing through A(-2; 4) and D(0; 6):

Since D(0; 6) is the y-intercept, $c=6$.
The equation of $g$ is $\boxed{g(x) = x + 6}$.

The average gradient between points A and D is the slope of the line segment connecting them. Since A and D are points on the line $g(x)$, this is simply the gradient of $g(x)$.
From Step 4, the gradient is $m=1$.
The average gradient between A and D is $\boxed{1}$.

Leave your answer in the form $h(x) = a(x+p)^2 + q$.
First, write $f(x)$ in the form $f(x) = a(x+p)^2 + q$. The turning point of $f(x)$ is $(-\frac{1}{2}, \frac{25}{4})$ and $a=-1$.

Reflection about the x-axis means $y \to -y$, so $f(x) \to -f(x)$:

Translation 3 units to the right means replacing $x$ with $(x-3)$:
\begin{align*} h(x) &= ((x-3) + \frac{1}{2})^2 - \frac{25}{4} \ &= (x - \frac{6}{2} + \frac{1}{2})^2 - \frac{25}{4} \ &= (x - \frac{5}{2})^2 - \frac{25}{4} \end{align*}
The equation of $h$ is $\boxed{h(x) = (x - \frac{5}{2})^2 - \frac{25}{4}}$.

From Step 1, the x-intercepts of $f(x)$ are $x = -3$ and $x = 2$.
Since $f(x) = -x^2 - x + 6$ is a parabola opening downwards, $f(x) > 0$ for the x-values between its x-intercepts.
The values of $x$ for which $f(x) > 0$ are $\boxed{-3 < x < 2}$.

Set $f(x) = 4$:
\begin{align*} -x^2 - x + 6 &= 4 \ -x^2 - x + 2 &= 0 \ x^2 + x - 2 &= 0 \ (x+2)(x-1) &= 0 \end{align*}
The solutions are $x = -2$ or $x = 1$.
We are given $f(p) = 4$ and $f(r) = 4$, so $p$ and $r$ are these two x-values.
We are also given that $r < 0$. Therefore, $r = -2$.
This means $p$ must be the other value