Show that g(x) = -x2 + 4x + 5. Calculate the average gradient of g between A(0; 5) and C(2; 9). Find the length of BD, given that B is the negative x-intercept of g(x) and D is the negative x-intercep

The general form of a parabola with a turning point $(p, q)$ is $g(x) = a(x-p)^2 + q$.
Given the turning point C(2; 9), we substitute $p=2$ and $q=9$:

Given the y-intercept A(0; 5), we substitute $x=0$ and $g(x)=5$ to find $a$:
\begin{align*} 5 &= a(0-2)^2 + 9 \ 5 &= a(-2)^2 + 9 \ 5 &= 4a + 9 \ 4a &= 5 - 9 \ 4a &= -4 \ a &= -1 \end{align*}
Now substitute $a=-1$ back into the equation:
\begin{align*} g(x) &= -1(x-2)^2 + 9 \ g(x) &= -(x^2 - 4x + 4) + 9 \ g(x) &= -x^2 + 4x - 4 + 9 \ g(x) &= -x^2 + 4x + 5 \end{align*}
This matches the required form.

The coordinates are A(0; 5) and C(2; 9).
The formula for the average gradient (slope) between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

The average gradient is $\boxed{2}$.

First, find the x-intercepts of $f(x) = \frac{1}{2}x^2 - 8$. Set $f(x) = 0$:
\begin{align*} \frac{1}{2}x^2 - 8 &= 0 \ \frac{1}{2}x^2 &= 8 \ x^2 &= 16 \ x &= \pm 4 \end{align*}
From the graph, B is the negative x-intercept, so B(-4; 0).

Next, find the x-intercepts of $g(x) = -x^2 + 4x + 5$. Set $g(x) = 0$:
\begin{align*} -x^2 + 4x + 5 &= 0 \ x^2 - 4x - 5 &= 0 \ (x-5)(x+1) &= 0 \ x = 5 \quad &or \quad x = -1 \end{align*}
From the graph, D is the positive x-intercept, so D(5; 0).

The length of BD is the distance between B(-4; 0) and D(5; 0):

The length of BD is $\boxed{9 \text{ units}}$.

The x-intercepts of $f(x)$ are $x = -4$ and $x = 4$.
Since $f(x) = \frac{1}{2}x^2 - 8$ is a parabola opening upwards, $f(x) \ge 0$ when $x$ is outside or on the x-intercepts.
Therefore, $x \le -4$ or $x \ge 4$.
The solution is $\boxed{x \le -4 \text{ or } x \ge 4}$.

For $f(x) = \frac{1}{2}x^2 - 8$: This parabola opens upwards and its turning point is at $x=0$. So, $f(x)$ is strictly increasing for $x > 0$.
For $g(x) = -x^2 + 4x + 5$: This parabola opens downwards and its turning point is at $x = -\frac{4}{2(-1)} = 2$. So, $g(x)$ is strictly increasing for $x < 2$.
For both functions to be strictly increasing, we need the intersection of these two intervals:
$x > 0$ AND $x < 2$.
The solution is $\boxed{0 < x < 2}$.