A parabola is defined by the equation f(x) = (x−4)2 − 9. Find the turning point E, the x-intercept A (with the smaller x-value), the y-intercept C, and the reflection M of C in the axis of symmetry. T

4.1
The equation of the parabola is $f(x) = (x-4)^2 - 9$.
This is in the form $y = a(x-p)^2 + q$, where $(p, q)$ is the turning point.
Comparing the given equation, the turning point E is $(4, -9)$.
The coordinates of E are $\boxed{(4, -9)}$.

4.2
A is an x-intercept of $f(x)$. To find the x-intercepts, set $f(x) = 0$.

$$(x-4)^2 - 9 = 0$$ $$(x-4)^2 = 9$$

Take the square root of both sides:

$$x-4 = \pm\sqrt{9}$$ $$x-4 = \pm 3$$

Solve for $x$:

$$x-4 = 3 \implies x = 7$$ $$x-4 = -3 \implies x = 1$$

From the graph, A is the x-intercept with the smaller x-value.
The coordinates of A are $\boxed{(1, 0)}$.

4.3
First, find the coordinates of C. C is the y-intercept of $f(x)$. Set $x=0$.

$$f(0) = (0-4)^2 - 9$$ $$f(0) = (-4)^2 - 9$$ $$f(0) = 16 - 9$$ $$f(0) = 7$$

So, C is $(0, 7)$.
The axis of symmetry of $f(x) = (x-4)^2 - 9$ is the vertical line $x=4$ (the x-coordinate of the turning point).
M is the reflection of C$(0, 7)$ in the line $x=4$.
The x-coordinate of C is 0. The distance from C to the axis of symmetry ($x=4$) is $4-0=4$ units.
The x-coordinate of M will be $4+4=8$.
The y-coordinate remains the same during reflection across a vertical line.
The coordinates of M are $\boxed{(8, 7)}$.

4.4
The equation of line $g$ is in the form $y = mx+c$.
C is the y-intercept of $f$ and $g$. From 4.3, C is $(0, 7)$. So, $c=7$.
The equation of $g$ is $y = mx+7$.
D is the x-intercept of $g$.
DE is parallel to the y-axis, which means D and E have the same x-coordinate.
From 4.1, E is $(4, -9)$. So, the x-coordinate of D is 4.
Since D is an x-intercept, its y-coordinate is 0.
So, D is $(4, 0)$.
Now, calculate the gradient $m$ using points C$(0, 7)$ and D$(4, 0)$:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{0 - 7}{4 - 0} = \frac{-7}{4}$$

The equation of $g$ is $\boxed{y = -\frac{7}{4}x + 7}$.

4.5
To find the inverse function $g^{-1}$, swap $x$ and $y$ in the equation for $g$ and solve for $y$.
The equation of $g$ is $y = -\frac{7}{4}x + 7$.
Swap $x$ and $y$:

$$x = -\frac{7}{4}y + 7$$

Subtract 7 from both sides:

$$x - 7 = -\frac{7}{4}y$$

Multiply by $-\frac{4}{7}$:

$$-\frac{4}{7}(x - 7) = y$$ $$y = -\frac{4}{7}x + \left(-\frac{4}{7}\right)(-7)$$ $$y = -\frac{4}{7}x + 4$$

The equation of $g^{-1}$ is $\boxed{y = -\frac{4}{7}x + 4}$.

4.6
We need to find the values of $x$ for which $x(f(x)) \le 0$.
This inequality holds true if:

1. $x \ge 0$ AND $f(x) \le 0$
2. $x \le 0$ AND $f(x) \ge 0$

From 4.2, the x-intercepts of $f(x)$ are $x=1$ (point A) and $x=7$ (point B).
Since $f(x)$ is an upward-opening parabola, $f(x) \le 0$ when $x$ is between or equal to the x-intercepts.
So, $f(x) \le 0$ for $1 \le x \le 7$.
And $f(x) \ge 0$ for $x \le 1$ or $x \ge 7$.

Now, let's apply these to the two cases:

Case 1: $x \ge 0$ AND $f(x) \le 0$
This means $x \in [0, \infty)$ AND $x \in [1, 7]$.
The intersection of these two conditions is $1 \le x \le 7$.

Case 2: $x \le 0$ AND $f(x) \ge 0$
This means $x \in (-\infty, 0]$ AND ($x \in (-\infty, 1]$ or $x \in [7, \infty)$).
The intersection of these two conditions is $x \le 0$.

Combining the results from Case 1 and Case 2, the values of $x$ for which $x(f(x)) \le 0$ are:
$\boxed{x \le 0 \text{ or } 1 \le x \le 7}$.