The function is given by f(x) = (4)/(x-3) + 4. Calculate the coordinates of M, D and the value of t if y=x+t is a line of symmetry.

Here are the solutions to the questions:

4.1 Write down the coordinates of M.
The function is given by $f(x) = \frac{4}{x-3} + 4$.
For a function of the form $y = \frac{a}{x-p} + q$, the vertical asymptote is $x=p$ and the horizontal asymptote is $y=q$. The point of intersection of the asymptotes is $(p, q)$.
In this case, $p=3$ and $q=4$.
So, the vertical asymptote is $x=3$ and the horizontal asymptote is $y=4$.
The point M is the intersection of these asymptotes.

$$M = (3, 4)$$

The coordinates of M are $\boxed{(3, 4)}$.

4.2 Calculate the coordinates of D.
D is the y-intercept of $f(x)$. To find the y-intercept, set $x=0$.

$$f(0) = \frac{4}{0-3} + 4$$ $$f(0) = \frac{4}{-3} + 4$$ $$f(0) = -\frac{4}{3} + \frac{12}{3}$$ $$f(0) = \frac{8}{3}$$

The coordinates of D are $\boxed{\left(0, \frac{8}{3}\right)}$.

4.3 If $y=x+t$ is the equation of a line of symmetry of $f$, calculate the value of $t$.
The lines of symmetry for a hyperbola of the form $y = \frac{a}{x-p} + q$ pass through the point of intersection of the asymptotes $(p, q)$ and have gradients 1 and $-1$.
From 4.1, the point of intersection of the asymptotes is M $(3, 4)$.
The given line of symmetry is $y=x+t$, which has a gradient of 1.
Substitute the coordinates of M $(3, 4)$ into the equation $y=x+t$:

$$4 = 3 + t$$ $$t = 4 - 3$$ $$t = 1$$

The value of $t$ is $\boxed{1}$.

4.4 Determine the values of $x$ for which $f(x) \le 0$.
We need to solve the inequality $\frac{4}{x-3} + 4 \le 0$.

$$\frac{4}{x-3} \le -4$$

Divide by 4:

$$\frac{1}{x-3} \le -1$$

Consider two cases for $x-3$:

• Case 1: $x-3 > 0 \implies x > 3$ Multiply both sides by $(x-3)$ (the inequality sign does not change):

  $$1 \le -1(x-3)$$ $$1 \le -x + 3$$ $$x \le 3 - 1$$ $$x \le 2$$

  This contradicts our assumption $x > 3$. So, there are no solutions in this case.

• Case 2: $x-3 < 0 \implies x < 3$ Multiply both sides by $(x-3)$ (the inequality sign reverses):

  $$1 \ge -1(x-3)$$ $$1 \ge -x + 3$$ $$x \ge 3 - 1$$ $$x \ge 2$$

  Combining with our assumption $x < 3$, we get $2 \le x < 3$.

The values of $x$ for which $f(x) \le 0$ are $\boxed{2 \le x < 3}$.

4.5 Calculate the coordinates of A.
A is the point on $f$ closest to M. These points lie on the lines of symmetry. From the graph, A is in the first quadrant, so it lies on the line of symmetry with a positive gradient, which is $y=x+1$ (from 4.3).
To find the coordinates of A, we set $f(x) = x+1$:

$$\frac{4}{x-3} + 4 = x+1$$ $$\frac{4}{x-3} = x+1-4$$ $$\frac{4}{x-3} = x-3$$

Multiply both sides by $(x-3)$:

$$4 = (x-3)^2$$

Take the square root of both sides:

$$\pm\sqrt{4} = x-3$$ $$\pm 2 = x-3$$

This gives two possible values for $x$:

1. $2 = x-3 \implies x = 5$
2. $-2 = x-3 \implies x = 1$

If $x=5$, then $y = 5+1 = 6$. So, $(5, 6)$.
If $x=1$, then $y = 1+1 = 2$. So, $(1, 2)$.
From the graph,