Let's solve each part of the problem step by step.

Part 1: $f(x) \geq 0$

The function is $f(x) = \frac{x(x+3)}{x-1}$.

To find where $f(x) \geq 0$, we need to find the intervals where the function is non-negative. This requires analyzing the sign of the numerator and denominator.

1. Numerator: $x(x + 3)$

   • $x(x + 3) = 0$ at $x = 0$ or $x = -3$
   • For $x < -3$, $x(x + 3) > 0$ (positive).
   • For $-3 < x < 0$, $x(x + 3) < 0$ (negative).
   • For $x > 0$, $x(x + 3) > 0$ (positive).

2. Denominator: $x - 1$

   • $x - 1 = 0$ at $x = 1$
   • For $x < 1$, $x - 1 < 0$ (negative).
   • For $x > 1$, $x - 1 > 0$ (positive).

3. Combined Sign:

   • For $x \in (-\infty, -3) \cup (0, 1)$, the function is positive.
   • For $x \in (-3, 0) \cup (1, \infty)$, the function is negative.
   • At $x = 0$ and $x = -3$, the function equals zero.

Therefore, $f(x) \geq 0$ on the intervals $(-\infty, -3] \cup [0, 1]$.

So the correct answer is: $\textbf{]-\infty, -3] \cup [0, 1[}$.

Part 2: Horizontal/Oblique Asymptotes

To find the horizontal or oblique asymptotes of $f(x) = \frac{x(x+3)}{x-1}$, we consider the degrees of the polynomials:

• The degree of the numerator is 2.
• The degree of the denominator is 1.

Since the degree of the numerator is greater than the denominator, the function has an oblique asymptote found by polynomial long division.

$\frac{x^2 + 3x}{x - 1} = x + 4 + \frac{4}{x - 1}$

As $x \to \infty$, $\frac{4}{x - 1} \to 0$. Thus, the oblique asymptote is $y = x + 4$.

So the correct answer is: $\textbf{y = x + 4}$.

Part 3: Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (where the function is undefined):

$x - 1 = 0 \implies x = 1$

Therefore, the vertical asymptote is at $x = 1$.

So the correct answer is: $\textbf{only x = 1}$.

Part 4: Extreme Points

To find extreme points, we take the derivative and set it to zero:

$f(x) = \frac{x(x+3)}{x-1}$

$f'(x) = \frac{(x-1)(2x+3) - x(x+3)}{(x-1)^2} = \frac{2x^2 + 3x - 2x - 3 - x^2 - 3x}{(x-1)^2} = \frac{x^2 - 3}{(x-1)^2}$

Setting the numerator to zero:

[ x^2 - 3 =