The given function is h(x) = (6)/(x-2) - 1.

The denominator cannot be zero, so $x-2 \neq 0$.

The vertical asymptote is $x=2$.

The constant term is $-1$.
The horizontal asymptote is $y=-1$.

The equations of the asymptotes are:
Vertical asymptote: $\boxed{x=2}$
Horizontal asymptote: $\boxed{y=-1}$

Set $h(x) = 0$:

The x-intercept is $\boxed{(8, 0)}$.

Set $x = 0$:

The y-intercept is $\boxed{(0, -4)}$.

The numerator is zero when $8-x=0 \implies x=8$.
The denominator is zero when $x-2=0 \implies x=2$.
These points divide the number line into intervals: $(-\infty, 2)$, $(2, 8)$, $(8, \infty)$.

• For $x < 2$ (e.g., $x=0$): $\frac{8-0}{0-2} = \frac{8}{-2} = -4$. Since $-4 < 0$, the inequality holds for $x < 2$.
• For $2 < x < 8$ (e.g., $x=3$): $\frac{8-3}{3-2} = \frac{5}{1} = 5$. Since $5 > 0$, the inequality does not hold for $2 < x < 8$.
• For $x > 8$ (e.g., $x=9$): $\frac{8-9}{9-2} = \frac{-1}{7}$. Since $\frac{-1}{7} < 0$, the inequality holds for $x > 8$.

So, $h(x) < 0$ when $x < 2$ or $x > 8$.

We need to find the intersection of $(-\infty, 2) \cup (8, \infty)$ with $[0, \infty)$.

• For $x < 2$: The part within $[0, \infty)$ is $[0, 2)$.
• For $x > 8$: The part within $[0, \infty)$ is $(8, \infty)$.

The values of $x$ for which $h(x) < 0$ for $x \in [0; \infty)$ are $\boxed{x \in [0, 2) \cup (8, \infty)}$.