Just to elaborate on Michael Hardy's answer a little bit...
Suppose we have a rational function written like $f(x)=\frac{p(x)}{q(x)}$. Performing the long division $p(x)\div q(x)$ gives you a different way of writing the same function, $$f(x)=a(x)+\frac{r(x)}{q(x)},$$ as a polynomial plus a rational function whose numerator's degree is smaller than its denominator's degree. (To be more specific, $a(x)$ has degree equal to the degree of the original numerator $p(x)$ minus the degree of the denominator $q(x)$. The degree of $r(x)$ is smaller than that of $q(x)$ because $r(x)$ is the remainder of the long division.)
As you already understand, $\frac{r(x)}{q(x)}$ tends to zero as $x$ tends to infinity, because the degree of the numerator is smaller than the degree of its denominator. That means that as $x\rightarrow\infty$, the function $f(x)$ tends towards the polynomial $a(x)$.
If $p(x)$ has degree exactly one more than that of $q(x)$, then $a(x)$ has degree one: it's a line. We call this line a slant or oblique asymptote of $f(x)$. If the degree of $p(x)$ is more than one higher than that of $q(x)$, then $a(x)$ is a quadratic, or a cubic, or whatever, but it's not a line. So $f(x)$ doesn't have a slant asymptote in this direction. (We still might say something like $f(x)$ asymptotically approaches a(x).)
Hopefully this explains why asymptotes only occur when the degree of the numerator is exactly one more than that of the denominator. It also might give you a hint for how you can find slant asymptotes of functions that aren't rational: if you can rewrite your function as a line plus something that goes to zero, you've got yourself an asymptote!
It always helps to divide/factor by $x$ raised to the highest power to see why the rules work the way they do. The important (yet obvious) thing to keep in mind is that for any constant $c$, $\dfrac{c}{x} \to 0$ as $x \to \infty$.
For the first case, you have the right idea. For instance, let $f(x) = \dfrac{x^2+5x+100}{x^3-5}$. You need to find $\lim_\limits{x \to \infty} f(x)$:
$$\lim_{x \to \infty}\frac{x^2+5x+100}{x^3-5} = \lim_{x \to \infty}\frac{x^2\left(\frac{5}{x}+\frac{100}{x^2}\right)}{x^3\left(1-\frac{5}{x^3}\right)} = \lim_{x \to \infty}\frac{1}{x} = 0$$
This is just another way of saying that the greatest powers will outgrow the other terms of the function as $x \to \infty$, and if the degree of the denominator is higher, clearly this means it tends to $0$. On the other hand, for the opposite case, the limit clearly tends to $\pm \infty$ (sign depends on the sign of the leading coefficients of the numerator and denominator).
For the second case, you can once again show this through factoring. For instance, say you have a function in the form $f(x) = \dfrac{ax^2+bx+c}{dx^2+ex+f}$. You once again need to find $\lim_\limits{x \to \infty} f(x)$:
$$\lim_{x \to \infty} \frac{ax^2+bx+c}{dx^2+ex+f} = \lim_{x \to \infty} \frac{x^2\left(a+\frac{b}{x}+\frac{c}{x^2}\right)}{x^2\left(d+\frac{e}{x}+\frac{f}{x^2}\right)} = \frac{a}{d}$$
You can think of this as the $ax^2$ and $dx^2$ terms eventually outgrowing all the other terms as $x \to \infty$, so all that “remains” is $\dfrac{ax^2}{dx^2} = \dfrac{a}{d}$.
Can you use similar arguments to show the third case (such as when the numerator is a cubic and the denominator is a quadratic)? As a start, you can notice that the terms with lower powers will eventually vanish in both the numerator and denominator, and you can try to justify this with a limit.
Addition: You got the idea. Using the limit, it becomes apparent that due to the numerator begin a single power higher, all that will “remain” as $x \to \infty$ is $mx$. As you saw, however, the limit gives only the slope $m$ and not the $y$-intercept ($b$) of the slanted asymptote, which is in the form $y = mx+b$. For this purpose, you would have to calculate another limit as well:
$$\lim_{x \to \infty} f(x)-mx = b$$
This gives the difference between $f(x)$ and $mx$ as $x \to \infty$, which will give $b$:
$$\implies \lim_{x \to \infty} \frac{3x^3+6x^2+4x+2}{x^2+5x+2}-3x = \lim_{x \to \infty} \frac{3x^3+6x^2+4x+2-3x^3-15x^2-6x}{x^2+5x+2}$$
Simplifying and using case $(2)$, you get
$$\lim_{x \to \infty} \frac{-9x^2-2x+2}{x^2+5x+2} = -9$$
Putting it all together, you get $y = 3x-9$, which is the same result obtained via polynomial division.
Best Answer
Your work is correct. Note that your solutions are the ''more simple'' rational functions that satisfies the requests. Obviously you can find infinitely many other rational functions that do the same, but have some other property.