Question 583086
a) Vertical asymptotes. like x=-2, happen only when a denominator is zero. 
{{{f(x)=1/(x+2)}}} has a vertical asymptote at x=-2
As x approaches -2, the denominator approaches zero and the absolute value of f(x) grows without bounds. The graph of {{{f(x)=1/(x+2)}}}  looks like this:
{{{graph(300,300,-5,1,-50,50,1/(x+2))}}} Coincidentally that function also has {{{y=0}}} as an asymptote, as you can see from the graph.
As the absolute value of x (and consequently of x+2) grows larger, and larger, f(x) grows  closer and closer to zero. A {{{y=0}}} horizontal asymptote happens when your rational function is a quotient and the denominator polynomial has a higher degree than the numerator.
CAUTION: Not every time a denominator is zero, you have a vertical asymptote.
If you make sure that the denominator, and only the denominator is zero at x=-2, you can be sure that the function will have an x=-2 asymptote.
If the numerator and denominator are zero at the same time, the function can be equivalent to another function that does not have a vertical asymptote.
For example, {{{p(x)=(x+2)/(x+2)}}} is equivalent to {{{q(x)=1}}} for all values of x except x=-2, and you know that {{{q(x)=1}}}  graphs as a horizontal line with {{{y=1}}}, and does not have a vertical asymptote. The graph for {{{p(x)=(x+2)/(x+2)}}} looks just like the same horizontal y=1 line, except for a hole at x=-2, where p(x) does not exist. 
b) From what I said above, you must realize that for a vertical {{{x=4}}} asymptote, you need the denominator to be zero for {{{x=4}}}.
{{{g(x)=1/(x-4)}}} would work. It also has a {{{y=0}}} asymptote, because the denominator, x-4 has degree 1, and the numerator, 1, has degree zero.
c) {{{h(x)=1/((x-1)(x-2))}}} has asymptotes {{{x=1}}} and {{{x=2}}} because those are zeros of the denominator.
The only horizontal asymptote for {{{h(x)=1/((x-1)(x-2))}}} is {{{y=0}}} and we need {{{y=1}}}, but that is easy to fix: we just add 1.
{{{m(x)=1+h(x)=1+1/((x-1)(x-2))}}}  has {{{x=1}}}, {{{x=2}}} and {{{y=1}}} asymptotes.
You can make it look fancier:
{{{m(x)=1+1/((x-1)(x-2))}}}={{{(x-1)(x-2)/((x-1)(x-2))}}}+{{{1/((x-1)(x-2))}}}={{{(x^2-3x+2)/(x^2-3x+2)}}}+{{{1/(x^2-3x+2)}}}={{{(x^2-3x+3)/(x^2-3x+2)}}}
d) For an {{{x=0}}} asymptote we want x as a factor in the denominator, but not tin the numerator. A {{{y=0}}} asymptote would be easier, but you saw in part c) how you can get a horizontal asymptote at a different y value
{{{r(x)=-1+1/x}}}={{{(1-x)/x}}} has {{{x=0}}} and {{{y=-1}}} asymptotes.
{{{s(x)=-1}}}+{{{1/x^2}}}={{{(1-x^2)/x^2}}} would work too.