Horizontal Asymptote of $Z(x)= \frac{1}{Q(x)}=\frac{1}{2x^4-12x^3-67x^2-14x+9}$ I just had a lecture about reciprocal functions. The horizontal asymptote rule is based on the degree of the function in the numerator and the degree of the function in the denominator.
The polynomial and the reciprocal functions that I am analyzing are $$Q(x)=2x^4-12x^3-67x^2-14x+9$$
$$Z(x)= \frac{1}{Q(x)}=\frac{1}{2x^4-12x^3-67x^2-14x+9}$$
The horizontal asymptote rule is based on the degree of the polynomial.
1.) If the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator then the horizontal asymptote is y=0.
2.) If the degree of the polynomial is equal the the degree of the polynomial in the denominator then the horizontal asymptote is a ratio of lead coefficients.
3.)  If the degree of the polynomial in the numerator is greater than the degree of the polynomial is greater than the degree of the polynomial in the denominator then there is no horizontal asymptote.
The reciprocal function that I am looking at seems to follow the first rule.  The numerator is not really a polynomial or a power function.  It is just a 1.  The denominator has a degree of 4.  The numerator is less than the denominator, therefore I would conclude the first rule applies.  Also when I look at the factored form of the reciprocal function I can see that zero is possible in the denominator in four different ways.  This to me suggests that there are asymptotes at y=0 as well.
Here is the factored form of the reciprocal function $$Z(x)= \frac{1}{2(x-0.27209)(x+.51993)(x+3.3235)(x-9.5713)}$$ The graph between factors x=0.27209 and x=-0.51993 dips down toward zero but then shoots back up.  How close does the function have to approach the horizontal asymptote to be considered an asymptote?  Is anybody else concerned about this topic?  I am trying to test the reciprocal function between these points by putting numbers into the equation.  I may have to wait until tomorrow to look closer at my equations but I am having a hard time trying to see how close the reciprocal function gets to zero between x=-0.51993 and x=0.27209.  The graph on the right is the one I am looking at.

 A: 
How close does the function have to approach the horizontal asymptote to be considered an asymptote?

Technically, infinitely close. This idea is summed up by a limit, and it is how we define a horizontal asymptote:
Given a function $f(x)$, if
$$
\lim_{x\to -\infty}f(x)=c\quad\text{or}\quad\lim_{x\to +\infty}f(x)=c,
$$
we say that the line $y=c$ is a horizontal asymptote.
In other words: we have a horizontal asymptote if the value of $f(x)$ tends towards some value number $c$ as the value of $x$ becomes infinitely large (in either of the positive or negative directions).
The horizontal asymptote rules you state are really describing the behaviour of rational functions as their input values get larger and larger:

*

*

If the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator then the horizontal asymptote is $y=0$.

This is true because larger-degreed polynomials grow much faster than lower-degreed ones. Eventually, the denominator will out-grow the numerator to an extent that makes the quotient effectively $0$.


*

If the degree of the polynomial is equal the degree of the polynomial in the denominator then the horizontal asymptote is a ratio of lead coefficients.

We can reason that the growth of two equally-degreed polynomials will be about the same. That is, the lower-degreed terms will be negligible for large values of $x$ and so we approach a value that is the ratio of the highest-degree terms in both polynomials.


*

If the degree of the polynomial in the numerator is greater than the degree of the polynomial is greater than the degree of the polynomial in the denominator then there is no horizontal asymptote.

The numerator will eventually outgrow the denominator, so the quotient will tend to infinity instead of approaching some value.
That being said, the horizontal asymptote of a graph $y=Z(x)$ in question is the line $y=0$. Does this answer your question?
