Week of October 21, 2019

Asymptotes

In general, you will be given a rational (fractional) function, and you will need to find the domain and any asymptotes. You'll need to find the vertical asymptotes, if any, and then figure out whether you've got a horizontal or slant asymptote, and what it is. To make sure you arrive at the correct (and complete) answer, you will need to know what steps to take and how to recognize the different types of asymptotes.

Let's get some practice:

• Find the domain and all asymptotes of the following function:

$\mathbf{\color{green}{\mathit{y} = \dfrac{\mathit{x}^2 + 3\mathit{x} + 1}{4\mathit{x}^2 - 9}}}$

I'll start with the vertical asymptotes.

They (and any restrictions on the domain) will be generated by the zeroes of the denominator, so I'll set the denominator equal to zero and solve.

$4x^2 - 9 = 0$

$4x^2 = 9$

$x^2 = \dfrac{9}{4}$

$x = \pm \dfrac{3}{2}$

Then the domain is all x-values other than $\pm \frac{3}{2}$, and the two vertical asymptotes are at $x = \pm \frac{3}{2}$.

Next I'll turn to the issue of horizontal or slant asymptotes.

Since the degrees of the numerator and the denominator are the same (each being 2), then this rational has a non-zero (that is, a non-x-axis) horizontal asymptote, and does not have a slant asymptote. The horizontal asymptote is found by dividing the leading terms:

$y = \dfrac{x^2}{4x^2} = \dfrac{1}{4}$

Then the full answer is:

domain: $\mathbf{\color{purple}{ \mathit{x} \neq \pm \frac{3}{2} }}$

vertical asymptotes: $\mathbf{\color{purple}{ \mathit{x} = \pm \frac{3}{2} }}$

horizontal asymptote: $\mathbf{\color{purple}{ \mathit{y} = \frac{1}{4} }}$

slant asymptote: none

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A given rational function may or may not have a vertical asymptote (depending upon whether the denominator ever equals zero), but (at this level of study) it will always have either a horizontal or else a slant asymptote.

Note, however, that the function will only have one of these two; you will have either a horizontal asymptote or else a slant asymptote, but not both. As soon as you see that you have one of them, don't bother looking for the other one.

• Find the domain and all asymptotes of the following function:

$\mathbf{\color{green}{\mathit{y} = \dfrac{\mathit{x} + 3}{\mathit{x}^2 + 9}}}$

The vertical asymptotes come from the zeroes of the denominator, so I'll set the denominator equal to zero and solve.

x2 + 9 = 0

x2 = –9

Oops! This has no solution. (Duh! The denominator is a sum of squares, not a difference. So of course it doesn't factor and it can't have real zeroes. I should remember to look out for this, and save myself some time in the future.)

Since the denominator has no zeroes, then there are no vertical asymptotes and the domain is "all x".

Since the degree is greater in the denominator than in the numerator, the y-values will be dragged down to the x-axis and the horizontal asymptote is therefore "y = 0". Since I have found a horizontal asymptote, I don't have to look for a slant asymptote.

My full answer is:

domain: all x

vertical asymptotes: none

horizontal asymptote: y = 0 (the x-axis)

slant asymptote: none

The Special Case with the "Hole"

We've dealt with various sorts of rational functions. When you were first introduced to rational expressions, you likely learned how to simplify them. You'd factor the polynomials top and bottom, if you could, and then you'd see if anything cancelled off.

What if you've found the zeroes of the denominator of a rational function (so you've found the spots disallowed in the domain), but one or another of the factors cancels off? Let's look at an example of exactly that situation:

• Find the domain and all asymptotes of the following function:

$\mathbf{\color{green}{\mathit{y} = \dfrac{\mathit{x}^2 - \mathit{x} - 2}{\mathit{x} - 2}}}$

It so happens that this function can be simplified as:

$y = \dfrac{x^2 - x - 2}{x - 2}$

$= \dfrac{(x - 2)(x + 1)}{x - 2}$

$= \dfrac{\color{blue}{(x - 2)}(x + 1)}{\color{blue}{(x - 2)}}$

$= x + 1$

So the entire rational function simplifies to a linear function. Clearly, the original rational function is at least nearly equal to y = x + 1 — though I need to keep in mind that, in the original function, x couldn't take on the value of 2. But what about the vertical asymptote? Is there one at x = 2, or isn't there?

If there is a vertical asymptote, then the graph must climb up or down it when I use x-values close to the restricted value of x = 2. I'll try a few x-values to see if that's what's going on.

x = 1.5, y = 2.5

x = 1.9, y = 2.9

x = 1.95, y = 2.95

x = 1.99, y = 2.99