FINDING VERTICAL ASYMPTOTES

As introduced in the earlier section, Introduction to Asymptotes:

DEFINITION asymptote
An ‘asymptote’ (pronounced AS-sim-tote) is a curve (usually a line)
that another curve gets arbitrarily close to as $\,x\,$ approaches $\,+\infty\,$, $\,-\infty\,$, or a finite number.

In particular, here is the definition of a vertical asymptote:

DEFINITION vertical asymptote
A ‘vertical asymptote’ is a vertical line that another curve gets arbitrarily close to as $\,x\,$ approaches a finite number.

Specifically, the vertical line $\,x = c\,$ is a vertical asymptote for a function $\,f\,$
if and only if at least one of the following conditions is true:
  • as $\,x\rightarrow c^+\,$, $\,f(x)\rightarrow\pm\infty\,$
  • as $\,x\rightarrow c^-\,$, $\,f(x)\rightarrow\pm\infty\,$

The tangent function
has infinitely many vertical asymptotes.

For example,
the vertical line $\,x = \frac{\pi}{2}\,$ is a vertical asymptote.

as $\,x\rightarrow {\frac{\pi}{2}}^{+}\,$, $\,\tan x\rightarrow -\infty\,$

as $\,x\rightarrow {\frac{\pi}{2}}^{-}\,$, $\,\tan x\rightarrow \infty\,$

NOTES ABOUT VERTICAL ASYMPTOTES

VERTICAL ASYMPTOTES FOR RATIONAL FUNCTIONS

How can a rational function (a ratio of polynomials) ‘blow up’ near a finite input?
When you're ‘trying’ to divide by zero!
Remember—dividing by a very small number gives a very big number.

VERTICAL ASYMPTOTES for RATIONAL FUNCTIONS
A vertical asymptote for a rational function occurs where the denominator is zero,
and the numerator is nonzero.

In other words, for polynomials $\,N\,$ and $\,D\,$, the rational function $\displaystyle \,\frac{N(x)}{D(x)}\,$
has a vertical asymptote at $\,c\,$ wherever $\,D(c) = 0\,$ and $\,N(c)\ne 0\,$.

CHECKING THE BEHAVIOR OF A RATIONAL FUNCTION
NEAR A VERTICAL ASYMPTOTE

Given a vertical asymptote, you usually want to know how the function behaves nearby.
Are the outputs going to infinity? To negative infinity?
Although you could certainly use a graphing calculator or WolframAlpha to see this behavior,
you should also be able to determine it algebraically, as shown next.

EXAMPLE:
The function $\displaystyle\,f(x) = \frac{-3}{2x + 1}\,$ has a vertical asymptote at $\,x = -\frac 12\,$.

Why?   The denominator, $\,2x + 1\,$, is zero when $\,x = -\frac 12\,$, and the numerator is (always) nonzero.

Here is the thought process for determining what the outputs from $\,f\,$ look like, close to $\,-\frac 12\,$:

Consider values of $\,x\,$ a bit less than $\,-\frac 12\,$ (i.e., just to the left of $\,-\frac 12\,$):
  • the denominator is small and negative (see the graph of $\,y = 2x+1\,$)
  • the numerator is negative
Thus:   as $\,x\rightarrow {-\frac 12}^{-}\,$, $\,f(x) \approx \frac{\color{green}{(-)}}{\color{red}{(\text{small }-)}} \rightarrow +\infty\,$.

Observe that the notation   $\,\frac{(-)}{(\text{small }-)}\,$   is being used to denote a (normal-sized) negative number, divided by a small negative number—which produces a large positive number.

Next, consider values of $\,x\,$ a bit more than $\,-\frac 12\,$ (i.e., just to the right of $\,-\frac 12\,$):
  • the denominator is small and positive (see the graph of $\,y = 2x+1\,$)
  • the numerator is negative
Thus:   as $\,x\rightarrow {-\frac 12}^{+}\,$, $\,f(x) \approx \frac{\color{green}{(-)}}{\color{blue}{(\text{small }+)}} \rightarrow -\infty\,$.

Observe that we did not compute actual values of the output near $\,x = -\frac 12\,$!
There's no need—it would be working too hard.
All we need to know is:
  • the sign (positive or negative) of the numerator and denominator
  • the denominator is small (close to zero)
  • dividing by increasingly small numbers give increasingly big numbers
Master the ideas from this section
by practicing the exercise at the bottom of this page.

When you're done practicing, move on to:
puncture points (holes)
On this exercise, you will not key in your answer.
However, you can check to see if your answer is correct.
PROBLEM TYPES:
1 2 3 4 5 6 7 8 9 10 11 12
AVAILABLE MASTERED IN PROGRESS

(MAX is 12; there are 12 different problem types.)