PreCalculus

12.1. Vertical asymptotes

Given the rational function in which P(x) and Q(x) are polynomial functions:

        P(x)
f(x) = ------
        Q(x)

If f(x) -> +∞ or f(x) -> -∞ as x -> c then the line x = c is a vertical asymptote. Furthermore, the line x = c is a vertical asymptote of the graph of f(x) if and only if the denominator Q(c) = 0 and the numerator P(c) ≠ 0. That is, vertical asymptotes are located at the zeroes of the denominator Q(x). A graph may approach a vertical asymptote from either side but will never cross over it. There are four cases:

As x approaches c from the left f(x) increases without bound:
```
lim f(x) = +∞
^{x -> c-}
```
As x approaches c from the left f(x) decreases without bound.
```
lim f(x) = -∞
^{x -> c-}
```
As x approaches c from the right f(x) increases without bound.
```
lim f(x) = +∞
^{x -> c+}
```
As x approaches c from the right f(x) decreases without bound.
```
lim f(x) = -∞
^{x -> c+}
```

For example, find the vertical asymptotes of:

f(x) = 2x/(x² - 4)

Factor the denominator.

         2x          2x
f(x) = ------ = --------------
       x² - 4   (x + 2)(x - 2)

Then the vertical asymptotes are located at the zeroes of the denominator:

x = -2 and x = +2

Another example, graph the function:

          x² - 4
f(x) = -----------
       x² + 3x + 2

Factor the numerator and denominator and simplify:

       (x + 2)(x - 2)   (x - 2)
f(x) = -------------- = -------
       (x + 1)(x + 2)   (x + 1)

Then the vertical asymptote is given by:

x = -1

The x-intercept is found by solving:

0 = (x - 2)/(x + 1)

x = 2

The y intercept is:

f(0) = (0 - 2)/(0 + 1) = -2

Note that this function is also undefined at x = -2; however there is no asymptote at the corresponding point on the graph. An undefined point is usually represented as an open circle.

Figure 12-1: Vertical asymptote

From the graph of f(x) we see that for c = -1:

lim f(x) = +∞, 
^{x -> c-}

and,

lim f(x) = -∞, 
^{x -> c+}