Finding Points of Discontinuity in Rational Functions
What Is Discontinuity in Rational Functions?
A rational function is a fraction where both the numerator and denominator are polynomials. Discontinuity happens when the function breaks — when the output doesn't exist at certain input values.
In math terms, you're looking for x-values that make the denominator equal to zero. Those are your problem spots.
But not all discontinuities are the same. You need to know the difference, because your textbook or professor will definitely ask.
Two Types of Discontinuity You Must Know
Removable Discontinuity (Holes)
This happens when a factor cancels out in both the numerator and denominator. The function would work if you plugged in that x-value — it's just missing that one point.
Example: (x² - 4)/(x - 2)
Factor the numerator: (x - 2)(x + 2)/(x - 2)
The (x - 2) cancels. At x = 2, the original function is undefined. But after canceling, you'd get x + 2 = 4 if you plugged in 2.
That's a hole at (2, 4).
Non-Removable Discontinuity (Vertical Asymptotes)
This happens when a factor in the denominator cannot be canceled. The function blows up to infinity — or doesn't exist at all — at that x-value.
Example: 1/(x - 3)
No factor cancels. At x = 3, you're dividing by zero. The function has a vertical asymptote at x = 3.
The graph goes straight up (or down) and never touches that line.
How to Find All Points of Discontinuity
Here's the process. Memorize it.
Step 1: Set the denominator equal to zero
Take your rational function f(x) = P(x)/Q(x). Write Q(x) = 0.
Solve for x. These are your candidates for discontinuity.
Step 2: Factor both numerator and denominator
Factor P(x) and Q(x) completely. Look for matching factors.
Step 3: Cancel any common factors
If a factor appears in both numerator and denominator, cancel it. Those x-values give you removable discontinuities (holes).
The x-values that remain in the denominator after canceling give you vertical asymptotes.
Step 4: List your answers
State each x-value and classify it. Hole or vertical asymptote? That's it.
Worked Example
Find all points of discontinuity for:
f(x) = (x² - 9)/(x² - x - 6)
Step 1: Denominator = 0
x² - x - 6 = 0
Factor: (x - 3)(x + 2) = 0
Candidates: x = 3, x = -2
Step 2: Factor numerator
x² - 9 = (x - 3)(x + 3)
Step 3: Cancel common factors
f(x) = (x - 3)(x + 3)/(x - 3)(x + 2)
The (x - 3) cancels. So at x = 3, we have a hole.
After canceling, the denominator still has (x + 2). So at x = -2, we have a vertical asymptote.
Step 4: Results
x = 3: Removable discontinuity (hole)
x = -2: Non-removable discontinuity (vertical asymptote)
To find the hole coordinates, plug x = 3 into the simplified function (after canceling):
f(x) = (x + 3)/(x + 2)
f(3) = (3 + 3)/(3 + 2) = 6/5
Hole is at (3, 6/5).
Quick Reference: Holes vs. Vertical Asymptotes
| Feature | Hole (Removable) | Vertical Asymptote |
|---|---|---|
| What causes it | Factor cancels in numerator and denominator | Factor cannot be canceled |
| Function value | Undefined at that point, but limit exists | Undefined, limit goes to ±∞ |
| Graph appearance | Single missing point | Graph shoots up or down |
| Finding coordinates | Simplify, then plug in the x-value | Only need the x-value |
Common Mistakes That Will Cost You Points
- Forgetting to factor first. You cannot find holes without factoring. Ever.
- Canceling when nothing cancels. If no factor appears in both numerator and denominator, you have no holes. Just vertical asymptotes.
- Misclassifying the discontinuity. A hole is not the same as an asymptote. Know the difference.
- Not checking for hidden cancellations. Sometimes you need to factor trinomials or difference of squares to see the common factor.
- Finding asymptotes in the wrong direction. Test values on each side of the asymptote to see if the function goes to +∞ or -∞.
Practice Problems to Try
Find all points of discontinuity for each function:
- f(x) = (x + 1)/(x² - 4)
- f(x) = (x² - 1)/(x - 1)
- f(x) = (x² + 5x + 6)/(x² - 9)
- f(x) = 3/(x + 2)
Answers
- x = 2, x = -2 (both vertical asymptotes — neither factor cancels)
- x = 1 (hole — (x² - 1) = (x - 1)(x + 1), which cancels)
- x = 3 (hole — numerator factors to (x + 2)(x + 3), denominator to (x - 3)(x + 3), so (x + 3) cancels)
- x = -2 (vertical asymptote — no cancellation possible)
The Bottom Line
Finding points of discontinuity comes down to one question: What makes the denominator zero, and can it be canceled?
If yes → hole. If no → vertical asymptote. Factor everything, cancel what you can, and classify the rest.
That's the whole process. No magic, no shortcuts. Factor and check.