10 Holes — AP Precalculus | The Complete Equation

AP PRECALCULUS — UNIT 1B · POLYNOMIAL & RATIONAL FUNCTIONS

1.10 — Rational Functions and Holes

Notes — Removable Discontinuities from Cancellation

💡 Learning Objectives

By the end of this lesson you will be able to (AP CED 1.10.A):

Identify holes (removable discontinuities) in the graph of a rational function.
Determine the exact location (x, y) of a hole.
Distinguish a hole from a vertical asymptote.
Describe the limit of r(x) as x approaches the hole.

1. What Is a Hole?

A hole in the graph of a rational function is a single missing point — a place where the function is undefined, but the surrounding graph behaves smoothly on either side. Holes arise when a common factor cancels from the numerator and the denominator of r(x).

💡 Definitions

Let r(x) = N(x) / D(x). Suppose (x − a) is a factor of BOTH N(x) and D(x), and after cancellation the simplified form r̃(x) is defined at x = a.

Then x = a is NOT in the domain of r (the original expression is undefined there).
The graph has a HOLE at (a, r̃(a)): the point is missing but the curve approaches it from both sides.
The hole is a removable discontinuity — the function could be made continuous there by defining r(a) = r̃(a).

Left: a hole at (2, 3) — the graph looks like the line y = x + 1 but with one missing point. Right: a vertical asymptote at x = 2 — the graph shoots off to infinity.

💡 Quick Check

For r(x) = (x − 4)(x + 1) / (x − 4), predict: hole or asymptote at x = 4?

Both numerator and denominator have the factor (x − 4) — it cancels.
After cancellation: r̃(x) = x + 1, which is defined at x = 4 with value 5.

So the graph has a HOLE at (4, 5), not a vertical asymptote.

2. Finding the Location of a Hole

To find a hole, factor numerator and denominator completely. Identify any factor common to both. For each common factor (x − a), the hole is at x = a. Its y-coordinate is the value of the SIMPLIFIED rational function at x = a.

📘 Example

Example 1 — Finding a hole

Find any holes in the graph of r(x) = (x² − 9) / (x² − 2x − 3).

Factor: N = (x − 3)(x + 3), D = (x − 3)(x + 1). Common factor: (x − 3).

Simplify: r̃(x) = (x + 3) / (x + 1), for x ≠ 3.

Hole at x = 3. y-coordinate: r̃(3) = 6 / 4 = 3/2.

So the hole is at the point (3, 3/2). Also x = −1 is a vertical asymptote (uncancelled).

⚠️ Common Mistake

A common error is to compute the y-coordinate by plugging into the ORIGINAL r(x) — that gives 0/0, undefined. Always plug into the SIMPLIFIED expression r̃(x).

3. Holes and Limits

Even though r(x) is undefined at a hole, the LIMIT of r(x) as x approaches that input exists. The limit equals the y-coordinate of the hole — because the curve is smooth right up to that missing point.

📘 Example

Example 2 — The limit at a hole

For r(x) = (x² − 4) / (x − 2), find lim r(x) as x → 2.

Factor: (x − 2)(x + 2) / (x − 2) = x + 2 for x ≠ 2.

lim r(x) as x → 2 = lim (x + 2) as x → 2 = 4.

The limit is 4 even though r(2) itself is undefined. The graph has a hole at (2, 4).

🎯 AP Tip

On the AP Exam, you may be asked to 'justify that the discontinuity at x = a is removable.' A complete justification names the common factor in numerator and denominator, shows the cancellation, and computes the limit by evaluating the simplified expression at x = a.

📘 Try This

For r(x) = (x² − 5x + 6) / (x² − 4), find all holes and their coordinates.

N = (x − 2)(x − 3); D = (x − 2)(x + 2); common factor (x − 2).
Simplified r̃(x) = (x − 3)/(x + 2).
r̃(2) = (−1)/(4) = −1/4.

Hole at (2, −1/4). Vertical asymptote at x = −2.

4. Summary

A hole occurs at x = a when (x − a) is a common factor of numerator and denominator that cancels.
The y-coordinate of the hole is the value of the SIMPLIFIED expression at x = a.
The limit of r(x) at a hole equals the y-coordinate of the hole; the function itself is still undefined there.
A hole is a removable discontinuity; a vertical asymptote is NOT removable.

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