How To Find The Coordinate Of A Hole: Step-by-Step Guide

How to Find the Coordinate of a Hole in a Graph

You're graphing a rational function, everything looks smooth, and then — wait. Not a vertical asymptote where the graph shoots off to infinity. Practically speaking, there's a single point missing. Just a single point, floating there alone, excluded from the curve.

That's a hole.

And if you've ever wondered how to find exactly where that hole sits on the coordinate plane, you're in the right place. It's one of those skills that shows up in algebra and precalculus, and once you see the pattern, it's actually pretty straightforward.

What Is a Hole in a Graph?

A hole (sometimes called a removable discontinuity) is a single point on a rational function where the graph is undefined — but unlike a vertical asymptote, the function doesn't blow up to infinity nearby. The graph approaches that point from both sides, gets really close, but never actually touches it.

Here's why holes happen: they occur when a factor in the numerator and a factor in the denominator are the same. That common factor cancels out, which simplifies the function — but the original function still can't use that value because you'd be dividing by zero.

Let me show you what I mean.

Take this function:

$f(x) = \frac{x^2 - 9}{x^2 - x - 6}$

At first glance, this looks like any rational function. But if you factor both parts:

• Numerator: $(x - 3)(x + 3)$
• Denominator: $(x - 3)(x + 2)$

There's a common factor of $(x - 3)$. When you cancel it, you get the simplified function $g(x) = \frac{x + 3}{x + 2}$.

But here's the thing — the original function $f(x)$ is undefined when $x = 3$, because that makes the denominator zero. The cancellation doesn't magically bring that point back to life. It just tells us that everywhere except at $x = 3$, the function behaves like $\frac{x + 3}{x + 2}$.

That missing point at $x = 3$ is the hole.

How a Hole Differs from a Vertical Asymptote

This is worth pausing on, because mixing these up is one of the most common mistakes students make.

A vertical asymptote happens when the denominator equals zero and there's no matching factor in the numerator to cancel it. The graph shoots off toward infinity as it approaches that x-value. Think of $f(x) = \frac{1}{x}$ — as x gets closer to zero from either side, the function values grow without bound.

A hole happens when there's a matching factor in the numerator. The graph approaches the point smoothly from both sides and just... stops. Even so, there's no dramatic shooting off to infinity. Just a single point where the function doesn't exist.

Why Finding the Coordinate of a Hole Matters

Here's the real-world version of why this matters: understanding holes teaches you how to read a function's behavior, not just its algebraic form.

In calculus, holes become important when you're learning about limits. You'll see functions where the limit exists at a point even though the function isn't defined there — and that's exactly what a hole is. The limit approaches a specific y-value, but the function itself can't deliver it.

Beyond that, if you're graphing rational functions by hand (or checking your work on a calculator), knowing how to find holes helps you create an accurate graph. Missing that single point might not seem like a big deal, but on a test, it's the difference between a correct answer and a careless error.

And honestly? That said, once you get comfortable finding holes, you'll start recognizing them instantly. It becomes one of those skills that just clicks.

How to Find the Coordinate of a Hole

Alright, let's get into the actual process. Here's the step-by-step method:

Step 1: Factor Both the Numerator and Denominator

Start with your rational function in simplest form — meaning it's already written as one fraction, numerator over denominator. Factor each part completely using whatever methods work: factoring by grouping, the quadratic formula, recognizing difference of squares, and so on.

As an example, with $f(x) = \frac{x^2 - 4x + 3}{x^2 - 1}$:

• Numerator factors to $(x - 1)(x - 3)$
• Denominator factors to $(x - 1)(x + 1)$

Step 2: Look for Common Factors

Scan through your factored numerator and denominator. If you see the same factor appearing in both, that's your hole-causing factor.

In our example, $(x - 1)$ appears in both. That's the one.

Step 3: Cancel the Common Factor

Cancel the common factor just like you would in any fraction simplification. This gives you the "simplified" function that describes the graph everywhere except at the hole.

After canceling $(x - 1)$, we get $g(x) = \frac{x - 3}{x + 1}$.

Step 4: Find the x-Coordinate

Set the canceled factor equal to zero and solve. This x-value is where the hole lives — it's the value that makes the original denominator zero.

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For our canceled factor $(x - 1)$: $x - 1 = 0$ $x = 1$

So the hole's x-coordinate is 1.

Step 5: Find the y-Coordinate

This is the step people sometimes forget. Because of that, you don't plug the x-value into the original function — it's undefined there. Instead, plug it into the simplified function you got after canceling.

Using $g(x) = \frac{x - 3}{x + 1}$ and plugging in $x = 1$: $g(1) = \frac{1 - 3}{1 + 1} = \frac{-2}{2} = -1$

So the y-coordinate is -1.

Step 6: Write the Coordinate Pair

The hole is at $(1, -1)$.

That's it. Factor, cancel, solve for the x-value that makes the canceled factor zero, then plug that x into the simplified function to get the y-value.

Common Mistakes People Make

Let me save you some pain by pointing out the errors I see most often:

Using the original function to find the y-coordinate. This is the big one. Students plug the x-value into the original function, get a division by zero error, and panic. Remember — you need the simplified function for the y-coordinate. The original function is undefined at that point by definition.

Forgetting to include the hole entirely. Some students cancel the factor and then act like the hole doesn't exist. They graph the simplified function and call it a day. But that single missing point is still part of the function's graph. Always plot it.

Confusing holes with vertical asymptotes. If the factor doesn't cancel — if it's only in the denominator — you have a vertical asymptote, not a hole. The graph behaves completely differently near each.

Not factoring completely. If you don't factor down to linear factors, you might miss a common factor hiding in a quadratic. Always factor completely before looking for cancellations.

Practical Tips That Actually Help

A few things that will make your life easier:

Check for holes before graphing. Whenever you're given a rational function, factor it first. Even if the problem doesn't ask you to find holes, spotting them early prevents mistakes later.

Use the simplified function for everything except the hole. Once you've canceled the common factor, use the simplified version to find the y-intercept, x-intercepts, and behavior at infinity. Just remember to add the hole back in as a single point. Turns out it matters.

Double-check by substitution. After you think you've found a hole at $(a, b)$, verify by checking that the original function equals $b$ when you plug in $a$ to the simplified version — and that the original function is undefined at $x = a$.

Watch for multiple holes. Yes, you can have more than one hole. If the numerator and denominator share multiple common factors, each one creates a hole. Apply the same process to each canceled factor.

Frequently Asked Questions

Can a hole be at any x-value? A hole can only occur at an x-value that makes both the numerator and denominator zero in the original function — specifically, at a value that creates a common factor. If only the denominator is zero, you have a vertical asymptote instead.

How do I know if there's a hole or a vertical asymptote? Factor the function. If a factor cancels out completely, you have a hole at the value that makes that factor zero. If a factor in the denominator doesn't cancel, you have a vertical asymptote at that value.

What if the common factor is squared? The process is the same. As an example, if you have $\frac{(x-2)^2}{(x-2)}$ in the original function, you can cancel to get $x-2$, and the hole is at $x = 2$. The squared factor doesn't change the location of the hole — it just means the graph approaches the hole from the same side on both directions (touching but not crossing).

Do I need to plot the hole on the graph? Yes. A hole is part of the function's graph — it's just a single point that's excluded from the curve. Always plot it as an open circle at the coordinate you found.

Can holes occur at the y-intercept? Absolutely. If the hole's x-coordinate is 0, then the hole is on the y-axis. You'd plot it as $(0, y)$ with an open circle.

The Bottom Line

Finding the coordinate of a hole comes down to a simple sequence: factor, cancel, solve for the excluded x-value, then plug that x into the simplified function to get the y-value. The coordinate pair tells you exactly where to draw that open circle on your graph.

It's one of those concepts that feels tricky the first time, but once you work through a few examples, the pattern becomes automatic. The key is remembering that the original function and the simplified function tell you different things — one tells you where the graph doesn't exist, and the other tells you what the graph would be if that point weren't missing.

Now you've got the method. A few practice problems and it'll be second nature.