Rational Functions and Holes
Why this matters
Imagine you're using a GPS to drive from Boston to a friend's house in a nearby town. The route is a perfectly straight road, except for one small problem: a single, tiny bridge is out at mile marker 2. The road exists right up to the edge of the gap and picks up immediately on the other side.
Your GPS is smart. It knows the road should be there. To properly describe the route, it can't just say "the road is y = x + 2." It has to add a note: "Warning: Bridge out at x=2."
That's exactly what a hole in a rational function is. It's a single, missing point on a graph that is otherwise perfectly predictable. In this lesson, we'll learn how to spot these "missing bridges" algebraically and find their exact coordinates.
Concept overview
flowchart TD
A[Start with r(x) = P(x)/Q(x)] --> B{Factor P(x) and Q(x)};
B --> C{Is there a common factor (x-c)?};
C -- No --> D[No hole from this factor. Check for asymptote.];
C -- Yes --> E[Hole exists at x = c];
E --> F[Create simplified function r_s(x) by canceling (x-c)];
F --> G[Calculate y-coordinate L = r_s(c)];
G --> H[Hole is at point (c, L)];
Core explanation
Hello! I'm Saavi, and today we're tackling a concept that can feel a little strange at first: holes in rational functions. But I promise, once you see the logic, it's very straightforward.
First, let's remember what a rational function is. It's any function that is a fraction of two polynomials, like r(x) = P(x) / Q(x). The key to understanding these functions is looking at the denominator, Q(x). Any x-value that makes the denominator zero is a point of discontinuity—a place where the graph has some kind of break.
There are two main types of discontinuities: vertical asymptotes and holes. We'll cover asymptotes in another lesson, but the big difference is this:
- A hole is a removable discontinuity.
- A vertical asymptote is a non-removable discontinuity.
Think of it like a tiny glitch in a computer program. If it's a hole, you can "patch" the code by removing a faulty line that appears in two places. If it's a vertical asymptote, it's a fundamental break in the logic that you can't patch.
How to Find a Hole
A hole exists when a specific x-value makes both the numerator and the denominator zero. This happens when they share a common factor.
Let's use our classic example:
r(x) = (x^2 - 4) / (x - 2)
Step 1: Factor everything. The first step, always, is to factor the numerator and the denominator completely.
- Numerator:
x^2 - 4is a difference of squares, which factors to(x - 2)(x + 2). - Denominator:
x - 2is already factored.
So, our function becomes:
r(x) = [(x - 2)(x + 2)] / (x - 2)
Step 2: Identify common factors.
Do you see a factor that appears on both the top and the bottom? Yes! It's (x - 2).
Because this factor exists in both, it creates a hole at the x-value that makes that factor zero.
x - 2 = 0
x = 2
So, we know there's a hole in the graph at x = 2.
This is the crucial part that addresses the official AP standard (EK 1.10.A.1): A hole occurs when a factor (x-c) appears in the denominator, and it also appears in the numerator with an equal or greater multiplicity. In our case, (x-2) has a multiplicity of 1 in the numerator and 1 in the denominator, so we get a hole.
Step 3: Find the coordinates of the hole.
A hole isn't just an x-value; it's a point with (x, y) coordinates. We have the x-coordinate: x = 2. How do we find the y-coordinate?
This is an indeterminate form. It doesn't give us an answer. It just tells us that something interesting is happening at x=2.
Instead, we look at the simplified function. Since the (x - 2) factors can be canceled out, we can create a new, simpler function that is identical to our original one everywhere except at the hole.
r(x) = [(x - 2)(x + 2)] / (x - 2)
Simplified function: y = x + 2
This simplified function, y = x + 2, tells us what the graph looks like. It's a simple straight line. To find the y-coordinate of the hole, we plug our x-value, x=2, into this simplified function:
y = 2 + 2 = 4
So, the hole is located at the point (2, 4).
The Formal Definition: Limits
In calculus, we use a special notation to describe this behavior. We say that the limit of r(x) as x approaches 2 is 4. We write it like this:
lim_{x→2} r(x) = 4
This notation (EK 1.10.A.2) is just a formal way of saying, "As you get infinitely close to x=2 from either the left or the right side, the y-value of the function gets infinitely close to 4." The function itself is undefined at x=2, but the values all around it are pointing to y=4.
So, the graph of r(x) = (x^2 - 4) / (x - 2) looks exactly like the graph of the line y = x + 2, but with a small open circle at the point (2, 4) to show that this single point is missing.
Worked examples
Let's walk through a few examples together to make sure this process is crystal clear.
A Standard Case
Problem: Find the location of the hole in the graph of f(x) = (x^2 + x - 6) / (x + 3).
Solution:
- 1Factor the numerator and denominatorThe denominator,
(x + 3), is already as simple as it gets. For the numerator,x^2 + x - 6, we need two numbers that multiply to -6 and add to +1. Those numbers are +3 and -2. So, the factored function is:f(x) = [(x + 3)(x - 2)] / (x + 3) - 2Identify the common factor and the x-coordinate of the holeThe common factor is
(x + 3). This tells us there is a hole wherex + 3 = 0, which means atx = -3. - 3Create the simplified functionWe cancel the common
(x + 3)factor from the top and bottom: Simplified function:y = x - 2 - 4Find the y-coordinate of the holeNow, we plug the x-coordinate of the hole (
x = -3) into our new simplified function:y = (-3) - 2 = -5
Distinguishing Holes and Asymptotes
Problem: Identify all discontinuities for g(x) = (x^2 - 3x + 2) / (x^2 - 1). Are they holes or vertical asymptotes? If there is a hole, find its coordinates.
Solution:
- 1Factor everything
- Numerator:
x^2 - 3x + 2factors to(x - 1)(x - 2). - Denominator:
x^2 - 1is a difference of squares, factoring to(x - 1)(x + 1). Our function is:g(x) = [(x - 1)(x - 2)] / [(x - 1)(x + 1)]
- Numerator:
- 2Identify all potential discontinuitiesThe denominator is zero when
x=1orx=-1. So, our discontinuities are atx=1andx=-1. - 3Distinguish between holes and vertical asymptotes
- Look at the factor
(x - 1). It appears in both the numerator and the denominator. This meansx=1is a hole. - Look at the factor
(x + 1). It appears in the denominator but not in the numerator. This meansx=-1is a vertical asymptote.
- Look at the factor
- Find the coordinates of the hole.
To find the hole's location, we first create the simplified function by canceling the
(x-1)factor: Simplified function:y = (x - 2) / (x + 1)Now, plug the hole's x-coordinate (x=1) into this simplified function:y = (1 - 2) / (1 + 1) = -1 / 2
Try it yourself
Ready to try a couple on your own? Don't worry about getting it perfect, just focus on applying the steps we just covered.
Problem 1:
Find the coordinates of the hole in the graph of r(x) = (x^2 - 9) / (x + 3).
Hint: The numerator is a difference of squares. What does a^2 - b^2 factor into?
Problem 2:
Find the coordinates of the hole in the graph of g(x) = (2x^2 + 5x - 3) / (x + 3).
Hint: Factoring the numerator 2x^2 + 5x - 3 is the trickiest part. You're looking for two numbers that multiply to `2 -3 = -6and add to+5`.*
Take your time, work through the steps, and check your work. You've got this.
Practice — 8 questions
In simple terms, this lesson explains how to find and graph "holes" in rational functions. These are like tiny, specific gaps in an otherwise continuous line or curve.
- 1.10.A: Determine holes in graphs of rational functions.
- 1.10.A.1
- If the multiplicity of a real zero in the numerator is greater than or equal to its multiplicity in the denominator, then the graph of the rational function has a hole at the corresponding input value.
- 1.10.A.2
- If the graph of a rational function r has a hole at x = c, then the location of the hole can be determined by examining the output values corresponding to input values sufficiently close to c. If input values sufficiently close to c correspond to output values arbitrarily close to L, then the hole is located at the point with coordinates (c, L). The corresponding mathematical notation is lim_{x→c} r(x) = L. It should be noted that lim_{x→c-} r(x) = lim_{x→c+} r(x) = lim_{x→c} r(x) = L.
flowchart TD
A[Start with r(x) = P(x)/Q(x)] --> B{Factor P(x) and Q(x)};
B --> C{Is there a common factor (x-c)?};
C -- No --> D[No hole from this factor. Check for asymptote.];
C -- Yes --> E[Hole exists at x = c];
E --> F[Create simplified function r_s(x) by canceling (x-c)];
F --> G[Calculate y-coordinate L = r_s(c)];
G --> H[Hole is at point (c, L)];
Read what Saavi narrates
Hello everyone, it's Saavi from Shrutam. Let's talk about something that sounds a bit weird... holes in graphs.
Imagine you're using a GPS for a road trip. The route is a perfectly straight road, but there's a tiny bridge out at mile marker 2. The road exists right up to the gap and picks up immediately on the other side. That's a hole! It's a single, missing point on a graph that is otherwise completely normal. Today, we'll learn how to find the exact coordinates of these "missing bridges."
A hole happens in a rational function when a factor in the numerator cancels out with an identical factor in the denominator. Let's see it in action with a classic example: the function r of x equals... the quantity x-squared minus 4... divided by... the quantity x minus 2.
First, we factor everything. The numerator, x-squared minus 4, becomes (x minus 2) times (x plus 2). The denominator is just (x minus 2).
So now we have... (x minus 2)(x plus 2) all over (x minus 2).
See that common factor of (x minus 2)? That's our clue. It tells us there's a hole where x minus 2 equals zero... which is at x equals 2.
But a hole is a point, so we need a y-coordinate. Here's a common mistake I see every year: students plug x equals 2 back into the original function. You'll just get zero divided by zero, which is undefined. Don't do that.
Instead, we use the simplified function. We cancel out the (x minus 2) factors, and we're left with a much friendlier function: y equals x plus 2.
Now, plug x equals 2 into *this* simplified function. y equals 2 plus 2, which is 4.
So, the hole is at the point (2, 4). The graph looks exactly like the line y equals x plus 2, but with a tiny open circle at the point (2, 4) to show that point is missing.
It's all about factoring, canceling, and then plugging into the simplified version. Once you get that rhythm down, you'll be able to spot these holes with confidence. Keep practicing, and you'll master it in no time.
This will always result in `0/0`, which is undefined and doesn't give you the y-coordinate.
Always plug the x-value into the **simplified** function (the one that's left after you cancel the common factors).
A hole is a missing *point* on the graph, which requires both an x and a y coordinate. An incomplete answer may not get full credit.
Always express the hole as a coordinate pair `(x, y)`.
They represent different behaviors. A hole is a single missing point. An asymptote is a line the function approaches but never touches.
Remember the rule: if the factor from the denominator *cancels* with a factor in the numerator, it's a hole. If it *doesn't* cancel, it's a vertical asymptote.
Factoring is the very first step. If you get it wrong, every subsequent step will be incorrect, even if your method is right.
After you factor, quickly multiply the factors back together (using FOIL or distribution) in your head or on scratch paper to confirm they match the original polynomial.