AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/1/1-10)
Unit 1 · Topic 1.10
1.10 Rational Functions and Holes
A hole is a single missing point on a graph. It shows up in a rational function when a factor of the denominator is completely canceled by the same factor in the numerator. You find its height from the simplified function, or with a limit.
Key terms
- hole
- removable discontinuity
- common factor
- limit
When a hole appears
Suppose x = c makes both the numerator and the denominator zero. Compare how many times (x − c) appears in each.
If (x − c) shows up at least as many times on top as on the bottom, every copy in the denominator cancels, and the graph has a hole at x = c. If the bottom has more copies, one is left over after canceling, and there's a vertical asymptote instead.
The original function is still undefined at x = c, because its denominator is zero there. That's why the graph has a gap: one point is missing.
Finding where the hole is
Cancel the common factors to get a simplified function. It agrees with r everywhere except at x = c.
Plug c into the simplified function to get L. The hole is at the point (c, L).
In limit language: as x gets close to c from either side, r(x) gets close to L. You write lim (x→c) r(x) = L, and both one-sided limits equal L too. The function approaches a value it never actually reaches at that input.
A full picture of a rational function
Factored form gives you almost everything at once:
- Holes: factors that cancel completely from the denominator.
- Vertical asymptotes: factors left in the denominator after canceling.
- Zeros: factors left in the numerator after canceling (if they aren't also at a hole).
- Horizontal or slant asymptote: from the leading terms (topic 1.7).
- Domain: all real numbers except zeros of the original denominator, both holes and asymptotes.
Why canceling is allowed, and its limits
(x − 3)/(x − 3) equals 1 for every input except x = 3, where it is 0/0 and undefined. So canceling gives a function that matches the original everywhere except at that one input. That single difference is exactly the hole.
Holes are easy to miss on a calculator. A graph on the screen usually looks unbroken because a single missing point is too small to show, and a table may show ERROR only if 3 happens to be one of the listed inputs. Find holes with algebra, not by looking.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Locating a hole and other features
Let r(x) = (x² − 9)/(x² − x − 6). Find any holes, vertical asymptotes, zeros and horizontal asymptote.
Show the solutionHide the solution
- Step 1: Factor: r(x) = (x − 3)(x + 3) / ((x − 3)(x + 2)).
- Step 2: (x − 3) appears once on top and once on the bottom, so it cancels completely: hole at x = 3. The simplified function is (x + 3)/(x + 2).
- Step 3: Height of the hole: (3 + 3)/(3 + 2) = 6/5. So lim (x→3) r(x) = 6/5 and the hole is at (3, 6/5).
- Step 4: Left in the denominator: (x + 2), so there's a vertical asymptote at x = −2.
- Step 5: Left in the numerator: (x + 3), so the zero is x = −3.
- Step 6: Leading terms x²/x² = 1 give the horizontal asymptote y = 1.
Answer: Hole at (3, 6/5); vertical asymptote x = −2; zero at x = −3; horizontal asymptote y = 1.
- Example 2
Trap: a hole on the x-axis
Let g(x) = (x − 1)²(x + 2) / ((x − 1)(x + 3)). Is x = 1 a zero of g?
Show the solutionHide the solution
- Step 1: x = 1 makes both top and bottom zero. Its multiplicity is 2 on top and 1 on the bottom, so the factor cancels completely from the bottom: there's a hole at x = 1.
- Step 2: Simplified: (x − 1)(x + 2)/(x + 3). At x = 1 this gives 0 · 3/4 = 0, so lim (x→1) g(x) = 0.
- Step 3: The hole sits at (1, 0), right on the x-axis. But g(1) is undefined, so 1 is not a zero. It's tempting to call it one because the simplified function equals 0 there.
Answer: No. g has a hole at (1, 0); 1 is not in the domain, so it is not a zero.
- Example 3
Building a function with given features
Write a rational function with a hole at x = 2, a vertical asymptote at x = −1, and a horizontal asymptote y = 3.
Show the solutionHide the solution
- Step 1: A hole at x = 2 needs a factor (x − 2) on top and bottom.
- Step 2: A vertical asymptote at x = −1 needs (x + 1) in the denominator only.
- Step 3: A horizontal asymptote y = 3 needs equal degrees with leading coefficients in the ratio 3:1. Put 3x on top to match the degree of (x + 1).
- Step 4: Try r(x) = 3x(x − 2) / ((x + 1)(x − 2)). Check: simplified 3x/(x + 1); asymptote x = −1; leading terms 3x/x = 3; the hole is at (2, 3·2/3) = (2, 2).
Answer: One answer: r(x) = 3x(x − 2) / ((x + 1)(x − 2)), which has a hole at (2, 2). Many other answers work.
Common mistakes
- Plugging c into the original function to find the hole's height. That gives 0/0. Use the simplified function.
- Treating a hole on the x-axis as a zero. The function is undefined at a hole, so it can't equal zero there.
- Calling every canceled factor a hole. If the denominator had more copies than the numerator, there's a vertical asymptote.
On the exam
- Expect to find the coordinates of a hole, often alongside asymptotes and zeros. Give both coordinates.
- Limit questions may ask for lim (x→c) r(x) at a hole. The answer is the y-coordinate of the hole, even though r(c) is undefined.
Connected topics
Videos
Check yourself
4 questions on 1.10 Rational Functions and Holes. Pick an answer to see if you got it, and why.
Let r(x) = ((x − 1)²(x + 2))/((x − 1)(x + 5)). Which of the following describes the graph of r at x = 1?
Let r(x) = (x − 3)/((x − 3)²(x + 1)). Which of the following describes the graph of r at x = 3?
Let f(x) = (x² + kx − 6)/(x − 2), where k is a constant. The graph of f has a hole at x = 2. What are the coordinates of the hole?
Let r(x) = (x − 4)/(x² − 5x + k), where k is a constant. For which of the following values of k does the graph of r have exactly one vertical asymptote?
0 of 4 answered