Does f(x) = ((x - 1)(x + 4))/((x + 4)) have a hole or an asymptote at x = -4? [1 point]

Find the y-coordinate of the hole of f(x) = (x^2 - 1)/(x - 1). [1 point]

College Board AP 2023 (style)-style practice: Section I, Part A (multiple choice, no calculator). The graph of f(x) = (x^2 - 4)/(x - 2) has which feature at x = 2? (A) A vertical asymptote (B) A zero (C) A hole (D) A horizontal asymptote

The correct answer is (C), a hole. Factor the numerator: x^2 - 4 = (x - 2)(x + 2). The factor (x - 2) cancels with the denominator, so f(x) = x + 2 for x ≠ 2. The value x = 2 is excluded from the domain even though the simplified form is defined there, which is exactly a removable discontinuity, or hole.

United StatesPrecalculusSyllabus dot point

What creates a hole in the graph of a rational function, and how do you find its location?

Topic 1.10 Rational Functions and Holes: identify removable discontinuities (holes) of a rational function from factors common to numerator and denominator, and find the coordinates of each hole.

A focused answer to AP Precalculus Topic 1.10, covering how common factors create removable discontinuities (holes), how to find a hole's coordinates by cancelling and substituting, and how holes differ from asymptotes.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
What makes a hole
Finding the coordinates of a hole
Holes versus asymptotes
Try this

What this topic is asking

The College Board (Topic 1.10) wants you to identify holes (removable discontinuities) in the graph of a rational function. A hole appears at any input where a factor is common to numerator and denominator, so it cancels. You must find the coordinates of each hole by cancelling the common factor and substituting into the simplified expression, and you must distinguish a hole from a vertical asymptote.

What makes a hole

The defining contrast with Topic 1.9 is cancellation: a denominator factor that cancels gives a hole, while a denominator factor that survives gives a vertical asymptote.

Finding the coordinates of a hole

You substitute into the simplified form, not the original, because the original is undefined at $a$ . The simplified form agrees with the original everywhere except at the hole, so it gives the height the curve "wants" at that point.

Locating a hole

Find the coordinates of the hole in $f(x) = \frac{x^2 - x - 6}{x^2 - 9}$ .

step 1 Factor numerator and denominator

$x^2 - x - 6 = (x - 3)(x + 2)$ and $x^2 - 9 = (x - 3)(x + 3)$ .

step 2 Identify and cancel the common factor

Both share $(x - 3)$ . Cancelling gives $f(x) = \frac{x + 2}{x + 3}$ for $x \neq 3$ . The cancelled factor marks a hole at $x = 3$ .

step 3 Substitute into the simplified form

At $x = 3$ : $\frac{3 + 2}{3 + 3} = \frac{5}{6}$ .

step 4 State the hole and the remaining feature

The hole is at $\left(3, \frac{5}{6}\right)$ . The remaining denominator factor $(x + 3)$ does not cancel, so $x = -3$ is a vertical asymptote, not a hole.

Holes versus asymptotes

The exam routinely puts a hole and a vertical asymptote in the same function to test whether you can tell them apart. The deciding question is always: does the denominator factor cancel? If yes, hole; if no, vertical asymptote. After cancelling, the surviving denominator factors are the asymptotes and the cancelled ones are the holes. Listing a hole as an asymptote, or vice versa, is the single most common error in rational-function graphing, and it comes entirely from skipping the factor-and-cancel step.

A finer point is that the hole's $x$ -value is still excluded from the domain even though the simplified function is perfectly happy there. The simplification changes the formula but not the domain: the original function was never defined at the cancelled value, so the graph must show a hole rather than filling the point in. Keeping the domain restriction attached to the original function, even after you simplify, is what justifies drawing the open circle and is the conceptual heart of a removable discontinuity.

Try this

Q1. Does $f(x) = \frac{(x - 1)(x + 4)}{(x + 4)}$ have a hole or an asymptote at $x = -4$ ? [1 point]

Cue. The $(x + 4)$ factor cancels, so there is a hole at $x = -4$ , not an asymptote.

Q2. Find the $y$ -coordinate of the hole of $f(x) = \frac{x^2 - 1}{x - 1}$ . [1 point]

Cue. Cancel to $x + 1$ ; at $x = 1$ the value is $2$ , so the hole is at $(1, 2)$ .

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)1 marksSection I, Part A (multiple choice, no calculator). The graph of

f(x) = \frac{x^2 - 4}{x - 2}

has which feature at

x = 2

? (A) A vertical asymptote (B) A zero (C) A hole (D) A horizontal asymptote

Show worked answer →

The correct answer is (C), a hole.

Factor the numerator: $x^2 - 4 = (x - 2)(x + 2)$ . The factor $(x - 2)$ cancels with the denominator, so $f(x) = x + 2$ for $x \neq 2$ . The value $x = 2$ is excluded from the domain even though the simplified form is defined there, which is exactly a removable discontinuity, or hole.

AP 2024 (style)3 marksSection II (free response, calculator allowed). Let

g(x) = \frac{(x - 5)(x + 1)}{(x - 5)(x - 3)}

. (a) Identify any holes and any vertical asymptotes of

g

. (b) Find the exact coordinates of each hole.

Show worked answer →

A 3-point question separating holes from asymptotes.

(a) Features (1 point): the factor $(x - 5)$ cancels, giving a hole at $x = 5$ . The remaining denominator factor $(x - 3)$ does not cancel, so $x = 3$ is a vertical asymptote.
(b) Hole coordinates (2 points): after cancelling, $g(x) = \frac{x + 1}{x - 3}$ for $x \neq 5$ . Substitute $x = 5$ into the simplified form: $\frac{5 + 1}{5 - 3} = \frac{6}{2} = 3$ . So the hole is at $(5, 3)$ .

Related dot points

Sources & how we know this

AP Precalculus Course and Exam Description — College Board (2023)