Find the vertical asymptotes of f(x) = (x + 1)/((x - 4)(x + 2)). [1 point]

For f(x) = (1)/((x - 3)^2), what are the one-sided limits as x → 3? [1 point]

United StatesPrecalculusSyllabus dot point

Where does a rational function shoot off to infinity, and how do you describe that behavior with limits?

Topic 1.9 Rational Functions and Vertical Asymptotes: locate the vertical asymptotes of a rational function from the zeros of the denominator that do not cancel, and describe the behavior with one-sided limits.

A focused answer to AP Precalculus Topic 1.9, covering how denominator zeros that do not cancel give vertical asymptotes, how to do sign analysis for one-sided behavior, and limit notation.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

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

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What this topic is asking
Where vertical asymptotes occur
One-sided behavior by sign analysis
Assembling the full graph
Try this

What this topic is asking

The College Board (Topic 1.9) wants you to find the vertical asymptotes of a rational function and describe the behavior near them. A vertical asymptote sits at any input where the denominator is zero but the numerator is not (after cancelling common factors). You must use sign analysis to decide whether the graph goes to $+\infty$ or $-\infty$ on each side, and write the result with one-sided limits.

Where vertical asymptotes occur

So the routine is: factor everything, cancel shared factors (those are holes), and the denominator factors that remain give the vertical asymptotes.

One-sided behavior by sign analysis

Near a vertical asymptote the graph goes to one of $\pm\infty$ from each side. Determine which by examining signs:

The numerator's sign near $a$ is roughly constant (just evaluate it at $a$ ).
The denominator is a tiny number near $a$ ; its sign just to the left and just to the right of $a$ can differ.

Matching signs (numerator and denominator both positive, or both negative) give $+\infty$ ; opposite signs give $-\infty$ .

One-sided limits at a vertical asymptote

Find the vertical asymptote of $f(x) = \frac{x + 4}{x - 1}$ and determine the behavior on each side.

step 1 Locate the asymptote

The denominator $x - 1$ is zero at $x = 1$ , and the numerator $x + 4 = 5$ there is nonzero, so $x = 1$ is a vertical asymptote.

step 2 Sign of the numerator near x = 1

Near $x = 1$ the numerator $x + 4 \approx 5$ is positive.

step 3 Sign of the denominator on each side

Just left of $1$ (say $x = 0.9$ ), $x - 1 = -0.1$ is a small negative. Just right of $1$ (say $x = 1.1$ ), $x - 1 = 0.1$ is a small positive.

step 4 Combine the signs and write the limits

Left side: $\frac{+}{-} \to -\infty$ , so $\lim_{x \to 1^-} f(x) = -\infty$ . Right side: $\frac{+}{+} \to +\infty$ , so $\lim_{x \to 1^+} f(x) = +\infty$ . The odd-multiplicity factor flips the sign across the asymptote, as expected.

Assembling the full graph

Vertical asymptotes combine with zeros (Topic 1.8), holes (Topic 1.10) and end behavior (Topic 1.7) to give the complete graph. The clean order is: factor numerator and denominator, cancel to find holes, read the remaining denominator factors as vertical asymptotes, read the remaining numerator factors as zeros, and use the degree comparison for the horizontal or slant asymptote. With those four features and a couple of test points, you can sketch any rational function in Unit 1.

A subtle but examined point is the role of multiplicity at the asymptote. Many students remember that vertical asymptotes exist but forget that an even-power factor sends the graph to the same infinity on both sides, producing a graph that hugs the asymptote like two upward (or two downward) branches rather than one up and one down. Tying the side-by-side behavior to the parity of the denominator factor, exactly as multiplicity governs crossing for zeros, makes the sketch reliable and explains the contrast the exam often highlights between $\frac{1}{x - 2}$ and $\frac{1}{(x - 2)^2}$ .

Try this

Q1. Find the vertical asymptotes of $f(x) = \frac{x + 1}{(x - 4)(x + 2)}$ . [1 point]

Cue. Denominator zeros at $x = 4$ and $x = -2$ ; neither cancels, so both are vertical asymptotes.

Q2. For $f(x) = \frac{1}{(x - 3)^2}$ , what are the one-sided limits as $x \to 3$ ? [1 point]

Cue. Even-power denominator stays positive both sides; with positive numerator, $\lim_{x \to 3^-} = \lim_{x \to 3^+} = +\infty$ .

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). Where is the vertical asymptote of

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

? (A)

x = -1

(B)

x = 1

(C)

x = 3

(D)

y = 1

Show worked answer →

The correct answer is (C), $x = 3$ .

A vertical asymptote occurs where the denominator is zero and the numerator is not. The denominator $x - 3$ is zero at $x = 3$ , and the numerator $x + 1 = 4$ there is nonzero, so $x = 3$ is a vertical asymptote. Choice (A) is the zero of the function, not the asymptote.

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

g(x) = \frac{x}{(x - 2)^2}

. (a) Identify the vertical asymptote. (b) Determine the one-sided behavior of

g

x \to 2^-

and as

x \to 2^+

, and write each with limit notation.

Show worked answer →

A 3-point question on one-sided behavior at a vertical asymptote.

(a) Vertical asymptote (1 point): the denominator $(x - 2)^2$ is zero at $x = 2$ and the numerator $x = 2$ is nonzero, so $x = 2$ is a vertical asymptote.
(b) One-sided limits (1 point each side): near $x = 2$ the numerator is positive (about $2$ ) and the denominator $(x - 2)^2$ is a small positive number on both sides (a square is never negative). So $\lim_{x \to 2^-} g(x) = +\infty$ and $\lim_{x \to 2^+} g(x) = +\infty$ ; the graph rises to $+\infty$ on both sides.

Related dot points

Sources & how we know this

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