Difficulty: Introductory to Intermediate | Prerequisites: Factoring polynomials, basic knowledge of trigonometric functions (sec, cos), comfort with limits from Part 1 notes.
Vertical asymptotes mark the x-values where a function's output escapes to infinity. They show up whenever you divide by something that can equal zero, and understanding them is essential for sketching rational functions and for later work with improper integrals. This topic builds on the limit concepts from Section 1.5: you already know how to evaluate limits from a graph, and now you are finding where those limits blow up. If you are comfortable factoring quadratics and simplifying rational expressions, you are ready.
Vertical asymptotes occur at x-values that make a function's denominator zero without also making the numerator zero. Factor the numerator and denominator, cancel any common factors (those give holes, not asymptotes), and the remaining zeros of the denominator are your vertical asymptotes. For trigonometric functions like x³ sec(x), asymptotes appear where the trig function itself is undefined.
Vertical asymptote
A vertical line x = a where at least one of lim (x→a⁻) f(x) or lim (x→a⁺) f(x) is +∞ or −∞. The graph of f shoots upward or downward without bound near x = a.
Think of it as: a wall the graph can never cross, because the function's output is exploding.
Rational function
A function of the form p(x)/q(x), where p and q are polynomials. Vertical asymptotes can only occur at x-values where q(x) = 0.
In simple terms, this means: a fraction of polynomials, and the trouble spots are wherever the bottom equals zero.
Removable discontinuity (hole)
A point where the numerator and denominator share a common factor that cancels. After cancellation, the function is well-behaved at that x-value, so it is a hole in the graph rather than an asymptote.
Think of it as: the zero in the denominator is "cancelled out" by the same zero in the numerator, so the function does not blow up there, it just has a missing point.
sec(x)
The secant function, defined as 1/cos(x). It is undefined wherever cos(x) = 0, which is at x = π/2, 3π/2, and so on (odd multiples of π/2).
In simple terms, this means: sec(x) inherits all the trouble spots of cos(x), because dividing by zero makes sec(x) blow up.
Step 1: Set the denominator equal to zero
Find all x-values where the denominator is zero. These are the candidates for vertical asymptotes.
Step 2: Factor both numerator and denominator
Factor completely. This reveals any common factors.
Step 3: Cancel common factors
A factor that appears in both numerator and denominator cancels. The x-value from that factor is a hole, not a vertical asymptote.
Step 4: The remaining denominator zeros are your vertical asymptotes
After cancellation, any x-value that still makes the denominator zero is a vertical asymptote.
Worked example: f(x) = (3x − 3) / (x² − 3x)
Factor: 3(x − 1) / [x(x − 3)].
Denominator zeros: x = 0 and x = 3.
No common factors with the numerator (the numerator has the factor x − 1, which is different).
Vertical asymptotes at x = 0 and x = 3.
Worked example: f(x) = (3x + 3) / (x² − x − 2)
Factor: 3(x + 1) / [(x − 2)(x + 1)].
Common factor: (x + 1) cancels.
After cancellation: 3 / (x − 2).
x = −1 is a hole (removable discontinuity), not an asymptote.
Vertical asymptote at x = 2 only.
Worked example: f(x) = x / (x² + 1)
Denominator: x² + 1 = 0 has no real solutions (x² = −1 is impossible for real numbers).
No vertical asymptotes. Answer: NONE.
The critical distinction
If a factor cancels between numerator and denominator, the function has a hole at that x-value, not a vertical asymptote. The function can be "repaired" at a hole by defining a single value there.
If the factor does not cancel, the function genuinely blows up, producing a vertical asymptote.
Worked example: f(x) = (2x − 6) / (x² − 8x + 15)
Factor numerator: 2(x − 3).
Factor denominator: (x − 3)(x − 5).
Common factor: (x − 3) cancels.
After cancellation: 2 / (x − 5).
x = 3 is a hole. x = 5 is the vertical asymptote.
Worked example: f(x) = 3x / (x + 1)²
Numerator: 3x. Denominator: (x + 1)².
No common factors.
Denominator zero: x = −1 (with multiplicity 2).
Vertical asymptote at x = −1.
Worked example: f(x) = 9 / (x − 7)
Already simplified. Denominator zero at x = 7.
Vertical asymptote at x = 7.
Where trig asymptotes come from
Functions like sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x) blow up wherever cos(x) = 0.
cos(x) = 0 at x = π/2 + nπ for any integer n. On the interval [0, π], that means x = π/2.
Worked example: f(x) = x³ sec(x) on [0, π]
sec(x) = 1/cos(x), so f(x) = x³/cos(x).
cos(x) = 0 at x = π/2 within [0, π].
The numerator x³ at x = π/2 is (π/2)³, which is not zero.
Vertical asymptote at x = π/2.
General approach for trig functions
Rewrite sec, csc, tan, or cot in terms of sin and cos.
Find where the denominator (cos or sin) equals zero within the given interval.
Check that the numerator is not also zero at that point (if it is, further analysis is needed).
Students often report every denominator zero as a vertical asymptote. It is not. If the numerator and denominator share a common factor at that zero, the factor cancels and the result is a hole, not an asymptote. Always factor first.
Students sometimes think x² + 1 = 0 has real solutions and report vertical asymptotes for functions like x / (x² + 1). It does not. x² + 1 is always positive for real x, so no asymptote exists.
Students often confuse vertical and horizontal asymptotes. Vertical asymptotes are about what happens as x approaches a finite value (the output blows up). Horizontal asymptotes are about what happens as x goes to ±∞ (the output levels off). They are separate questions.
Students sometimes forget to restrict the interval for trig problems. If the problem says "on [0, π]," only list asymptotes within that interval, not every place the trig function is undefined.
⚠️ "Find the vertical asymptotes" is one of the most common question types in Calculus I. It appears on WeBWorK, quizzes, and exams.
⚠️ The hole-vs-asymptote distinction is a favourite exam trap. Expect at least one problem where a common factor cancels and the "obvious" answer is wrong.
⚠️ Problems 9 through 12 in this assignment have multiple parts: identify asymptotes, evaluate limits near them, and match the graph. On the exam, you may need to determine the sign of the function on each side of the asymptote (positive infinity vs negative infinity) to pick the correct graph.
⚠️ Trig-based asymptote questions (like x³ sec(x) on an interval) test whether you can connect trig identities to the rational-function framework. Rewrite everything in terms of sin and cos.
True or False: If the denominator of a rational function equals zero at x = 4, there is always a vertical asymptote at x = 4. (False, it could be a hole if the numerator also has the factor (x − 4).)
Fill in the blank: f(x) = (3x − 3) / (x² − 3x) has vertical asymptotes at x = ____ and x = ____. (0 and 3)
True or False: f(x) = x / (x² + 1) has no vertical asymptotes. (True)
Fill in the blank: For f(x) = (3x + 3) / (x² − x − 2), after cancelling the common factor, the only vertical asymptote is at x = ____. (2)
True or False: sec(x) has a vertical asymptote wherever sin(x) = 0. (False, sec(x) = 1/cos(x), so its asymptotes are where cos(x) = 0.)
Q: Find all vertical asymptotes of f(x) = (2x − 6) / (x² − 8x + 15).
A: Factor: 2(x − 3) / [(x − 3)(x − 5)]. The (x − 3) cancels, leaving 2 / (x − 5). x = 3 is a hole. The vertical asymptote is x = 5.
Q: Find all vertical asymptotes of f(x) = 9 / (x − 7).
A: The denominator is zero at x = 7. The numerator is 9 (never zero). Vertical asymptote at x = 7.
Q: Find all vertical asymptotes of f(x) = 3x / (x + 1)².
A: Denominator zero at x = −1. Numerator at x = −1 is 3(−1) = −3, which is not zero. No cancellation. Vertical asymptote at x = −1.
Q: Does f(x) = x / (x² + 1) have any vertical asymptotes?
A: x² + 1 = 0 has no real solutions. No vertical asymptotes. Answer: NONE.
Q: Find all vertical asymptotes of f(x) = x³ sec(x) on [0, π].
A: Rewrite as x³ / cos(x). cos(x) = 0 at x = π/2 within [0, π]. The numerator at x = π/2 is (π/2)³ ≠ 0. Vertical asymptote at x = π/2.
Vertical asymptotes are closely tied to continuity. A function cannot be continuous at a vertical asymptote, so identifying asymptotes helps you map out a function's discontinuities, which is the focus of the next section in most Calculus I courses.
Later, when you study curve sketching (typically around Chapter 4), you will combine vertical asymptotes with first- and second-derivative information to draw accurate graphs of rational functions.
In Calculus II, improper integrals sometimes involve vertical asymptotes within the interval of integration. Recognising them is essential for deciding whether such an integral converges or diverges.
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