AP® Precalculus review sheet from Aim for Five (aimforfive.com/precalc/units/1/1-9)
Unit 1 · Topic 1.9
1.9 Rational Functions and Vertical Asymptotes
A vertical asymptote is an input where a rational function's outputs blow up toward ∞ or −∞. It happens where the denominator is zero and that zero isn't fully canceled by the numerator.
Key terms
- vertical asymptote
- unbounded
- factor
- limit notation
When there is a vertical asymptote
Let r(x) = p(x)/q(x). If a is a real zero of q but not a zero of p, the graph of r has a vertical asymptote at x = a.
If a is a zero of both, compare multiplicities. When a appears more times as a zero of the denominator than of the numerator, there is still a vertical asymptote at x = a. When the numerator's multiplicity is the same or greater, there's a hole instead (topic 1.10).
In practice: factor both parts, cancel common factors as far as they go, and any factor still left in the denominator gives a vertical asymptote.
What happens near the asymptote
Near x = a, the denominator gets very close to 0 while the numerator does not, so the fraction gets huge in size. The outputs increase or decrease without bound.
You describe each side with a one-sided limit. The notation x → a⁺ means x approaches a from the right (inputs greater than a), and x → a⁻ means from the left.
For example, lim (x→a⁺) r(x) = ∞ means the graph shoots up just to the right of x = a, and lim (x→a⁻) r(x) = −∞ means it drops down just to the left.
Finding the direction
To decide between ∞ and −∞ on each side, look at the signs of the factors for an input just to that side of a. The factor that becomes 0 is a tiny positive or tiny negative number, and the other factors keep their signs.
A shortcut: if the leftover factor (x − a) in the denominator has odd multiplicity, the two sides go in opposite directions. If it has even multiplicity, both sides go the same direction.
A graph never crosses its vertical asymptote, because the function is undefined at x = a.
Asymptotes in tables and contexts
In a table, a vertical asymptote shows up as outputs that get larger and larger in size as the inputs approach a. If r(2.9) = 100, r(2.99) = 1,000 and r(2.999) = 10,000, the outputs are growing without bound as x → 3⁻.
In a context, a vertical asymptote usually marks a limit that can't be reached. If the cost, in thousands of dollars, of removing p percent of a pollutant is C(p) = 80p/(100 − p), there's a vertical asymptote at p = 100: lim (p→100⁻) C(p) = ∞. Removing every last bit of the pollutant would cost an unlimited amount.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Finding all vertical asymptotes
Find the vertical asymptotes of r(x) = (x − 1)(x + 4) / ((x + 4)²(x − 5)).
Show the solutionHide the solution
- Step 1: The denominator is zero at x = −4 and x = 5.
- Step 2: x = 5 is not a zero of the numerator, so x = 5 is a vertical asymptote.
- Step 3: x = −4 is a zero of both. Its multiplicity is 2 in the denominator and 1 in the numerator. Canceling one factor leaves r(x) = (x − 1)/((x + 4)(x − 5)) for x ≠ −4, with (x + 4) still in the denominator.
- Step 4: Since 2 > 1, x = −4 is also a vertical asymptote, not a hole.
Answer: Vertical asymptotes at x = −4 and x = 5.
- Example 2
One-sided limits at the asymptotes
For the same function, r(x) = (x − 1)/((x + 4)(x − 5)) when x ≠ −4, find the one-sided limits at x = 5 and x = −4.
Show the solutionHide the solution
- Step 1: Just right of 5 (like x = 5.01): numerator ≈ 4 (positive), x + 4 ≈ 9 (positive), x − 5 is a tiny positive number. Positive over tiny positive gives a huge positive output, so lim (x→5⁺) r(x) = ∞.
- Step 2: Just left of 5: x − 5 is a tiny negative number, so lim (x→5⁻) r(x) = −∞.
- Step 3: Just right of −4 (like x = −3.99): numerator ≈ −5 (negative), x + 4 is tiny positive, x − 5 ≈ −9 (negative). The denominator (tiny positive)(−9) is a tiny negative number. Negative over tiny negative is a huge positive number, so lim (x→−4⁺) r(x) = ∞.
- Step 4: Just left of −4: x + 4 is tiny negative, so the denominator (tiny negative)(−9) is tiny positive. Negative over tiny positive gives lim (x→−4⁻) r(x) = −∞.
Answer: lim (x→5⁺) r(x) = ∞, lim (x→5⁻) r(x) = −∞, lim (x→−4⁺) r(x) = ∞, lim (x→−4⁻) r(x) = −∞.
- Example 3
An even-multiplicity asymptote
Describe the behavior of f(x) = 3/(x − 2)² near x = 2.
Show the solutionHide the solution
- Step 1: The denominator is zero at x = 2 and the numerator is 3, so there is a vertical asymptote at x = 2.
- Step 2: (x − 2)² is positive on both sides of 2, and it's tiny near 2. So 3 divided by a tiny positive number is a huge positive number on both sides.
- Step 3: Even multiplicity means both sides go the same direction.
Answer: lim (x→2⁻) f(x) = ∞ and lim (x→2⁺) f(x) = ∞: the graph shoots up on both sides of x = 2.
Common mistakes
- Calling x = a a hole just because a factor cancels. If a copy of the factor is left in the denominator after canceling, it's still a vertical asymptote.
- Guessing the direction of the one-sided limits. Check the sign of every factor just to each side of a.
- Writing a vertical asymptote as y = a. Vertical asymptotes are vertical lines: x = a.
On the exam
- Multiple-choice questions often give a factored rational function and ask for its vertical asymptotes, holes or one-sided limits. Factor and cancel first.
- When asked to justify an asymptote, give the limit statement, such as lim (x→5⁺) r(x) = ∞.
Connected topics
Videos
Check yourself
4 questions on 1.9 Rational Functions and Vertical Asymptotes. Pick an answer to see if you got it, and why.
The graph of which of the following functions has vertical asymptotes at x = −1 and x = 4 and no other vertical asymptotes?
Let r(x) = (x + 1)/(x − 2)². Which of the following is true?
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?
The function r is given by r(x) = (x² − 9)/(x² + x − 6).
Which of the following statements about r is true?
0 of 4 answered