Rational Functions — Complete Guide
Domain, vertical asymptotes, holes, horizontal asymptotes, oblique asymptotes, and graphing. Master every type of rational function behavior with worked examples.
Quick Reference — Asymptote Rules
Vertical Asymptotes & Holes
- Step 1: Factor numerator and denominator
- Common factors cancel → holes (removable discontinuities)
- Remaining denominator zeros → vertical asymptotes
- Hole coordinate: plug x into simplified form
Horizontal Asymptotes
- deg(num) < deg(den) → y = 0
- deg(num) = deg(den) → y = leading coeff ratio
- deg(num) > deg(den) → no HA
- deg(num) = deg(den) + 1 → oblique asymptote
What Is a Rational Function?
A rational function is a ratio of two polynomials:
f(x) = p(x) / q(x)
where p(x) and q(x) are polynomials and q(x) ≠ 0
The domain excludes any x-values that make the denominator zero. Always set q(x) = 0 and solve — those solutions are not in the domain.
Example: Find the domain of f(x) = (x + 3) / (x² − 9)
Set denominator = 0: x² − 9 = 0 → (x − 3)(x + 3) = 0 → x = 3 or x = −3
Domain: all real numbers except x = 3 and x = −3
Write as: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞)
Vertical Asymptotes
A vertical asymptote is a vertical line x = a where the function grows without bound. It occurs where the denominator equals zero and the numerator does not. You must factor and cancel common factors first — those give holes, not asymptotes.
Factor both numerator and denominator completely
Cancel any common factors — each canceled factor produces a hole
Set the remaining denominator equal to zero and solve — each solution is a vertical asymptote
Critical reminder
Always factor and simplify before identifying asymptotes. A canceled factor is a hole, not a vertical asymptote. Skipping this step is the most common mistake on rational function problems.
Holes (Removable Discontinuities)
A hole appears when the same factor cancels from both numerator and denominator. The function is undefined at that single x-value, but the graph looks continuous everywhere else — picture a curve with one invisible pinhole missing.
How to Find the Hole Coordinate
Identify the common factor that cancels, e.g. (x − a). The x-coordinate of the hole is x = a.
Plug x = a into the simplified (reduced) form of the rational function — not the original. This gives the y-coordinate.
Write the hole as a coordinate pair: (a, y). Plot it as an open circle on the graph.
Example: f(x) = (x² − 4) / (x − 2)
Factor: (x − 2)(x + 2) / (x − 2)
Cancel (x − 2) → simplified form: x + 2, with x ≠ 2
Hole at x = 2: plug into simplified form → 2 + 2 = 4 → hole at (2, 4)
No vertical asymptote — denominator has no remaining factors after cancellation.
Horizontal Asymptotes
A horizontal asymptote is a horizontal line y = k that the function approaches as x → ±∞. It is determined entirely by comparing the degrees of the numerator (n) and denominator (d).
| Degree Comparison | Horizontal Asymptote | Example | Note |
|---|---|---|---|
| deg(num) < deg(den) | y = 0 | (x + 1) / (x² + 4) | Numerator degree 1 < denominator degree 2 |
| deg(num) = deg(den) | y = a/b (ratio of leading coefficients) | (2x² + 3) / (5x² − 1) | y = 2/5 |
| deg(num) > deg(den) | None (check for oblique if deg(num) = deg(den) + 1) | (x² − 1) / (x − 2) | Do polynomial long division |
Important distinction
Unlike vertical asymptotes, a rational function can cross its horizontal asymptote for finite x-values. The horizontal asymptote only describes behavior as x → ±∞.
Oblique (Slant) Asymptotes
When the degree of the numerator is exactly one more than the degree of the denominator, there is no horizontal asymptote but there is an oblique asymptote — a slanted line y = mx + b that the graph approaches at the extremes.
How to Find the Oblique Asymptote
Perform polynomial long division of p(x) ÷ q(x). The quotient (ignoring the remainder) is the equation of the oblique asymptote.
Example: f(x) = (x² − 1) / (x − 2). Divide x² − 1 by x − 2.
x² ÷ x = x
x · (x − 2) = x² − 2x
Subtract: (x² − 1) − (x² − 2x) = 2x − 1
2x ÷ x = 2
2 · (x − 2) = 2x − 4
Subtract: (2x − 1) − (2x − 4) = 3 (remainder)
Result: x + 2 remainder 3 → f(x) = (x + 2) + 3/(x − 2)
Oblique asymptote: y = x + 2 (the remainder vanishes as x → ±∞)
Graphing Rational Functions — Step-by-Step
Follow these steps in order to produce an accurate sketch of any rational function.
Find the domain
Set the denominator equal to zero and solve. Every solution is excluded from the domain.
Factor and cancel
Factor numerator and denominator completely. Cancel common factors — each canceled factor is a hole, not an asymptote.
Find vertical asymptotes
Set the remaining denominator equal to zero. Each solution gives a vertical asymptote x = a.
Find holes
For each canceled factor (x − a), find the y-coordinate by plugging a into the simplified form. Plot as an open circle.
Find horizontal or oblique asymptote
Compare degrees of numerator and denominator. Use the three-case rule for HA, or do long division for oblique.
Find intercepts
y-intercept: evaluate f(0). x-intercepts: set numerator of simplified form equal to zero and solve.
Sign chart and sketch
Test the sign of f(x) in each interval created by vertical asymptotes and x-intercepts. Use sign info to determine whether branches go up or down near asymptotes. Sketch with smooth curves.
Worked Examples
f(x) = (x + 2) / (x² − 4)
Find all asymptotes, holes, intercepts, and domain.
Step 1 — Factor
Denominator: x² − 4 = (x + 2)(x − 2)
f(x) = (x + 2) / [(x + 2)(x − 2)]
Step 2 — Cancel common factor (x + 2)
Simplified: f(x) = 1 / (x − 2), x ≠ −2
Hole at x = −2: plug into simplified → 1/(−2 − 2) = −1/4
Hole at (−2, −1/4)
Step 3 — Vertical asymptote
Remaining denominator: x − 2 = 0 → x = 2 (vertical asymptote)
Step 4 — Horizontal asymptote
Simplified has deg(num) = 0 < deg(den) = 1 → y = 0
Step 5 — Intercepts
y-intercept: f(0) = 1/(0 − 2) = −1/2 → point (0, −1/2)
x-intercept: numerator of simplified = 1 ≠ 0 → no x-intercept
f(x) = (2x² + 3x − 2) / (x² − 1)
Find all asymptotes, holes, and the horizontal asymptote.
Step 1 — Factor both
Numerator: 2x² + 3x − 2 = (2x − 1)(x + 2)
Denominator: x² − 1 = (x − 1)(x + 1)
No common factors → no holes
Step 2 — Vertical asymptotes
Denominator zeros: x − 1 = 0 → x = 1; x + 1 = 0 → x = −1
Vertical asymptotes at x = 1 and x = −1
Step 3 — Horizontal asymptote
deg(num) = 2 = deg(den) = 2 → y = leading coeff ratio
y = 2/1 = 2 (horizontal asymptote)
Step 4 — Intercepts
y-intercept: f(0) = (0 + 0 − 2)/(0 − 1) = −2/−1 = 2 → point (0, 2)
x-intercepts: set numerator = 0 → (2x − 1)(x + 2) = 0 → x = 1/2 or x = −2
Both are in the domain → x-intercepts at (1/2, 0) and (−2, 0)
f(x) = (x² − 1) / (x − 2)
Find the oblique asymptote using polynomial long division.
Step 1 — Check degrees
deg(num) = 2, deg(den) = 1. Since 2 = 1 + 1, there is an oblique asymptote.
Step 2 — No common factors, so vertical asymptote at x = 2
Domain: all real numbers except x = 2
Step 3 — Polynomial long division
Divide x² − 1 by x − 2:
x² ÷ x = x → x(x − 2) = x² − 2x
Subtract: (x² − 1) − (x² − 2x) = 2x − 1
2x ÷ x = 2 → 2(x − 2) = 2x − 4
Subtract: (2x − 1) − (2x − 4) = 3
f(x) = x + 2 + 3/(x − 2)
Result
As x → ±∞, the term 3/(x − 2) → 0, so f(x) ≈ x + 2
Oblique asymptote: y = x + 2
Intercepts
y-intercept: f(0) = (0 − 1)/(0 − 2) = 1/2 → point (0, 1/2)
x-intercepts: x² − 1 = 0 → (x − 1)(x + 1) = 0 → x = 1 or x = −1 → points (1, 0) and (−1, 0)
Frequently Asked Questions
How do you find vertical asymptotes of a rational function?
First factor both the numerator and denominator completely. Cancel any common factors — each canceled factor gives a hole, not an asymptote. The remaining values that make the denominator equal zero are the vertical asymptotes. For example, in f(x) = (x+2)/(x²-4), factor the denominator to get (x+2)(x-2). The factor (x+2) cancels, giving a hole at x=-2. The remaining denominator factor (x-2) gives a vertical asymptote at x=2.
What are the three cases for horizontal asymptotes?
Horizontal asymptotes depend on comparing the degrees of the numerator n and denominator d. Case 1: if n < d, the horizontal asymptote is y = 0 (the x-axis). Case 2: if n = d, the horizontal asymptote is y = (leading coefficient of numerator) / (leading coefficient of denominator). Case 3: if n > d, there is no horizontal asymptote — instead there may be an oblique (slant) asymptote when n = d + 1.
What is a hole in a rational function?
A hole (removable discontinuity) occurs when the same factor cancels from both the numerator and denominator. The function is undefined at that x-value, but the graph looks like a continuous curve with a single missing point. To find the hole: set the canceled factor equal to zero to get the x-coordinate, then plug that x-value into the simplified (reduced) rational function to get the y-coordinate. For example, if (x+2) cancels and the simplified form is 1/(x-2), plug in x=-2 to get y = 1/(-2-2) = -1/4, so the hole is at (-2, -1/4).
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