Understanding functions with polynomial ratios and their asymptotic behavior
Reserve & Extensions • K-12
A rational function is a ratio of two polynomials:
These functions have fascinating behavior near values where the denominator equals zero, producing vertical lines the graph approaches but never crosses, and horizontal lines the graph settles toward for very large x.
Since division by zero is undefined, the domain of a rational function excludes all x-values where q(x) = 0.
Find the domain of f(x) = (x + 3) / (x2 − 4).
Set the denominator equal to zero: x2 − 4 = 0, so x2 = 4, giving x = 2 or x = −2.
Domain: all real numbers except x = 2 and x = −2.
A vertical asymptote occurs at x = a if the denominator is zero at x = a but the numerator is not. The graph shoots toward positive or negative infinity near x = a.
If both the numerator and denominator are zero at x = a, the common factor cancels, leaving a hole -- a single missing point in the graph.
Analyze f(x) = (x2 − 1) / (x2 − x − 2).
Factor: f(x) = (x − 1)(x + 1) / [(x − 2)(x + 1)]
The factor (x + 1) appears in both numerator and denominator, so x = −1 produces a hole, not an asymptote. Cancel to get f(x) = (x − 1)/(x − 2), with the note that x ≠ −1.
The remaining denominator zero is x = 2 (numerator is 2 − 1 = 1 ≠ 0), so x = 2 is a vertical asymptote.
The hole is at x = −1. Plugging into the simplified form: y = (−1 − 1)/(−1 − 2) = −2/−3 = 2/3. The hole is at (−1, 2/3).
The horizontal asymptote describes the end behavior -- what happens as x grows very large (positive or negative).
| Degree of Numerator vs. Denominator | Horizontal Asymptote |
|---|---|
| Numerator degree < Denominator degree | y = 0 |
| Numerator degree = Denominator degree | y = (leading coefficient of p) / (leading coefficient of q) |
| Numerator degree > Denominator degree | No horizontal asymptote (slant or other end behavior) |
Analyze and sketch key features of f(x) = 2x / (x − 3).
The graph approaches x = 3 vertically and settles toward y = 2 horizontally.
A vertical asymptote is NOT just "denominator = 0." You must check that the numerator is non-zero at that point. If both are zero, you have a hole, not an asymptote. Always factor first.
To sketch a rational function: (1) Factor fully. (2) Find domain restrictions. (3) Identify asymptotes and holes. (4) Find intercepts. (5) Test a point in each region between vertical asymptotes. (6) Draw the curve approaching asymptotes.
1. Find the vertical asymptote(s) and horizontal asymptote of f(x) = 5 / (x + 2).
Vertical asymptote: x = −2 (denominator zero). Horizontal asymptote: degree of numerator (0) < degree of denominator (1), so y = 0.
2. Find and classify all discontinuities of f(x) = (x2 − 9) / (x − 3).
Factor: (x − 3)(x + 3) / (x − 3). Both are zero at x = 3, so there is a hole at x = 3. Simplified: f(x) = x + 3 (x ≠ 3). The hole is at (3, 6). No vertical asymptotes.
3. Find the horizontal asymptote of f(x) = (3x2 + 1) / (5x2 − 2).
Same degree (both 2). Horizontal asymptote: y = 3/5.
4. Determine all asymptotes and intercepts of f(x) = (x + 1) / (x2 − x − 6).
Factor denominator: (x − 3)(x + 2). No common factors with numerator. Vertical asymptotes: x = 3 and x = −2. Horizontal asymptote: degree 1 < degree 2, so y = 0. x-intercept: x = −1 (set numerator = 0). y-intercept: f(0) = 1/(−6) = −1/6.
5. Does f(x) = x2 / (x + 1) have a horizontal asymptote? Explain.
No. The degree of the numerator (2) is greater than the degree of the denominator (1). There is no horizontal asymptote. Instead, there is a slant (oblique) asymptote, found by polynomial long division: y = x − 1.