Domain and Graph Transformations of Rational Functions
The graph of a rational function such as f(x) = a/(x − h) + k is the hyperbola y = a/x shifted h units horizontally and k units vertically, with the domain x ≠ h and the asymptotes x = h and y = k. In brief: for f(x) = 2/(x − 3) + 1, the domain is x ≠ 3, the vertical asymptote is x = 3, the horizontal asymptote is y = 1 and the range is y ≠ 1. The domain of any rational function excludes the zeros of its denominator.
Transformations let you sketch many rational functions quickly from the basic graph of y = 1/x, without a table of values. A function given as a quotient of linear expressions can be rewritten in the shifted form by division, which makes its asymptotes visible. This guide explains how to find the domain, how the parameters a, h and k transform the graph, how to rewrite (ax + b)/(cx + d) and how to sketch the graph, with a table of examples.
How do you find the domain of a rational function?
Set the denominator equal to zero and exclude its zeros from the real numbers. The domain is determined by the formula as given, before any simplification.
- f(x) = 3/(x + 5): x ≠ −5.
- f(x) = (x − 1)/(x2 − 4): x ≠ ±2.
- f(x) = x/(x2 + 1): all real numbers.
Using interval notation, the first domain is (−∞, −5) ∪ (−5, ∞).
How do transformations change the graph?
Start with y = 1/x, which has the asymptotes x = 0 and y = 0 and passes through (1, 1) and (−1, −1). For g(x) = a/(x − h) + k:
- h: shifts the graph right (h > 0) or left (h < 0). The vertical asymptote moves to x = h.
- k: shifts the graph up (k > 0) or down (k < 0). The horizontal asymptote moves to y = k.
- a: stretches the graph vertically by |a| (the branches move away from the asymptotes for |a| > 1). If a < 0, the graph is reflected so that the branches lie in quadrants II and IV relative to the asymptotes’ crossing point.
The center of symmetry of the hyperbola is the point (h, k), the intersection of the asymptotes. Domain: x ≠ h. Range: y ≠ k.
Example 1: f(x) = 1/(x + 2) − 3
h = −2 and k = −3. The asymptotes are x = −2 and y = −3. The center is (−2, −3). Points: x = −1 gives 1 − 3 = −2, so (−1, −2); x = −3 gives −1 − 3 = −4, so (−3, −4). The y-intercept: f(0) = 1/2 − 3 = −2.5. The zero: 1/(x + 2) = 3, x = −2 + 1/3 = −5/3.
Example 2: f(x) = −2/(x − 1) + 4
Asymptotes x = 1 and y = 4. Since a = −2 < 0, the branches are in the upper-left and lower-right parts relative to (1, 4). Points: x = 2 gives −2 + 4 = 2; x = 0 gives 2 + 4 = 6; x = 3 gives −1 + 4 = 3. The zero: −2/(x − 1) = −4, x − 1 = 1/2, so x = 1.5.
How do you rewrite (ax + b)/(cx + d)?
A quotient of two linear expressions can be written as a constant plus a fraction by polynomial division. The horizontal asymptote is y = a/c and the vertical asymptote is x = −d/c.
Example 3: f(x) = (2x + 1)/(x − 3)
Divide: 2x + 1 = 2(x − 3) + 7. So f(x) = 2 + 7/(x − 3). The center is (3, 2). The asymptotes are x = 3 and y = 2. The y-intercept is f(0) = 1/(−3) = −1/3, and the zero is x = −1/2. Check: 2 + 7/(x − 3) at x = 0: 2 − 7/3 = −1/3 ✓.
Example 4: f(x) = (3x − 5)/(2x + 4)
Write 3x − 5 = (3/2)(2x + 4) − 11. So f(x) = 3/2 − 11/(2x + 4) = 3/2 − 11/(2(x + 2)) = 3/2 − (11/2)/(x + 2). The asymptotes are x = −2 and y = 3/2.
Example 5: horizontal shift and vertical stretch together
f(x) = (4 − x)/(x + 1) = −1 + 5/(x + 1). The center is (−1, −1). The graph passes through (0, 4) and (4, 0) and (−2, −6).
How do you sketch the graph step by step?
- Find the domain and draw the vertical asymptote(s) as dashed lines.
- Find the horizontal asymptote and draw it.
- Find the intercepts: f(0) and the zeros of the numerator.
- Plot a few extra points on each side of the vertical asymptote.
- Draw each branch approaching the asymptotes, without crossing the vertical asymptote.
For the hyperbola y = a/(x − h) + k, the two branches are in opposite “corners” formed by the asymptotes. The sign of a decides which corners.
Table of examples
| Function | Vertical asymptote | Horizontal asymptote | Domain | Range |
|---|---|---|---|---|
| 1/x | x = 0 | y = 0 | x ≠ 0 | y ≠ 0 |
| 1/(x + 2) − 3 | x = −2 | y = −3 | x ≠ −2 | y ≠ −3 |
| −2/(x − 1) + 4 | x = 1 | y = 4 | x ≠ 1 | y ≠ 4 |
| (2x + 1)/(x − 3) | x = 3 | y = 2 | x ≠ 3 | y ≠ 2 |
| (3x − 5)/(2x + 4) | x = −2 | y = 3/2 | x ≠ −2 | y ≠ 3/2 |
| (4 − x)/(x + 1) | x = −1 | y = −1 | x ≠ −1 | y ≠ −1 |
Where do these ideas come from?
The study of the graphs of rational functions started with the coordinate geometry of Descartes and Fermat. The classification of curves defined by equations xy + ax + by + c = 0 as hyperbolas goes back to the conic sections of Apollonius. Asymptotes were studied by Apollonius for the hyperbola and later extended to general curves in the seventeenth and eighteenth centuries.
Where are they used today?
The average cost per item C(n) = (500 + 3n)/n = 3 + 500/n is a shifted reciprocal function: the horizontal asymptote y = 3 is the variable cost per item, which the average cost approaches as production grows. Concentration of a solution after adding water, the combined work rate and the focal length formulas are all of this type. Rational functions of a higher degree are used in engineering as transfer functions.
Which mistakes are common?
Reading the shift with the wrong sign
In 1/(x + 2), the vertical asymptote is x = −2, not x = 2.
Mixing up the asymptotes
The vertical asymptote comes from the denominator (x = h), and the horizontal asymptote is the added constant k.
Drawing the graph through the asymptote
The curve never meets the vertical asymptote and approaches the horizontal asymptote without being equal to it for finite x.
Forgetting that the horizontal asymptote is not in the range
For a/(x − h) + k, the value y = k is never attained.