This note covers the reciprocal function \(f(x) = \dfrac{1}{x}\) and rational functions of the form \(\dfrac{ax + b}{cx + d}\). You will learn how their graphs look, how to find their asymptotes and intercepts, and why the reciprocal function is its own inverse. These graphs model situations where a quantity levels off, such as an average cost.
Key ideas
The reciprocal function. \(f(x) = \dfrac{1}{x}\) is defined for all \(x\) except \(x = 0\). Its graph has two separate branches, one in the top-right and one in the bottom-left of the axes.
- As \(x\) gets very large (positive or negative), \(\dfrac{1}{x}\) gets close to 0. So the \(x\)-axis, \(y = 0\), is a horizontal asymptote.
- As \(x\) gets close to 0, \(\dfrac{1}{x}\) becomes very large in size. So the \(y\)-axis, \(x = 0\), is a vertical asymptote.
- There are no intercepts with either axis.
- The domain is \(x \neq 0\) and the range is \(f(x) \neq 0\).
- The graph is symmetrical about the line \(y = x\). This means that \(f\) is self-inverse: \(f^{-1}(x) = \dfrac{1}{x}\).
Rational functions of the form \(f(x) = \dfrac{ax + b}{cx + d}\) (with \(c \neq 0\)). Their graphs are the reciprocal graph, stretched, moved sideways and moved up or down.
- Vertical asymptote: where the denominator is 0, so \(x = -\dfrac{d}{c}\). The function is not defined there.
- Horizontal asymptote: as \(x\) becomes very large, the constants \(b\) and \(d\) matter less, and \(f(x)\) approaches \(\dfrac{a}{c}\). So the line is \(y = \dfrac{a}{c}\).
- \(x\)-intercept: set the numerator to 0: \(x = -\dfrac{b}{a}\) (if \(a \neq 0\)).
- \(y\)-intercept: put \(x = 0\): \(y = \dfrac{b}{d}\) (if \(d \neq 0\)).
- Domain: \(x \neq -\dfrac{d}{c}\). Range: \(f(x) \neq \dfrac{a}{c}\).
Another helpful method is to divide the numerator by the denominator and rewrite the function as \(f(x) = \dfrac{k}{x - p} + q\). Then the vertical asymptote is \(x = p\) and the horizontal asymptote is \(y = q\). The sign of \(k\) tells you the shape: if \(k \gt 0\), the branches are in the top-right and bottom-left relative to the asymptotes' crossing point. If \(k \lt 0\), they are in the top-left and bottom-right.
Sketching. Draw the asymptotes as dashed lines and label their equations. Mark the intercepts. Then draw each branch so that it passes through the intercepts and approaches the asymptotes without touching or crossing them.
Solving equations. To solve \(\dfrac{ax + b}{cx + d} = k\), multiply both sides by the denominator and solve the linear equation. Check that your answer does not make the denominator 0. You can also use a GDC to graph the function and the line \(y = k\), then find the intersection.
Self-inverse rational functions (HL). If \(f(x) = \dfrac{ax + b}{cx - a}\), that is, if the number in the denominator is the opposite of the \(x\) coefficient in the numerator, then \(f\) is its own inverse, so \(f(f(x)) = x\).
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(f(x) = \dfrac{1}{x}\) | Reciprocal function. Domain \(x \neq 0\). Asymptotes \(x = 0\) and \(y = 0\). |
| \(f(x) = \dfrac{ax + b}{cx + d}\) | Rational function with a linear numerator and denominator. |
| Vertical asymptote | \(x = -\dfrac{d}{c}\), where the denominator is 0. |
| Horizontal asymptote | \(y = \dfrac{a}{c}\), the value \(f(x)\) approaches for large \(|x|\). |
| \(f(x) = \dfrac{k}{x - p} + q\) | Shifted form. Asymptotes \(x = p\) and \(y = q\). |
| Self-inverse | \(f^{-1} = f\). True for \(\dfrac{1}{x}\) (SL and HL) and for \(\dfrac{ax + b}{cx - a}\) (HL). |
Worked examples
Example 1 For \(f(x) = \dfrac{2x + 1}{x - 3}\), find the equations of the asymptotes and the coordinates of the axis intercepts, and state the domain and range.
Step 1: vertical asymptote. The denominator is 0 when \(x = 3\). So \(x = 3\).
Step 2: horizontal asymptote. Compare the coefficients of \(x\) on top and bottom: \(\dfrac{2}{1} = 2\). So \(y = 2\).
Step 3: \(x\)-intercept. The numerator is 0 when \(2x + 1 = 0\), so \(x = -\dfrac{1}{2}\). The point is \(\left(-\dfrac{1}{2}, 0\right)\).
Step 4: \(y\)-intercept. When \(x = 0\), \(f(0) = \dfrac{1}{-3} = -\dfrac{1}{3}\). The point is \(\left(0, -\dfrac{1}{3}\right)\).
Step 5: domain and range. Domain: \(x \neq 3\). Range: \(f(x) \neq 2\).
Answer: asymptotes \(x = 3\) and \(y = 2\); intercepts \(\left(-\dfrac{1}{2}, 0\right)\) and \(\left(0, -\dfrac{1}{3}\right)\); domain \(x \neq 3\); range \(f(x) \neq 2\).
Example 2 Sketch \(g(x) = \dfrac{3}{x + 2} - 1\), showing the asymptotes and the intercepts.
Step 1: the function is in the form \(\dfrac{k}{x - p} + q\) with \(p = -2\) and \(q = -1\). The vertical asymptote is \(x = -2\) and the horizontal asymptote is \(y = -1\).
Step 2: \(y\)-intercept. \(g(0) = \dfrac{3}{2} - 1 = 0.5\). The point is \((0, 0.5)\).
Step 3: \(x\)-intercept. Solve \(\dfrac{3}{x + 2} - 1 = 0\), so \(\dfrac{3}{x + 2} = 1\), giving \(x + 2 = 3\) and \(x = 1\). The point is \((1, 0)\).
Step 4: shape. Since \(k = 3 \gt 0\), the branches are on the upper right and the lower left of the point where the asymptotes cross, which is \((-2, -1)\). The right branch passes through \((0, 0.5)\) and \((1, 0)\) and flattens towards \(y = -1\). The left branch is below \(y = -1\) and falls steeply towards the line \(x = -2\).
Answer: a two-branch curve with asymptotes \(x = -2\) and \(y = -1\), passing through \((0, 0.5)\) and \((1, 0)\).
Example 3 A workshop has a fixed cost of 500 euros and each item costs 8 euros to make. The average cost per item for \(n\) items is \(C(n) = \dfrac{500 + 8n}{n}\). (a) Write \(C(n)\) in the form \(a + \dfrac{b}{n}\) and find the horizontal asymptote. (b) Find the number of items for which the average cost is 12 euros.
(a) Split the fraction: \(C(n) = \dfrac{500}{n} + \dfrac{8n}{n} = 8 + \dfrac{500}{n}\). As \(n\) becomes very large, \(\dfrac{500}{n}\) approaches 0, so \(C\) approaches 8. The horizontal asymptote is \(C = 8\). This means the average cost can never fall below 8 euros.
(b) Set \(8 + \dfrac{500}{n} = 12\). Then \(\dfrac{500}{n} = 4\), so \(n = \dfrac{500}{4} = 125\).
Answer: (a) \(C(n) = 8 + \dfrac{500}{n}\), asymptote \(C = 8\); (b) 125 items.
Common errors
- Common error: finding the vertical asymptote at \(x = d\) instead of \(x = -\dfrac{d}{c}\). Set the whole denominator equal to 0 and solve.
- Common error: stating the horizontal asymptote as \(y = \dfrac{b}{d}\). That is the \(y\)-intercept. The asymptote uses the coefficients of \(x\): \(y = \dfrac{a}{c}\).
- Common error: letting the curve cross or touch an asymptote in a sketch. Each branch should get closer and closer without meeting the line.
- Common error: connecting the two branches into one curve. The graph has a break at the vertical asymptote.
- Common error: forgetting to check that a solution does not make the denominator 0.
Exam tip: write the equations of both asymptotes on your sketch, and label each intercept with its coordinates. These are the details that show you understand the shape.
Check yourself
- Find the equations of the asymptotes of \(f(x) = \dfrac{4x - 1}{2x + 5}\).
Answer
The denominator is 0 when \(x = -2.5\), so the vertical asymptote is \(x = -2.5\). The horizontal asymptote is \(y = \dfrac{4}{2} = 2\). - (HL) Show that \(f(x) = \dfrac{2x + 3}{x - 2}\) is self-inverse.
Answer
\(f(f(x)) = \dfrac{2\left(\dfrac{2x + 3}{x - 2}\right) + 3}{\dfrac{2x + 3}{x - 2} - 2}\). Multiply top and bottom by \(x - 2\): numerator \(4x + 6 + 3x - 6 = 7x\); denominator \(2x + 3 - 2x + 4 = 7\). So \(f(f(x)) = \dfrac{7x}{7} = x\), which means \(f^{-1} = f\). - State the domain and range of \(h(x) = \dfrac{5x}{x + 1}\).
Answer
The denominator is 0 when \(x = -1\), so the domain is \(x \neq -1\). The horizontal asymptote is \(y = \dfrac{5}{1} = 5\), so the range is \(h(x) \neq 5\). - Solve \(\dfrac{6}{x - 1} = 3\).
Answer
Multiply both sides by \(x - 1\): \(6 = 3(x - 1)\), so \(2 = x - 1\) and \(x = 3\). This does not make the denominator 0, so it is valid.