1The idea in one paragraph

The reciprocal function f(x) = 1/x sends each number to one divided by it. It cannot accept 0, and it never outputs 0, so its graph has two separate pieces, called branches, that approach the axes without ever touching them: the axes are its asymptotes. Swapping x and y leaves y = 1/x unchanged, so it is its own inverse. The rational functions of this course, f(x) = (ax + b)/(cx + d), are the same curve stretched and moved. Each has one vertical asymptote, where the denominator is zero, x = −d/c, and one horizontal asymptote, the value approached for large x, y = a/c. With those two lines and the intercepts with the axes, you can sketch any of them.

2The reciprocal function 1/x

Build a table and the shape appears.

Two things happen. As x gets close to 0, 1/x becomes huge: 1/0.001 = 1000, and 1/(−0.001) = −1000. So near x = 0 the graph shoots up on the right and down on the left, and the line x = 0 (the y-axis) is a vertical asymptote. As x gets large, positive or negative, 1/x gets close to 0 without reaching it: 1/1000 = 0.001. So the line y = 0 (the x-axis) is a horizontal asymptote.

Figure 1 shows the result, a curve called a rectangular hyperbola. Its key facts:

The branches lie in the first quadrant (x and y both positive) and the third (both negative), because a positive number has a positive reciprocal and a negative one a negative reciprocal.

3The reciprocal function is its own inverse

A function is self-inverse if f⁻¹ = f: doing it twice gets you back where you started. The reciprocal function has this property, and there are three ways to see it.

By composition. f(f(x)) = 1/(1/x) = x, for every x ≠ 0. Since f ∘ f is the identity, f undoes itself.

By the swap method of 2.5. Write y = 1/x and swap x and y: x = 1/y. Rearranging gives y = 1/x, the same function.

By the graph. The graph of f⁻¹ is the reflection of the graph of f in y = x. Figure 1 is symmetric in y = x (the point (4, ¼) reflects to (¼, 4), which is also on the curve), so the reflection is the same curve.

The same is true of y = k/x for any constant k ≠ 0: swap x and y in y = k/x and you get x = k/y, so y = k/x again. Section 6 shows which other rational functions are self-inverse.

1/x is self-inverse: f(f(x)) = x, so f⁻¹(x) = 1/x. Its graph is symmetric in the line y = x.

4Stretching and moving 1/x

Every graph on this page is y = 1/x after a stretch and a translation, which 2.11 treats in full. Two cases are worth recognising now; Figure 2 shows both.

y = k/x is a vertical stretch of 1/x by factor k. The asymptotes are still the axes, but the curve sits further out: y = 4/x passes through (2, 2) instead of (1, 1). If k < 0, the branches move into the second and fourth quadrants.

y = 1/(x − h) + k is 1/x translated h to the right and k up. The asymptotes move with it, to x = h and y = k. For y = 1/(x − 2) + 3, the asymptotes are x = 2 and y = 3. The intercepts come from the usual rules: at x = 0, y = 1/(−2) + 3 = 2.5; and y = 0 when 1/(x − 2) = −3, so x − 2 = −⅓ and x = 5/3.

Put over one denominator, 1/(x − 2) + 3 = (1 + 3(x − 2))/(x − 2) = (3x − 5)/(x − 2). That is a function of the form (ax + b)/(cx + d), which is where the next section starts.

5Rational functions (ax + b)/(cx + d)

A rational function in this course is one of the form

f(x) = (ax + b)/(cx + d), with c ≠ 0, x ≠ −d/c.

Its graph is always a hyperbola like the one in Figure 1, moved and stretched, and you can find everything you need to sketch it from a, b, c and d.

The vertical asymptote is x = −d/c. The denominator cx + d is zero when x = −d/c. The function is undefined there, and near it the denominator is tiny, so the output is huge.

The horizontal asymptote is y = a/c. For very large x, the constants b and d hardly matter: (ax + b)/(cx + d) behaves like ax/cx = a/c. You can make this precise by dividing top and bottom by x: (a + b/x)/(c + d/x), and as x → ±∞ both b/x and d/x go to 0, leaving a/c.

The y-intercept is (0, b/d), found by putting x = 0 (if d ≠ 0).

The x-intercept is (−b/a, 0), found by setting the numerator to zero, since a fraction is zero only when its top is zero (if a ≠ 0).

For f(x) = (ax + b)/(cx + d): vertical asymptote x = −d/c, horizontal asymptote y = a/c.

Learn both, or rederive them in the exam in two lines as above: denominator zero for the vertical one, large x for the horizontal one.

The domain is x ≠ −d/c and the range is y ≠ a/c: the function takes every value except the one it approaches forever.

Worked example. Sketch y = (4x + 2)/(2x − 3). Here a = 4, b = 2, c = 2, d = −3.

Now place the branches. Draw the two asymptotes as dashed lines; they cross at (3/2, 2) and divide the plane into four regions. The two intercepts are both in the bottom-left region, so one branch lives there. The other branch is in the opposite region, top right. A test point confirms it: at x = 3, y = 14/3 ≈ 4.67, which is above y = 2 and to the right of x = 3/2. Figure 3 shows the finished sketch.

Why the branches go where they do. Dividing out gives (4x + 2)/(2x − 3) = 2 + 8/(2x − 3). Check: 2(2x − 3) + 8 = 4x + 2. So the graph is 2 plus a reciprocal-type term. When 2x − 3 > 0 (to the right of the vertical asymptote), the extra term is positive and the curve is above y = 2; to the left it is negative and the curve is below. That one line of algebra tells you which regions the branches are in without a test point.

Special cases.

• If a = 0, the function is b/(cx + d). Then the horizontal asymptote is y = 0 and there is no x-intercept, because the numerator is never zero. For example 6/(x + 2) has asymptotes x = −2 and y = 0 and y-intercept (0, 3).
• If d = 0, the vertical asymptote is x = 0, the y-axis, and there is no y-intercept. For example (4x + 600)/x has asymptotes x = 0 and y = 4.
• If the top is a multiple of the bottom, as in (2x + 4)/(x + 2) = 2, the fraction cancels to a constant and there is no hyperbola at all, only the line y = 2 with a gap at x = −2. Exam questions avoid this case, but it explains why the curve always needs ad ≠ bc.

6The inverse of a rational function

Find the inverse of f(x) = (4x + 2)/(2x − 3) by the method of 2.5, with one extra algebraic step: the new y appears twice, so collect the y terms and factorise.

The inverse is also a rational function. Its asymptotes are x = 2 and y = 3/2: exactly those of f, with x and y swapped. That is no coincidence. The graph of f⁻¹ is the reflection of f in y = x, so a vertical asymptote of f becomes a horizontal asymptote of f⁻¹ and the other way round. Figure 4 shows both graphs. This also gives a quick check on the domain and range: the range of f, y ≠ 2, is the domain of f⁻¹, x ≠ 2.

Which rational functions are self-inverse? A graph that is its own reflection in y = x must have its asymptotes cross on the line y = x. The crossing point is (−d/c, a/c), so this needs −d/c = a/c, that is d = −a. For example g(x) = (2x + 5)/(x − 2) has a = 2 and d = −2, and indeed:

To show a function is self-inverse in an exam, do exactly that: find f⁻¹ and show it equals f, or show that f(f(x)) simplifies to x.

7Rational functions in context

Rational functions model anything shared out: a fixed cost spread over more and more items, a fixed distance covered at different speeds, a dose spread through a growing volume. The horizontal asymptote is then the long-run value, and it usually has a meaning worth stating.

A small firm prints T-shirts. It pays €600 to set up a design, then €4 for each shirt. The average cost per shirt for n shirts is

A(n) = (4n + 600)/n = 4 + 600/n.

Figure 5 shows the graph for n > 0. At n = 100 the average cost is €10; at n = 600 it is €5. The horizontal asymptote is A = 4: however many shirts are printed, the setup cost shared among them shrinks towards nothing, and the average cost falls towards the €4 each shirt costs to make, without ever reaching it. The vertical asymptote n = 0 has no practical meaning, since you cannot print zero shirts, and the domain is n ∈ ℤ⁺.

A question might ask how many shirts are needed for an average cost below €5. Then 4 + 600/n < 5, so 600/n < 1, so n > 600: at least 601 shirts. Multiplying through by n was allowed because n is positive; with a variable that could be negative, you would have to be careful, since multiplying by a negative flips an inequality.