A rational function has its variable in the denominator, and it starts from inverse proportion y = k/x, where the product of two quantities stays fixed, like the time it takes to fill a bath.
Its graph is a hyperbola in two branches that gets ever closer to the x-axis and y-axis without ever touching them; such lines are called asymptotes.
y = k/(x − a) + b is y = k/x translated, so the point (a, b) where the two asymptotes meet becomes the centre of the curve, and even a complicated general form tidies into this shape when you divide the numerator by the denominator.
In the first picture the quadrants the hyperbola lies in and a rectangle of area k change with k; in the second, the asymptotes and the curve move together with a and b.
Rational functions start from inverse proportion
💧 A tap and a bath
①Suppose you fill a bath with 12 L of water
②At 1 L a minute it takes 12 minutes, at 2 L a minute 6 minutes, at 3 L a minute 4 minutes
③As the flow per minute x goes up, the time y goes down, and the product xy is always 12
④Writing this inverse proportion as a formula gives y = k/x
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The hyperbola y = k/x as k changes, with a rectangle of area k whose opposite corners are the origin and a point on the curve
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The basic rational function
y = kx (x ≠ 0)
xy = k: the product of the two variables is always the same
📐 Properties of the basic form
①f(−x) = −f(x), so the graph is symmetric about the origin
②If k > 0 the graph lies in the first and third quadrants; if k < 0, in the second and fourth
③The larger |k| is, the further the graph sits from the origin
④The asymptotes are the x-axis (y = 0) and the y-axis (x = 0): the graph gets ever closer to both but never meets them
⑤The rectangle in the picture always has area k, showing that for any point on the curve the product of its coordinates is k
Translation and asymptotes
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y = k/x (blue) translated by a in the x-direction and b in the y-direction (gold), with the two asymptotes
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Standard form
y = kx − a + b
Asymptotes x = a and y = b
🔑 Why the meeting point of the asymptotes matters
①The graph of y = k/x is symmetric about the origin (0, 0)
②y = k/(x − a) + b is that graph moved by (a, b), so it is symmetric about the point (a, b)
③(a, b) is where the asymptotes x = a and y = b meet
④The domain is all real x except x = a, and the range is all real y except y = b
From general form to standard form
General form
y = cx + dex + f
Divide the numerator by the denominator to get quotient + remainder/denominator
🧮 Converting y = (2x + 5)/(x + 1)
①2x + 5 = 2(x + 1) + 3, so dividing the numerator by the denominator gives quotient 2 and remainder 3
②So y = 2 + 3/(x + 1) = 3/(x + 1) + 2
③In standard form, k = 3, a = −1, b = 2
④The asymptotes are x = −1 and y = 2
⑤The graph is symmetric about the point (−1, 2)
Shortcuts
Reading straight from the general form
Horizontal asymptote y = b
e.g. for (2x + 5)/(x + 1), b = 2/1 = 2
(coefficient of x on top) ÷ (coefficient of x underneath)
Vertical asymptote x = a
e.g. x + 1 = 0 gives a = −1
The x that makes the denominator 0
Value of k
e.g. 2x + 5 = 2(x + 1) + 3 gives k = 3
When the denominator's x-coefficient is 1, the remainder on dividing top by bottom
Inverses of rational functions
Inverse function
y = kx − a + b ↔ y = kx − b + a
The roles of the asymptotes swap (a ↔ b); the two graphs are symmetric about the line y = x
🪞 A function that is its own inverse
①Swapping x and y in y = k/x gives x = k/y, that is y = k/x, the same as before
②So y = k/x is its own inverse
③The inverse of y = k/(x − a) + b is y = k/(x − b) + a: a and b trade places
④If a = b, the function is again its own inverse, and its graph is symmetric about the line y = x
Worked examples and unit test
Example 1
Find the equations of the asymptotes of the graph of y = 2/(x − 3) + 1.
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For y = k/(x − p) + q, the asymptotes are x = p and y = q.
y = 2/(x − 3) + 1
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Read off p = 3 and q = 1.
vertical asymptote x = 3, horizontal asymptote y = 1
▸ x = 3, y = 1
Set k = 2, a = 3, b = 1 in the second picture to see this graph.
Example 2
Write y = (2x + 1)/(x − 1) in the form y = k/(x − p) + q and find the equations of its asymptotes.
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Divide the numerator by the denominator: 2x + 1 = 2(x − 1) + 3.
y = (2(x − 1) + 3)/(x − 1) = 2 + 3/(x − 1)
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Read the asymptotes from the standard form.
y = 3/(x − 1) + 2, asymptotes x = 1, y = 2
▸ x = 1, y = 2
The horizontal asymptote y = 2 equals the ratio 2/1 of the x-coefficients, so you can guess it even before dividing.
Summary
Standard form of a rational function
y = kx − a + b
Asymptotes x = a, y = b; domain x ≠ a; range y ≠ b
Grade-10 school exam type
The graph of y = 3/(x − 2) + 1 is symmetric about the point (a, b). What is a + b?
①1
②2
③3
④4
⑤5
▸ ③ 3
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A rational function y = k/(x − p) + q is symmetric about the intersection of its asymptotes, (p, q).
Center of symmetry = (p, q) = (2, 1)
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Add a = 2 and b = 1.
a + b = 2 + 1 = 3
🎯 Points that often come up
①The vertical asymptote is the x that makes the denominator 0; the horizontal asymptote comes from the ratio of the x-coefficients on top and bottom
②If k > 0 the graph lies towards the first and third quadrants; if k < 0, towards the second and fourth
③Convert the general form (cx + d)/(ex + f) to standard form by division
④x = a is left out of the domain and y = b out of the range
⑤The inverse swaps a and b, and if a = b the function is its own inverse