Mathematics, five units · Grade 12

How do you find the asymptotes of a rational function?

A vertical asymptote sits where the denominator is zero and the numerator is not — solve the denominator for zero and check each root. A horizontal asymptote is found by comparing the degrees of numerator and denominator: equal degrees give the ratio of the leading coefficients, a smaller numerator gives zero, and a larger numerator gives none.

Learning objectives

Vertical asymptotes

Set the denominator equal to zero and solve. Each solution is a candidate, because dividing by something approaching zero sends the function off without bound.

But check the numerator at each one. If the numerator is zero there too, the factor cancels and there is a hole in the graph rather than an asymptote — a single missing point, not a wall.

That check is the whole difference between a correct answer and a confident wrong one, and it costs one substitution per candidate.

Horizontal asymptotes

Compare the degrees. If the numerator has the lower degree, the function tends to zero far out in both directions, so the horizontal axis is the asymptote.

If the degrees are equal, the function tends to the ratio of the leading coefficients — not to one, and not to the ratio of the constant terms.

If the numerator has the higher degree, there is no horizontal asymptote at all. The function grows without bound, and at this level that is a complete answer.

What to do with them in a full investigation

The asymptotes are the frame of the sketch. Draw them as dashed lines first and the shape of the curve is largely determined before a single point is plotted.

They also split the domain. Intervals of increase and decrease are reported between asymptotes, never across one, because the function does not exist at the asymptote itself.

Worked examples

  1. Find the asymptotes of f(x) = 1 over (x − 2)

    1. Denominator zero: x equals 2
    2. The numerator there is 1, which is not zero, so it is a genuine asymptote
    3. Degrees: 0 on top, 1 below, so the numerator is smaller

    Answer: Vertical at x = 2, horizontal at y = 0

  2. Find the horizontal asymptote of f(x) = (3x squared + 1) over (x squared − 4)

    1. Compare degrees: 2 on top and 2 below, so they are equal
    2. Take the ratio of the leading coefficients
    3. That is 3 over 1

    Answer: Horizontal at y = 3

  3. Does f(x) = (x squared − 1) over (x − 1) have an asymptote at x = 1?

    1. The denominator is zero at x equals 1
    2. Check the numerator there: 1 minus 1, which is also zero
    3. The factor cancels, leaving x plus 1

    Answer: No. There is a hole at x = 1, not an asymptote

Common mistakes

Declaring an asymptote wherever the denominator vanishes
If the numerator vanishes at the same point the factor cancels, giving a hole rather than an asymptote. One substitution per candidate settles it.
Taking the ratio of the constant terms for a horizontal asymptote
Far from the origin the leading terms dominate and the constants are irrelevant. The asymptote is the ratio of the leading coefficients.
Reporting an interval of increase that crosses an asymptote
The function does not exist at the asymptote, so an interval cannot span it. Intervals are reported on each side separately.

What to remember

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