Topic 2.8 · AA Standard Level

Two asymptotes, and which coefficient each comes from

For (2x + 3)/(x − 4) the vertical asymptote is x = 4 and the horizontal one is y = 2. Swap them and you get x = 2 and y = 4, two lines the curve goes straight through.

the two asymptotes
x = 4vertical
y = 2horizontal
never crossedverdict

An asymptote is a line the curve approaches and does not reach. A line the curve crosses has failed the only test there is.

Start with 1/x

f(x) = 1/x, defined for every x except 0. Two branches, one in each of the first and third quadrants, and:

  1. Vertical asymptote x = 0, because that is the one input the function refuses.
  2. Horizontal asymptote y = 0, because 1/x shrinks towards nothing as x grows.
  3. No intercepts at all, which is unusual and worth noticing: 1/x is never zero, and x = 0 is not allowed.

And it is self-inverse: 1/(1/4) = 4. Apply it twice and you are back. Graphically that means it is its own reflection in y = x, which is the cleanest example of the reflection idea from 2.2.

The general form, and where each asymptote comes from

For f(x) = (ax + b)/(cx + d):

  1. Vertical asymptote: x = −d/c, from the bottom. It is the x the denominator forbids.
  2. Horizontal asymptote: y = a/c, from the leading coefficients. For large x the constants stop mattering and the ratio settles on a/c.

So for (2x + 3)/(x − 4): the bottom is zero at x = 4, and the ratio of the leading coefficients is 2/1, so y = 2.

Check the horizontal one by substituting something large. f(100) = 2.115 and f(10 000) = 2.00. It is closing on 2 from above, and it never arrives.

Why this curve never crosses y = 2. Set it equal and see:

(2x + 3)/(x − 4) = 2  →  2x + 3 = 2x − 8  →  3 = −8

which is false for every x, so there is no solution and the curve misses the line everywhere. That is a proof, not a description, and it takes one line. For this family of functions the horizontal asymptote is never crossed, which is special: other kinds of curve cross theirs happily. The one thing to check is that the numerator is not a multiple of the denominator: (2x + 4)/(x + 2) is just the number 2 with a hole in it, and there is no curve to speak of.

The swap is invisible unless you test it. x = 2 and y = 4 look like a perfectly reasonable pair of answers. But the curve passes through (2, −3.5), which is on the line x = 2, and through (9.5, 4), which is on the line y = 4, so it crosses both of them, and a line the curve crosses cannot be an asymptote. One substitution kills the swapped answer.

The rest of the sketch

A sketch of a rational function has to show both asymptotes and any intercepts. For (2x + 3)/(x − 4):

  1. y-intercept: put x = 0, giving 3/(−4) = −0.75.
  2. x-intercept: the fraction is zero when the top is, so 2x + 3 = 0 and x = −1.5.
  3. Two branches, one either side of x = 4. Both fall as x increases. On the left the curve comes in just under y = 2 from far left, passes the y-intercept at −0.75 and drops away towards the asymptote; on the right it falls from enormous values towards 2 from above.

f(3) = −9 and f(5) = 13, which is the jump across the asymptote and the quickest way to get the two branches the right way up.

The inverse swaps the asymptotes, and that is the check. Rearranging y = (2x + 3)/(x − 4) gives

f⁻¹(x) = (4x + 3)/(x − 2)

whose vertical asymptote is x = 2 and whose horizontal one is y = 4: the original pair, exchanged. That has to happen, because reflecting in y = x turns a vertical line into a horizontal one. If your inverse's asymptotes are not the swap of the original's, one of the two is wrong. And f⁻¹(13) = 5, which is where f(5) = 13 started.

On the GDC: asymptotes are not drawn for you

Neither machine draws an asymptote, and both will join the two branches with a near-vertical line if the window is wrong, which looks exactly like a curve crossing its own asymptote. Knowing where the lines are is how you read the screen.

When you may use it. Analysis Paper 1 is non-calculator, and sketching rational functions with their asymptotes labelled is Paper 1 work. Use the machine to confirm a sketch, and expect it to need help with the window.

TI-Nspire CX II

  1. A Graphs page with f1(x)=(2x+3)/(x-4)
  2. menu → Window / Zoom → Window Settings, x from -10 to 18, y from -12 to 16
  3. Add the asymptotes yourself as f2(x)=2 and, for the vertical one, menu → Graph Entry/Edit → Relation, then type x=4. The Line entry only offers fixed templates like y = mx + b and will not accept it
  4. Analyze Graph → Zero gives -1.5, and the y-intercept reads off at x = 0 as -0.75

Casio fx-CG50

  1. MENU → Graph, with Y1=(2X+3)÷(X-4), then F6 DRAW
  2. SHIFT F3 V-Window and widen it; F3 STD is too small here and the two branches merge
  3. SHIFT MENU SET UP and set Draw Type to Plot rather than Connect, which stops the machine joining the branches across the asymptote. Set it back to Connect afterwards: it is a global setting and every graph you draw until you do will come out as dots
  4. SHIFT F5 G-SOLVE → ROOT gives -1.5, and Y-ICEPT gives -0.75

The mark people lose. Giving the asymptotes the wrong way round. x = 2 and y = 4 is an answer with the right numbers in the wrong roles, and nothing in the algebra objects. The habit that prevents it: the vertical one comes from the bottom of the fraction, because that is the x the function is not allowed to have. Say "bottom, so vertical" every time. Then test it: substitute x = 2 and see that the curve is quite happily at −3.5.

Your turn

Throughout: f(x) = (2x + 3)/(x − 4).

1. Give the x value of the vertical asymptote.

2. Give the y value of the horizontal asymptote.

3. Find the y-intercept.

4. Find f(5).

5. A student gives the asymptotes as x = 2 and y = 4. What settles it in one step?

Question 5. A student gives the asymptotes as x equals 2 and y equals 4. What settles it in one step?
Where the marks go

1 markThe vertical asymptote, from the denominator.

1 markThe horizontal asymptote, from the leading coefficients.

1 markBoth intercepts.

1 markA sketch with both asymptotes drawn and labelled.

The last mark is for the labels. An unlabelled dashed line is not an asymptote as far as a markscheme is concerned, and writing x = 4 beside it costs four characters.

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