For (x² + 1)/(x − 1) the leading coefficients are both 1, so the old rule says y = 1. At x = 101 the curve is at 102.02, which is 101 away from that line and 0.02 from y = x + 1.
One gap shrinks and the other grows. That is what makes one of these lines an asymptote and the other a line that happens to be on the diagram.
At 2.8 the top and the bottom were both linear, and the horizontal asymptote was the ratio of the leading coefficients. That rule only works when the two degrees are equal. Here the top is quadratic and the bottom is linear, so the top wins and the curve has nowhere horizontal to go.
Divide. x² + 1 divided by x − 1 gives x + 1 with remainder 2, so:
(x² + 1)/(x − 1) = x + 1 + 2/(x − 1)
Now read it. As x grows, the 2/(x − 1) term goes to zero and what is left is x + 1. So the curve closes on the line y = x + 1, which is the oblique asymptote, and the remainder term is exactly the gap.
Check it: at x = 3 the remainder term is 2/2 = 1, and f(3) = 5 against the line's 4. At x = 11 it is 2/10 = 0.2, and f(11) = 12.2 against 12. The algebra predicted the gap before the arithmetic confirmed it.
The degree rule, all four cases. Compare the degree of the top with the degree of the bottom:
| Degrees | What you get |
|---|---|
| top < bottom | Horizontal asymptote y = 0. The bottom wins. |
| top = bottom | Horizontal asymptote y = a/c, the leading coefficients. This is the 2.8 case. |
| top = bottom + 1 | Oblique asymptote, found by division. |
| top > bottom + 1 | No linear asymptote at all. The curve runs away faster than any line. |
Count the degrees before you do anything else. One glance decides which of the four you are in, and three of the four need no work.
What the wrong rule looks like. Applying y = a/c here gives y = 1/1 = 1, and the figure draws that line so you can watch it fail. At x = 3 the curve is 4 above it; at x = 11, 11.2 above; at x = 101, 101.02 above. The gap is not closing, it is growing roughly as fast as x.
That is worth a moment, because a wrong asymptote is a special kind of error: the sketch still has a line on it, in a plausible place, and nothing about the picture announces the problem. The only defence is counting the degrees first.
Everything else comes off the two forms:
Two points worth knowing because they catch a sketching error: f(0) and f(−1) are both −1. If your left branch passes through only one of them, it is the wrong shape.
And the contrast, so the rule has two sides. (2x² + 1)/(x² − 1) has equal degrees, so it does have a horizontal asymptote, at y = 2/1 = 2. g(10) = 2.0303 and g(1000) = 2 to three decimals. It also has two vertical asymptotes, at x = 1 and x = −1, because the bottom is zero twice.
Same sub-topic, different case, and the only thing you did differently was count.
An oblique asymptote is invisible in a small window, because near the origin the curve does not look like a line at all. The calculator earns its place here by letting you zoom out far enough that the curve and the line become indistinguishable.
When you may use it. Analysis Paper 1 is non-calculator, and the division is Paper 1 work. On Paper 2 plot the curve and the line together as a check that your division was right, which is the fastest way to catch a sign slip.
The mark people lose. Giving a horizontal asymptote because the previous sub-topic always had one. It is the single commonest error here and it is a habit rather than a misunderstanding. The habit to replace it with: write down the two degrees before you write anything about an asymptote. Two numbers, and they tell you which of four cases you are in.
Throughout: f(x) = (x² + 1)/(x − 1).
1. Find f(3).
2. The oblique asymptote is y = x + c. Give c.
3. How far is the curve above that asymptote at x = 11? Give it to 1 decimal place.
4. Find the y-intercept.
5. Why does this curve have no horizontal asymptote?
1 markThe division, carried out correctly.
1 markThe oblique asymptote stated as an equation.
1 markThe vertical asymptote.
1 markIntercepts, or a statement that there is no x-intercept.
On a "sketch the curve" question the asymptotes are drawn as dashed lines and labelled with their equations. An unlabelled dashed line does not earn the mark, and neither does a labelled line in the wrong place.
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