Two numbers, written before anything else, decide which of four cases the question is. Students arrive from 2.8 with one case memorised and apply it to all four, and the resulting sketch looks entirely reasonable.
Leave the slider at 3 and ask which of the two straight lines is the asymptote. At that x the curve is 1 above the slant line and 4 above the horizontal, so the horizontal is not obviously absurd. Let the class argue briefly.
Then drag it to 12 and read the two numbers aloud: 0.182 and 12.182. One is heading for zero and the other is tracking x. That is what an asymptote is, and it is a much better definition than "a line the curve gets close to", because both lines are close to the curve somewhere.
The useful follow-up question is at what point does y = 1 stop looking plausible. The honest answer is that it never stops looking plausible on a small sketch, which is why the method has to be counting degrees and not looking.
| Question | Answer |
|---|---|
| 1. f(3) | 10/2 = 5. |
| 2. The c in y = x + c | 1, so y = x + 1. |
| 3. The gap at x = 11 | 2/10 = 0.2. |
| 4. The y-intercept | −1. |
| 5. Why no horizontal asymptote | B. The numerator outranks the denominator. |
Question 3 is the one worth marking carefully. The gap is the remainder divided by (x − 1), and a student who answers 2 has understood the division and not the consequence.
1 markThe division, carried out correctly.
1 markThe oblique asymptote as an equation.
1 markThe vertical asymptote.
1 markIntercepts, or a statement that there is no x-intercept.
On a sketch, both asymptotes must be dashed and labelled with their equations. An unlabelled line does not earn it. Say this explicitly, because students who have drawn a beautiful curve are often surprised to lose marks on two dashed lines.
| They wrote | What happened |
|---|---|
| y = 1 as an asymptote | The error this page exists for. The 2.8 rule applied where the degrees differ. Drag the slider to 12 rather than explaining. |
| y = x − 1 | Sign slip in the division. Easy to self-check by multiplying back: (x − 1)(x − 1) is x² − 2x + 1, not x² + 1. |
| y = x | Divided only the leading terms and stopped. x²/x is x, and then the next step produces the + 1. |
| 2 on question 3 | Gave the remainder rather than the gap. The remainder is 2; the gap is 2/(x − 1). |
| 11.2 on question 3 | Measured to y = 1. Two errors compounding, and worth separating: which line, then how far. |
| 4 on question 1 | Evaluated the asymptote instead of the curve. Sometimes a genuine misreading, sometimes a belief that the curve IS the line once you have divided. |
| +1 on question 4 | Sign. The denominator at x = 0 is −1. Worth a reminder that the y-intercept is just f(0) and nothing special. |
| An x-intercept | Solved x² + 1 = 0 and got ±i, or got confused and wrote 1. There is no real x-intercept and saying so is the mark. |
| A monotonic sketch | Carried over the shape from 2.8, where each branch really is monotonic. Here each branch turns, at x = 1 ± √2, which is 2.414 and −0.414, where the curve reaches 4.828 and −0.828. |
"Can the curve cross the oblique asymptote?" Yes, and it is a good question because the answer differs from 2.8. The gap here is 2/(x − 1), which is never zero, so this particular curve does not. But (x² − 1)/x = x − 1/x crosses its asymptote y = x nowhere either, while (x³ + x)/(x² + 1) is exactly x and equals its "asymptote" everywhere. The general answer is that an oblique asymptote can be crossed, unlike the 2.8 horizontals, and the remainder term tells you whether.
"Do I have to use long division?" No, and many students are faster at the alternative: write x² + 1 = (x − 1)(x + a) + b and compare coefficients. For this one, expanding gives x² + (a − 1)x + (b − a), so a = 1 and b = 2. It is the same work in a shape that feels less error-prone, and it is worth offering both.
"Which way does the curve approach?" Read the sign of the remainder term. 2/(x − 1) is positive for x > 1, so the right branch sits above the line, and negative for x < 1, so the left branch sits below it. That is a mark on a sketch question and it costs one line of thought.
"What about top degree two more than the bottom?" Then there is no linear asymptote at all, and the fourth row of the table says so. x³/x = x², which is not a line. Students who have learnt "divide to find the asymptote" will divide anyway and produce a parabola, then call it an asymptote. It is worth one counter-example.
| Demonstrate | Plot the curve, the slant line and y = 1 together, then set the window to −100 to 100 both ways. The curve and the slant line become one line and y = 1 is nowhere near either. Then come back in to −6 to 12 and the real shape returns. Zooming out is the demonstration; nothing said at the board does it as well. |
|---|---|
| Where they stick | The connecting line across x = 1, which they report as part of the graph. Same artefact as 2.8 and the same fix: Draw Type to Plot on the Casio, or a wider window on the Nspire. Remind them to set it back. |
| The check | Graph the original and the divided form, x + 1 + 2/(x − 1), as two functions. If the division was right they are indistinguishable. That catches a sign slip instantly and is faster than checking the algebra. |
The superscript box catches people here: typing (x^2+1)/(x-1) puts the whole denominator into the exponent. Use the squared key, or the right arrow to leave the box.
| Step | What |
|---|---|
| 1 | Recall 2.8's rule and ask whether it can be the whole story. Write both degrees for (x² + 1)/(x − 1). |
| 2 | Evaluate f(11) and f(101) numerically. 12.2 and 102.02. Ask what line those are approaching. |
| 3 | Somebody will say y = x. Divide, and get x + 1 with remainder 2. |
| 4 | The divided form read aloud: the remainder term vanishes, the line is what is left. |
| 5 | The figure. Both lines, both gaps, slider from 3 to 12. |
| 6 | The four-case table, derived by comparing degrees rather than given. |
| 7 | The full sketch: both asymptotes, y-intercept, no x-intercept, both turning points. |
| 8 | The equal-degree contrast, (2x² + 1)/(x² − 1), with its two vertical asymptotes. |
Do not say "the asymptote is the ratio of the leading coefficients". It is the sentence students bring from 2.8 and it is the direct cause of y = 1 here. The accurate version has a condition attached: when the degrees are equal, the horizontal asymptote is that ratio. Say the condition first and the rule second.
Do not let "oblique" and "slant" drift apart. Both words appear in textbooks and the guide uses oblique. Use one in class and mention the other once, because a student who meets "slant asymptote" in a revision book should not think it is a different thing.