Analysis only. The guide prints both formulae in the guidance column, x = −d/c and y = a/c, so the content is knowing which is which rather than deriving them.
Start on view 2, the swapped pair, and ask whether those lines are asymptotes. The two crossings are marked on the curve, at (2, −3.5), which sits on x = 2, and (9.5, 4), which sits on y = 4. A line the graph goes through is not a line the graph approaches, and here it goes through both.
Then view 1, where the real pair sits with the curve closing on both and touching neither. The definition does all the work, and going wrong-answer-first means the definition arrives as a tool rather than as a sentence to copy down.
View 3 is worth the extra minute: the inverse has the two asymptotes exchanged, because reflecting in y = x turns vertical into horizontal. That is a free check on every inverse they will ever find in this sub-topic.
| Question | Answer |
|---|---|
| 1. Vertical asymptote | x = 4, where x − 4 is zero. |
| 2. Horizontal asymptote | y = 2, the ratio 2/1. |
| 3. y-intercept | 3/(−4) = −0.75. |
| 4. f(5) | 13/1 = 13. |
| 5. What settles the swapped pair | B. f(2) = −3.5, so the curve crosses x = 2. |
The x-intercept is −1.5, from the numerator. It is not asked for directly but it belongs on any sketch, and a class that has found it is less likely to confuse it with the asymptote at a later date.
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 label is the mark. A dashed line with no equation beside it earns nothing, and students who have drawn a perfectly good sketch lose a mark they did the work for. Make labelling part of the drawing, not a check at the end.
| They wrote | What happened |
|---|---|
| x = 2 and y = 4 | The pair swapped. Right numbers, wrong roles, and nothing in the algebra objects. Send them to f(2). |
| x = −4 | Read −d/c as "the d with a minus in front" rather than solving cx + d = 0. With d = −4 the answer is +4. |
| x = −1.5 | The x-intercept. They have used the top where the bottom belongs. |
| y = 0 | The horizontal asymptote of 1/x, carried across. It is y = 0 only when the top has no x in it. |
| y = 3 | The constant on the top rather than the coefficient of x. |
| 0.75 | The sign on the y-intercept. The bottom at x = 0 is −4. |
| −1.33 | −4/3: the fraction inverted. Worth a word about reading top over bottom in the order written. |
| 2.6 on question 4 | 13/5, dividing by x instead of x − 4. The bottom at x = 5 is 1, which is what makes the value so large. |
| −9 on question 4 | f(3), the other side of the asymptote. Useful rather than careless: the two together give the branches. |
"Can a graph cross its asymptote?" In general yes, and the Applications page at 2.4 is built on an example that does. For this family it cannot, and the reason is worth showing: setting (2x + 3)/(x − 4) = 2 gives 3 = −8, which has no solutions. So the right answer is "not this one, and here is the proof", which is more useful than a rule either way.
"Why is 1/x self-inverse?" Because the operation is its own undoing: the reciprocal of the reciprocal is the original. Graphically it is symmetric about y = x. Worth asking whether any other function they know does this; most classes will not think of one, and (x + 1)/(x − 1) at AHL 2.14 is the pay-off.
"How do I know which branch is which?" Substitute one value either side of the vertical asymptote. f(3) = −9 and f(5) = 13, so the left branch is below and the right branch is above. Two substitutions and the sketch is determined; guessing from the shape of the formula is how students end up with the branches swapped.
"Does the domain matter?" Yes, and it is the honest way to state the vertical asymptote: the domain is every real number except 4, and the asymptote is at the excluded value. Framing it that way links back to 2.2 and forward to the restricted-domain inverses at AHL 2.14.
| Demonstrate | Graph it in the default window first and let the machine draw the vertical line joining the two branches. Ask whether that line is part of the graph. It is not; it is the plotter connecting two points it should not have joined. Then set Draw Type to Plot on the Casio, or widen the window on the Nspire, and watch it disappear. That is a minute well spent, because the same artefact appears in every rational-function question they will meet. |
|---|---|
| Where they stick | Neither machine draws an asymptote, so students report that the calculator "did not show it". They have to add y = 2 as a second function, and the vertical one needs the Nspire's Graph Entry/Edit → Equation → Line, which is not where anyone looks. On the Casio it is F3 TYPE → X= in the graph list, which is no easier to find. |
| The check | Substitute a large x. f(100) = 2.115 and f(10000) = 2.00. If the claimed horizontal asymptote is not what those values are closing on, it is wrong, and it takes one entry. |
Analysis Paper 1 has no calculator and sketching these with labelled asymptotes is Paper 1 work. The machine is for confirming a sketch and for showing the connecting-line artefact.
| Step | What |
|---|---|
| 1 | 1/x from a table of values, including x = 0.1 and x = 10. Two branches, two asymptotes, no intercepts. |
| 2 | Self-inverse, by applying it twice and by the symmetry in y = x. |
| 3 | Now (2x + 3)/(x − 4). Ask for the asymptotes before giving the formulae. |
| 4 | Figure view 2: the swapped pair, with the crossings marked. Collect the objection. |
| 5 | Name both formulae, with "bottom, so vertical" said out loud. |
| 6 | The 3 = −8 proof that y = 2 is never reached. |
| 7 | Both intercepts, then f(3) and f(5) to orient the branches. |
| 8 | The inverse, and the asymptotes swapping, as the closing check. |
Do not say "a graph never crosses an asymptote". It is false in general, your own Topic 2.4 page is built on a counterexample, and a student who believes it will reject a correct answer later. Say that for (ax + b)/(cx + d) the horizontal asymptote is never reached, and show the one-line reason.
Do not say "the asymptote is where the graph breaks". It describes the picture and gives no method. The usable version is that the domain excludes one value and the asymptote sits at it, which is a sentence they can act on with a denominator in front of them.