WitCH 8: Oblique Reasoning

February 13, 2019

5:58 pm

79 Comments on WitCH 8: Oblique Reasoning

analysis, asymptotes, Specialist Mathematics, textbooks, WitCH

A reminder, WitCH 2, WitCH 3 and WitCH 7 are also open for business. Our new WitCH comes courtesy of John the Merciless. Once again, it is from Cambridge’s text Specialist Mathematics VCE Units 3 & 4 (2019). The text provides a general definition and some instruction, followed by a number of examples, one of which we have included below. Have fun.

Update

With John the Impatient’s permission, I’ve removed John’s comments for now, to create a clean slate. It’s up for other readers to do the work here, and (the royal) we are prepared to wait (as is the continuing case for WitCh 2 and Witch 3).

This WitCH is probably difficult for a Specialist teacher (and much more so for other teachers). But it is also important: the instruction and the example, and the subsequent exercises, are deeply flawed. (If anybody can confirm that exercise 6G 17(f) exists in a current electronic or hard copy version, please indicate so in the comments.)

Update (05/02/20)

It is obviously long, long past time to sort out this godawful mess. We apologise to all those industrious commenters, who nailed the essential wrongness, and whose hard work was left hanging. This will also be a very long update; you should pour yourself a stiff drink, grab the bottle and get comfy. For the benefit of the Twitter addicts, however, who now find it difficult to concentrate for more than two paragraphs, here’s the punchline:

Cambridge‘s notion of “non-vertical asymptote” is so vague, falling so short of a proper definition, it is close to meaningless and it is pointless. This leads to the claim $\color{red}\boldsymbol{y \to \frac{x}{\sqrt{x}}}}$ in Example 31 being flawed in three distinct ways. In particular, it follows from Cambridge that the “curve” $\color{red}\boldsymbol{y = \frac{x+1}{\sqrt{x-1}}}$ is a “non-vertical asymptote” to the function $\color{red}\boldsymbol{F(x) = \frac{x+1}{\sqrt{x-1}}}$ .

The source of Cambridge‘s confusion is also easy to state:

More complicated asymptotes cannot be defined, interpreted or computed in the manner possible for simpler asymptotes.

Unfortunately, the discussion in Cambridge is so sparse and so far from coherent that directly critiquing the excerpt above would probably be incomprehensible. So, we’ll first try to make very careful sense of “non-vertical asymptotes”, taking some minor whacks at Cambridge along the way. Then, with that sense as foundation, it will be easy work to hammer the excerpt above. To simplify the discussion, we’ll only consider the asymptotes of a function $\boldsymbol{F(x)}$ as $\boldsymbol{x \to \infty}$ . Obviously, the case where $\boldsymbol{x \to -\infty}$ can be handled in like manner, and vertical asymptotes are pretty straightforward.

OK, take a swig and let’s go.

1. Horizontal Asymptotes

We’ll begin with the familiar and demon-free case, for asymptotes as horizontal lines. Consider, for example, the function

$\color{blue}\boldsymbol{g(x) = \frac{2x + 1}{x-3}\,.}$

We can write $\boldsymbol{g(x) = 2 \ + \ 7/(x-3)}$

, noting the second term is tiny when $\boldsymbol{x}$

huge. So, we declare that $\boldsymbol{g(x)}$

has the horizontal asymptote $\boldsymbol{y = 2}$

Formally, the asymptote in this example is captured with limits. The underlying functional behaviour is $\boldsymbol{g(x)\to 2}$ or, more officially, $\boldsymbol{\lim\limits_{x\to\infty}g(x) = 2}$ . The limit formalism is of no benefit here, however, and is merely likely to confuse. The informal manner of thinking about and writing limits, as is standard in schools, suffices for the understanding of and computation of horizontal asymptotes.

Note also that horizontal asymptotes can easily be spotted with essentially no calculation. Consider the function

$\color{blue}\boldsymbol{h(x) = \frac{4x^2 + 2}{x^2 + 1}\,,}$

which is Example 29 in Section 6G of Cambridge. Here, it is only the highest powers, the $\boldsymbol{x^2}$

terms in the numerator and denominator, that matter; the $\boldsymbol{x^2}$

s cancel, and so the function has horizontal asymptote $\boldsymbol{y=4}$

. This simplification also applies to suitable rooty (algebraic) functions. The function

$\color{blue}\boldsymbol{k(x) = \frac{4x + 2}{\sqrt{x^2 + 1}}\,,}$

for example, also has horizontal asymptote $\boldsymbol{y=4}$

, for the same reason.

2. Linear Asymptotes

We’ll now consider general linear asymptotes, which are still reasonably straight-forward (and which include horizontal asymptotes as a special case). There are, however, two demons lurking, one conceptual and one computational.

Cambridge begins with the example

$\color{blue}\boldsymbol{f(x) = \frac{8x^2 -3x+2}{x}\,.}$

The function can be rewritten as $\boldsymbol{f(x) = 8x -3 \ + \ 2/x}$

, and it seems simple enough: a similar “when $\boldsymbol{x}$

is huge, the leftovers are tiny” intuition leads us to declare that $\boldsymbol{f(x)}$

has the linear asymptote $\boldsymbol{y = 8x -3}$

. The trouble is that the example, and the discussion that follows, are too simple, so that the demons remain hidden.

3. A Conceptual Demon

How do we formally (or semi-formally) express that a function has a linear asymptote? For the example above, Cambridge writes “f(x) will approach the line y = 8x – 3”. This is natural, intuitive and sufficiently clear. It is considering the asymptote to be a geometric object, as the line, which is fine at the school level.

What, however, if we want to think of the asymptote as the function $\boldsymbol{g(x) = 8x -3}$ ? The problem is that both $\boldsymbol{f(x)}$ and $\boldsymbol{8x-3}$ are zooming off to infinity, which means that writing $\boldsymbol{f(x) \to 8x -3}$ , or anything similar, is essentially meaningless. Those arrows are shorthand for limits and, fundamentally, the limit of a function must be a number, not another function. (For vertical asymptotes we consider infinity to be an honorary number, which may seem dodgy, but which can be justified.)

The obvious way around this problem is simply to ignore it. As long as we stick to asymptotes of reasonably simple functions – rational functions and carefully chosen others – Cambridge’s intuitive approach is fine. But in the, um, unlikely event that our situation is not so simple, then we have to carefully consider the limiting behaviour of functions. So, we’ll continue.

Suppose that the function $\boldsymbol{F(x)}$ appears to have the linear function $\boldsymbol{L(x)}$ as asymptote. To capture this idea, the simple trick is to consider the difference of the two functions. If $\boldsymbol{x}$ is huge then this difference should be tiny, and so $\boldsymbol{L(x)}$ being the asymptote to $\boldsymbol{F(x)}$ is captured by the limit statement

$\color{blue}\boxed{\boldsymbol{F(x)- L(x)\ \longrightarrow \ 0}}$

Again, none of this is really necessary for capturing linear asymptotes of simple functions. But it is necessary, at least as underlying guidance, if we wish to consider less simple functions and/or more general asymptotes.

WARNING: It may seem as if our boxed definition of linear asymptote $\boldsymbol{L(x)}$ would work just as well for $\boldsymbol{L(x)}$ non-linear, but there is a trap. There is another, third demon to deal with. Before that, however, our second demon.

4. A Computational Demon

It is very import to understand that the simple, “highest powers” trick we indicated above for spotting horizontal asymptotes does not work for general linear asymptotes. Cambridge‘s introductory example $\boldsymbol{f(x)}$ illustrates the issue, albeit poorly and with no subsequent examples. For a better illustration, we have Exercise 12 in Section 6G, which presents us with the function

$\color{blue}\boldsymbol{m(x) = \frac{4x^2 + 8}{2x +1}\,.}$

Here, it would be invalid to divide the $\boldsymbol{4x^2}$

by the $\boldsymbol{2x}$

and then declare the linear asymptote to be $\boldsymbol{y =2x}$

. The problem is the $\boldsymbol{+1}$

in the denominator has an effect, and we are forced to perform long division or something similar to determine that effect. So, $\boldsymbol{m(x) = 2x -1 \ + \ 10/\left(2x +1\right)}$

, and then it is clear that the linear asymptote is $\boldsymbol{y =2x -1}$

All that is fine, as far as it goes. Cambridge routinely performs long division to correctly determine linear asymptotes (including horizontal asymptotes, for which the simpler highest powers trick would have sufficed). What, however, if the function is not rational? Consider, for example, Example 31 above. Can we safely ignore the $\boldsymbol{-1}$ in the denominator of the function, as Cambridge has done? And, if so, why?

In order to stay in the linear world for now, we’ll leave Example 31 and instead consider two other examples:

$\color{blue}\boldsymbol{p(x) = \frac{x^2}{\sqrt{x^2 -1}} \qquad\qquad q(x) = \frac{x^2}{\sqrt{x^2 -x}}\,.}$

So, are we permitted to ignore the $\boldsymbol{-1}$

and the $\boldsymbol{-x}$

in the denominators of $\boldsymbol{p(x)}$

and $\boldsymbol{q(x)}$

? It turns out that the answers are “Yes” and “No”: the function $\boldsymbol{p(x)}$

has linear asymptote $\boldsymbol{y =x}$

, but $\boldsymbol{q(x)}$

has linear asymptote $\boldsymbol{y =x +\frac12}$

The behaviour of $\boldsymbol{p(x)}$ and $\boldsymbol{q(x)}$ is, of course, anything but obvious. In particular, the “highest powers” and long division tricks are of no assistance here. Moreover, similar difficulties arise with the analysis of non-horizontal asymptotes of pretty much all rooty functions. It is essentially impossible for a Victorian school text to cover non-horizontal asymptotes of non-rational functions. The text must either work very hard, or it must cheat very hard; Cambridge takes the latter approach.

5. A Nonlinear Demon

Almost there. We have non-linear “asymptotes” left to consider, which are fundamentally demonic. This is demonstrated by Cambridge‘s very first example, Example 26, which considers the function

$\color{blue}\boldsymbol{r(x) = \frac{x^4 +2}{x^2}\,.}$

Dividing through by $\boldsymbol{x^2}$

gives $\boldsymbol{r(x) = x^2 + 2/{x^2}}$

, after which Cambridge baldly declares “The non-vertical asymptote has equation $\boldsymbol{y = x^2}$

“. But why, exactly? All we have to go on is Cambridge‘s intuitive approach to $\boldsymbol{f(x)}$

above. Intuitively, the function $\boldsymbol{r(x)}$

approaches the parabola $\boldsymbol{y = x^2}$

. That’s fine, and we can formalise it by setting $\boldsymbol{Q(x)=x^2}$

. Then guided by our boxed definition above, we can write $\boldsymbol{r(x) - Q(x) \ \to \ 0}$

, and we seem to have our asymptote.

The problem is that there are a zillion functions that will fit into that blue box. It is also true, for example, that $\boldsymbol{r(x) - r(x) \ \to \ 0}$ . This means that $\boldsymbol{r(x)}$ itself just as good an asymptote as $\boldsymbol{Q(x)}$ . So, yes, one can reasonably declare that $\boldsymbol{r(x)}$ is asymptotic to $\boldsymbol{Q(x)}$ , but we cannot declare $\boldsymbol{Q(x)}$ to be THE asymptote to $\boldsymbol{r(x)}$ .

Of course we want asymptotes to be unique, so what do we do? There are two ways out of this mess, the first solution being to restrict the type of function that we’ll permit to be an asymptote. That’s intrinsically what we did when considering horizontal and linear asymptotes, and it is exactly why the multiple asymptote problem didn’t arise in those contexts: linear functions only had to compete with other linear functions. Now, for general rational functions, we must broaden the notion of asymptote, but it’s critical to not overdo it: we now permit polynomial asymptotes, but nothing more general. Then, every rational function will have a unique polynomial (possibly linear, possibly horizontal) asymptote.

But now, what do we do for non-rational functions? These functions may naturally have rooty asymptotes, as illustrated by Example 31 above. Such functions can also be dealt with, by suitably and carefully generalising “polynomial asymptote”, but it’s all starting to get fussy and annoying. It is making a second solution much more attractive.

That second way out of this mess is to just give up on unique nonlinear “asymptotes”. Then, our blue box becomes a definition of two functions being “asymptotically equivalent”. We don’t get unique asymptotes, but we do get a clear way to think about the asymptotic behaviour of functions. (You can’t always get what you want, But if you try sometimes, you just might find …)

One final comment. The computational demon we mentioned above is of course still around to bedevil non-linear asymptotes. This is well illustrated by Exercise 17, which considers the function

$\color{blue}\boldsymbol{s(x) = \frac{x^2 + x + 7}{\sqrt{2x+1}}\,.}$

As discussed by commenters below, this exercise used to have a part (f), asking for the vertical asymptote. That part was deleted in later editions, and it is easy to guess why; the $\boldsymbol{+1}$

in the denominator of $\boldsymbol{s(x)}$

turns out to matter, so that the “highest powers” technique that Cambridge invalidly employs on such examples gives the wrong answer.

6. The Devil in Cambridge‘s detail

Finally. We can now deal with the Cambridge excerpt above. There is no need to write at length here, since everything follows from the discussion above, as indicated. We begin with a clarification of the punchline:

The claim $\color{red}y \to \frac{x}{\sqrt{x}}}$ in Example 31 is triply flawed: it is meaningless (conceptual demon); the calculation is invalid (computational demon); and the conclusion is wrong (nonlinear demon).

To be fair, it turns out that the -1 in the denominator of the function in Example 31 can be ignored, but not because of the highest powers trick that Cambridge appears to employ. This also means y = √x will be the non-linear asymptote to y = (x + 1)/√(x-1) if we suitably generalise the notion of “polynomial asymptote”.

That’s plenty wrong, of course. But to round off, here are the other problems with the excerpt:

The “graph of y = f(x)” doesn’t approach anything. It is the function that approaches the asymptote.
Underlying the punchline, the definition of “non-vertical asymptote” is hopelessly vague and does not remotely mean what Cambridge thinks it means.
We do not “require √(x – 1) > 0; we require x – 1 > 0.

And, we’re done. Thank Christ.

79 Replies to “WitCH 8: Oblique Reasoning”

Number 8 says:

February 16, 2019 at 1:49 pm

Part (f) is not present in the current online version of the text. It is also not present in the current student book PDF available from the same website.

I will however confirm that the Specialist Mathematics Units 3&4 students I spoke with last year (and also at least one teacher of the subject) were unable to tell me whether or not the non-linear asymptotes required labelling on graphs.

My follow-up question to said teacher was whether or not students would be penalised for leaving these “construction lines” in their graphs to produce sketches by addition of ordinates.

The advice I was eventually given was to have students draw and label the non-linear asymptotes and hope for the best.

Would definitely like to know (for genuine curiosity more than anything else) what the convention is here and (less importantly) what VCAA is and isn’t prepared to mark as correct.

Reply
1. marty says:
  
  February 16, 2019 at 3:23 pm
  
  Thanks Number 8. Still curious about current hard copies. It is impossible to tell from the Study Design what is expected from students. Whatever is expected, the above excerpt from the text is a mess.
  
  Reply
  1. Number 8 says:
    
    March 22, 2019 at 10:13 am
    
    17f is in my hard copy (printed 2016).
    
    Reply
2. John Friend says:
  
  January 5, 2021 at 10:52 am
  
  I’ve recently learned that in answer to the question:
  
  What types of asymptotes are required in Specialist Maths? Only linear asymptotes (vertical, horizontal and diagonal/slant/oblique)? Or are curvilinear asymptotes also required?
  
  VCAA gave the following answer:
  
  In secondary mathematics the term asymptote without any qualification is taken to refer to vertical, horizontal, or oblique straight-line asymptotes. Curvilinear asymptotes are not required.
  
  I’m astounded such a crystal clear answer was given. Now all we need is for such a clear statement to be included in the new Stupid Design so that all teachers (and textbook writers) know where things stand.
  
  By the way, publishers could do worse things than pay Marty to polish up his 05/02/2020 update for inclusion in their new editions. I have not seen a more clear and accessible discussion anywhere. The only thing I *might* add would be something like:
  
  The way in which the concept is used is that if one is given a function $F(x)$ , it is interesting to study other functions $g(x)$ that are “asymptotic to $F(x)$ ” in various ways. One meaning of this phrase (discussed by Marty in his update) is
  $\displaystyle F(x) - g(x) \rightarrow 0$ .
  Another somewhat different meaning is
  $\displsyatyle \frac{F(x)}{g(x)} \rightarrow 1}$
  and there are many other variations on this concept.
  
  Reply
  1. marty says:
    
    January 5, 2021 at 11:12 am
    
    Thanks, JF. It is, of course, sensible to restrict to linear asymptotes in VCE, and to have the word “asymptote” defined/understood to mean that. As for the publishers, in general they simply don’t give a fuck.
    
    Reply
    1. John Friend says:
      
      January 6, 2021 at 2:00 pm
      
      For what it’s worth, I’ve attached some calculations to supplement Marty’s 05/02/2020 update. Enjoy (and please feel free to delight in pointing out any errors). Some of you might consider writing a SAC of shit based on this content.
      
      Non-rational functions and non-linear asymptotes
      
      Reply
      1. John Friend says:
        
        January 6, 2021 at 8:17 pm
        
        Revised copy with screenshots from Cambridge included.
        
        Non-rational functions and non-linear asymptotes
        
        Reply
        
        marty says:
        
        January 10, 2021 at 5:14 pm
        
        Thanks, JF. Nice work. I guess if VCAA is declaring non-linear asymptotes are verboten then this may all now be academic.
        
        Reply
        
        John Friend says:
        
        January 10, 2021 at 7:32 pm
        
        Verboten one minute, un-verboten the next …
        
        And even if verboten:
        
        1) I think anyone teaching Specialist Maths should make an effort to have a deeper understanding of this. All Specialist teachers know how to use polynomial division to find an oblique asymptote for rational functions such as $\displaystyle f(x) = \frac{x^2 + 1}{x-1}$ , but how many could explain to the curious student why $\displaystyle f(x) = \frac{x^2 + 1}{x-1} \approx \frac{x^2}{x} = x$ gives the wrong answer and is therefore invalid?
        
        As a Yr 12 student learning this stuff, the latter ‘method’ would make perfect sense to me given that you’re only interested $x \rightarrow \infty$ .
        
        How many teachers could explain or why a definition like the blue box is needed?
        
        How many could explain why the Cambridge method delivers a correct asymptote for $\displaystyle f(x) = \frac{x+1}{\sqrt{x-1}}$ but not for $\displaystyle f(x) = \frac{x^2 + x + 7}{\sqrt{2x+1}}$ ?
        
        2) Teachers should take pride in knowing more mathematics than what they have to teach. How can students get a true appreciation of what they’re learning if they can’t be given some greater context. And how can the curious student be encouraged to remain curious …?
        
        So verboten or not I think the ideas discussed in this post and comments are very important.
        
        What makes things really difficult is the sloppy teaching of asymptotes in Methods Units 1/2. For example, I hear shit like “You can never cross an asymptote” all the time.
        
        Reply
        
        marty says:
        
        January 10, 2021 at 8:15 pm
        
        Hi, JF. Of course I agree that teachers need to understand safely beyond the curriculum in order to teach the curriculum with confidence and clarity and efficiency. And of course the mantra approach to more difficult topics is a recipe for disaster. I’d be happy if teachers understood this post, and I think it would have value if they did so. But I also wouldn’t be too fussed if teachers simply had a proper understanding of linear asymptotes. Many, I am sure, do not.
        
        The reason for my post, of course, was not to argue what teachers should or do understand. The point was to argue what textbook writers should or do (not) understand. Teachers have some excuse; Cambridge has none.
        
        Reply
John says:

February 16, 2019 at 4:25 pm

My hard copy is 2016 – admittedly old – and has part(f). It’s possible the digital platform my school uses only has digital copies of the older text. But I have had students from this year (2019) using pdf copies asking about part(f). Whatever, it’s a mess – incompetent meddling in things outside the scope of the course that leaves a trail of destruction for years to come.

Reply
1. Number 8 says:
  
  February 16, 2019 at 8:03 pm
  
  I can’t find copies printed since 2016, but a lot of the students I work with are using digital copies which means I haven’t been asked about this…
  
  …yet.
  
  Reply
Number 8 says:

February 17, 2019 at 1:32 pm

OK – I’ll attempt to make a start (be gentle with me…)

But as x approaches infinity, surely y also approaches infinity?

The fact that y approaches sqrt(x) is true, but since sqrt(x) approaches infinity, isn’t this a bit redundant?

Reply
John says:

February 17, 2019 at 2:43 pm

I understand your point, but then why is a big deal made of the diagonal asymptotes of something like x^2 – y^2 = 1?

Reply
1. Number 8 says:
  
  February 17, 2019 at 4:52 pm
  
  Because straight line asymptotes on graphs of non functions are a key feature of ALL graphs of this type?
  There also seems to be a (perhaps implicit) idea that asymptotes are straight lines – in that the graph of y=1/x has horizontal and vertical asymptotes and then a rotation by Pi/4 clockwise creates the two diagonal asymptotes.
  
  No combination of dilations, reflections, rotations make straight asymptotes into non-straight line asymptotes and because sheers are (as far as I can tell) absent from the VCE curriculum…
  
  …but I really don’t know.
  
  Reply
  1. John says:
    
    February 18, 2019 at 5:09 pm
    
    I’d add that the general hyperbola (x-h)^2/a^- (y-k)^2/b^2 = 1 can be obtained from y = 1/x using not just rotations but also translations and dilations. So the argument could be made that the general diagonal asymptotes y – k = +/- b/a (x – h) emerge naturally via transformations from the vertical and horizontal asymptotes of y = 1/x and this validates them.
    
    But it would appear that a diagonal asymptote is also called an oblique asymptote ….
    
    Thomas and Finney (Calculus, 9th ed. pp 227 – 228, 231) says:
    
    “If the degree of numerator of a rational function is one greater than the degree of the denominator, the graph has an oblique asymptote, that is, a linear asymptote that is neither vertical nor horizontal.”
    
    and talks about graphing with asymptote and dominant terms. For example, it talks about the dominant terms of y = (x^3 + 1)/x and says “For |x| large, y approx x^2. For x near zero, y approx 1/x …. [and y has] a vertical asymptote at x = 0 …” It’s exercises include graphing of rational functions where ” … the graphs and equations of the asymptotes and dominant terms” are required. Thomas and Finney says dominant terms “… are the key to predicting a function’s behavior.”
    
    So Thomas and Finney appear to define oblique asymptotes as linear asymptotes that are neither vertical or horizontal. So something like f(x) = x + 1/x would have an oblique asymptote y = x, whereas f(x) = x^2 + 1/x would have the dominant term x^2 for large |x| and the graph of y = x^2 would be included as a feature in a sketch graph of y = f(x) but not referred to as any sort of asymptote.
    
    Stewart also talks about curves that have “… asymptotes that are oblique, that is, neither horizontal nor vertical.” and actually gives a mathematical definition in terms of limits. There is no mention of, for want for a better word, curved asymptotes. However, one of the exercises talks about f(x) = (x^3 + 1)/x, asks the reader to show that
    
    lim[f(x) – x^2, x –> +/- oo] = 0
    
    and then says that y = f(x) is asymptotic to the parabola y = x^2 ….
    
    So Evans has opened a real can of worms, as has past questions on VCAA exams and the current Study Design ….
    
    Reply
marty says:

February 17, 2019 at 5:42 pm

Number 8, you are spot on with both remarks. Whatever the text is attempting to do in Example 31, to write y → √x is at best vague, and at worst meaningless. And, you are correct that there is a fundamental distinction one should make between linear and non-linear “asymptotes”. The text not only fails to make that distinction, it frames the discussion so that this distinction is almost impossible to make.

Reply
1. John says:
  
  February 17, 2019 at 8:42 pm
  
  I understand what Number 8 is saying and it makes a lot of sense (I too think you’re on the money).
  
  So what do we do with, say, f(x) = x + 1/x? Do we talk about a straight line asymptote in this case (even though there is no the same motivation as there is for rotated hyperbolas)? And if we do, what do we say about the asymptotes of, say, f(x) = x^2 + 1/x … What’s so special about straight line asymptotes? Is it OK for asymptotes to be polynomial but not non-polynomial (in which case, what do we say about f(x) = e^x + e^-x, say?)
  
  The crazy thing is that we’re all in the dark.
  
  The Study Design is singularly uninformative on this (as it is on many things) and as far as I know there has been no advice given in any VCAA Bulletin. In such cases one might look to past exams and Reports for additional insight. So it’s extremely interesting to note the 2000 Exam 1 Q1 Part 2 and subsequent comments in the Examiners Report (I don’t think the Study Design has changed all that much over 15 years with this part of the course), as well as the 2008 Exam 2 Q1.
  
  Reply
  1. Number 8 says:
    
    February 18, 2019 at 5:08 pm
    
    That, Sir John, is the right question to be asking, because it is (a) common enough a type of question and (b) rare enough (non existent) in official VCAA publications. I would assume even raising it at a “meet the examiners” session may prove pointless? I don’t get to go to these, so cannot offer to raise it.
    
    Reply
2. Damo says:
  
  February 22, 2019 at 2:59 pm
  
  I will confess that teaching this part of the course prompts some anxiety in me. I can only imagine what it does to the students. I feel that it all stems from the study design, which offers very, very little in the way of direction. I am less anxious about rational functions, because I feel on (fairly) safe ground with assuming that both the numerator and denominator are polynomial functions. As curvilinear asymptotes have been marked in on previous exam questions (as noted by John) I feel fairly confident about telling students that they should include these on their graph along with the equation. However, at the bottom of the Functions and Graphs section of the Study Design sits the dot point: graphs of simple quotient functions. I do not feel that it is at all obvious what a ‘simple quotient function’ actually is, and the level of detail that students would be expected to include with these graphs. My anxiety is not helped by there being no mention of ‘quotient functions’ in any textbooks that I have looked at. I think that Example 31 is an attempt to address this dot point, although I can’t be sure. In any case, I now feel sick, so I am going to take whatever I can scrounge up in my drawer and go and lie down in the storage cupboard.
  
  Reply
  1. Damo says:
    
    February 22, 2019 at 3:35 pm
    
    Also, with Example 31, and going by how curvilinear asymptotes have been calculated earlier in the text, for this question should it not be y=sqrt(x-1)?
    
    Reply
    1. marty says:
      
      February 22, 2019 at 8:19 pm
      
      Thanks very much, Damo. Your question is excellent and goes to the heart of this tricky WitCH: what, if anything, is wrong with the answer y = √(x-1) to 31(c)?
      
      Underlying your question, what precisely does one need to show to prove that a certain “line or curve” is a “non-vertical asymptote”?
      
      Reply
      1. Damo says:
        
        February 22, 2019 at 10:51 pm
        
        I’m not sure, but let me have a go. With polynomial functions a(x) and b(x), if a(x)/b(x)=q(x)+r(x)/b(x) then the graph of y=a(x)/b(x) will have a “non-vertical asymptote” at y=q(x) because as x goes to infinity (or neg infinity) r(x)/b(x) will go to 0 and the distance between the curve and q(x) will thus go towards 0. With the above example, we can do something similar: (x+1)/√(x-1)=(x+1)√(x-1)/(x-1)=(x-1+2)√(x-1)/(x-1)=√(x-1)+2√(x-1)/(x-1)
        
        The second part of this equation goes towards 0 as x goes towards infinity, so the graph goes towards √(x-1). I feel that this is a more consistent extension of the approach taken with polynomial functions, as opposed to the logic that the textbook uses to give an asymptote of y=√x. But, well, really?
        
        It leaves me feeling deeply uneasy, and if students really can expect to get this type of question, then I don’t know where it stops, and I don’t know how to prepare them adequately to tackle such questions. In particular, Number 8 refers below to “quotient functions, or rational functions as some IB textbooks have taken to calling them”, but my question is: Are quotient functions the same thing as rational functions? Because in the Functions and Graphs section of the Study Design, one dot point refers specifically to graphs of rational functions (with mention of asymptotes and stationary points) and another dot point refers to “graphs of simple quotient functions” (the dot point in its entirety). I don’t know what they mean by that. I’m just going to have a quick lie down in my son’s pillow fort. It feels safe in there.
        
        Reply
    2. John says:
      
      February 22, 2019 at 11:36 pm
      
      That’s very good, Damo! I hadn’t thought of that: y = Sqrt[x-1] + 2/Sqrt[x-1] so y –> Sqrt[x-1] as x –> +oo.
      
      So there’s at least *two* ‘plausible’ non-linear asymptotes.
      
      What about re-writing as y = (Sqrt[x] + 1/Sqrtx])/Sqrt[1 – 1/x] and then taking x –> oo. What, if anything, is wrong with that (it also gives y = Sqrt[x]).
      
      Or expanding y = (x+1)/Sqrt[x-1] at infinity (ask WolframAlpha: y = Sqrt[x] + (x to a negative power terms)) and then taking the limit?
      
      I totally agree with Marty’s question:
      
      “what precisely does one need to show to prove that a certain “line or curve” is a “non-vertical asymptote”?”
      
      And just to throw the cat among the pigeons, could y = Sqrt[x-a] where a e R+ represent plausible equations for the non-linear asymptote? An infinite number of non-linear asymptotes …?!
      
      Is the equation for the non-linear asymptote unique? (This question can only be answered after answering Marty’s question). And if it’s NOT unique, does it even make sense to talk about a non-linear asymptote? Under what circumstances *is* it unique? Does f(x) = x^2 + 1/x have a unique non-linear asymptote?
      
      So many questions! The only things I feel sure about right now are:
      
      An oblique (diagonal/slant) asymptote is a linear asymptote that is not parallel or vertical to the axes.
      A curvilinear asymptote is an asymptote that is not linear.
      
      Cambridge originally had a question 17: f(x) = (x^2 + x + 7)/Sqrt[2x+1] with a part (f) that asked for the equation of the “… other asymptote.” (the previous part asked for the vertical asymptote). This question appears to have been removed some time after 2016 ….
      
      Reply
      1. Damo says:
        
        February 23, 2019 at 12:05 am
        
        John, I have been trying to craft a reply for a while now (I actually have already made one attempt at posting a reply, but for some reason it didn’t appear. Boo). However, you have summed up much of what I’m thinking. So many questions, precious few answers. To add to the uncertainty, Number 8 commented that:
        
        Quotient functions, or “rational functions” as some IB textbooks have taken to calling them…
        
        However, in the Study Design under Functions and Graphs there is one dot point that says:
        
        • graphs of rational functions of low degree, their asymptotic behaviour and nature and location of stationary points
        
        And another dot point which says simply
        
        • graphs of simple quotient functions.
        
        So I am assuming that VCAA intend for quotient functions to be different to rational functions (so they contain non-polynomial functions?). However, there is no elaboration on what these functions may look like, no direction to how detailed such graphs would need to be and no explicit reference to them in any textbooks I’ve seen. I’m assuming that it means that questions like Example 31 are fair game, but, well. Yeah.
        
        I am now going to crawl into my son’s pillow fort. It feels safe in there.
        
        Reply
  2. Number 8 says:
    
    February 22, 2019 at 5:34 pm
    
    Damo, I hear and share your anxiety with this part of the SM34 course. Quotient functions, or “rational functions” as some IB textbooks have taken to calling them bring with them all sorts of questions about what to label – not helped by a number of the issues you raise above. But the worst of them all (in my opinion anyway) is the example you mentioned of an examiners report where non linear asymptotes are labelled but there is no help from the report nor the study design about whether this is required.
    
    Students ask (rightfully so) about this every year and I am still at a loss about what examiners want to see. Which, unfortunately is the driving motivation here, not education.
    
    Reply
Number 8 says:

February 17, 2019 at 7:08 pm

Was the author trying to follow a VCAA guide perhaps? (Some sarcasm intended, but not sure where it is directed)

Reply
1. marty says:
  
  February 19, 2019 at 12:14 pm
  
  The author was trying, and failing, to impress.
  
  Reply
  1. Number 8 says:
    
    February 19, 2019 at 5:52 pm
    
    Trying to impress who? The general consensus (from this website and other places) is that no-one really reads anything except the exercises in these books – possibly because it doesn’t make sense to a lot of students (whom it is presumably written for…?)
    
    Reply
    1. marty says:
      
      February 20, 2019 at 11:41 am
      
      Whom, not who, but good points. I was just being snarky. There’s no question the authors try to impress on occasion, but here they’re just faking it. However, the fact that teachers don’t read this text doesn’t mean they’re not unduly “impressed” by it. Yes the text is unreadable for many teachers (and most students), but my sense is that teachers typically blame their own perceived shortcomings for this, rather than placing the blame where it belongs, with the authors.
      
      Reply
Damo says:

February 23, 2019 at 12:21 am

Apologies for the double up comments above, I think my posting today reflects my feelings about this topic. I don’t know what I’m doing.

Reply
marty says:

February 23, 2019 at 12:30 am

Damo, you appear to know MUCH more of what you’re doing than the authors of the textbook. You have given a perfect argument that the function y = (x+1)√(x-1) is asymptotic to the “curve” y = √(x-1).

(Sorry about the delay: your comment was hiding in spam.)

“Rational function” means a ratio of polynomials (analogous to “rational number”). The term “quotient function”, similar to “hybrid function”, doesn’t have any precise meaning. Still, one can get a sense of “simple quotient” as a ratio of polynomials, rooty guys and the like.

Reply
1. John says:
  
  February 23, 2019 at 7:45 am
  
  Yes Marty, rational function means exactly what you say. But graphs of simple quotient functions could mean anything – from what I’ve seen in textbooks it has been interpreted as graphs of reciprocal quadratic functions, which is dumb because these are just rational functions of low degree ….
  
  I don’t disagree with your interpretation, but it’s not what immediately springs to my mind. My interpretation (that I use in the classroom) is that it means graphs of things like f(x) = 1/ln(x), f(x) = 1/(e^x -1), f(x) = 1/Sqrt[2x-1] etc. In other words, graphs of reciprocals of ‘familiar’ non-polynomial functions.
  
  I real problem would be if you’re right and therefore something like f(x) = (x^2 + 1)/Sqrt[x-1] is fair game. Because this function has all sorts of potential pitfalls when it comes to its asymptotic behaviour (however that is meant to be defined) …. I’d love to see what Cambridge thinks the equation of the curvilinear asymptote, if one exists, is ….
  
  What adds to the mayhem is that the Study Design specifically mentions graphs of the reciprocal circular functions, which I would have categorised within my above interpretation of simple quotient functions. It’s good that they get a specific mention, but it’s bad that you have to guess about the rest.
  
  What’s interesting is that asymptotic behaviour (whatever that’s specifically meant to mean) is only explicitly mentioned with the dot point on rational functions. But clearly this is intended to also apply to graphs of reciprocal circular functions.
  
  And of course an issue with my interpretation above is that it assumes quotient means reciprocal.
  
  Reply
  1. Damo says:
    
    February 23, 2019 at 8:37 am
    
    Adding further to the mayhem is that in the introduction to this section, we get “In this area of study students cover inverse circular functions, reciprocal functions, rational functions and other
    simple quotient functions…”. Which seems to indicate that they consider quotient functions to be different (or broader in scope) to reciprocal (and rational) functions, except that they then only refer to reciprocal circular functions in the dot point. My concern is that the last dot point is so open that an examiner could conceivably put in a question like Ex31 and justify it by pointing to this dot point.
    
    Reply
    1. marty says:
      
      February 23, 2019 at 10:01 am
      
      Thanks, John and Damo. It is worth noting that “reciprocal function” also has no meaning that is not trivial: ignoring zeroes, every function is a reciprocal function.
      
      I almost have sympathy for the VCAA on this issue. But, I don’t.
      
      It is close to impossible and is definitely unworkable to state precisely what a curriculum will cover, what functions can be expected and what the student is expected to do with them. So, though “simple quotient function” and “reciprocal function” are not mathematical terms, there is some intuition for what the terms mean. If this intuition is then nailed down with clear examples, good communication and general good faith, then things are ok.
      
      Alas, clear examples, good communication and general good faith are not VCAA’s strong suits.
      
      Reply
Damo says:

February 23, 2019 at 9:06 am

Returning to the work at hand, I also question finding the stationary point before considering the asymptotic behaviour. If you do this the other way around, then it is obvious that the stationary point has to be a minimum and alleviates the need to find the second derivative.

Reply
1. marty says:
  
  February 23, 2019 at 10:03 am
  
  Excellent point.
  
  Reply
John says:

February 28, 2019 at 4:50 pm

The key is to have a definition of either “behaviour as x –> oo” or “curvilinear” asymptote.

The sources I’ve consulted all seem to give the following definition:

y = g(x) is a horizontal, oblique or curvilinear asymptote of the function y = f(x) if lim[f(x) – g(x), |x| –> oo] = 0, where g(x) neq f(x).

This defines a unique horizontal, oblique or curvilinear asymptote for all rational functions.

But not for things like Cambridge’s y = (x+1)/Sqrt[x-1] where all functions of the form y = Sqrt[x-a], a e R+, satisfy the condition, implying an infinite number of asymptotes to be labelled on any graph that’s drawn. And for a function like y = (x^2 + 1)/Sqrt[x – 1], although it would be natural to think like Cambridge and claim h(x) = x^(3/2) as a curvilinear asymptote, this is not the case. It fails the above definition. However, a function that *does* satisfy the above definition is h(x) = x^(3/2) + Sqrt[x]/2 (another one is h(x) = (x + 1)Sqrt[x-1]). But how you get h(x) = x^(3/2) + Sqrt[x]/2 is not within the scope of the course, nor is long division of a radical into a polynomial ….

So a tighter definition is needed if the curvilinear asymptote is to be unique in general, and I haven’t come across one.

I suppose you could arbitrarily decide that it had to be a power series containing positive powers (not necessarily integer) of x (so that it corresponded to the leading terms in the expansion at infinity) …. I note that Sqrt[x] is always the leading term in such a power series for Sqrt[x – a] …. But this is all way past Yr 12.

I have no problem with the dot point “simple quotient functions”, *provided* identification of non-linear asymptotes is not required for non-rational functions. But the Study Design does not say one way or the other. I’d love to see it on the exam, purely for the huge sh*t storm it would generate.

It appears the the curriculum documents have never been clear on exactly what functions and features are required. In the past, rational functions of the type f(x) = ax^m + bx^-n + c (m,n = 1,2) have been explicitly stated but nothing said about asymptotes. Even the old VCAB Mathematics A and B Handbook (a much better curriculum document than all the Study Designs, containing substance rather than waffle) says nothing. And going further back to VUSEB the rational functions were all of the type where the degree of the numerator was never greater than the degree of the denominator by more than 1 (so that asymptotes were always linear).

In summary, it is my opinion that in Specialist Maths Units 3&4 you can reasonably ask for linear asymptotes (vertical, horizontal, diagonol/slant/oblique) of any function but you can only ask for non-linear asymptotes when the function is rational. It would be nice if the Study design made such clarification, with examples, as well as gave a glossary of notation, definitions and terms (which the old VCAB Handbook did).

Reply
1. marty says:
  
  March 1, 2019 at 1:06 am
  
  John, I don’t think the definition of asymptote you quote/semiquote is natural, and I don’t think it gives a unique asymptote for rational functions.
  
  Reply
John says:

March 1, 2019 at 7:51 am

Well, that’s a revolting development because in that case I don’t see the way forward through this mess.

Do you (or anyone else) have a counter example of a rational function where the definition does *not* give a unique horizontal, diagonal/slant/oblique or curvilinear asymptote?

Reply
1. marty says:
  
  March 1, 2019 at 9:41 am
  
  John, maybe I misunderstood what you wrote or intended to write. As concrete examples, what is the asymptote(s) to each of the following functions: 1) y = x + 1/x, 2) y = x, 3) y = 2 + 1/x, 4) y = 2? The answer can be either according to a) the Cambridge definition, b) your definition, c) what we would *like* the answer to be. So, twelve questions in all.
  
  Reply
  1. John says:
    
    March 1, 2019 at 11:43 am
    
    By my definition (and Cambridge):
    
    1) VA x = 0, OA y = x.
    2) No asymptotes. (NB f(x) neq g(x) in the definition I’ve written).
    3) VA x = 0, HA y = 2.
    4) No asymptotes. (NB f(x) neq g(x) in the definition I’ve written).
    
    I should say that I haven’t used the definition to *find* the equations of the oblique and horizontal asymptotes, I’ve considered what happens when |x| –> oo (like Cambridge does). But what I did do is use the definition to *confirm* the equations were ‘correct’ (well, correct under the definition).
    
    A bit like *finding* the limit of x^2 as x –> w, say, by substituting x = 2, getting 4 and then *confirming* the limit is 4 using the epsilon-delta criterion (the confirming part doesn’t happen in Yr 12).
    
    Which is what’s needed with an example like (x^2 + 1)/Sqrt[x-1] to *confirm* that something like h(x) = Sqrt[x] is actually *wrong*.
    
    Reply
    1. marty says:
      
      March 1, 2019 at 12:01 pm
      
      John, sticking to 1), is y = x + 1/x + 1/x^2 an asymptote to y = x + 1/x, in the sense of a) and/or b) and/or c)?
      
      Reply
      1. John says:
        
        March 1, 2019 at 1:26 pm
        
        OK, I see what you mean. I’m assuming the powers in the equation of the horizontal, diagonal/slant/oblique or curvilinear asymptote are always non-negative. Otherwise it’s total anarchy:
        
        Because why stop with y = x + 1/x + 1/x^2 being an asymptote to y = x + 1/x …. Let’s have y = x + a/x + b/x^2 where a,b e R+. And let’s add y = x + a/x + b/x^2 + c/x^3 etc.
        
        Complete anarchy.
        
        So the assumption that asymptotes have non-negative powers needs to be an explicit requirement. I guess I’ve always assumed this because I’ve never see y = x + 1/x in *any* textbook having asymptotes additional to x = 0 and y = x ….
        
        Reply
marty says:

March 1, 2019 at 1:28 pm

Yep, that’s the point. The Cambridge definition leads to “anarchy”. So, how does one avoid anarchy?

Reply
1. John says:
  
  March 1, 2019 at 5:22 pm
  
  Cambridge’s ‘definition’ is more a technique: Let x –> oo and see what function y approaches. The technique works for rational functions (just like finding the value of some limits is OK by substituting the value in) but will fail in general for non-rational quotient functions eg. (x^2 + 1)/Sqrt[x – 1] if not done carefully.
  
  Some of the anarchy can be avoided using a definition like:
  
  y = g(x) is a horizontal, oblique or curvilinear asymptote of the function y = f(x) if lim[f(x) – g(x), |x| –> oo] = 0, where g(x) neq f(x) and g(x) has no negative powers of x.
  
  This definition also deals with all the curvilinear asymptotes h(x) = Sqrt[x – a], a e R+, of (x+1)/Sqrt[x-1]: Take the series at infinity, let x –> oo, and keep the sqrt[x] term.
  
  So it’s not complicated for rational functions (it’s just like calculating limits by substituting the value of x) but is potentially tricky for non-rational quotient functions.
  
  Reply
marty says:

March 2, 2019 at 8:55 am

Cambridge gives both a technique and a general definition. As John notes, the technique is best thought as applying to rational functions, but even in this context the technique conflicts with the general definition.

I think the Cambridge definition can be tightened in the manner John suggests, but it’s probably easier to directly state the type of function permitted to be an asymptote, rather than the piece-by-piece ruling out of undesired behaviour.

Reply
John says:

March 9, 2019 at 8:03 am

Well, I would propose that the type of function permitted to be a curvilinear asymptote is a sum (either finite or infinite) of powers of x where all powers of x are non-negative (and not necessarily integers).

Marty, what definition of function would you propose?

Reply
1. marty says:
  
  March 9, 2019 at 2:17 pm
  
  John, I don’t think there’s one correct answer. It depends upon which functions you want to have asymptotes, and whether you want the asymptotes to be unique. The question is fundamentally tricky. A blithe approach, as in the excerpt above, is asking for trouble.
  
  Reply
  1. John says:
    
    March 9, 2019 at 5:39 pm
    
    So we can just say we want polynomial non-linear asymptotes for rational functions. That works fine, particularly at the VCE level. And we can all agree that hyperbolas have straight line asymptotes (vertical and horizontal, or diagonal/slant/oblique).
    
    But it gets trickier for non-rational functions eg. the hyperbolic functions sinh and cosh. I do want them to have unique curvilinear asymptotes, or at least to be able to talk about how they behave as |x| –> oo.
    
    Do we simply say that things like y = (x+1)/Sqrt[x-1] or (x^2 + 1)/Sqrt[x-1] has no non-linear asymptotes? Ruling out functions would seem to be no better (or worse) than ruling out undesired behaviour …?
    
    It boils down to definitions – but I haven’t seen definitive definitions anywhere that address the issues in this wiTCH (certainly none that you could easily present at the VCE level). In fact, this wiTCH is the possibly the only place where these issues have actually been raised, discussed and some attempt made at a resolution …?
    
    So, what’s our criteria? If there’s no unique curvilinear asymptote in the sense of lim[f(x) – g(x), |x| –> oo] = 0 then do we simply say that there’s no non-linear asymptote …?
    
    Reply
    1. marty says:
      
      March 10, 2019 at 3:43 pm
      
      Yes, to all of that. It is natural enough to restrict oneself to polynomial asymptotes. Or, more generally and as you suggested earlier, sums of powers of x. But then plenty of functions will not have asymptotes. Such is mathematical life.
      
      Reply
Damo says:

March 11, 2019 at 11:34 am

I’m not sure that I have anything new to add here, but this post has been rattling around my head a bit and I thought I’d get my thoughts down. Specifically, there has been a lot of discussion on here about what an asymptote is, but it really got me thinking, why are asymptotes important? What was the motivation around finding and defining asymptotes? And I realised that I don’t know the answer to that, that I have never really thought about it. However, I suspect that it was primarily to aid graphing (and to give us an understanding of how an unfamiliar function behaves through comparison to a familiar function). In which case, as soon as a curvilinear asymptote becomes overly difficult to sketch, it has lost its usefulness and, well, what’s the point. I think that this is central to the above discussion between John and Marty, but for my own benefit I thought that I’d make it explicit, and be corrected if my thinking is wrong. If I am right with this, however, then are we even being too open by restricting asymptotes to polynomials or sums of powers of x? Certainly, some of these asymptotes could be tricky enough to sketch on their own that we could reasonably ask, what’s the point? While on the other hand, an asymptote of y=sqrt(x) could be entirely reasonable. At this level, I would think that it would make sense for the study design to be very explicit about what asymptotes students could be expected to sketch, and, if they do want students to sketch curvilinear asymptotes, I don’t see any reason why they need to go beyond quadratic asymptotes.

Reply
marty says:

March 11, 2019 at 2:31 pm

Damo, all excellent thoughts. You are correct, the idea of “asymptote” is to give a simple guide to the behaviour of a function at infinity. And yes, the more elaborate we permit “asymptote” to be, the less use it is. It’s like Lewis Carrol’s map with a 1:1 scale; if the model is too specific you may as well just use the real thing as the model.

So, what if we only permit an asymptote to have a single power of x? Is there still a way that we can talk about the asymptote for a function? What is the definition? What about, for example, the function

(x³ + 4x²)/(x + 9) .

What is the “asymptote” by John’s suggested multi-power definition? How do we find it? What is the “asymptote” by Damo’s suggested single-power definition? How do we find it? Which definition is “better”?

As to what this means for sorting out the VCE curriculum, I don’t know and I don’t much care. My hobby is pointing out crap, and pointing out that the educational authorities and the textbook writers are either unaware or don’t give a stuff that it’s crap. Or, of course, both.

But, until some powers that be indicate that they are aware of the school crap and are concerned to fix it, discussing how one might fix it seems to me an exercise in futility.

Reply
Damo says:

March 11, 2019 at 4:46 pm

To start with, I would say that the asymptote of (x^3 + 4x^2)/(x + 9) would be y=x^2 – 5x +45. However, I would query the usefulness of this asymptote, at least when it comes to sketching the function.

What do we know about the graph of this function?
– We can easily ascertain the behaviour of the function as x-> -9.
– It has x-int at (-4,0) and (0,0)
– Using the first derivative, we can see that (0,0) is also a TP. Less easily, we can find the other two turning points.
– Finally, we know that as x -> +/-inf, the graph goes towards +inf.

With this information, we can get a pretty good sketch of the function. So how important is it to sketch the asymptote? How important is accuracy when sketching it? How much does it actually add to our understanding of the function? Is this a reasonable question for Year 12 students?

This last question is, I think, the key. Because, this topic does prove stressful for Year 12 students. Also,I don’t know the answers to the other questions that I have raised above, but I would like to be able to give as much clarity to my students as possible. And while the initial post identified some crap in a textbook, I think that it does raise some important questions for VCAA. Not that I would hold my breath, waiting for an answer.

Finally, an interesting (to me) aside. The question in the original post actually first appeared in Galbraith and Fitzpatrick’s Pure Mathematics (mostly as presented here). So, this has been rattling around Victoria for a while, and Evans might not have to bear full responsibility for this particular can of worms. Although it does raise another question – Is this plagiarism? And fairly mindless plagiarising at that?

Reply
1. John says:
  
  March 11, 2019 at 8:09 pm
  
  Hi Damo, and thanks for your latest contributions to this topic. I’m really glad to see I’m not alone in mulling over this issue. Everything you’ve said makes really good sense. Particularly your questions:
  
  How important is it to sketch the asymptote? How much does it actually add to our understanding of the function?
  
  Yes, it’s good to wonder how reasonable a question this is for Year 12 students. But what about 1st year maths students? 2nd year …?
  
  Yes, it’d be great to get clarification from VCAA on what asymptotes *they* expect, as well as clarification on what’s meant by reciprocal functions as opposed to quotient functions. But you’re right about the breath ….
  
  You could argue that nothing beyond linear (vertical, horizontal, diagonal) asymptotes are needed. And I think you could well argue that only vertical and horizontal asymptotes are really needed when it comes to drawing graphs ….
  
  Now, if anyone can point to a reference that gives a definitive definition or formalises the issues raised in this wiTCH, I’d love to know!
  
  Yes, the question in the original post has been rattling around at least since the late 70’s (and maybe earlier in whatever textbook preceded Galbraith ….). In my edition (a loyal servant that got me and then my sister through Pure and Applied Maths):
  
  p150 Eg. 9*: Sketch the graph of the function defined by f(x) = (x+1)/Sqrtx-1]. (The * means “May be omitted”, I assume because it wasn’t part of the course).
  
  It starts by discussing domain, the lack of axes-intercepts, vertical asymptote and then:
  
  “When x is very large, f(x) approx x/Sqrt[x]; in fact f(x) is slightly greater than x^1/2 as x increases without bound. Hence g(x) = x^1/2 is an asymptote whch f(x) approaches from above as x –> oo.”
  
  It then continues with finding the turning point and then shows the graph.
  
  It would be interesting to know what the Study Design equivalent of the time has to say about asymptotes, although I suspect it’s very little if the subsequent VCAB syllabus is anything to go by (which by the way is infinitely better than the current Study Design) ….
  
  Yes, I’ve noticed a lot of similarities between questions in Cambridge and questions in Galbraith and Fitzpatrick ….. The publisher of G&F was Jacaranda Press, acquired in 1976 by Wiley (which I don’t believe has any relationship with Cambridge ….)
  
  Anyway, the other leading Year 12 textbook of the time was Lucas and James ‘Sixth Form Pure Math’ (Lloyd O’Neil publisher), which has little to add this discussion and did not (perhaps wisely so) go near examples like this.
  
  Reply
  1. John Friend says:
    
    January 7, 2021 at 11:13 am
    
    Exercise 5(b) *Q34 of F&G gives the function $\displaystyle f(x) = \frac{x}{\sqrt{x+2}}$ .
    
    What’s strange about this question is that the answer says that $y = \sqrt{x+2}$ is an asymptote …
    
    So clearly they’ve used the method
    
    $\displaystyle f(x) = \frac{x}{\sqrt{x+2}} = \frac{x+2-2}{\sqrt{x+2}} = \frac{x+2}{\sqrt{x+2}} - \frac{2}{\sqrt{x+2}} = \sqrt{x+2} - \frac{2}{\sqrt{x+2}}$
    
    rather than their Example 9 (invalid) method, which would have given
    
    $\displaystyle f(x) = \frac{x}{\sqrt{x+2}} \approx \frac{x}{\sqrt{x}} = \sqrt{x}$ .
    
    I don’t know why F&G have done this, particularly since $f(x)$ is asymptotic to $y = \sqrt{x}$ …
    
    (And as an aside, unless a restriction is placed on the type of function that’s permitted to be an asymptote, there are an infinite number of asymptotes of the form $y = \sqrt{x - a}$ where $a \in R$ ).
    
    The “mindless plagiarising” of Cambridge did not extend to including this question in its exercises. Instead it chose Ex 6G Q17: $\displaystyle f(x) = \frac{x^2+x+7}{\sqrt{2x+1}}$ .
    
    This question is obviously a disaster:
    The invalid approach
    
    $\displaystyle f(x) = \frac{x^2+x+7}{\sqrt{2x+1}} \approx \frac{x^2 + x}{\sqrt{2x}} = \frac{x^{3/2} + x^{1/2}}{\sqrt{2}}$
    
    gives the wrong answer (easily checked using the definition in Marty’s blue box) but what’s comical is that Cambridge gives $y = \sqrt{x}$ as an asymptote, which is not even the correct wrong answer!!
    
    The bottom line is that the textbooks are useless and the VCAA Stupid Design is useless. The proposed Stupid Design is currently useless, but there’s an opportunity to remedy this with strong and overwhelming feedback. Fix the Stupid Design and textbooks such as Cambridge might also get fixed.
    
    Reply
    1. marty says:
      
      January 10, 2021 at 5:12 pm
      
      Thanks, JF. It’s not great that F & G tricked themselves although I don’t necessarily whack people/texts too hard for getting this stuff wrong. Getting this stuff wrong, however, for 50 odd years is pretty damning on Victorian maths culture as a whole.
      
      Reply
      1. John Friend says:
        
        January 10, 2021 at 8:08 pm
        
        “50 odd years” *or more* …!!
        
        The problem is that, like the stupid intersection of a function and its inverse question, one source from way back gets it wrong, but then everyone else blindly copies/plagiarises that source … F & G got it wrong. OK, shit happens. But then the people who write later textbooks (like the Cambridge text) just blindly copy/plagiarise the error. And then teachers come along and just blindly regurgitate from the textbook.
        
        No-one bothers to check what they’re plagiarising!
        
        And then the lepers without their bells that wrote those textbooks end up on panels for writing exams and Stupid Designs …
        
        I will look to see if I can find any textbook between F&G and Cambridge that has the same example and method.
        
        (As an aside: In Cambridge’s earlier editions they had an example that said (incorrectly) that the solution to $|z - 1| + |z + 1| = 1$ was the set of all points lying on a hyperbola. I know teachers in commercial revision programs who copied that example into their ‘lecture’ notes. One such teacher refused to believe me but went ashen after choosing several simple points on the hyperbola and discovering they didn’t work. I wrote to Cambridge pointing out this error, never got a reply but the next edition correctly showed the example had no solutions. The example was then deleted in subsequent editions (just like the asymptote example under discussion).
        
        Reply
        
        marty says:
        
        January 10, 2021 at 8:40 pm
        
        Yes, and I really, really don’t understand this. Of course there have always been teachers who notice conceptual errors, but it then seems to just stop there, unreported and uncorrected. There seems to be no expectation that the perpetuation of these errors is unnecessary and inexcusable. There isn’t the mathematical culture to shame such errors into non-existence. In brief, pretty much no one gives a shit, or at least enough of a shit.
        
        Reply
        
        John Friend says:
        
        January 11, 2021 at 9:00 am
        
        Marty, it’s possible that errors ARE being reported but the hubris-ridden big-heads who write this stuff refuse to accept they’ve made mistakes in all but the most totally blatant of cases. We witness this all the time with dodgy trial exam questions, dodgy trial exam solutions (eg. https://mathematicalcrap.com/2020/09/25/mavs-trials-and-tribulations/), dodgy VCAA Examination Reports, dodgy Stupid Designs etc. It all gets reported and what happens …? (In fairness to VCAA, a new broom seems to be taking feedback on dodgy Examination Reports more seriously than has previously been the case). So I doubt anything’s different when it comes to textbooks.
        
        I think there IS a culture of giving a shit, but this culture is consistently subverted by an influential inbred enclave of protected hubris-ridden big-heads who are not even 10% as smart as they think they are. That is the real problem.
        
        Reply
marty says:

March 12, 2019 at 11:31 pm

Damo (and John), thanks again. I didn’t mean to deny the worth of clarifying the nature and extent of this topic for VCE students and their beleaguered teachers. But, coming to a sense on this blog of how the topic could or should be framed gives essentially no insight into VCAA thought (or lack thereof). It’s the latter that is of practical importance, and I see no way of gaining that insight without the use of a cattle prod.

What I do like to do is to try to tease out the real mathematical ideas and issues and questions and worth. Whether and how that all fits in with VCE is secondary (for me). To that extent, I think you’re approaching the questions in a perfectly understandable but somewhat constrained, VCE-ish manner. So, you ask how much the “asymptote” helps with graphing, but the underlying question is, what is the behaviour of the function? It is the algebraic/analytic characterisation of this behaviour that is powerful, not the picture.

I asked the questions about the function (x³ + 4x²)/(x + 9) with the hope of clarifying such characterisations. Then one can more clearly consider more general functions. The Cambridge text is plain wrong on rational functions, so doesn’t have a hope of talking sense about the harder example above. (F & G is not great on this topic, but it’s on much solider ground than Cambridge.)

Reply
Red Five says:

August 26, 2019 at 7:15 pm

Some final thoughts from me (it has really been bugging me but more because VCAA is so quiet on the matter, I don’t know what to draw on graphs and what is time wasting!):

1. For a function, the vertical asymptotes are important because they affect differentiability, continuity. Horizontal asymptotes (or lack thereof) give a sense of what happens for x large (both directions) and are therefore important. Other types of asymptote are really (in my limited opinion) not so useful and are not really required except perhaps as construction lines to assist sketching.

2. For non-functions it gets trickier as each family of graphs has its own standard features (such as the x^2-y^2=1 hyperbola and related graphs) which are used in identifying the graph as belonging to that family.

3. Everything else I am putting in the “too hard, hope VCAA doesn’t ask” basket.

Reply
John Friend says:

August 26, 2019 at 8:51 pm

I generally agree with all your thoughts. Although I think one could argue that oblique/diagonal/slant asymptotes (however you want to name them) are important. And I still wonder about the importance (or not) of non-linear asymptotes associated with the graphs of things like the hyperbolic functions sinh(x) and cosh(x) ….

In particular, the lack of clarity by VCAA is appalling. I really wonder about the credentials of the people who wrote the Study Design. Back in the day, the equivalent to the Study Design had much more clarity, detail and included clarifying examples (however, there was still a lack of clarity with regards to asymptotes).

Nowadays, the Study Design is just a pompous self-serving paperweight. More often or not, you have to rely on exam questions to get clarity (my current example for this is the NHT 2019 Exam 2 Section A Q18 ….)

This lack of clarity seems to be a real deficiency in the Victorian mathematics curriculum documentation, going back many decades. I’m not sure what it’s been like or is in other states.

I find it interesting that I haven’t yet found a textbook book (including typical 1st yr university level mathematics textbooks) that discusses non-linear asymptotes and argues the case for or against them ….

Reply
marty says:

August 27, 2019 at 9:50 am

Thanks, RF and JF, for your latest comments, and thanks to everyone who has given this very tricky WitCH a go. I will update the post soon, and will try to summarise and to clarify all that is here.

In brief, “asymptotes” are important because they indicate the limiting behaviour of the function. But it is the function behaviour that is actually of importance, and the asymptotes are only of importance to the extent they give some intuition for that behaviour. The trouble in VCE is that asymptotes become the end-game, and so the proper and meaningful consideration of function behaviour degenerates into a formulaic ritual.

In general, this ritual is merely silly. When considering oblique and non-linear asymptotes, however, the function behaviour can be too complicated to be captured by the ritual. And then, we have a WitCH.

Reply
1. John Friend says:
  
  August 27, 2019 at 6:57 pm
  
  You closing paragraph: Very nicely put.
  
  Reply
John Friend says:

February 5, 2020 at 7:25 pm

Re: Update (05/02/20).
Amazing!! So much work. Thankyou, Marty.

So (at the risk of looking like I’ve missed the point) the pragmatic conclusions are:

The asymptotes of a rational function (whether they be linear or non-linear) are unique and worth finding.

If a non-rational function f(x) has a non-linear asymptote h(x) in the sense that lim[(f(x) – h(x), x –> oo) = 0 (excluding h(x) = f(x)) then h(x) is not unique in general.

Discussion of non-linear asymptotes of non-rational functions at the secondary school level is not a clever thing to do, particularly for a textbook (whose motivation is probably to want to look clever) because (a) they are not unique, and (b) their computation is non-trivial.

Reply
1. marty says:
  
  February 5, 2020 at 10:50 pm
  
  You’re welcome, John. Here’s my answer:
  
  i) If you want to talk about unique asymptotes then you must restrict the type of function that is allowed to be an asymptote. People may have differing conventions, and be more or less explicit, but the restriction must be there somehow.
  
  ii) This is implicitly understood when you’re considering horizontal asymptotes or linear asymptotes: the name tells you what is permitted. Maybe “asymptotes” to rational functions are a little less standard, but probably still standard enough: they will be naturally polynomial and are readily computed.
  
  iii) So, (i) and (ii) confirm your first statement.
  
  iv) Your second statement is (pragmatically) correct, unless the type of permissible asymptote has somehow been restricted. (But also note, there is no a priori reason to exclude h(x) = f(x).) One could argue that a polynomial asymptote to a non-rational function should implicitly be regarded as unique, and one could declare that to be the case, a la points (i) and (ii). Personally, I think that once you’re out of the linear world, you gotta carefully declare the rules.
  
  v) Your third point is correct, but I’d go further. Even linear asymptotes to non-rational functions are a problem, because of (b), unless the functions are very carefully chosen. For example, I don’t see how a school student could be expected to determine the linear asymptotes of the functions p(x) and q(x) above. One could carefully choose non-rational functions where the linear asymptotes can be determined (i.e. without cheating). But it’d be a pretty Swiss cheesy topic.
  
  Reply
  1. John Friend says:
    
    February 6, 2020 at 8:10 am
    
    i) is a really key point that you make – it is completely lacking in the VCAA Study Design (among many other things).
    
    Reply
    1. marty says:
      
      February 6, 2020 at 12:34 pm
      
      Yes. If you say “horizontal asymptote” or ‘linear asymptote” or (less commonly) “polynomial asymptote” then it is clear what you mean and the asymptote, if it exists, is unique. If you say “asymptote” then it will usually be interpreted to mean “linear asymptote”, and again that’s fine. In any other context, or with any other meaning, the obligation is to say exactly what you mean.
      
      Reply
SRK says:

February 8, 2020 at 3:43 pm

Many thanks for the update Marty. I have recently finished teaching this topic to my class, and (with some trepidation) my approach was: (i) if you don’t have CAS, you’ll only be asked to sketch rational functions where the asymptotes are lines or parabolas, so just find the quotient (ii) if you have a CAS, just copy the fucking screen on to the page.

Reply
1. marty says:
  
  February 8, 2020 at 3:53 pm
  
  Thanks, SRK, and you’re welcome. That also raises an interesting question: are asymptotes of non-rational functions ever actually examined?
  
  Reply
  1. John Friend says:
    
    February 8, 2020 at 6:29 pm
    
    Good question. The answer is yes, they are routinely examined via the reciprocal trig functions sec, cosec and cot. But if we’re talking about non-rational functions of the ilk discussed in this blog, then the answer is no.
    
    As an aside, I have only ever seen non-linear asymptotes examined twice in the last couple of decades. Both times it was in the context of a rational function.
    
    At the university level you have the hyperbolic functions sinh and cosh with their exponential asymptotes ….
    
    Essential was trying to be cute and clever, but only succeeded in looking dumb. At least the other textbooks know their limitations ….
    
    Reply
    1. marty says:
      
      February 8, 2020 at 7:05 pm
      
      Thanks, JF. Yes, I should have been more specific. I meant asymptotes to (non-rational) algebraic functions: rooty guys.
      
      Reply
John Friend says:

January 11, 2021 at 10:42 am

I’d like to add two things to Update 05/02/2020 point 4 with regards to the functions p(x) and q(x):

1) p(x) and q(x) also have linear asymptotes $y = -x$ and $y = -x - \frac{1}{2}$ . This doesn’t affect the argument point 4 makes, it’s stated simply for completeness for those who are still watching the game.

2) A simple way (and easier than using a series expansion at infinity) of understanding why the $-x$ in the denominator of q(x) can’t be ignored is to write:

$\displaystyle q(x) = \frac{x^2}{\sqrt{x^2 - x}} = \frac{x^2}{\sqrt{(x - \frac{1}{2})^2 - \frac{1}{4}}}$ . The $-\frac{1}{4}$ in the denominator can be ignored in the same way that the $-1$ in the denominator of $p(x)$ can be ignored. But ‘obviously’ the $-\frac{1}{2}$ which originates from the $-x$ in the process of completing the square cannot be ignored. It also shows where the other linear asymptote stated in 1) comes from. (I should point out that this ‘method’ is not quite kosher).

Reply
John Friend says:

October 11, 2022 at 11:18 am

The new edition of Cambridge Specialist Maths for Units 3-4 (for the new Study Design) is now available. I have looked at Chapter 8 Section 8G (Rational Functions). It makes all the same mistakes with asymptotes as the previous edition. In particular, it has retained Example 31 together with the invalid calculation of the ‘curved’ (sqrt) asymptote.

It is very evident that the authors do not know or understand that:

1) Their calculation of the ‘curved’ asymptote in Example 31 is invalid.

2) Ex 8G Q16 requires students to use the same invalid approach as Example 31 to get a sqrt asymptote.

3) Functions such as that in Example 31 and Q16 are not rational functions and are not part of the new (or old) Study Design.

4) Their Example 26 and Example 27 (and Ex 8G Q1 (b) and (g)) require students to find ‘curved’ asymptotes (parabolic) despite VCAA stating very clearly that only straight line asymptotes are in the Study Design.

Who proof reads this stuff?? (*)

@Marty: Maybe you could mention all this to David Treeby (one of the authors). I’m sure he’s not responsible for this mess, but his name will have more influence with Cambridge in getting those dodgy examples and questions deleted than a Friendly name.

* None of these examples or questions have been marked as not being part of the Study Design.

Reply
1. marty says:
  
  October 11, 2022 at 4:32 pm
  
  Thanks, John. In the authors’ (limited) defense, VCAA”s rushing out the new Curriculum presumably means the textbook companies also rushed out the new books. I didn’t know David was part of Cambridge SM34. If so, I think that’s new, and his involvement would probably have been minor.
  
  Reply
  1. John Friend says:
    
    October 11, 2022 at 4:48 pm
    
    That’s what I figured. David (Treeby) and Gareth (Ainsworth) are both new additions to the Evans et al. authorship. (By the way, for those who don’t know, both have a PhD in mathematics and both are quality mathematics teachers).
    Cambridge has had 6 years to fix these errors (by the simple expedient of deleting them!!) I can only assume that the person who wrote this section for Edition 1 also wrote/vetted it for Edition 2 and said “All good, nothing to change here, go to print.”
    
    My main reason for posting is that I think there will be a lot of new and inexperienced Specialist Maths teachers next year and they will think that those examples and questions reflect the Study Design in some subtle way … So I’ll be explicit:
    
    Ignore those examples and questions. They are a combination of cagel (*) and outside-the-scope-of-the-course. Someone’s $\displaystyle still$ trying to show off (after 6 years!), but all they’re showing off is their foolishness and ignorance.
    
    * With a nod to Harry Harrison and the Stainless Steel Rat. (Added post-Marty follow-up comment)
    
    Reply
    1. marty says:
      
      October 11, 2022 at 4:50 pm
      
      Language … But yes, most textbook editing I’m sure is token at best.
      
      Reply
      1. John Friend says:
        
        October 11, 2022 at 5:20 pm
        
        Noted.
        
        I’m also amazed that this was not brought to Cambridge’s attention in the last 6 years (but maybe it was and the writer simply said “Ignore the crank. All good.”)
        
        Obviously no-one at Cambridge reads your blog! (At least VCAA reads it. I think …)
        
        Reply

Related posts:

79 Replies to “WitCH 8: Oblique Reasoning”

Leave a Reply Cancel reply