This question may not be the worst ever but, as discussed here, it’s pretty bad. I think there’s also more than the obvious to say about it. The question is from the 2023 Specialist Mathematics Exam 2 (not yet online).

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# WitCH 114: Rational Dysfunction

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This question may not be the worst ever but, as discussed here, it’s pretty bad. I think there’s also more than the obvious to say about it. The question is from the 2023 Specialist Mathematics Exam 2 (not yet online).

On my first read, I thought that you could long divide to find the values of and , then find afterwards. That way, you could avoid having to think about the other vertical asymptote not listed in the question. But then, the wording also implies that there are only two asymptotes, and you’d have to assume is a double root for the quadratic in the denominator, which isn’t the case. So, the wording could lead you astray if you think too much about it. That’s what I can see, at least.

Concur. My initial approach to tackling this question was generating a system of equations through first equation the discriminant to 0, as I had assumed ax^2+bx+c had one real root, and the others through long division as you had suggested. I reached a solution only through realising that ax^2+bx+c did in fact factor into two distinct factors, and applied the other equations. Confusing for a completely unnecessary reason.

See the WitCH.

EDIT: Doh! This is the WitCH. I am way too tired.

I wonder if anyone cares that by this definition of asymptote, the straight line also “has asymptote “. According to this example question, linear functions have asymptotes!

Then immediately following with the same wording and using is quite problematic. One of these is a function, the other not. They should not be treated the same way.

I would say instead that the function *is asymptotic to* as approaches plus or minus infinity (or both). This can be defined as the absolute value of the difference of the function values approaching zero. This would be mathematically sound, and allows for easy generalisation later on.

For a function on the reals, an asymptote is when the application of a given function is asymptotic to plus or minus infinity as approaches . This can be defined as the limit of the values of the function diverging to plus or minus infinity respectively. We can be ambiguous and say that has an asymptote at .

This annoyed me.

I have three issues with this question, two have been mentioned.

1. Is an asymptote? Maybe, but this is multiple choice so there is no time to think about it.

2. When the question says there are asymptotes and names two, does that mean these are the only two? Maybe, but this is multiple choice so there is no time to think.

3. And I should have led with this: THIS IS MULTIPLE CHOICE AND THERE IS NO TIME TO THINK. Just push some buttons, find an answer that works and move on. That is not even remotely “doing” mathematics.

Thanks, RF. Obviously 2 is the main issue. Why might y = 2x + 1 not be an asymptote?

There was a very long (and amazingly educational, for those prepared to read it all…) WiTCH on this some time ago.

From memory, the definition of an asymptote is not clear from VCAA’s many examples and could be thought of more as a construction line than an asymptote.

For instance, does the graph of the function approach the graph of as gets really large/small? Why? Is this a key feature of the graph of the function?

If not, calling an asymptote is not really appropriate.

It was WitCH 8: Oblique Reasoning (https://mathematicalcrap.com/2019/02/13/witch-8-oblique-reasoning/).

Thanks, Sai.

Thanks, RF. I’m not sure there’s anything in the WitCH to suggest that y = 2x + 1 isn’t an asymptote.

My issue is that there is nothing in the question to suggest that there are three asymptotes.

There is nothing to say there are only two either, but I’m not sure how valid that argument is in this case.

Thanks, RF. Of course the “two asymptotes” thing is by far the worst aspect of the question. But I do want to say more.

I’ll admit, when posting this WitCH, I had forgotten that CAS was here. And of course, CAS poisons everything.

Glen, this is high school. I think you’re being too pure.

Glen, this is off the point of the WitCH, but I’m trying to figure out why the language of the question is so annoying to you. There seem to be two aspects to your complaint, and I’m not sure which is the driver. (See below.)

To be sure, the language of “asymptotes” in VCE is a mess, and this mess can lead to mathematical error. The purpose of my book-length WitCH 8 was to try to lay out the ideas as clearly as possible.

Now to your objections.

1) You seem to want asymptotes to be the geometric objects, the lines, rather than the functions (or relations). Is that correct?

I see the point, but honestly I don’t care much. Vertical lines will always be a little peculiar to deal with, since they’re geometrically very simple, but are not the graphs of functions (of x). In WitCH 8 I simply ducked “vertical asymptotes” and stuck to functions as asymptotes (and then eventually functions being “asymptotically equivalent”).

2) You seem to object to the function y = 2x + 1 having y = 2x + 1 as an asymptote.

I agree it’s a confusing trivial case. But what language/definition are you suggesting to rule it out?

(1) Yes, (2) See below…

My basic motivation is to eliminate confusion in students by presenting the material logically. Asymptotes should denote singular behaviour by functions. These are things like the vertical axis for . Only functions can have asymptotes (or be asymptotic to something), and asymptotes themselves should not be functions. In high school, asymptotes would I expect be vertical lines.

A separate idea is the behaviour of functions defined on as their parameter approaches plus or minus infinity. While it is standard terminology to use the words “asymptotic to blah” to describe properties of these functions, I believe it is a critical distinction to stop precisely there and go no further. Using the word “asymptote” goes further.

You might wonder if these two ideas should be so strictly made distinct. The answer is unequivocally yes, but students are not expected to fully grasp why. This is the reason that I said in another comment that people putting the curriculum together need a broader view. They need to be experts. (Briefly: While plus and minus infinity are boundary points just like the points and for a finite interval , the study of functions that have asymptotes at and is very different to the study of functions that are asymptotic to something at plus or minus infinity. They go in different spaces, and depending on the details, can be used for different purposes. It is a big deal to sort this out precisely later, and that should be facilitated by the curriculum design (not hindered).)

My example is supposed to demonstrate that it is not reasonable for an asymptote to be its own asymptote or asymptotic to anything, so it is supposed to be my shortcut to generate a reaction similar to mine from someone mathematically minded. This is because asymptotes should not be functions in the first place. If they were functions, then there is no singular behaviour as they should be defined at every point in their domain. So the original function didn’t have an asymptote there after all.

Thanks, Glen. I agree that it is generally preferable to think of “asymptote” as the geometric thing rather than the function. My purpose in the previous WitCH was to try to explain non-linear “asymptotes”, which is where the Cambridge textbook strayed.

But I still think you’re worrying way too much about this here. And I *really* don’t want to hint to VCAA that they should make their questions even wordier. They’re already fusspots.

The question reads “… asymptotes given by y = 2x +1 and x = 1”. How would you reword this?

“The graph of … has asymptote x = 1 and is asymptotic to y = 2x + 1.”

Caveat: With the understanding that when unspecified, “asymptotic to” means at plus and minus infinity, not just one of them. If not, then more words are needed. Which makes sense, as the behaviour of the function is more complicated.

A swing and a miss.

I’m not attempting to address the badness of the question. The rewording I suggest is not to fix the question, just to illustrate the point I’m making.

I know. I’m disagreeing with the point you’re making.

Does the Folium of Descartes have an asymptote?

Thanks, BiB. For the less informed (like me), see here. The folium of Descartes is given by the equation . Picture below for a = 1.

No. On (and elsewhere where it defines a function) it is asymptotic to .

Glen, forget it.

For those pondering this, I’m not suggesting anything deeply wrong, in the nature of WitCH 8. I’m just thinking about how it sucks as an exam question, and i think it sucks for more than just the “has two [but maybe more] asymptotes”.

What is the optimal strategy for doing this question in an exam?

Long Division or other methods of doing it by hand would probably take too long for a one mark question.

The low-skill way of doing this question would be just to input all of these on the graphing calculator and look at the graphs, no need for mathematics at all. And the more skilled students can notice that a must be 1/2, and that the denominator must be 0 when x = 1, narrowing it down to B and C.

Or, if graphing is too slow, you could use the “expand” function on the CAS as well on the various options, which makes it trivial to identify asymptotes.

This question would be a much better question if it were tech-free, and worth three or four marks.

This is the kind of pondering that lead me to post the WitCH. I think it could have been a good question, and it very much isn’t.

Here is my approach as a student sitting the exam. The function does have an oblique asymptote, therefore I “Propfrac” it in the CAS, writing the denominator as (x-1)(mx+n) and get the remainder and equate it to 2x+1 to find m and n. At this stage I simply don’t care if the denominator is a perfect square or not as if it does, it will reflect via the value of m and n. I do not have a huge issue with this question simply because it’s MC and I do not give a hood much about it. The trick for doing MC is “one can’t overthink”. You barely have 1.5 min to do one questions, so go by the instinct (that’s what I tell my student.) But I agree the wording is not nice. But it’s VCAA, what do you expect?

Sigh. I hate the modern world.

It really is fundamentally broken. At which point in that process did someone do some mathematics?

I know Marty you think I’m too formal or whatever, but much more thought needs to be a big part of whatever solution could ever get these subjects out of the quagmire they are languishing in currently. While the students and the curriculum do not need to assess pure or formal ideas (and IMO shouldn’t), those setting the curriculum and designing the assessment do need to be familiar with how everything ties together and where things naturally go. They should be trained in mathematics. At least in my opinion.

It’s broken by CAS here. The formalness of “asymptote” is a very distant second.

You can do partial fractions in 1 line on the CAS. From there look at what does go to 0 in the limit as x goes to +- infinity. This gives you a and b (based on what is left over). Your last step is to solve for c given you know one of the roots of the quadratic.

2 lines on the cas and a couple scratchy lines of working.

So, even though the wording of the question is stuffed, it doesn’t matter that it’s stuffed because CAS. What does that tell us?

(ps: Please use a more distinct name than “anon”.)

I believe the term we used at school was, ‘CAS bashing’.

Now, it takes a while to relinquish preconceived notions of mathematics and adopt a robotic CAS attack, but that’s what VCAA rewards.

So far as what it tells us, VCAA sees the CAS as an entire solution method rather than a solution aid.

I will add that this question wasn’t the worst I’ve seen so far as unintuitive CAS garbage, and I did appreciate the need to consider limiting behaviour (however obvious)

Thanks, James. I have my own personal style of CAS bashing. I highly recommend it.

It is possible to do SM MCQ2 by hand in the time restriction – as you don’t need the remainder of the long division – but not easy to see it in the time pressure of a MCQ in an exam, after being spun about by the 2 or 3 asymptote issue, with the temptation of the CAS hammer sitting next to you, etc…

Yes, as a tech-free question, this would probably be quite nice.

As for the optimal strategy as a MCQ CAS active question? Pretty much what others have said. Long division with CAS and match coefficients.

CAS bashing. Sketch the graph with each of the MC options, panic that one option looks right seems to have an extra asymptote, then realise VCAA are daft, and move on.

Hi,

I think Marti and/or JF may have written about “oblique asymptotes” a few years ago in a witch?!

Presumably the question was expecting the candidate to divide the denominator into the numerator and equate the coefficients to obtain values for a and b which equate to y= 2x+1 and then calculate c =-a-b with x = 1

Steve R

Click to access Slantsymptotes.pdf

Sai linked the WitCH above. I don’t think it’s particularly relevant, not at least to my concerns with this question.

To summarise my concerns with this question (the wording is atrocious, not just my post):

– The wording implies that the given asymptotes are the ONLY asymptotes, which is not true and painfully confusing.

– The question positions students to recognise that a vertical asymptote of x=1 implies ax^2+bx+c equals zero when x=1. However, as of the point above, this can mislead students to assume that the denominator can be factorised into (x-1)^2, or that the discriminant is equal to zero, both creating incorrect systems of equations.

– Nitpick, but this question asks for the values of a, b, and c ‘respectively’, yet MCQ17 lists solutions in the form ‘a=…, b=…,’ etc. Annoying inconsistency, makes questions hard to read.

Overall, while this question may not have a major error (in the sense of the word), it provides a serious obstacle for students in completing this exam for their ‘CAS-push-button-less-than-1.5-minutes’ exercise, more so than a lot of more blunderous VCAA blunders. Uncomfy.

To summarise my concerns with this question (the wording is atrocious, not just my post):

– The wording implies that the given asymptotes are the ONLY asymptotes, which is not true and painfully confusing.

– The question positions students to recognise that a vertical asymptote of x=1 implies ax^2+bx+c equals zero when x=1. However, as of the point above, this can mislead students to assume that the denominator can be factorised into (x-1)^2, or that the discriminant is equal to zero, both creating incorrect systems of equations.

– Nitpick, but this question asks for the values of a, b, and c ‘respectively’, yet MCQ17 lists solutions in the form ‘a=…, b=…,’ etc. Annoying inconsistency, makes questions hard to read.

Overall, while this question may not have a major error (in the sense of the word), it provides a serious obstacle for students in completing this exam for their ‘CAS-push-button-less-than-1.5-minutes’ exercise, more so than a lot of more blunderous VCAA blunders. Uncomfy.

I think this would have been a better question if it did only have one x-intercept since that would reward students who recognize that the denominator must be a perfect square and used the discriminant to find the solution. As it is I think most students will just type it into their CAS and cycle through the options with ‘prop frac’ until they get the desired result. I don’t think they would even do the minor bits of algebra suggested here by other commentators. I think if you can do an exam question using CAS and trial and error there should be something in it that means that thinking about the question and understanding something will save you time – but here those of us who thought about it lost time. I got the wrong answer thinking there was only one x asymptote, and when I checked it on CAS realised it was wrong because I had the wrong oblique asymptote, then reread the question several times and thought about whether another asymptote was permitted while the clock ticked on.

Thanks very much, Amber. This is getting to why the question really irritates me.

It looks like a question you can readily think about. So, the “two asymptotes” thing really punished the students who think you should think. But even if the language had been properly precise, it’s still a nasty question to do via brain rather than via zombie.

And, once again: thank you David and thank you Kaye. What you’ve contributed to Victorian maths ed will never be forgotten.

This what was I did at first shot on the basis that it’s multiple choice and must be quick and I neither teach SM nor do CAS voluntarily.

It’s NOT correct by any methodology but it does get the right answer and perhaps some students may have done this – so it’s not a great question.

x-1 cannot = 0 and 2x+1 cannot = 0 due to asymptotes (yes I know 2x+1 is not correct but if I’m in rush, this was what auto-pilot gave me and I couldn’t think of anything better). Hence bottom line = d(x-1)(x+1/2) = d(x2 – 1/2x – 1/2). Hence b = c, hence B as the only one like that.

Hi, JJ. Am I correct that your method is completely invalid but, by dumb luck, gives the correct answer?

Yes – but you’d think it’s instinctively what some people might do.

OK, I’m going to take another swing at this question:

It starts out nice enough, is an asymptote, so . So far so good.

Next, you can work out fairly easily that .

Unfortunately, realising this means there is not a repeated factor in the denominator…

Then it gets problematic. The way I proceeded from here (sans CAS) was to guess that there was another vertical asymptote of the form and then work backwards.

Coming at the question again, to see if the alternative MCQ strategy worked – you can narrow it down to options B and C pretty quickly then just graph them both. Total time a lot less, but what is the question really testing in that case?

Not much.

If I have to do it teach free, I would use the remainder theorem and equating the coefficient as seen in the attached file, it costs me 5 mins.

Thanks for posting this. My own thinking was slightly different.

is pretty easy to see, and since must be a factor of the denominator, must be a solution to .

This means that

At this point, I will admit I stopped, grabbed the CAS and did it the way VCAA intended.

Thanks, Victoria. That’s more or less how I would have approached it, but without so much expanding and detail. So, the 2x + 1 asymptote (or whatever Greg wants to call it) gives

.

Obviously a = -1/2. But then you have to work. The term gives you b = -a/2, but then you

stillhave to work. Then x = 1 (asymptote) gives you c = – a – b.One version or the other of this is straight-forward enough. But, with the second hand of the clock ticking, it’s almost impossible to see a quick path and to not stuff up the fractions and negatives.

A truly horrid problem.

For what it’s worth:

x = 1 is a vertical asymptote therefore

… (1)

Using the appropriate command on your CAS of choice:

(Irrelevant Stuff)

Compare with :

and .

Substitute these values into (1) and solve for c: .

Option B.

I’m not condoning the question but I think it can be done efficiently using a combination of CAS and a small pinch of mathematical acumen.

The CAS makes getting (Irrelevant Stuff) a lot easier. I haven’t checked how much time it takes to get the leading terms ‘by hand’.

I will echo the comments made by other commentators:

I think the wording is very misleading (borderline duplicitous) to the point of being wrong.

This is really similar to what I did, I’ll post in a separate comment.

Here’s my attempt for what its worth. I don’t think there is any need for hard calculation or some computer assistance.

From the fact that is undefined as , we know that when , which gives . Call this (1).

We also know that as so similarly

which implies the coefficients of the cubic and quadratic terms must vanish. That is, , then , and finally plugging into (1) we see that .

We disagree on the words being used, and it seems we may also disagree on the complexity of the question. The arithmetic is just multiplying and dividing by 2, and the simultaneous equations basically solve themselves. Doing the long division at the start with the aid of a computer just makes the question harder than it needs to be, in my opinion.

Nonsense. You have 90 seconds to decide whether to use CAS or brain, the approach to use with either, and to reliably get the answer.

The question punishes thinking, and thus is horrid.

I agree, it undoubtedly punished thinking this year. (In any year, most MCQ’s punish thinking – any dirty way of finding the correct option is the name of the game).

However, some good may come from it in the future:

Glen’s solution exploits , that is, Glen is exploiting the definition of a diagonal (oblique, slant) asymptote.

I do not believe this gets taught in most classrooms. I believe that students are told to do the polynomial long division and then write down the asymptote ‘by inspection’ – they are never taught why this technique gives the asymptote. Perhaps MCQ 2 will encourage deeper teaching of diagonal asymptotes in the future, particularly since Glen’s approach saves a lot of time if doing a question like this ‘by hand’. And many students and teachers like to anticipate MCQ’s getting asked in Exam 1.

It’s an ill wind that blows no good.

@Glen: Nice solution.

I’m not doubting for a minute that several of the MCQs on this paper could have been made into decent Paper 1 questions.

This raises the question… why is Paper 1 only 1 hour and why does Paper 2 contain MCQs???

It all just seems like quite an easy problem to solve…

Because VCAA loves CAS and will not, ever, listen to reason.

Even if I accept that argument, surely a decent, CAS-active paper can be written that doesn’t have multiple choice?

Shorten Paper 2 by 30 minutes, add a couple more decent questions to Paper 1…

Yeah. Not likely.

No.

I’ve tried to find the approach to this question that would require the least understanding by a student, while still getting the correct answer.

propFrac(x^3/(a*x^2+b*x+c))|{a=2, b=-4, c=2} (for option A)

You can then just cycle through the options until you find one that has the correct oblique asymptote and a denominator=0 when x=1.

Hi Damo, long time no comment (or see).

“I’ve tried to find the approach to this question that would require the least understanding by a student, while still getting the correct answer.”

Just for the perverse pleasure?

Partly for the perverse pleasure. But also, I need to try and help my students pick up whatever marks they can, even if they don’t really deserve it.

And also, I think this question is even more horrid than commenters above have realised. Even when using the CAS, they have tried to solve the question. But you don’t have to. Just plug in the values and see what works. Which trivialises the question to: Do you know what asymptotes are? And many Multiple Choice questions can be trivialised in this way.

Thanks, Damo. Point taken.