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How many marks roughly would this question be worth (AQA A level Maths)?

Find the coodinates of the stationary point(s) of the curve: y = 2+3x^2/3x-1 and prove the nature of the stationary point(s).
Reply 1
Original post by lanky_giraffe
Find the coodinates of the stationary point(s) of the curve: y = 2+3x^2/3x-1 and prove the nature of the stationary point(s).

Not sure, but its reasonably straightforward so solving a quadratic to get the stationary points and the second derivative simplfies nicely though its a few lines of algebra. Maybe ~6/7, but take that with a pinch of salt.

Fwiw, the problem is easier to understand by writing it as
y = x + 1/3 + 7/[3(3x-1)]
So shifted reciprocal with asymptotes y~x. So a min in Q1 and a max in Q3.
(edited 4 months ago)
Original post by mqb2766
Not sure, but its reasonably straightforward so solving a quadratic to get the stationary points and the second derivative simplfies nicely though its a few lines of algebra. Maybe ~6/7, but take that with a pinch of salt.

Fwiw, the problem is easier to understand by writing it as
y = x + 1/3 + 7/[3(3x-1)]
So shifted reciprocal with asymptotes y~x. So a min in Q1 and a max in Q4.

how is it easier to understand by writing it as that?

Surely I would just use the quotient rule twice, no?
Reply 3
Original post by lanky_giraffe
how is it easier to understand by writing it as that?

Surely I would just use the quotient rule twice, no?

Sure to chug through the algebra, its "easiest" to do the quotient rule twice. Though that form simplifies the second derivative part.

For sketching the function and understanding/validating the nature of the stationary points though, its useful. It says the function is roughly (a couple of translations/scaling)
y ~ x + 1/x
which is easy to sketch.
(edited 4 months ago)
Original post by mqb2766
Sure to chug through the algebra, its easiest to do the quotient rule twice.

For sketching the function and understanding/validating the nature of the stationary, its useful.

Okay, thank you.

Also, how did you turn it into y = x + 1/3 + 7/[3(3x-1)]?? I’m a bit confused
Reply 5
Original post by lanky_giraffe
Okay, thank you.

Also, how did you turn it into y = x + 1/3 + 7/[3(3x-1)]?? I’m a bit confused

Either polynomial division or the add zero trick (twice) so the numerator can be written as
[3x^2 - x + x - 1/3 + 1/3 + 2]/(3x-1) = (3x^2-x)/(3x-1) + (x-1/3)/(3x-1) + (7/3)/(3x-1)
Really the add zero trick (borrow remainders) is just doing polynomial division on the fly.

Its really just making the fraction proper, which is a good habit to get into.
(edited 4 months ago)
Original post by mqb2766
Either polynomial division or the add zero trick (twice) so the numerator can be written as
[3x^2 - x + x - 1/3 + 1/3 + 2]/(3x-1) = (3x^2-x)/(3x-1) + (x-1/3)/(3x-1) + (7/3)/(3x-1)
Really the add zero trick (borrow remainders) is just doing polynomial division on the fly.

Its really just making the fraction proper, which is a good habit to get into.

Honestly, I think this may just confuse me at the minute, but thank you very much for your help.

I dont even know how to do polynomial divison (I dont think its required for my course), but i will likely come back to this once ive finished the course in about a months time (I'm teaching myself the course in my gap year).
Reply 7
Original post by lanky_giraffe
Honestly, I think this may just confuse me at the minute, but thank you very much for your help.

I dont even know how to do polynomial divison (I dont think its required for my course), but i will likely come back to this once ive finished the course in about a months time (I'm teaching myself the course in my gap year).

Sure, if youve not covered it, just ignore it. The standard way would be to quotient rule twice and chug through the algebra.
Original post by mqb2766
Sure, if youve not covered it, just ignore it. The standard way would be to quotient rule twice and chug through the algebra.

Yeah, that's how I worked it out; it was about two pages of workings.
Original post by mqb2766
Sure, if youve not covered it, just ignore it. The standard way would be to quotient rule twice and chug through the algebra.

I think it might come up later in the course, so thank you for the heads-up and the help!
Reply 10
Original post by lanky_giraffe
Yeah, that's how I worked it out; it was about two pages of workings.

It shouldnt be that much but it depends how youve written it etc and 6/7 marks at the start may be a bit underestimated but Id be surprised if it was more than 8.

Tbh, this is one of the things where having a rough idea (sketch/guestimate) about the function/solution is really worthwhile as its easy to make algebraic slips with that amount of working.
Original post by mqb2766
It shouldnt be that much but it depends how youve written it etc and 6/7 marks at the start may be a bit underestimated but Id be surprised if it was more than 8.

Tbh, this is one of the things where having a rough idea (sketch/guestimate) about the function/solution is really worthwhile as its easy to make algebraic slips with that amount of working.

Yeah, that's very true, thank you for the advice. I will try and sketch out how I imagine the graph would look at the start and then I’ll see if my answer looks close for future questions.

Me saying 2 pages was a bit of an exaggeration, it was more like 1 and a bit.
Original post by mqb2766
Either polynomial division or the add zero trick (twice) so the numerator can be written as
[3x^2 - x + x - 1/3 + 1/3 + 2]/(3x-1) = (3x^2-x)/(3x-1) + (x-1/3)/(3x-1) + (7/3)/(3x-1)
Really the add zero trick (borrow remainders) is just doing polynomial division on the fly.

Its really just making the fraction proper, which is a good habit to get into.

Polynomial division is not something I would recommend ... many current students were not taught long division ever.

You can avoid using it ... it does not have the prominance it once had and is never asked for explicitly in a question,
Original post by Muttley79
Polynomial division is not something I would recommend ... many current students were not taught long division ever.

You can avoid using it ... it does not have the prominance it once had and is never asked for explicitly in a question,

Thank you for the contribution; I really don’t want to do polynomial division. I can’t even do it with numbers, let alone algebra 🤣.
Original post by mqb2766
Not sure, but its reasonably straightforward so solving a quadratic to get the stationary points and the second derivative simplfies nicely though its a few lines of algebra. Maybe ~6/7, but take that with a pinch of salt.

Fwiw, the problem is easier to understand by writing it as
y = x + 1/3 + 7/[3(3x-1)]
So shifted reciprocal with asymptotes y~x. So a min in Q1 and a max in Q3.

When you diff, you get y(x)=quadratic(3x1)2y'(x) = \dfrac{quadratic}{(3x-1)^2}, so the sign of y' will be the same as the sign of {quadratic}. So you don't really need to look at the 2nd derivative.
(edited 4 months ago)
Original post by lanky_giraffe
Thank you for the contribution; I really don’t want to do polynomial division. I can’t even do it with numbers, let alone algebra 🤣.

It can always be avoided ....
(edited 4 months ago)

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