# A level further maths question

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f(x)=(ax^2+bx+c)/(x^2+dx+e) a is not 0

f(x) has no roots or vertical asymptotes, has a maximum at 2, minimum at 0.5, y intercept at (0,1.5)

show that (b-0.5d)^2=(b-2d)^2=2e

I am really not sure how to approach this question, I have tried making f(x)=k and find an equation for the discriminant but this has not given the right answer, how would I go about answering this question?

f(x) has no roots or vertical asymptotes, has a maximum at 2, minimum at 0.5, y intercept at (0,1.5)

show that (b-0.5d)^2=(b-2d)^2=2e

I am really not sure how to approach this question, I have tried making f(x)=k and find an equation for the discriminant but this has not given the right answer, how would I go about answering this question?

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#2

(Original post by

f(x)=(ax^2+bx+c)/(x^2+dx+e) a is not 0

f(x) has no roots or vertical asymptotes, has a maximum at 2, minimum at 0.5, y intercept at (0,1.5)

show that (b-0.5d)^2=(b-2d)^2=2e

I am really not sure how to approach this question, I have tried making f(x)=k and find an equation for the discriminant but this has not given the right answer, how would I go about answering this question?

**Emily~3695**)f(x)=(ax^2+bx+c)/(x^2+dx+e) a is not 0

f(x) has no roots or vertical asymptotes, has a maximum at 2, minimum at 0.5, y intercept at (0,1.5)

show that (b-0.5d)^2=(b-2d)^2=2e

I am really not sure how to approach this question, I have tried making f(x)=k and find an equation for the discriminant but this has not given the right answer, how would I go about answering this question?

You could try making the fraction proper, but id guess this may not be necessary?

Last edited by mqb2766; 1 month ago

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(Original post by

I guess you've got to at least differentiate and consider the stationary points at x=2 and 0.5?

You could try making the fraction proper, but id guess this may not be necessary?

**mqb2766**)I guess you've got to at least differentiate and consider the stationary points at x=2 and 0.5?

You could try making the fraction proper, but id guess this may not be necessary?

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#4

you just need to simplify it bring the bottom of the fraction up and (making the powers negative so mulitplying by (x^-2 dx^-1 e^-1) basic laws of indices,

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#5

(Original post by

This is year 12 Further maths so I don’t know how to differentiate fractions like this yet, in class we have found the stationary points by using the discriminant but I’m not sure how to apply that to this question

**Emily~3695**)This is year 12 Further maths so I don’t know how to differentiate fractions like this yet, in class we have found the stationary points by using the discriminant but I’m not sure how to apply that to this question

The intersection of that horizontal line with the function would have a single solution.

Cross multiply the quadratic fraction and max value and "solve" the resulting quadratic - single solution or discriminant.

Last edited by mqb2766; 1 month ago

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#6

**Emily~3695**)

f(x)=(ax^2+bx+c)/(x^2+dx+e) a is not 0

f(x) has no roots or vertical asymptotes, has a maximum at 2, minimum at 0.5, y intercept at (0,1.5)

show that (b-0.5d)^2=(b-2d)^2=2e

I am really not sure how to approach this question, I have tried making f(x)=k and find an equation for the discriminant but this has not given the right answer, how would I go about answering this question?

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#7

Perhaps completing the square on both the numerator and denominator could help.

EDIT. Don't do this

EDIT. Don't do this

Last edited by B_9710; 1 month ago

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#8

**Emily~3695**)

f(x)=(ax^2+bx+c)/(x^2+dx+e) a is not 0

f(x) has no roots or vertical asymptotes, has a maximum at 2, minimum at 0.5, y intercept at (0,1.5)

show that (b-0.5d)^2=(b-2d)^2=2e

I am really not sure how to approach this question, I have tried making f(x)=k and find an equation for the discriminant but this has not given the right answer, how would I go about answering this question?

When you say "maximum at 2, minimum at 0.5", I presume you mean for all x, and that it achieves its bounds.

Your idea of using the discriminant for the maximum and minimum values will give you two equations.

Using the information for the y-intercept will give you a relationship between "c" and "e" - allowing you to eliminate "c".

**Edit:**See following.

Putting these bits of information together will allow you to derive the desired equation - "(b-0.5d)^2=(b-2d)^2=2e"

**Note:**No differentiaton is required.

Last edited by ghostwalker; 1 month ago

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#9

(Original post by

I think people have been misinterpreting your question.

When you say "maximum at 2, minimum at 0.5", I presume you mean for all x, and that it achieves its bounds.

**ghostwalker**)I think people have been misinterpreting your question.

When you say "maximum at 2, minimum at 0.5", I presume you mean for all x, and that it achieves its bounds.

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#10

(Original post by

I was wondering why this question didn't seem to make much sense as phrased

**davros**)I was wondering why this question didn't seem to make much sense as phrased

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(Original post by

Can't recall seeing one like it on here before - doesn't mean there hasn't been one mind - and the "at 2" etc., does naturally lend itself to mean x=2, so somewhat misleading.

**ghostwalker**)Can't recall seeing one like it on here before - doesn't mean there hasn't been one mind - and the "at 2" etc., does naturally lend itself to mean x=2, so somewhat misleading.

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#12

(Original post by

Thank you for your help it makes more sense now the wording was definitely quite misleading which is why I think I was confused.

**Emily~3695**)Thank you for your help it makes more sense now the wording was definitely quite misleading which is why I think I was confused.

I realised this morning that my solution is flawed.

The desired equation can be obtained if we have a=1, but just because the function's range is [0.5,2] (viz. bounded), this does not imply that a=1.

Last edited by ghostwalker; 1 month ago

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#13

**Emily~3695**)

Thank you for your help it makes more sense now the wording was definitely quite misleading which is why I think I was confused.

Is it taken from a book? Can you upload an image?

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#14

**Emily~3695**)

Thank you for your help it makes more sense now the wording was definitely quite misleading which is why I think I was confused.

If you have a=3/2, b=7/3, c=3/2, d=2/3, e=1 (for example), then this set of values meets all the given criteria, but: (b-0.5d)^2=4e and (b-2d)^2=e

As davros asked, an image of the original question would be useful, if not a necessity, now.

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#15

(Original post by

Further checking shows that the desired equation is not necessarily true.

If you have a=3/2, b=7/3, c=3/2, d=2/3, e=1 (for example), then this set of values meets all the given criteria, but: (b-0.5d)^2=4e and (b-2d)^2=e

As davros asked, an image of the original question would be useful, if not a necessity, now.

**ghostwalker**)Further checking shows that the desired equation is not necessarily true.

If you have a=3/2, b=7/3, c=3/2, d=2/3, e=1 (for example), then this set of values meets all the given criteria, but: (b-0.5d)^2=4e and (b-2d)^2=e

As davros asked, an image of the original question would be useful, if not a necessity, now.

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**ghostwalker**)

Further checking shows that the desired equation is not necessarily true.

If you have a=3/2, b=7/3, c=3/2, d=2/3, e=1 (for example), then this set of values meets all the given criteria, but: (b-0.5d)^2=4e and (b-2d)^2=e

As davros asked, an image of the original question would be useful, if not a necessity, now.

This is the original question it’s pretty much exactly what I put

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#17

Perhaps post an image of the question, next time.

Last edited by ghostwalker; 1 month ago

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(Original post by

Your missed one vital fact from your post "a horizontal asymptote at y=1". This implies a=1 and the rest follows.

Perhaps post an image of the question, next time.

**ghostwalker**)Your missed one vital fact from your post "a horizontal asymptote at y=1". This implies a=1 and the rest follows.

Perhaps post an image of the question, next time.

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#19

(Original post by

Ah ok I didn’t even realise that when I was answering the question, thanks for your help

**Emily~3695**)Ah ok I didn’t even realise that when I was answering the question, thanks for your help

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(Original post by

Without it you cannot deduce that a=1. I trust you understand why it means a=1: Divide numerator and denominator by x^2, and consider the limit as x goes to infinity.

**ghostwalker**)Without it you cannot deduce that a=1. I trust you understand why it means a=1: Divide numerator and denominator by x^2, and consider the limit as x goes to infinity.

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