KieRigby
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If anyone could help with this FP1 question, I would be very grateful!

A curve has an equation y=(x^2+2)/(x^2-4x)

a)Write down the asymptotes (Done this part)
b)i) Prove no part of the graph exists in the region -1 <= y <= 1/2 (part i'm having trouble with)

ii) Find the co-ordinates of the turning point on (may become easier after I know how to do part i)

c) Sketch the curve (I can do this too)

Any help would be great thanks
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Farhan.Hanif93
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(Original post by KieRigby)
A curve has an equation y=(x^2+2)/(x^2-4x)

b)i) Prove no part of the graph exists in the region -1 <= y <= 1/2 (part i'm having trouble with)
Imagine a graph of some curve y=f(x), say, and imagine a bunch of horizontal lines drawn on the same set of axes. Can you see why "no part of the curve exists in this region..." is the same thing as saying that "horizontal lines in this region do not intersect the curve"?

If yes, notice that horizontal lines have equation y=c, for any given real value of c. So you want to find the values of c, where f(x)=c has no solutions as this will give you the "horizontal lines" that do not meet the curve, which in turn tells you that the curve doesn't exists there.

Try rewriting f(x) = c as a quadratic equation and consider it's discriminant (recall that a quadratic equation has no solutions if the discriminant is negative).

ii) Find the co-ordinates of the turning point on (may become easier after I know how to do part i)
The horizontal line method works here too, in this case - notice that turning points are places where the horizontal lines have exactly one intersection with the curve. This is equivalent to considering the discriminant of the above quadratic and finding the values of c that make it 0.
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KieRigby
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Try rewriting Image as a quadratic equation and consider it's discriminant (recall that a quadratic equation has no solutions if the discriminant is negative).
Thanks for your help

I understand you're explanation, however the question asks prove no part of the graph exists in region -1 <= y <= 1/2. How can this be possible? The line cannot intersect with the curve and not intersect with it at all? Perhaps the question is wrong?

Either way, your explanation was perfect! Thank you
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Farhan.Hanif93
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(Original post by KieRigby)
Thanks for your help

I understand you're explanation, however the question asks prove no part of the graph exists in region -1 <= y <= 1/2. How can this be possible? The line cannot intersect with the curve and not intersect with it at all? Perhaps the question is wrong?
Take another read of the explanation, it may help. Think of it this way:

"No part of the graph existing in the region -1\leq y \leq 1/2" is the same thing as saying "any horizontal line y=c drawn (with -1\leq c \leq  1/2) will not intersect the curve". The natural thing to do next is to consider intersections of the curve with the arbitrary horizontal line y=c, and showing that c = \dfrac{x^2+2}{x^2-4x} has no solutions for the range -1\leq c \leq 1/2. As explained above, this is done by considering the discriminant.

it would be a lot easier to explain with a picture, but it's late and I'm tired. There's nothing wrong with the question.
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KieRigby
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(Original post by Farhan.Hanif93)
Take another read of the explanation, it may help. Think of it this way:

"No part of the graph existing in the region -1\leq y \leq 1/2" is the same thing as saying "any horizontal line y=c drawn (with -1\leq c \leq  1/2) will not intersect the curve". The natural thing to do next is to consider intersections of the curve with the arbitrary horizontal line y=c, and showing that c = \dfrac{x^2+2}{x^2-4x} has no solutions for the range -1\leq c \leq 1/2. As explained above, this is done by considering the discriminant.

it would be a lot easier to explain with a picture, but it's late and I'm tired. There's nothing wrong with the question.
Funny story. The question was wrong, my teacher made a mistake when producing the work sheet. So there is something wrong with the question.
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Notnek
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(Original post by KieRigby)
Funny story. The question was wrong, my teacher made a mistake when producing the work sheet. So there is something wrong with the question.
The inequality -1 <= y <= 1/2 in the question should have been -1 < y < 1/2.

Is that the mistake you are referring to?
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KieRigby
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(Original post by notnek)
The inequality -1 <= y <= 1/2 in the question should have been -1 < y < 1/2.

Is that the mistake you are referring to?
Yeah, it stressed me out so much before the exam and then to just be told 'it's a mistake' was very frustrating after being convinced so many times that there was 'nothing wrong with the question'.
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MathMeister
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What is a rational function... it is not on the FP1 edexcel syllabus. We only have complex numbers, iteration, parametric stuff, matrices, series and induction
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davros
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(Original post by MathMeister)
What is a rational function... it is not on the FP1 edexcel syllabus. We only have complex numbers, iteration, parametric stuff, matrices, series and induction
Basically just an expression that is one polynomial divided by another.

It's nothing special - you may even study them without knowing what they're called
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Notnek
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(Original post by MathMeister)
What is a rational function... it is not on the FP1 edexcel syllabus. We only have complex numbers, iteration, parametric stuff, matrices, series and induction
A rational function is a function of the form \displaystyle \frac{p(x)}{q(x)}, where p and q are polynomials.

It's not part of Edexcel, but for AQA FP1 you need to know about rational functions where p and q have degree 1 or 2 e.g. \frac{2x+3}{x^2+2x+1}. They study things like asymptotes, maxima, minima, intersection points etc.
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physicsmaths
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consider y(x^2-4x)=x^2+2
then use discriminant to find range of values of y hence solving it. Takes a few minutes.


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KieRigby
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(Original post by physicsmaths)
consider y(x^2-4x)=x^2+2
then use discriminant to find range of values of y hence solving it. Takes a few minutes.


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The question is incorrect though, it is unsolvable.
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physicsmaths
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(Original post by KieRigby)
The question is incorrect though, it is unsolvable.
just saying, thats how you would find the actual range of y


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physicsmaths
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1/2 n 1


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