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FP2 Rational Graphs

Prove that if y=(x-k)/(x^2-4x-k) can take all values as x varies then 0<k<5.

Right so I rearranged to form the equation:

(y)x^2-(4y+1)x+(k-ky)=0
Then we know for there to be real points on the graph then the discriminant of that equation must be greater than or equal to 0.
So finding the discriminant gives (-4y-1)^2-4y(k-ky) >0(or equal to)
Where do I go from here??
Original post by Mathematicus65
Prove that if y=(x-k)/(x^2-4x-k) can take all values as x varies then 0<k<5.

Right so I rearranged to form the equation:

(y)x^2-(4y+1)x+(k-ky)=0
Then we know for there to be real points on the graph then the discriminant of that equation must be greater than or equal to 0.
So finding the discriminant gives (-4y-1)^2-4y(k-ky) >0(or equal to)
Where do I go from here??


You need this to be true for all y. So, any thoughts on how you can show a quadratic is always greater than or equal to zero?
Original post by ghostwalker
You need this to be true for all y. So, any thoughts on how you can show a quadratic is always greater than or equal to zero?


No clue ..
Original post by Mathematicus65
No clue ..


Here's a clue: can the quadratic be this shaped, roughly speaking? \cap
Original post by Mpagtches
Here's a clue: can the quadratic be this shaped, roughly speaking? \cap


I don't know how I would determine that???
I put the information into the discriminant to get (16+4k)y^2 + (8-4k)y+1>0

But don't see how this helps me
Original post by Mathematicus65
I put the information into the discriminant to get (16+4k)y^2 + (8-4k)y+1>0

But don't see how this helps me


This is your new quadratic, and yes it requires quite a bit of algebra to work out.

Forget about it for now. In general, how do you show a quadratic is always greater than or equal to zero?

One method in spoiler

Spoiler

Original post by ghostwalker
This is your new quadratic, and yes it requires quite a bit of algebra to work out.

Forget about it for now. In general, how do you show a quadratic is always greater than or equal to zero?

One method in spoilers

Spoiler



Well if I complete the square for the original quadratic in the denominator - (x^2-4x-k) then you get (x-2)^2 - (k+4)..
Original post by Mathematicus65
I put the information into the discriminant to get (16+4k)y^2 + (8-4k)y+1>0

But don't see how this helps me


This is the closest you got. You have to calculate a discriminant a second time.
Original post by EricPiphany
This is the closest you got. You have to calculate a discriminant a second time.


image.jpg
Okay so ive managed to work it out now but my solution is still incorrect due to the Incorrect signs. I have found that k needs to be <= 0 and >=5 not 0<k<5 so what have I done wrong??
Original post by Mathematicus65
Well if I complete the square for the original quadratic in the denominator - (x^2-4x-k) then you get (x-2)^2 - (k+4)..


Not in the original equation, in your new one, involving y and k.

Original post by Mathematicus65
image.jpg
Okay so ive managed to work it out now but my solution is still incorrect due to the Incorrect signs. I have found that k needs to be <= 0 and >=5 not 0<k<5 so what have I done wrong??


You don't want the discriminant to be >=0. This implies that there are roots, and so your inequality, (16+4k)y^2 + (8-4k)y+1>=0 , won't be true for all y.

You require the discriminant to be <=0, which is why you have the signs the wrong way round.

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