x Turn on thread page Beta
 You are Here: Home >< Maths

# FP2 Rational Graphs watch

1. 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??
2. (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?
3. (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 ..
4. (Original post by Mathematicus65)
No clue ..
Here's a clue: can the quadratic be this shaped, roughly speaking?
5. (Original post by Mpagtches)
Here's a clue: can the quadratic be this shaped, roughly speaking?
I don't know how I would determine that???
6. 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
7. (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:
Show

Complete the square to get:

a(x+b)^2 +c

This is always >=0 if c>=o
8. (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:
Show

Complete the square to get:

a(x+b)^2 +c

This is always >=0 if c>=o
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)..
9. (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.
10. (Original post by EricPiphany)
This is the closest you got. You have to calculate a discriminant a second time.

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??
11. (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)

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.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: October 17, 2015
Today on TSR

### Complete university guide 2019 rankings

Find out the top ten here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams