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??