SOLUTION: y={{{(2-(8/x^2))/(x-(8/x^2))}}} For this equation I need to give the domain, asymptotes, and removable discontinuities. I get both of the asymptotes but I just have no idea what

Algebra ->  Rational-functions -> SOLUTION: y={{{(2-(8/x^2))/(x-(8/x^2))}}} For this equation I need to give the domain, asymptotes, and removable discontinuities. I get both of the asymptotes but I just have no idea what      Log On


   



Question 430397: y=%282-%288%2Fx%5E2%29%29%2F%28x-%288%2Fx%5E2%29%29
For this equation I need to give the domain, asymptotes, and removable discontinuities. I get both of the asymptotes but I just have no idea what the domain and removable discontinuities are. :( Please help me! Thank you and have a great day!!

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
One restriction on the value of x is that it should not be equal to 0. Another restriction is that x-8%2Fx%5E2 should not be equal to 0. This is the same as saying that %28x%5E3+-+8%29%2Fx%5E2 should not be equal to 0, x%5E3+-+8 is not equal to 0, or x is not equal to 2.
The DOMAIN is then all real numbers not equal to 0 or 2. This said,
.
As x goes to infinity, the expression goes to y = 0, because the degree of the numerator is less than the degree of the denominator. (Because the expression will have no real roots.)
Hence y = 0 is a HORIZONTAL ASYMPTOTE.
Since the denominator is a quadratic irreducible over the real numbers, the expression has NO VERTICAL ASYMPTOTES.
There are REMOVABLE DISCONTINUITIES at x = 0 and x = 2, because they cancel out in the process of simplification. They correspond to "holes" at the graph at the point (0,1) and (2, 2/3).
graph%28400%2C+400%2C+-10%2C10%2C+-10%2C10%2C+%282-8%2Fx%5E2%29%2F%28x-8%2Fx%5E2%29%29