document.write( "Question 430397: y=\"%282-%288%2Fx%5E2%29%29%2F%28x-%288%2Fx%5E2%29%29\"\r
\n" ); document.write( "\n" ); document.write( "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!!
\n" ); document.write( "

Algebra.Com's Answer #298948 by robertb(5830)\"\" \"About 
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. \r
\n" ); document.write( "\n" ); document.write( "The DOMAIN is then all real numbers not equal to 0 or 2. This said, \r
\n" ); document.write( "\n" ); document.write( ".\r
\n" ); document.write( "\n" ); document.write( "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.) \r
\n" ); document.write( "\n" ); document.write( "Hence y = 0 is a HORIZONTAL ASYMPTOTE. \r
\n" ); document.write( "\n" ); document.write( "Since the denominator is a quadratic irreducible over the real numbers, the expression has NO VERTICAL ASYMPTOTES.\r
\n" ); document.write( "\n" ); document.write( "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).
\n" ); document.write( "\"graph%28400%2C+400%2C+-10%2C10%2C+-10%2C10%2C+%282-8%2Fx%5E2%29%2F%28x-8%2Fx%5E2%29%29\"
\n" ); document.write( "
\n" );