document.write( "Question 389522: For the function f(x)=(x^4-625)/((x^2-25)(x+5)(x^2+25))\r
\n" ); document.write( "\n" ); document.write( "a) How do you find the holes in a function?\r
\n" ); document.write( "\n" ); document.write( "b) Identify any vertical asymptotes\r
\n" ); document.write( "\n" ); document.write( "c) Identify any holes
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Algebra.Com's Answer #276018 by robertb(5830)\"\" \"About 
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\n" ); document.write( "a) If there are any cancellations, then the roots of the cancelled terms are candidates for the occurence of holes. The domain of f is all real numbers except -5 and 5. So it's POSSIBLE for the function to have holes there. \r
\n" ); document.write( "\n" ); document.write( "b) The function becomes, after cancellation, . So there is a vertical asymptote at x = -5.\r
\n" ); document.write( "\n" ); document.write( "c) \"x%5E2+-+25\" was cancelled from the top and the bottom. Its roots are
\n" ); document.write( "-5 and 5. Since there is a vertical asymptote at x = -5, there is a \"hole\" at the graph only at x = 5. The expression \"x%5E2+%2B+25\"is never zero for real values of x. (The roots are imaginary.)
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