document.write( "Question 62456: Find the vertical and horizontal asymptotes of Y=(5x^2-8x+1)/(2x^2-5x-12)\r
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Algebra.Com's Answer #43205 by uma(370)\"\" \"About 
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Given y = (5x^2 - 8x + 1)/ (2x^2 - 5x - 12)\r
\n" ); document.write( "\n" ); document.write( "To find the vertical asymptotes , equate the denominator to zero.\r
\n" ); document.write( "\n" ); document.write( "==> 2x^2 - 5x - 12 = 0
\n" ); document.write( "==> 2x^2 - 8x + 3x - 12 = 0
\n" ); document.write( "==> 2x(x-4) + 3(x-4) = 0
\n" ); document.write( "==> (x-4)(2x + 3) = 0
\n" ); document.write( "==> x - 4 = 0 or 2x + 3 = 0
\n" ); document.write( "==> x = 4 or x = - 3/2\r
\n" ); document.write( "\n" ); document.write( "x = 4 and x = -3/2 are the two vertical asymptotes.\r
\n" ); document.write( "\n" ); document.write( "To find the horizontal asymptotes, \r
\n" ); document.write( "\n" ); document.write( "Since the degrees of the numerator and the denominator are the same (each being 2), then this rational has a non-zero (non-x-axis) horizontal asymptote, not a slant asymptote, and the horizontal asymptote is found by dividing the leading terms:
\n" ); document.write( "= 5/2\r
\n" ); document.write( "\n" ); document.write( "Thus y = 5/2 is the horizontal asymptote and x = 4 and x = -3/2 are the vertical asymptotes.\r
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