document.write( "Question 59578: -2(x+1)(x+4)^2
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Algebra.Com's Answer #40843 by Edwin McCravy(20056)\"\" \"About 
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document.write( " -2(x+1)(x+4)²\r\n" );
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document.write( "  (x+2)²(x-3)\r\n" );
document.write( "What are the vertical and horizontal asymptotes.\r\n" );
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document.write( "If the numerator and denominator are relatively\r\n" );
document.write( "prime polynomials, i.e., have no common factor\r\n" );
document.write( "other than 1, then:\r\n" );
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document.write( "1. The vertical asymptotes are found by setting the\r\n" );
document.write( "denominator = 0.  For each value obtatined a\r\n" );
document.write( "vertical asymptote will be obtained whose equation\r\n" );
document.write( "is x = that value.\r\n" );
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document.write( "denominator = (x+2)²(x-3) = 0\r\n" );
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document.write( "have solutions x = -2, x = 3\r\n" );
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document.write( "So these are the equations of the two horizontal\r\n" );
document.write( "asymptotes:\r\n" );
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document.write( "x = -2 and x = 3\r\n" );
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document.write( "2. There will be a horizontal asymptote if and only\r\n" );
document.write( "if the degree of the numerator is not larger than\r\n" );
document.write( "the degree of the denominator.\r\n" );
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document.write( "A. If the degree of the numerator is less than the\r\n" );
document.write( "   degree of the denominator, the horizontal asymptote\r\n" );
document.write( "   is always the x-axis, which has equation y = 0\r\n" );
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document.write( "B. If the degrees of the numerator and denominator are\r\n" );
document.write( "   equal, then the horizontal asymptote is y = a/b,\r\n" );
document.write( "   where a and b are the leading coefficients of the\r\n" );
document.write( "   numerator and denominator respectively, when the\r\n" );
document.write( "   numerator and denominator are multiplied out.\r\n" );
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document.write( "Now in this case, if we were to multiply the numerator\r\n" );
document.write( "and denominator out, we would find that both would\r\n" );
document.write( "have degree 3.  So the rule B holds.\r\n" );
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document.write( "We don't need to multiply out the top and bottom,\r\n" );
document.write( "since all we need is the leading term of each.\r\n" );
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document.write( "We can see that if we were to multiply the numerator\r\n" );
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document.write( "-2(x+1)(x+4)²\r\n" );
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document.write( "out, that the leading term would be -2x³, and the\r\n" );
document.write( "leading coefficient would be a = -2\r\n" );
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document.write( "We can see that if we were to multiply the denominator\r\n" );
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document.write( "(x+2)²(x-3)\r\n" );
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document.write( "out, that the leading term would be x³, and the\r\n" );
document.write( "leading coefficient would be b = 1.\r\n" );
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document.write( "Therefore the horizontal asymptote has equation\r\n" );
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document.write( "y = -2/1 or y = -2\r\n" );
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document.write( "The green and blue vertical lines are the vertical\r\n" );
document.write( "asymptotes and the red horizontal line is the \r\n" );
document.write( "horizontal asymptote.  If we were to extend the graph\r\n" );
document.write( "farther to the left and to the right, the blue curve\r\n" );
document.write( "would get closer to the horizontal asymptote, and if\r\n" );
document.write( "we extended the graph farther upward and downward,\r\n" );
document.write( "the blue curve would get closer and closer to the\r\n" );
document.write( "two vertical asymptotes.\r\n" );
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document.write( "Edwin
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