document.write( "Question 1039476: Determine the equations of any horizontal and vertical asymptotes of
\n" ); document.write( "f(x)=x^2+8x+15/x^3+5x^2-x-5 <---The first equation is over the second.\r
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Algebra.Com's Answer #654218 by Boreal(15235) $\"\"$ $\"About$

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The horizontal asymptote requires looking the degree of the first/degree of the second, here 2/3. The horizontal asymptote is 0. As x gets very large, the leading term drives both the numerator and the denominator. This becomes x^2/x^2, or 1/x, and as x gets large, this polynomial goes to 0.
\n" ); document.write( "Vertical asymptotes occur where the denominator = 0
\n" ); document.write( "x^3+5x^2-x-5
\n" ); document.write( "1===+5===-1===-5, synthetic division
\n" ); document.write( "1===6=====5====0
\n" ); document.write( "x=-1 is a root, so 1 is a potential vertical asymptote.
\n" ); document.write( "The quadratic quotient is x^2+6x+5, and that factors into (x+5)(x+1). Set those equal to zero, and -1 and -5 are potential asymptotes. We need to be sure the numerator isn't 0 also at any of these points.
\n" ); document.write( "factor numerator and denominator
\n" ); document.write( "(x+5)(x+3)/(x-1)(x+5)(x+1)
\n" ); document.write( "The x+5 divide out, so the function is (x+3)(x-1)(x+1), and there are asymptotes at +/- 1 and a hole at x=-5\r
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