document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=984101>Question 984101</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I am trying to work with an oblique asymptote and I am at a delimma.  The problem is -x^3+2x^2+2x-5 /x^2+x-12.   I have gotten to this for the denominator but I am stumped.   x^2+x-12  as (x+4)(x-3).   Can you help me? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #604900 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=604900\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> When you do long division of the denominator into the numerator, you get (-x+3) in the quotient with remainder of -12x +31\r<BR>\n" );
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document.write( "If you multiply the denominator, unfactored, by (-x+3) and then add -12x +31, you will get the original function back.\r<BR>\n" );
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document.write( "The slant asymptote is -x+3<BR>\n" );
document.write( "There are vertical asymptotes at x=-4 and x+3.\r<BR>\n" );
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document.write( "Note, in order to show the minus 1x slope, you need larger numbers on the calculator for x, like at least 50 or preferably 100.\r<BR>\n" );
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