document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1160149>Question 1160149</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the end behavior of f(x) = 2^ -x  + 5 as x goes to infinity? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #783340 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=783340\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If this is f(x)=2^(-x)+5<BR>\n" );
document.write( "as x goes to infinity, 2^(-x), or 1/2^x, approaches 0 as x gets large, and the function value is 5.<BR>\n" );
document.write( "Note: negative exponents do not change the overall sign.\r<BR>\n" );
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