document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=915987>Question 915987</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Please explain why I am getting this wrong. The graph of an exponential function given by f (x) = A(b^x)+c is shown below:\r<BR>\n" );
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document.write( "http://i.imgur.com/GAywHDn.png </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #555878 by </B><A HREF=/tutors/aboutme.mpl?userid=josh_jordan><B>josh_jordan(263)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=josh_jordan><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=josh_jordan><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=555878\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The points listed are at (-2,12) and (0,4) and the asymptote is y = 3.  You are also given the form of A(b^x) + c.  If you plug in the first point you'd get A(b^-2) + 3 = 12, which would give you A(b^-2) = 9.  Using your second point, you'd get A(b^0) + 3 = 4, which would give you A + 3 = 4, leaving you with A = 1.  So, you can now substitute 1 for your first function: 1(b^-2) = 9.  Solving for b would give you 1/3.  You can now substitute 1/3 for b, and 3 for c in A(b^x) + c:\r<BR>\n" );
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document.write( "f(X) = (1/3)^x + 3 </SPAN>\n" );
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