document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=65621>Question 65621</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I need help to find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4).\r<BR>\n" );
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document.write( "Does anyone grasp this? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #46264 by </B><A HREF=/tutors/aboutme.mpl?userid=venugopalramana><B>venugopalramana(3286)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=venugopalramana><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.geocities.com/venugopalramana/FREE_MATHS_ASSISTANCE.html valign=middle><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=46264\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I need help to find an exponential function of the form f(x)=ba^x+c with y-intercept 2, horizontal asymptote y=-2, that passes through the point P(1,4). <BR>\n" );
document.write( "Does anyone grasp this?<BR>\n" );
document.write( "Y=B*A^X+C<BR>\n" );
document.write( "IT GOES THROUGH (1,4)...SO<BR>\n" );
document.write( "4=B*A^1+C=AB+C............................1<BR>\n" );
document.write( "HORIZINTAL ASYMPTOTE IS Y=-2...THAT IS AS X TENDS TO + OR - INFINITY Y TENDS TO -2.<BR>\n" );
document.write( "CASE 1 ...AS X TENDS TO + INFINITY Y TENDS TO -2 <BR>\n" );
document.write( "-2=B*A^(INFINITY)+C......HENCE |A| MUST BE <1 AND THEN A^(INFINITY)=0<BR>\n" );
document.write( "-2=C.............................2<BR>\n" );
document.write( "Y INTERCEPT IS 2..THAT IS AT X=0,Y=2<BR>\n" );
document.write( "2=B*A^0+C=B+C.............3<BR>\n" );
document.write( "2=B-2<BR>\n" );
document.write( "B=4<BR>\n" );
document.write( "FROM 1...<BR>\n" );
document.write( "4=4A-2<BR>\n" );
document.write( "A=6/4..............NOT POSSIBLE AS IT IS CONTRARY TO OUR BASIC REQUIREMENT IN CASE 1 THAT |A|<1.SO <BR>\n" );
document.write( "CASE 2.....AS X TENDS TO - INFINTY, Y TENDS TO -2.<BR>\n" );
document.write( "PROCEEDING AS ABOVE<BR>\n" );
document.write( "-2=B*A^(-INFINITY)+C......<BR>\n" );
document.write( "-2=C.............................2<BR>\n" );
document.write( "Y INTERCEPT IS 2..THAT IS AT X=0,Y=2<BR>\n" );
document.write( "2=B*A^0+C=B+C.............3<BR>\n" );
document.write( "2=B-2<BR>\n" );
document.write( "B=4<BR>\n" );
document.write( "FROM 1...<BR>\n" );
document.write( "4=4A-2<BR>\n" );
document.write( "A=6/4 =1.5.......<BR>\n" );
document.write( "HENCE EQN. IS <BR>\n" );
document.write( "Y=4(1.5)^X-2<BR>\n" );
document.write( " </SPAN>\n" );
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