document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=750896>Question 750896</A>: <I><SPAN STYLE=\"word-break: break-all;\"> An investment was worth $2500 in 1995. By 2000 it had grown to $3500. Let y be the value of the investment in year x, where x = 0 represents 1995. Assuming that the value of the investment increases in a linear fashion, write an equation that models the value of the investment in year x. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #456868 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=456868\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> An investment was worth $2500 in 1995. By 2000 it had grown to $3500. Let y be the value of the investment in year x, where x = 0 represents 1995. Assuming that the value of the investment increases in a linear fashion, write an equation that models the value of the investment in year x.<BR>\n" );
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document.write( "You have 2 points: (0,2500) and (5,3500)<BR>\n" );
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document.write( "slope = (3500-2500)/(5-0) = 1000/5 = 200<BR>\n" );
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document.write( "intercept = f(0) = 2500<BR>\n" );
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document.write( "Equation:<BR>\n" );
document.write( "f(x) = 200x + 2500<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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document.write( " </SPAN>\n" );
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