document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1124504>Question 1124504</A>: <I><SPAN STYLE=\"word-break: break-all;\"> At the beginning of 2010 Monique's car was worth $10,000, but the value of her car decreases exponentially. She notices that the value of her car decreases by 20% every 3 years. \r<BR>\n" );
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document.write( "a. What is the 3-year growth factor for the value of Monique's car?\r<BR>\n" );
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document.write( "b. What is the 1-year growth factor for the value of Monique's car?\r<BR>\n" );
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document.write( "c. Write a function f that determines the value of Monique's car (in dollars) in terms of the number of years t since the beginning of 2010. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #740860 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=740860\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> f = p * (1 + r) ^ n\r<BR>\n" );
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document.write( "f is the future value<BR>\n" );
document.write( "p is the present value<BR>\n" );
document.write( "r is the interest rate per time period.<BR>\n" );
document.write( "n is the number of time periods.\r<BR>\n" );
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document.write( "that's the general formula for the future value of something when compound interest is concerned.\r<BR>\n" );
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document.write( "compound interest is a form of exponential growth or decay.\r<BR>\n" );
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document.write( "if the car cost 10,000 and decreases 20% in value every 3 years, then:\r<BR>\n" );
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document.write( "you can solve as follows:\r<BR>\n" );
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document.write( "time period is every 3 years.<BR>\n" );
document.write( "interest rate is interest rate per every 3 years.\r<BR>\n" );
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document.write( "f = p * (1 + r) ^ n becomes f = 10,000 * (1 - .20) ^ n\r<BR>\n" );
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document.write( "simplify this to get f = 10,000 * .80 ^ n\r<BR>\n" );
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document.write( "after 3 years, the value of the car is 10,000 * .8 ^ 1 = 8000.<BR>\n" );
document.write( "after 6 years, the value of the car is 10,000 * .8 ^ 2 = 6400.<BR>\n" );
document.write( "after 9 years, the value of the car is 10,000 * .8 ^ 3 = 5121.536\r<BR>\n" );
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document.write( "the average annual growth rate would be calculated as follows:\r<BR>\n" );
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document.write( "find the future value after one 3 year period.\r<BR>\n" );
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document.write( "assuming the present value is one, the formula would become.\r<BR>\n" );
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document.write( "f = 1 * (1 - .2) ^ 1 = 1 * .8 = .8\r<BR>\n" );
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document.write( "so, the present value becomes .8 after 3 years.\r<BR>\n" );
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document.write( "to find the annual growth rate, you would do the following.\r<BR>\n" );
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document.write( "f = p * (1 + r) ^ n becomes .8 = 1 * (1 + r) ^ 3\r<BR>\n" );
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document.write( "divide both sides of the equation by 1 to get .8 = (1 + r) ^ 3\r<BR>\n" );
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document.write( "take the cube root of both sides of the eqution to get .8 ^ (1/3) = 1 + r\r<BR>\n" );
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document.write( "subtract 1 from both sides of the equation to get .8 ^ (1/3) - 1 = r\r<BR>\n" );
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document.write( "solve for r to get r = -.0716822333.\r<BR>\n" );
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document.write( "that's the annual growth rate.\r<BR>\n" );
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document.write( "to confirm, use this rate in the original equation of .8 = 1 * (1 + r) ^ 3\r<BR>\n" );
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document.write( "you will get .8 = 1 * (1 - .0716822333) ^ 3\r<BR>\n" );
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document.write( "simplify to get .8 = .8\r<BR>\n" );
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document.write( "if you graph this equation, it will look like this.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" src = \"http://theo.x10hosting.com/2018/092801.jpg\" alt = \"$$$\" >\r<BR>\n" );
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document.write( "from the graph, you can see that the loss in value is 20% every 3 years.\r<BR>\n" );
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document.write( "in year 0, the value is 10,000<BR>\n" );
document.write( "in year 3, the value is 8000<BR>\n" );
document.write( "in year 6, the value is 6400<BR>\n" );
document.write( "in year 9, the value is 5120\r<BR>\n" );
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document.write( "the every 3 year growth rate is (1 - .2) = .8\r<BR>\n" );
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document.write( "the every year growth rate is (1 - .0716822333) = .9283177667\r<BR>\n" );
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document.write( " </SPAN>\n" );
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