document.write( "Question 459333: #5 on homework.\r
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document.write( "sketch graph of function f(x)=log2(x+1) by transforming y=log2x\r
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document.write( "Find domain and vertical asympote \n" );
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Algebra.Com's Answer #316076 by lwsshak3(11628)![]() ![]() ![]() You can put this solution on YOUR website! sketch graph of function f(x)=log2(x+1) by transforming y=log2x \n" ); document.write( ".. \n" ); document.write( "I don't have the means to draw a graph for you, but I can show you how to do it. \n" ); document.write( ".. \n" ); document.write( "Start with the basic log function: y=logx \n" ); document.write( "The first point on the graph to note is (1,0), that is, when x=1, y=0. \n" ); document.write( "When x<1, approaching asymptote, y-axis or x=0, y becomes increasingly negative. \n" ); document.write( "When x>1, y becomes increasingly positive but at a slower rate. \n" ); document.write( ".. \n" ); document.write( "y=log2x \n" ); document.write( "This moves the (1,0) point to the (1/2,0) point. \n" ); document.write( "The asymptote, x=0, does not change, but the point at which y goes from negative to positive is at x=1/2 instead at x=1 as in the basic function. Both curves have similar shapes with the log2x curve sitting above the logx curve \n" ); document.write( ".. \n" ); document.write( "log2(x+1) \n" ); document.write( "This just moves previous curve one unit to the left, placing the asymptote at x=-1 instead of x=0, and moving the (1/2,0) point to the (-1/2,0) point. Curve also sits above previous curve. \n" ); document.write( ".. \n" ); document.write( "Hope this helps you understand the log curves. \n" ); document.write( " |