document.write( "Question 637132: Graph the function: g(x) = 2/3 log_2(x-3) + 2 \r
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document.write( "Plot two points and the asymptotes (if any) of the graph of the function. \r
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document.write( "Help please! Thank you very much! Can anyone please walk me through the steps to achieve this answer?! \n" );
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Algebra.Com's Answer #401553 by lwsshak3(11628) ![]() You can put this solution on YOUR website! \n" ); document.write( "Graph the function: g(x) = 2/3 log_2(x-3) + 2 \n" ); document.write( "Plot two points and the asymptotes (if any) of the graph of the function. \n" ); document.write( "I don't have the means to graph it for you, but I can describe how to do it. \n" ); document.write( "** \n" ); document.write( "start with the basic log curve: y=logx (base 10) \n" ); document.write( "like all logx curves, regardless of base, they have the y-axis is an asymptote, and x-intercept=1. \n" ); document.write( "So this basic curve starts from (0,-∞), crosses the x-axis at 1 and gradually increases to ∞. \n" ); document.write( ".. \n" ); document.write( "For given log curve: g(x) = 2/3 log_2(x-3) + 2 \n" ); document.write( "Since it is a lower base(2), the curve is above the basic curve when <1, and below the basic curve when x>1.There is no change to the y-axis asymptote or x-intercept at 1. \n" ); document.write( "The 2/3 multiplier stretches the curve, making it more negative below the x-axis and more positive above the x-axis. \n" ); document.write( "(x-3) moves the asymptote and x-intercept 3 units to the right at x=3 and x=4 respectively \n" ); document.write( "Finally, +2 bumps the entire curve 2 units up \n" ); document.write( ".. \n" ); document.write( "note:If you have a graphing calculator, it could greatly help you follow the steps above \n" ); document.write( " |