document.write( "Question 1063800: Graph ofy=|x+1|+|2x-1| \n" ); document.write( "

Algebra.Com's Answer #678943 by Edwin McCravy(20060) $\"\"$ $\"About$

You can put this solution on YOUR website!
Graph of y=|x+1|+|2x-1|
\n" ); document.write( "

\r\n" );
document.write( "Here's what you do, step by step:\r\n" );
document.write( "\r\n" );
document.write( "Set the expressions inside the absolute value bars equal 0,\r\n" );
document.write( "x+1 = 0        2x-1 = 0\r\n" );
document.write( "  x = -1         2x = 1\r\n" );
document.write( "                  x =   \r\n" );
document.write( "These define three intervals: x < -1, -1 < x < 1/2, and x > 1/2\r\n" );
document.write( "\r\n" );
document.write( "For x < -1, |x+1| = -(x+1) and |2x-1| = -(2x-1)\r\n" );
document.write( "so y = -(x+1)-(2x-1) = -x-1-2x+1 = -3x\r\n" );
document.write( "\r\n" );
document.write( "For -1 < x < 1/2, |x+1| = x+1 and |2x-1| = -(2x-1)\r\n" );
document.write( "so y = x+1-(2x-1) = x+1-2x+1 = -x+2\r\n" );
document.write( "\r\n" );
document.write( "For x > 1/2, |x+1| = x+1 and |2x-1| = 2x-1\r\n" );
document.write( "so y = x+1+2x-1 = 3x-1-2x+1 = 3x\r\n" );
document.write( "\r\n" );
document.write( "So the graph is this piecewise graph:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "So draw the three lines:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Then erase the parts of the three lines that extend \r\n" );
document.write( "past their part of the domain:\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Edwin

\n" ); document.write( "

\n" );