document.write( "Question 1117647: sketch the graph of y=3+|x-2|clearly stating your arguments. what is the range of the function? \n" ); document.write( "

Algebra.Com's Answer #732808 by greenestamps(13203) $\"\"$ $\"About$

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\n" ); document.write( "The response from the other tutor was rather brief, and may not be much help to you if you are having trouble understanding the problem....

\n" ); document.write( "An absolute value is always 0 or positive. So the minimum value of $\"abs%28x-2%29\"$ is 0; and that value is obtained when $\"x-2=0\"$ --> $\"x+=+2\"$ .

\n" ); document.write( "So in your function, $\"3%2Babs%28x-2%29\"$ , the minimum value is when x=2; and at x=2 the function value is $\"3%2Babs%282-2%29+=+3%2Babs%280%29+=+3%2B0+=+3\"$ .

\n" ); document.write( "So the turning point (\"vertex\") of the graph is at (2,3).

\n" ); document.write( "For values of x less than 2, x-2 is negative, so $\"abs%28x-2%29+=+2-x\"$ , and the equation is $\"y+=+3%2B2-x\"$ or $\"y+=+-x%2B5\"$ ; that line has a slope of -1.

\n" ); document.write( "For values of x greater than 2, x-2 is positive, so $\"abs%28x-2%29+=+x-2\"$ , and the equation is $\"y+=+3%2Bx-2\"$ or $\"y+=+x%2B1\"$ ; that line has a slope of +1.

\n" ); document.write( "So the minimum value of the function is at (2,3) and from that point the graph to the right has a slope of +1 and the graph to the left has a slope of -1.

\n" ); document.write( "Clearly the lower end of the range of the function is y=3. And there is nothing to keep the value of y getting larger \"forever\"; so the range of the function is from 3 to infinity. \n" ); document.write( "

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