Question 1117647
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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....<br>
An absolute value is always 0 or positive.  So the minimum value of {{{abs(x-2)}}} is 0; and that value is obtained when {{{x-2=0}}} -->  {{{x = 2}}}.<br>
So in your function, {{{3+abs(x-2)}}}, the minimum value is when x=2; and at x=2 the function value is {{{3+abs(2-2) = 3+abs(0) = 3+0 = 3}}}.<br>
So the turning point ("vertex") of the graph is at (2,3).<br>
For values of x less than 2, x-2 is negative, so {{{abs(x-2) = 2-x}}}, and the equation is {{{y = 3+2-x}}} or {{{y = -x+5}}}; that line has a slope of -1.<br>
For values of x greater than 2, x-2 is positive, so {{{abs(x-2) = x-2}}}, and the equation is {{{y = 3+x-2}}} or {{{y = x+1}}}; that line has a slope of +1.<br>
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.<br>
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.