is this right?
27. h(x)= |x-3| ; x=0,1,2,3,4,5,6
h(x); |0-3| =3
h(x); |1-3|=2
h(x); |2-3|=1
h(x); |4-3|=1
h(x); |5-3|=2
h(x); |6-3|=3
correct, but you left out h(3) = 0
rather than h(x);|1-3| it is more customary to write
h(1) = |1-3| = 2
since h(x) = |x-3|
and you just replace all the x's with 1's