# Algebra2

posted by .

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

• Algebra2 -

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

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