(Homework) Converting an NFA to a DFA with an epsilon transition, but no transition for some a in the alphabet

Question

If my alphabet is {a,b} and my nfa has the following transitions:

State    |      a        b        epsilon        
--------------------------------------------
q0              q1      null        q1
q1              q2       q1         none
q2              q2       q1         none

Is this table wrong? should delta(q0, b) = q1 because q0 can move on epsilon to state q1?

Answer 1

As olydis stated in the comments section:

"If from q1 (or any state you reach from q1 via epsilon-transition) you reach q1 when b is read, then delta(q0, b) = q1"