Converting NFA to DFA

I've converted an NFA to a DFA. But even after checking over it a few times, it still doesn't feel right. I'm sure this is trivial, but I'd like someone to give me an idea where I went totally wrong on this.

NFA:

DFA:

Well, of course you have to merge the two empty states and there should be two transitions $(\{q_5\}, a, \emptyset)$ and $(\{q_5\}, b, \emptyset)$ if you want a complete automaton, but otherwise, it looks right and I agree with Subhayan: both automata accept $ab^+ \cup ab^+a$.

StackExchange Computer Science Q#14458, answer score: 3