MIAMI GARDENS, Fla. — Miami Dolphins quarterback Tua Tagovailoa has been named the AFC's offensive player of the week for his season-opening performance against the Los Angeles Chargers.
The NFL announced Wednesday that Tagovailoa won his second AFC offensive player of the week honors after completing 28-of-45 passes for 466 yards, three touchdowns and one interception in a 36-34 road win.
His 466 passing yards were the fifth-most in a game in team history and fourth-most by any player in an opening game in NFL history.
Tagovailoa's final pass to Tyreek Hill with 1:45 left proved to be the game-winner.
The recognition could have easily gone to Hill, whose 215 receiving yards were the third-most in an opening game in NFL history.
Tagovailoa also won the award for his performance in a 42-38 win at the Baltimore Ravens last season.
Despite the accolades, Tagovailoa told reporters he's not letting it go to his head.
"You're either going to win the Super Bowl after you win your first game or you're the worst team ever and everything that comes with that," Tagovailoa said. "So, I would say for myself, that's one of the reasons I don't pay too much attention to the outside noise."
The Dolphins will travel to New England to face the Patriots this Sunday night in a game that will be televised on WPTV.
"We got 16 more games to play against 16 really, really good teams," Tagovailoa said. "And it starts with New England this week, and that's our main focus."