Tee Higgins Claims AFC's Weekly Offensive Honor for the First Time

On November 4, 2026, Cincinnati Bengals receiver Tee Higgins was named the AFC's offensive standout of the week for the first time after a standout performance against the Tennessee Titans. Higgins caught 10 passes for 155 yards and two touchdowns in the Bengals’ 43-36 victory over the weekend.

The award extends a strong season for the 27-year-old. Higgins has 726 receiving yards, second most in the league, and four touchdowns. Among qualified receivers, he ranks third for yards gained per route, with only Jaxson Smith-Njigba and Zay Flowers ahead.

Coach Zac Taylor said Cincinnati kept throwing to Higgins and Ja’Marr Chase as the offense gained momentum in the high-scoring game. He noted that short passes became bigger gains as the receivers found room to make plays.

Higgins said the performance reflected the team’s emphasis throughout the season and training camp on continuing to fight. He also called the win a team effort, despite the game not unfolding as Cincinnati had hoped.

Higgins is the fourth Bengal to receive a weekly league honor this season. Kicker Evan McPherson was recognized in Week 1, while quarterback Joe Burrow and special-teams player Ryan Rehkow earned awards in Week 7. Cincinnati next faces Atlanta in a game scheduled for Spain.