Australia’s Alex de Minaur defeated hometown hero Jenson Brooksby in the final of the Atlanta Open on Sunday, winning in straight sets 6-3 6-3.
It is the second time De Minaur has won the Atlanta Open – also getting the job done in 2019 when he met American Taylor Fritz in the final.
This time around, De Minaur feasted on Brooksby’s second serve, winning 81 per cent (13-of-16) of those opportunities, compared to just 29 per cent when the American was able to land his first serve.
It was even more pronounced in the second set as the fast-finishing De Minaur continued to strangle the life out of Brooksby, as he was only able to win the point twice from 12 second serves.
Despite his relatively comfortable victory, it was De Minaur who had to face adversity first as he so often does before fighting back, facing the first two break points of the game, but he saved both before capitalising on his only break point opportunity in the opening set.
The Aussie threatened to run away with the match when he broke again in the opening game of the second frame, but Brooksby instantly snatched back a break of his own.
Ultimately, De Minaur was just better both on serve and in his return game, winning 59 per cent of his points on serve compared to 52 per cent for Brooksby.
He also saved four of Brooksby’s five break point opportunities, while winning all four of his own.