The expression under the square root is \(ax + b\), which is \(-x + 9\).
Since the square root of a negative number is not defined in the set of real numbers, we require \(-x + 9 \geq 0\).
To find the domain, we solve the inequality \(-x + 9 \geq 0\). Given \(a = -1\), we solve $x <= \frac{-9}{-1} = 9$.
Based on the solution to the inequality, the domain in interval notation is $(-∞, 9]$.
The domain of the function \(f(x) = \sqrt{-x + 9} + 2\) is $(-∞, 9]$.
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