Set $f(x) = 0$ and solve:

$-9(x + 1)^2 + 81 = 0$

$-9(x + 1)^2 = -81$

$(x + 1)^2 = 9$

$x + 1 = \pm 3$

$x = -1 + 3 = 2 \quad \text{or} \quad x = -1 - 3 = -4$

Since $a > 0$, we have $a = 2$.