(a) This equation has a quadratic form so its acceleration is constant. Comparing this equation with standard constant acceleration kinematic equation, \(x=\frac{1}{2} a t^{2}+v_{0} t+x_{0}\), we will find its magnitude as

\[

\frac{1}{2} a=-1 \Rightarrow a=-2 \mathrm{~m} / \mathrm{s}^{2}

\]

So this choice is incorrect.

(b) "Object's velocity at the initial time" means its initial velocity. Comparing the two equations below reveals that initial velocity is \(v_{0}=+6 \mathrm{~m} / \mathrm{s}\).

\[

\begin{gathered}

x=\frac{1}{2} a t^{2}+v_{0} t+x_{0} \\

x=-t^{2}+6 t-9

\end{gathered}

\]

The positive sign indicates that the initial velocity is toward the positive \(x\)-axis. So, this choice is also incorrect.

(c) By setting \(t=0\) in the position-time equation, its initial position is obtained. So, \(x_{0}=\) \(-9 \mathrm{~m}\). Negative indicates that the object is on the negative side of the \(x\)-axis initially. Thus, this choice is correct.