Solution: We recall that for any random variable $X$, the ejection of $X$ is given by \begin{align*} E[X] & = \sum_{x} x \cdot P(X = x) \\ & = -3 \times \frac{1}{6} + 6 \times \frac{1}{2} + 9 \times \frac{1}{3} = \frac{11}{2}. \end{align*} Thus, \[ \textcolor{teal}{ \boxed{ E[X] = \frac{11}{2} } } \]

We also have \begin{align*} E[X^2] & = \sum_{x} x^2 \cdot P(X = x) \\ & = (-3)^2 \times \frac{1}{6}+ 6^2 \times \frac{1}{2} + 9^2 \times \frac{1}{3} = \frac{93}{2}. \end{align*} So, \[ \textcolor{teal}{ \boxed{ E[X^2] = \frac{93}{2} } } \]

Now to find the expectation of $(2x + 1)^2$, we recall some properties of the expectation. For any random variables $X,X_1, X_2, \dots, X_n$ \begin{align*} & E \left[ \sum_{i=1}^{n} X_i \right] = \sum_{i=1}^{n} E[X_i] \\[1ex] & E[\alpha X] = \alpha E[X] \text{ and } E[X + \alpha ] = E[X] + \alpha \end{align*} for any constant $\alpha $. So, \begin{align*} E[(2x + 1)^2] & = E \left[ 4X^2 + 4X + 1 \right] \\ & = 4 E[X^2] + 4 E[X] + 1 \\ & = 4 \times \frac{93}{2} + 4 \times \frac{11}{2} + 1 \\ & = 209. \end{align*} Hence, \[ \textcolor{teal}{ \boxed{ E[(2x + 1)^2] = 209 } } \]