\[ \text { If } f(x) \text { is defined as follows, find (a) } f(-3) \text {, (b) } f(0) \text {, and }(c) f(3) \] \( f f(x)=\left\{\begin{array}{ll}x^{2} & \text { if } x<0 \\ 2 & \text { if } x=0 \\ 3 x+1 & \text { if } x>0\end{array}\right. \) \( \begin{array}{l}\text { (a) } f(-3)=9 \text { (Simplify your answer.) } \\ \text { (b) } f(0)=\square \text { (Simplify your answer.) }\end{array} \)

Let's analyze the piecewise function step-by-step! For \( f(-3) \): Since \(-3 < 0\), we use the first case of the function: \[ f(-3) = (-3)^2 = 9. \] For \( f(0) \): Since \(x = 0\), we use the second case of the function: \[ f(0) = 2. \] Now for \( f(3) \): Here, \(3 > 0\), so we use the third case of the function: \[ f(3) = 3(3) + 1 = 9 + 1 = 10. \] In summary: - (a) \( f(-3) = 9 \) - (b) \( f(0) = 2 \) - (c) \( f(3) = 10 \)