\( 1 \leftarrow \quad \begin{array}{l}\text { Sketch a graph of } f \\ f(x)=-3 x^{2}-2\end{array} \)

The graph of the function \( f(x) = -3x^2 - 2 \) is a downward-opening parabola because of the negative coefficient in front of the \( x^2 \) term. Its vertex can be found at the point \( (0, -2) \), as there are no linear or constant terms affecting the x-coordinate. The parabola will be symmetric around the y-axis, and as you move away from the vertex, the value of \( f(x) \) decreases rapidly due to the factor of -3. To sketch this, plot the vertex at \( (0, -2) \) and then mark points on both sides; for example, when \( x = 1 \), \( f(1) = -3(1)^2 - 2 = -5 \), and when \( x = -1 \), \( f(-1) = -3(-1)^2 - 2 = -5 \). This gives you points \( (1, -5) \) and \( (-1, -5) \). The graph will reach toward negative infinity as \( x \) moves away from the vertex in either direction. Don't forget to label your axes and add arrows to show that the graph continues on indefinitely!