Find (a) the slope of the curve at the given point $P$, and (b) an equation of the tangent line at $P$.
\[
y=-2-4 x^{2}, \quad P(3,-38)
\]

Step 1 ：Given the function \(y = -2 - 4x^2\) and the point \(P(3,-38)\).

Step 2 ：To find the slope of the curve at the given point, we need to take the derivative of the function.

Step 3 ：The derivative of \(y = -2 - 4x^2\) is \(y' = -8x\).

Step 4 ：Evaluating this derivative at the point \(x = 3\) gives us the slope of the curve at that point, which is \(-8*3 = -24\).

Step 5 ：So, the slope of the curve at the point \(P(3,-38)\) is \(-24\).

Step 6 ：To find the equation of the tangent line at \(P\), we can use the point-slope form of a line, which is \(y - y_1 = m(x - x_1)\), where \(m\) is the slope and \((x_1, y_1)\) is the point.

Step 7 ：Substituting \(m = -24\), \(x_1 = 3\), and \(y_1 = -38\) into the point-slope form gives us the equation of the tangent line: \(y - (-38) = -24(x - 3)\).

Step 8 ：Simplifying this equation gives us \(y = -24x + 72\).

Step 9 ：Final Answer: The slope of the curve at the point \(P(3,-38)\) is \(\boxed{-24}\). The equation of the tangent line at \(P\) is \(\boxed{y = -24x + 72}\).