(a) Find the slope of the curve $y=x^{2}-3 x-4$ at the point $P(2,-6)$ by finding the limit of the secant slopes through point $P$. (b) Find an equation of the tangent line to the curve at $P(2,-6)$.

Step 1 ：To find the slope of the curve at a given point, we need to find the derivative of the function at that point. The derivative of a function at a point gives the slope of the tangent line at that point. The derivative of a function \(f(x)\) is given by \(f'(x)\).

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

Step 3 ：To find the slope of the curve at the point \(P(2,-6)\), we substitute \(x=2\) into the derivative. The slope of the curve at the point \(P(2,-6)\) is \(1\).

Step 4 ：To find the equation of the tangent line to the curve at \(P(2,-6)\), we can use the point-slope form of a line, which is \(y - y_1 = m(x - x_1)\), where \((x_1, y_1)\) is a point on the line and \(m\) is the slope of the line. In this case, \((x_1, y_1) = (2, -6)\) and \(m = 1\).

Step 5 ：The equation of the tangent line to the curve at \(P(2,-6)\) is \(y = x - 2\).

Step 6 ：Final Answer: The slope of the curve \(y=x^{2}-3 x-4\) at the point \(P(2,-6)\) is \(\boxed{1}\). The equation of the tangent line to the curve at \(P(2,-6)\) is \(\boxed{y = x - 2}\).