Differentiate the function and find the slope of the tangent line at the given value of the independent variable \[ s=t^{3}-t^{2}, t=-6 \]

Step 1 ：Given the function \(s=t^{3}-t^{2}\) and \(t=-6\)

Step 2 ：Differentiate the function to find \(ds/dt\). The derivative of \(t^{3}\) is \(3t^{2}\) and the derivative of \(t^{2}\) is \(2t\). So, \(ds/dt = 3t^{2} - 2t\)

Step 3 ：Substitute \(t=-6\) into \(ds/dt\) to find the slope of the tangent line at that point. The slope is \(3(-6)^{2} - 2(-6) = 120\)

Step 4 ：Final Answer: The slope of the tangent line at \(t = -6\) is \(\boxed{120}\)