Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises - Page 124: 15

$s'(t) = 3t^{2} -2t$ $slope =s'(-1) = 5$ Equation of tangent line: $y = 5x+3$

$s(t) = t^{3} - t^{2}$ $s(-1) = -1-1 = -2$ Now we know the point on the graph is: (-1,-2) So, $s'(t) = 3t^{3-1} -(2)t^{2-1}$ $s'(t) = 3t^{2} -2t$ Slope at t= -1: $slope = s'(-1) = 5$ Equation of tangent line: $\frac{y+2}{x+1} = 5$ $y= 5(x+1) -2$ $y = 5x+3$