1. **Problem:** Find the slope of the tangent line to $f(x) = 3x^2 - 2x$ at the point $(-1,5)$ and then find the equation of the tangent line in the form $y = mx + b$. 2. **Formula:** The slope of the tangent line at a point is given by the derivative $f'(x)$ evaluated at that point. 3. **Find the derivative:** $$f'(x) = \frac{d}{dx}(3x^2 - 2x) = 6x - 2$$ 4. **Evaluate the slope at $x = -1$:** $$m = f'(-1) = 6(-1) - 2 = -6 - 2 = -8$$ 5. **Use point-slope form to find the equation:** The point-slope form is: $$y - y_1 = m(x - x_1)$$ Substitute $m = -8$, $x_1 = -1$, and $y_1 = 5$: $$y - 5 = -8(x + 1)$$ 6. **Simplify the equation:** $$y - 5 = -8x - 8$$ $$y = -8x - 8 + 5$$ $$y = -8x - 3$$ **Final answer:** The slope of the tangent line is $-8$ and the equation of the tangent line is $$y = -8x - 3$$.