1. **State the problem:** We want to analyze the function $y = -x^3 + 6x^2$ and understand its graph using the slope of the tangent line. 2. **Formula for slope of tangent line:** The slope of the tangent line at any point $x$ on the curve is given by the derivative $y' = \frac{dy}{dx}$. 3. **Find the derivative:** $$y = -x^3 + 6x^2$$ Using power rule: $$y' = -3x^2 + 12x$$ 4. **Interpretation:** The slope $y' = -3x^2 + 12x$ tells us how steep the curve is at any $x$. 5. **Find critical points (where slope = 0):** $$-3x^2 + 12x = 0$$ Factor out $-3x$: $$-3x(x - 4) = 0$$ So, $$x = 0 \quad \text{or} \quad x = 4$$ 6. **Determine nature of critical points:** Second derivative: $$y'' = \frac{d}{dx}(-3x^2 + 12x) = -6x + 12$$ Evaluate at $x=0$: $$y''(0) = 12 > 0$$ (local minimum) Evaluate at $x=4$: $$y''(4) = -6(4) + 12 = -24 + 12 = -12 < 0$$ (local maximum) 7. **Find function values at critical points:** $$y(0) = 0$$ $$y(4) = -(4)^3 + 6(4)^2 = -64 + 96 = 32$$ 8. **Summary:** - The graph has a local minimum at $(0,0)$. - The graph has a local maximum at $(4,32)$. - The slope of the tangent line changes sign at these points, indicating peaks and valleys. This analysis helps sketch the graph and understand its shape using tangent slopes.