1. The problem is to find the points on the curve $y = x^3 + 3x^2 - 9x + 4$ where the tangent line is horizontal. 2. A tangent line is horizontal where the derivative $y'$ is zero. 3. Compute the derivative: $$y' = \frac{d}{dx}(x^3 + 3x^2 - 9x + 4) = 3x^2 + 6x - 9$$ 4. Set $y' = 0$ to find the critical points: $$3x^2 + 6x - 9 = 0$$ 5. Divide both sides by 3: $$x^2 + 2x - 3 = 0$$ 6. Factor the quadratic: $$(x + 3)(x - 1) = 0$$ 7. Solve for $x$: $$x = -3 \quad \text{or} \quad x = 1$$ 8. Find the corresponding $y$ values: - For $x = -3$: $$y = (-3)^3 + 3(-3)^2 - 9(-3) + 4 = -27 + 27 + 27 + 4 = 31$$ - For $x = 1$: $$y = 1^3 + 3(1)^2 - 9(1) + 4 = 1 + 3 - 9 + 4 = -1$$ 9. The points where the tangent is horizontal are: - Smaller $x$ value: $(-3, 31)$ - Larger $x$ value: $(1, -1)$