1. **State the problem:** Find all points on the curve defined by $$x^2 + y^2 + 4x - 2y = -1$$ where the tangent lines are (a) horizontal and (b) vertical. 2. **Rewrite the curve equation:** The curve is implicitly defined by $$F(x,y) = x^2 + y^2 + 4x - 2y + 1 = 0$$. 3. **Find the derivative \( \frac{dy}{dx} \):** Using implicit differentiation, $$\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) + \frac{d}{dx}(4x) - \frac{d}{dx}(2y) + \frac{d}{dx}(1) = 0$$ which gives $$2x + 2y \frac{dy}{dx} + 4 - 2 \frac{dy}{dx} = 0$$ 4. **Solve for \( \frac{dy}{dx} \):** Group terms with \( \frac{dy}{dx} \), $$2y \frac{dy}{dx} - 2 \frac{dy}{dx} = -2x - 4$$ $$\frac{dy}{dx}(2y - 2) = -2x - 4$$ $$\frac{dy}{dx} = \frac{-2x - 4}{2y - 2} = \frac{-2(x + 2)}{2(y - 1)} = \frac{-(x + 2)}{y - 1}$$ 5. **Horizontal tangents:** Occur when \( \frac{dy}{dx} = 0 \), so numerator must be zero: $$-(x + 2) = 0 \implies x = -2$$ 6. **Find corresponding \(y\) values for \(x = -2\):** Substitute into original equation: $$(-2)^2 + y^2 + 4(-2) - 2y = -1$$ $$4 + y^2 - 8 - 2y = -1$$ $$y^2 - 2y - 4 = -1$$ $$y^2 - 2y - 3 = 0$$ 7. **Solve quadratic for \(y\):** $$y = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-3)}}{2} = \frac{2 \pm \sqrt{4 + 12}}{2} = \frac{2 \pm 4}{2}$$ So, $$y = 3 \quad \text{or} \quad y = -1$$ 8. **Points with horizontal tangents:** $$(-2, 3)$$ and $$(-2, -1)$$. 9. **Vertical tangents:** Occur when \( \frac{dy}{dx} \) is undefined, i.e., denominator zero: $$y - 1 = 0 \implies y = 1$$ 10. **Find corresponding \(x\) values for \(y = 1\):** Substitute into original equation: $$x^2 + (1)^2 + 4x - 2(1) = -1$$ $$x^2 + 1 + 4x - 2 = -1$$ $$x^2 + 4x - 1 = -1$$ $$x^2 + 4x = 0$$ 11. **Solve quadratic for \(x\):** $$x(x + 4) = 0$$ $$x = 0 \quad \text{or} \quad x = -4$$ 12. **Points with vertical tangents:** $$(0, 1)$$ and $$(-4, 1)$$. **Final answers:** - Horizontal tangents at $$(-2, 3)$$ and $$(-2, -1)$$. - Vertical tangents at $$(0, 1)$$ and $$(-4, 1)$$.