1. **State the problem:** Find the vertex and x-intercepts of the quadratic function $f(x) = -x^2 - 3x - 4$. 2. **Formula for vertex:** The vertex of a parabola given by $f(x) = ax^2 + bx + c$ is at $\left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)$. 3. **Calculate the x-coordinate of the vertex:** $$x_v = -\frac{b}{2a} = -\frac{-3}{2 \times -1} = -\frac{-3}{-2} = -\frac{3}{2} = -1.5$$ 4. **Calculate the y-coordinate of the vertex:** $$f(-1.5) = -(-1.5)^2 - 3(-1.5) - 4 = -2.25 + 4.5 - 4 = -1.75$$ 5. **Vertex:** The vertex is at $(-1.5, -1.75)$. 6. **Find x-intercepts:** Solve $f(x) = 0$: $$-x^2 - 3x - 4 = 0$$ Multiply both sides by $-1$ to simplify: $$\cancel{-}x^2 \cancel{-} 3x \cancel{-} 4 = \cancel{-}0$$ $$x^2 + 3x + 4 = 0$$ 7. **Calculate discriminant:** $$\Delta = b^2 - 4ac = 3^2 - 4 \times 1 \times 4 = 9 - 16 = -7$$ 8. Since $\Delta < 0$, there are no real x-intercepts. **Final answers:** - Vertex: $(-1.5, -1.75)$ - x-intercepts: None