1. **State the problem:** We are given the quadratic function $$h(x) = -\frac{1}{2}x^2 + 4x - 10$$ and need to find its vertex, line of symmetry, x-intercepts, and y-intercept. 2. **Formula for vertex:** The vertex of a parabola given by $$ax^2 + bx + c$$ is at $$x = -\frac{b}{2a}$$. 3. **Calculate the vertex x-coordinate:** Here, $$a = -\frac{1}{2}$$ and $$b = 4$$. $$x = -\frac{4}{2 \times -\frac{1}{2}} = -\frac{4}{-1} = 4$$ 4. **Calculate the vertex y-coordinate:** Substitute $$x=4$$ into $$h(x)$$: $$h(4) = -\frac{1}{2}(4)^2 + 4(4) - 10 = -\frac{1}{2} \times 16 + 16 - 10 = -8 + 16 - 10 = -2$$ 5. **Vertex:** The vertex is at $$\boxed{(4, -2)}$$. 6. **Line of symmetry:** The line of symmetry is the vertical line through the vertex's x-coordinate: $$x = 4$$ 7. **Determine if parabola opens up or down:** Since $$a = -\frac{1}{2} < 0$$, the parabola opens downward. 8. **Find x-intercepts:** Solve $$h(x) = 0$$: $$-\frac{1}{2}x^2 + 4x - 10 = 0$$ Multiply both sides by $$-2$$ to clear fraction: $$\cancel{-2} \times \left(-\frac{1}{2}x^2 + 4x - 10\right) = \cancel{-2} \times 0$$ $$x^2 - 8x + 20 = 0$$ Use quadratic formula: $$x = \frac{8 \pm \sqrt{(-8)^2 - 4 \times 1 \times 20}}{2} = \frac{8 \pm \sqrt{64 - 80}}{2} = \frac{8 \pm \sqrt{-16}}{2}$$ Since discriminant is negative, no real x-intercepts. 9. **Find y-intercept:** Set $$x=0$$: $$h(0) = -\frac{1}{2} \times 0 + 4 \times 0 - 10 = -10$$ So y-intercept is $$\boxed{-10}$$. **Final answers:** - Vertex: $$(4, -2)$$ - Line of symmetry: $$x = 4$$ - Parabola opens downward - No real x-intercepts - y-intercept: $$-10$$