1. The problem is to analyze the quadratic function $$y = -6(x - 4)^2 + 3$$ and understand its key features. 2. The general form of a quadratic function in vertex form is $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. 3. Here, $a = -6$, $h = 4$, and $k = 3$. The vertex is at $(4, 3)$. 4. Since $a = -6$ is negative, the parabola opens downward. 5. The vertex is the maximum point of the parabola because it opens downward. 6. The axis of symmetry is the vertical line $x = h = 4$. 7. To find the y-intercept, set $x = 0$: $$y = -6(0 - 4)^2 + 3 = -6(16) + 3 = -96 + 3 = -93$$ 8. To find the x-intercepts, set $y = 0$ and solve for $x$: $$0 = -6(x - 4)^2 + 3$$ $$-6(x - 4)^2 = -3$$ $$(x - 4)^2 = \frac{1}{2}$$ $$x - 4 = \pm \sqrt{\frac{1}{2}} = \pm \frac{\sqrt{2}}{2}$$ $$x = 4 \pm \frac{\sqrt{2}}{2}$$ 9. So the x-intercepts are at $$x = 4 + \frac{\sqrt{2}}{2}$$ and $$x = 4 - \frac{\sqrt{2}}{2}$$. 10. Summary: - Vertex: $(4, 3)$ - Opens downward - Axis of symmetry: $x = 4$ - Y-intercept: $(0, -93)$ - X-intercepts: $\left(4 + \frac{\sqrt{2}}{2}, 0\right)$ and $\left(4 - \frac{\sqrt{2}}{2}, 0\right)$