1. The problem is to analyze the quadratic function $$y = -3(x - 1)^2 + 12$$ and understand its graph. 2. This is a quadratic function in vertex form: $$y = a(x - h)^2 + k$$ where $(h, k)$ is the vertex and $a$ determines the direction and width of the parabola. 3. Here, $a = -3$, $h = 1$, and $k = 12$. Since $a$ is negative, the parabola opens downward. 4. The vertex is at the point $(1, 12)$, which is the maximum point of the parabola because it opens downward. 5. The parabola is narrower than the standard parabola $y = (x - 1)^2$ because $|a| = 3 > 1$, meaning it is vertically stretched. 6. The axis of symmetry is the vertical line $x = 1$. 7. To find the y-intercept, set $x = 0$: $$y = -3(0 - 1)^2 + 12 = -3(1) + 12 = 9$$ So the y-intercept is at $(0, 9)$. 8. To find the x-intercepts, set $y = 0$ and solve: $$0 = -3(x - 1)^2 + 12$$ $$-3(x - 1)^2 = -12$$ $$(x - 1)^2 = 4$$ $$x - 1 = \pm 2$$ $$x = 1 \pm 2$$ So the x-intercepts are at $x = -1$ and $x = 3$, points $(-1, 0)$ and $(3, 0)$. Final answer: The parabola opens downward with vertex at $(1, 12)$, y-intercept at $(0, 9)$, and x-intercepts at $(-1, 0)$ and $(3, 0)$.