1. **Stating the problem:** We are given a parabola that opens downwards with x-intercepts near $-6$ and $6$, and a vertex near $(0,7)$. We want to find the equation of this parabola. 2. **Formula used:** The standard form of a parabola with vertex at $(h,k)$ is $$y = a(x - h)^2 + k$$. Since the parabola opens downwards, $a < 0$. 3. **Identify vertex:** From the graph description, vertex is at $(0,7)$, so $h=0$ and $k=7$. The equation becomes $$y = a x^2 + 7$$. 4. **Use x-intercepts to find $a$:** The parabola crosses the x-axis at $x = -6$ and $x = 6$, so $y=0$ at these points. 5. Substitute $x=6$, $y=0$ into the equation: $$0 = a(6)^2 + 7$$ $$0 = 36a + 7$$ $$36a = -7$$ $$a = -\frac{7}{36}$$ 6. **Final equation:** $$y = -\frac{7}{36} x^2 + 7$$ This equation matches the given parabola with vertex at $(0,7)$ and x-intercepts near $-6$ and $6$. **Answer:** $$\boxed{y = -\frac{7}{36} x^2 + 7}$$