1. **State the problem:** We are given a downward-opening parabola with vertex near $(0,20)$ and x-intercepts near $-5$ and $5$. We want to find the equation of this parabola. 2. **Formula and rules:** The standard form of a parabola with vertex $(h,k)$ is $$y = a(x - h)^2 + k$$ where $a$ determines the direction and width of the parabola. Since it opens downward, $a < 0$. 3. **Identify vertex:** Given vertex is approximately $(0,20)$, so $h=0$ and $k=20$. The equation simplifies to $$y = a x^2 + 20$$ 4. **Use x-intercepts:** The parabola crosses the x-axis near $x = -5$ and $x = 5$, so $y=0$ when $x = \\pm 5$. 5. **Plug in an x-intercept to find $a$:** Using $x=5$, $y=0$: $$0 = a(5)^2 + 20$$ $$0 = 25a + 20$$ $$25a = -20$$ $$a = \frac{-20}{25} = -\frac{4}{5}$$ 6. **Write final equation:** $$y = -\frac{4}{5} x^2 + 20$$ This equation matches the vertex and x-intercepts described. **Final answer:** $$y = -\frac{4}{5} x^2 + 20$$