1. The problem asks to find the equation of a parabola given its vertex and x-intercepts. 2. The vertex is at $ (2, -1) $ and the x-intercepts are at $ x = -2 $ and $ x = 6 $. 3. The standard form of a parabola with vertex $ (h, k) $ is: $$ y = a(x - h)^2 + k $$ 4. Here, $ h = 2 $ and $ k = -1 $, so: $$ y = a(x - 2)^2 - 1 $$ 5. Since the parabola passes through the x-intercepts, substitute $ x = -2 $ and $ y = 0 $ to find $ a $: $$ 0 = a(-2 - 2)^2 - 1 $$ $$ 0 = a(-4)^2 - 1 $$ $$ 0 = 16a - 1 $$ $$ 16a = 1 $$ $$ a = \frac{1}{16} $$ 6. Therefore, the equation of the parabola is: $$ y = \frac{1}{16}(x - 2)^2 - 1 $$ 7. This equation matches the given vertex and x-intercepts.