Answer: the x-intercepts are (5, 0) and (-3, 0).Step-by-step explanation:The standard equation of a vertical parabola with vertex at (h, k) isy = a(x - h)^2 + k. Here we are told that the vertex is at (1, -16), which means that h = 1 and y = -16. Thus, we have:y = a(x - 1)^2 + (-16).We are told also that the graph passes through (0, -15). Substituting -15 for y and 0 for x, we get:-15 = a(0 - 1)^2 - 16, or-15 = a - 16Then a must be 1, and the equation of the parabola isy = (x - 1)^2 - 16.Now to find the x-intercepts: Set y = 0 and solve for x:0 = (x - 1)^2 - 16, or(x - 1)^2 = 16, orx - 1 = ±√16 = ±4Then x = 1 ± 4, or x = 5 or x = -3, so the x-intercepts are (5, 0) and (-3, 0).