Anonymous

Let the equation be y-4=a(x+2)^2, where a=constant. (1,1)=> 1-4=a(1+2)^2=> -3=9a=> a=-1/3 Thus, the equation is y-4=(-1/3)(x+2)^2 or 3y=-x^2-4x+8

Anonymous

y = a(x - h)^2 + k y = a(x + 2)^2 + 4 for some a Substitute (1, 1) for (x, y) and solve for a. 1 = a(1 + 2)^2 + 4 1 = 9a + 4 -3 = 9a a = -1/3 y = -(x + 2)^2/3 + 4

Anonymous

Since you don't specify, I'll assume the parabola is vertical. The vertex is (-2,4), so the vertex form of the equation is  y = a(x+2)² + 4 Plug in (1,1) and solve for a 1 = a(1+2)² + 4 = 9a+4 a = -⅓ y = -⅓(x+2)² + 4