Find the x-intercepts of the parabola with the vertex (1,1) and the y intercept (0,-3). write your answer in this form (x1,,y2).

We start with the equation y-k = a(x-h)^2.  Since the vertex is at (1,1), this equation becomes y-1 = a(x-1)^2.  Since the curve passes thru (0, -3), we get:

-3-1 = a(0-1)^2, or -4 = a.  Thus, the eqn of the parabola is

y-1 = -4(x-1)^2.  To find the x-intercepts, set y = 0 and solve the resulting equation for x:

-1 = -4(x-1)^2, or 1 = 4(x-2)^2, or (1/4) = (x-2)^2.

Taking the sqrt of both sides, (1/2) = plus or minus x-2. 

Then x -2 = 1/2, or 2x - 4 = 1, or x = 5/2

Also, -(x-2) = 1/2, or  -2(x-2) = 1, or -2x + 4 = 1, or -2x = -3, or x=3/2.

Thus, the x-intercepts are (3/2, 0) and (5/2, 0).

Is there more to the question?