just for reference, here's the vertex form of a parabola equation:
f(x) = a(x - h)^2 + k
because this is in vertex form, finding the vertex is a little easier -- it's just (2, -1), bc the vertex is found as (h, k)
the domain is all real numbers, (-∞, ∞), because there aren't any constraints on the domain in this case
to determine the range, you have to figure out whether or not this parabola curves upwards or down, and in this case it curves up!! you know this because your A value (the implied coefficient of 1) is positive, so the range will be from the y value of your vertex to POSITIVE infinity, so [-1, ∞)
then your x and y intercepts are found by plugging in zeros for the values:
x-intercept: (plug in 0 for y)
0 = (x - 2)^2 - 1
1 = (x - 2)^2
1 = x - 2
x = 3
y-intercept: (plug in 0 for x)
y = (0 - 2)^2 - 1
y = 4 - 1
y = 3
