How to find equation of a parabola with vertex and y-int

http://img848.imageshack.us/img848/3707/…

I'm not sure how to go about solving this problem. The two equations of a parabola my teacher taught me is:
y=ax^2+bx+c
and
y=a(x-h)^2+k

Thanks!

Vertex is at (h,k) so
y = a(x + 2)^2 - 1
Replace (x,y) with (0,-3), the y-intercept, and solve for a.
-3 = a(0 + 2)^2 - 1
-3 = 4a - 1
-2 = 4a
a = -1/2
y = -1/2(x + 2)^2 - 1

Vertex (- 2, - 1):

h = - 2
k = - 1

y-int. (0, - 3):

x = 0
y = - 3

y = a(x - h)² + k
- 3 = a[0 - (- 2)]² + (- 1)
- 3 = a(2)² - 1
- 3 = a(4) - 1
- 3 = 4a - 1
4a = - 3 + 1
4a = - 2
a = - 2/4
a = - 1/2

y = - 1/2(x + 2)² - 1
y = - 1/2(x² + 4x + 4) - 1
y = - 1/2 x² - 2x - 2 - 1
y = - 1/2 x² - 2x - 3
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