The vertex must be at x = -3......
And the form would be :
y = a(x +3)^2 - 15a = ax^2 + 6ax + 9a - 15a = ax^2 + 6ax - 6a.......and since a is positive, let 6a = b so b is also positive......so the final form is
y = ax^2 + bx - 6a = ax^2 + 6ax - 6a
A) Notice that (0, -3) could be on the graph if a = 1/2....
B) (0,3) cannot be on the graph because "a" would have to be -1/2.... but "a" is positive
C) (-3, 0) ...we must have 0 = 9a - 18a - 6a.....and "a" would have to be 0.....again.....this isn't possible because "a" is positive
D) (3,3) .....we must have 3 = 9a + 18a - 6a → 3 = 21a → a = 1/7 .....so this point could exist
E) (-3, -3).......we must have -3 = 9a - 18a - 6a → -3 = -15a → a = 1/5....so this point could exist
In summary (A), (D) and (E) are possible, but (B) and (C) are not
Here's a graph of the possibilites.......https://www.desmos.com/calculator/rsqelub4qd
