1. I'm assumig that the paths are perfect parabolas
this means that their general forms can be written in 
it's easier to find vertex form first then expand to get general form
vertex form is
where the vertex is (h,k) and a is a constant
firework #1
vertex is (10,50), so (h,k)=(10,50) and h=10, k=50

to find the value of
, subsitute another point
(0,0)



so the equation in vertex form is 
expand to get general form



2.
same as last time
vertex is (10,72) so (h,k)=(10,72) so h=10 and k=72
equation is 
find 
use another point
(0,22)




so the equation in vertex form is 
3.
range is the numbers that h is allowed to be
think about what h represents. it represents the height of the rocket
from the graph, we can see that the lowest possible height is 0yd and the highest height is 50yd
so range is 0 to 50 or 0≤h≤50
domain is the numbers that t is allowed to be
think about what t represents. it represents how long the rocket has been flying
it will stop flying when it hits the ground or at t=20
it starts flying at t=0
so domain is from 0 to 20 or 0≤t≤20