WILL MARK BRAINLIEST PLEASE HELP

  1) h = -1/2t^2 +10t

  2) h = -1/2(t -10)^2 +72

  3) domain: [0, 20]; range: [0, 50]

Step-by-step explanation:

1.) I find it easiest to start with the vertex form when the vertex is given. The equation of the presumed parabolic path for Firework 1 is ...

  h = a(t -10)^2 +50

To find the value of "a", we must use another point on the graph. (0, 0) works nicely:

  0 = a(0 -10)^2 +50

  -100a = 50 . . . . . . subtract 100a

  a = -1/2 . . . . . . . . . divide by -100

Then the standard-form equation is ...

  h = (-1/2)(t^2 -20t +100) +50

  h = -1/2t^2 +10t

__

2.) The path of Firework 2 is translated upward by 22 units from that of Firework 1.

  h = -1/2(t -10)^2 +72

__

3.) The horizontal extent of the graph for Firework 1 is ...

  domain: 0 ≤ t ≤ 20

The vertical extent of the graph for Firework 1 is ...

  range: 0 ≤ h ≤ 50

step-by-step explanation:

c. (3,2,6) is the answer.