Answer:
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
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2.) The path of Firework 2 is translated upward by 22 units from that of Firework 1.
h = -1/2(t -10)^2 +72
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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
