Question

the flight path of a firework is modelled by the function h(t)=-5(t-5)^2+127

Answer 1

Question (B)

h= _5 (t - 5)^2 +127.

t = 0

h = -5(0 - 5)^2 +127

h= -5(-5)^2 + 127

h=-5(25) +127

h = - 125 + 127

h= 2

Answer 2

h(t)=-5(t-5)^2+127 solve for max height
b) the vertex coordinates are right there (5,127) which makes the max height 127m

c) plug in the max height 127 to solve for time
127=-5(t-5)^2+127
127-127=-5(t-5)^2+127-127
0=-5(t-5)^2
0 divided by -5={-5(t-5)^2}divided by -5
0=(t-5)^2 factor
0=(t-5)(t+5) using the zeros rules
0=t-5+5
5=t
there for 5 second is how long it took the rocket to reach max height 127m

Answer 3

How would you solve the exact same equation, but if it was h=-4(t-5)^2+102?