WILL MARK AS BRAINLIEST. THIS IS TIME SENSITIVE1.] Use technology to determine an appropriate model of the data. (-1,0), (1,-4), (2,-3), (4,5), (5,12)a.] f(x) = – (x+1)^2 + 4b.] f(x) = (x+1)^2 + 4c.] f(x) = – (x-1)^2 - 4d.] f(x) = (x-1)^2 - 42.] Use technology to determine an appropriate model of the data. (0,1), (1,4), (2,5), (3,4), (4,1)a.] f(x) = – (x+2)^2 + 5b.] f(x) = (x+2)^2 + 5c.] f(x) = – (x-2)^2 + 5d.] f(x) = (x-2)^2 - 53.] The height of a free falling object at time t can be found using the function, h(t) = - 12t^2 + 36t. Where h(t) is the height in feet and t is the in seconds. Find the time when the object hits the grounda.] 1 secb.] 2 secc.] 3 secd.] 4 sec4.] The height of a soccer ball can be modeled by the function h(t) = - 8t^2 + 32t. Where h(t) is the height in feet and t is the in seconds. Find the time when the soccer ball reaches its maximum heighta.] 1 secb.] 2 secc.] 3 secd.] 4 sec

(1) Attached figure 1 is the solution
By graphing the model of the data ⇒⇒⇒ red points
(-1,0), (1,-4), (2,-3), (4,5), (5,12)
and graphing the quadratic equations
∴ the solution will be choice (d)
d.] f(x) = (x-1)² - 4
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(2) Attached figure 2 is the solution
By graphing the model of the data ⇒⇒⇒ red points
(0,1), (1,4), (2,5), (3,4), (4,1)
and graphing the quadratic equations
∴ the solution will be choice (c)
c.] f(x) = – (x-2)² + 5
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problems (1),(2) can be solved by substituting with the model of data
at the general form the quadratic function:
F(x) = a (x+b)² + c
and solve to find a, b, c
OR, It is easier to use the graph to find the appropriate quadratic equation
as  we did previously
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(3)
      h(t) = - 12t² + 36t.
      when the object hits the ground ⇒⇒ h = 0
∴   - 12t² + 36t = 0              ⇒⇒ factor t
∴  t ( -12t +36 ) = 0
t = 0 (unacceptable)    OR   t = 36/12 = 3 sec.

∴ the solution will be choice (c)
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(4)
    h(t) = - 8t² + 32t
At maximum height  ⇒⇒⇒      dh/dt = 0
∴ dh/dt = -16t +32 = 0
∴ t = 32/16 = 2 sec.

∴ the solution will be choice (b)