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
Accepted Solution
A:
(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 ======================================================
(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 ======================================================
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 =========================================================
(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) =======================================================
(4) h(t) = - 8t² + 32t At maximum height ⇒⇒⇒ dh/dt = 0 ∴ dh/dt = -16t +32 = 0 ∴ t = 32/16 = 2 sec.