A rock is thrown upward from a bridge that is 57 feet above a road. The rock reaches its maximum height above the road 0.76 seconds after it is thrown and contacts the road 3.15 seconds after it was thrown.
(a) Write a Function (f) that determines the rock's height above the road (in feet) in terms of the number of seconds t since the rock was thrown.

On this question we see that we are given two points on a certain graph that has a maximum point at 57 feet and in 0.76 seconds after it is thrown, we know can say this point is a turning point of a graph of the rock that is thrown as we are told that the function f determines the rocks height above the road (in feet) in terms of the number of seconds t since the rock was thrown therefore this turning point coordinate can be written as (0.76, 57) as we are told the height represents y and x is represented by time in seconds. We are further given another point on the graph where the height is now 0 feet on the road then at this point its after 3.15 seconds in which the rock is thrown in therefore this coordinate is (3.15,0).

now we know if a rock is thrown it moves in a shape of a parabola which we see this equation is quadratic. Now we will use the turning point equation for a quadratic equation to get a equation for the height which the format is [tex]f(x)= a(x-p)^2 +q[/tex] , where (p,q) is the turning point. now we substitute the turning point

[tex]f(t) = a(t-0.76)^2 + 57[/tex], now we will substitute the other point on the graph or on the function that we found which is (3.15, 0) then solve for a.

0 = a(3.15 - 0.76)^2 + 57

-57 =a(2.39)^2

-57 = a(5.7121)

-57/5.7121 =a

-9.9788169 = a then we substitute a to get the quadratic equation therefore f is

[tex]f(t) = -9.9788169(t-0.76)^2 +57[/tex]