# A rock is thrown upward from a bridge that is 73 feet above a road. The rock reaches its maximum height above the road 0.98 seconds after it is thrown and contacts the road 3.21 seconds after it was thrown. Write a function ff that determines the rock's height above the road (in feet) in terms of the number of seconds tt since the rock was thrown. f(t)=f(t)= Hint: the function ff can be written in the form f(t)=c*(t−t_1)(t−t_2) f(t)=c*(t-t_1)(t-t_2) for fixed numbers cc, t_1t_1, and t_2t_2.

A rock is thrown upward from a bridge that is 73 feet above a road. The rock reaches its maximum height above the road 0.98 seconds after it is thrown and contacts the road 3.21 seconds after it was thrown.
Write a function ff that determines the rock's height above the road (in feet) in terms of the number of seconds tt since the rock was thrown.
$f\left(t\right)=f\left(t\right)=$
Hint: the function ff can be written in the form for fixed numbers cc, t1t1,
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skarvama
Let the function f(t) be
$f\left(t\right)=c\left(t-{t}_{1}\right)\left(t-{t}_{2}\right)$
At $t=3.21sec$, rock's height = 0 feet
so, $t=3.21$ sec is one of the zeroes of the function f(t)
Therefore, $f\left(t\right)=c\left(t-{t}_{1}\right)\left(t-3.21\right)$
At t= 0.98 seconds, rock is at the maximum height i.e. it represents the vertex of the quadratic function f(t).
Since the zeroes are equidistant from the vertex,
$\left(3.21-0.98\right)=\left(0.98-{t}_{1}\right)$
$⇒2.23=\left(0.98-{t}_{1}\right)$
$⇒{t}_{1}=-1.25$
So, $f\left(t\right)=c\left(t+1.25\right)\left(t-3.21\right)$
At t = 0 seconds, f(t) = 73
$73=c\left(1.25\right)\left(-3.21\right)$
$⇒c=-18.1931$
so, $f\left(t\right)=-18.1931\left(t+1.25\right)\left(t-3.21\right)$