Question 323983
The height in feet of an object thrown upward is given by the equation h = 12t + 9 – ½t2, where h is the height of the object after t seconds. 
What is the initial position? 
What is the height after ½ second? 
At what time does it reach its maximum height? 
What is the maximum height? 
--------------------------
h = 12t + 9 – ½t2  A strange function, but could be true on a different planet.
{{{h(t) = -(1/2)t^2 + 12t + 9}}}
-----------------------------------
What is the initial position? 
@ t=0: h(t) = 9  (no units were specified, 9 something)
---------------------
What is the height after ½ second?
{{{h(1/2) = -(1/2)*(1/4) + 6 + 9 = 14.875}}}
-------------------------------------------
At what time does it reach its maximum height?
At the vertex of the parabola, t = -b/2a
t = -12/(-1) = 12 seconds
---------------------------
What is the maximum height?
h(12) = (-1/2)*144 + 12*12 + 9
h(12) = 81