an object thrown directly upward is at a height of h feet after t seconds , where h = - 12 ( t - 3 ) ^ 2 + 120 . at what height , in feet , is the object 2 seconds after it reaches its maximum height ?
"we see that h will be a maximum h = 120 when t - 3 = 0 , that is when t = 3 . at t = 5 , h = - 12 * ( 5 - 3 ) ^ 2 + 120 = - 12 ( 4 ) + 120 = 72 the answer is e ."
add(multiply(negate(12), power(subtract(add(3, 2), 3), 2)), 120)