Aball is thrown into the air with an upward velocity of 32 ft/s. its height (h) in feet after t seconds is given by the function h= -16t^2 + 32t + 6. how long does it take the ball to reach its maximum height? what is the ball’s maximum height? roun

For this case we have the following equation:
 h = -16t ^ 2 + 32t + 6
 Deriving we have:
 h '= -32t + 32
 We equal zero and clear t:
 -32t + 32 = 0
 32t = 32
 t = 32/32
 t = 1 s
 Then, the maximum height is given by:
 h (1) = -16 * (1) ^ 2 + 32 * (1) + 6
 h (1) = 22 feet
 
 It takes the ball to reach its maximum height about:
 t = 1 s
 The ball's maximum height is:
 h (1) = 22 feet

You're doing something i dont know how to do but search up on google slope calculator maybe that will get you there