H (t) = - 16t^2+65t+3 how many seconds to reach max height. nearest tenth

Two ways to answer:

A. Using calculus:

H (t) = - 16t^2+65t+3

To find maximum of H (t), equate H' (t) = 0, or

H' (t) = - 32t+65=0 = > t=65/32=2.03125, or

t=2.0 s. to the nearest 10

B, Without using calculus, i. e. by completing the square

H (t)

=-16t^2+65t+3

=-16 (t^2-65/16t) + 3

=-16 (t^2-2 * (65/32) t + (65/32) ^2) + 3 + (65/32) ^2

=-16 (t-65/32) ^2+3 + (65/32) ^2

meaning the maximum occurs at (t-65/32) = 0, or t=65/32=2.03125

Answer: t=2.0 seconds (to the nearest tenth)