The height of an object moving vertically is s(t)=16t^2+80t+245 (feet) at time t seconds. Find the object's maximum height and the time at which it occurs. Discuss domain, find the critical point, and use the first or second derivative test to confirm your answer.

Question

anonymous · Answer

$$dy/dx = 32t+80$$$$t=-80/32$$$$d²y/dx²=32 $$$$y=145$$

that is the minimum height, as u can see the a in 'ax²' of ur function is positive value