How do I answer each part of this question step by step

Question

The height 𝑦 of an object thrown straight upward in feet is given by 𝑦 = −16𝑡² + 256𝑡 + 25.

where t is the time in seconds.

a.) At what time is the object at its maximum height?

b.) What is the maximum height of the object?

Answer 1

This is a quadratic function, so it is a parabola. The parabola opens downward, so the maximum height will be at the vertex. For parabolas in the form of at² + bt + c, the vertex is found using

t (seconds) = -b / 2a

t = -256 / 2(-16)

t = 8

So, the object reaches its maximum height (t) in 8 seconds. Now you can plug it back in and solve for y (height)

y = -16(8)² + 256(8) + 25

y = -1,024 + 2,048 + 25

y = 1,049

Its maximum height is 1,049.

Answer 2

a) To get maximum we differentiate our function: y' = - 32t + 256 = 0, t = 256/32 = 8 sec.

Now because second derivative y'' = - 32 < 0 by second derivative test we have maximum at t = 8 sec;

b) y_max= y(8) = - 16·64 + 256·8 + 25 = 1049 ft