Answer: a. 16.9 m b. 4.082 seconds c. no d. 2.041 seconds e. 21.41 mStep-by-step explanation:A graphing calculator can help you answer these questions quickly and easily. The graph is attached.__a.Evaluate the function at t=3. h(t) = -4.9t^2 +20t +1 h(3) = (-4.9×3 +20)×3 +1 = 5.3×3 +2 = 16.9The baton will be at a height of 16.9 meters after 3 seconds.__b.The axis of symmetry of quadratic ax^2+bx+c is located at x=-b/(2a). The point of release and the later point at the same height will be symmetrically located about the axis of symmetry. The peak height will be on the axis of symmetry of the graph. That value of time is ... tpeak = -20/(2(-4.9)) = 20/9.8 ≈ 2.041 . . . secondsThe baton will be at the release height again after 2×2.041 = 4.082 seconds.__c.As we saw in part (b), the maximum height will be when t=2.041 seconds. That height is found by evaluating the function for that value of t. h(2.041) = (-4.9(2.041) +20)(2.041) +1 = 21.41 . . . metersThe baton will not reach a height of 25 meters.__d.We found this answer in part (b): 2.041 seconds to reach the maximum height.__e.We found this answer in part (c): 21.41 meters is the maximum height._____Additional commentThe time at maximum height is 20/9.8 seconds = 100/49 = 2 2/49 seconds. In general, the answer values shown here are rounded to 4 significant figures. The graphing calculator rounds to 3 decimal places.