h(t) = -16t^2 + 60t + 2.

The above Eq is a parabola. Since the

coefficient of X^2 is negative, the

parabola opens downward. The coordinate

of the vertex is the maximum point on

the graph:

h'(t) = -32t + 60 = 0,

-32t + 60 = 0,

-32t = -60,

t = 1.875. = The X coordinate of the

vertex.

Substitute 1.875 for t in the given Eq:

h(1.875)=-16*(1.875)^2 + 60*1.875 + 2,

h(1.875) = -56.25 + 112.5 + 2 = 58.25. = The Y coordinate of the vertex.

V(X,Y) = V(t,h) = V(1.875,58.25) = The maximum point on the graph.

The graph will be zero at the points where the graph crosses the X axis:

Solve the given Eq using the Quadratic

Formula and get:

X = -0.03304, and X = 3.783.

Or (-0.03304,0), (3.783,0).