From the function rule h(t)=60t-16t^2+2. How do we tell from the graph and the table whe h'(t) will be positive, where it will be negative, and where it is zero.
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).