An object is thrown upward with an initial velocity of 32 feet per second. the objects height is modeled by the function h(t) = - 16t^2 + 32t
where t is the time of the at height, h(t). what is the maximum height of the object?

Using calculus...

dh/dt=-32t+32

When velocity, dh/dt, is equal to zero, the object is at its maximum height.

dh/dt=0 only when 32t=32, t=1 second, so the maximum height is h(1)

h(1)=-16+32=16ft

Using algebra

The maximum height will occur at the midpoint between the two zeros of the height function because of a quadratics symmetry...

h(t)=0 when

-16t^2+32t=0

-16t(t-2)=0

so t=0 and 2

The midpoint is t=1

And of course this will give you the same as we found earlier.  The maximum height is h(1) ft

35=5×77=7×1when we simplify 7 by 7we will get 5 and 1, 5 divided by 1 is 5