Question 10369
{{{h = -16t^2 + 46t}}} This is the equation for a parabola that opens downward. The quadratic equation is already in standard form: {{{ax^2 + bx + c}}} where x = t, a = -16, b = 48, and c = 0.

The maximum height of the thrown object will be at the vertex (the maximumum value) (h, t) of the parabola.

The t-coordinate (equivalent to the x-coordnate) of the vertex (this is the time at which the object reaches its maximum height) is given by {{{(-b/2a)}}} where a = -16 and b = 48.

{{{(-48)/2(-16) = 3/2}}} so, at t= 3/2 secs, the object is at its maximum height.
To find the value of the maximum height at t = 3/2 secs, substitute t = 3/2 into the original quadratic equation and solve for h.

{{{h = -16(3/2)^2 + 48(3/2)}}}
{{{h = -16(9/4) + 48(3/2)}}}
{{{h = -36 + 72}}}
{{{h = 36 ft}}}

As a check, you could take the first derivative of the quadratic equation and set it to zero to find the value of t at the maximum height. 
{{{h = -16t^2 + 48t}}}
{{{dh/dt = -32t + 48}}}
{{{-32t + 48 = 0}}}
{{{-32t = -48}}}
{{{t = -48/(-32)}}} = {{{3/2}}}secs, same as previous answer for the time the object reaches maximum height.