Question

A rocket is launched from a tower. The height of the

rocket, y in feet, is related to the time after launch, x


in seconds, by the given equation. Using this


equation, find the maximum height reached by the


rocket, to the nearest tenth of a foot.


y = –16x2 + 246x + 100


=

Answer 1

1045.6 feet

Explanation:

A graphing calculator easily shows you the vertex of the parabola. The y-coordinate there is the rocket's maximum height: 1045.6 feet.

__

Quadratic function ax^2 +bx +c has its vertex at x=-b/(2a). Here, that x-value is ...

x = -(246)/(2(-16)) = 7 11/16 . . . seconds

The corresponding height of the rocket is ...

y = (-16(7 11/16) +246)(7 11/16) +100 = 123(7 11/16) +100 = 1045 9/16 ≈ 1045.6

The maximum height reached by the rocket is 1045.6 feet.