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

=

step-by-step explanation:

i dont get the question

step-by-step explanation:

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.