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 out the time at which the rocket will reach its max, to the nearest 100th of a second. y=−16x 2 +171x+87

Answer 1

To find the time at which the rocket will reach its max height, we need to find the x-coordinate of the vertex of the quadratic equation.

The x-coordinate of the vertex of a quadratic equation in the form y = ax^2 + bx + c is given by the formula x = -b / 2a.

In this case, our quadratic equation is y = -16x^2 + 171x + 87.
So, a = -16 and b = 171.

Using the formula x = -b / 2a, we can calculate the x-coordinate of the vertex:

x = -171 / (2 * -16)
x = -171 / -32
x ≈ 5.34

Therefore, the rocket will reach its max height at approximately 5.34 seconds after launch.