A hovercraft takes off from a platform. It's height(in meters), x seconds after takeoff, is modeled by:
What is the height of the hovercraft at me the time of take off?

14s

Step-by-step explanation:

1. Since the point in this question is considering the moment the hovercraft has took off, h=0 till it lands on the ground. We can do this algebraic calculation. To find the roots of the quadratic equation.

Looking at the equation:

Also, these points mark the initial and ending point of the hovercraft path. As 11 and -3 are the roots, the time between those points is the elapsed time by the hovercraft. Considering also, the x-axis, the time.

h(x)=(-x+11)(x+3)\\h(x)=-x^{2}-3x+11x+33\\0=-x^{2}+8x+33x'=t_1=11, and\: x''=t_2=-3\\

2. Graphically, we can see that movement of the hovercraft, since a< 0. The concavity is downward.

As 11 and -3 are the roots, the time between the those points, is the elapsed  time by the hovercraft

11-(-3)=14s

Its height(in meters),x seconds after takeoff, is modeled by h(x)= -(x-11)(x+3) how many seconds aft

Its height(in meters),x seconds after takeoff, is modeled by h(x)= -(x-11)(x+3) how many seconds aft

Step-by-step explanation:

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step-by-step explanation: