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Its height(in meters),x seconds after takeoff, is modeled by h(x)= -(x-11)(x+3) how many seconds after takeoff will the hovercraft land on the ground

User Iivannov
by
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2 Answers

6 votes

Answer:

14s

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+33\\\\x'=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-example-1
Its height(in meters),x seconds after takeoff, is modeled by h(x)= -(x-11)(x+3) how-example-2
User ItsGeorge
by
4.4k points
3 votes

Answer:


x = 11\,s

Explanation:

The instants when hovercraft lands on the ground are those when h(x) = 0. Given that height function is a second-order polynomial in its factorized form, such instants are related to the roots of the function.

Time is a positive real number and the only possible solution:


x = 11\,s

User Aksoy
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4.7k points