For this discussion, you will work in groups to answer the questions. In a video game, airplanes move from left to right along the path described by LaTeX: y=2+\frac{1}{x} y = 2 + 1 x . They can shoot rockets tangent to the direction of flight at targets on the x-axis. Where will a rocket fired from LaTeX: \left(1,3\right) ( 1 , 3 ) hit the target? Where will a rocket fired from LaTeX: \left(2,2.5\right) ( 2 , 2.5 ) hit the target? Where will a rocket fired from LaTeX: \left(2.5,2.4\right) ( 2.5 , 2.4 ) hit the target? Where will a rocket fired from LaTeX: \left(4,\:2.25\right) ( 4 , 2.25 ) hit the target?

When fired from (1,3) the rocket will hit the target at (4,0)

When fired from (2, 2.5) the rocket will hit the target at (12,0)

When fired from (2.5, 2.4) the rocket will hit the target at

When fired from (4,2.25) the rocket will hit the target at (40,0)

Explanation:

All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).

Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.

In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:

so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:

m=y'=-1

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

and solve for x

x=4

so, when fired from (1,3) the rocket will hit the target at (4,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)

so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

and solve for x

x=12

so, when fired from (2, 2.5) the rocket will hit the target at (12,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)

so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

and solve for x

so, when fired from (2.5, 2.4) the rocket will hit the target at

Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)

so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

and solve for x

so, when fired from (4,2.25) the rocket will hit the target at (40,0)

I uploaded a graph that represents each case.

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

hope this ! this is not copied! search it up, it's free of plagiarism!