A toy rocket is launched from the ground and follows a parabolic path represented by the equation y=−12x2+8x

Question

. At the same time, a paper airplane is thrown from a rooftop at a height of 30 feet and follows a straight path to the ground represented by the equation y=−2x+30
. Let x=
time, in seconds and y=
height, in feet. Which of the following statements is true?(1 point)
Responses

The time in seconds, x, must be positive values, and the height in feet, y, can be any value therefore the viable solutions will be in the first and fourth quadrants.
The time in seconds, x , must be positive values, and the height in feet, y , can be any value therefore the viable solutions will be in the first and fourth quadrants.

The time in seconds, x, and the height in feet, y, must be positive values, therefore the viable solutions will be in the first quadrant.
The time in seconds, x , and the height in feet, y , must be positive values, therefore the viable solutions will be in the first quadrant.

The time in seconds, x, can be any value and the height in feet, y, must be positive values, therefore the viable solutions will be in the first and second quadrant.
The time in seconds, x , can be any value and the height in feet, y , must be positive values, therefore the viable solutions will be in the first and second quadrant.

The time in seconds, x, and the height in feet, y, have no constraints, therefore the viable solution can be in any quadrant.

Answer 1

To analyze the height equations of the toy rocket and the paper airplane, let’s break it down:

1. Toy Rocket: The height \( y \) is given by the equation \( y = -12x^2 + 8x \).

   • This is a parabolic equation that opens downwards (as indicated by the negative coefficient of \( x^2 \)).
   • The rocket starts from the ground, which means when \( x = 0 \), \( y = 0 \) (it is launched from height 0).
   • The rocket will reach a maximum height and then fall back to the ground, which will eventually lead to negative heights as time goes on after it descends.

2. Paper Airplane: The height \( y \) is given by the straight line equation \( y = -2x + 30 \).

   • This is a linear equation where the airplane is thrown from a height of 30 feet.
   • As time \( x \) (in seconds) increases, the height \( y \) decreases, and it will hit the ground when \( y = 0 \).

Constraints on Time and Height:

• Time \( x \):

  • In the context of both the rocket and the airplane, \( x \) must be non-negative (i.e., \( x \geq 0 \)), since negative time doesn’t apply in this physical scenario.

• Height \( y \):

  • For the rocket: The maximum height occurs at a certain point and can then become negative as it falls back to the ground after reaching its peak height.
  • For the airplane: The height starts from 30 feet and decreases until it hits the ground (0 feet), at which point it will not go negative in a physical sense.

Quadrants:

• The first quadrant of the Cartesian plane contains points where \( x \) (time) and \( y \) (height) are both positive.
• The fourth quadrant contains points where \( x \) is positive and \( y \) is negative.
• Therefore, the valid solutions for the toy rocket will be in the first quadrant (when it is ascending) and the fourth quadrant (as it descends past ground level). The paper airplane will stay strictly within the first quadrant until just before it hits the ground.

Conclusion:

The statement matching this analysis is:

"The time in seconds, x, must be positive values, and the height in feet, y, can be any value therefore the viable solutions will be in the first and fourth quadrants."