A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y =-0.04x2+8.3x + 4.3, where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land?

Question

(1 point)
• 208.02 m
• 416.03 m
• 0.52 m
• 208.19 m

Answer 1

To find where the rocket lands, we need to find the x-value when y = 0 (the height is 0 when it lands).

y = -0.04x^2 + 8.3x + 4.3
0 = -0.04x^2 + 8.3x + 4.3

This is a quadratic equation that can be solved using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, a = -0.04, b = 8.3, and c = 4.3.

Plugging in the values:

x = (-8.3 ± sqrt(8.3^2 - 4(-0.04)(4.3))) / 2(-0.04)
x = (-8.3 ± sqrt(68.89 + 0.688)) / -0.08
x = (-8.3 ± sqrt(69.578)) / -0.08
x = (-8.3 ± 8.34) / -0.08

x = (-16.64) / -0.08 or x = (-0.04) / -0.08

x ≈ 208 m or x ≈ 0.5 m

Therefore, the rocket will land approximately 208 meters horizontally from its starting point. The closest option is 208.19 m.