Amodel rocket is launched from a roof into a large field. the path of the rocket can be modeled by the equation y=-0.04x^2+9.1x+11.9, 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 it’s starting point will the rocket land?

228.8 meters

Step-by-step explanation:

y = -0.04 x² + 9.1 x + 11.9

When the rocket lands, y = 0:

0 = -0.04 x² + 9.1 x + 11.9

Solving for x, start by multiplying by -1:

0 = 0.04 x² - 9.1 x - 11.9

Multiply by 100:

0 = 4x² - 910 x - 1190

Solve with quadratic formula:

x = [ -b ± √(b² - 4ac) ] / 2a

x = [ 910 ± √((-910)² - 4(4)(-1190)) ] / 2(4)

x = [ 910 ± √(847140) ] / 8

x ≈ -1.3, 228.8

Since x can't be negative, x = 228.8 meters.

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.

answer: 40.95

step-by-step explanation: