Question

A football player punts a ball. The path of the ball can be modeled by the equation y = –0.004x2 + x + 2.5, where x is the horizontal distance, in feet, the ball travels and y is the height, in feet, of the ball. How far from the football player will the ball land? Round to the nearest tenth of a foot.
125.0 ft

127.5 ft

247.5 ft

252.5 ft

Answer 1

The answer is 252.5 ft

y = –0.004x² + x + 2.5
Let's take y = 0 and we'll get the quadratic equation:

–0.004x² + x + 2.5 = 0
The general formula for quadratic equation is:
ax² + bx + c = 0

a = -0.004
b = 1
c = 2.5

$x_{1,2} = \frac{-b+/- \sqrt{ b^{2}-4ac} }{2a} =\frac{-1+/- \sqrt{ 1^{2}-4*(-0.004)*2.5} }{2*(-0.004)} = \frac{-1+/- \sqrt{1+0.04} }{-0.008} = \\ =\frac{-1+/- \sqrt{1.04} }{-0.008} = \frac{-1+/-1.02}{-0.008}$

$x_1=\frac{-1+1.02}{-0.008} = \frac{0.02}{-0.008} = -2.5 \\ x_2=\frac{-1-1.02}{-0.008} = \frac{2.02}{-0.008} = 252.5$

Since distance cannot be negative (x1), the correct answer is x2 = 252.5 ft

Answer 2

Answer:the answer is 252.5

Step-by-step explanation: