Answer:
[tex]y = -0.1x^2 + 250 ft[/tex]
Step-by-step explanation:
Because this quadratic equation would have the curve-down form of:
[tex]y = -ax^2 + b[/tex]
where a and b are positive coefficient.
If we let the peak (250 ft) of the curve be at x = 0. Then
[tex]y = -a0^2 + b = 250[/tex]
[tex]b = 250[/tex]
Also at the begins and ends, thats where y = 0, the 2 points are separated by 100 ft. So let the begin at -50 ft and the end at 50ft. We have
[tex]-a(\pm 50)^2 + 250 = 0[/tex]
[tex]-a2500 = -250[/tex]
[tex]a = 250/2500 = 0.1[/tex]
Therefore, the model quadratic equation of our path would be
[tex]y = -0.1x^2 + 250 ft[/tex]