3 point solution

Another choice for the "best" parabola might be the unique parabola that shares three important points with the catenary. These points are the apex and the two intersections with the ground (the x-axis). The equation of this parabola is again of the form g(x) = p1 x^2 + p2 due to the symmetry about the vertical axis. Furthermore, since the vertex is on the vertical axis, we know that p2 = fc which yields the functional form

Input := 

g = p1 x^2 + fc

Output =

               2
625.0925 + p1 x

We choose p1 so that the x-intercept of the parabola is at (L, 0) -- the x-intercept of the catenary. Substituting L into g and setting it equal to zero yields

Input := 

p1star = p1 /. Flatten[Solve[(g /. x -> L) == 0, p1]]

Output =

-0.00698155

Input := 

bestg = g /. {p1 -> p1star, p2 -> p2star}

Output =

                       2
625.0925 - 0.00698155 x

Input := 

Plot[{f[x], bestg}, {x, -L, L},
 AspectRatio -> Automatic,
 PlotStyle -> {{AbsoluteThickness[2]},
               {AbsoluteThickness[3],
                AbsoluteDashing[{3, 6}]}},
 AxesLabel -> {"[ft]", "height [ft]"}];