Linear Equation matches polynom

Questions
1

Hi, i have a Problem finding a nice solution with mathematica. The Problem: I want to find all coordinates where a linear equation matches a polynom. I tried it with findroot, but that didnt bring the result i need. My Solution so far is shown below. The problem with findroot is that i have to give a starting point and that i only get one point close to the starting point. Maybe somebody can give me a hint what to do. Thank You! Remove[“Global`*”]
Points = {{0,
    0}, {1, -1}, {2, -2}, {3, -4}, {4, -3}, {5, -4}, {6, -1}, {7, \
-4}, {8, 2}, {9, 3}, {10, -1}, {11, -4}};
f = Interpolation[Points,
InterpolationOrder → 1](*Polynom from Points*);

\ = 45 °; (*Angle of Linear*)
fg[x_, Pos_] := -x*Tan[\] + Points[][[2]] +
Points[][[1]]*
Tan[\] (*Linear equation throug point (Points), gradient \
from \ *)
Plot[{fg[x, 9], f}, {x, 0, 11}, Epilog → Map[Point, Points],
ImageSize → Large, AspectRatio → 1] (*Plot*)

sol = FindRoot[fg[x, 9] == f, {x, 8}, AccuracyGoal → 8,
PrecisionGoal → 8];
solution = x /. sol
(*Check*)
f == fg[solution, 9]
f

2

I hope this helps, I am rewriting the problem a little bit.

First the list of points:

points = {{0, 0}, {1, -1}, {2, -2}, {3, -4}, {4, -3}, {5, -4}, {6, -1}, {7, -4}, {8, 2}, {9, 3}, {10, -1}, {11, -4}};

Next the piecewise linear interpolation function:

f = Interpolation[points, InterpolationOrder -> 1]

Visualize them together in a plot:

Plot[f[x], {x, 0, 11}, Epilog -> {AbsolutePointSize[6], Red, Point[points]}]

out.png
out.png360×223 3.31 KB

Next the ‘fg’ function:

fg[x_, pos_] := -x*Tan[Pi/4] + points[[pos, 2]] + points[[pos, 1]]*Tan[Pi/4]

And visualize again, everything together:

Plot[Join[{f[x]}, Table[fg[x, i], {i, 1, 12}]], {x, 0, 11}, 
 Epilog -> {AbsolutePointSize[6], Red, Point[points]}]

out2.png
out2.png360×216 6.92 KB

Now solve for one of the intersections:

FindRoot[fg[x, 9] == f[x], {x, 8.5}]  (* returns {x->8.} *)