How to make a Line[] with no end?

Question

I'm trying to do this:

In this graph, the secant points are aproximated in order to become the tangent, it seems I need some kind of function which plots a line based on two points and it's points do not limit the end of the line.

This is what I could do, as you can evaluate (see?), you'll see the line ends at the proposed points.

Show[{
  Plot[{15 - 2 x^2}, {x, -1, 3}],
  Graphics[Line[{{1, 13}, {2, 7}}]]
  }]

I've read the reference on line but I still have no idea on how to do it, I imagine there are two ways of doing it, (1) the way I did, with Line; (2) With the Plot function.

Answer 1

A little bit more general (for any function, interval, starting and final points...) and implementing most of your labels:

appDeriv[f_, x0_, xinit_, pRange_List] := Module[{xseq, tan, tan1, l, plot, absPR},
   xseq      = Table[x0 + Exp[-n/2] (xinit - x0), {n, 0, 10}]; (* exploration points *)
   tan       = D[f@x, x] /. x -> x0;                           (* real derivative *)
   tan1[x_] := (f[x] - f[x0])/(x - x0);                        (* approx derivative  *)
   l[t_]    := Line[Transpose[{pRange, pRange t + (-t x0 + f@x0)}]]; (*line w/given angle*)
   plot = Plot[f@x, {x, pRange[[1]], pRange[[2]]}, PlotStyle->Thick, PlotLabel->InputForm@f];
   absPR = AbsoluteOptions[plot, PlotRange][[1, 2]];     (*get actual prange for the plot*)
 Animate[Show[
  plot,
  Graphics[{
    Thick, PointSize[Large],
    Green, l[tan], Red, l[tan1[p]], Black,  (* The two lines *)
    Point[{x0, f@x0}],Point[{p,AbsoluteOptions[plot, AxesOrigin][[1,2,2]]}], Point[{p, f@p}],
    Inset[ "x =" <> ToString@N@p, {p, (absPR[[2, 1]] 9 + absPR[[2, 2]])/10}],
    Inset[Framed@Grid[{{"Exact ", tan}, {"Calculated", N@tan1[p]}}], {Left,Center}, 
          {Left, Center}, Background -> Yellow]}]], 
 {p, xseq}] ];
GraphicsGrid[{{appDeriv[Sin@# + 1 &, 0, 2, {-2, 2}],appDeriv[2 # + 2 # # &, 1, 2, {-2, 2}]}}]

Answer 2

line[x_] := Solve[{a + b == 13, a x + b == 15 - 2 x^2}, {a, b}] // Quiet
f[x_, x0_] := {15 - 2 x^2, (a x + b) /. line[x0]}

Animate[ Plot[{f[x, x0], -4 x + 17}, {x, -2, 3}, PlotRange -> {0, 18}, PlotStyle -> Thick,
               Evaluated -> True, Epilog -> {PointSize[0.025], 
               Point[{{1, 13}, {x0, 15 - 2 x0^2}}]}],           {x0, -2, 2.5}]

Adding any formulae has been left up to you.

Answer 3

f[x_] := 15 - 2 x^2

x0 = 1;
tangentLine[x_] = Normal[Series[f[x], {x, x0, 1}]]

secantLine[h_, x_] := InterpolatingPolynomial[{{1, f[1]}, {h, f[h]}}, x]

Animate[Plot[{f[x], tangentLine[x], secantLine[h, x]}, {x, -1, 3}, 
  AxesLabel -> {"x", "y"}, 
  Epilog -> {AbsolutePointSize[7], Point[{{1, 0}, {1, f[1]}, {h, 0}, {h, f[h]}}]}, 
  PlotRange -> {-4, 20}, PlotStyle -> {Red, Blue, Green}],
  {h, Range[2., 1.1, -.1] ~Join~ Range[1.09, 1.01, -.01] ~Join~ Range[1.009, 1.001, -.001]}]

Adding the labels is up to you (hint: modify the setting of Epilog by adding Text[] primitives).

Answer 4

It seems to me that you are asking for a way to extend a line segment specified by two points. Here is one method you might use.

extend[x:{a_, b_}, d_] := (d/Norm[a - b]) (x - {b, a}) + x

This will extend a line segment a b by distance d in either direction. Example:

DynamicModule[{pt = {{1, 1}, {2, 1}}},
 LocatorPane[Dynamic @ pt,
  Graphics[
   {Red, Line @ Dynamic @ extend[pt, 5]},
   PlotRange -> {{-10, 10}, {-10, 10}},
   Axes -> True,
   GridLines -> Automatic
  ]
 ]
]