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family function x^2 - 2*(m - 2)*x + m - 2

a) Create a Manipulate to explore the behaviour of the functions of this family for m ∈ [-10,10]. Mark the minimum value of the parabolas with a red point. What do you observe about these points? Use the interval [-20,20] for x.

b) Collect/create the coordinates of the minimum value for the 21 values of m (integer values from -5 to 5). Find the coefficients a,b and c such that the points are on the curve of equation $ax^2+bx+c=0$.

I couldn't solve the question b

Here what I wrote:

f1[m_, x_] = x^2 - 2*(m - 2)*x + m - 2; 
Assuming[-10 <= m <= 10, 
  Minimize[{f1[m, x], -20 <= x <= 20, -10 <= m <= 10}, x] // Simplify]

It gives:

{-6 + 5 m - m^2, {x -> -2 + m}}

Please help.

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  • $\begingroup$ What is the family of functions that are of concern? Are they given by f[m_][ x_] := x^2 - 2*(m - 2)*x + m - 2; $\endgroup$
    – m_goldberg
    Commented Oct 17, 2020 at 4:28
  • $\begingroup$ yes the function is x^2 - 2*(m - 2)*x + m - 2 $\endgroup$
    – hihograss
    Commented Oct 17, 2020 at 5:11

2 Answers 2

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Part b of the question has two errors. For integer values of m from -5 to 5 you get 11 values, not 21. Second, the formula in the b question must be a x^2 + b x + c == y . Tell your teacher, that he asked nonsense. With this corrections you get

f1[m_, x_] = x^2 - 2*(m - 2)*x + m - 2;
    Assuming[-10 <= m <= 10, 
    min = Minimize[{f1[m, x], -20 <= x <= 20, -10 <= m <= 10}, x] // 
    Simplify; {ymin[m_], xmin[m_]} = {min[[1]], x /. min[[2]]}]

sol = Solve[
  Table[a xmin[m]^2 + b xmin[m] + c == ymin[m], {m, -10, 10}], {a, b, 
   c}, Reals]

(*   {{a -> -1, b -> 1, c -> 0}}   *)

pl1 = Plot[a x^2 + b x + c /. First@sol, {x, -10, 10}];
pl2 = ListPlot@Table[{xmin[m], ymin[m]}, {m, -10, 10}];
Show[pl1, pl2]
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You are asked to find the interpolating polynomial through the coordinates of the minima of the family of functions for different values of m.

Using interpolation

One solution using interpolation as prescribed is

minimaCoordinates = {x /. Last[#], First[#]} &@
     Minimize[f1[#, x], x] & /@ Range[-10, 10];
interpol[x_] = ExpandAll@InterpolatingPolynomial[minimaCoordinates, x]
(* x - x^2 *)

You can see that the solution is $a=-1$, $b=1$, $c=0$.

You can also visualize the result

Plot[interpol[x], {x, Sequence @@ MinMax[minimaCoordinates[[All, 1]]]},
 Epilog -> Point[minimaCoordinates]]

Visualization of the minima and the interpolating polynomial

Without interpolation

Of course, this can also be obtained along the lines of your attempt. You are missing the crucial final step of identifying your result with a polynomial.

(# /. First@Solve[First[#] == x, m]) &@ 
  With[{x = x /. First@Solve[D[f1[m, x], x] == 0, x]},
   {x, f1[m, x]}
   ] // ExpandAll
(* {x, x - x^2} *)
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