Finding missing coordinates in a non-linear graph?

Question

If I have the y value for the x values of 1, 2, 3, 7, and 9 in a graph, but I need all 1 through 12, how would I figure out the missing points?

 x  |  y
---------
 1  |  5
 2  |  6
 3  |  7
 7  | 10
 9  | 11

Answer 1

959137
Set up the following equation, then substitute into it whatever $x$ you desire to get the corresponding $y$.
$\begin{vmatrix} x^4&x^3&x^2&x&1&y\\ 9^4&9^3&9^2&9&1&11\\ 7^4&7^3&7^2&7&1&10\\ 3^4&3^3&3^2&3&1&7\\ 2^4&2^3&2^2&2&1&6\\ 1&1&1&1&1&5 \end{vmatrix}= \begin{vmatrix} x^4&x^3&x^2&x&1&y\\ 6561&729&81&9&1&11\\ 2401&343&49&7&1&10\\ 81&27&9&3&1&7\\ 16&8&4&2&1&6\\ 1&1&1&1&1&5 \end{vmatrix}=0$

$y=\frac{\begin{vmatrix} x^4&x^3&x^2&x&1&0\\ 6561&729&81&9&1&11\\ 2401&343&49&7&1&10\\ 81&27&9&3&1&7\\ 16&8&4&2&1&6\\ 1&1&1&1&1&5 \end{vmatrix}} {\begin{vmatrix} 6561&729&81&9&1\\ 2401&343&49&7&1\\ 81&27&9&3&1\\ 16&8&4&2&1\\ 1&1&1&1&1 \end{vmatrix}}$

The above is my first naïve attempt, for which I apologize. The following is my second, better attempt.

First of all, there are three collinear points among the five given points. They are $(1\mid 5);(2\mid 6);(3\mid 7).$
There are also two more extraneous points. They are $(7\mid 10);(9\mid 11).$

The two groups can be considered separately as straight lines, with common point $(5\mid 9)$. The function is
$y=\left\{\begin{array}{lcl} \frac{x+13}2&&(x\ge 5)\\ x+4&&(x\le 5) \end{array}\right.$

In my conception, the entire locus consists of those portions of the intersecting lines at, or south of, the intersection.

Answer 2

It depends on the exact nature of the non-linear graph you expect. Is it a polynomial? Exponential? Rational? You need to inter/extrapolate the values in between and right after the data points you know, so you can use the general form of the type of function you need. Plug in the known data points, and then you'll get a system of equations you can solve for the unknown constants' values in the general formula.

Word/PowerPoint are good for this kind of work; just go onto insert -> chart -> XY (scatter) and start imputing the data points. Then go onto the treadline option in the design tab and choose the type of function you want. It will give you the best fit for the function type (choosing a 5th degree polynomial in your case will guarantee you an exact fit, although sometimes it can be exact for lower degree polynomials.) as well as the function's equation (by ticking the appropriate boxes). You can then use the function to extrapolate/interpolate the values to the points you want by simply plugging in the values you want.