Fitting points to parabola-like shapes

Question

I have 5 points which I would like to fit onto a curve. Mostly this is just for fun, but trying to do it is hopefully teaching me and my daughter some mathematics. We're using shape-building to learn about the properties of equations.

We would like the curve to be concave to the left like a parabola and we would like it to hit each point.

We tried a linear regression of a quadratic function and the software did not hit all points. Is that a limitation of the software or are we simply trying to do something that a parabola cannot do?

The points are:

1, 2
1.5, 1.9
2, 1.5
1.5, 1.1
1, 1

Essentially a sideways, symmetrical parabola-like shape.

I tried higher order functions but as you might expect, the shape that the software produced was not concave to the left.

I'd like to know both whether there exists a perfect-fit equation which is concave forever, and also whether there exists a perfect-fit equation which is simply concave in the region 1,1

In general I'd like to understand whether every concave curve can be described as a single equation and if so, how.

Answer 1

The points given are symmetrical about $y=1.5$ appear to form a curve with a stationary point at $(2,1.5)$.

You would get a pretty good fit with a $2^{\text{nd}}$ order curve $(R^2=0.96)$ using $$2-x=3.7457(y-1.5)^2$$

and an even better one with a $4^{\text{th}}$ order curve $(R^2\approx 1)$ using
$$2-x=9.7222(y-1.5)^4+1.5694(y-1.5)^2$$

as shown in the diagram below.

Answer 2

A parabola opening to the left has the equation $$x = Ay^2 + By + C, \quad A < 0.$$

This gives you 3 parameters from which to try to vary the curve, but you would like to pass through 5 points. You can either

1. reduce the number of points, e.g. leaving points 1,3,5 or 2,3,4 to get the shape you want exactly (but it won't pass through the other 2 points you left out); or
2. find values of $A,B,C$ to make the curve as close as possible to all 5, but then quite likely, it won't pass through any of them altogether...

Answer 3

Five points determine a conic,its equation has 5 constants.

$$ ax^2+ 2 h xy + by^2 +dx+ ey =1$$

Four points determine a parabola, its equation has 4 constants:

$$ ax^2+\sqrt{ab } xy + by^2 +dx+ ey =1$$

By Cramer's rule (arranged in a matrix) we can find the parabola passing through four given points ( as also the conics through 5 points).

In the figure below(drawn using Geogebra) 4 fixed points$(A,B,C,D)$ are placed on a known convenient parabola $ y = x^2/4+1 $ shown.

A fifth variable point placed on a slider (p1,p2,P,p3,p4) forms ellipses when placed outside parabola and when placed inside it, form a hyperbolas.

The locus of a fifth point to remain on a parabola set up through 4 prescribed points is the same parabola.

So from the data you gave there are 5 such combinations/cases. In each case place the fifth point on a slider (with a reasonable slider domain) and see all possibilities as above.

I believe all possible convex/concave shapes can be obtained this way.

Next, I took the 5 points you gave and it is seen to makes a narrow ellipse, but not narrow enough to make a parabola with arms going to $ \infty $ to the left. So I perturbed the second point on a circle with first point as center. When the second point is placed a bit below the location you have given, (all the other four points being fixed ) a parabola can be captured. Exact value for symmetric case is given below.

EDIT1:

After what is mentioned above if you wish to have a parabola opening to left side with its axis displaced parallel to x-axis $k=1.5, $ passing through $( 1,1), (1,2), (2,1.5) $ the one and only possibilitycorresponding to $x=1.5$ y-coordinates should be

$$ (y_1= 1.85355, y_2= 1.14645) $$

symmetric to $y= 1.5$