Saturday, 16 August 2008

How To Find Minimum Point In An Equation

Given a mathematics equation, sometimes we need to know the minimum point or values of its lowest co-ordinates. There are 2 methods to locate this lowest point in the equation.

The 2 methods are:

  1. graphical

  2. mathematical re-expression of equation

I shall deal with both methods here.

Example: y = x2 - 2x + 3 (a quadratic equation)

1) Graphical method

By plotting the curve generated by the equation, we are able to visually identify the lowest point and get the co-ordinates directly.

The demerit of this method is having to plot the graph out and the co-ordinate values are as accurate as plotting accuracy.

The graph is drawn as shown in diagram 1.

Diagram 1 Equation plotted to identify the lowest point on the curve

From diagram 1, you are able to see that the lowest point is at x = 1 and y = 2. (The point marked with the red cross). This graphical method is straight forward but needs effort in sketching the curve accurately.

2) Mathematical re-expression method

This method involves more manipulation of the equation rather than sketching graph.

It makes use of the "Completing the Square" method in factorisation to extract out the lowest point. For a review of the "Completing the Square" method, please refer to this link.

y = x2 - 2x + 3 can be re-written as

y = (x2 - 2x + c2) -c2 + 3

Note: The "-c2" term outside the closing bracket is to retain the originality of the equation.
The purpose is to form a y expression with an order of 2 or (x -c)2 for reason which will be clear later.

What is this "c" to be ?

By following the principle of the "Completing the Square" technique, "c" can be seen to be (2/2).

Therefore the new equation becomes
y = [x2 - 2x + (2/2)2] - (2/2)2 + 3

Now, here comes the important concept of this method to locate the lowest point.

From the new y expression, you see that to make the y value the lowest possible, the only thing you can do is to make the x value equal to the "c" value or the (2/2) value.
That is, [x - (2/2)] = 0 ==> x = (2/2) for lowest value of y.

When x = (2/2) = 1, y must then be -(2/2) + 3 = +2.

Therefore, the answer to the lowest co-ordinates is x = 1 and y = 2.

This is the same as for the graphical method. However, it does not involve tedious plotting, beside more accurate numerical values is ensured.


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