Quadratic Functions of Two Variables
Notes for Math 210 - by M. Flashman

Suppose f(x,y) = A x^{2} + B xy + C y^{2}
.

To begin assume that A is not 0

We reorganize this function by completing the square
on the x's.

So f(x,y) = A( x^{2} + B/A xy +B^{2}
/4A^{2} y^{2} ) + (C - B^{2} /4A)y^{2}
or
**f(x,y) = A( x + B/(2A) y)**^{2}
) + ((4AC - B^{2} )/4A^{2})y^{2} .

Now assume 4AC-B^{2} >0.

If **A >0
then f(x,y) >0** for all (x,y) but (0,0). So (0,0) is a minimum for f.

If **A <
0 then f(x,y) < 0 **for all (x,y) but (0,0). So (0,0)
is a maximum for f.

Now assume that 4AC-B^{2} <0.

If **y
= 0 and x>0 **then **f(x,y) has its sign the same as A**.

But when **x
= -B/2A y** and **y>0**, **f(x,y)
has it sign opposite to that of A**.

Thus no matter how close we consider (x,y) to (0,0),
there will always be points where f(x,y) > 0 and where f(x,y)<0.

Thus (0,0) is neither a maximum nor a minimum for f(x,y)
... it is a saddle.

If A = 0 and B is not 0 then 4AC-B^{2} <
0 and

** f(x,y) = y (Bx +
Cy).**
The line Bx+Cy
= 0 passes through (0,0) and divides the plane into two regions- one where
Bx+Cy is positive and one where Bx+Cy is negative. Since B is not 0, the
line Bx+Cy = 0 is not horizontal, so there are points in the plane with
y>0 in each of the two regions.

If **y>0
and Bx+Cy is positive**, then f(x,y) is positive.

If **y>0
and Bx+Cy is negative **then f(x,y) is negative.

Thus, no matter how close we consider (x,y) to (0,0),
there will always be points where f(x,y) > 0 and where f(x,y)<0.

Thus (0,0) is neither a maximum nor a minimum for f(x,y)
... it is a saddle.

**Summary: If 4AC-B**^{2}
< 0, then (0,0) is saddle point for f.

**If 4AC-B**^{2} >
0, and A>0 the (0,0) is a minimum for f.

**If 4AC-B**^{2} >
0, and A<0 the (0,0) is a maximum for f.