# General quadratic functions in two variables

Consider the quadratic function $$2x^2-4xy+y^2-3x+4y$$. This can be expressed as $$2(x-5/4)^2+(y-1/2)^2-4(x-5/4)(y-1/2)-7/8$$

Is there any advantage of expressing in the latter form? Are there some features of the function that become apparent by looking at the second expression? In other words, why would we ever want to write the function in the other manner?

• The second form makes it quite obvious that both derivatives vanish at $x={5\over4},y={1\over2}$. Feb 7, 2019 at 15:29
• Thanks, that makes sense! Feb 7, 2019 at 15:35
• Related to that, the level curves of this function are a family conics centered at $(5/4,1/2)$.
– amd
Feb 7, 2019 at 19:33

$$f(x,y) = 2(x-m)^2 + (y-n)^2 - 4(x-m)(y-n)+p$$
Conveniently, $$S = (m,n,p)$$ is the saddle point.
You can easily check this on any graphing software, such as GeoGebra Classic. Simply create three sliders $$m$$, $$n$$, and $$p$$, then create the point $$S$$ as I defined it above, and define $$f$$ as above. Fiddle with the sliders and all will make sense.
• Define $S$ as above using the same parameters. What you should get is: as you change parameters, $S$ will move and will drag the curve with it. The curve does not change, but will be translated along the motion of $S$. Feb 7, 2019 at 16:04