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?

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

1 Answer 1


Taking a parametrized version of your function:

$$ f(x,y) = 2(x-m)^2 + (y-n)^2 - 4(x-m)(y-n)+p $$

Your function has a 3D surface curve called a hyperbolic paraboloid, which admits a saddle point.

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.

  • $\begingroup$ Thanks, I did use GeoGebra to plot it (before posting) but didn't quite get the role of the parametrized version. Will play around with the parameters further. $\endgroup$
    – PGupta
    Feb 7, 2019 at 16:02
  • $\begingroup$ 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 $S$. $\endgroup$
    – ex.nihil
    Feb 7, 2019 at 16:04

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