## Function

f(x,y) = px^{2}+ qxy + ry^{2}

## Domain

xandyare elements of the set of allReal numbers

## Graphs

## Derivative

With respect tox:2px + qy

With respect toy:qx + 2ryCritical points are formed where the two derivatives equal 0. These critical points may be minimums, maximums, or saddle points. In the first graph, for

x+^{2}xy+y, 0 =^{2}2x + ywhich happens only whenxandyare both 0. Fory, 0 =x + 2yonly whenxandyare both 0. Therefore the only critical point is (0,0) which is the lowest point, 0.

## Integral

With respect tox:(+^{1}/_{3})px^{3}+ (^{1}/_{2})px^{2}y + ry^{2}xc

With respect toy:px+^{2}y + (^{1}/_{2})qxy^{2}+ (^{1}/_{3})rq^{3}c

Double Integral:(^{1}/_{3})px^{3}y + (^{1}/_{4})qx^{2}y^{2}+ (^{1}/_{3})ry^{3}x

Break down, first with respect tox:(^{1}/_{3})px^{3}+ (^{1}/_{2})qx^{2}y + ry^{2}xThis is taken then with respect to

yand the final double integral is returned. The double integral allows us to find the volume of an area under a surface and thexyplane. Looking at two examples, we want to find the area bound byxfrom -1 to 1 andyfrom -1 to 1 in the first graph. We first evaluate the first breakdown from -1 to 1. The second integral is then taken and evaluated from -1 to 1. For our first graph, the volume fromx=-1..1 andy=-1..1 is (^{8}/_{3}). If we look at the volume fromx=0..1 andy=0..1, the process is repeated, and the volume is (^{11}/_{12}).

## Interesting Features

When working with this function, there are many different items to look at. The general formula forms a curved surface that gets the lowest at the origin (0,0). Since the values are squared, the negatives become positive and the curve appears symmetric. The

pscales one end of the curve causing it to rise. Theqvalue scales the values in a symmetric manner effectingxandy. Therfunctions exactly the same asponly on the other axis.

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