An Exploration of xy = ax + by + c

a=b=1a=5/2; b = -4

Notice that the graph goes intersects both axis at (1, 0) and (0, 1). For other values of a and b the graph of the function will be two lines, perpendicular, whose intersection with the axis will be at the points (b, 0) and (0, a). So we have made the graph of the following function:

which is a particular case of the general equation that was posed.

Observe that the graph of this expression IS different from the graph of the expression that results after the following transformations:

so for x different from b

.

The graph of this expression has to exclude the value at which the denominator becomes 0. So the grphs have to be different:

In the next step, we are going to consider the function (x - b)(y - a) - K = 0. Let's consider a = b = 1, K =1:

The next graph shows some functions of the family (x - 1)(y - 1) - K = 0, for K > 1:

(Although only one 'arm' of the graph is pointed by the arrow, the compete graph should have two 'arms')

As K increases, the graph seems to be further to the origin. The 'last' one corresponds to K =10.
For K positive but less than 1, the graphs seem to approach to the origin:

Note that as K goes to 0, we tend to get our first function. For K negative, I would expect a reflection on the x-axis, and a similar behavior (in absolute value!):

The conjecture seems to be true. Why does it happen? Consider the following graphs, for K a number of the form 1/n, n integer:

Observe that there is a relationship between 1/n and the x- and y- intercepts of these graphs. Note that for K = 1, the graph crosses at (0,0). But for K = 1/2, the graph intersects the x-axis at (1/2,0) and the y axis at (0, 1/2); similarly, for K=1/10, the graph intersects the x-axis at (1/10, 0) and the y-axis at (0, 1/10). So the expected behavior for K an integer, would be similar: the intersection with the x-axis will be at (K, 0) and the intersection with the y-axis will be at (0, K):

We need a closer view in order to determine if the conjecture was true:

Well it does not seem to be true! Observe that the graph for K=2 intersects the x- and y- axis at -1! and the graph for K = 10 intersects the axis at -9. A third graph, for K = 8 is shown in the second graph. The intersections are at -7. So it seems that what is happening for K positive is different, the intersections are at (0, -K+1) for the y-axis and (-K+1, 0) for the y axis.
Let's see what would be the bahavior for K negative:

For K = -1, the graph intersects the axis at 2; and for K = -2, it intersects the axis at 3. Let's take a closer view for K = -10:

Well, the graph for K= -10 intersects the axis at 10. It seems that the expression for negative integer values behave, with respect to the intersections, in the same way os for K positive, so the intersections ar at (0, -K+1) for the y-axis and (-K+1, 0) for the y-axis. Can we expect then a similar behavior for values of K in (-1, 0)?:

It seems that follows the same pattern as in the positive case, except in that the intersections are translated 1 unit. For K =-1/2, the intersections are at (1/2) + 1, and for K = -1/10, they are at (1/10) + 1. Why is 1? The 1 is probably related to the values of a and b. I would expect that if, for example a = 1 and b = 6, then the intersects for the y axis will translate, vertically 5 units (6 = 1 + 5). Let's see what happens. The next graph shows the expression (x-1)(y-6) - K = 0, for K = 1, 2, and 10. The y-intersections are at 5, 4, and -4 respectively. (Recall that for (x-1)(y-1) - K = 0, they were at 0, -1, and -9); so, the conjecture seems to be true.

And then if we change x instead of y, we will get a horizontal translation, and the x-intersects will change. For a positive the translation will be to the right (we will have still a factor of the form (x - a). For a negative the graph will translate to the left (the factor is then (x + a)):

The next graph shows the expression (x-6)(y-1) - K = 0, for K = 1, 2, and 10. The x-intersections are at 5, 4, and -4 respectively. (Recall that for (x-1)(y-1) - K = 0, they were at 0, -1, and -9); so, the conjecture seems to be true.
This highlights the fact that the numbers a and b are the values of the asymptotes of the graph. Recall that for K = 0, or the case in which in the general expression xy = ax + by + c, c = ab, the graph corresponds to two perpendicular lines, one intersecting the x-axis at a, and the other intersecting the y-axis at b. Since the right hand side of the expression xy = ax + by + c can always be written as

we can take K = c - ab, and so the analysis done provides full information about the resulting graphs.

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Summary

would be, the particular case in which c = ab was analyzed. Doing this, the expression factored to (x- a)(y - b) = 0 and its graph showed to be two perpendicular lines, and a and b represented the x- and y- intercepts respectively. Changing the expression to (x- a)(y - b) + K= 0, the a and b showed to be the asymptotes of a hyperbola, and K determined the x- and y- intercepts of the graph. In any case, the x-intercepts are given by (-K+ a , 0) and the y-intercepts are (0, -K + b).
For K < -1. For -1< K < 0 and for 0 < K < 1. Finally the general case follows by writting the general expresion xy = ax + by + c as (x- a)(x - b) + K = 0, where K = c - ab.

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Posible Extensions

(y - mx- a)(y - nx - b) = 0

So different hyperbolas would be obtained. For example for , the graph corresponds to:

And for, the graph is (the asymptotes are drawn also):

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