Final Project

by

Jackson Huckaby

Consider the graphs of the equation

xy=ax+by+c

for various substitutions of real number coefficients a, b, and c.

To explore this equation we will fix a and b, and create a slider value on c.

An interactive file can be found here.

As we set c equal to -2,-1,0,1,2,3 and overlay the graphs we get the following image.

we see that the lines are pulling farther and farther away from the origin as n increases. We also see that when n becomes negative a reflection over the line x=1 occurs.

How else can we manipulate xy=ax+by+c?

Lets try setting x=0. So this becomes 0y=a0+by+c, or 0=y+c, using the same values of c we get the following set of horizontal lines:

If we make y=0 instead of x, we get 0=x+c. This will result in a set of vertical lines.

Lets observe the graphs of x-b=0 and y-a=0

Rewriting x-b=0 first, we get x=b.

So therefore by setting b equal to -3,-2,-1,0,1,2,3 we get:

Lets try the same thing with y=a, we can predict that this time we will get vertical lines instead.

And we do!

So now let us explore the affect of graphing (x-b)(y-a)=0

And then let us graph our original equation xy=ax+by+c on the same plane:

So let us vary our c values and overlay again and see what we get:

By looking at this graph we can clearly see that the graph of (x-a)(x-b)=0 gives us our asymptotes for the equation xy=ax+by+c.

So let us examine why this could be true. If we distribute:

(x-b)(y-a)=n(for any n)

xy-ax-by+ab=n

xy=ax+by-ab+n

And compare this to our equation in question: xy=ax+by+c

So: as long as c=-ab+n then we are getting the same equation.

Lets verify.

If we let a=5 and b=-2 and n=1

(x+2)(y-5)=1

And then let us graph xy=5x-2y+11:

And what we realize is...

They are the exact same graph!