Final Assignment

Part 1


Allison McNeece

1. Consider the graphs of the equation: xy=ax+by+c for various substitutions of real number coefficients a,b, and c.

Give a comprehensive description of the families of curves, with illustrations, that result.

Let's start by varying the values for a while keeping b and c fixed at 1:

With a positive we have:

equations with positive a


-the graph is a hyperbola with asymptotes at y=a and x=1

With a negative we have:

equations with negative a values

graph with negative values of a


-the graph is a hyperbola with asymptotes y=a and x=1 as before but the graph has inverted across the x axis


Now let's vary b but keep a and c fixed at 1:

With b positive:

equations with positive b

graph with positive b


-again we have a hyperbola but the asymptotes are now at y=1 and x=b

With b negative:

equations with negative b

graph with negative values for b


-same graph as previous (for the values of b=-5,-4,-3,-2)with asymptotes at y=1 and x=b but inverted across the y axis.

-note the graph of xy=ax-y+c. This ends up being just two straight lines. hmmmm.....


And lastly we vary c keeping a and b fixed at 1:

With c positive:

equations with positive c values

graph with positive c values


-We again have a hyperbola. The asymptotes are at y=1, x=1

-How is c changing the graph? It seems that as |c| increases the vertex of the hyperbola move further away from the center of the hyperbola

With c negative:

equations with negative c values

graph with negative c values


-same as previous graph but when c=-1 we have two intersecting lines and the hyperbolas are inverted across the y axis

What have we learned here?

The graph of xy=ax+by+c is a hyperbola with asymptotes of x=b and y=a

rewritten asymptotes are where (x-b)(y-a)=0

If we were to rewrite this equation we would have:


hmm.. that looks a lot like our original equation: xy=ax+by+c

What does this tell us about the graph that we had above where a=1,b=1,c=-1?


What if we were to look at (x-b)(y-a)=k for any integer k?

Then we would have xy=xa+by-ab+k

So when we were looking at a=1 and b=1 with varying values for c we had this equations:


Thus c=k.



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