**Assignment 3: Problem 1**

**Jason P. Pickhardt**

__Problem:__

Consider the equation in the *xb* plane:

__Exploration:__

In order to graph the given equation in the *xb* plane we must
first understand what the *xb* plane is. This is the case where are *x*-values are still on the horizontal axis and the *b*-values are on the vertical axis. The
following picture is the graph of this equation in the *xb* plane. What results is a graph
of a hyperbola that has two asymptotes. We can find these asymptotes by knowing
that there are some values for which the given equation is false. This happens
when:

x = 0,

b = -x,

Now we can graph different values of *b* for the equation. The following is an example of the equation for
*b *= 3. The result is a horizontal
line that crosses the original graph at two points. These represent two real
roots for the original equation at the value *b *= 3.

We can make a conjecture from this. Notice that for values
of b from -2 < *b *< 2 there are
no real solutions. When *b* > 2 we
have two negative real solutions and when *b*
< -2 there are two positive real solutions. Let’s see what happens when *b* = -2 and *b* = 2.

Hence, when *b* = -2,
we have the positive real solution of *x*
= 1. When *b* = 2, we have a negative
real solution of *x* = -1. Another
interesting thing to notice is what happens when we have the equation:

The new graph is a hyperbola. If we were to graph any value of b we would end up with two real solutions to the equation.

__Conclusion:__

When graphing the equation in the *xb* plane the
result is a hyperbola with asymptotes of *x*
= 0 and –*x* =* b*. Also, the equation has two real solutions for all values *b* > 2 and *b* < -2 and one real solution when *b* = -2 and *b* = 2. When -2
< *b *< 2 there are no real
solutions that satisfy the equation.