**Kristin Ottofy**

Assignment 3

Investigation 1:

Consider again the equation

Now graph this relation in the xb plane. We get this graph.

Similarly we can overlay multiple graphs for the constant term, c.

Here we have graphs for c = 1, 3, 5, or 7.

If we take any particular value of b, say b = 5, and overlay this equation on the graph we add a line parallel to the x-axis. If it intersects the curve in the xb plane the intersection points correspond to the roots of the original equation for that value of b. We have the following graph.

For each value of b we select, we get a horizontal line. It is clear on a single graph that we get two negative real roots of the original equation when b > 2, one negative real root when b = 2, no real roots for -2 < b < 2, One positive real root when b = -2, and two positive real roots when b < -2.

Consider the case when c = - 1 rather than + 1.

Graph other values of

con the same axes.

We have a family of hyperbolas.

The equation for c = 0 will yield the asymptotes, x = -b and x = 0

Each horizontal line representing a value of b will cross each hyperbola 0, 1, or 2 times.

Exploration:

If I let b=y, then I can graph as follows:

When c = 0, the asymptote for the hyberbola forms. This is also written as

x

^{2}+ yx = 0 where y = -x^{2}/x = -x. So, when c = 1, a linear line is formed. Using the original notation, we can say b = -x when c = 0.

Below, we can see the horizontal line formed by the line y = 2.

The line, y=2, intersects x

^{2}+ yx + 1 = 0 in one place where c = 1. It also intersects when c=0 in one place.

Well, the solution to the system of equations {y = 2 and x

^{2}+ yx + 1 = 0} is as follows:x

^{2}+ 2x + 1 = 0(x+1)(x+1) = 0

So, x = -1.

Thus, giving a root at x = -1.

Similarly, we can find where y = 2 and x

^{2}+ yx - 1 = 0 intersect atand .

Here is a graph of when y=3:

We can calculate that two intersections occur for x

^{2}+ yx + 2 = 0 and y = 3 at(-2, 3) and (-1, 3). The midpoint between these two points is at (-(3/2), 3).

Two intersections also form for x

^{2}+ yx - 2 = 0 and y = 2 atThe midpoint between these two points occurs at (-(3/2), 3)).

The midpoint between the intersections can be generalized as (-(b/2), b).

Thus, the midpoint can be found at x = -(y/2).

This happens to be the same line as 2x + y = 0 by:

(-1)(x) = (-(y/2))(-1)

-x = y/2

(2)(-x) = (y/2)(2)

-2x = y

-2x+2x = y + 2x

0 = y + 2x = 2x + y.

By Investigation 2, the vertex of the hyperbola when c > 0 lies on the line

2x + y = 0. This is a linear equation which can be easily found by finding the vertices of a hyperbola and then calculating the slope.

We can set 2x + y = x

^{2}+ yx + c to find an intersection:

2x - x

^{2}- c = yx - y2x - x

^{2}- c = y(x - 1)y = (2x - x

^{2}- c) / (x - 1)

Now I plug that back into 2x+y=0:

2x + ((2x - x

^{2}- c) / (x - 1))=0

This gives me the x value of the point of intersection.

If I pick an equation such as b = y= p, where p is any real number greater than the y value of the vertex of the hyperbola, then I should get 2 real value solutions and the line 2x+y=0 should intersect halfway between the 2 real values.

Example:

Let c = 1,

Then, I can let y = 3 to find the two real values and the intersection with 2x+y=0.

If I plug y = 3 into 2x+y=0, I get 2x+3=0, 2x=-3,

x=-3/2.

Now what are the two real values? The intersection of

where c = 1 with y=3, plugging in y=3 gives

.

The intersection of

where c = 1 and y=3, gives

.

Thus, x = -3/2 is equidistant between the 2 real values.