Jackie Gammaro

Assignment 3 - Graphs in the xb plane

If we graph y = x² + bx + 1, from b = -3, -2, -1, 0, 1, 2, 3, the following picture is obtained.

We can see from this picture, the y axis is a line of reflection for opposite values of b.

Also, clearly the y-intercept is -1 for all equations.

The chart below summarizes the nature of the roots for varying values of b.

 b Number of Roots Nature of Roots Vertex -3 two real, irrational (1.5, -1.25) -2 one real, rational, x = 1 (1,0) -1 two complex, imaginary (0.5, 0.75) 0 one complex, imaginary, x = i (0,1) 1 two complex, imgainary (-0.5 , 0.75) 2 one real, rational, x = -1 (-1,0) 3 two real, irrational (-1.5, -1.25)

To determine the locus of vertices, one should notice the curve appears to be a parabola.

Step 1: To determine the equation of the quadratic curve, I chose three vertices (1,0), (0,1) and (-1,0) along the desired path.

Step 2: Using these three verices and the equation y = ax²+ bx + c, one obtain the resulting equations:

0 = a + b + c

1 = c

0 = a - b + c

Step 3: Thus letting c = 1, the two resulting equations are:

0 = a + b + 1

0 = a - b + 1

Step 4: Thus adding these two equations, you result with a value of a = -1. If a = -1, thus b = 0.

Step 5: One obtains the equation y = -x²+1 for the locus of vertices.

The following graphic shows the path of the parabolas from b = -3 to 3 and it's locus of vertices: y = -x²+1.

Now consider the xb plane. In this case we let y = b.

The following is the graph of x² + yx + 1 = 0 and the intersection of it with graphs of y =b, from b = -3 to 3

If we look at the table above for different values of b, we can see a pattern in the number of intersections each line makes with the equation 0 = x² + yx + 1

 b Number of Roots Nature of Roots Number of Intersection y = b with equation 0 = x² + yx + 1 -3 two real, irrational two -2 one real, rational, x = 1 one (1, -2) -1 two complex, imaginary none 0 one complex, imaginary, x = i none 1 two complex, imgainary none 2 one real, rational, x = -1 one (-1, 2) 3 two real, irrational two

We notice when b = -2, the root is 1, which is also the x value for the intersection of y = -2 with the equation 0 = x² + yx + 1.

We notice when b = 2, the root is -1, which is also the x value for the intersection of y = 2 with the equation 0 = x² + yx + 1.

We notice when b = -1, 0, or 1, the roots are imaginary and y = b does not intersect with the equation 0 = x² + yx + 1.

As we look to b = -3 and 3 one notices there are two roots for the original y = x² + bx + 1. I conjecture that the value of the roots for each value of b will be the x value of the intersection of the graph y = b and 0 = x² + yx + 1. When solving for the roots of the equation y = x² -3x + 1, using the quadratic formula, one obtains the values x = (3 ±√5)/2, and for y = x² +3x + 1, one obtains the values

x = (-3 ±√5)/2, which are the values of the intersection with the graph y = b.

Part 2 - Now lets consider the case when c = -1, thus y = x² + bx - 1, from b = -3 to 3.

Notice all graphs have two real roots.

It appears that the locus of vertices is now y = -x² -1. The following is a the graph of y = x² + bx -1 moving along the graph y = -x² -1.

When looking in the xb plane, we obtain the following graph. One can see the points of intersections 0 = x² + yx -1 and the graph of

y =b are the roots of the equation y = x² + bx -1.

Part 3 - Graph other values of c on the same axes.

Thus one notices when b = 1, the locus of vertices for the equation y = x² + x + c, is now a line, x = -½.

Thus holding b constant, and now changing c makes the locus of vertices a line.