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 yintercept 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.