by Courrey J. Alexander
First, letís consider the graph of the equation y=ax2+x+1. Since we are looking at graphs in the xa plane, we need to consider the graph of the equation ax2+x+1=0. This way we can identify all roots for a . Hereís the graph for the equation.
Here the equation will have one, real negative root at a=0. When a is between 0 and approximately 0.25, the equation will have two, real negative roots. When a is less that 0, then the equation will have two roots (a positive and a negative). There arenít any real roots when a is greater than 0.25.
Letís look at the graph of y=x2+bx+1 for -3£b£3.
The graph of the parabola is translated upward from the right until the axis of symmetry is at 1. Then the graph is translated downward. The parabola always passes through (0,1) on the y-axis. When b=2,-2, the equation has one real negative and one real positive root, respectively. When b<-2, the equation has two real, positive roots. When b>-2, the equation has two real, negative roots. For any value of b between 2 and –2, the equation has no real roots.
If we consider the locus of the vertices of the graph above, we can see that it forms a parabola that opens downward (y=-x2+1).