Frater Elijah

It has now become a standard exercise, with the availability of certain technologies, to construct graphs of the equation:

Drawing several graphs of the function f(x) = ax² + bx + c for different values of the parameters a, b, and c, can be used to illustrate the different effects of these parameters. The graphs being drawn while holding two of these parameters constant, and varying the third. From these resulting graphs one can explore any patterns that may arrise between these graphs and roots of the equation ax² + bx + c = 0.

For example, if we set b = {-3, -2, -1, 0, 1, 2, 3} while keeping a = 1 = c and overlay these graphs, then this corresponds to the following equation and picture for the mentioned parameters:

From this picture we can discuss the behavior of a parabola as the parameter b changes. The parabola always passes through the same point on the y-axis, this is called the y -intercept and correpsonds to the point (0, 1) of the equation f*(x) above.

For b < -2 it can be seen that the parabola will intersect the x-axis in two places on the positive side of the x-axis, that is, the original equation will have two positive real roots. Similarly, for b > 2 the parabola will intersect the x-axis twice on the negative side, that is, the original equation will have two negative real roots.

For b = -2 the graph of f*(x) is tangent to the x-axis (intersects at one point only) on the positive side and thus f*(x) has only one positive real root. When b = 2 this corresponds to one negative real root as seen in the picture (the graph is tangent to the x-axis in one point on the negative side). For -2 < b < 2, f*(x) does not intersect the x-axis at all, and so there do not exist any real roots for f*(x) [note: there exist solutions in the complex plane]. Now let us consider the following. Consider the locus of vertices of the set of parabola’s graphed above. That is define a set

P_1* = {vertices of f*(x)| f*(x) = x² + bx +1 and b = -3, -2, -1, 0, 1, 2, 3}

We have the following picture, which corresponds to P_1*:

When overlapped with f*(x) we have:

Notice that P_1* passes through the vertices of f*(x) as it should.

If we let b tend to positive and negative infinity we will arrive at the following set of points.

P₁ = {vertices of f(x)| f(x) = x² + bx +1 and b R}

Now this corresponds to the equation of:

g(x) = -x² + 1

The negative sign will reflect the parabola y = x² over the x-axis and the +1 will shift the graph 1 unit up in the positive y direction. The vertices can be found as discussed in X-ploration #2. Try some other variations and see what comparisons you can make.

Let us start with the equation x² + bx + 1 = 0 now graph this in relation to the xb plane. To do this we need only solve for b (similar to solving for y) and then graph the function with our x-axis and then a b-axis in place of y. Look at the following simplification:

x² + bx + 1 = 0

ý bx = - (x² + 1)

ý b = - (x² + 1)/x

If we take any particular value of b, say b = 3 and overlay this equation on the graph we add a line parallel to the x-axis.

If this line intersects the curve in the xb plane, the intersection points correspond to the roots of the original equation for that value of b. It is clear from the above graph that we will have two negative real roots (negative side of x-axis).

Let us check this using the quadratic formula for finding roots, x = [-b±(b^2-4ac)^.5]/2a. This gives us

x₁ = (-3 + (5)^.5)/2 which is about -.381966 and x₂ = (-3 - (5)^.5)/2 which is about -2.61803

for the particular case when b =3.

In the picture below these roots are illustrated by where the vertical lines intersect the graph.

For each different b we select, we get a horizontal line. It is clear also from the above graph that we arrive at the following correspondences for various b.

Let us look at a variance of the above case. Let c = -1 rather than +1. This gives rise to the following picture (it is graphed on top of the case when c = +1 to see the relationship).

Using the quadratic formula first (just to check) we arrive at the following roots,

x = (-3±(13)^.5)/2. Now is this where the horizontal line of b = 3 intersects? Let’s see…

The x roots correspond to x = .302776 and x = -3.3.278

We got it going now!

So we have it that graphs of ax² + bx + c = 0 in the xb plane, gives rise to the roots of the particular quadratic for that particular b (when plotting the horizontal line with b).

Let us start with the following example: x² + 5x + c = 0

ý c = - ( x² + 5x)

Now graphing this in the xc plane gives rise to a parabola (of course!)

For each value of c considered, it’s graph will be a line crossing the parabola in 0, 1, or 2 points, with the intersections being at the roots of the original equation at that value of c. The following graph corresponds to c=1, that is, x² + 5x + 1 = 0.

With the roots corresponding to x Å -.208712 and x Å -4.79129. Now by looking at the graph we can see that the equation will have only one real root at c = 6.25. This corresponds also to x = 2.5 (check this!):

For c > 6.25 the equation will not have any real roots and for c < 6.25 the original equation will have two roots. Look on the graph and ask yourself when these roots will be positive and when they will be negative (or both).

For 0 < c < 6.25, both roots will be negative, and for c < 0 one root will be positive and the other root will be negative. Try more and see what you can come up with.

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