Figure 2.1

^{}

Situation 3: This situation caused me to hypothesize the conjecture that, where A is the A value from the equation. In addition, g(x) was tangent to f(x) at the point (10,10). This meant that x2 and x3 were the same (10), and showed that the conjecture held true for tangents.

Situation 6: A concave down parabola in the 1st quadrant still holds the conjecture, provided that the conjecture is changed to

Situation 7: Irrational A values work for the conjecture.

Situation 8: A concave up parabola in the third quadrant works.

Situation 9: Anthat is concave down in the third quadrant works.

Situation 10: Intersections in both 1st and 3rd quadrants work as long as the intersections of one line are x2 and x3, and the intersections of the other line are x1 and x4.^{}

After situation 10, I proved my conjecture that when the lines g(x)=x and h(x)=2x intersect the parabola, the value of D is

Using the quadratic formula , I found the intersections of lines f(x) and g(x) : and .

^{}

Using the same method, and could also be found.

Here, g(x) and h(x) were changed to see if this changed the conjecture. The situations tested appear algebraically below in Figure 3.1, and graphically in Figure 3.2

Figure 3.1

Situation 11: Here, I changed my conjecture to tentatively be, where m is from and n is from.

Situation 14: Here I attempted to determine whether the b value in affected the D value. It did not, most likely because the part of the intersection point this problem is concerned with is the x value, and with parabolas, the x value increases proportionally on both sides of the axis of symmetry.

After trying all of these parabolas, I again proved my conjecture, the only difference being that now and .

^{}

After proving this conjecture, I investigated similar situations with cubic polynomials. Shown below is the graph of, , and (Figure 4.1).

The intersections, similar to the quadratic examples, are labeled as x1, x2, x3, x4, x5, and x6 from left to right as shown.

In this case, I again used my calculator to determine the values of these intersections. Once these were calculated, the values of , , and were determined:

After rearranging these values I found that, and then tested this conjecture on other cubic functions, shown below algebraically in Figure 5.1, and graphically in Figure 5.2

Figure 5.1

Example 3 = Cubics with negative roots work with the conjecture.

Example 3 = The SM, SL, and SR values of this example were about half of those in example 1, leading me to change my conjecture to ^{}

Example 5 = Here, as in the quadratic, I changed the equation to 3x to observe its affect on the SM, SL, and SR values. When this change showed a change in the S values I formed the conjecture that.^{}

This is an interesting way to show the variety of parabolas that you considered.

Nicely done. Stating each of your situations illustrates your reasoning.

Nice job pointing out x2 and x3.

Well done. You basically came up with one of the missing pieces for D, which is, when you distribute and simplify, get x2+x3+x6 – (x1+x4+x5)

Why didn’t you get rid of the 1/A since the other factor, Sm-Sl-Sr = 0 ?