- Level: International Baccalaureate
- Subject: Maths
- Word count: 815
Ib math SL portfolio parabola investigation
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Introduction
___________
IBHL 2 Math
Parabola Investigation
In this investigation, relationships between the points of intersection of a parabola and two different lines were examined.^{[a]}
First, the parabola and the lines and were used as an example.
As seen on the adjacent graph, the points of intersection were labeled left to right as x1, x2, x3, and x4.
Using a graphing calculator, these values were found:
x1=1.764, x2=2.382, x3=4.618, and x4=6.234.
At this point, x1 was subtracted from x2, and x3 from x4, and the resulting numbers were labeled SL and SR respectively:
After this, a value D was found:
To further investigate this relationship, I followed this same process with many different parabolas and lines.
Middle
^{[b]}
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.^{[c]}
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 .
^{[d]}
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 | x | 3x | 2 | |
12 | 2x | 4x | 1.2 | |
13 | 4x | 3.6 | ||
14 | 2x | 1 | ||
15 | 1 |
Conclusion
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 ^{[f]}
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.^{[g]}
[a]Adequate, but not ideal.
[b]This is an interesting way to show the variety of parabolas that you considered.
[c]Nicely done. Stating each of your situations illustrates your reasoning.
[d]Nice job pointing out x2 and x3.
[e]Great job!
[f]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 ?
[g]Conclusion!
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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