= 0.18
X4-X3 = SR
= 2.85-2.33
= 0.52
D= 0.52 – 0.18
=0.33
X1= 0.75
X2= 1
X3= 1.5
X4 =2
SL = 0.25
SR= 0.50
D= 0.50 – 0.25
= 0.25
Evaluation
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For a = 1 ,D value is equal to 1.00
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For a = 2 , D value is equal to 0.50
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For a= 3 , D value is equal to 0.33
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For a= 4 , D value is equal to 0.25
If the pattern of the D values are examined it can be concluded that it is directly proportional with 1/a
3) Investigate your conjecture for any real value of a , and any placement of the vertex. Refine your conjecture as necessary, and prove it.
At part two , I have examined the a values grater than 1. But at this part as a can take any reel value I will be examining the a values less then 0 to prove whether my conjecture is true or not.
For a = -1
Assume that b= 8, c = -7
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x1 – x2 = 1.27 – 1 = 0.27 = SL
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x3 – x4 = 6 - 4.73 = 1.27 = SR
D =1.27-0.27
=1
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The conjecture I made previously was D= 1 /a. However it doesn’t seem to work for a= -1 as 1/a is equal to -1 in this case but D is equal to 1. So, I will test this conjecture with another reel value of “a”.
- As I have to test my conjecture for any placement of vertex, I will be testing it for a parabola with a vertex in the 2nd quadrant:
X1:-4.5
X2: -5.12
X3:-0.88
X4: -1
SL= -5.12-(-4.5) = -0.62
SR= -1-(-0.88) = -0.12
D= -0.12-(-0.62) =0.50
According to my conjecture, D value found is not correct as it must have been 1/a which is -0.50 in this case. It can be concluded that there is a problem with the sign of a in my conjecture as despite “a”’s having a negative sign D is always positive . If I modify my conjecture as D = 1/ lal the problem will be solved.
4)Does your conjecture hold if the intersecting lines are changed? Modify your conjecture, if necessary , and prove it.
-
Previously I have made a conjecture about the D values of a parabolas intersection of lines y=x and y=2x. To test my cojecture I have worked with several parabolas and modified my conjecture accordingly as D = 1/ lal . To generalize this conjecture I must test it’s validity by changing the intersecting lines.
- Let’s test the conjecture for y= 3x and y=4x
I will continue working on my first parabola which was x2 -6x + 11=y
X1 : 1.26
X2: 1.46
1.46-1.26 = 0.2 = SL
X3: 7.54
X4: 8.74
8.74 - 7.54 = 1.2 = SR
D= 1.2- 0.2 = 1 = 1/ lal
- As it can be seen above, even tough I have changed the intersecting lines into y=3x and y= 4x, my conjecture still worked.
-
To be sure, I will test my conjecture for the intersecting lines y=5x , y=6x and the parabola 5x2 -5x + 4
X1: 0.46
X2: 0.55
0.55- 0.46 = 0.09 = SL
X3: 1.45
X4 : 1.74
1.74-1.45 = 0.29 = SR
D = 0.29-0.09 = 0.20
As 1/ 5 is equal to 0.20. I proved my conjecture to be working with different intersecting lines than y=x and y=2x.
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Both two trials were made with intersecting lines that have 1 difference between their slopes. Now I will test my conjecture for the intersecting lines y= 2x and y= 4x and the parabola x2 -6x + 11.
X2: 1.76
1.76- 1.26 =0.5= SL
X3: 6.24
X4: 8.72
8.74 – 6.24 = 2.5 = SR
D = 2.5- 0.5 = 2.0
- According to my conjecture D must have been equal to 1 but it is 2 in this case . When I worked with lines which had one difference in their slopes, my conjecture worked, however it doesn’t seem to be working with the lines y= 2x and y=4x. It is obvious that I must modify my conjecture in order to involve slope differences of the lines.
M2- M1 = 4-2 = 2
Interestingly I found D value to be 2 .
I think if I modify my conjecture as
D= (m2 – m1) / lal it will become more accurate.
Conjecture :
y=b 1x (these are the equations of the lines)
y=b 2 x
D=|b2-b1| /| a|
Proof:
ax2+(b− b1)x+c= 0
ax2+(b− b2)x+c= 0
x2,3= −( b− b1)±√((b− b1)2−4ac)/ 2a
x1,4= −( b− b2)±√((b− b2)2−4ac)/2a
D=|SL−SR| = |(x2−x1) – (x4−x3)|= |x2−x1− x4+x3| = |(x2+x3)−(x1+x4)|
x2+x3=[−(b− b1)/2a]2= −(b− b1)/a
x1+x4=[−(b− b2)/2a]2= −(b− b2)/a
D=|(−(b− b1)/a)−( −(b− b2)/a)|= |b2− b1|/| a|
To test my new conjecture I will work on two other intersecting lines which have negative slope values.
-
y= -3x , y= -6x, y = x2 +5x + 8
X1: -0.78
X2: -1.17
SL : -0.39
X3: -6.83
X4 : -10.22
SR : -3.39
l-3.39+0.39l =3.00
(m2 – m1) = 6-3 = 3
A =1
3/1 = 3.
So ,the conjecture is proved
5)Determine whether a similar conjecture can be made for cubic polinomials
According to the fundamental theorem of algebra:
ax3+bx2+cx+d= a(x−x1)(x−x2)(x−x3)
(x−x1)(x−x2)(x−x3)= x2−xx1−xx2+x1x2 (xx3)
=x3−x2x1−x2x2−xx1x2− x2x3+xx1x3+xx2x3−x1x2x3
= a(x3−(x1+x2+x3)x2+(x1x2+x2x3+x1x3)x−(x1x2x3))
= ax3−a(x1+x2+x3)x2+a(x1x2+x2x3+x1x3)x−a(x1x2x3)
From the proof we can see what each of the coefficient equals:
a=a
b= −a(x1+x2+x3)
c= a(x1x2+x2x3+x1x3)
d= a(x1x2x3)
The sum of the roots can be obtained by b:
b= −a(x1+x2+x3)
x1+x2+x3= b/−a = −b/a
According to the formula
D=|(x2+x3)−(x1+x4)|
The sum of the roots for a cubic polynomial is –b/a so
D= |(−b/a)−( −b/a)|=0
6)The conjecture obtained by the cubic polynomials can be applied to higher order polinomials as the roots will cancel out so D=0 will always be true.
Conclusion
- By this investigation, I had a chance to apply the fundamental formulae we have learned in lesson and this, I think played a very important role in making my knowledge about parabolas more permenant. I also learned how to make a conjecture and how to empower it by several different methods. This investigation , I think improved my ability to think critically and independently while providing a practical base for the theoretical knowledge I have learned in the lesson.
