And Y= x²+4x+5 (a=1, b= 4, c=5)
The Vertex of the first quadratic lies in the First Quadrant
You can complete the square to convert ax2 + bx + c to vertex form, but it's simpler to just use a formula (derived from the completing-the-square process) to find the vertex.
For a given quadratic y = ax2 + bx + c, the vertex (h, k) is found by computing h = –b/2a, and then evaluating y at h to find k where k = (4ac – b2) / 4a.
So the vertex for the first one is is h = =
K = =
So the Vertex is (2,1)
The Vertex of the second quadratic does not lie in the first quadrant
Y= x²+4x+5 (a=1, b= 4, c=5)
h = =
K = =
That the vertex is not in the first quadrant can also be proved by using technology (autograph) which is shown in the graph below:
3. Investigate your conjecture for any real value of a and any placement of the vertex. Refine your conjecture as necessary and prove it. Maintain the labelling convention used in parts 1 and 2 by having the intersections of the first line to be and , and the intersections with the second line to be and .
This Question implies that I have to investigate and modify my conjecture for any real value for “a” and any placement of the vertex. This means that I am going to change not only “a”, but also the variables “b” and “c” in the parabola. The lines of y = x and y = 2x are still going to be the same. Similarly, I will note the intersections of those lines with the parabola and also calculate D again.
I used several parabolas to derive to a modified conjecture to the case of a vertex in different quadrants.
Y=−x²+4x+5
= {-1,449}
= {-1.193}
= {4,193}
= {3.449}
D= - =( - ) - (- ) = 0.256 – (-0.744)= 1
= -1 D-1
My previously conjecture has to be modified from = D
To = D
In this case it would be = -1 = 1 = D
The second parabola I used in the form of y=ax²+bx+c is
y=-13x²-4x+7
= {-1}
= {-0,9509}
= {0,5663}
= {0,5385}
D= - =( - ) - (- ) = 0.0491 – (-0.0278)= 0.0769
= = 0.0769 = D
Y = log(6)x² + 3x – 2
= -2.3697
= -3.3398
= 0.7696
= 1.0846
D = - =( - ) - (- ) = | (-0.9701) – (0.3150) | = | (-1.2851) | = 1.2851
D= = = 1.2851
Y = 5x² - 3x – 2
= -0.3062
= -0.3483
= 1.1483
= 1.3062
D = - =( - ) - (- ) = | (-0.0421) – (0.1579) | = | -0.2000 | = 0.2
D= = = 0.2
Y = -18x² + 5x +6
= -0.5
= -0.4768
= 0.6991
= 0.6667
D = - =( - ) - (- ) = | 0.0232 – (-0.0324) | = | -0.0556 | = 0.0556
D= = = 0.0556 regarding only 4 significant figures.
My modified conjecture works so far which is shown in table of results below:
To prove my conjecture I picked a random parabola:
y=−√12 (x²)−12x+2
= -4.1796
=-3.9008
= 0.14802
= 0.13814
D = - =( - ) - (- ) = | 0.2788 – (-0.0988) | = | 0.28868 | = 0.28868
D= = = 0.28868
Hereby my conjecture is proved.
4. Does your conjecture hold if the lines are changed? Modify your conjecture if necessary and prove it.
The questions asks if there is a change in the conjecture if the two lines y=x and y=2x are changed. Therefore, I am going to change the lines and see if any modification to the conjecture has to be made. In the beginning I will keep the parabola the same and only change the lines. Afterwards, I will change both the lines and the conjecture. If necessary, I will modify my conjecture and at the end prove it.
The Parabola for the first two example is going to stay the same:
Y= x²-6x+11
I changed the lines to y=x +2 and y= 3x
Now I am going to use my conjecture from question 3 and check for validity:
=1.4586
=1.6972
=5.3028
=7.5414
D = - =( - ) - (- ) = | 0.2386 – (2.2386) | = | -2 | = 2
D= = = 0.5
My conjecture is not working if the lines are changed !!!
Now I am trying to modify my conjecture by analysing patterns which are occurring when the Parabola and/or the lines are changed.
Second Example
The Parabola is: y = (x-5)2+2
The lines that I chose are:
y=x
y=0.5x
The x-coordinates of these intersections as they appear from left to right on the x-axis are: x1, x2, x3, and x4, respectively.
The x-values as they appear on the graph are as follows:
x1 = 3.6972
x2 = 4.5
x3 = 6
x4 = 7.3028
The values of x2 - x1 is named SL and the values of x4 – x3 is named SR.
SL = x2 - x1 = 4.5 - 3.6972 = 0.8028
SR = x4 – x3 = 7.3028 – 6 = 1.3028
D = | SL - SR| = |0.8028 - 1.3028| = 0.5
Third Example
The Parabola is: y = (x-5)2+2
The lines that I chose are:
y=x+1
y=x-1
The x-coordinates of these intersections as they appear from left to right on the x-axis are: x1, x2, x3, and x4, respectively.
The x-values as they appear on the graph are as follows:
x1 = 3.4384
x2 = 4
x3 = 7
x4 = 7.5616
The values of x2 - x1 is named SL and the values of x4 – x3 is named SR.
SL = x2 - x1 = 4 - 3.4384 = 0.5616
SR = x4 – x3 = 7.5616 – 7 = 0.5616
D = | SL - SR| = |0.5616 - 0.5616| = 0
Fourth Example
The Parabola is: y = (x-5)2+2
The lines that I chose are:
y=2x+1
y=2x-1
The x-coordinates of these intersections as they appear from left to right on the x-axis are: x1, x2, x3, and x4, respectively.
The x-values as they appear on the graph are as follows:
x1 = 2.8377
x2 = 3.1716
x3 = 8.8284
x4 = 9.1623
The values of x2 - x1 is named SL and the values of x4 – x3 is named SR.
SL = x2 - x1 = 3.1716 - 2.8377 = 0.3339
SR = x4 – x3 = 9.1623 - 8.8284 = 0.3339
D = | SL - SR| = |0.3339 - 0.3339| = 0
After now having investigated four different parabolas; I found a relation that might not seem obvious in the first place but if you realize it than you might get the hint.
I analyzed the patterns and I asked myself what actually a line is and what its components are. So I figured y=x + d is actually not everything there is this one part that I did not take in consideration in the conclusion before. This can best be explained by using a general parabola: y=a(x+b)2+c and general lines: y=d1x+e and y=d2x+f. None of the values of “b”, “c”, “e” or “f” have any impact on the value of ”D”.
The conjecture which I made from the previous Question was: D=1/|a|
However, this conjecture is not valid anymore as the lines are changed and therefore has to be modified: The critical values which have to be paid attention on are: “d1”, “d2” and “a” because they determine the gradient of the lines and the steepness of the parabola.
Through calculating and experimenting with these values I found out that the new conjecture will have to be:
D = |d1 – d2|/|a|
In order to prove this I will use the formula to see if my value for “D” equals the value I got from graphing the parabolas
5. Determine whether a similar conjecture can be made for cubicpolynomials.
This question wants me to create a similar conjecture for cubic polynomials. Therefore I am going to keep the lines constant for the first few trials and only change the cubic polynomial and then I will change both the lines and the cubic polynomial. The difficulty hereby is that the cubic polynomial cuts the two lines more than four times. Therefore list all the intersections and try to find some kind of a pattern by considering both the consecutive terms and the ones following it.
In order to be able to find a conjecture, I am going to keep the cubic polynomial as simple as possible.
First Example
Y=(x-1)(x-3)(x-5)
Y=x and y = 0.5 x
The x-coordinates of these intersections as they appear from left to right on the x-axis are: x1, x2, x3, x4, x5 and x6 respectively.
The x-values as they appear on the graph are as follows:
x1 = 1.0706
x2 = 1.1658
x3 = 2.3434
x4 = 2.6578
x5 = 5.2716
x6 = 5.4909
The values of x2 - x1 is named SL and the values of x4 – x3 is named SR.
SL = x2 - x1 = 1.1658 - 1.0706 = 0.0952
SR = x4 – x3 = 2.6578 - 2.3434 = 0.3144
D = | SL - SR| = |0.0952 - 0.3144| = 0.2192
Basically I said that “D” is the first 4 Terms of the polynomial !
However, I am also interested in seeing how the last two points of intersections are related to “D”, therefore I subtracted them from each other and found out that the last point of intersections subtracted from the point of intersections just before is also equal to D.
x6 - x5 = 5.4909 - 5.2716 = 0.2193
D = x6 - x5 = 5.4909 - 5.2716 = 0.2193
Therefore I can generalize my findings to:
D = | x6 - x5 | = | x4 – x3| - | x2 - x1|
Second Example
Y=2(x-1)(x-3)(x-5)
Y=x and Y=0.5x
The x-coordinates of these intersections as they appear from left to right on the x-axis are: x1, x2, x3, x4, x5 and x6 respectively.
The x-values as they appear on the graph are as follows:
x1 = 1.0331
x2 = 1.0706
x3 = 2.6578
x4 = 2.8222
x5 = 5.1447
x6 = 5.2716
The values of x2 - x1 is named SL and the values of x4 – x3 is named SR.
SL = x2 - x1 = 1.0706 - 1.0331 = 0.0375
SR = x4 – x3 = 2.8222 - 2.6578 = 0.1644
D = | SL - SR| = |0.0375 - 0.1644| = 0.1269
Again, I want to see whether x6 - x5 gives me the same value as D, therefore I calculated:
x6 - x5 = 5.2716 - 5.1447 = 0.1269
As it can be seen, my findings have been repeated to be true:
D = | x6 - x5 | = | x4 – x3| - | x2 - x1|
Third Example
Y=4(x-1)(x-3)(x-5)
Y=x and Y=0.5x
The x-coordinates of these intersections as they appear from left to right on the x-axis are: x1, x2, x3, x4, x5 and x6 respectively.
The x-values as they appear on the graph are as follows:
x1 = 1.0161
x2 = 1.0331
x3 = 2.8222
x4 = 2.9089
x5 = 5.075
x6 = 5.1447
The values of x2 - x1 is named SL and the values of x4 – x3 is named SR.
SL = x2 - x1 = 1.0331 - 1.0161 = 0.017
SR = x4 – x3 = 2.9089 - 2.8222 = 0.0867
D = | SL - SR| = |0.017 - 0.0867| = 0.0697
Again, I want to see whether x6 - x5 gives me the same value as D, therefore I calculated:
x6 - x5 = 5.1447 – 5.075 = 0.0697
As it can be seen, my findings have been repeated to be true:
D = | x6 - x5 | = | x4 – x3| - | x2 - x1|
Fourth Example
Y=(x-2)(x-4)(x-7)
Y=x and Y=0.5x
The x-coordinates of these intersections as they appear from left to right on the x-axis are: x1, x2, x3, x4, x5 and x6 respectively.
The x-values as they appear on the graph are as follows:
x1 = 2.1148
x2 = 2.2813
x3 = 3.316
x4 = 3.67
x5 = 7.2152
x6 = 7.4027
The values of x2 - x1 is named SL and the values of x4 – x3 is named SR.
SL = x2 - x1 = 2.2813 - 2.1148 = 0.1665
SR = x4 – x3 = 3.67 - 3.316 = 0.354
D = | SL - SR| = |0.1665 - 0.354| = 0.1875
Again, I want to see whether x6 - x5 gives me the same value as D, therefore I calculated:
x6 - x5 = 7.4027 - 7.2152 = 0.1875
As it can be see, my findings has been repeated to be true:
D = | x6 - x5 | = | x4 – x3| - | x2 - x1|
Now I am going to change both the lines and the cubic polynomial
Fifth Example
The polynomial is: Y=3(x-1)(x-5)(x-7)
The lines are: Y=2x and Y=0.5x
The x-coordinates of these intersections as they appear from left to right on the x-axis are: x1, x2, x3, x4, x5 and x6 respectively.
The x-values as they appear on the graph are as follows:
x1 = 1.007
x2 = 1.0289
x3 = 4.6399
x4 = 4.9003
x5 = 7.0927
x6 = 7.3312
The values of x2 - x1 is named SL and the values of x4 – x3 is named SR.
SL = x2 - x1 = 1.0289 - 1.007 = 0.0219
SR = x4 – x3 = 4.9003 - 4.6399 = 0.2604
D = | SL - SR| = |0.0219 - 0.2604| = 0.2385
Again, I want to see whether x6 - x5 gives me the same value as D, therefore I calculated:
x6 - x5 = 7.3312 - 7.0927 = 0.2385
As it can be see, my findings has been repeated to be true:
D = | x6 - x5 | = | x4 – x3| - | x2 - x1|
Conclusion
To find a conjecture I realized that D = | x6 - x5 | but how is this related to the principle of trigonometry so
First of all, I wanted to find the gradient of the curve:
Gradient of the curve (a) = rise/run = y6 – y5 / x5 – x6 = y6 – y5 / D
y6 and y5 are the y-coordinates of the points of intersection x6 and x5 respectively-
In order to now find “D” I am going to rearrange the equation to:
D = y6 – y5 / a
Now I want to have a look at the gradients of the lines intersecting the polynomial:
m1 = y5 / x5 So: y5 = m1 * x5
m2 = y6 / x6 So: y6 = m2 * x6
Furthermore it can be said that x5 and x6 have about the same value and so I want to set them equal, in order to make a simpler conjecture. Therefore x6 = x5 = w
Now I am going back to my original equation and insert my new findings:
D = ((d2 * w)– (d1 * w)) / a
D = (w*(d2-d1)) / a
D = (d2 – d1) / (a/w)
I found my conjecture to be the following:
D = |d2-d1| / |2a|
In order to see if my conjecture is valid, I am going to create a table to check my results:
My value of “D” unfourtanetly is only close to the real value of my derived conjecture but still this can give you an approximate idea of the concept of cubicpoynomials.
6. Consider whether the conjecture might be modified to include higher order polynomials.
I am trying to find a conjecture for 4th degree polynomials first using the GDC.
First example
Polynomial: y = (x-1)(x-3)(x-4)(x-5)
The Lines: y = x-6 and y = 0.5x-6
The x-coordinates of these intersections as they appear from left to right on the x-axis are: x1, x2, x3, and x4, respectively.
The x-values as they appear on the graph are as follows:
x1 = 1.2679
x2 = 1.3236
x3 = 2.1896
x4 = 2.382
The values of x2 - x1 is named SL and the values of x4 – x3 is named SR.
SL = x2 - x1 = 1.3236 - 1.2679 = 0.0557
SR = x4 – x3 = 2.382 - 2.1896 = 0.1924
D = | SL - SR| = |0.0557 - 0.1924| = 0.1367
Second Example
Polynomial: y = 2(x-1)(x-3)(x-4)(x-5)
Lines: y= 4x – 19 and 3x - 12
The x-coordinates of these intersections as they appear from left to right on the x-axis are: x1, x2, x3, and x4, respectively.
The x-values as they appear on the graph are as follows:
x1 = 1.2236
x2 = 1.4902
x3 = 2.1389
x4 = 2.6102
The values of x2 - x1 is named SL and the values of x4 – x3 is named SR.
SL = x2 - x1 = 1.4902 - 1.2236 = 0.2666
SR = x4 – x3 = 2.6102 - 2.1389 = 0.4713
D = | SL - SR| = |0.2666 - 0.4713| = 0.2047
Third Example
Polynomial: y = (x-1)(x-3)(x-4)(x-5)
Lines: y= 4x – 19 and 3x - 12
The x-coordinates of these intersections as they appear from left to right on the x-axis are: x1, x2, x3, and x4, respectively.
The x-values as they appear on the graph are as follows:
x1 = 1.1285
x2 = 1.4009
x3 = 2.2886
x4 = 3
The values of x2 - x1 is named SL and the values of x4 – x3 is named SR.
SL = x2 - x1 = 1.4009 - 1.1285 = 0.2724
SR = x4 – x3 = 3 - 2.2886 = 0.7114
D = | SL - SR| = |0.2724 - 0.7114| = 0.439
Fourth Example
Polynomial: y = (x-1)(x-3)(x-4)(x-5)(x-6)
Lines: y = 3x and y= 5x
The x-coordinates of these intersections as they appear from left to right on the x-axis are: x1, x2, x3, and x4, respectively.
The x-values as they appear on the graph are as follows:
x1 = 1.0266
x2 = 1.0463
x3 = 2.4175
x4 = 2.5821
The values of x2 - x1 is named SL and the values of x4 – x3 is named SR.
SL = x2 - x1 = 1.0463 - 1.0266 = 0.0197
SR = x4 – x3 = 2.5821 - 2.4175 = 0.1646
D = | SL - SR| = |0.0197 - 0.1646| = 0.1449
Conclusion
Each time the order of the polynomial has changed, the conjecture has changed as well. Therefore I assume that the power plays an important role in the conjecture.
The conjecture that I found in question 5 was: D = |d2-d1| / |2a|
At that case, the polynomial was to the power of 3. Therefore is the coefficient of a is (power-1).
Therefore my new general conjecture for any polynomial is: D = |d2-d1| / |(power-1)*a|,
Where d1 and d2 are the gradients of the lines intersecting the polynomial and a is the gradient of the polynomial and the power is the order of the polynomial.
Now I want to summarize my findings form question 6 and check whether my conjecture is true.
Again the calculated value for D did not come out to be exactly the same as the value for D that I got from drawing the polynomials and intersecting it with the lines. This is due to the fact that the gradient of the curve does not stay constant and so is only approximated. Therefore, my conjecture can only come out to be close, but not exactly the same.
Conclusion
After now having investigated in the patterns of the intersections of parabolas with two lines, I want to conclude that the general conjecture for any polynomial is:
D = (d2-d1) / ((power-1)*a), where d1 and d2 are the gradients of the lines intersecting the polynomial and a is the gradient of the polynomial and the power is the order of the polynomial.
The difficulties in this Project were that it was impossible to find an exact conjecture for D. It was always only an approximate. This is due to the fact that the gradient of the line, a, is approximated and changed throughout the whole curve. Therefore the calculated values for D will always vary slightly from the actual value for D.
However, the conjecture that I found comes out to be very close.
Parabola Investigation
Patrick Vollmer
Math Higher Level
