The vertex of this equation is (3, 2) and the value, a = 1. Thus, this parabola has its turning point in the first quadrant.
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Using the software ‘Fx Graph4’ I have plotted these 3 functions on a graph.
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Then I have found the four intersection points of these three functions and labeled them as X1, X2, X3, X4 respectively from left to right. (see Graph 1)
Graph 1: Value of a=1
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Then I found out the values of (X2 – X1) and (X4 – X3) and called them SL and SR respectively.
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I finally calculated the value of D = |SL – SR|
These results are illustrated in table 1.
(software used: “Microsoft Excel”)
I am now going to repeat steps 1 to 5 for different values of a, all being greater than 0 and all belonging to Z+.
Graph 2: Value of a=2
Graph 3: Value of a=3
Graph 4: Value of a=4
Graph 5 : Value of a=5
Graph 6 : Value of a=6
Table 2: Showing results obtained from a=1 to a=6
Forming my conjecture1:
I can see that the value of D is related to the value of a. Accordingly, I can form a hypothesis that D =
D(1) = = 1 → TRUE
D(2) = = 0.5 →TRUE
D(3) = = →TRUE
D(4) = = 0.25 →TRUE
D(5) = = 0.2→TRUE
D(6) = = 0.
Proof for my conjecture1
I will now prove this conjecture by using mathematical deduction.
D = , a>0 and a Z+
Let the equation of any parabola be y = ax2 + bx + c
Since this parabola intersects the lines y = x and y = 2x the points of Intersection of the parabola with the lines can be obtained from the Roots of the equation by equating ax2 + bx + c = x and ax2 + bx + c = 2x
D = |SL - SR|
= |x2 – x1 – x4 + x3|
= |() – () – () +() |
= || = || =
D= since, a > 0; ||= Hence this Conjecture is valid and has been proved for all a>0.
VALIDATION and TESTING:
Now I will test my conjecture1 for all real values of ‘a’ in quadrant 1 starting from decimal values, then irrational numbers and then negative integers and decimals.
(Note: For ALL parabolas where a R+ I have used the eqation y=a(x-3)2+2. But for all parabolas where a R- I have used the eqation y= a(x-2)2+5. This is because when a > 0 the parabola must intersect y=x hence h>k and when a < 0 the parabola must intersect y=2x and hence h<2k.)
Graph 5: Value of a=0.5
Graph 6: Value of a = 2.5
Graph 7: Value of a=
Graph 8: Value of a=
Graph 9: Value of a=-1
Graph 10: Value of a=-2
Graph 11: Value of a=-3
Graph 12: Value of a=-4
Graph 13: Value of a =-0.5
Graph 14: Value of a=-
Table 3: Showing the testing for all real values of a.
Hence I can modify my conjecture1:
D = , a R.
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Where D is defined as |(x2 - x1) – (x4 – x3)| where x1 and x4 are the points of intersection of lines y=2x and x2 and x3 are the points of intersection of lines y=x with the parabola y=a(x-h)2 + k
- h and k > 0 ( Vertex in the first quadrant)
(The proof for this modified conjecture is the same as the proof given above on page 7 where we got D = )
PARABOLAS IN OTHER QUADRANTS INTERSECTING WITH LINES y=x and y=2x.
I will now see if my conjecture holds true for parabolas having their vertex in other quadrants.
Graph 15: Value of a =-4 and turning point is in the 2nd quadrant.
Graph 16: Value of a= 2 and turning point is in the 3rd quadrant.
Graph 17: Value of a= 3 and turning point is in the 4th quadrant.
Table 4: Showing the testing of my conjecture for turning points in different quadrants.
Hence I can modify my conjecture1:
D = , a R.
-
Where D is defined as |(x2 - x1) – (x4 – x3)| where x1 and x4 are the points of intersection of lines y=2x and x2 and x3 are the points of intersection of lines y=x with the parabola y=a(x-h)2 + k
(The proof for this modified conjecture is the same as the proof given above on page 7 where we got D = and we know that the value of D only depends on the coefficient of x2 not the values of b and c. Hence the ONLY condition for my conjecture is that the parabola MUST have four intersection points with the two lines.)
ANY PARABOLA INTERSECTING WITH ANY 2 LINES
In the previous part of my portfolio I investigated the patterns formed in the intersection of two lines y=x and y=2x with any parabola and found D.
Now I will further expand my portfolio by investigating patterns formed in with any line.
Let us assume that:
- P, N are the x coefficients of straight lines y= Px + p and y=Nx + n.
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P, N
The basic quadratic equation y = would intersect the two lines at four distinct points and I have found these four points, labeling the intersections with y=Px+p as X2 and X3 and the intersections with y=Nx+n as X1 and X4. Using these four points I will again find the value of D = | (X2 – X1) – (X4 – X3) |.
D = | (X2 – X1) – (X4 – X3) |.
D = | X2 – X1 – X4 + X3|.
D =| |
D = |
D = |
D= |
This is the proof by deduction for the value of D. It holds true for the conjecture1 in the previous part of my portfolio as D = | | =
Now from this proof we can generalize that D= |, where P and N and are the x coefficients of 2 straight lines and a and is the coefficient of x2 in a parabola. The value of D is defined as | (X2-X1) - (X4-X3) | where X1 and X4 are the points of intersection of y=Nx +n with any parabola and X2 and X3 are the points of intersection of y=Px + p with the same parabola. NOTE here we have assumed P < N to define our 4 intersection points clearly and the value of D is dependent ONLY on the coefficients of x in the two lines and x2 in the parabola.
EXAMPLE:
Graph 18: The three lines are y=1-x, y=3x+2 and y=3(x-2)2-5 and are used to test my conjecture.
Table 5: Showing how D is true for any intersecting line.
D =| | = 4/3 = 1.3333 → Hence my conjecture is valid and true.
FOR CUBIC POLYNOMIALS
Before going on to see what happens with cubic polynomials I would like to define the value of D in other ways. We know that X2 and X3 are the roots of the intersection of the polynomial with the first line and X1 and X4 are the roots of the intersection of the polynomial with the second line. Thus:
D = | (X2 – X1) – (X4 – X3) |.
D = | X2 – X1 – X4 + X3|.
D = | X2 + X3 – X4 – X1|.
D = | (X2 + X3) – (X4 + X1)|.
D =| (sum of the roots of the first line) – (sum of the roots of the second line) |.
(Where the gradient of the first line is smaller than the gradient of the second)
Now for a cubic equation, we have 6 roots when it intersects with two straight lines. These six roots are X1, X2, X3, X4, X5, and X6.
We can express any cubic equation in its factored form as:
= a(x r1) (x2 r2x r3x + r2r3)
= a(x-r1) (x2 (r2+r3)x + r2r3)
= a(x3 - (r2+r3)x2 + r2r3x r1 x2 + r1(r2+r3)x r1r2r3)
= a(x3 - (r1+ r2+r3)x2 + (r1+r2 + r1r3 + r2r3)x - r1r2r3)
= ax3 - a(r1+ r2+r3)x2 + a(r1+r2 + r1r3 + r2r3)x - a(r1r2r3)
If we compare this equation to the standard cubic equation y = ax3+bx2+cx+d then:
b= - a (r1+ r2+r3)
r1+ r2+r3 =
Hence we can see the sum of the roots of any cubic equation = where ‘a’ is the coefficient of x3 and ‘b’ is the coefficient of x2 and this is proved above by mathematical deduction.
Now for our cubic equation we have: X2, X3, and X6 as the roots of the first line intersecting with the curve and X1, X4, and X5, as the roots of the second line intersecting with the curve.
=| (sum of the roots of the first line) – (sum of the roots of the second line) |
= | (X1 +X4 + X5) - (X2 +X3 + X6) |
We know that (X1 +X4 + X5) = and (X2 +X3 + X6) =
- (|
= +
=0
Thus D = 0.
We know this is true as when a straight line y= Nx+n intersects with a cubic curve y= ax3+bx2+cx+d only the value of ‘c’ and‘d’ are changed. Hence the value of D is zero for any cubic polynomial as the intersecting lines do not have an influence on the value of ‘a’ and ‘b’ that determine D.
EXAMPLE:
Graph 19 and Table6: Showing how D = 0 for a cubic polynomial.
Table 7: Showing how the value of D = 0 for a cubic polynomial.
FOR HIGER ORDER POLYNOMIALS:
- We can express any polynomial of degree ‘n’ in its factored form as:
y = ]
Where ‘a’ is the coefficient of xn and r1, r2, r3, …rn-1, and rn are all roots of the polynomial.
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This can also be expressed as (observed from the cubic equation on page 18):
f(x) = axn - a (r1+ r2+r3…rn) xn-1 +…. + (-1)n a (r1r2r3…rn)
I am now going to use mathematical induction and try and prove that the sum of the roots of a polynomial of degree n is
Let us assume that p (k) statement is true.
Now if we multiply both sides by (x - rk+1) →
Hence p (k+1) statement is true.
p(3),p(k) and p(k+1) statement is true and hence we have proved that the sum of the roots of a polynomial with degree ‘n’ is where ‘a’ is the coefficient of xn and b is the coefficient of xn-1.
Now that we know that the sum of the roots of any polynomial is we can find the value of D for any polynomial.
We already know that value of D for a parabola intersecting with two lines and for a cubic curve intersecting with two lines.
For a polynomial higher than that of degree 3, the coefficients of Xn and xn-1 always remain ‘a’ and ‘b’ respectively. The two lines y=Px+p and y=Nx+n do not have an effect on the value of ‘a’ and ‘b.
Thus the value of D is →
=| (sum of the roots of the first line) – (sum of the roots of the second line) |
We know that (sum of the roots of first line) = and (sum of the roots of the second line) =
- (|
= +
=0
Thus D = 0 for any polynomial higher than order 2.
CONCLUSION
I can conclude my portfolio by saying that there was a pattern found with the intersection of 2 lines and a polynomial. The value of D which is defined as the modulus of the sum of the roots of the first intersecting line minus the sum of the roots of the second intersecting line was found for each polynomial.
For Polynomial of degree 2 → D= |, where P and N and are the x coefficients of any 2 straight lines and a and is the coefficient of x2 in a parabola.
For a Polynomial of degree 3 or higher → D = 0.
These are the patterns that I found and proved in my portfolio. This portfolio has further scope and the intersections of a cubic curve with a parabola can also give interesting patterns.
