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Find values of D for other parabolas of the form y= ax2+bx+c, a>0, with vertices in quadrant 1, intersected by the lines y=x and y=2x. Consider various values of a, beginning with a=1. Make a conjecture about the value of D for these parabolas.
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I’m going to consider three different parabolas with vertices in quadrant 1 in the form of form y= ax2+bx+c, a>0.
y= (x−4)2+2 = x2−8x+18
y= 2x2−8x+9
y= 4x2−20x+26
To summarize, the results are listed in the chart below:
Conjecture:
The relationship of D and a looks like it should be:
D=|−1/a|
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Investigating this conjecture further for any real value of a and any placement of the vertex, I try different values. The labeling convention used in parts 1 and 2 by having the intersections of the first line to be x2 and x1 and the intersections with the second line to be x1 and x4 will be maintained.
y= −x2−2x+6
y= −x2−5x−8
y=−2x2+3x+5
The results of investigating different real values of a and placement of the vertex is shown below:
The conjecture still holds and the results of D fit the conjecture found in part 2.
Proof:
y= ax2−bx+c
y= x
ax2−(b−1)x+c=0
x= (1−b)±√((b−1)2−4ac)
2a
x2,3=(1−b)±√((b−1)2−4ac)
2a
y= ax2−bx+c
y= 2x
ax2−(b−2)x+c=0
x= (2−b)±√((b−2)2−4ac)
2a
X1,4= (2−b)±√((b−2)2−4ac)
2a
SL= x2−x1
SR= x4−x3
D=|SL−SR| = |(x2−x1) – (x4−x3)|= |(x2+x3)−(x1+x4)|= |(1–b)2/2a – (2–b)2/2a|
= 1/a|1−b−2+b|=1/a
- To prove that the conjecture will still hold with the lines are changed, I will use the same examples from part 2. Now, the two intersecting lines will be y=1.5x and y=3x.
y= x2−8x+18
y= 2x2−8x+9
y= 4x2−20x+26
Some modifications had to be made to the conjecture. The new conjecture is:
y=b1x (these are the equations of the lines)
y=b2x
D=|b2− b1| /a
So D is actually the absolute value of the difference of the slopes of the intersecting lines divided by the a of the parabola.
Proof:
ax2+(b− b1)x+c= 0
ax2+(b− b2)x+c= 0
x= −( b− b1)±√((b− b1)2−4ac)
2a
x= −( b− b2)±√((b− b2)2−4ac)
2a
D=|SL−SR| = |(x2−x1) – (x4−x3)|= |x2−x1− x4+x3|= |x2+x3−x1− x4|= |(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
- A similar conjecture can be made for cubic polynomials.
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)
From the expression for b, we can find the sum of the roots:
b= −a(x1+x2+x3)
x1+x2+x3= b/−a = −b/a
From the conjecture and proof for the parabola,
We know that:
D=|(x2+x3)−(x1+x4)|
For a cubic polynomial, we’ve found that the sum of the roots is –b/a so
D= |(−b/a)−( −b/a)|=0
- The conjecture can be modified to include higher order polynomials and it would very similar to the cubic one. For higher order polynomials, the roots will cancel out so D=0 will always be true.
anxn+an-1xn-1+…+a1x+a0= an(x−xn)(x−xn-1)…(x−x1)
anx3−an(xn+xn-1+…+x1)+ an(xnxn-1+xn-1xn-2+…+xnx1)+an(xnxn-1…x1)
xn+xn-1+…x1=(an−1)/-an
