Parabola Investigation

Type 1- Math Portfolio

Parabola Investigation

Introduction

Parabolas are a conic sections , they are formed by the intersection of a right circular conical surface and a plane parallel to a generating straight line of the surface. The parabola has many important applications like for example in the field of Physics, engineering and in many other areas. They are used from automobile headlight reflectors to the design of ballistic missiles. In physics, it is best known as the trajectory path in projectile motion. It is also used in the body in motion under the influence of a uniform gravitational field without air resistance. Another instance in which parabolic motion occurs in nature is in two-body orbits, for example a body under the influence of the gravity of the sun. Shapes of parabola are also found in main cables on a typical suspension bridge

Equation: x2-x+11, y=x and y=2x

x1=1.764

x2= 2.382

x3=4.618

x4=6.236

x2-x1= 0.618 = SL

x4-x3 = 1.618 = SR

D = | SL-SR|

= |0.618 – 1.618|

=|-1|

= 1

Equation 2: 2x2-14x+27, y=x, y=2x

x1=2.419

x2= 3

x3=4.5

x4=5.581

x2-x1= 0.581 = SL

x4-x3 = 1.081 = SR

D = | SL-SR|

= |0.581 – 1.081|

=|-0.5|

= 0.5

Equation 2: 3x2-12x+13, y=x, y=2x

x1=1.279

x2= 1.566

x3=2.768

x4=3.387

x2-x1= 0.287 = SL

x4-x3 = 0.619 = SR

D = | SL-SR|

= |0.287 – 0.619|

=|-0.33|

= 0.33

Equation 3: 4x2-22x+29, y=x, y=2x

x1=1.677

x2= 1.867

x3=3.883

x4=4.323

x2-x1= 0.19 = SL

x4-x3 = 0.44 = SR

D = | SL-SR|

= |0.44 – 0.19|

=|-0.25|

= 0.25

Equation 4: 5x2-26x+35, y=x, y=2x

x1=1.883

x2= 2.161

x3=3.239

x4=3.717

x2-x1= 0.278 = SL

x4-x3 = 0.478 = SR

D = | SL-SR|

= |0.278 – 0.478|

=|-0.2|

= 0.2

Equation 5 :6x2-30x+39, y=x, y=2x

x1=2.167

x2= 1.885

x3=3

x4=3.448

x2-x1= 0.282 = SL

x4-x3 = 0.448 = SR

D = | SL-SR|

= |0.282 – 0.448|

=|-0.166|

= 0.166

Equation

Value of ‘a'

Value of D

x2-6x+11

1

1

2x2-14x+27

2

0.5

3x2-12x+13

3

0.33

4x2-22x+29

4

0.25

5x2-26x+35

5

0.2

6x2-30x+39

6

0.166

We can see from all the above worked examples and from the above table that D varies inversely with a. So the conjecture is

Equation 1 : x2+8x-18, y=x, y=2x

x1=-9

x2= -8.196

x3=2.196

x4=2

x2-x1= 0.804= SL

x4-x3 = -0.196 = SR

D = | SL-SR|

= |0.804 – (-0.196)|

=|1|

= 1

Equation 2: 2x2+6x-5, y=x, y=2x

x1=-3.266

x2= -2.871

x3=0.871

x4=0.766

x2-x1= 0.395= SL

x4-x3 = -0.105 = SR

D = | SL-SR|

= |0.395 – (-0.105)|

=|0.5|

= 0.5

Equation 3: -3x2-6x+22, y=x, y=2x

x1=-4.115

x2= -4.352

x3=1.685

x4=1.782

x2-x1= -0.237= SL

x4-x3 = 0.097= SR

D = | SL-SR|

= |-0.237 – 0.097|

=|-0.33|

= 0.33

Equation 4: 4x2+18x+11, y=x, y=2x

x1=-3.454

x2= -3.118

x3=-0.882

x4=-0.796

x2-x1= 0.336= SL

x4-x3 = 0.086 = SR

D = | SL-SR|

= |0.336 – 0.086|

=|0.25|

= 0.25

Equation 5: -5x2-26x+35, y=x, y=2x

x1=-6.480

x2= -6.652

x3=1.052

x4=1.080

x2-x1= -0.172= SL

x4-x3 = 0.028 = SR

D = | SL-SR|

= |-0.172 –0.028|

=|-0.2|

= 0.2

Equation

Value of ‘a'

Value of D

X2+8x-18

1

1

2x2-14x+27

2

0.5

-3x2-6x+22

-3

0.33

4x2+18x+11

4

0.25

-5x2-26x+35

-5

0.2

From the above table we can observe that D still varies with the inverse of ‘a'

So,

Proof for the above conjecture is as follows

The roots of a quadratic equation are given by

When a equation of the form ax2+bx+c is intersected by the line y=x, the points of intersection are can be determined using algebraic methods.

y= ax2+bx+c

y=x

After subtracting both the above equations we get,

ax2+(b-1)x+c=0

Therefore, the roots of the above equation will be the intersecting points for the line and the parabola

The points are labelled from left to right, so,

and

Similarly, the intersecting points of the parabola and the second line that is x1 and x4 are found as follows

Y= ax2+bx+c

Y=2x

Subtracting both the equations we get,

ax2+(b-2)x+c=0

Therefore, the roots of the above equation will be the intersecting points

The points are labelled from left to right

and

SL =x2-x1

=

SR=x4-x3

y= x2-6x+11

y= x

y=3x

x1=1.459

x2=2.382

x3=4.618

x4=7.541

x2-x1= 0.923= SL

x4-x3 = 2.923 = SR

D = | SL-SR|

= |0.923 – 2.923|

=|-2|

= 2

Equation 2:

y= x2-6x+11

y= x

y=4x

x1=1.258

x2=2.382

x3=4.618

x4=8.742

x2-x1= 1.124= SL

x4-x3 = 4.124 = SR

D = | SL-SR|

= |1.124 – 4.124|

=|-3|

= 3

Equation 3:

y= x2-6x+11

y= x

y=5x

x1=1.113

x2=2.382

x3=4.618

x4=9.887

x2-x1= 1.269= SL

x4-x3 = 5.269 = SR

D = | SL-SR|

= |1.269 – 5.269|

=|-4|

= 4

From the above examples we can see that the D contradicts the conjecture.

The proof for the above conjecture is as follows,

y= mx+n and y=px+q intersect the parabola with equation in the form ax2+bx+c

y= ax2+bx+c

y= mx+n

Subtracting both the equations, we get,

ax2 +(b-m)x (c-n) = 0

The solution to the above equation is the intersecting points of the line and the parabola

Similarly, the points of intersection between the quadratic equation and the line y = px+q can be found out by,

0= ax2+(b-p)x +(c-q)

Using the quadratic formula, we determine the points of intersection

, ,, and

Unlike a quadratic polynomial that has 4 intersecting points a cubic polynomial has 6 intersecting points. Therefore, the D for a cubic polynomial can be defined

Let us find D for the below equation( for the same intersecting lines)

Equation 1: y=x3-3x2-x+3, y=x, y=2x

x1 = -1.128

x2 =-1.262

x3 = 0.660

x4 =0.798

x5 =3.602

x6 =3.330

x2-x1 = -0.134

x4-x3=0.138

x6-x5=-0.272

D = |(-0.134)-(0.138)-(-0.272)|

D = |0|

D=0

Equation 2: y=2x3-3x2-3x+2, y=x, y=2x

x1 = -1.111

x2 =-1.220

x3 = 0.345

x4 =0.409

x5 =2.375

x6 =2.202

x2-x1 = -0.109

x4-x3=0.064

x6-x5=-0.173

D = |(-0.109)-(0.064)-(-0.173)|

D = |0|

D=0

From the above examples, we see that in all cases the D=0

Let us see how the value of D will change if the order of the degree is greater than 3

Let the polynomial equation be y= anxn + an-1xn-1+…+a1x+a0

Let the the lines y=cx+d & y=mx+n

If R1, R2,…,Rn are the roots of anxn + an-1xn-1+…+a1x+a0=0, then

R1+R2+R3+…+Rn = -an-1/an

R1.R2+R1.R3+…+Rn-1.Rn= an-2/an

.

.

.

R1.R2.R3…Rn = (-1)n.a0/an

Now, after the introduction of the 2 lines, y= cx+d and y=mx+n

SL =sum of the roots of anxn + an-1xn-1+…+a1x+a0 - (cx+d) = 0 ------------Equation (i)

SR = sum of the roots of anxn + an-1xn-1+…+a1x+a0 - (mx+n) = 0 -----------Equation (ii)

From the above, we can write,

anxn + an-1xn-1+…++(a1 - c)x+(a0-d) = 0-----------Equation (i)

anxn + an-1xn-1+…++(a1 - m)x+(a0-n) = 0----------Equation (ii)

From above, we can write

SL= -an-1/an and also SR= -an-1/an,

Because the roots are not affected by the introduction of the lines cx+d and mx+n

Hence D= SL - SR =0

Therefore, for any higher order polynomial , D =0

Let's consider the polynomial equation y= anxn + an-1xn-1+…+a1x+a0 and the lines y=lx+m & y=px+q

We know that if R1, R2,…,Rn are the roots of anxn + an-1xn-1+…+a1x+a0=0, then

R1+R2+R3+…+Rn = -an-1/an

R1.R2+R1.R3+…+Rn-1.Rn= an-2/an

.

.

.

R1.R2.R3…Rn = (-1)n.a0/an

Here we have:

SL =sum of the roots of anxn + an-1xn-1+…+a1x+a0 - (lx+m) = 0 (call this equation as Eq. 1) and

SR = sum of the roots of anxn + an-1xn-1+…+a1x+a0 - (lx+m) = 0 (call this equation as Eq. 2)

We can re write:

(1) as anxn + an-1xn-1+…++(a1 - l)x+(a0-m) = 0 &

(2) as anxn + an-1xn-1+…++(a1 - p)x+(a0-q) = 0

From the principles explained above,

SL= -an-1/an and also SR= -an-1/an, i.e., they are unaffected by the introduction of lx+m or px+q, as compared to the sum of the roots of the original polynomial.

Hence D= SL - SR =0

Thus the conclusion is that D is 0 for all polynomials of order higher than 2.

Conclusion:

From the above investigation, I can conclude the following:

1. The value of D varies inversely with |a| for the same intersecting lines, that is =, y=x and y=2x

2. The value of D becomes different for the polynomial when it is intersected by different intersecting lines. The formula that is derived in

3. The value of D of a polynomial of degree greater than 2 is always zero

Please be aware that the free essay that you were just reading was not written by us. This essay, and all of the others available to view on the website, were provided to us by students in exchange for services that we offer. This relationship helps our students to get an even better deal while also contributing to the biggest free essay resource in the UK!