• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month
Page
1. 1
1
2. 2
2
3. 3
3
4. 4
4
5. 5
5
6. 6
6
7. 7
7
8. 8
8
9. 9
9
10. 10
10
11. 11
11
12. 12
12
13. 13
13
14. 14
14
15. 15
15
16. 16
16

# Maths Portfolio Type I Parabola Investigation

Extracts from this document...

Introduction

Ken Chen

Consider the parabola y = (x – 3)2 + 2 = x2 – 6x + 11, which is drawn together with the lines y = x and y = 2x on the same axes.

Using the program called Qax Grapher, which has functions same as a GDC, the parabola y = (x – 3)2 + 2 and the two lines y = x and y = 2x are graphed as illustrated below.

There are 4 intersections in the axes which have the x-values labeled as x1, x2, x3 and x4 as they appear from left to right on the x-axis.

x1and x4 – the x-values of the intersections between the line y = 2x and the parabola y = x2 – 6 x + 11 on the left and right hand side of the graph respectively.

x2and x3 – the x-values of the intersections between the line y = x and the parabola y = x2 – 6 x + 11 on the left and right hand side of the graph respectively.

Using GDC, these x-values can be found.

x1 = 1.764        x2 = 2.382        x3 = 4.618        x4 = 6.236

Call SL as the difference between x1 and x2 and SR as the difference between x3 and x4.

SL = x2 – x1 = 2.382 – 1.764 = 0.618

and

SR = x4 – x3 = 6.236 – 4.618 = 1.618

D is the difference between the two differences SL and SR.

D = |SL – SR| = 1.618 – 0.618 = 1.

In order to find out the pattern of the D-value, other parabolas which have same characteristics as the parabola above (the parabolas need to be in the form y = ax2 + bx + c (a > 0) with vertices in quadrant 1) are considered.

• Consider the parabolas with a = 1

When the parabola y = (x – 4)2 + 1 = x2 – 8x + 17

Middle

→ It can be seen that when D-value is 3 when the parabolas have a = Error! Reference source not found. and there are four intersections between the parabola and the lines.

From these examples, the conjecture can be obtained.

“For any four intersections of the two lines y = x, y = 2x and the parabola ax2 + bx + c (a > 0 and its vertex is in quadrant 1); the value of D which is calculated by D = |(x2 – x1) – (x4 – x3)| equals to Error! Reference source not found..”

Testing the Validity of the Statement

To test for the validity of the conjecture above, other parabolas which have a as a real number and have vertices place in any quadrants are investigated.

Firstly, the pattern in intersections of any parabolas that have real value of a and the lines y = x and y = 2x is investigated out to see whether the conjecture above still holds true.

• a = - 4 (a Error! Reference source not found. Z)

When the parabola y = - 4x2 + 32x - 45 and the lines y = x and y = 2x are graphed, four values of x are obtained.

x1 = 1.934        x2 = 2.073        x3 = 5.427        x4 = 5.816

SL = x2 – x1 = 2.073 – 1.934 = 0.139

SR = x4 – x3 = 5.816 – 5.427 = 0.389

D = |SL – SR| = 0.389 – 0.139 = 0.25 = Error! Reference source not found.

When the parabola y = - ½ x2 + 3x + 5 and the lines y = x and y = 2x are graphed, four values of x are obtained.

x1 = -2.317        x2 = -1.742        x3 = 5.742        x4 = 4.317

SL = x2 – x1 = -1.742 + 2.317 = 0.575

SR = x4

Conclusion

Consider y = x4 – 6x3 + 9x2 – 2. There are 8 intersections.

x1 = -0.377        x2 = 0.726        x3 = 2.000        x4 = 3.651        x5 = -0.343        x6 = 1.000

x7 = 1.529        x8 = 3.814

x1, x2, x3, x4 are the x-values of the intersections of the quadric polynomial and the y = x

x5, x6, x7, x8 are the x-values of the intersections of the quadric polynomial and the y = 2x

D = (x1 + x2 + x3 + x4) - (x5 + x6 + x7 + x8) = 6 – 6 = 0.

• Polynomial of order 5

Consider y = 2x5 – 6x4 – 2x3 + 7x2 + 2x – 1. There are 10 intersections.

x1 = -0.914        x2 = -0.484        x3 = 0.340        x4 = 1.139        x5 = 2.919        x6 = -1.000

x7 = -0.384        x8 = 0.442        x9 = 1.000        x10 = 2.942

x1, x2, x3, x4, x5 are the x-values of the intersections of the quadric polynomial and the y = x

x6, x7, x8, x9, x10 are the x-values of the intersections of the quadric polynomial and the y = 2x

D = (x1 + x2 + x3 + x4 + x5) - (x6 + x7 + x8 + x9 + x10) = 3 – 3 = 0

The conjecture can be made: “The difference between the sum of all the x-values of the intersections of the polynomial and the line y = x and the sum of all those of the intersections of the polynomial the lines y = 2x equals to 0”

Justification

The justification is same as the justification for cubic polynomial. According to the Viete’s theorem, it can be seen that the sum of all the roots is independent of the coefficient of x. So the sum always equals to Error! Reference source not found. which is a constant for any intersections of the polynomial (with order higher than 3) and any changing lines.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

## Found what you're looking for?

• Start learning 29% faster today
• 150,000+ documents available
• Just £6.99 a month

Not the one? Search for your essay title...
• Join over 1.2 million students every month
• Accelerate your learning by 29%
• Unlimited access from just £6.99 per month

# Related International Baccalaureate Maths essays

1. ## Investigating Parabolas

y = x2-4x+5 Process - x2 - x1 = 1.39 - 1 = 0.39 = SL - x4 - x3 = 5 - 3.61 = 1.39 = SR - So, the ?SL - SR ? = ?0.39 - 1.39? = ?-1?= 1, in other words, D = 1.

2. ## Ib math SL portfolio parabola investigation

The intersections, similar to the quadratic examples, are labeled as x1, x2, x3, x4, x5, and x6 from left to right as shown. In this case, I again used my calculator to determine the values of these intersections. Once these were calculated, the values of , , and were determined:

1. ## Ib math HL portfolio parabola investigation

= = 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

2. ## Parabola investigation. In this task, we will investigate the patterns in the intersections of ...

Secondly, we consider the parabola, the lines and. By using the manual calculation, we can calculate the four intersections. Calculate the intersections between f(x) and h(x): (1) (2) Substitute (2) into (1): or . Using calculator to obtain the approximate values of x or * Substitute into (2): (x, y)

1. ## Shady Areas. In this investigation you will attempt to find a rule to approximate ...

= 0.4 ?10 (2/(x2 + 4)) dx = (1/8) (0.5 + 0.4 + 2[0.4923076923+ 0.4705882353+ 0.4383561644] = 0.462813023 Calculator: math-> 9. fnInt ( enter 2/(x2 + 4) ,X ,1 ,3 ) ENTER OR Y= enter y = 2/(x2 + 4) 2nd TRACE 7 ENTER graph displayed 0 ENTER 1 S ENTER S ?f(x)dx = 0.463647609 4.

2. ## Math Investigation - Properties of Quartics

= 0 Solving this with the quadratic formula gives us the two remaining roots. Roots of the function are therefore written below: P: = Q: = 5 R: = 3 S: = As it is a straight line, we can directly find out the ratios of x values.

1. ## Analysis of Functions. The factors of decreasing and decreasing intervals (in the y ...

Power functions are not periodic because a periodic function repeats function values after regular intervals. It is defined as a function for which f(x+a) = f(x), where T is the period of the function. In the case of power functions, it can't be a periodic function is that there is no definite or constant period like "a".

2. ## Music and Maths Investigation. Sine waves and harmony on the piano.

We can begin with the first term of the harmonic series, 1. The largest power of less than or equal to 1 is 0. Hence, the first term of our new series is 1. The second term of the harmonic series is .

• Over 160,000 pieces
of student written work
• Annotated by
experienced teachers
• Ideas and feedback to