• 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
  17. 17
    17
  18. 18
    18
  19. 19
    19
  20. 20
    20
  21. 21
    21
  22. 22
    22
  23. 23
    23
  24. 24
    24
  25. 25
    25
  26. 26
    26
  27. 27
    27
  28. 28
    28
  29. 29
    29
  30. 30
    30
  31. 31
    31
  32. 32
    32
  33. 33
    33
  34. 34
    34
  35. 35
    35
  36. 36
    36
  37. 37
    37
  38. 38
    38
  39. 39
    39
  40. 40
    40
  41. 41
    41
  42. 42
    42
  43. 43
    43
  44. 44
    44
  45. 45
    45
  46. 46
    46

Maths HL Type 1 Portfolio Parabolas

Extracts from this document...

Introduction

Maths Portfolio 1 HL Type 1 Parabola Investigation 1. Consider the function To find the four intersections in the graph shown above using the GDC, i. Press the 2nd button and then the TRACE button to select the CALC function. Select the intersect function by pressing button 5. ii. Select the first curve of intersection and press ENTER. iii. Select the second curve of intersection and press ENTER. iv. Select the area of estimation of the intersection point and press ENTER. v. The first intersection point between Repeat the above steps i-iv, to obtain the other intersection points between . The second intersection point between The first intersection point of The second intersection point of The To find the values of , To calculate the value of , 2. To find other values of D for other parabolas of the form ,with vertices in quadrant 1, intersected by the lines Consider the parabola and the lines , The intersections of the parabola with the lines can be calculated using both the GDC and manual calculation. By manual calculation, To calculate the intersection between , Sub (2) into (1), Sub Sub The intersections between the parabola are (3,3) and (6,6). By using the GDC, To calculate the intersection between i. Key in equations of into the GDC, ii. Press the TRACE button to plot the graph on the GDC, iii. Press the 2nd button and the TRACE button to select the CALC function. Select the intersect function by pressing 5 to calculate the intersection between the parabola and the linear line. iv. Select the first curve which is the parabola by pressing ENTER and then the second curve which is the line by pressing ENTER, then estimate the location of the intersection by moving the cursor using the left and right directional buttons and then press ENTER. v. Hence, the first intersection of the parabola is . ...read more.

Middle

into (1), Let be , To find the imaginary intersection between the parabola and the line Substitute (2) into (1), Let Hence, the x-values are: Calculation of D, The conjecture still holds true when the intersections are not real numbers, for real values of a and a>0. To prove the conjecture using the general equation of , Let the roots of the general equation be . Hence, it can be deduced that, Since , where To find the x-values of intersections between the parabola and the lines Substitute (2) into (1), Hence, since the roots of the equation are The sum of the roots of the equation, i.e. Substitute (3) into (1), Hence, since the roots of the equation are The sum of the roots of the equation, i.e. , Hence, the conjecture is proven for all real values of a, . 4. Investigating the conjecture when the intersecting lines are changed. To investigate whether the conjecture still works when the intersecting lines are changed, I will be using the same parabola while varying the intersecting lines. To vary intersecting lines, the intersectings lines all follow the general equation of , where m is the gradient and c is the constant. Hence, for the two intersecting lines, I will be varying the m value and the c value. The equation of the two intersecting lines will be as follows: Line equation 1: Line equation 2: The values of will be varied. Consider the parabola and the intersecting lines of , The intersections between the parabola and the intersecting lines can then be found via Autograph software: The intersections between the parabola and the line are (-2,-8) and (3,12) The intersections between the parabola and the line are (-5.162,5.162) and (1.162,-1.162) Let the x-values of the intersections between the parabola and the line be . Let the x-values of the intersections between the parabola and the line be Hence, the x-values: Calculation of D: The conjecture does not hold when the intersecting lines are changed. ...read more.

Conclusion

To find the intersections between the cubic function and the first quadratic curve, Substitute (2) into (1), Since the roots of the equation are The sum of roots,i.e. To find the intersections between the cubic function and the first quadratic curve, Substitute (3) into (1), Since the roots of the equation are The sum of roots, i.e. Since This can be proven graphically, Consider the cubic function and the quadratic equations The x-values: Calculation of D: |3-3| Alternatively, , where Hence, the conjecture that can be made for cubic polynomials is that for all cubic polynomials that are intersected with linear lines, however, when the cubic polynomial is intersected with quadratic equations, , where 6. Consider whether the conjecture might be modifired to include higher order polynomials. To consider higher order polynomials, The general equation would be , where n is the highest degree. Alternatively, it can be written as . The roots of the equation would be Hence, The sum of roots will be as proven earlier. When the polynomial intersects with a line that is at least two degrees lower than the polynomial,i.e. , the two equations can then be equated to find the points of intersection which will be the roots of the new equation. Hence the sum of roots will then be Hence when the parabola is intersected with two linear lines, the value of D will be Therefore, the conjecture that will hold as long as the polynomial is intersected with a line that is at least two degrees lower than the polynomial. However, when the polynomial is intersected with a line that is one degree lower,i.e. Equation of polynomial: Equation of intersecting line: , The sum of roots of the new equation will be , as according to examples above. When the polynomial is intersected by two lines that are one degree lower than the polynomial, the value of D will be Hence when the higher order polynomials of degree n is intersected with lines that are one degree less, the conjecture will be . ?? ?? ?? ?? 2 ...read more.

The above preview is unformatted text

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

See related essaysSee related essays

Related International Baccalaureate Maths essays

  1. Math IA Type 1 In this task I will investigate the patterns in the ...

    There seems to be a pattern developing that D and a are inversely proportional but to affirm this, I will look at another parabola with the value of a as 8.

  2. Ib math HL portfolio parabola investigation

    This means that the parabola MUST intersect with y=x. If a parabola cuts y=x, then it also cuts y=2x. For a parabola to cut y=x at two points, its turning point should be below the line. For the turning point to be below the line the x-coordinate of the vertex should be greater than the y-coordinate. This implies h>k.

  1. Math Portfolio Type II Gold Medal heights

    on the curve of the graph must also be taken into consideration. The graph attained by this calculation does fit these points very well as the graph on the following page proves. The axis' have again been shifted to provide a better visualization of the graph.

  2. Math IA type 2. In this task I will be investigating Probabilities and investigating ...

    Then I set up column 5 as the odds function, which is defined as Since the Probability of Player C losing and the probability of Player D winning are equal I will now substitute the values into the equation to find the expression for Odds.

  1. MAths HL portfolio Type 1 - Koch snowflake

    General formula for An The general formula can be verified by putting in values from the table. Stage 4 For the diagram above I have used a software called Dr. Bill's software of the Von Koch snowflake simulation which I found online.

  2. Shadow Functions Maths IB HL Portfolio

    As we know the roots, we can determine the equation for the shadow function: With this equation, we can now find the points of intersection between and , by equating both functions. if or where Thus, the two points of intersection are and From this, we can calculate the shadow-generating

  1. Math IA Type 1 Circles. The aim of this task is to investigate ...

    Another important thing to note is that throughout this investigation, all lengths will be measured in Units, since no other unit measure was provided in the parameters for the investigation. Part I. For Part One, let r =1, and an analytic approach will be used to find OP?, when OP=2, OP = 3, and OP = 4.

  2. MATH IA: investigate the position of points in intersecting circles

    AP?=OA ?OAP is an isosceles triangle. AG is perpendicular to OP so as a result of this, AOP=AP?O. With all this being said OG=GP = OP?. Assume that OG=x, this will make GP=3-x. OAG and ?AGP are both right triangles that share the side AG. To solve for x we will use the Pythagorean triangle and

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