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miah

  • one year ago

ohkay so I REALLY need help.

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  1. miah
    • one year ago
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    Using the direction, size and vertex of your parabola, think of a quadratic equation in general form, y = a(x - h)2 + k, that would closely match your parabola. Remember (h, k) represents the vertex of the parabola. Enter that equation in the Input bar.

  2. miah
    • one year ago
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  3. miah
    • one year ago
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    @amistre64

  4. miah
    • one year ago
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    @KyrstanMiller24

  5. KyrstanMiller24
    • one year ago
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    what grade of math are you in?

  6. miah
    • one year ago
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    12th

  7. miah
    • one year ago
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    @AirPotta

  8. miah
    • one year ago
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    @Mertsj

  9. miah
    • one year ago
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    Solving Quadratic Equations

  10. miah
    • one year ago
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    Im doing a project thing. and I was going thru the steps then I go to that one and I stopped cause I didnt know what to do

  11. miah
    • one year ago
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    @Mertsj

  12. miah
    • one year ago
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    its the pic in the attatchment

  13. miah
    • one year ago
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    If you have access to a digital camera, take a picture of a parabola existing in the real world. Upload that picture onto your computer's hard drive. If you don't have access to a digital camera, search the internet for an image of a parabola existing in the real world. Be sure to follow the guidelines and safety precautions for completing internet searches. In a new GeoGebra window, select the axes button or the grid button underneath the toolbar if you cannot see either in your GeoGebra window. Deactivate the Algebra window. Go to View in the Menu bar. Deselect Algebra. Confirm there is a check mark next to Input Bar, also under View in the Menu bar. Insert the image from your digital camera or internet search into GeoGebra using the Insert Image button . Remember that this tool might be under one of the other buttons. Select the tiny triangle in the bottom right corner of each button until you find it. Once it’s activated, insert the picture by selecting the coordinate grid. You may need to adjust the placement of your image on the screen. To do this, right-click on the image and select Object Properties… Under the Position tab, enter the coordinates of two or three corners of your picture.

  14. miah
    • one year ago
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    Select the point tool , then place a point at the vertex of the parabola in your image. Using the direction, size and vertex of your parabola, think of a quadratic equation in general form, y = a(x - h)2 + k, that would closely match your parabola. Remember (h, k) represents the vertex of the parabola. Enter that equation in the Input bar. See how closely the graphed parabola matches your image parabola. You may need to try different values of a to make the parabola fit. To change the values of the equation, right-click on the parabola, choose Object Properties…, and change the equation under the Basic tab. Close the window. Once the graphed parabola matches as close as possible to the image, make note of the quadratic equation, both in general and standard form (y = ax2 + bx + c). When completed, the graphed parabola should lie on top of the parabola in your image. If necessary, right-click on the image and go to the Color tab to make the image more transparent so the graphed parabola can be seen over the image.

  15. miah
    • one year ago
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    Identify the vertex, axis of symmetry, domain and range of the graphed parabola by inspecting your graph. Find the x- and y-intercepts of the parabola using the "Intersect Two Objects" icon . Remember this tool may be hidden below other buttons. To find the x-intercepts, select the parabola, then the x-axis. To find the y-intercepts, select the parabola, then the y-axis. Find the discriminant and explain the best method of finding the x-intercepts algebraically. Find the x-intercepts, or solutions to the quadratic equation, algebraically using each of the following methods: factoring, the quadratic formula, and completing the square. If a method cannot be used, explain why. Show all work involved with each method. Write a paragraph or two describing the picture you took. Provide details such as where it was found, how you found it, and why you chose this parabola in your activity. Send the activity to your instructor with the following information: Title (1 point) Materials Used (1 point) Procedure (1 point) Data(20 points) Include: A screenshot of GeoGebra with the picture you uploaded in the background and the graphed parabola on top. This can be done using the menu at the top of the GeoGebra screen. Go to "File," "Export," "Graphics View to Clipboard." Right-click in your document and choose "Paste." (2 points) The quadratic equation matching your picture in general form and in standard form. (2 points) Vertex, axis of symmetry, domain and range of the parabola. (4 points) The x- and y-intercepts found using GeoGebra. (2 points) A complete sentence explaining which method of solving would be best used based on the value of the discriminant. (2 points) The solutions, or x-intercepts, of the parabola found algebraically by factoring, using the quadratic formula and completing the square. If one of these methods cannot be used, a complete sentence explaining why. (3 points) 1-2 paragraphs describing the parabolic image chosen. Provide details such as where it was found, how you found it, and why you chose this parabola in your activity. (5 points)

  16. amistre64
    • one year ago
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    well, a parabola can be defined by its x intercepts, called zeroes or roots. For what x values does your parabola cross the x axis at?

  17. miah
    • one year ago
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    2,8 ?

  18. amistre64
    • one year ago
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    i cant tell that part, pictures not that detailed for me to see, |dw:1365103947488:dw|

  19. amistre64
    • one year ago
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    are those x values 2 and 8 on the picture you posted?

  20. amistre64
    • one year ago
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    you have 2 arches ... which one are you using?

  21. amistre64
    • one year ago
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    if its the front one, im guessing its a 2.5 and a 12.5 but im not sure what you are looking at

  22. miah
    • one year ago
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    the back 1

  23. amistre64
    • one year ago
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    ...why? front one seems simpler; 3 and 13, while the back one might be iffier

  24. miah
    • one year ago
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    ohkay front

  25. amistre64
    • one year ago
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    lol, do you agree that 3 and 13 look about right for those x values?

  26. miah
    • one year ago
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    yes

  27. amistre64
    • one year ago
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    we will also want to know the value of x that is in the center of those, that will be our "line of symmetry".

  28. amistre64
    • one year ago
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    what value is the center between 3 and 13?

  29. miah
    • one year ago
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    uhm ?

  30. amistre64
    • one year ago
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    if you cant find the middle between 3 and 13, then you are definantly not prepared to do quadratics ....

  31. amistre64
    • one year ago
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    we want some value such that 3+n = 13-n can you solve for n?

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