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habbster Group Title

PARABOLA how to find the vertex and points y^2=9x

  • one year ago
  • one year ago

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  1. terenzreignz Group Title
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    Isolate x first.

    • one year ago
  2. habbster Group Title
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    done with that x=(1/9)y^2

    • one year ago
  3. terenzreignz Group Title
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    Okay, now you have your quadratic equation in standard form \[ay^{2}+by+c\] Now the vertex (h,k) is given by \[\huge h=-\frac{b}{2a}\]\[\huge k=\frac{4ac-b^{2}}{4a}\] Alternatively, you can just find h, and then evaluate the function at y = h.

    • one year ago
  4. habbster Group Title
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    example please I still dont get it

    • one year ago
  5. terenzreignz Group Title
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    Well, I'm used to them being functions of x, so let's use y = f(x) for now, ok? It's practically the same for y, anyway... so suppose y = x² + 4x + 8 then for this case your a = 1 b = 4 c = 8 Plugging them in your formula gives you h = -2 k = 4 So the vertex of this parabola is at (-2 , 4)

    • one year ago
  6. habbster Group Title
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    \[h=-0/\frac{ 2}{ 9 }\] i come up with this

    • one year ago
  7. terenzreignz Group Title
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    Yes, indeed, and finally, you can see that h = ?

    • one year ago
  8. habbster Group Title
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    h=0 ?

    • one year ago
  9. terenzreignz Group Title
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    Correct. Now you can either compute for k, or just evaluate x when y = 0.

    • one year ago
  10. habbster Group Title
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    x is also 0?

    • one year ago
  11. terenzreignz Group Title
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    @habbster SOMETHING REALLY IMPORTANT When you have x², then after computing for h and k, your vertex is at (h,k) BUT when you have a y², then after computing for h and k, your vertex is at (k,h) DON'T FORGET!

    • one year ago
  12. terenzreignz Group Title
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    And yes, x = 0

    • one year ago
  13. habbster Group Title
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    oh so if the y is the one with a power the vertex is (k,h)

    • one year ago
  14. terenzreignz Group Title
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    I had to give you the more difficult, yet general formula to prepare you for more complicated parabolas in the future. However, now, you must remember that whenever your quadratic equation is of the form y = kx² or x = ky² in other words, b = 0 and c = 0 Then your vertex is always, ALWAYS at (0,0)

    • one year ago
  15. terenzreignz Group Title
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    And yes, you're right.

    • one year ago
  16. habbster Group Title
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    how about the points? where the parabola will pass thru?

    • one year ago
  17. terenzreignz Group Title
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    Well, just give a random value for y, say m, and solve for x(m). (which is x evaluated when y = m) in that case, the point (x(m), m) is a point on the parabola. You can choose any real number value for y.

    • one year ago
  18. terenzreignz Group Title
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    For example, I like the number 6. Then x = (1/9)y² x = (1/9)(6)² x = (1/9)(36) x = 4 Hence, the point (4,6) is on the parabola.

    • one year ago
  19. habbster Group Title
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    if i replace the value of y is equal to 1 then the value of x is 1/9?

    • one year ago
  20. terenzreignz Group Title
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    That is correct, and hence, the point (1/9 , 1) is on the parabola as well.

    • one year ago
  21. habbster Group Title
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    ok i get it thank you!! for your time :)

    • one year ago
  22. terenzreignz Group Title
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    No problem :) Terence out ...

    • one year ago
  23. habbster Group Title
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    w8

    • one year ago
  24. habbster Group Title
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    @terenzreignz what is the focus and directx?

    • one year ago
  25. terenzreignz Group Title
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    I'm really sorry I left too soon... Another form for a parabola is this equation: \[\huge (x - x_{o})^{2} = 4p(y - y_{o})\]or alternatively \[\huge (y - y_{o})^{2} = 4p(x - x_{o})\] First express the equation in this form... (this can usually be done by completing the square, etc...)

    • one year ago
  26. terenzreignz Group Title
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    That p-value there is key. Anyway, your equation, written like this, takes the form \[\huge (y - 0)^{2} = 4\left( \frac{9}{4} \right)(x - 0)\]or just simply\[\huge y^{2}=4\left( \frac{9}{4} \right)x\] You can see that the value for p is (9/4). Then since this is a horizontal parabola (you have a y²), then your focus and directrix are both (9/4) units away from your vertex along the horizontal. While that may not sound too appealing, it simply means you take the x-value of your vertex, and consider adding and subtracting (9/4) from it. Since your x-value is 0, then adding (9/4) gives you (9/4) and subtracting (9/4) gives you -(9/4). So, copy the y value, you get two points: \[\large \left( \frac{9}{4},0 \right) \ and \ \left( -\frac{9}{4},0 \right)\]. One of these is your focus.

    • one year ago
  27. terenzreignz Group Title
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    To find out which one, it is necessary to find out in which direction your parabola opens. This particular parabola opens to the right, as the coefficient of y² is positive. This means that your focus is to the RIGHT of your vertex, hence, \[\left( \frac{9}{4},0 \right)\]is your focus.

    • one year ago
  28. terenzreignz Group Title
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    What about the directrix? It is the line parallel to the y axis passing through the other point we got earlier. So you can see that it is the line \[x = -\frac{9}{4}\]and now you're done.

    • one year ago
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