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anonymous

  • 5 years ago

find the x intercept of a parabola with vertex (3,-2) and y intercept (0,7)

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  1. amistre64
    • 5 years ago
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    (y+2) = (x-3)^2 +7 y = (x-3)^2 +5

  2. amistre64
    • 5 years ago
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    1,-6,14 3 +- sqrt(36 - 4(14))/2 ... aint see it happening

  3. amistre64
    • 5 years ago
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    yes I do lol.... its right there

  4. anonymous
    • 5 years ago
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    write it in this form (x1,y1)(x2,y2)

  5. amistre64
    • 5 years ago
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    (y+2) = (x-3)^2 +C y = (x-3)^2 +C -2 C-2 = 7; C =9 ........................................ y = (x-3)^2 +9 y = x^2 -6x +9 +9; ug, scratch that y = x^2 -6x +7 .............................. 6 +- sqrt(36 -28) sqrt(8)...2sqrt(2) ---------------- = 2 (3 + sqrt(2), 0) (3 - sqrt(2), 0) maybe lol

  6. anonymous
    • 5 years ago
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    \[y=a(x-3)^2-2\] and \[7=a(0-3)^2+2\] \[7=9a+2\] \[5=9a\] \[a=\frac{5}{9}\]

  7. anonymous
    • 5 years ago
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    yes?

  8. anonymous
    • 5 years ago
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    giving \[y=\frac{5}{9}(x-3)^2-2\]

  9. anonymous
    • 5 years ago
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    in the (x1,y1) and (x2,y2) form please, i need two intercepts

  10. anonymous
    • 5 years ago
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    ok we have the equation for the parabola. finding x- intercepts means setting y = 0 and solve for x.

  11. amistre64
    • 5 years ago
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    x^2 -6x + 7 is good

  12. anonymous
    • 5 years ago
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    in other words love \[0=\frac{5}{9}(x-3)^2-2\]

  13. amistre64
    • 5 years ago
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  14. amistre64
    • 5 years ago
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    i used 0,7 ; 3,-2 ; 6,7 lol

  15. anonymous
    • 5 years ago
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    in the (x1,y1) and (x2,y2) form

  16. anonymous
    • 5 years ago
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    please

  17. anonymous
    • 5 years ago
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    well x-intercept means \[(x_1,0)\]

  18. amistre64
    • 5 years ago
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    already did; (3-sqrt(2),0) and (3+sqrt(2),0)

  19. anonymous
    • 5 years ago
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    what equation did you use?

  20. amistre64
    • 5 years ago
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    x^2 -6x +7

  21. anonymous
    • 5 years ago
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    (y-k)=a(x-h)^2

  22. amistre64
    • 5 years ago
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    (y-7)+9 = (x-3)^2 (y+2) = (x-3)^2

  23. anonymous
    • 5 years ago
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    boy am i dumb. i wrote \[y=a(x-3)^2+2\] when it should have been \[y=a(x-3)^2-2\] and \[7=a(-3)^2-2\] \[7=9a-2\] \[9=9a\] \[a=1\]

  24. amistre64
    • 5 years ago
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    :) its ok lol

  25. anonymous
    • 5 years ago
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    so equation is \[y=(x-3)^2-2\]

  26. anonymous
    • 5 years ago
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    idk how to change it to the two points on the graph, please type it that way

  27. anonymous
    • 5 years ago
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    or if you prefer \[y+2=(x-3)^2\]

  28. amistre64
    • 5 years ago
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    i posted the points 2 times already...

  29. amistre64
    • 5 years ago
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    (3+sqrt(2),0) (3-sqrt(2),0) right?

  30. anonymous
    • 5 years ago
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    zeros at \[3\pm \sqrt{2}\]

  31. anonymous
    • 5 years ago
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    with the parenthesis? and two different points?

  32. anonymous
    • 5 years ago
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    with the parenthesis? and two different points?

  33. anonymous
    • 5 years ago
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    by inspection

  34. amistre64
    • 5 years ago
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    now that 3 and four times lol

  35. anonymous
    • 5 years ago
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    \[(2 -\sqrt{3},0)\] \[(2+\sqrt{3},0)\]

  36. amistre64
    • 5 years ago
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    sat lol; that was backwards

  37. anonymous
    • 5 years ago
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    are the x-intercepts sorry i slowed you up

  38. anonymous
    • 5 years ago
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    oh yes it was! \[(3-\sqrt{2},0)\] \[(3+\sqrt{2},0)\]

  39. anonymous
    • 5 years ago
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    please in the form of (x1,y1) and (x2,y2) idk what (2-sqrt of 3, 0( means...

  40. anonymous
    • 5 years ago
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    means \[x_1=3-\sqrt{2}\] \[y_1=0\]

  41. amistre64
    • 5 years ago
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    it means the the sqrt(2) is an EXACT form of an irrational number that can only be expressed approximately in decimal form; of which we have no way of determining the requirements of since you have given us no parameters to establish them by

  42. anonymous
    • 5 years ago
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    and \[x_2=3+\sqrt{2}\] \[y_2=0\]

  43. amistre64
    • 5 years ago
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    y = 0 exactly at: x = 3-sqrt(2) AND x= 3-sqrt(2) ; otherwise it doesnt...

  44. amistre64
    • 5 years ago
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    lol... make one of those a +sqrt(2)

  45. anonymous
    • 5 years ago
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    whats x1 and y1?

  46. amistre64
    • 5 years ago
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    x1 = 3-sqrt(2) y1 = 0

  47. anonymous
    • 5 years ago
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    its wrong

  48. amistre64
    • 5 years ago
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    x1 = 1.414213562373095048801688724209.... no, its right; you simply havent given us a means to approximate the answer.

  49. amistre64
    • 5 years ago
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    its like saying; what color is the sky? blue. wrong, its a shade of blue that is determined by whatever the criteria is for saying 'blue' in a way that is acceptable...

  50. amistre64
    • 5 years ago
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    how old am I ? the right answer depends on how accurate of an answer you want right?

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