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idontgetconicsections

  • 2 years ago

Select all of the quadrants that the parabola whose equation is y=sqrt(x-4) occupies. I II III IV So the basic rules are x is always positive to the right so: x=ay^2 opens to the right x is always negative to the left; so x=-ay^2 opens to the left My problem is: I got rid of the radical by squaring, which gave me y^2 = x-4, then I subtracted x and y^2 from both sides to get -x=-y^2-4. Am I supposed to multiply all numbers by -1 to get rid of the negative numbers or do I just leave it like that?

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  1. idontgetconicsections
    • 2 years ago
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    @amistre64

  2. idontgetconicsections
    • 2 years ago
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    |dw:1342621312995:dw|

  3. mukushla
    • 2 years ago
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    multiply by -1 u will have \( x=y^2+4 \) can u draw it?

  4. idontgetconicsections
    • 2 years ago
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    |dw:1342621412614:dw|

  5. idontgetconicsections
    • 2 years ago
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    Or is it only in quadrant 1?

  6. mukushla
    • 2 years ago
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    u should move it to right

  7. idontgetconicsections
    • 2 years ago
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    |dw:1342621507272:dw|

  8. slaaibak
    • 2 years ago
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    Well, the answer of a square root is nonnegative.. so definitely can't be 3 or 4

  9. idontgetconicsections
    • 2 years ago
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    Like that?

  10. idontgetconicsections
    • 2 years ago
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    Thanks

  11. slaaibak
    • 2 years ago
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    No it's not

  12. idontgetconicsections
    • 2 years ago
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    So it's I and IV

  13. idontgetconicsections
    • 2 years ago
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    Why?

  14. idontgetconicsections
    • 2 years ago
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    Why is it wrong slaaiback?

  15. slaaibak
    • 2 years ago
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    I just told you. "Well, the answer of a square root is nonnegative.. so definitely can't be 3 or 4"

  16. idontgetconicsections
    • 2 years ago
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    It's only I then?

  17. slaaibak
    • 2 years ago
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    |dw:1342621638765:dw|

  18. slaaibak
    • 2 years ago
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    Yes. Only 1

  19. idontgetconicsections
    • 2 years ago
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    Uhmm... why doesn't it look like a parabola?

  20. slaaibak
    • 2 years ago
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    because I freehanded the graph. Just to illustrate where it goes. You can make it parabola-like yourself

  21. idontgetconicsections
    • 2 years ago
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    Haha. Well, I have another question: Score: Select all of the quadrants that the parabola whose equation is y = x² - 4 occupies I chose all quadrants... is this correct or is it only 1 and 2 because it's nonnegative?

  22. mukushla
    • 2 years ago
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    sorry u must consider ristrictions for x ,y from \( y=\sqrt{x-4} \) u have y>=0 and x>=4

  23. slaaibak
    • 2 years ago
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    It's all of them. x^2 is nonnegative yes, but x^2 - 4 is not always nonnegative

  24. idontgetconicsections
    • 2 years ago
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    This is really confusing.. which quadrants does x = y² + 1 occupy? I chose 1 because it's nonnegtaive, but 1 is not negative? So... ?

  25. idontgetconicsections
    • 2 years ago
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    Oh wait, that's a square root, so it's quadrant 1 and 4??

  26. myko
    • 2 years ago
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    omg, it's 1 and 4. The quadrants where x is positive

  27. slaaibak
    • 2 years ago
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    ye 1 and four looks correct to me.

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