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anonymous

  • 5 years ago

Find the quadratic equation from the following data The intercepts are (-3,0) and (0,18)

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  1. myininaya
    • 5 years ago
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    quadratic has the form f(x)=ax^2+bx+c right?

  2. myininaya
    • 5 years ago
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    we are given that those two points lie on the quatric so we have f(-3)=a(-3)^2+b(-3)-c=0 and we have f(0)=c=18 so f(-3)=9a-3b-18=0

  3. myininaya
    • 5 years ago
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    so we have 9a-3b=18 or 3a-b=6 if we divide both sides by 3

  4. myininaya
    • 5 years ago
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    but we have two unknowns so now i'm stuck maybe we should try the form f(x)=a(x-h)^2+k

  5. anonymous
    • 5 years ago
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    I don't know much, but if I am correct, the equation must be of the form ay^2+by+c=0

  6. anonymous
    • 5 years ago
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    As you see there is only one x intercept

  7. myininaya
    • 5 years ago
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    you can use either form

  8. anonymous
    • 5 years ago
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    I don't think so, as you see while you use the x form, you are putting the value of x in the equation, and that I think will not work

  9. anonymous
    • 5 years ago
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    Though I am not sure

  10. myininaya
    • 5 years ago
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    we know the y-intercept which is (0,18) so we have f(x)=a(x-h)^2+18 now we need to find h and a

  11. myininaya
    • 5 years ago
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    0=f(-3)=a(0-h)^2+18=ah^2+18 so we have ah^2=-18

  12. myininaya
    • 5 years ago
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    so either way we still have two unknowns

  13. myininaya
    • 5 years ago
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    oh i see what you are saying no y=f(x) not x=y

  14. anonymous
    • 5 years ago
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    Now you got it

  15. myininaya
    • 5 years ago
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    its f(x)=ax^2+bx+c or you can use f(x)=a(x-h)^2+k

  16. myininaya
    • 5 years ago
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    we don't know anything else about the parabola?

  17. anonymous
    • 5 years ago
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    I beg your pardon?

  18. anonymous
    • 5 years ago
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    If you find any solution please post it here, I will have to leave now, as there is a black out here

  19. myininaya
    • 5 years ago
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    i was just asking if we know anything else about the parabola?

  20. myininaya
    • 5 years ago
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    go lokisan!

  21. anonymous
    • 5 years ago
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    Iam, I may be able to look at it a bit later. My first impressions of it are to find the slope of the line that joins the two points, use that to perform a rotation back to the x-axis, solve the equation normally and then transform back. Or you could transform the general equation by the same angle and solve straight from the points. Like I said, this is a cursory glance. I don't have time available at the moment to work through it properly, but I will when I can...promise.

  22. anonymous
    • 5 years ago
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    haha, just saw what you wrote, myininaya...

  23. myininaya
    • 5 years ago
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    lol .

  24. anonymous
    • 5 years ago
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    And I forgot to say, you may want to translate your points to the right 3 units before rotating...so your points would be (-3,0) --> (0,0) and (0,18) --> (3,18). Hope this is making sense.

  25. myininaya
    • 5 years ago
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    ignorant, do we know anything else about the parabola?

  26. myininaya
    • 5 years ago
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    like the vertex or something else?

  27. anonymous
    • 5 years ago
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    No that is all the information I have about it

  28. myininaya
    • 5 years ago
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    can you make sense out what lokisan is saying? it looks like he has a way to find the equation of the parabola

  29. anonymous
    • 5 years ago
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    Yes, I can get the idea of what he is saying. And I am trying to work it out on pen and paper. If you have any more clue, please post it

  30. myininaya
    • 5 years ago
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    do you know derivatives?

  31. anonymous
    • 5 years ago
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    Yes sure

  32. anonymous
    • 5 years ago
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    http://en.wikipedia.org/wiki/Rotation_matrix

  33. myininaya
    • 5 years ago
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    ok awesome

  34. myininaya
    • 5 years ago
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    f'(x)=2ax+b

  35. anonymous
    • 5 years ago
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    Forget the matrix stuff if you haven't done it. You can use the equations.

  36. anonymous
    • 5 years ago
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    Actually I happen to have learnt that rotation part of matrix

  37. anonymous
    • 5 years ago
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    Good

  38. anonymous
    • 5 years ago
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    Your problem is: (3,18)=R(x,y) where the ( , ) are vectors and R is the rotation matrix. You need to invert R to find (x,y)=R^{-1}(3,18). You'll then have two points, (0,0) and (x,0) (since both points will now be on the x-axis). I'm rushing through this...I've just thought of a 'better' method, but I'm rushed...if I can sort it out later, I will.

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