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Express y as a function of x

  1. Oct 7, 2014 #1
    I'm slightly embarrassed to ask such an easy question but how do you rearrange the following equation such that y=f(x)?

    x2+2xy-3y2=0

    I always have trouble with these types of problems when there are multiple "y"'s in an equation especially when they have different powers. Can someone please help?
     
    Last edited: Oct 7, 2014
  2. jcsd
  3. Oct 7, 2014 #2

    Mark44

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    The left side factors very easily. You will get two relationships with x as a function of y.
     
  4. Oct 7, 2014 #3
    Can you clarify what you mean? If I factor the equation I get (x-y)(x+3y)=0 but I want it in a y=f(x) form.
     
  5. Oct 7, 2014 #4
    It does not satisfy the definition of a function, why would you need to express it as one?
    At any rate, I would go about expressing y by either completing the square or factorising and then expressing y.
     
  6. Oct 7, 2014 #5
    Can you show how its done? I want to express it in a y=f(x) format since there is another function that intersects with this one.
    Do you know any other way to find the coordinate that it intersects without having to express it in a y=f(x) format?
     
  7. Oct 7, 2014 #6
    Yes, you have 2 canonic equations. You can express y or x from either one, plug it into the other equation and solve the resulting equation. Assuming those 2 sets intersect, the result will not be an empty set.
    Again, the first is not a function, it's some sort of a relation between the two variables. It is a set, there can be intersection between sets.
     
    Last edited: Oct 7, 2014
  8. Oct 7, 2014 #7

    PeroK

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    First, consider x as fixed, then you have a quadratic equation in y. Simply solve that quadratic and you have y expressed in terms of x. If there is more than one value of y for a given x, then this will not be a function, but you could make it a function by restricting the values of y to be unique for each x.
     
  9. Oct 7, 2014 #8
    I'm not sure if I understand you. The problem still persist even if you try to rearrange the equation such that x=f(y).

    Also, if you want to rearrange the equation Vi(t)+1/2g(t)^2=D for t such that t equals a function of Vi and g, how do you do it?

    Can someone actually demonstrate it?
     
    Last edited: Oct 7, 2014
  10. Oct 7, 2014 #9

    Mark44

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    Your notation is confusing, especially Vi(t), which could be interpreted as Vi being a function of t, or V times i, which is a function of t.
    I'm going to assume you mean this: vit + (1/2)gt2 = D. This equation is quadratic in t, so you can use the Quadratic Formula to solve for t. To make it a bit easier, write the equation in this form: gt2 + 2vit - D = 0.

    In this form, a = g, b = 2vi, and c = -D. Just plug these into the Quadratic Formula to get t.
     
  11. Oct 7, 2014 #10

    WWGD

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    It may be possible to rotate the plane to get rid of the mixed term.
     
  12. Oct 7, 2014 #11
    Alright, thanks.

    What do do you do if you can't use the quadratic equation?

    For example, how do you rearrange the following equation:

    y=x^5+x^3+x^2+x+4 so that x=f(y)
     
    Last edited: Oct 7, 2014
  13. Oct 7, 2014 #12

    PeroK

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    It's really not very difficult. ##x^2 +2xy - 3y^2 = 0 \ \ iff \ \ (x-y)(x+3y)=0 \ \ iff \ \ y = x \ \ or -\frac{x}{3}##

    And that's y expressed in terms of x. As explained above, this is not a function of x.

    If you want y = f(x), where (x, f(x)) is a solution to the equation, then you have two obvious options:

    ## y = f(x) = x## and ##y = f(x) = -\frac{x}{3}##
     
  14. Oct 7, 2014 #13

    Mark44

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    I don't believe it's possible for this one. Did you make it up?
     
  15. Oct 7, 2014 #14
    Yes I made it up. Is it really impossible to find the inverse of this equation?
     
  16. Oct 7, 2014 #15

    WWGD

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    The inverse/implicit function theorems give conditions under which a local inverse (may)exist(s).
     
  17. Oct 7, 2014 #16
    I see. What about other equations that has an inverse function?

    For example, y=x^3+x^2 has an inverse function according to wolfram but how the heck do you solve it? Is there a rule/method you can use to solve these types of problems? What are these problems called anyway?
     
  18. Oct 8, 2014 #17

    HallsofIvy

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    There are formulas for solving quadratic, cubic, and quartic equations. It has been proven that there cannot be a formula, involving only roots, addition, multiplication, addition, and subtraction, for solving polynomials of degree 5 or higher.
     
  19. Oct 8, 2014 #18

    WWGD

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    Do you mean for _all_ quintics (sorry if you do, I am no expert in this area)? I think if the Galois group of the polynomial is solvable, then you can use those operations only.
     
  20. Oct 8, 2014 #19

    HallsofIvy

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    I meant a general formula.. Yes, of course, there exist solutions to some polynomials of any degree.
     
  21. Oct 8, 2014 #20

    WWGD

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    Ah, sorry, I am kind of new to that area.
     
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