1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Staff: Mentor

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    Staff: Mentor

    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

    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    Staff: Mentor

    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

    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    User Avatar
    Science Advisor
    Gold Member

    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

    User Avatar
    Staff Emeritus
    Science Advisor

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

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Ah, sorry, I am kind of new to that area.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook