# Express y as a function of x

1. Oct 7, 2014

### beamthegreat

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. Oct 7, 2014

### Staff: Mentor

The left side factors very easily. You will get two relationships with x as a function of y.

3. Oct 7, 2014

### beamthegreat

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.

4. Oct 7, 2014

### nuuskur

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.

5. Oct 7, 2014

### beamthegreat

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?

6. Oct 7, 2014

### nuuskur

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
7. Oct 7, 2014

### PeroK

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.

8. Oct 7, 2014

### beamthegreat

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
9. Oct 7, 2014

### 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.

10. Oct 7, 2014

### WWGD

It may be possible to rotate the plane to get rid of the mixed term.

11. Oct 7, 2014

### beamthegreat

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
12. Oct 7, 2014

### PeroK

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 $y = f(x) = -\frac{x}{3}$

13. Oct 7, 2014

### Staff: Mentor

I don't believe it's possible for this one. Did you make it up?

14. Oct 7, 2014

### beamthegreat

Yes I made it up. Is it really impossible to find the inverse of this equation?

15. Oct 7, 2014

### WWGD

The inverse/implicit function theorems give conditions under which a local inverse (may)exist(s).

16. Oct 7, 2014

### beamthegreat

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?

17. Oct 8, 2014

### HallsofIvy

Staff Emeritus
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.

18. Oct 8, 2014

### WWGD

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.

19. Oct 8, 2014

### HallsofIvy

Staff Emeritus
I meant a general formula.. Yes, of course, there exist solutions to some polynomials of any degree.

20. Oct 8, 2014

### WWGD

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