# Inverse of polynomial in dy^2 + ey + f form

1. Sep 10, 2010

### stevec

Hello,

I'm trying to find the inverse polynomial of y = ax^2 + bx + c in the form of x = dy^2 + ey + f.

I'm able to approximate this using Excel, but would prefer a more elegant solution. Any suggestions?

Steve

2. Sep 10, 2010

### Yuqing

Isolate x in terms of y using the quadratic formula should give you the inverse.

3. Sep 10, 2010

### Mentallic

I don't think I'm understanding your problem properly because the inverse of $$y=ax^2+bx+c$$ IS $$x=ay^2+by+c$$ and so d=a, e=b, f=c

4. Sep 10, 2010

### Gerenuk

5. Sep 10, 2010

### stevec

Thanks for the replies, I apologize for not being more descriptive in my question.

Gerenuk's reply is close to what I am looking for, although I think I may need to increase the terms - a quick set of data against the equation was off.

6. Sep 11, 2010

### HallsofIvy

Staff Emeritus
Mentallic is correct: the "inverse function" to $y= ax^2+ bx+ c$ is just $x= ay^2+ by+ c$. Now use the quadratic formula to solve for y:
$$y= \frac{-b\pm\sqrt{b^2- 4ac}}{2a}$$

But notice the "$\pm$". The quadratic function is not one-to-one and so does not have a true "inverse". You could restrict x to one side or the other of the vertex of the parabolic graph, thus using either the "+" or the "-".

7. Sep 12, 2010

### Gerenuk

Neither of you is correct, because you don't understand the question. He might not have used the proper wording, but it's not hard to guess what he is really looking for.

@SteveC: Maybe you want to look at Chebychev Polynomials and their Approximation theory. They provide a method to vaguely minimize the maximum total error. Whereas the series expansion I wrote down only aims to be best a y=0.

8. Sep 13, 2010

### Mentallic

Yep, I already acknowledged that I might not be understanding it properly.