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

[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

 

[Yuqing]

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

 

[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

 

[Gerenuk]

Maybe you mean
http://www.wolframalpha.com/input/?i=Series[%28-b%2Bsqrt[b^2-4*a*%28c-y%29]%29%2F%282*a%29%2C{y%2C0%2C2}]

 

[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.

 

[HallsofIvy]

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 "-".

 

[Gerenuk]

Mentallic is correct

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.

 

[Mentallic]

Neither of you is correct, because you don't understand the question.

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

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

The way he asked it, my answer is correct. I was only sceptical about my answer because it was too simple.