# How to Offset a polynomial

1. Feb 25, 2013

### zzinfinity

How to "Offset" a polynomial

Suppose I have a function for a curve, for example y=x2. I want to find a function to "offsets" it by 2 units. That is, I want a larger parabola that is exactly 2 units away from my original parabola. What I have in mind is the offset command in AutoCAD. Is there a simple transformation that can be done to my function to do this? And if its not possible, is it possible to do with functions other than polynomials? I've been thinking about this for a while, and I feel like its not possible, but I was wondering if anyone had ever encountered an algebraic way to do this.

2. Feb 25, 2013

### pwsnafu

Re: How to "Offset" a polynomial

What do you mean? What does "exactly 2 units away" mean? What do you mean by a "a larger parabola"?

3. Feb 25, 2013

### micromass

Re: How to "Offset" a polynomial

Maybe you should include a drawing or picture of what you want to do?

4. Feb 25, 2013

### Simon Bridge

Re: How to "Offset" a polynomial

I think he means something like this:

For $y=x^2$ ... the outer curve would have roots $x = \pm 2$ and intersect the y axis at $y=-2$. With those three points, a parabola can be determined.

Maybe more like: the outer boundary of the set of all points within 2 units of the curve.

5. Feb 25, 2013

### JJacquelin

6. Feb 25, 2013

### dodo

Re: How to "Offset" a polynomial

These formulas are producing offsets in directions normal to the curve. The reference to CAD software makes me think that the OP simply wants to translate the parabola. (Guesswork here.)

In which case, a direction of translation should be specified. For example, to translate along the X-axis, you would add/subtract some constant "c" to "x":$$y = (x \mp c)^2$$
To translate along the Y-axis, you add/subtract to "y":$$y = x^2 \pm c$$
To translate in an arbitrary (straight-line) direction, you would do a combination of the above. If you want to translate "c" units in a direction determined by an angle "theta" (with respect to the positive X-axis), then the horizontal translation is $c \cos \theta$ and the vertical translation is $c \sin \theta$, leaving the formula as$$y = (x - c \cos \theta)^2 + c \sin \theta$$
But yes, I agree with micromass than a picture would help to avoid guessing.

7. Feb 25, 2013

### fortissimo

Re: How to "Offset" a polynomial

If you want to have your parabola precisely two units away from your orignal one, at every point, then you will not obtain a parabola. This would be equivalent to having a ball (with a given radius) rolling around at the perimeter of the parabola, noting the curve it traces out. For instance, for y1 = x^2 and y2 = x^2 + c the Euclidian distance between these two curves approaches zero as x tends to infinity, while the distance at x=0 is c. Scaling and translating the parabola will not produce the desired result either, since the parabolas will move farther apart as x tends to infinity.

Last edited: Feb 25, 2013
8. Feb 25, 2013

### zzinfinity

9. Aug 27, 2014

### mathmut

Hello,

mine is a similar question.

Suppose we make a co-ordinate system with y and x axes.locate a point 5 units above the vertex which is at the ORIGIN.this point is the focus.now, we make a dashed line 5 units below parallel to y=0 and this is our directrix.considering this origin we sketch a parabola according to the equation y2=4fx where f is the focus.lets call this parabola 1

NOW,the question is.i intend to OFFSET the parabola 1 curve 2 units BELOW.such that the new curve(i am keeping it un named and un-defined and ambigious by calling it a curve) i get is 2 units exactly below the parabola 1 with the same curvature(similar to offseting a line or a semi-circle in a CAD modelling software such as pro-e or SW).WILL THIS NEW CURVE BE A PARABOLA (as per its definition of equal distances from directrix and focus)???

10. Aug 27, 2014

### Simon Bridge

Welcome to PF;
Did you follow the link in posts #7 and #8?

Do you mean that each point on the second curve has a y value that is 2 units less than the y value of the first, for the same value of x? (Sketch what you mean and check.)

11. Aug 27, 2014

### mathmut

Dear Simon,

thank you for your reply..

i actually want to know if this NEW curve is a parabola or not? will this NEW parabola take on a NEW EQUATION? or will it have the SAME equation as the parabola 1 ? because we have just offsetted the whole parabola 1 2 units downwards.

12. Aug 28, 2014

### Simon Bridge

Don't know for sure but it sounds like the new curve will also be a parabola with a different equation. To be sure, I need you to answer my questions.

If you follow suggestions, you should be able to work it out for yourself anyway.

Note - shifting a curve given by f(y)=g(x) x0 a units to the right and b units up gives a new equation: f(y-b)=g(x-a), but it is the same curve in a new position.

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