# How to control a graph of f(x) = y

1. Mar 21, 2012

### Xcrypt

I'm programming a camera for a game, and to achieve some effect I think the best solution would be some kind of function.

What I need to achieve, is a function that:

- 'starts' at x1, with a y value of y1.
- 'ends' at x2, with a y value of y2.
- I also need some parameter to control the 'curve amount' of the graph in between x1 and x2

('starts' and 'ends' mean that only the solution for the set of x values ['start' - 'end'] matter)

Unfortunately my calculus knowledge is very limited so I hope that someone here could explain to me what the correct solution is, and/or how to get there.

2. Mar 21, 2012

### Robert1986

Well, one idea might be to pick an x value in between x1 and x2 and force the graph to pass through there to enforce the "curve amount" requirement. From there, I could show you how to construct a quadratic polynomial that will fit all three points.

3. Mar 22, 2012

### Xcrypt

Well, please go on then :)

4. Mar 22, 2012

### chiro

Hey Xcrypt and welcome to the forums.

The subject you are looking for is known as interpolation. For your problem, there are many different forms of interpolation procedures that produce a variety of different effects.

If you want to just get the techniques then any book that covers hermite interpolation, b-splines, NURBS, catmull-clark splines amongst others will give you the formulas and the techniques.

I know that this book: http://books.google.com.au/books/about/Computer_animation.html?id=lP5uw516QXIC&redir_esc=y has examples in terms of code that work so you might be interested in getting this for your own bookshelf but it's fairly old and there's probably a lot of better books since that came out.

5. Mar 22, 2012

### Xcrypt

Thanks, just the techniques would do, the code (should) come naturally when you understand the techniques :)

Are these techniques you are naming covered in a calculus book?
I have heard many of them by name, but I haven't looked into them yet

6. Mar 22, 2012

### chiro

They are applications of calculus, but I doubt you'll find them in a standard calculus book. You usually find them in specialized books for things like computer graphics, engineering and areas where these are used and where specific information about the application is also talked about.

7. Mar 22, 2012

### Robert1986

I don't know what your exact situation is, but it seems that a quadratic polynomial might be the best way to go. If it doesn't then look up some of the stuff chiro mentioned.

So, we are going to pick three points on this curve and find a quadratic polynomial that fits these data. A quadratic has the form:

$$p(x)=a_2x^2 + a_1x + a_0$$

and we want to find a quadratic such that:

$$p(x_i) = y_i, i = 1,2,3$$

that is, we have three equatations in three unknowns. The unknowns are the $a_i, i = 1,2,3$. Now, just solve this system for the three unknows, and you will have your polynomial. Also, you can pick a 4th point and construct a cubic polynomial in a similar manner. In general, if you pick n points, you can construct an (n-1)th degree polynomial that fits those points.

8. Mar 22, 2012

### Xcrypt

Thanks, Robert1986 and chiro :)

9. Apr 5, 2012

### Xcrypt

update:

I've found a 2nd degree polynomial that goes through 3 points.

x=0.0, y=1.0
x=0.3, y=0.3
x=1.0, y=0.0

Solved: ((40/21)*((x)^(2)))-((61/21)*(x))+1
(check http://www.freemathhelp.com/equation-grapher.html )

But this is not what I want because it's going under the x-axis instead of making a smooth curve. I think I'll use a bezier curve instead :)

10. Apr 5, 2012

### genericusrnme

Lagrange polynomials are pretty handy
You could set up three points, x1, x2 and a new x3 in the middle, shift the coefficient of the x3 polynomial up and down and you should be able to change the 'curve ammount'

Last edited: Apr 5, 2012
11. Apr 5, 2012

### Office_Shredder

Staff Emeritus
I think the problem as stated is slightly ill-defined. Do you only want that shape on a bounded subset, or does the function need to have a horizontal asymptote as x goes to infinity?

12. Apr 5, 2012

### Xcrypt

I'm not sure what a bounded subset is, but I certainly don't want an asymptote. The problem is slightly ill defined because I don't have much knowledge about calc + I'm not used to expressing math terms in English. But I hope you can get a 'feel' for what I want to achieve?

Anyway, I solved the problem with bezier curve now, still it would be nice to know how to do it with 'normal' polynomials.

I'll try to define better how I want to define the curve:

There should be three points making up the curve:
P1 (starting point)
P3 (end point)
P2, Length(P1,P2) = Length(P2,P3) (mid point)

You should be able to 'move' P2 around on a line that is perpendicular to [P1,P3], and goes through the point .5*(P1+P3)

13. Apr 5, 2012

### genericusrnme

When he said asymptote I think he meant f(x)->0 as x -> infinity, like what's happening on your graph.
The bounded subset just means that you're only interested in the values of x between p1 and p3.

My idea about the lagrange polynomials would be able to do that!

The lagrange polynomials are handy little polynomials.
Let x1,x2,x3 be the x coordinates of p1,p2,p3 and y1,y2,y3 the y coordinates, then you can make 3 poynomials, which give zero at two of the three points and the value 1 at the other, like this;
$P_1 (x1) = 1$ , $P_1 (x2) = P_1 (x3)=0$
$P_2 (x2) = 1$ , $P_2 (x1) = P_2 (x3)=0$
$P_3 (x3) = 1$ , $P_3 (x1) = P_3 (x2)=0$

To make this work you'd just have

$P_1(x) = \frac{(x-x2)(x-x3)}{(x1-x2)(x1-x3)}$
$P_2(x) = \frac{(x-x1)(x-x3)}{(x2-x1)(x2-x3)}$
$P_3(x) = \frac{(x-x1)(x-x2)}{(x3-x1)(x3-x2)}$

I hope you can see that these will give the values I gave earlier, so to be able to do what you want (what I think you want at least) you define your f(x,a), where a defines the y value at x2;

$f(x,a) = y_1 P_1(x) + a P_2(x) + y_3 P_3(x)$

Using the properties of the polynomials you can easily verify that this will give you the desired results.

Is this the kind of thing you are looking for?

14. Apr 5, 2012

### Xcrypt

I think so, yes. Thanks