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I have been pondering this question for a little while, and unfortunately the Google has not given me an answer.

I understand that for example you had a table or graph that crosses or touches the x axis at say x= -2, 0, 3 you could form an equation as

f(x) = ax(x+2)(x-3)

Then solve for a and you have your function.

I have considered transforming a graph to force zeros, and in the couple of attempts I made, it was successful, but I am unsure if this would be the mathmatically proper way to do so.

So my question is if you have a graph or table of coordinates similar to my example above, but the points never cross zerI, what would be the proper Mathmatics procedure to find the equation.

UPDATE

y = x^2. Vertex = 0,0 and zero = 0

In comparison to:

y = (x-1)^2+1 vertex = 1,1 and zero = null

Its the same form but in a different position. In this situation the functions were provided, but for clarification of what I am looking for, I thought this would help.

Thank you.

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  • $\begingroup$ You may be asking about Lagrange interpolation: en.wikipedia.org/wiki/Lagrange_polynomial $\endgroup$ – Ethan Bolker Oct 1 '18 at 20:53
  • $\begingroup$ Your question does not make sense. The "table of coordinates similar to my example above" lists zeroes, and you want a solution without zeroes. So what ? $\endgroup$ – Yves Daoust Oct 1 '18 at 20:59
  • $\begingroup$ Sorry for being confusing. I mean to say you could graph a table of cordinates that would appear to be the same as the one in my example, but the points land in such a way that they are either to far up or down to cross zero. $\endgroup$ – SynchroDynamic Oct 1 '18 at 21:02
  • $\begingroup$ I think you at least pointed me in the right direction @Ethan Bolker. Thank you $\endgroup$ – SynchroDynamic Oct 1 '18 at 21:03
  • $\begingroup$ "but the points land in such a way that they are either to far up or down to cross zero." If you put it that way that makes no sense. If they don't touch at $x=3,0,-2$ then why do you even mention $x=3,0,-2$ $3,0,-2$ aren't any different than any other values of $x$ where the function doesn't. That's saying animal 1 eats pickles, and jelly beans when you feed it. But what animals don't eat anything when you feed it pickles and jelly beans. Well if it doesn't eat any thing, why feed it pickles and jelly beans? $\endgroup$ – fleablood Oct 1 '18 at 22:22
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Touching the $x $ axis at $x=a $ means the function has a root at $x=a $ and not touching means it doesn't.

If we restrict ourselves only to polynomials (a very small class of functions) then having a root/touching the axis at $x=a$ means $x-a$ is a factor of the polynomial $p (x) $.

If $(x-a) $ is a factor but $(x-a)^2$ is not than $x=a $ is a single root and the graph crosses at at angle. If $(x-a)^k $ divides the polynomial and $k $ is the largest integer that does so, then $x=a $ is a multiple root and the graph goes flat as it touches the $x$ axis. If $k$ is even it touches but doesn't cross and if $k $ is odd it crosses.

So if your graph touches at $x=3,x=0,x=-1$ but never crosses than the SIMPLEST polynomial is $(x-3)^2(x+1)^2x^2$. But that certainly is not the ONLY polynomial. Any $(x-3)^{2k}x^{2j}(x+1)^{2m}p (x) $ for any polynomial $p (x) $ that has no real roots (Example: $p (x)=x^2+1$ or $p (x)=9 (-x^6-5x^2-3)$)

In you example $f (x)=x(x-3)(x+2) $ is obviously not the ONLY function that crosses at $x=3,0,-2$. It isn't even the only polynomial that does. Any $x^{2k+1}(x+2)^{2j+1}(x-3)^{2m+1}p (x) $ where $p (x) $ is a polynomial with no real root.

Food for thought: for a polynomial to never touch the $x $ axis then the leading power must be even (do you see why). And any even power coefficient can be made to never touch the $x $ axis if you add or subtract a large enough constant to "push" it up or down far enough.

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  • $\begingroup$ Thank you for your answer. It helped me figure out what I was looking for. $\endgroup$ – SynchroDynamic Oct 6 '18 at 6:04
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I ended up figuring this out with the help of @Ethan Bolker. He pointed me in the direction of LaGrange interpolation. After learning a bit, I came across another fantastic work of Newton. Newton's divided difference is simple to learn and easy to use. It gives a general form as I was looking for, and there is no need for any transforming or nonsense like I was trying to do.

Newton's Divided Difference:

How Newton Sequence Works

Once you make these computations, you take the top values of the sequence. The first y value is by itself, then the second contains the factor of the of the first x, the third contains the factors of the first and second x values and so on in that pattern.

An Example

$x = {-2, -1, 0, 1, 2}$

$y = {4, 1, 0, 1, 4}$

$-2_1 , 4_1$

$\frac{1-4}{-1_2-(-2_1)} = -3$

$-1_2 , 1_2$ ---------------- $\frac{1-(-3)}{0_3-(-2_1)} = 2$

$\frac{0-1}{0_3-(-1_2)} = -1$ ----------------------- $\frac{1-2}{1_4-(-2_1)} = 1$

$ 0_3 , 0_3$ ---------------- $\frac{1-(-1)}{1_4-(-1_2)} = 1$ ----------------------- $\frac{1-1}{2_5-(-2_1)} = 0$

$\frac{1-0}{1_4-0_3)} = 1$ --------------------------- $\frac{2-1}{1_4-(-1_2)} = 1$

$ 1_4 , 1_4$ ---------------- $\frac{3-1}{2_5-0_3} = 2$

$\frac{4-1}{2_5-1_4} = 3$

$ 2_5, 4_5$

Then once you have your sequence completed it a pretty simple:

start with $y_1$, Factors of x values will be marked as $x_1$'s factor $= (x + b)_1$

$$f(x) = 4 + -3(x + 2)_1 + 2(x + 2)_1(x + 1)_2 + 1(x + 2)_1(x + 1)(x)_3 + 0(x + 2)_1(x + 1)_2(x)_3(x - 1)_4$$

If you FOIL all of this out and combine like terms, you are left with:

$$f(x) = x^2$$

Obviously, for this particular example their are far simpler ways to get this solution, however, this works with any data you may have. The only rule that I was able to find is that the x values need to be in order. They can be descending or ascending, but they just need to be in order.

I hope this is helpful to others, and I thank everyone for helping me find this answer.

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