How to find the equation / function that does not cross through or touch the x axis

Question

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.

Answer 1

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.

Answer 2

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.