Need help finding the equation of a line for something I am

[Sylus]

Hello everyone. I can't go into too much detail, but I am working on something in which I have points on a line and I need to know what the formula for the points are. I do not remember how to do this. It is a curved line and I know some points. I can also get any points if that helps anyone find the equation.

Here are 3 points I used to find an eqation

X Y
35 1
100 0.35
200 0.175
I got

y= 0.00005000x2 −0.01675x+1.525

However, this was not the answer as for when I put in X = 150, I got a result of Y = 0.137.

I was looking for X = 150 to give the result of Y = 0.234

I quite honestly am not really certain at all how to find this equation. I also am not able to go into too much detail, but what I can do is give any points that someone may need to help me find this formula.

Please help. Thank you.

[HallsofIvy]

Given a finite number of points, there are an infinite number of curves that pass through the given points!

In a situation in which you have three points you can choose a type of function (trig, exponential, polynomial) that has three parameters and use the points to get three equations to solve for those parameters.

In particular, given three points, such as you start with, then there exist a unique quadratic, [tex]y= ax^2+ bx+ c[/tex], that passes through those three points. That appears to be what you did. But there is NO guarantee that the quadratic that passes through those three points will pass through this fourth point!

If you want a polynomial that passes through (35, 1), (100, 0.35), (200, 0.175), and (150, 0.234) you will need to start with a cubic, [tex]y= ax^3+ bx^2+ cx+ d[/tex] and solve the four equations
[tex]1= a(35)^2+ b(35)^2+ d(35)+ d[/tex]
[tex]0.35= a(100)^2+ b(100)^2+ d(100)+ d[/tex]
[tex]0.175= a(200)^2+ b(200)^2+ d(200)+ d[/tex]
[tex]0.234= a(150)^2+ b(150)^2+ d(150)+ d[/tex]
for the four numbers a, b, c, and d.

[Baltuilhe]

Good morning!

If you don't mind having a 'generic' equation, you don't need to solve a system.

A sample to solve your second degree equation... and you'll try third degree, ok?

Known points:
(35, 1)
(100, 0.35)
(200, 0.175)

'Create' the following:
[tex]y=\frac{(x-100)(x-200)}{(35-100)(35-200)}\cdot 1+\frac{(x-35)(x-200)}{(100-35)(100-200)}\cdot 0.35+\frac{(x-35)(x-100)}{(200-35)(200-100)}\cdot 0.175[/tex]

This is the solution!

Have fun!