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I'm trying to figure this out and I need help.

I have a set of points and I know they are quadratic because the differences between the y points aren't the same.

But how do I find the equation of a nonlinear equation?

X | Y

-2 | -1

-1 | -4

3 | 4

4 | 11

dev4life
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    Welcome to Math Stack Exchange! A linear equation looks like $y = ax + b$ for some numbers $a$ and $b$ and a quadratic equation looks like $y = ax^2 + bx + c$ for some numbers $a$, $b$ and $c$. You are right that quadratic equations are non-linear, but there are other kinds of non-linear equations too, that are not quadratic (for example cubic or exponential equations.) Do you have reason to believe that this equation is quadratic, besides the fact that it is non-linear? – Matthew van Eerde Dec 02 '20 at 20:54
  • Yes. Because I plugged the points into a few websites and a quadratic equation was returned in all the sites. But it didn't explain how it was done. So I was looking for help. – dev4life Dec 03 '20 at 03:31

1 Answers1

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First you posit a quadratic equation in unknowns a, b, c thus: Y = a$X^2$ + b$X$ + c. Now plug in your five X and Y values, yielding five (linear!) equations in a, b, c. Solve for a, b, c (I assume you know how to solve linear equations in three unknowns).

PMar
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  • I did try that but I was getting some odd values. I'll have to try it again. Thanks! – dev4life Dec 03 '20 at 03:30
  • You should end up with $(a, b, c) = (1, 0, -5)$ which is to say $y = x^2 - 5$. $(x, y) = (-2, -1) \implies a(-2)^2 + b(-2) + c = 0 \iff 4a - 2b + c = -1$. $(x, y) = (-1, -4) \implies a(-1)^2 + b(-1) + c = -4 \iff a - b + c = -4$. $(x, y) = (3, 4) \implies a3^2 + b3 + c = 4 \iff 9a + 3b + c = 4$. You can use these three points to solve for $(a, b, c)$. As a check you can use $(x, y) = (4, 11) \implies a4^2 + b4 + c = 11 \iff 16a + 4b + c = 11$ – Matthew van Eerde Dec 04 '20 at 12:16