# Is a model with xs squared a good fit for a parabola? [closed]

It should be, but the fact that the xs and not the parameters is squared confuses me.

• why not just use linear regression to fit the parabolic data? ie Y ~ X + X^2 May 6, 2022 at 20:12
– whuber
May 7, 2022 at 13:35

$$y=ax^2+bx+c$$
If you use the features $$x$$ and $$x^2$$ then $$y$$ is linear with respect to $$a,b,c$$ and you can simply run a linear regression that will estimates $$a,b,c$$.
It can be surprising to see that a parabola is linear in some sense, but the trick is that the word "linear" in linear regression means linear with respect to the coefficients (here $$a,b,c$$) not the feature (here $$x$$).
• Maybe you should look at how a (multiple) linear regression model is defined. The features are your observations/data and the parameters are what are actually estimated. The model assumes that the predicted value $y$ can be written (on average, conditionally to the data) as a linear combination of features, and the coefficients of the linear combination are the $\beta_i$. Maybe this link can be useful too : en.wikipedia.org/wiki/Polynomial_regression May 7, 2022 at 12:37