# How to interpret coefficients of $x$ and $x^2$ in same regression

If I have the below functional form for an OLS regression, how do I interpret the $x$ and $x^2$? I cannot interpret them separately, correct? Do I interpret them as a summation of the two coefficients, so the effect of a one unit change of $x$ on $y$ is $0.5 + 0.3 = 0.8$?

$$y = a + 0.5x + 0.3x^2 + e$$

Update: Now with 2 squared terms, and one that is an interaction term with the other and a binary variable, is there no rule for how to interpret the coefficients for all values of $x$? Or does it depend on the value of $x$?
\begin{align} y = 0.04\; &+\; 0.05\times {\rm drop}\; +\; 0.008\times {\rm season}\; -\; 0.0004\times {\rm season}^2 \\ &+\; 0.002\times {\rm dropseason}\; -\; 0.0003\times {\rm dropseason}^2\; -\; 0.029\times p \\ &-\; 0.01\times r_1\; +\; 0.003\times r_2\; +\; 0.01\times r_3\; +\; 0.003\times v_5\; +\; e \end{align}

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There are very good answers below. It may also help you to read my answers here: Does it make sense to add a quadratic term but not the linear term to a model?, & here: Why is polynomial regression considered a special case of multiple linear regression? –  gung Jul 21 '14 at 3:22
I think it would be more accurate to rewrite your question as: "How do I interpret the coefficients of x and x^2". (Since your actual interpretation of "x" and "x^2" would be independent of the regression itself) –  Steve S Jul 21 '14 at 5:23
That belongs as part of the question, not as a comment, @Rebecca. –  gung Jul 21 '14 at 14:37
In addition, nothing about the answer changes in this situation. You should still consider the terms together as a unit & the effect of a 1-unit change in x depends on the level of x from which the change takes place. –  gung Jul 21 '14 at 14:46

Such an equation describes a curved relationship between $y$ and $x$ - a parabola:

(This particular set of parameters correspond to a minimum at $x= -\frac{_5}{^6}$, just off the left margin of this plot.)

Consequently, you should keep all terms in the same x-variable together, since they describe the way $y$ is related to $x$.

Do I interpret them as a summation of the two coefficients, so the effect of a one unit change of x on y is 0.5 + 0.3 = 0.8.

No. The effect of the $x^2$ term on a one-unit change in $x$ is not constant.

Consider increasing $x$ from 0 to 1 and then from 10 to 11:

At $0$, the expected value of $y$ is $a$ (a=7 in my plot)
At $1$, the expected value of $y$ is $a+0.5\times 1+0.3\times 1^2$

The average increase in $y$ when $x$ increases from 0 to 1 is $0.5\times 1+0.3\times 1^2 = 0.8$

At $10$, the expected value of $y$ is $a+0.5\times 10+0.3\times 10^2$ (a=7 in my plot)
At $11$, the expected value of $y$ is $a+0.5\times 11+0.3\times 11^2$

The average increase in $y$ when $x$ increases from 10 to 11 is $0.5\times (11-10)+0.3\times (11^2-10^2) = 6.8$

So there's not one single number -- it depends on which $x$ you look at.

It may be useful to describe the effect of a unit change at some low value, some high value and somewhere in between.

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It doesn't pay to interpret them separately. They are connected. The formula for the vertex in a quadratic equation $y = a x^{2} + b x + c$ is $-\frac{b}{2a}$. The effect of changing $x$ from $s$ to $t$ is $a (t^{2} - s^{2}) + b (t - s)$. In a regression setting I often set $s$ to the first quartile of $x$ and $t$ to the $3^{\textrm{rd}}$ quartile, so as to estimate the inter-quartile-range $x$ effect.

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The most straightforward way to interpret is through multivariate Taylor expansion. If you don't know what is it, then forget what I just wrote.

If you take a derivative of the model specification, you'll see that your coefficients are the Taylor series coefficients.

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