Slopes Along a Quadratic Curve

Why a Quadratic Curve Has Different Slopes

A straight regression line has the same slope at every predictor value. A quadratic regression curve can rise quickly in one part of its range and more slowly in another, or its direction of change can differ across the range. Consequently, saying only that a quadratic model has a particular “slope” is incomplete: we must specify where along the curve we want the slope.

The derivative provides that local slope. For a fitted quadratic mean function, it gives the rate at which the model’s expected response changes at a particular predictor value. This is a description of the fitted curve, not by itself evidence that changing the predictor causes the response to change.

The Quadratic Mean Function and Its Local Slope

Let $X$ be a quantitative predictor and $Y$ a quantitative response. The quadratic regression model represents the conditional mean response as a quadratic function of $x$:

E(Y\mid X=x)=\beta_0+\beta_1x+\beta_2x^2.

Here, $\beta_0$, $\beta_1$, and $\beta_2$ are population coefficients. In a sample, least squares estimates them with $\hat{\beta}_0$, $\hat{\beta}_1$, and $\hat{\beta}_2$. The fitted mean response at predictor value $x$ is therefore

\hat{y}(x)=\hat{\beta}_0+\hat{\beta}_1x+\hat{\beta}_2x^2.

The local slope is the derivative of this fitted function with respect to $x$. Differentiating each term gives

\frac{d\hat{y}(x)}{dx} = \hat{\beta}_1+2\hat{\beta}_2x.

Definition: The fitted local slope at $x=x_0$ in a quadratic regression is the derivative of the fitted mean function evaluated at $x_0$: $$ \hat{\beta}_1+2\hat{\beta}_2x_0. $$ It describes the fitted change in mean response per one-unit increase in the predictor, locally around $x_0$.

The coefficients and the slope have units. If $Y$ is measured in response units and $X$ in predictor units, $\hat{\beta}_1$ has units of response units per predictor unit, while $\hat{\beta}_2$ has units of response units per predictor unit squared. Multiplying $\hat{\beta}_2$ by $x$ gives the same units as $\hat{\beta}_1$, so the local slope is measured in response units per predictor unit.

The derivative describes the rate of change at a point, not an exact change over a large interval. For a small increase in $x$, the slope is a useful approximation to the change in fitted mean response per unit increase. An exact change across an interval can be found by subtracting the fitted values at its endpoints.

Why the Derivative Gives the Local Rate

The difference quotient compares the fitted response at $x_0+h$ with the fitted response at $x_0$, divided by the predictor change $h$. Substituting the quadratic fitted function and expanding its terms shows what happens as $h$ becomes small:

\begin{aligned} \frac{\hat{y}(x_0+h)-\hat{y}(x_0)}{h} &= \frac{ \hat{\beta}_1h+\hat{\beta}_2\big((x_0+h)^2-x_0^2\big) }{h}\\ &= \frac{ \hat{\beta}_1h+\hat{\beta}_2(2x_0h+h^2) }{h}\\ &= \hat{\beta}_1+2\hat{\beta}_2x_0+\hat{\beta}_2h. \end{aligned}

As $h$ approaches zero, the last term $\hat{\beta}_2h$ approaches zero. The difference quotient therefore approaches $\hat{\beta}_1+2\hat{\beta}_2x_0$, which is the derivative at $x_0$. This justifies interpreting the derivative as the fitted rate of change at that predictor value.

For a finite increase $\Delta$, the exact change in fitted mean response is instead

\hat{y}(x+\Delta)-\hat{y}(x) = (\hat{\beta}_1+2\hat{\beta}_2x)\Delta+\hat{\beta}_2\Delta^2.

The first term is the local slope at $x$ multiplied by the change in the predictor; the second term accounts for the fact that the slope changes along the quadratic curve. Thus the local slope times $\Delta$ is generally an approximation, not the exact change, unless $\Delta$ is sufficiently small or $\hat{\beta}_2=0$.

Interpreting a Slope from a Fitted Model

A local-slope calculation is straightforward, but its interpretation depends on the model being used appropriately. The derivative can always be calculated from the fitted equation; its statistical interpretation as a feature of the mean response relies on the regression setup being reasonable.

The model assumes that the conditional mean response is adequately represented by a quadratic function over the predictor range of interest. For standard regression interpretation, errors are taken to have mean zero conditional on the predictor values; the usual least-squares inferential framework also commonly assumes independent errors with constant variance. Normal errors are an additional assumption for exact small-sample inference, but are not needed to differentiate the fitted function. A slope calculated outside the range represented by the data is an extrapolation and may not describe the actual mean relationship there.

In context, state the predictor value, direction, approximate amount of change, and units. For example, “At a fertilizer dose of 6 dose units, the fitted mean yield is increasing by about 1.8 tons per hectare for each additional dose unit” is more informative than “the slope is 1.8.” This interpretation concerns the fitted conditional mean. It does not say that every observation changes by that amount, and it does not establish a causal effect.

Worked Example: Yield at a Particular Fertilizer Dose

Suppose a hypothetical analysis of crop plots produces the fitted quadratic mean function

\hat{y}(x)=12+3.6x-0.15x^2,

where $x$ is fertilizer dose measured in units of 10 kilograms per hectare and $\hat{y}(x)$ is fitted mean yield in tons per hectare. The observed doses in this example range from 0 to 10 dose units. Assume the quadratic mean shape is adequate over that range, and that observations are independent with mean-zero errors and roughly constant variance. These assumptions support the regression interpretation; the calculation below does not require a significance test.

Differentiate the fitted mean function:

\frac{d\hat{y}(x)}{dx}=3.6-0.30x.

At $x=6$ dose units, substitute $x=6$:

$$ 3.6-0.30(6)=3.6-1.8=1.8. $$

Thus, at a dose of 60 kilograms per hectare, the fitted mean yield is increasing locally by about 1.8 tons per hectare for each additional dose unit. Since one dose unit is 10 kilograms per hectare, this is equivalent to about 0.18 tons per hectare for each additional kilogram per hectare, locally. The statement is about the fitted mean near that dose, not every plot’s yield.

Worked Example: A Local Decline in a Fitted Symptom Score

Consider a hypothetical fitted model relating treatment dose $d$, in milligrams, to a symptom score $S$, in score points:

\hat{S}(d)=42-5d+0.25d^2.

Suppose doses in the data range from 0 to 10 milligrams. For this illustration, assume the conditional mean is reasonably represented by the quadratic model in that range, and observations meet the independence and mean-zero error conditions used for regression interpretation. The fitted local slope is

\frac{d\hat{S}(d)}{dd}=-5+0.50d.

At $d=6$ milligrams, the calculation is

$$ -5+0.50(6)=-5+3=-2 $$

score points per milligram. In context, the fitted mean symptom score is decreasing locally by about 2 points for each additional milligram of dose near 6 milligrams. This is a model-based association; the equation alone does not show that dose caused the score to change. It also does not imply that the decrease is exactly 2 points across every full one-milligram interval.

Worked Example: Comparing a Local Slope with an Exact Change

A hypothetical model relates elapsed time $t$, in minutes, to the mean temperature $T$, in degrees Celsius, of a cooling object:

\hat{T}(t)=80-6t+0.20t^2.

Suppose the observed time range is 0 to 10 minutes and the quadratic mean form is adequate over that range. Assume independent observations and mean-zero errors for the regression interpretation. The fitted local slope is

\frac{d\hat{T}(t)}{dt}=-6+0.40t.

At $t=5$ minutes, the slope is

$$ -6+0.40(5)=-6+2=-4 $$

degrees Celsius per minute. So near 5 minutes, the fitted mean temperature is decreasing at about 4 degrees Celsius per minute. Now calculate the exact fitted change from 5 to 6 minutes using the model:

\begin{aligned} \hat{T}(6)-\hat{T}(5) &=(80-6(6)+0.20(6^2))-(80-6(5)+0.20(5^2))\\ &=(80-36+7.2)-(80-30+5)\\ &=51.2-55=-3.8 \end{aligned}

The exact fitted mean change over that one-minute interval is a decrease of 3.8 degrees Celsius. This is close to the local-slope approximation of a 4-degree decrease, but it is not identical. The slope is a rate at one point; the endpoint subtraction gives the exact model-based change across the interval.

Why the Linear Coefficient Is Not Always the Slope

A common mistake is to interpret $\hat{\beta}_1$ as the slope everywhere in a quadratic regression. That is incorrect unless the quadratic term is zero. The derivative is

\hat{\beta}_1+2\hat{\beta}_2x,

so the local slope depends on $x$ whenever $\hat{\beta}_2\ne 0$. The coefficient $\hat{\beta}_1$ is the slope specifically at $x=0$, because substituting zero gives $\hat{\beta}_1+2\hat{\beta}_2(0)=\hat{\beta}_1$. If zero is outside the observed range, this interpretation at zero may not be practically relevant.

Another pitfall is to report a derivative without its units or context. A numerical value such as $-2$ is not a complete interpretation: it must be tied to a predictor value and expressed in response units per predictor unit. Finally, avoid turning a fitted association into a causal statement unless the study design and assumptions support that conclusion.

Key takeaway: For a fitted quadratic mean function $\hat{y}(x)=\hat{\beta}_0+\hat{\beta}_1x+\hat{\beta}_2x^2$, the local slope at $x_0$ is $\hat{\beta}_1+2\hat{\beta}_2x_0$. Report it at a specified predictor value, with response-per-predictor units, and interpret it as a local change in the fitted mean.

Check Your Understanding

Use the quadratic derivative rule and keep the units and context in each interpretation.

For $\hat{y}(x)=7+4x-0.5x^2$, calculate the fitted local slope at $x=3$.
A fitted curve is $\hat{R}(u)=20-2u+0.1u^2$, where $u$ is measured in hours and $R$ in liters. What are the units of its local slope, and what is the slope at $u=5$?
Why is the linear coefficient in a quadratic model not generally the slope across the entire predictor range?
For a quadratic fitted curve, is the derivative at $x=4$ necessarily the exact change in fitted response from $x=4$ to $x=5$? Explain the distinction.
What information should be included when interpreting a calculated local slope in an applied setting?

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