Why Interpret the Fitted Curve?
A regression coefficient in a polynomial model does not always describe a constant change in the response. For a quadratic or cubic curve, the fitted relationship can bend, flatten, or reverse direction as the predictor changes. The useful questions are therefore local and structural: What response does the model fit at a particular predictor value? What is the local rate of change there? Where, if anywhere, does the fitted curve turn?
This tutorial considers polynomial regression with one quantitative predictor \(x\). Polynomial models remain linear regression models because they are linear in their unknown coefficients, even though their fitted curves are not straight lines. The interpretation here concerns the fitted curve; a derivative alone does not establish a causal effect.
The calculations assume the fitted polynomial is the curve being interpreted over the predictor values of interest. The usual regression assumptions concern the errors around the conditional mean, such as mean zero and independence; constant variance and normality are used for familiar inference procedures, not to take the derivative of a fitted curve. A turning point outside the observed predictor range is an algebraic feature of the fitted polynomial, not necessarily a credible feature of the relationship in the data.
Quadratic Curves: Slope and Turning Point
A fitted quadratic curve has the form
Its marginal effect depends on \(x\):
If \(\hat{\beta}_2\ne 0\), the stationary point occurs at \(x^*=-\hat{\beta}_1/(2\hat{\beta}_2)\). A positive quadratic coefficient gives a convex curve and a minimum; a negative quadratic coefficient gives a concave curve and a maximum.
The squared term is smallest when \(x=-\hat{\beta}_1/(2\hat{\beta}_2)\). If \(\hat{\beta}_2>0\), its coefficient is positive, so this is the curve's minimum; if \(\hat{\beta}_2<0\), the curve has a maximum there. Differentiation gives the same result by setting \(\hat{\beta}_1+2\hat{\beta}_2x=0\). If \(\hat{\beta}_2=0\), the model is linear and this quadratic turning-point formula does not apply.
Worked Example: A Peak in a Hypothetical Service-Use Curve
Suppose a hypothetical insurer models an annual service-use index \(y\) against a customer engagement score \(x\), using observations with \(0\le x\le7\). The fitted curve is \(\hat{y}(x)=30+8x-x^2\). Find the fitted value and marginal effect at \(x=2\), and locate and interpret the turning point.
Method. Differentiate the fitted curve to obtain the marginal effect: \(\frac{d\hat{y}}{dx}=8-2x\). At \(x=2\), the marginal effect is \(8-2(2)=4\) index units per score point. The fitted value at \(x=2\) is \(30+8(2)-2^2=30+16-4=42\).
Turning point. Set the derivative to zero: \(8-2x=0\), so \(x=4\). The second derivative is \(-2\), which is negative, so the curve has a maximum there. Its fitted value is \(\hat{y}(4)=30+8(4)-4^2=30+32-16=46\). The value \(x=4\) lies inside the observed range from 0 to 7, so the fitted peak is not an extrapolation beyond that range.
Check and trap avoided. At \(x=2\), the positive slope of 4 is consistent with the curve still rising toward its maximum at 4. The coefficient 8 is the slope only at \(x=0\); treating it as the slope everywhere would ignore the quadratic term.
Cubic Curves: Several Changes in Direction
A fitted cubic curve has the form
Its marginal effect is a quadratic function, so it can have zero, one, or two distinct roots. The second derivative helps classify those roots. A cubic with \(\hat{\beta}_3\ne0\) also has an inflection point where its second derivative is zero; that point marks a change in curvature, not necessarily a maximum or minimum.
At a stationary point where the second derivative is positive, the curve has a local minimum; if it is negative, the curve has a local maximum. If both the first and second derivatives are zero, the second-derivative test is inconclusive: inspect the derivative's sign on either side or use the curve's higher-order behavior. Always compare candidate points with the observed \(x\)-range before interpreting them as features supported by the data.
Worked Example: Classifying Two Stationary Points of a Cubic
A hypothetical fitted response curve, based on observations from \(x=0\) to \(x=8\), is \(\hat{y}(x)=5+\frac{1}{3}x^3-4x^2+12x\). Find its stationary points, classify them, calculate their fitted responses, and identify the inflection point.
Step 1: Find stationary points. The first derivative is \(\frac{d\hat{y}}{dx}=x^2-8x+12=(x-2)(x-6)\). Thus the stationary points are \(x=2\) and \(x=6\).
Step 2: Classify and evaluate them. The second derivative is \(2x-8\). At \(x=2\), it is \(4-8=-4<0\), so there is a local maximum. The fitted response is \(\hat{y}(2)=5+\frac{8}{3}-16+24=13+\frac{8}{3}=\frac{47}{3}\). At \(x=6\), the second derivative is \(12-8=4>0\), so there is a local minimum. Its fitted response is \(\hat{y}(6)=5+72-144+72=5\).
Step 3: Locate the inflection point. Set the second derivative to zero: \(2x-8=0\), giving \(x=4\). The fitted response there is \(\hat{y}(4)=5+\frac{64}{3}-64+48=5+\frac{16}{3}=\frac{31}{3}\). Since the second derivative changes from negative to positive at 4, this is an inflection point, not a stationary point: the slope there is \(16-32+12=-4\), not zero.
Check and trap avoided. Both stationary points and the inflection point lie within the observed range. The derivative factors exactly as \((x-2)(x-6)\), and its sign is positive before 2, negative between 2 and 6, and positive after 6. This confirms the maximum-then-minimum pattern. A positive cubic coefficient does not mean the curve is increasing everywhere.
Reading Coefficients Without Misreading the Curve
For a polynomial model, coefficients work together to determine the slope at each predictor value. In a quadratic, \(\hat{\beta}_1\) is the marginal effect at \(x=0\), not a universal effect; \(\hat{\beta}_2\) determines how quickly that slope changes. In a cubic, \(\hat{\beta}_1\) is again the slope at zero, while the quadratic and cubic terms cause the slope itself to vary with \(x\). The sign of one coefficient alone does not tell you whether the fitted curve is increasing or decreasing throughout the data range.
Units matter. If \(y\) is measured in thousands of dollars and \(x\) in years, then the derivative is in thousands of dollars per year. A second derivative is in response units per squared predictor unit. The derivative is a local rate; for a finite increase from \(x_1\) to \(x_2\), the fitted change is \(\hat{y}(x_2)-\hat{y}(x_1)\), which need not equal the derivative at either endpoint.
Worked Example: A Minimum and a Zero Marginal Effect
A hypothetical operating-cost model, in cost units, is fitted against a process setting \(x\), with data observed for \(0\le x\le5\): \(\hat{C}(x)=30-4x+x^2\). Find the minimum fitted cost and the marginal effect at \(x=2\). State how the marginal effect changes on either side of the minimum.
Method. Differentiate: \(\frac{d\hat{C}}{dx}=-4+2x\). Set this equal to zero: \(-4+2x=0\), so \(x=2\). The second derivative is \(2>0\), confirming a minimum. The fitted cost is \(\hat{C}(2)=30-4(2)+2^2=30-8+4=26\) cost units.
At \(x=2\), the marginal effect is \(-4+2(2)=-4+4=0\) cost units per setting unit. For \(x<2\), \(-4+2x<0\), so fitted cost is decreasing; for \(x>2\), the derivative is positive, so fitted cost is increasing. The stationary point lies within the observed range.
Check and trap avoided. The derivative of \(x^2\) is \(2x\), so the derivative is \(-4+2x\), not a constant \(-4\). Substitution at \(x=2\) gives zero, not a nonzero slope. Also, the minimum response is 26; substituting \(x=2\) into the original curve, rather than the derivative, is necessary to obtain the fitted cost.
A Reliable Interpretation Sequence
For an exam-style calculation, keep the curve, its derivatives, and the predictor range distinct. A useful sequence is:
Keep each coefficient and power visible so that the derivative can be checked term by term.
Substitute the requested predictor value into the first derivative to calculate its marginal effect.
Set the first derivative equal to zero and solve; do not infer turning points from coefficient signs alone.
Use the second derivative or a derivative sign check to classify each point, then substitute into the original fitted curve for its response.
State whether the point is within the observed predictor range and report the marginal effect in response units per predictor unit.
A frequent exam trap is to solve the derivative correctly but report that derivative's value as the fitted response. The response at a turning point must be calculated from \(\hat{y}(x)\), not from \(\frac{d\hat{y}}{dx}\). Another is to announce an in-range maximum or minimum from a polynomial root that lies outside the range actually observed. Both mistakes can be avoided by keeping the original curve, derivative, and data range explicit.
These interpretations are aligned with the SOA Exam SRM (Statistics for Risk Modeling) syllabus. They describe a fitted association under the chosen model; they do not, by themselves, establish that changing \(x\) would cause the response to change by the calculated amount.
Check Your Understanding
For each problem, use the fitted curve and the stated predictor range. Answers are included for checking.
- A hypothetical fitted curve is \(\hat{y}(x)=12+6x-\frac{1}{2}x^2\), with \(0\le x\le10\). Find its turning point, classify it, and calculate the marginal effect at \(x=8\). (Answer: maximum at \(x=6\), fitted value \(30\); marginal effect at \(x=8\) is \(-2\) response units per predictor unit.)
- A quadratic operating-cost curve is \(\hat{C}(x)=20-8x+2x^2\), with \(0\le x\le5\). Find the minimum and the marginal effect at \(x=3\). (Answer: minimum at \(x=2\), fitted cost \(12\); marginal effect at \(x=3\) is \(4\) cost units per predictor unit.)
- A cubic fitted curve is \(\hat{y}(x)=2+x^3-6x^2+9x\), with \(0\le x\le4\). Find and classify its stationary points, then calculate the inflection point's predictor and fitted response. (Answer: local maximum at \(x=1\), fitted value \(6\); local minimum at \(x=3\), fitted value \(2\); inflection point at \(x=2\), fitted value \(4\).)
- For \(\hat{y}(x)=10+2x-x^2+\frac{1}{3}x^3\), calculate the marginal effect at \(x=2\) and state whether the curve is locally concave up or down there. (Answer: marginal effect \(2\) response units per predictor unit; concave up, since the second derivative at \(x=2\) is \(2>0\).)
- A fitted quadratic curve is \(\hat{y}(x)=7-3x+\frac{1}{2}x^2\), observed for \(0\le x\le5\). Find the stationary point and the marginal effect at \(x=4\). (Answer: minimum at \(x=3\), fitted value \(2.5\); marginal effect at \(x=4\) is \(1\) response unit per predictor unit.)