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Taylor Polynomials · Tutorial 805 of 998

Quadratic Taylor Models and Curvature

Learn to build a quadratic Taylor model, interpret its second-derivative term, and compare its local approximation with the tangent line.

Beginner 9 min read

What You'll Learn

  • Construct a degree-two Taylor polynomial at a specified center
  • Explain how the quadratic coefficient is determined by the second derivative
  • Compare a quadratic Taylor model with its tangent-line approximation
  • Use the sign of the quadratic term to describe the model’s position relative to the tangent
  • Distinguish curvature information at one point from concavity on a neighborhood
  • Recognize conditions that justify the quadratic model as a local approximation

From a Tangent Line to a Quadratic Model

A tangent line gives a function’s value and slope at a chosen point. It is the simplest local model that matches both. But a graph can bend away from its tangent line, and a straight line cannot represent that bending. A quadratic polynomial can: its squared term records the function’s second-derivative information at the center.

This tutorial develops the degree-two Taylor polynomial and compares it with the tangent-line approximation. The focus is local: the polynomial is built to match the function’s value, first derivative, and second derivative at one point. How well it approximates the function nearby depends on appropriate smoothness conditions.

Definition of the Degree-Two Taylor Polynomial

Let \(f\) be a function with a first and second derivative at a point \(a\). The degree-two Taylor polynomial of \(f\) centered at \(a\) is the quadratic polynomial whose value and first two derivatives agree with those of \(f\) at \(a\).

Definition: The degree-two Taylor polynomial of \(f\) at \(a\) is \(P_2(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2}(x-a)^2\).

The terms have distinct roles. The constant term makes \(P_2(a)=f(a)\). The linear term supplies the slope \(f'(a)\). The quadratic term supplies the second derivative \(f''(a)\), because differentiating \(\frac{f''(a)}{2}(x-a)^2\) twice gives \(f''(a)\).

The tangent-line approximation, also called the linearization, is \(L(x)=f(a)+f'(a)(x-a)\). Thus the degree-two polynomial keeps the tangent line and adds one correction:

$$ P_2(x)=L(x)+\frac{f''(a)}{2}(x-a)^2. $$

This comparison is the main point: the quadratic model and the tangent line have the same value and slope at the center, while the added term describes the bending captured by the second derivative there.

Worked Example: A Quadratic Function

Worked Example: A Quadratic Function

Find the degree-two Taylor polynomial of \(f(x)=3-2x+5x^2\) centered at \(a=1\), and compare it with the tangent line.

First evaluate the function at the center:

$$f(1)=3-2(1)+5(1)^2=3-2+5=6.$$

Differentiate and evaluate the first derivative:

$$f'(x)=-2+10x,\qquad f'(1)=-2+10(1)=8.$$

Differentiate again and evaluate the second derivative:

$$f''(x)=10,\qquad f''(1)=10.$$

Substitute these values into the definition:

$$P_2(x)=6+8(x-1)+\frac{10}{2}(x-1)^2=6+8(x-1)+5(x-1)^2.$$

The tangent line is \(L(x)=6+8(x-1)\). To check the polynomial, differentiate it:

$$P_2(1)=6,\qquad P_2'(x)=8+10(x-1),\qquad P_2'(1)=8,\qquad P_2''(x)=10.$$

These match \(f(1)=6\), \(f'(1)=8\), and \(f''(1)=10\). In fact, expanding \(P_2\) gives \(6+8x-8+5x^2-10x+5=3-2x+5x^2=f(x)\). Here the quadratic model reproduces the function exactly.

What the Quadratic Term Says

Since \((x-a)^2\) is positive whenever \(x\ne a\), the sign of the quadratic correction is determined by \(f''(a)\). If \(f''(a)>0\), then \(P_2(x)\) lies above its tangent line for every \(x\ne a\). If \(f''(a)<0\), then \(P_2(x)\) lies below its tangent line. If \(f''(a)=0\), the degree-two polynomial is the tangent line; that fact alone does not tell us what the function does beyond its second-order information at \(a\).

This comparison describes the polynomial model. To draw conclusions about the function’s concavity throughout a neighborhood, we need information about the second derivative throughout that neighborhood. For example, if \(f''(x)>0\) throughout an interval, then \(f'\) is increasing there; if \(f''(x)<0\) throughout an interval, then \(f'\) is decreasing there. A sign at just one point does not establish either neighborhood claim. If \(f''\) is continuous at \(a\) and \(f''(a)>0\), continuity does ensure \(f''(x)>0\) on some interval around \(a\); the corresponding statement holds when \(f''(a)<0\).

Here, “curvature” refers to the bending indicated by the second derivative in the quadratic model. It is not a claim that \(f''(a)\) is always the geometric curvature of the graph: for a graph \(y=f(x)\), geometric curvature involves both \(f''(x)\) and \(f'(x)\).

Why the Formula Works

The coefficients are chosen so that the polynomial and the function agree in their first three pieces of derivative data at the center. This gives a direct justification of the formula.

1
Match the value.
At \(x=a\), both \((x-a)\) and \((x-a)^2\) vanish, so \(P_2(a)=f(a)\).
2
Match the first derivative.
Differentiating gives \(P_2'(x)=f'(a)+f''(a)(x-a)\), so \(P_2'(a)=f'(a)\).
3
Match the second derivative.
Differentiating once more gives \(P_2''(x)=f''(a)\), hence \(P_2''(a)=f''(a)\).

These three requirements also determine the coefficients uniquely. Suppose a quadratic polynomial has the form \(Q(x)=A+B(x-a)+C(x-a)^2\) and matches \(f\), \(f'\), and \(f''\) at \(a\). Then \(Q(a)=A=f(a)\), \(Q'(x)=B+2C(x-a)\) gives \(B=Q'(a)=f'(a)\), and \(Q''(x)=2C\) gives \(C=f''(a)/2\). Therefore \(Q=P_2\).

Matching derivatives explains how to construct the polynomial. To justify its use as an approximation, a further condition is helpful. Suppose \(f''\) is continuous on an interval containing \(a\) and \(x\). The Fundamental Theorem of Calculus, applied twice, gives

$$ f(x)=f(a)+f'(a)(x-a)+\int_a^x (x-t)f''(t)\,dt. $$

Subtract the proposed quadratic polynomial from both sides. Since \(\int_a^x(x-t)\,dt=(x-a)^2/2\), this gives the exact error identity

$$ f(x)-P_2(x)=\int_a^x (x-t)\bigl(f''(t)-f''(a)\bigr)\,dt. $$

To see what this says near \(a\), let \(M\) be the largest value of \(\lvert f''(t)-f''(a)\rvert\) between \(a\) and \(x\). Taking absolute values and integrating the absolute value of the linear factor gives

$$ |f(x)-P_2(x)|\leq \frac{M}{2}|x-a|^2. $$

As \(x\) approaches \(a\), continuity of \(f''\) makes \(M\) approach zero. Thus the error divided by \((x-a)^2\) approaches zero: the error is smaller than a fixed multiple of \((x-a)^2\) sufficiently close to \(a\). This argument explains both the role of the quadratic term and a sufficient condition for \(P_2\) to be a valid local approximation. Merely matching derivative values at a point should not be confused with establishing a particular error bound.

Worked Example: A Negative Second Derivative

Worked Example: A Negative Second Derivative

Find the degree-two Taylor polynomial for \(f(x)=\sqrt{x}\) at \(a=4\), and compare its estimate at \(x=4.2\) with the tangent-line estimate.

The function value is \(f(4)=2\). Its derivatives are \(f'(x)=\frac{1}{2\sqrt{x}}\) and \(f''(x)=-\frac{1}{4x^{3/2}}\). Therefore

$$f'(4)=\frac14,\qquad f''(4)=-\frac{1}{4\cdot 8}=-\frac{1}{32}.$$

Substitution gives

$$P_2(x)=2+\frac14(x-4)-\frac{1}{64}(x-4)^2.$$

The tangent line is \(L(x)=2+\frac14(x-4)\). At \(x=4.2\), \(x-4=0.2\), so

$$L(4.2)=2+\frac14(0.2)=2.05,$$
$$P_2(4.2)=2+0.05-\frac{0.04}{64}=2.05-0.000625=2.049375.$$

Check the derivative matches at the center: \(P_2'(x)=\frac14-\frac{1}{32}(x-4)\) and \(P_2''(x)=-\frac{1}{32}\), so \(P_2(4)=2\), \(P_2'(4)=\frac14\), and \(P_2''(4)=-\frac{1}{32}\), as required. The actual value is \(\sqrt{4.2}\approx2.049390\). The quadratic estimate is closer than the tangent-line estimate in this case. Its negative quadratic correction places the model below its tangent, consistent with the negative second derivative.

Worked Example: Comparing Models for a Reciprocal

Worked Example: Comparing Models for a Reciprocal

Find the degree-two Taylor polynomial and tangent line for \(f(x)=1/x\) at \(a=2\), then estimate \(f(2.2)\).

Evaluate the function and its first two derivatives at the center:

$$f(2)=\frac12,\qquad f'(x)=-\frac{1}{x^2},\qquad f'(2)=-\frac14.$$
$$f''(x)=\frac{2}{x^3},\qquad f''(2)=\frac{2}{8}=\frac14.$$

Hence

$$P_2(x)=\frac12-\frac14(x-2)+\frac18(x-2)^2,\qquad L(x)=\frac12-\frac14(x-2).$$

At \(x=2.2\), the difference from the center is \(0.2\). The tangent estimate is

$$L(2.2)=\frac12-\frac14(0.2)=0.5-0.05=0.45.$$

The quadratic estimate is

$$P_2(2.2)=0.5-0.05+\frac18(0.04)=0.45+0.005=0.455.$$

For a derivative check, \(P_2'(x)=-\frac14+\frac14(x-2)\) and \(P_2''(x)=\frac14\). Thus \(P_2(2)=\frac12\), \(P_2'(2)=-\frac14\), and \(P_2''(2)=\frac14\), matching the function data. The exact value at \(2.2=11/5\) is \(f(2.2)=5/11\approx0.454545\). The quadratic estimate is closer than \(0.45\), the tangent estimate.

Worked Example: A Cubic Function

Worked Example: A Cubic Function

Find the degree-two Taylor polynomial for \(f(x)=x^3\) at \(a=1\), and use it to estimate \(f(1.1)\).

The function value and derivatives at the center are

$$f(1)=1,\qquad f'(x)=3x^2,\qquad f'(1)=3,$$
$$f''(x)=6x,\qquad f''(1)=6.$$

Therefore

$$P_2(x)=1+3(x-1)+3(x-1)^2.$$

At \(x=1.1\), \(x-1=0.1\), so

$$P_2(1.1)=1+3(0.1)+3(0.01)=1+0.3+0.03=1.33.$$

The tangent estimate is \(L(1.1)=1+3(0.1)=1.3\), while the exact value is \(f(1.1)=(1.1)^3=1.331\). The quadratic estimate has error \(0.001\), compared with \(0.031\) for the tangent estimate. Finally, \(P_2'(x)=3+6(x-1)\) and \(P_2''(x)=6\), giving \(P_2(1)=1\), \(P_2'(1)=3\), and \(P_2''(1)=6\). All three required matches hold.

A Common Mistake: Overstating What One Value Tells You

A common error is to infer the behavior of \(f'\) on an interval from the sign of \(f''\) at a single point. For example, the calculation \(f''(a)>0\) does not, by itself, prove that \(f'\) is increasing throughout a neighborhood of \(a\). The second derivative might be negative at other points arbitrarily close to \(a\).

Correction: To conclude that \(f'\) is increasing on an interval, establish that \(f''(x)>0\) throughout that interval. If \(f''\) is continuous at \(a\) and \(f''(a)>0\), then continuity guarantees positivity on some neighborhood of \(a\). Without that additional information, interpret \(f''(a)\) as the coefficient data for the quadratic model at the center, not as a neighborhood-wide sign claim.

Another useful distinction is between a model’s position relative to its tangent and the accuracy of that model. The sign of \(f''(a)\) tells us whether the quadratic correction raises or lowers the tangent-line model. It does not, on its own, give a numerical error bound for the actual function. The error identity above supplies such control when \(f''\) is continuous.

When the Quadratic Model Is Useful

The degree-two Taylor polynomial improves on a tangent line when second-derivative information matters over a small range around the center. It can capture a local bend that a straight-line model necessarily misses. Its usefulness in a specific calculation depends on how close \(x\) is to \(a\), the behavior of \(f''\) between those points, and the accuracy required.

The essential comparison is simple: both models share the function’s value and slope at the center, and the quadratic model adds \(\frac{f''(a)}{2}(x-a)^2\). That correction is zero at the center and grows with the square of the distance from it. Its sign describes the quadratic model’s bending relative to the tangent; additional conditions are needed for stronger claims about the function nearby.

Check Your Understanding

Use the definitions and comparisons from this tutorial to answer the following.

  1. Write the degree-two Taylor polynomial of \(f\) at \(a\), and state what value each of its three terms encodes.
  2. How does the degree-two polynomial differ from the tangent-line approximation?
  3. If \(f''(a)<0\), where does the quadratic model lie relative to its tangent line for \(x\ne a\)?
  4. What condition on \(f''\) throughout an interval allows you to conclude that \(f'\) is increasing there?
  5. Why does matching \(f(a)\), \(f'(a)\), and \(f''(a)\) not automatically provide a numerical error bound?

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