Back to Taylor polynomials

Exercise 8

Avg difficulty 2.5

In Exercise, we found that the degree 2 Taylor polynomial centered at a=0a = 0 of a quadratic function is the quadratic function itself. In this exercise, we explore how changing the center of the approximation offers additional insight into the function.

Let f(x)=12x2−2x+5f(x) = \frac{1}{2}x^2 - 2x + 5, and let a=2a = 2 be the center at which we will find a degree 22 Taylor polynomial approximation of ff.

  1. By finding f′(x)f'(x), f′′(x)f''(x), f(2)f(2), f′(2)f'(2), and f′′(2)f''(2), determine T2(x)T_2(x), the degree 22 Taylor polynomial approximation of ff that is centered at a=2a = 2.

  2. Plot both f(x)f(x) and T2(x)T_2(x) on the same axes. What do you observe?

  3. What does the algebraic form of T2(x)T_2(x) tell you about the original function f(x)f(x)?

  4. What is the tangent line approximation to f(x)f(x) at a=2a = 2? What is special about the function's behavior at this input value?

Boelkins, Active Calculus, §8.2 Taylor polynomials, Exercise 8

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Adapted from Active Calculus © 2012–2025 Matthew Boelkins. Licensed under CC BY-SA 4.0. Source: Active Calculus. Imported at revision dc86ba3.