To analyze the first-order Taylor approximation of a quadratic function $f(x) = ax^2 + bx + c$ at a point $x_0$, we first need to derive the Taylor approximation and then compare it to $f(x)$.

1. First-Order Taylor Approximation:

The first-order Taylor approximation of $f(x)$ at $x = x_0$ is given by: $T_1(x) = f(x_0) + f'(x_0)(x - x_0)$ where $f'(x)$ is the first derivative of $f(x)$.

For the function $f(x) = ax^2 + bx + c$, we have: $f'(x) = 2ax + b$

Evaluating $f(x)$ and $f'(x)$ at $x_0$: $f(x_0) = a x_0^2 + b x_0 + c$ $f'(x_0) = 2a x_0 + b$

Substituting into the Taylor approximation: $T_1(x) = a x_0^2 + b x_0 + c + (2a x_0 + b)(x - x_0)$

2. Comparing $T_1(x)$ and $f(x)$:

To compare $T_1(x)$ and $f(x)$, let's find the difference between $f(x)$ and $T_1(x)$: $f(x) - T_1(x) = (a x^2 + b x + c) - (a x_0^2 + b x_0 + c + (2a x_0 + b)(x - x_0))$

Simplify this expression: $f(x) - T_1(x) = ax^2 + bx + c - a x_0^2 - b x_0 - c - (2a x_0 x - 2a x_0^2 + bx - bx_0)$ $= ax^2 - a x_0^2 + bx - b x_0 - 2a x_0 x + 2a x_0^2 - bx + bx_0$

Combining like terms: $f(x) - T_1(x) = ax^2 - 2a x_0 x + a x_0^2$ $= a(x^2 - 2x_0 x + x_0^2)$

Notice that the expression $x^2 - 2x_0 x + x_0^2 = (x - x_0)^2$, so: $f(x) - T_1(x) = a(x - x_0)^2$

3. Determining the Sign of the Difference:

The difference $f(x) - T_1(x) = a(x - x_0)^2$ depends on the coefficient $a$ and the squared term $(x - x_0)^2$:

• $(x - x_0)^2 \geq 0$ for all $x$ (it is always non-negative since it is a square).
• If $a > 0$, then $f(x) - T_1(x) = a(x - x_0)^2 \geq 0$. Thus, $f(x) \geq T_1(x)$ for all $x$, meaning the first-order Taylor approximation is always less than or equal to $f(x)$.
• If $a < 0$, then $f(x) - T_1(x) = a(x - x_0)^2 \leq 0$. Thus, $f(x) \leq T_1(x)$ for all $x$, meaning the first-order Taylor approximation is always greater than or equal to $f(x)$.

4. Conclusion:

By examining the coefficient $a$ of the quadratic function $f(x) = ax^2 + bx + c$:

• If $a > 0$, the parabola opens upwards, and the first-order Taylor approximation $T_1(x)$ is always less than or equal to $f(x)$.
• If $a < 0$, the parabola opens downwards, and the first-order Taylor approximation $T_1(x)$ is always greater than or equal to $f(x)$.

Thus, the sign of the coefficient $a$ determines whether the first-order Taylor approximation is an underestimate or an overestimate of the function $f(x)$.

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Would you like more details on any part of this explanation, or do you have any other questions?

Here are some related questions you might find interesting:

1. How does the second-order Taylor approximation compare to the first-order Taylor approximation?
2. What happens to the Taylor approximation when $x_0 = 0$?
3. How can the Taylor series be used to approximate non-polynomial functions?
4. Can the first-order Taylor approximation be applied to a cubic function?
5. How does the concavity of a function relate to the accuracy of the Taylor approximation?

Tip: When approximating a function using the Taylor series, the more terms you include, the closer the approximation will be to the actual function, especially near the point of expansion $x_0$.