To prove that the relative minimum or maximum of the quadratic function $f(x) = ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$, let's go through the steps using calculus.

Step 1: Find the first derivative of $f(x)$

We need to find the critical points by first taking the derivative of the function $f(x) = ax^2 + bx + c$.

$f'(x) = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b$

Step 2: Set the derivative equal to zero

Critical points occur where the first derivative is equal to zero or where the derivative does not exist. Since $f'(x) = 2ax + b$ is a polynomial, it exists for all values of $x$. Thus, we only need to find where the derivative equals zero:

$2ax + b = 0$

Step 3: Solve for $x$

Solve the equation $2ax + b = 0$ for $x$:

$2ax = -b$

$x = -\frac{b}{2a}$

This value of $x$ represents the critical point.

Step 4: Determine if it's a minimum or maximum

To confirm whether this critical point corresponds to a minimum or maximum, we take the second derivative of $f(x)$.

$f''(x) = \frac{d}{dx}(2ax + b) = 2a$

• If $a > 0$, then $f''(x) = 2a > 0$, indicating the function is concave up at $x = -\frac{b}{2a}$, meaning this point is a relative minimum.
• If $a < 0$, then $f''(x) = 2a < 0$, indicating the function is concave down at $x = -\frac{b}{2a}$, meaning this point is a relative maximum.

Conclusion:

The relative minimum or maximum of the quadratic function $f(x) = ax^2 + bx + c$ occurs at $x = -\frac{b}{2a}$.

Would you like more details or have any questions?

Here are 5 related questions to expand this understanding:

1. How does the second derivative test confirm concavity and the type of critical point?
2. What happens if $a = 0$ in the function $f(x) = ax^2 + bx + c$?
3. Can you find the minimum or maximum value of $f(x)$ at $x = -\frac{b}{2a}$?
4. How does this result relate to the vertex of a parabola in geometry?
5. Can we generalize this process for higher-degree polynomials?

Tip: The formula $x = -\frac{b}{2a}$ also gives the x-coordinate of the vertex of a parabola.