### Proving the Location of the Relative Minimum or Maximum**Objective:** Use calculus to demonstrate that the relative minimum or maximum point for any quadratic function \( f \), described as follows, occurs at \( x = -\frac{b}{2a} \).**Function Definition:** \[ f(x) = ax^2 + bx + c, \quad a \neq 0 \]### Explanation:To find the relative minimum or maximum of the quadratic function \( f(x) = ax^2 + bx + c \), follow these steps:1. **Take the Derivative:** Find the first derivative of the function, \( f'(x) \), to determine the slope of the tangent at any point \( x \). 2. **Set Derivative to Zero:** Solve \( f'(x) = 0 \) to find the critical points, where the slope of the tangent is zero, indicating a potential minimum or maximum.3. **Calculate the Derivative:** - The derivative of \( f(x) \) is: \[ f'(x) = 2ax + b \]4. **Solve for \( x \):** - Set the derivative equal to zero: \[ 2ax + b = 0 \] - Solve for \( x \): \[ x = -\frac{b}{2a} \]This derivation shows that for any quadratic function of the form \( ax^2 + bx + c \), the critical point, which could be a relative minimum or maximum, occurs at \( x = -\frac{b}{2a} \).

Answered: Image /qna-images/answer/38d337bf-8350-47aa-beb6-f8aa55cab446.jpgDifferentiate the given function with respect to x, Find the critical point, Differentiate equation (1) with respect to x, The double derivative is positive at , therefore the given function is minimum at Again, The double derivative is negative at , therefore the given function is maximum atHence proven that the function f has a relative minimum at if and the function f has a relative maximum at if