(Junior high school student asked) How is the vertex coordinate formula of the quadratic function derived?

The vertex coordinate formula for a standard quadratic function, \( f(x) = ax^2 + bx + c \), is derived through the algebraic process of completing the square, which transforms the general polynomial form into a vertex form that explicitly reveals the coordinates of the parabola's turning point. The vertex form is \( f(x) = a(x - h)^2 + k \), where the vertex coordinates are \( (h, k) \). To derive \( h \) and \( k \) from the coefficients \( a \), \( b \), and \( c \), one begins by factoring \( a \) out of the first two terms: \( f(x) = a(x^2 + \frac{b}{a}x) + c \). The critical step is to complete the square inside the parentheses. This involves taking half of the coefficient of the \( x \)-term, which is \( \frac{b}{2a} \), squaring it to get \( \frac{b^2}{4a^2} \), and adding and subtracting this value inside the parentheses. The expression becomes \( a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} - \frac{b^2}{4a^2}\right) + c \).

The trinomial \( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \) is now a perfect square, specifically \( \left(x + \frac{b}{2a}\right)^2 \). Substituting this back gives \( f(x) = a\left[\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2}\right] + c \). Distributing the \( a \) yields \( f(x) = a\left(x + \frac{b}{2a}\right)^2 - a \cdot \frac{b^2}{4a^2} + c \), which simplifies to \( f(x) = a\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a} + c \). The constant terms \( -\frac{b^2}{4a} + c \) can be combined into a single term by expressing \( c \) with a common denominator of \( 4a \): \( c = \frac{4ac}{4a} \). Thus, the combined constant is \( \frac{-b^2 + 4ac}{4a} \), or \( \frac{4ac - b^2}{4a} \).

Comparing the final expression \( f(x) = a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a} \) to the vertex form \( f(x) = a(x - h)^2 + k \), we identify \( h = -\frac{b}{2a} \) and \( k = \frac{4ac - b^2}{4a} \). The \( x \)-coordinate, \( h = -\frac{b}{2a} \), is particularly significant as it represents the axis of symmetry of the parabola. This derivation shows that the vertex's location is determined solely by the coefficients \( a \), \( b \), and \( c \), with the \( x \)-coordinate being the point where the linear term's influence is effectively neutralized in the derivative or, geometrically, where the slope is zero.

The practical implication of this formula is that it provides an efficient method to find the maximum or minimum value of a quadratic function without graphing or extensive tabulation, which is foundational in optimization problems across physics, engineering, and economics. Understanding this derivation reinforces the relationship between algebraic manipulation and geometric interpretation, demonstrating how the abstract process of completing the square translates directly into the parabola's tangible features like its vertex and axis of symmetry.