What is the Vertex of the Quadratic Function f(x) = x² - 6x?
Decoding the Vertex of f(x) = x² - 6x
In the realm of analytic geometry and algebra, the quadratic function $f(x) = x^2 - 6x$ describes a parabolic curve opening upward. To find the vertex—the turning point or the apex of the parabola—we can employ multiple mathematical lenses, ranging from algebraic formulas to calculus and completing the square.
Method 1: The Algebraic Formula ($x = -b / 2a$)
A standard quadratic function is written in the form $f(x) = ax^2 + bx + c$. For our specific function, $f(x) = x^2 - 6x$, the coefficients are:
- $a = 1$
- $b = -6$
- $c = 0$
The x-coordinate of the vertex is given by the formula $x = \frac{-b}{2a}$. Substituting our values in:
$$x = \frac{-(-6)}{2(1)} = \frac{6}{2} = 3$$
To find the corresponding y-coordinate, we substitute $x = 3$ back into the original function:
$$f(3) = (3)^2 - 6(3) = 9 - 18 = -9$$
Thus, the vertex is (3, -9).
Method 2: Completing the Square
Historically rooted in ancient Babylonian and Islamic mathematics (pioneered by scholars like Al-Khwarizmi), completing the square transforms the quadratic into vertex form, $f(x) = a(x - h)^2 + k$, where $(h, k)$ is the vertex.
- Start with $x^2 - 6x$.
- Take half of the linear coefficient ($-6$), which is $-3$, and square it to get $9$.
- Add and subtract this value: $(x^2 - 6x + 9) - 9$.
- Rewrite as a squared binomial: $(x - 3)^2 - 9$.
From this vertex form, we immediately see that $h = 3$ and $k = -9$, confirming the vertex is at (3, -9) since $a = 1 > 0$, indicating a minimum.
Method 3: Calculus and Optimization
From a calculus perspective, the vertex of a parabola represents its critical point where the derivative equals zero (the slope of the tangent line is horizontal).
Taking the derivative of $f(x) = x^2 - 6x$ with respect to $x$:
$$f'(x) = 2x - 6$$
Setting the derivative to zero to find the extremum:
$$2x - 6 = 0 \implies 2x = 6 \implies x = 3$$
Evaluating $f(3)$ yields $-9$, reinforcing that the minimum vertex sits gracefully at $(3, -9)$.