Q.For what value of $x$ is $y = x(5-x)$ maximum or minimum? OR Use differentials to find the value of $\sqrt{49.5}$.

Quadratic Extrema: From Intuition to Precision

Toss a ball straight up: it rises, slows, stops for an instant at the top, then falls. Plot its height against time and you get a parabola with exactly one turning point — a peak (maximum) or a valley (minimum). That single highest or lowest point is what quadratic extrema are about.

The Intuition First

A quadratic is $f(x) = ax^2 + bx + c$, with $a \neq 0$; its graph is a parabola.

• If $a > 0$, it opens upward (a U) and has a minimum at the bottom.
• If $a < 0$, it opens downward and has a maximum at the top.

The turning point is the vertex. Every quadratic has exactly one vertex — that's the extremum.

The Precise Statement

For $f(x) = ax^2 + bx + c$ with $a \neq 0$:

1. Vertex (extremum) at

2. Extremum value

3. Nature: $a > 0$ → minimum; $a < 0$ → maximum.

Why That $x$? A Quick Derivation

Complete the square:

The squared term is always $\geq 0$. When $a > 0$, $f(x)$ is smallest when the square is zero — at $x = -\frac{b}{2a}$. When $a < 0$, the largest value occurs at the same $x$.

Common Mistake to Avoid

Example

Find the extremum of $f(x) = 2x^2 - 8x + 5$. …