Maximum and Minimum Values

Completing the square, $f(x)=ax^2+bx+c = a\Big(x+\dfrac{b}{2a}\Big)^2 + \dfrac{4ac-b^2}{4a}$, does more than settle the sign question — it also pins down the single most extreme value $f(x)$ ever takes, because a square is never negative.

• If $a>0$: the squared term is always $\geq 0$, so $f(x)$ is smallest exactly when the square vanishes, i.e. at $x=-\dfrac{b}{2a}$. There $f(x)$ attains its absolute minimum value, $\dfrac{4ac-b^2}{4a}$, and $f(x)$ has no maximum — it grows without bound as $x\to\pm\infty$.
• If $a<0$: the sign flips, so $f(x)$ is largest at $x=-\dfrac{b}{2a}$, giving an absolute maximum value of $\dfrac{4ac-b^2}{4a}$, and no minimum exists.

Geometrically this is just the vertex of the parabola $y=ax^2+bx+c$: an upward-opening parabola ($a>0$) has a lowest point (the minimum), a downward-opening one ($a<0$) has a highest point (the maximum), and in both cases that turning point sits at $x=-\dfrac{b}{2a}$ — the parabola is symmetric about the vertical line through this $x$-value.

A very useful application: this lets you find the range of a rational function whose numerator and denominator are both quadratics. Set $y_0$ equal to the function's value, clear denominators to get a quadratic in $x$ with $y_0$ as a parameter, and demand that this quadratic have a real root (i.e. its discriminant $\geq 0$). That turns into an inequality in $y_0$ alone, whose solution set is exactly the range of the function.

Worked example. Find the maximum or minimum of $f(x) = 3x^2 - 12x + 7$. …