Range of a Quadratic

The range of a quadratic is bounded on one side by its vertex, if the vertex is in the domain. That italicised condition is the entire difficulty of this subtopic, and 0606 tests it deliberately.

Unrestricted domain: the vertex rules

$f(x) = x^2 - 6x + 14$, $x \in \mathbb{R}$. Complete the square: $(x - 3)^2 + 5$. Upward parabola, vertex value $5$. Range: $f(x) \ge 5$.

For a downward parabola ($a < 0$), the inequality flips: $f(x) \le q$. Two marks: the completed square (M), the stated range in function notation (A). "$x \ge 5$" is wrong; "$y \ge 5$" is tolerated; "$f(x) \ge 5$" is the safe form.

Restricted domain: sketch before you state

$f(x) = x^2 - 6x + 14$ for $0 \le x \le 2$. Vertex at $x = 3$, outside the domain. On $[0, 2]$ the function is decreasing (left of the vertex). Endpoints: $f(0) = 14$, $f(2) = 6$. Range: $6 \le f(x) \le 14$.

The vertex value $5$ appears nowhere, quoting it is the planted error. The safe procedure: locate the vertex, check whether it’s inside the domain, then evaluate whichever of $\{\text{vertex, endpoints}\}$ actually bound the outputs. A ten-second sketch makes all of it visible.

When the vertex is inside a closed domain, the range runs from the vertex value to the larger endpoint value (upward parabola), both endpoints must be checked, not just one.

Why this subtopic matters beyond itself

Range-of-a-quadratic is the engine inside several composite questions: the domain of an inverse (domain of $f^{-1}$ = range of $f$), the one-one restriction follow-ups, and applied “find the possible values of…” parts. Get the sketch-first habit here and those parts inherit it.

Common mistakes

• Vertex value quoted when the vertex lies outside the domain
• Only one endpoint evaluated on a closed domain
• Range in $x$-notation
• Open/closed boundary symbols not matching the domain’s (a strict domain end gives a strict range end)
• Min/max direction flipped when $a < 0$

Full topic context: Quadratic Functions notes · the general method: domain & range.