Problem

Use \((x-6)^{2}\ge~0\) to analyze \(f(x)=(x-6)^{2}+2\). State when equality occurs, derive the bound on \(f\), and give the extremum type, value and input, vertex, and range.

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What this problem is really about

The nonnegative-square fact gives an algebraic bound without relying only on the graph. We’ll identify where the square reaches equality, shift its lower bound by the outside constant, and use the attained value to state the vertex, minimum, and closed endpoint of the range.

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The squared term is strictly positive, so the value approaches 2 but never reaches it. A square equals 0 at x = 6.

Equality occurs at x = 6, but the minimum is 1 and vertex is (6, 1). Substituting x = 6 gives 2.

Equality occurs at x = −6, giving vertex (−6, 2). The factor x − 6 equals 0 at x = 6.

(x − 6)² ≥ 0, with equality at x = 6. Therefore f has minimum value 2 at x = 6, vertex (6, 2), and range [2, ∞).

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Use \((x-6)^{2}\ge~0\) to analyze \(f(x)=(x-6)^{2}+2\). State when equality occurs, derive the bound on \(f\), and give the extremum type, value and input, vertex, and range.

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