Show why \(0.5(x-6)^{2}\) is nonnegative and use that fact to bound \(h(x)=0.5(x-6)^{2}-7\). Give the equality input, extremum, and range in inequality and interval notation.

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The scaled square term is nonpositive, so the function has maximum −7. The positive factor 0.5 preserves nonnegativity.

The squared term is strictly positive, so y > −7. At x = 6 the square equals 0, so −7 is included.

The range is x ≥ −7, or [−7, ∞). A range describes output y-values, not input x-values.

0.5(x − 6)² ≥ 0, so h(x) ≥ −7. The minimum −7 is included at x = 6; the range is y ≥ −7, or [−7, ∞).

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Curriculum context

Course: Math II
Standard: A-SSE.3.b
Category: Algebra
Domain: Seeing Structure in Expressions
Objective: Complete the square to reveal maximum or minimum values of quadratic functions.
Problem type: Determine range from vertex form