Which statement is true about f of x plus 2 is equal to 1 sixth start absolute value x minus 3 end absolute value ?

Subtract 2 from both sides to get f(x) = (1/6)|x − 3| − 2. The absolute value is smallest when x = 3, so the minimum value of f(x) is −2; its range is therefore f(x) ≥ −2. The graph opens upward and has vertex (3, −2), so the true statement is that its range is greater than or equal to negative 2.

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