Problem

Establish that \(-(x+2)^{2}\le~0\) and determine when equality holds. Apply this to \(g(x)=-(x+2)^{2}+10\) to give the extremum type, value and location, vertex, and range.

Answer choices

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The function has maximum −1 at x = −2, vertex (−2, −1), and range (−∞, −1]. Substituting x = −2 gives 10.

The function has maximum 10 at x = −2 and vertex (−2, 10), but range (−∞, 10). The endpoint 10 is attained and must be included.

The function has maximum 10 at x = 2, vertex (2, 10), and range (−∞, 10]. The equality input is x = −2.

−(x + 2)² ≤ 0, with equality at x = −2. The function has maximum 10 at x = −2, vertex (−2, 10), and range (−∞, 10].

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